CBSE - i
Science
Class-X Unit - 4
PHYSICS
Light – Reflection and Refraction
CHEMISTRY
Carbon and its Compounds
BIOLOGY
Heredity and Evolution
CENTRAL BOARD OF SECONDARY EDUCATION
Shiksha Kendra, 2, Community Centre, Preet Vihar,Delhi-110 092 India
CBSE - i
CLASS-X
Science
Unit - 4
PHYSICS
Light – Reflection and Refraction
CHEMISTRY
Carbon and its Compounds
BIOLOGY
Heredity and Evolution
CENTRAL BOARD OF SECONDARY EDUCATION
Shiksha Kendra, 2, Community Centre, Preet Vihar,Delhi-110 092 India
The CBSE-International is grateful for permission to reproduce
and/or translate copyright material used in this publication. The
acknowledgements have been included wherever appropriate and
sources from where the material may be taken are duly mentioned. In
case any thing has been missed out, the Board will be pleased to rectify
the error at the earliest possible opportunity.
All Rights of these documents are reserved. No part of this publication
may be reproduced, printed or transmitted in any form without the
prior permission of the CBSE-i. This material is meant for the use of
schools who are a part of the CBSE-International only.
PREFACE
The Curriculum initiated by Central Board of Secondary Education -International (CBSE-i) is a progressive step in
making the educational content and methodology more sensitive and responsive to the global needs. It signifies the
emergence of a fresh thought process in imparting a curriculum which would restore the independence of the learner
to pursue the learning process in harmony with the existing personal, social and cultural ethos.
The Central Board of Secondary Education has been providing support to the academic needs of the learners
worldwide. It has about 11500 schools affiliated to it and over 158 schools situated in more than 23 countries. The
Board has always been conscious of the varying needs of the learners in countries abroad and has been working
towards contextualizing certain elements of the learning process to the physical, geographical, social and cultural
environment in which they are engaged. The International Curriculum being designed by CBSE-i, has been visualized
and developed with these requirements in view.
The nucleus of the entire process of constructing the curricular structure is the learner. The objective of the curriculum
is to nurture the independence of the learner, given the fact that every learner is unique. The learner has to understand,
appreciate, protect and build on values, beliefs and traditional wisdom, make the necessary modifications,
improvisations and additions wherever and whenever necessary.
The recent scientific and technological advances have thrown open the gateways of knowledge at an astonishing pace.
The speed and methods of assimilating knowledge have put forth many challenges to the educators, forcing them to
rethink their approaches for knowledge processing by their learners. In this context, it has become imperative for
them to incorporate those skills which will enable the young learners to become 'life long learners'. The ability to stay
current, to upgrade skills with emerging technologies, to understand the nuances involved in change management
and the relevant life skills have to be a part of the learning domains of the global learners. The CBSE-i curriculum has
taken cognizance of these requirements.
The CBSE-i aims to carry forward the basic strength of the Indian system of education while promoting critical and
creative thinking skills, effective communication skills, interpersonal and collaborative skills along with information
and media skills. There is an inbuilt flexibility in the curriculum, as it provides a foundation and an extension
curriculum, in all subject areas to cater to the different pace of learners.
The CBSE has introduced the CBSE-i curriculum in schools affiliated to CBSE at the international level in 2010 and is
now introducing it to other affiliated schools who meet the requirements for introducing this curriculum. The focus of
CBSE-i is to ensure that the learner is stress-free and committed to active learning. The learner would be evaluated on
a continuous and comprehensive basis consequent to the mutual interactions between the teacher and the learner.
There are some non-evaluative components in the curriculum which would be commented upon by the teachers and
the school. The objective of this part or the core of the curriculum is to scaffold the learning experiences and to relate
tacit knowledge with formal knowledge. This would involve trans-disciplinary linkages that would form the core of
the learning process. Perspectives, SEWA (Social Empowerment through Work and Action), Life Skills and Research
would be the constituents of this 'Core'. The Core skills are the most significant aspects of a learner's holistic growth
and learning curve.
The International Curriculum has been designed keeping in view the foundations of the National Curricular
Framework (NCF 2005) NCERT and the experience gathered by the Board over the last seven decades in imparting
effective learning to millions of learners, many of whom are now global citizens.
The Board does not interpret this development as an alternative to other curricula existing at the international level,
but as an exercise in providing the much needed Indian leadership for global education at the school level. The
International Curriculum would evolve on its own, building on learning experiences inside the classroom over a
period of time. The Board while addressing the issues of empowerment with the help of the schools' administering
this system strongly recommends that practicing teachers become skillful learners on their own and also transfer their
learning experiences to their peers through the interactive platforms provided by the Board.
I profusely thank Shri G. Balasubramanian, former Director (Academics), CBSE, Ms. Abha Adams and her team and
Dr. Sadhana Parashar, Head (Innovations and Research) CBSE along with other Education Officers involved in the
development and implementation of this material.
The CBSE-i website has already started enabling all stakeholders to participate in this initiative through the discussion
forums provided on the portal. Any further suggestions are welcome.
Vineet Joshi
Chairman
ACKNOWLEDGEMENTS
Advisory
Shri Vineet Joshi, Chairman, CBSE
Shri N. Nagaraju, Director(Academic), CBSE
Ideators
Ms. Aditi Misra
Ms. Amita Mishra
Ms. Anita Sharma
Ms. Anita Makkar
Dr. Anju Srivastava
English :
Ms. Sarita Manuja
Ms. Renu Anand
Ms. Gayatri Khanna
Ms. P. Rajeshwary
Ms. Neha Sharma
Ms. Sarabjit Kaur
Ms. Ruchika Sachdev
Geography:
Ms. Deepa Kapoor
Ms. Bharti Dave
Ms. Bhagirathi
Ms. Archana Sagar
Ms. Manjari Rattan
Ms. Anuradha Sen
Ms. Archana Sagar
Ms. Geeta Varshney
Ms. Guneet Ohri
Dr. Indu Khetrapal
Conceptual Framework
Shri G. Balasubramanian, Former Director (Acad), CBSE
Ms. Abha Adams, Consultant, Step-by-Step School, Noida
Dr. Sadhana Parashar, Director (Training),CBSE
Ms. Jaishree Srivastava
Dr. Kamla Menon
Dr. Meena Dhami
Ms. Neelima Sharma
Dr. N. K. Sehgal
Material Production Groups: Classes IX-X
Mathematics :
Science :
Dr. K.P. Chinda
Ms. Charu Maini
Mr. J.C. Nijhawan
Ms. S. Anjum
Ms. Rashmi Kathuria
Ms. Meenambika Menon
Ms. Reemu Verma
Ms. Novita Chopra
Dr. Ram Avtar
Ms. Neeta Rastogi
Mr. Mahendra Shankar
Ms. Pooja Sareen
Political Science:
Ms. Sharmila Bakshi
Ms. Archana Soni
Ms. Srilekha
Dr. Rajesh Hassija
Ms. Rupa Chakravarty
Ms. Sarita Manuja
Ms. Himani Asija
Dr. Uma Chaudhry
History :
Ms. Jayshree Srivastava
Ms. M. Bose
Ms. A. Venkatachalam
Ms. Smita Bhattacharya
Economics:
Ms. Mridula Pant
Mr. Pankaj Bhanwani
Ms. Ambica Gulati
Material Production Groups: Classes VI-VIII
English :
Ms. Rachna Pandit
Ms. Neha Sharma
Ms. Sonia Jain
Ms. Dipinder Kaur
Ms. Sarita Ahuja
Dr. Indu Khetarpal
Ms. Vandana Kumar
Ms. Anju Chauhan
Ms. Deepti Verma
Ms. Ritu Batra
Science :
Dr. Meena Dhami
Mr. Saroj Kumar
Ms. Rashmi Ramsinghaney
Ms. Seema kapoor
Ms. Priyanka Sen
Dr. Kavita Khanna
Ms. Keya Gupta
Mathematics :
Ms. Seema Rawat
Ms. N. Vidya
Ms. Mamta Goyal
Ms. Chhavi Raheja
Political Science:
Ms. Kanu Chopra
Ms. Shilpi Anand
Material Production Group: Classes I-V
Ms. Rupa Chakravarty
Ms. Anita Makkar
Ms. Anuradha Mathur
Ms. Kalpana Mattoo
Ms. Savinder Kaur Rooprai
Ms. Monika Thakur
Ms. Seema Choudhary
Mr. Bijo Thomas
Ms. Kalyani Voleti
Dr. Sadhana Parashar,
Head (I and R)
Shri R. P. Sharma, Consultant
Ms. Seema Lakra, S O
Geography:
Ms. Suparna Sharma
Ms. Leela Grewal
History :
Ms. Leeza Dutta
Ms. Kalpana Pant
Ms. Nandita Mathur
Ms. Seema Chowdhary
Ms. Ruba Chakarvarty
Ms. Mahua Bhattacharya
Coordinators:
Ms. Sugandh Sharma,
Dr. Srijata Das,
Dr. Rashmi Sethi,
E O (Com)
E O (Maths)
E O (Science)
Ms. Ritu Narang, RO (Innovation) Ms. Sindhu Saxena, R O (Tech) Shri Al Hilal Ahmed, AEO
Ms. Preeti Hans, Proof Reader
Content
Physics
Light – Reflection and Refraction
1.
Syllabus Coverage
1
Unit 4 - Light Reflection and Refraction
Core and Extension
2.
Matrix
2
3.
Scope Document
5
4.
•
Learning Objectives
•
Cross Curricular Links
•
Suggested Activities
Teachers' Notes (TN)
•
Warm up : Nature of Light
•
Reflection of Light – Regular and Diffused
•
Plane Mirror – Properties
•
Refraction of Light
•
Spherical Lenses and their Image formation
•
Total Internal Reflection and its applications
•
Applications of Refraction of light in nature
•
Dispersion
•
Structure and Function of Human eye
•
Eye defects and their correction
•
Extension
6
5.
Teacher Student Support Material (TSSM)
11
6.
Rubrics of Assessment For Learning - Physics
81
Chemistry
Carbon and its Compounds
1.
Syllabus Coverage
83
Carbon and its Compounds
2.
Matrix
3.
Scope Document
4.
•
Learning Objectives
•
Cross Curricular Links
•
Suggested Activities
Teachers' Notes (TN)
•
Warm up
•
Pre content
•
Bonding in Carbon Compounds
•
Nomenclature of Organic Compounds
•
Homologous series
•
Fractional distillation of petroleum
•
The Physical and Chemical Properties of Hydrocarbons
•
Alcohols
•
Carboxylic Acid
•
Polymes-Nylon and Terylene
84
91
5.
Teacher Student Support Material (TSSM)
115
6.
Formative Assessment
7.
Rubrics of Assessment
173
8.
Suggested Videos and Resources
175
Bio logy
Heredity and Evolution
1.
Syllabus Coverage
176
Heredity and Evolution
2.
Matrix
177
3.
Scope Document
182
4.
•
Learning Objectives
•
Cross Curricular Links
•
Suggested Activities
Teachers' Notes (TN)
•
Structure of DNA
•
Traits
•
Gene, genotype, phenotype, homologous chromosomes
•
Rules for Inheritance
•
Genetic Variation
•
Chromosomal basis of sex determination in human beings
•
our solar system and life
•
origin of life
•
Evolution
•
A Long Time- a time line
•
evidences that help in tracing evolutionary relationships
•
Evolution and Classification
•
the steps in species formation
•
human evolutionz
184
5.
Teacher Student Support Material (TSSM)
199
6.
Formative Assessment
7.
Rubrics of Assessment
276
8.
Suggested Videos and Resources
277
PHYSICS
LIGHT – Reflection and Refraction
UNIT 4
SYLLABUS COVERAGE
Unit 4 – Light – Reflection and Refraction
CORE

Nature of Light

Regular and Diffused reflection

Laws of reflection

Image formation by plane mirror

Refraction

Refractive Index

Spherical lenses

Image formation by lenses
A

Practical Application of reflection and refraction
B

Dispersion

Scattering of light

Structure and function of human eye

Eye defects and their correction
S
Y
L
L
U
S
EXTENSION

Electromagnetic Waves
1
Matrix
CONTENT/
INTENDED LEARNING
SKILL
CORE
Nature of light
Reflection of light
Students will be able to understand the
Observation,
nature/properties of light by
Understanding
observing/performing a few activities.
and analysis
Students will be able to understand the
Observation, To
phenomenon of reflection of light and its
Differentiate
reasons.
Types of
They will be able to differentiate between
Identification,
Reflection
regular and diffused reflection and state the
Application,
two laws of reflection.
Learning by
doing.
Illustration.
Laws of reflection
Image formed by
Student will understand that a plane mirror
Understanding,
a plane mirror
is the most suitable example of regular
Application,
and its
reflection.
Observation,
characteristics.
Use of plane
Learning by
They will be able to explain the
mirrors in making characteristics of the image formed by a
kaleidoscope or
plane mirror. They will be able to construct
periscope etc.
kaleidoscope, periscope etc.
http://www.goo
d-science-fairprojects.com/
2
doing
Refraction
Students will understand the phenomenon of Observation.
refraction and will be able to explain/ reason
out the difference in real and apparent
positions of objects in water.
Refraction index
They may do some activities and watch
Learning by
videos from the link provided to strengthen
doing,
the concepts.
Application and
reasoning.
www.youtube.co
m/watch?v=kc2o
73FyN3
http://hyperphys
ics-astro.gsu.edu
Spherical lenses
Student will be able to understand and
Observation,
and the images
illustrate the images formed by convex and
Illustration,
formed by them
concave lenses with the help of ray
Reasoning
diagrams. They will understand the Lens
Numerical and
formula and its use through numerical They
analytical skills
will know the use of these lenses on the basis
of their properties.
Practical
Students will be able to apply the
Observation,
application of
phenomenon of reflection and refraction in
Application
reflection and
day to day activities.
refraction
Dispersion
Students will be able to understand the
Observation,
phenomenon of dispersion and its cause.
Learning by
3
Scattering of light
They will be able to depict the formation of
doing,
rainbow with the help of activities.
Illustration
Students will be able to explain the
Observation,
appearance of blue sky and reddish orange
reasoning and
sun(during the sunrise and the sunset) with
application
the help of the phenomenon of scattering.
http://hyperphys
ics-astro.gsu.edu
Structure and
Function of Human eye
Students will be able to gain the knowledge
Observation,
about the structure of the human eye. They
Analysis
will also understand the function of various
parts of the eye.
The concept can be explained with the help
of the Ppt provided herewith.
http://www.enc
hantedlearning.co
m/subjects/anato
my/eye/label/la
beleye.shtml
Eye defects and
Students will come to know the various
Observation,
their correction
types of defects of vision and the causes and
reasoning,
correction fot them. They will be able to
application and
illustrate The defective and corrective eyes
analysis.
with the help of ray diagrams.
Extension
Students will learn about properties and
types of Electromagnetic waves.
4
Scope Document
Intended Learning outcomes
At the end of this unit, students should be able to

Describe the phenomenon of reflection, refraction and dispersion and scattering.

State, explain and apply the laws of reflection and explain the practical
application of reflection, refraction, dispersion and scattering.

Illustrate the formation of real and virtual images by convex and concave lenses.

Describe and understand the structure and function of human eye.

Describe the causes and correction of defects of vision.

Construct devices like kaleidoscope etc which work on the phenomenon of
reflection and refraction.
Cross curricular links

Mathematics – Measuring angles with the help of protractor, measuring the size
of the cardboard for making kaleidoscope etc.

Biology – Explaining the structure of human eye.
5
Teachers’ Notes (TN)
The teacher would go through the links, video clips and PPT prior to teaching in the
class. She is expected to frame thought provoking questions based on the web links.
1. Teacher will demonstrate few warm up Activities to explain properties of light.
2. Using daily life examples teacher will define Reflection of light and differentiate
between Regular and Diffused Reflections.
3. Teacher may demonstrate’ Mirror Magic’ Activities to explain properties of plain
mirror
and
organize
‘hands-on
‘activity
in
the
classroom
to
build
a
kaleidoscope/Periscope in order to teach characteristics of image formation. Teacher
may find the link http://www.good-science-fair-projects.com useful.
4. Following video links and Activities mentioned in TSSM can be used by the teacher
to explain the phenomenon of Refraction of light.
www.youtube.com/watch?v=kc2o73FyN3 and
http://hyperphysics-astro.gsu.edu
5. Teacher may remember following points while teaching the application of
Refraction in forming images by Spherical lenses.

Refraction may be defined not as the bending of a ray of light in going from one
medium to another but as the collection of phenomenon associated with the
change in the speed of a ray of light as it goes from one medium to another. We
usually (but not always) observe the effect of this change in speed of light
through the bending of a ray of light.

All ray diagrams, associated with refraction phenomenon, are always drawn on
the basis of the law of refraction. This is so both for convex and concave lenses
and it would be interesting to show that a ray incident on a convex lens, bends
6
towards the principal axis on both the surfaces of the lens. For a concave lens, the
bending is away from the principal axis at both the surfaces. Application of
Snell's law may also be done to show the reversal in the behavior of the two
types of lenses when the surrounding medium has refractive index more than
that of the material of the lens. All the time, it has to be remembered that the
normal to a spherical surface (for a lens there are two such surfaces) is drawn by
joining the point on the lens surface to the centre of curvature of THAT surface. It
would be pertinent to point out the three special rays, used for drawing ray
diagrams, are all drawn on the basis of the law of refraction.

The importance of the sign convention has to be emphasized.(We would
otherwise get different lens formulae for the convex and concave lenses) The
terms u, v, and f in the(single) lens formula, are all algebraic terms and we have
to put both the sign and the magnitude when substituting their values for any
numerical calculation. It would be interesting to work out the power of the
normal eye lens when viewing objects at Infinity and at the normal distance of
distinct vision I.e.,25 cm. The image distance for both the cases would be the size
of the eyeball which may be roughly taken as 2.3 cm.The power values are quite
high ( of the order of 45 D) and the difference in the power values, for the two
cases,(of the order of 4 D) is an indicator of the power of accommodation of the
normal eye.

The students may be given sufficient practice in drawing ray diagrams for
different situations. Some students have difficulty understanding how the entire
image of an object can be deduced once a single point on the image has been
determined. If the object is merely a vertical object (such as the arrow object used
in the example below), then the process is easy. The image is merely a vertical
line. In theory, it would be necessary to pick each point on the object and draw a
separate ray diagram to determine the location of the image of that point. That
would require a lot of ray diagrams as illustrated in the diagram below.
7
If the object is a vertical line, then the image is also a vertical line. For our
purposes, we will only deal with the simpler situations in which the object is a
vertical line that has its bottom located upon the principal axis. For such
simplified situations, the image is a vertical line with the lower extremity located
upon the principal axis.)

The teacher may interest the students by telling them the relationship between
the object distance and object size and the image distance and image size of a
convex lens. Starting from a large value, as the object distance decreases (i.e., the
object is moved closer to the lens), the image distance increases; meanwhile, the
image height increases. At the 2F point, the object distance equals the image
distance and the object height equals the image height. As the object is brought
closer than 2F the size and distance of the image starts increasing reaching to
highly magnified image formed at infinity when the object is kept at the focus F.

In case of a concave lens, as the object distance
is decreased, the image distance is decreased
and the image size is increased. So as an object
approaches the lens, its virtual image on the
same side of the lens approaches the lens as
well; and at the same time, the image becomes
8
larger thoughthe size of the image will always remain smaller than the object.
Method used for convex lens can be used again to explain the students, how the
entire image of an object can be deduced once a single point on the image has
been determined. If the object is merely a vertical object (such as the arrow object
used in the example below), then the process is easy. The image is merely a
vertical line. This is illustrated in the diagram below. In theory, it would be
necessary to pick each point on the object and draw a separate ray diagram to
determine the location of the image of that point. That would require a lot of ray
diagrams as illustrated in the diagram below. Fortunately, a shortcut exists. If the
object is a vertical line, then the image is also a vertical line. For our purposes, we
will only deal with the simpler situations in which the object is a vertical line that
has its bottom located upon the principal axis. For such simplified situations, the
image is a vertical line with the lower extremity located upon the principal axis.
The ray diagram above illustrates that the image of an object in front of a double
concave lens will be located at a position behind the double concave lens.
6. Teaching practical applications of Reflection and Refraction of light can be made
very interesting by taking examples from daily life. Hands – on session for
building Pin-hole camera may be arranged in the classroom as part of the activity
session. Concepts of Dispersion and Scattering of light may be taught by
explaining formation of rainbow and blue color of sky. Teacher may use the link
http://hyperphysics-astro.gsu.edu to explain these phenomenons.
7. Teacher may use the PPT to teach structure, function and defects of human eye.
Some more Teaching Suggestions, Activities and Demonstrations
1. Construct a kaleidoscope which allows the angle between the mirrors to be
adjusted. Describe what happens to the patterns observed when the angle
between the mirrors changes.
9
2. Experimentally investigate the characteristics of images formed in a plane mirror
using a optical bench, mirror and a candle. Illustrate using ray diagrams or
explain why a plane mirror produces a virtual image.
3. Suggest practical applications which illustrate the lateral inversion of an image in
a plane mirror and differentiate between lateral and vertical inversions.
4. Straddle a full-length mirror sideways, so that students see one leg in front of the
mirror, while the other leg is behind the mirror out of their view. The mirror
should be tall enough so that it reaches from the floor to your crotch. While
balancing on the rear leg, out of the view of the class, slowly raise the leg that is
visible to them and lean forward slightly. The reflection in the mirror creates the
illusion that the rear leg is also being lifted, allowing you to "levitate" before their
very eyes. Wearing a cape, having a fan blowing over the cape to produce a
breeze, and holding your arms outstretched may create an illusion of flight. This
demonstration is likely to be a success if it is set up carefully beforehand.
A follow-up might include a discussion of how mirrors are used to produce theatrical
effects and optical illusions.
5. Show students how to use ray diagrams as a means of verification for numerical
solutions to problems.
6. Perform an activity to observe the number of images formed by two mirrors
placed at 60°, 45°, and 30° angles to one another. Confirm that the number of
images observed and calculated agree with one another.
10
TEACHER
STUDENT
SUPPORT
MATERIAL
(TSSM)
11
Nature of Light
We think of light as ‘an agent’ that enables us to see and observe the world around us.
This is just one of the functions of what we now refer to as visible light. In general, we
can say that

Light is a form of energy and it can be converted to other forms

Light travels in straight lines

Light forms shadows.
Activity 1
We know that ‘energy is the ability to do work’. Hence, to show
that light is a form of energy we must be able to show that it can
do work. Work is done when an object get moved. So, if we can
show that light can move something, we can demonstrate that it
is a form of energy
Demonstration
Connect a motor to a solar panel. It is observed that when light shines on the solar
panel, the motor turns. This is because the solar panel convert light energy into
electrical energy and this then gets converted to kinetic energy in the motor.
Activity 2
To Show that Light Travels in Straight Lines.
Demonstration
Take a bulb and three symmetric pieces of cardboards
with a hole in the centre of each one of them. Align them
using a piece of cord. Notice that you can only see light
from the bulb when the holes in the cards are lined up.
12
Activity 3
We all know that shadows get formed when light coming from a
source is blocked. The shadow is then similar in outline to the object
blocking the light. We can use different objects, and our imagination,
to form very many interesting shadows. Some of them are shown
below.
In nature also, we understand the phenomenon of Lunar and Solar eclipses through the
formation of shadows.
The size and nature of the shadow observed, depends on the nature and size of both the
source of light and the obstacle put in its way. When we place our fingers between a
candle and a screen we notice that as we bring our finger closer to the candle, more
light gets blocked out and the shadow gets bigger.
Activity 4
Objective: Make simple observations of light and shadows to demonstrate and
understand that
Light travels in straight lines.
Write your observations and answer questions for each of the following:
1. Go to a dark room and turn on a light kept in one corner of the room. Observe
the places, if any, in the room which gets lighted up or stay dark.
2. Stand in the room with your back towards the light. Observe the change in your
shadow as you move around the room.
13
3. Make a small hole of diameter about 1 cm in an opaque sheet. Use this sheet to
block the light from a bulb by keeping it about 10 cm from the light source.
Position your eye at a place where you can see the light passing through the hole.
Record the details about the position of your eye when you see the light from the
lamp.
4. Show the relationship between the positions of your eye, the hole, and the source
of light on a top view diagram. Hence state your conclusion, if any, about the
condition necessary for you to see the light through the hole in the paper.
5. We often speak of ray of light or a beam of light to describe the propagation of
light. Explain how we are justified in using these words to describe light.
Activity 5
Objective: Observe the nature of the changes in the shadow with a change in distance
between the light source and the object.
Procedure:
1. Go to a dark room with an opaque linear object (say a pencil) and keep it about
20 cm from the light source on a blank sheet of paper.
2. Turn on the lamp and observe the shadow on the table cast by the pencil.
3. Observe the shape of the shadow carefully and see whether the shadow is lighter
and darker at different places. On the white paper draw a picture of the shadow
and give the details of the description of your observations.
4. Take the object 40 cm and then to 80 cm away from the light source. Do you
observe any changes in the shadow?
14
5. Analyse your observations to predict what the shadow would look like when the
object is moved to 120 cm from the light source.
Use a paper with a small hole in the center between the light and your standing
object. The size of the light source can now be thought of as being closer to a
point source of light.
Repeat the above steps in the procedure and again above observe the effect of
this nearly point light source on the size, shape, brightness, and sharpness of the
shadow cast by the standing opaque object.
6. Write out the details of the changes you observe in the shadows with this point
light source and try to analyse and explain your observations.
REFLECTION OF LIGHT
Reflection is the bouncing of light from a surface. We can see objects because
some of the light which leaves the objects hits our eyes. Luminous objects are a
source of light while non-luminous objects are seen as a result of light
reflected from them.
Examples of luminous objects: the sun, a light-bulb, a candle.
Examples of non-luminous objects: everything else (not emitting their own light)
Reflection of light changes the direction of a wave at an interface (surface of
separation) between two different media so that the light wave returns into the
medium from which it originated. The law of reflection state that (i) the angle at
which the wave is incident on the surface equals the angle at which it is reflected
and (ii) the incident ray, reflected ray and the normal lie on the same plane.
15
Types of Reflection
Reflection of light is either regular (mirror-like) or diffused (retaining the energy, but
losing the image) depending on the nature of the interface. Incoming and reflected
lights have same angle with the surface. If the surface reflects most of the light then we
call such surfaces as mirrors. Examine the given pictures below. They show regular and
diffuse reflection of light from given surfaces.
Diffused Reflections
Regular Reflection
Regular Reflection
In regular reflection the reflected rays follow a set pattern as shown in the figure above.
A mirror provides the most common model for regular light reflection, and typically
consists of a glass sheet with a metallic coating where the reflection actually occurs.
Reflection also occurs at the surface of transparent media, such as water or glass.
Diagram of regular reflection
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In the given diagram, a light ray PO strikes a vertical mirror at point O, and the
reflected ray is OQ. By projecting an imaginary line through point O perpendicular to
the mirror, known as the normal, we can measure the angle of incidence { The angle which
the incident ray PO makes with the normal } θi and the angle of reflection { angle which
the reflected ray OQ makes with the normal } θr.
Regular reflection forms images.
Diffuse/ Irregular Reflection
When light strikes the surface of a (non-metallic) material it bounces off in all directions
due to multiple reflections by the microscopic irregularities inside the material and by its
surface, if it is rough. Thus, an 'image' is not formed. This is called diffuse or irregular
reflection. The exact form of the reflection depends on the structure of the material. Most
of the objects around us reflect light irregularly. The light sent to our eyes by most of
the objects we see is due to diffuse reflection from their surface. Diffuse reflection is
the reflection of light from a surface such that an incident rays do not follow a set
pattern and is reflected at many angles rather than at just one angle as in the case of
regular reflection.
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Figure - General mechanism of diffused reflection by a solid surface
A surface may also exhibit both regular and diffuse reflection, as is the case, for
example, of glossy paints as used in home painting, which give also a fraction of regular
reflection, while matte paints give almost exclusively diffuse reflection.
Laws of reflection
The laws of reflection are as follows:
1. The incident ray, the reflected ray and the normal to the reflection surface at the
point of the incidence lie in the same plane.
2. The angle which the incident ray makes with the normal is equal to the angle
which the reflected ray makes to the same normal.
3. The reflected ray and the incident ray are on the opposite sides of the normal.
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Plane Mirror and the Image Formed by it.
The plane mirror is a highly polished surface which is the best example of regular
reflection. It forms a virtual image of an object reflecting light on to it.
A virtual image is formed at the location in space where all the reflected light appears
to diverge from. Since light from the object appears to diverge from this location, a
person who sights along a line at this location will perceive a replica or likeness of the
actual object.
Characteristics of the image formed by a plane mirror
1. In the case of plane mirrors, the image is said to be a virtual image. Virtual images
are images that are formed in locations where light does not actually reach. Light
does not actually pass through the location on the other side of the mirror; it only
appears to an observer as though the light is coming from this location. Whenever a
mirror (whether a plane mirror or otherwise) creates an image that is virtual, it will
be located behind the mirror where light does not really come from.
2. Plane mirror images are laterally inverted there are several other characteristics that
are worth noting. There is an apparent left-right reversal of the image formed by a
plane mirror. That is, if you raise your left hand, you will notice that the image raises
what would seem to be its right hand. If you raise your right hand, the image raises
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what would seem to be its left hand. This is often termed left-right reversal or
lateral inversion. The word AMBULANCE is written in inverse form in front of the
vehicle. The person in a vehicle in front of the ambulance can read it properly in
his/her rear view mirror. While there is an apparent left-right reversal of the
orientation of the image, there is no top-bottom vertical reversal. The image is said
to be upright or
erect.
3. The third characteristic of plane mirror images pertains to the relationship between
the object's distance of the object in front of the mirror to the distance of the image
inside the mirror. For plane mirrors, the object distance is equal to the image
distance. That is the image is the same distance behind the mirror as the object is in
front of the mirror. If you stand a distance of 2 meters from a plane mirror, you must
focus at a location 2 meters behind the mirror in order to view your image.
4. The fourth and final characteristic of plane mirror images is that the dimensions of
the image are the same as the dimensions of the object. If a 1.6-meter tall person
stands in front of a mirror, he/she will see an image that is 1.6-meters tall. The ratio
of the image dimensions to the object dimensions is termed the magnification. Plane
mirrors produce images that have a magnification of 1.
To summarize, images formed by plane mirrors are virtual, erect, laterally inverted,
the same distance from the mirror as the object's distance, and the same size as the
object.
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Reflection Activities
Activity – 6
Mirror Magic
Fix a comb, with its teeth pointing upwards, just in front of
a board or box-lid. Aim the comb at the sun - but do not
look at the sun. What do we observe? We will see that white
light shine through the teeth, to form parallel (side-by-side)
rays, making white lines. This activity shows that light
usually travels in a straight line.
Now hold a plane mirror, with its edge touching the board, in the path of the rays as
shown in the figure. Observe how the mirror reflects them. Angles formed with the
mirror before and after reflection are always equal. This verifies the law of reflection.
Activity – 7
To Prove Laws of Reflection
Things required: a large mirror, two cardboard tubes of similar length and diameter, a
flashlight, some books.
Steps to be followed
1. Place the books to prop the mirror upright.
2. Hold one tube at an angle with one end touching the mirror.
3. Ask a friend to hold the second tube at a matching angle.
4. Shine the flashlight into the tube you are holding.
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Observation
When the tubes are at the correct angle, the light will bounce off the mirror and down to
the end of the second tube. If we hold our hand at the end of the second tube, we will
see a circle of reflected light. On a rough surface, light is not reflected like this. It is
scattered back in several different directions.
This also proves that the incident ray and the reflected ray lie on the same plane.
Activity - 8
Make a Kaleidoscope
The word "kaleidoscope" means "beautiful to look at. A lot of beautiful symmetrical
patterns can be seen through a Kaleidoscope.
Material Required: A stiff cardboard, a pencil, scissors, black paper or a thick black felt
pen, aluminum foil, glue, clear plastic, tracing paper, cellophane tape, small colored
shapes or beads, broken pieces of bangles etc.
Hoe to go about it:
1. Cut out a piece of cardboard about 22 cm by 15 cm { 9 inches by 6 inches }
2. With the pencil, divide the card into four equal strips. Each strip should be
1{inches wide }
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3. Stick foil over two of the panels as shown in the figure. Make sure it is as smooth
as possible.
4. Stick black paper over the third panel or color it black.
5. Leave the fourth panel blank.
6. Fold the cardboard to make a triangular shape and tape the side to hold it in
place.
8. Stick a piece of clear plastic over each end of the kaleidoscope.
9. Put the colored shapes or beads over one piece of plastic and stick some tracing
paper over the top. Leave enough room for the shapes to slide about.
10. Patterns can be seen when the Kaliedoscope is held towards bright light and
looked into.
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How did this happen?
The light bounces back and forth between the foil mirrors. The reflections of the colored
shapes or beads make interesting patterns. To change the pattern, shake your
kaleidoscope so the shapes or beads move into new positions
Activity – 9
Alternate Method to Build a Kaleidoscope.
Material required: 3 plane mirrors, rubber bands, cardboard, tracing or greaseproof
paper, small coloured objects, scissors.
Method:
Hold the mirrors with the reflective sides pointing inwards.
Wrap the mirrors in a cardboard and fix with a rubber band.
Seal one end with tracing or greaseproof paper.
Put some pieces of coloured objects inside the tube and view.
Additional Activity:
1. Place two mirrors at 90º.
2. Put a small object like a table tennis ball between the mirrors and check the
number of images Produced.
3. Slowly reduce the angle between the mirrors and check how many images are
produced.
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Activity – 10
Making of a Periscope
Things required: two plane mirrors, cardboard or long box, cello tape, scissors.
Hoe to go about it?
1. Arrange the things as shown in the figure.
Questions:
1.
Explain how a periscope works.
2.
Suggest two uses of a periscope.
REFRACTION OF LIGHT
Refraction is usually defined as the bending of light as it passes from one medium to
another.
Take a glass slab and illuminate it with a beam of
light. Observe that it bends on the way in and
also on the way out. Light travels in a straight
line in any homogenous medium. However,
when it passes from one medium to another
medium it changes its direction. This change in
the direction of light is called refraction. Since
the densities of the media are different, light
25
travels with different speed in different media. Speed of light in vacuum is 300.000.000
km per hour. The decrease in the speed of light, when it passes from a rarer medium to
a denser medium, bends the light ray toward the normal to the boundary between the
two media.
Amount of bending depends on the refractive index of the media and the angle
between the light ray and the line perpendicular (normal) to the surface separating the
two media (medium/medium interface)
Each medium has a different refractive
index. The angle between the incident
light ray and the normal to the
boundary, at the point of incidence, is
called the angle of incidence. The
angle between the refracted light ray
and the normal is called the angle of
refraction.
Index of Refraction
We find the amount of refraction by using the refractive indices of the media. Refractive
index is the ratio of the speed of light in vacuum to the speed of the light in given
medium.
Approximate values of the indices of refraction of some common substances are given
below.
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Snell's Law
Snell's Law relates the indices of refraction ‘n’ of the two media to the directions of
propagation in terms of the angles to the normal.
If the incident medium has the larger index of refraction, then the angle with the normal
is increased by refraction. The larger index medium is commonly called the "internal"
medium, since air with n=1 is usually the surrounding or "external" medium.
Lenses, spectacles, magnifying glass, microscope, binoculars, telescopes, camera lenses,
prisms, projectors, endoscope, periscope etc are some of the applications of refraction of
light.
Demonstration of Refraction of Light by Water
Refraction at the water surface gives the "broken pencil"
effect shown above. Submerged objects always appear to be
shallower than they are because the light bends downward
towards the water.
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Activity - 11
The Aim: To use refraction to make a pencil seem like it is broken.
Material Needed: A pencil {fairly long}, Water, A drinking glass.
Method:
• Put water in the glass so that it is half to three quarters full.
• Stand the pencil in the glass.
• Look at it from the side of the glass.
• Then look at the pencil from above the glass.
Results:
• From the side of the glass, the pencil seems broken (check out the photos below –
the first photo was actually a photo of the same experiment using a glass bowl
with a straight side. The second photo shows the pencil in the glass). Notice the
difference in the two photos? The curved glass and water act as a magnifying
glass!
• When looking from above, the pencil seems like it is bending at the surface of the
water.
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• If you move your eyes up and down from the side of the glass to the top you will
get to a point where you will see what seems to be another pencil in the water.
The Conclusion
We can see all the effects mentioned in the results in the
picture below. The different effects are seen because we are
looking at the pencil from different angles. From above the
bent light will make image of the pencil above the real pencil
and so gives the impression that the pencil is bent. From the
side, the bent light will lower the image of the pencil, but
because we are looking at it from the side, it will seem to be
broken. The
largeness of the pencil
is because
of
magnification caused by the curved glass.
Following Ray diagrams show formation of images due to Refraction of light in water.
The bent pencil...
And the broken pencil...
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Activity – 12
The Aim: To create a rainbow with all its colors using refraction.
Experiment 1:
Equipment Needed: A torch, A fairly shallow glass transparent rectangular bowl, A
small mirror, A white piece of cardboard or thin tissue paper, Water
Method:
• Pour the water into the glass bowl, until it is about half or three quarters full.
• Balance the mirror against the side of the bowl and in the water at an angle, with
the mirror pointing up out of the water.
• Shine the torch from above the water, through the water onto the mirror.
• Hold the white cardboard above the mirror until you have the reflection from the
mirror on it.
• Alternatively, make a frame with the cardboard and stick the tissue over the
cardboard, much like a screen. When you get the image onto the tissue screen
you will be able to see it from the top as it will shine through, rather than have to
look underneath to see the colors.
• HINT: If you have a sunny window and you look at the reflection of it in the
mirror in the water, you can also see the colors of the rainbow.
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Experiment
2:
Equipment needed: A prism, A strong light source, A piece of cardboard or plastic
with a small slit on one side, A white piece of cardboard
Method:
• Set up the light source and the cardboard with the slit so that the light shines
through the slit creating a thin beam of light on the surface you are working on.
• Place the white cardboard so that the beam of light shines onto it.
• Place the prism in-between the light source and the white cardboard so that the
beam of light passes through it.
• Move the prism around while watching the white cardboard.
Results
Experiment 1: The cardboard has the colors of the rainbow shining on it.
Experiment 2: While watching the cardboard and moving the prism, we will notice
rainbow colors.
The Conclusion
In both situations above, the white light was split into its seven colors. This happened
because of refraction of light in the water in experiment 1 and in the prism in
experiment 2.
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Spherical Lenses and their Image Formation
If a piece of glass or other transparent material takes an appropriate shape, it is possible
that parallel incident rays would either converge to a point or appear to be diverging
from a point. A piece of glass that has such a shape is called as a lens.
Most lenses are spherical lenses: their two surfaces are parts of the surfaces of spheres,
with the lens axis ideally perpendicular to both surfaces. Each surface can be convex
(bulging outwards from the lens), concave (depressed into the lens), or planar (flat). The
line joining the centers of the spheres making up the lens surfaces is called the principal
axis of the lens.
If the lens is biconvex or plano-convex, a beam of light travelling parallel to the lens
principal axis and passing through the lens will be converged (or focused) to a spot on
the axis, at a certain distance behind the lens (known as the focal length). In this case, the
lens is called a positive or converging lens.
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If the lens is biconcave or plano-concave a beam of light passing through the lens is
diverged (spread); the lens is thus called a negative or diverging lens. The beam after
passing through the lens appears to be emerging from a particular point on the axis in
front of the lens; the distance from this point to the lens is also known as the focal
length, although it is negative with respect to the focal length of a converging lens.
The Anatomy of a Lens
Lenses can be thought of as a series of tiny refracting prisms, each of which refracts
light to produce their own image. When these prisms act together, they produce a
bright image focused at a point.
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REFRACTION THROUGH LENSES
If a symmetrical lens were thought of as being a slice of a sphere, then there would be a
line passing through the center of the sphere and attaching to the mirror in the exact
center of the lens. This imaginary line is known as the principal axis. An imaginary
vertical axis bisects the symmetrical lens into halves. The imaginary centre of the lens is
termed as optical centre .Light rays incident towards either face of the lens and
traveling parallel to the principal axis will either converge or diverge at or from a point
on the principal axis. This point is known as the focal point of the lens. The focal point
is denoted by the letter F. Each lens has two focal points - one on each side of the lens.
Unlike mirrors, lenses can allow light to pass through either face, depending on where
the incident rays are coming from. Subsequently, every lens has two possible focal
points. The distance from the lens to the focal point is known as the focal length
(abbreviated by f). A lens have an imaginary point referred to as the 2F point. This is
the point on the principal axis that is twice as far from the vertical axis as the focal point
is.
Refraction Rules for a Converging Lens

Any incident ray travelling parallel to the principal axis of a converging lens will
refract through the lens and pass through the focal point on the opposite side of
the lens.

Any incident ray travelling through the focal point on the way to the lens will
refract through the lens and become parallel to the principal axis.

An incident ray that passes through the optical center of the lens will pass
through the lens without any deviation or negligible deviation.
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Refraction Rules for a Diverging Lens

Any incident ray travelling parallel to the principal axis of a diverging lens will
refract through the lens and travel in line with the focal point (i.e., in a direction
such that its extension will pass through the focal point). These rays after
refraction appear to be diverging from the focus.

Any incident ray travelling towards the focal point on the way to the lens will
refract through the lens and become parallel to the principal axis.

An incident ray that passes through the optical center of the lens will pass
through the lens without any deviation or negligible deviation.
The rules merely describe the behavior of three specific incident rays. While there is a
multitude of light rays being captured and refracted by a lens, only two rays are needed
in order to determine the image location.
Converging Lens Image Formation
Converging lenses can produce both real and virtual images. Images are formed at
locations where any observer is sighting as they view the image of the object through
the lens. So if the path of several light rays through a lens is traced, each of these light
rays will intersect at a point upon refraction through the lens. While different observers
will sight along different lines of sight, each line of sight intersects at the image location.
The diagram below shows several incident rays emanating from an object - a light bulb.
Three of these incident rays correspond to our three strategic and predictable light rays.
Each incident ray will refract through the lens and be detected by a different observer
(represented by the eyes). The point where the refracted rays are intersecting is the
location of the image.
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In this case, the image is a real image since the light rays are actually passing through
the image location. To each observer, it appears as though light is coming from this
location.
Diverging Lens Image Formation
Diverging lens create virtual images since the refracted rays do not actually converge to
a point. In the case of a diverging lens, the image location is located on the object's side
of the lens where the refracted rays would intersect if extended backwards. Every
observer would be sighting along a line in the direction of this image location in order
to see the image of the object. As the observer sights along this line of sight, a refracted
ray would come to the observer's eye. This refracted ray originates at the object, and
refracts through the lens. The diagram below shows several incident rays emanating
from an object - a light bulb. Three of these incident rays correspond to our three
strategic and predictable light rays. Each incident ray will refract through the lens and
be detected by a different observer (represented by the eyes). The location where the
refracted rays are intersecting is the image location. Since refracted light rays do not
actually exist at the image location, the image is said to be a virtual image. It would only
appear to an observer as though light were coming from this location to the observer's
eye.
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Images of Objects That Do Not Occupy a Single Point
The above case was the formation of an image by a "point object" - in this case, a small
light bulb. The same principles apply to objects that occupy more than one point in
space. For example, a person occupies a multitude of points in space. As you sight at a
person through a lens, light is being reflected from each individual point on that person
in all directions. Some of this light reaches the lens and refracts. All the light that
originates from one single point on the object will refract and intersect at one single
point on the image. This is true for all points on the object; light from each point
intersects to create an image of this point. The result is that a replica or image of the
object is created as we look at the object through the lens. This is depicted in the
diagram below.
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IMAGE FORMATION BY A LENS FOR OBJECTS PLACED AT DIFFERENT
POSITIONS IN FRONT OF IT
CONVERGING/CONVEX LENS
Step-by-Step Method for Drawing Ray Diagrams
Case 1. An object located beyond the 2F point of a double convex lens.
1. Pick a point on the top of the object and draw three
incident rays traveling towards the lens. Using a
scale, accurately draw one ray so that it passes
exactly through the focal point on the way to the
lens. Draw the second ray such that it travels
exactly parallel to the principal axis. Draw the third
incident ray such that it travels directly to the exact center of the lens. Place
arrowheads upon the rays to indicate their direction of travel.
2. Once these incident rays strike the lens, refract them according to the three rules of
refraction for converging lenses. The ray that passes through the focal point on the
way to the lens will refract and become parallel to the principal axis. Use a scale to
accurately draw its path. The ray that traveled parallel to the principal axis on the
way to the lens will refract and pass through the focal point. And the ray that
traveled to the exact center of the lens will continue in the same direction. Place
arrowheads upon the rays to indicate their direction of travel. Extend the rays past
their point of intersection.
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3.
Mark the image of the top of the object.
The image point of the top of the object is the point where the three refracted rays
intersect. All three rays should intersect at exactly the same point. This point is
merely the point where all light from the top of the object would intersect upon
refracting through the lens. Of course, the rest of the object has an image as well
and it can be found by applying the same three steps to another chosen point.
4.
After completing the first three steps, only the image location of the top extreme of
the object has been found. Thus, the process must be repeated for the point on the
bottom of the object. If the bottom of the object lies upon the principal axis (as it
does in this example), then the image of this point will also lie upon the principal
axis and be the same distance from the mirror as the image of the top of the object.
At this point the entire image can be filled in.
The ray diagram above illustrates that when the object is located at a position
beyond the 2F point, the image will be located at a position between the 2F point
and the focal point on the opposite side of the lens. The image will be inverted,
diminished (smaller than the object), and real.
Similarly ray diagrams for the image formation by a convex lens for other position
of the object can be drawn. It should be noted that the process of constructing a ray
diagram is the same regardless of where the object is located. While the result of
the ray diagram (image location, size, orientation, and type) is different, the same
three rays are always drawn. The three rules of refraction are applied in order to
determine the location where all refracted rays appear to diverge from (which for
real images, is also the location where the refracted rays intersect).
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An object is located at the 2F point (fig 1)
An object is located between the 2F and the focal point. (fig 2)
Fig 1
Fig 2
Ray Diagram for Object Located in Front of the Focal Point
In the three cases described above - the case of the object being located beyond 2F, the
case of the object being located at 2F, and the case of the object being located between 2F
and F - light rays are converging to a point after refracting through the lens. In such
cases, a real image is formed. As shown above, real images are formed when the object
is located a distance greater than the focal length from the convex lens. A virtual image
is formed if the object is placed between the optical centre and the focal point of the
convex lens.
40
A ray diagram for the case in which the object is located the focal point and the optical
centre is shown in the diagram at the left. Observe that in this case the light rays diverge
after refracting through the lens. When refracted rays diverge, a virtual image is
formed. The image location can be found by tracing all light rays backwards until they
intersect. For every observer, the refracted rays would seem to be diverging from this
point; thus, the point of intersection of the extended refracted rays is the image point.
Since light does not actually pass through this point, the image is referred to as a virtual
image. Observe that when the object in located in front of the focal point of the
converging lens, its image is an upright/erect and enlarged image that is located on the
object's side of the lens. In fact, one generalization that can be made about all virtual
images produced by lenses (both converging and diverging) is that they are always
upright/erect and always located on the object's side of the lens.
Ray Diagram for Object Located at the Focal Point
A convex lens produces a real image is when an object
is placed beyond the focus and a virtual image when
an object is placed within the focal point of the lens
(i.e., in front of F). But what happens when the object is
located at F? That is, what type of image is formed
when the object is located exactly at the focus of a
converging lens? The diagram below shows two
incident rays and their corresponding refracted rays.
For the case of the object located at the focal point (F), the light rays neither converge
nor diverge after refracting through the lens. As shown in the diagram above, the
refracted rays are traveling parallel to each other. Subsequently, the light rays may
converge beyond the range of the lens hence forming an enlarged, real, inverted image
at infinity.
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Converging Lenses - Object-Image Relations
Hence to Summarise
1. When the object is located at a location beyond the 2F point, the image will always
be real, inverted and smaller in size than the object and located somewhere in
between the 2F point and the focal point (F) on the other side of the lens.
2. When the object is located at the 2F point, the image will also be located at the 2F
point on the other side of the lens. In this case, the image will be inverted and of
the same size as the object.
3. When the object is located between F and 2F point, the image will be formed
beyond the 2F point on the other side of the lens. The image will be inverted and
magnified/enlarged.
4. When the object is located at the focal point, highly magnified real and inverted
image is formed at infinity.
5. When the object is located between the focal point and the optical centre, the image
will always be located somewhere on the same side of the lens as the object. It will
be virtual, erect and highly magnified.
It might be noted from the above descriptions that there is a relationship between the
object distance and object size and the image distance and image size. Starting from a
large value, as the object distance decreases (i.e., the object is moved closer to the lens),
the image distance increases; meanwhile, the image height increases. At the 2F point,
the object distance equals the image distance and the object height equals the image
height. Eight different object locations are drawn in red and labeled with a number; the
corresponding image locations are drawn in blue and labeled with the identical
number.
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REVISION EXERCISE
1. For which positions of an object in front of a converging lens a real image is
formed?
2. With the help of a ray diagram show the formation of a real image, of the same
size as the object, by a convex lens.
3. A converging lens is sometimes used as a magnifying glass. Depict with the help
of a ray diagram the position of the object in front of the lens to produce the
magnified effect.
Diverging /Concave Lens
Step-by-Step Method for Drawing Ray Diagrams
The method of drawing ray diagrams for a double concave lens is described below.
1. Pick a point on the top of the object and draw three
incident rays traveling towards the lens.
Using a scale, accurately draw one ray so that it
travels towards the focal point on the opposite side
of the lens; this ray will strike the lens before
43
reaching the focal point; stop the ray at the point of incidence with the lens. Draw
the second ray such that it travels exactly parallel to the principal axis. Draw the
third ray to the exact center of the lens. Place arrowheads upon the rays to indicate
their direction of travel.
2. Once these incident rays strike the lens, refract them according to the three rules of
refraction for double concave lenses.
The ray that travels towards the focal point
will refract through the lens and travel parallel
to the principal axis. Use a straight edge to
accurately draw its path. The ray that traveled
parallel to the principal axis on the way to the
lens will refract and travel in a direction such
that its extension passes through the focal
point on the object's side of the lens. Align a straight edge with the point of
incidence and the focal point, and draw the second refracted ray. The ray that
traveled to the exact center of the lens will continue to travel in the same direction.
Place arrowheads upon the rays to indicate their direction of travel. The three rays
should be diverging upon refraction.
3. Locate and mark the image of the top of the object.
The image point of the top of the object is the
point where the three refracted rays intersect.
Since the three refracted rays are diverging,
they must be extended behind the lens in
order to intersect. Using a scale, extend each of
the rays using dashed lines. Draw the
extensions until they intersect. All three
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extensions should intersect at the same location. The point of intersection is the
image point of the top of the object. The three refracted rays would appear to
diverge from this point. This is merely the point where all light from the top of the
object would appear to diverge from after refracting through the double concave
lens. Of course, the rest of the object has an image as well and it can be found by
applying the same three steps to another chosen point.
4. Repeat the process for the bottom of the object.
If the bottom of the object lies upon the principal axis (as
it does in this example), then the image of this point will
also lie upon the principal axis and be the same distance
from the lens as the image of the top of the object. At this
point the complete image can be filled in.
Diverging Lenses - Object-Image Relations
The diagrams above show that for any position of an object in front of a concave
lens, the image formed is

located on the object' side of the lens

a virtual image

an erect image

reduced in size (i.e., smaller than the object)
Another characteristic of the images of objects formed by diverging lenses pertains
to how a variation in object distance affects the image distance and size. The
45
diagram below shows five different object locations (drawn and labeled in red)
and their corresponding image locations (drawn and labeled in blue).
The diagram shows that as the object distance is decreased, the image distance is
decreased and the image size is increased. So as an object approaches the lens, its
virtual image on the same side of the lens approaches the lens as well; and at the
same time, the image becomes larger.
REVISION EXERCISE
1. Which of the following, a converging lens or a diverging lens can be used to
produce
(a) a real image that has the same size as the object
(b) a virtual and diminished image?
Support your answer with ray diagrams.
2. The image of an object is found to be
(a) real and reduced in size.
(b) real and magnified
(c) Virtual and magnified
(d) virtual and diminished.
What type of lens is used to produce each image?
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The Lens Formula
We have seen above how to find the details of the image of an object, formed by the lens
by drawing appropriate ray diagrams. This, however, is not the only method available
for this purpose. It is possible to find all the relevant details of the image by doing
calculations based on a simple formula called the lens formula.
The Lens Formula
1/v- 1/u = 1/f
Here v and u denotes the positions of the image and object and f denotes the focal
lengths of the lens.
All the terms in this formula are algebraic quantities i.e. we have to attach a plus or
minus sign to them, on the basis of standard accepted sign convention. The sign
convention is based on the following rules.
1. The optical centre of the lens is the reference point or origin for measuring all the
distances.
2. As per the standard (co-ordinate geometry ) sign convention :
I.
We draw all the ray diagrams with the light rays propagating from left to
right.
II.
All distances (measured from the optical centre of points, situated to the left
/right of the optical centre), are attached a minus/plus sign.
III.
The focal length of the convex lens if taken with a positive sign while that of
concave lens is taken with negative sign.
IV.
The size of an object/image ,situated above the principal axis is taken with a
plus(minus)
It implies that size of a (real) inverted image would be taken with a minus
sign while that of a (virtual) erect image would be taken with a plus sign.
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Magnification
The ray diagrams, drawn above have shown that the size of the image is in general
different from that of the object. We can call the ratio of the size of image (formed by a
lens ), to the size of the object ,as the magnification due to lens.
Thus Magnification, m= size of the image (I)/ /size of the object (O)
We can calculate this through a formula. The magnification formula, for a lens is
m=v/u.
Here v and u denote, as before, the positions of the image and the object respectively.
Magnification, m as this formula indicates would again be an algebraic quantity i.e. it
would have a plus or minus sign associated with it As per the sign convention, stated
above we can say :

For a real image inverted and enlarged image of an object, the magnification m
would be a negative quantity whose magnitude is greater than one.

For a real inverted and diminished image of an object, the magnification m
would be a negative quantity whose magnitude is less than 1 i.e is a fraction

For a virtual erect and enlarged of an object, the magnification m would be a
positive quantity whose magnitude is greater than 1

For a virtual, erect and diminished image of an object the magnification m,
would be a positive quantity whose magnitude is less than 1,i.e,a fraction
Power of a Lens
A convex lens, as we know, usually converges an incident parallel beam of light to a
point on its principal axis. The distance of this point (called the focus) from the optical
centre of the lens, is known as its focal length.
As per our common perception of the term, a lens would be more powerful if it does its
converging(diverging) ‘job’ more effectively or quickly. Thus, for more ‘powerful’ lens,
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the focal length would have a smaller magnitude and vice –versa. We, therefore, define
the power (p),of a lens as the reciprocal of its focal length(F).Thus
Power (p) =1/focal length (f)
i.e p=1/f
The focal length, in SI units is measured in meters. The corresponding SI unit of power
(metre⁻1)has been given the name DIOPTER (D)
the power of a convex lens ,of focal length 25cm (=0.25 ) would therefore, be
(1/+0.25d)=+4d.For a concave lens of focal length 20cm(0.2M),the power would be(1/0.20d)=-5d.
It follows that a lens of power +10d is a convex lens of focal length(1/10)m i.e,10cm
and.a lens of power -2d,would be a concave lens of focal length(1/2)m,ie,50cm
Try guessing the object positions, for a convex lens, for which m is a negative quantity
with a magnitude greater / less than 1
Can we have m as a positive fraction for a convex lens?
Is it possible for m to be negative for a concave lens? Can m be a positive quantity
greater than 1 for this lens?
It is time now to do some actual calculations based on these formulae.
Solved Examples
Example 1: an object is kept at a distance of 25 cm (ii) 45 cm from a convex lens of focal
length 20 cm. find the position , magnification and nature of image formed in each case
Solution: As per the standard sign convention, the object has to be kept to the left of the
lens.hence,in case (i) u = position of the object = -25 cm.The focal length of a convex lens
is taken with a positive sign . hence
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F = +20 cm
Substituting these values in the lens formula
1/v – 1/u = 1/f
We get
1/v – 1/-25 = 1/+20
Therefore, 1/v = 1/20 – 1/25 = {5-4} / 100 = 1/100
Therefore, v = +100 cm
The image is thus formed to the right of the lens at a distance of 100 cm from it
Therefore, magnification, m = v/u = +100 / -25 = -4
The image size is, therefore, 4 times as large as the object. The negative sign, with m,
implies an Inverted Image.
The image is, the therefore, a REAL IMAGE.
In case (ii) , we have u=45 cm, f= + 20cm
1/V - 1/-45 = 1/20
1/v =1/20 - 1/45 = 9-4/180= 5/180=1/36
V= +36 cm
The image is thus formed to the right of the lens at a distance of 36 cm from it.
Magnification, m=V/U = +36/-45= -4/5 = -0.8
The image formed is, therefore, a diminished image and its size is 0.8 times that of the
object.
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The negative sign, with m, implies an inverted image; the image is, therefore, a real
image.
Example2: A convex lens, of focal length 30 cm, forms an image of an object at a
distance of 30 cm from it. What is/ are the possible values of the distance of the object
from it?
Solution: Here it is not specified as to whether the image formed is to the right (real
image) or to the left (virtual image) the lens. We would, therefore, need to look at both
the possibilities.
Case I- Let the (real) image be formed to the right of the lens. We then have
V= + 30 cm
Also f= + 30 cm
Now 1/V- 1/U= 1/f
1/30- 1/u= 0 or u – D
The object, in this case, would, therefore, be to the left of the lens and at an infinite (very
large ) distance from it.
Case II – Let the (Virtual) image be formed to the left of the lens. We would then have
V= -30 cm and f= + 30 cm
1/-30 – 1/u = 1/30
-1/u= 1/30 + 1/30 = 2/30 = 1/15
U= -15cm
The object, in this case, would, therefore, be on the left of the lens and at a distance of 15
cm from it. Since this distance is less than the magnitude of the focal length, a convex
lens, as we know, forms a virtual (erect and enlarged) image to the left of the lens.
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Example: What is the power of concave lens of focal length 40cm?
This lens forms an image of the object whose magnification m=+0.4. Where is the object
located?
Solution: We have f= -40cm = - 0.4m
(For a concave lens, the focal length is taken with a negative sign)
Power = 1/f = 1/-0.4D = -2.5D
The power, of the given concave lens, is, therefore, -2.5D
Now m=+0.4 = v/u
(the positive value of m implies an erect an therefore, a virtual image)
V= +0.4u.
Substituting values in the lens formula
1/v-1/u= 1/f,
We get
1/+0.4u -1/u= -1/40
Or 1.5/u = -1/40
U= -60 cm
The object is, therefore, at a distance of 60 cm from the lens and to the left of it.
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Total Internal Reflection
We have already studied about refraction, the observed effect of light waves changing
direction when entering a new medium, due to a change in the speed of the wave. We
found that the change in direction of the wave can be quantified using the refractive
indexes of the two materials.
Now, imagine a ray of light entering an optically less dense material, from an optically
denser one. What happens? The light ray bends away from the normal. As the
following diagram shows, the farther the incident ray is from the normal, the farther the
refracted ray will be from it as well.
However, with a small angular change in the angle of incidence comes a bigger change
in angle of refraction (due to the refractive indexes of the two materials).
Let’s move on to an extreme case of this situation: when the ray exiting the optically
denser material is refracted to such an extent that it is bent to 900 from the normal.
The angle of incidence in this special case is called the critical angle because beyond
this point, there is a difference in the behavior of the light. When the critical angle for
53
the two substances is exceeded, a phenomenon known as Total Internal Reflection
or T.I.R. occurs. This means that instead of exiting the optically denser material and
being refracted, the incident ray is reflected inside the material (i.e. internally).
After this point, normal laws of reflection are followed, by the ray, off of the surface
between the two materials.
The critical angle of a boundary can be found quite simply, using Snell’s Law, which
states:
where 1 and 2 correspond to the first and second media entered respectively, and
therefore where
corresponds to the angle of incidence, and
corresponds to the
angle of refraction. In the position of the critical angle, we know that the angle of
refraction, , is 90 . Therefore, sin
is equal to 1. The angle of incidence is of course the
critical angle, so we now have:
The critical angle, c, can therefore be found simply by knowing the refractive indexes of
the two materials. It is also important to note that T.I.R. takes place only at the interface
of an optically denser material with one that is optically less dense, and not vice versa.
54
Applications of Total Internal Reflection
Total Internal Reflection can be used as a way to make diamonds sparkle and
information travel rapidly on light waves through a fiber optic.
The cut of the diamond favors total internal reflection. Most rays
entering the top of the diamond will internally reflect until they reach
the top face of the diamond where they exit. This gives diamonds their
bright sparkle.
A fiber optic is a glass "hair" which is so thin that once light enters one end, it can never
strike the inside walls at less than the critical angle. The light undergoes total internal
reflection each time it strikes the wall. Only when it reaches the other end is it allowed
to exit the fiber.
Actually an optical fiber has two layers: a core made of a material of with a high
refractive index, and a second, outer layer with lower refractive index. The light waves
transmitted by an optical fiber are reflected off of the boundary between these two
substances, as shown in the diagram of a cross-section of a fiber below.
The smaller the refractive index of the cladding is compared to the refractive index of
the core, the smaller the critical angle is, allowing T.I.R. to take place in more conditions
(as it can be more often exceeded).
Optical fibers are used in a growing number of fields. In communication they are used
for carrying signals precisely, and at the speed of light. This is faster than the speed of
energy transmission by electrons, and therefore faster than electric signals. In medicine,
optical fibers are used by operating doctors to view previously inaccessible places, such
55
as the inside of a lung. Optical fibers are helpful in that they allow the transmission of
light to or from places not usually possible. Because they are fibers, they can be bent,
allowing light to be bent easily and precisely around many corners, without the use of
more clumsy devices such as mirrors.
History of Total Internal Reflection
In 1854, British physicist named John Tyndall discovered the principle of optical fiber
by watching a stream of water flowing out of a barrel. His observation was that water
was carrying light, this illusion, of course, was due to total internal reflection, where the
light was bouncing off the sides of the water stream because the angle at which the light
was hitting the sides of the stream were larger than the critical angle between the two
media (air and water). This illusion is actually quite a popular act in magic shows where
the magician "pours light" using a physics principle.
Activity- 13
Light in a Test Tube
Materials

long test tube (the longer the better)

laser pointer

powdered milk or a few drops of liquid milk

water in container so water can be poured into test tube
Procedure
1. Take a clean test tube and put a small amount of powdered milk in it (only a
pinch). Fill the test tube with water and shake to mix up the powder and water.
2. Make sure the outside of the test tube is clean and dry.
3. You will be using the laser pointer to shine into the test tube. MAKE SURE
THAT YOU NEVER SHINE THE LASER INTO YOUR OWN OR ANYONE
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ELSE'S EYES! Hold the laser pointer at the bottom of the test tube and shine it up
through the test tube. The powder in the water has large enough particles that
the laser will scatter light into your eyes so that you can see the path of the laser
light through the test tube. If you have time try shining light through a test tube
of clear water. What do you think you will see?
4. Now rotate the laser so its light reflects off the side of the test tube. What do you
see? Have one of your group members put a finger near the test tube where the
light hits the test tube. Do you see laser light on the finger? If you do, try to shine
the laser at a smaller glancing angle onto the side of the test tube. Eventually all
(or almost all) the light will be reflected. THE INTENSITY OF LASER POINTERS
IS SMALL ENOUGH THAT IT WILL DO NO DAMAGE ON YOUR SKIN, BUT
DO NOT SHINE THEM IN ANYONE'S EYES.
5. Try to shine the light on the sides so that it reflects several times before coming
out of the top of the test tube. You have demonstrated total internal reflection.
6. If the light disappears inside the test tube, you have most likely added too much
powdered milk. Pour this out and start over. If you don't see the laser light inside
the test tube, you need to add a little more powdered milk.
Questions:
1. Can you conclude that light travels in a straight line inside the test tube?
2. What is the effect of the powdered milk? Why is it needed?
3. What do you think would happen if the test tube were ten times longer? Would
you still be able to see the zigzag path of the light? Why or why not?
4. Why would a thin, clear plastic fiber be better than the test tube with water and
powdered milk? Would we be able to see the path of the laser light in the plastic?
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5. Were you able to shine the laser inside the test tube at an angle such that a lot of
the light came outside the test tube? If so, draw a diagram showing the angle of
the laser light path through the test tube.
Activity – 14
To Make a Pin-Hole Camera.
Material required: a shoe box, cello tape, aluminum foil, tracing paper and pin.
Method:
1. Cut off one end of the box and seal it with tracing paper.
2. Cut out a small part in front of the box and paste a piece of aluminium foil over
the opening.
3. Make a hole in the aluminium foil with a pin.
Observations and results:
1. What happens to the size of the image as the object is:
(a) moved further away from the pin-hole.
(b) moved closer to the pin-hole.
Applications of Refraction of light in nature
1. Formation of rainbows
When there are a few showers with a bit of sunshine, a band of colours can be seen
in the sky. In the school laboratory, glass prisms can be used to produce the band of
colours by viewing sunlight through them. This band of colours is called a rainbow.
We are going to see how a rainbow can be produced.
a) Glass prism
When light travels from one medium to another, it refracts. In other words, it
changes direction and moves at different speeds in different mediums. We also
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know that white light is made up of different colours. Different colors of light
have different wavelength. Except in vacuum and air, they travel at different
speeds in different mediums.
When we allow a beam of light to pass through a simple prism, it bends as it
enters the prism and also as it leaves the prism. The different colours of the white
light travel with different speeds inside the prism. The slanted surfaces of the
prism makes the different colours emerge separately, thus in addition to bending
light as a whole, a prism separates white light into its component colors. This
splitting of light into its component colours is called dispersion, where prism is a
dispersing element.
(b) Internal reflection
When the sunlight strikes a raindrop, the light is refracted because the light is
moving from one medium (air) to another (water). Drops of rainwater can
disperse light in the same way as a prism. When the light comes out of the water
droplet, the sun's rays (white light) have now been split into their component
wavelengths (colors). In this way, each individual raindrop behaves just like a
prism, dispersing white sunlight into its seven component colors. The rainbow
we see is thus a result of many, many raindrops bending the sun's rays.
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2. Real and Apparent Depth
Refraction causes the object to appear above where it is actually positioned. This fact
is responsible for swimming pools appearing shallower than they really are. Light
from the bottom of the pool bends as it enters the air as shown in the diagram
below. Since eye cannot see in this bended way it makes the floor of the pool or an
object lying on the floor, appear raised.
3. Mirages
A mirage is an optical illusion. On a hot day, there seems to be a film of water on the
road. This is due to refraction of light. The air near the ground becomes hotter and
less dense than the air above. This difference in densities causes light rays from the
sun to bend.
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In desert, one may see images of distant trees during high temperature. Since in
nature, images of trees are formed only in water, therefore one is tricked into
believing that a tree is situated near some water body and its reflection is being seen.
However, on reaching the spot, it is found that there is no water.
When the sun is high in the sky, the sand gets heated first and then the layers of air
above it. The rays from the trees travel from an optically denser air layer to a rarer
layer and hence bend away from the normal. This bending continues and a stage is
reached where the angle of incidence becomes greater than the critical angle and
total internal reflection takes place. The totally reflected rays that reach the eyes
appear to come from a point on the ground where the image of the tree is formed.
Thus one sees an inverted image of the tree though there is no water around.
5. Blue Sky
The blue color of the sky is caused by the scattering of sunlight off the molecules of
the atmosphere. This scattering, called Rayleigh scattering, is more effective at short
wavelengths (the blue end of the visible spectrum). As light moves through the
atmosphere, most of the longer wavelengths, for example red, orange and yellow
light pass straight through. However, much of the shorter wavelength light is
absorbed by the atmosphere. The absorbed blue light is then radiated in different
directions. It gets scattered all around the sky. Whichever direction you look, some
of this scattered blue light reaches you. Since you see the blue light from everywhere
overhead, the sky looks blue.
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Activity 15
Experiment to Show the Recombination of White Light
Place a prism (P1) on a table and a screen behind it. Allow a narrow beam of light to be
incident on the prism (P1). The white light gets dispersed and we obtain a band of seven
colors on the screen.
Recombination of the Dispersed White Light
Now remove the screen and place another prism P2 of the same material in the opposite
direction. Place a white light screen behind P2. A spot of white light appears on the
screen. Thus the second prism has recombined the dispersed light.
The Human Eye and the Camera
The camera and the human eye have a lot in common. This might sound strange since
the camera is made by man while the human eye is the work of God. The table below
shows how different parts of the camera are similar, or perform functions similar to
those performed by the human eye.
Human eye
Camera
Function
converging lens
converging lens
bends and focuses the light
retina
Film
where the image is formed
Iris
shutter or diaphragm
controls the amount of light
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Picture of a camera
How does the human eye function?
How we see
The images we see are made up of light reflected from the objects we look at. This light
enters the eye through the cornea. Because this part of the eye is curved, it bends the
light, creating an upside-down image on the retina (this is eventually put the right way
up by the brain).
Focusing on a nearby object
Focusing on a distant object
What happens when light reaches the retina?
The retina is a complex part of the eye, but only the very back of it is light-sensitive.
This part of the retina has roughly the area of a 10p coin, and is packed with
photosensitive cells called rods and cones. These allow us to see images in colour and
detail, and to see at night.
63
Cones are the cells responsible for daylight vision. There are three kinds - each
responding to a different wavelength of light: red, green and blue. The cones allow us to
see in colour and detail.
Rods are responsible for night vision. They are sensitive to light but not to colour. In
darkness, the cones do not function at all.
Focusing the image
The lens focuses the image. It can do this because it is adjustable - using muscles to
change shape and help us focus on objects at different distances. The automatic focusing
of the lens is a reflex response and is not controlled by the brain.
Sending the image to the brain
Once the image is clearly focused on the sensitive part of the retina, energy in the light
that makes up that image creates an electrical signal. Nerve impulses can then carry
information about that image to the brain through the optic nerve.
When the iris opens, light from an object enters and passes through a converging lens.
The converging lens focuses the light on the retina, which acts as a screen where an
image is produced. In the eye, the lens does not move back and forth, but simply
becomes thinner or thicker. The image produced is real, inverted and smaller than the
object.
How does a camera function?
When the shutter is opened, light from an object enters and passes through a
converging lens. The converging lens focuses the light on the photographic film where
an image is produced. In a camera, the lens has a fixed shape and size. It is moved back
and forth in order to focus. The image produced is real, inverted and smaller than the
object.
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Images are formed in a camera by refraction in a manner similar to image formation in
the eye. However, accommodation to image closer objects is done differently in the eye
and camera.
There are four types of defect of the Eye: Myopia, Hypermetropia, Presbyopia and
Astigmatism. Below are given the nature of the defect, its causes and corrective
measures:Myopia:
Nearsightedness, also called myopia is common name for impaired vision in which a
person sees near objects clearly while distant objects appear blurred. In such a defective
eye, the image of a distant object is formed in front of the retina and not at the retina
itself. Consequently, a nearsighted person cannot focus clearly on an object farther
away than the far point for the defective eye.
Causes:
This defect arises because the power of the eye is too great due to the decrease in focal
length of the crystalline lens. This may arise due to either
(i)
Excessive curvature of the cornea, or
(ii)
Elongation of the eyeball.
65
Correction:This defect can be corrected by using a concave (diverging) lens. A concave lens of
appropriate power or focal length is able to bring the image of the object back on the
retina itself.
Method for calculating the power of the corrective lens: For calculating the required power of a corrective lens, we first find the power of the
eye at its far point. Then, we select a corrective lens of appropriate power to move the
far point to infinity. We then use the thin lens formula, written in terms of power P of
the lens as
The image distance v of the eye can be taken as 0.02 m approximately.
Myopia (Nearsightedness)
Hypermetropia:
Farsightedness, also called hypermetropia, common name for a defect in vision in which
a person sees near objects with blurred vision, while distant objects appear in sharp
focus. In this case, the image is formed behind the retina.
Causes:
This defect arises because either
(i) The focal length of the eye lens is too great, or
(ii) The eyeball becomes too short, so that light rays from the nearby object, say at
point N, cannot be brought to focus on the retina to give a distinct image.
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Correction:This defect can be corrected by using a convex (converging) lens of appropriate focal
length. When the object is at N’, the eye exerts its maximum power of accommodation.
Eyeglasses with converging lenses supply the additional focussing power required for
forming the image on the retina.
Hypermetropia (Farsightedness)
Astigmatism:
Astigmatism, a defect in the outer curvature on the surface of the eye that causes
distorted vision.In astigmatism, a person cannot simultaneously focus on both horizontal
and vertical lines.
Causes:
This defect is usually due to the cornea that is not perfectly spherical. Consequently, it
has different curvatures in different directions in vertical and horizontal planes. This
results in objects in one direction being well-focussed, while those in a perpendicular
direction not wellfocussed.
Correction:This defect can be corrected by using eyeglasses with cylindrical lenses oriented to
compensate for the irregularities in the cornea.
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Astigmatism
Presbyopia:
Presbyopia, progressive form of farsightedness that affects most people by their early
60s. The power of accommodation of the eye decreases with ageing. Most people find
that the near point gradually recedes.
Cause and Cure:
It arises due to the gradual weakening of the ciliary muscles and diminishing flexibility
of the crystalline lens. Simple reading eyeglasses with convex lenses correct most cases
of presbyopia. Sometimes, a person may suffer from both myopia and hypermetropia.
Such people often require bi-focal lenses. In the bi-focal lens, the upper portion of the bifocal lens is a concave lens, used for distant vision. The lower part of the bi-focal lens is
a convex lens, used for reading purposes.
Eye defects and how to correct them
Far (long) sightedness
A person who is far-sighted cannot focus on objects
that are fairly close to the eye. This is because the
lens cannot adjust to become thicker in the middle.
This causes the image to be formed beyond the
retina. This defect can be corrected by wearing a
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converging lens in the spectacles as shown in the diagram below.
Near (short) Sightedness
A person who is near-sighted cannot see distant
objects clearly. This is because the lens cannot adjust
to become thinner in the middle. This causes the
image to be formed in front of the retina. This defect
can be corrected by wearing a diverging lens in the
spectacles as shown in the diagram.
EXTENSION – Electromagnetic Waves
Do you listen to the radio, watch TV, or use
microwave oven? All these devices make use of
electromagnetic waves. Radio waves, microwaves,
infrared, visible light, ultraviolet, X rays and Gamma
rays are all examples of EM waves that differ from
each other in wavelength and frequency.
TIME VARYING ELECTRIC AND MAGNETIC FIELDS WHICH ARE MUTUALLY
PERPENDICULAR TO EACH OTHER AND ALSO TO THE DIRECTION OF
PROPAGATION ARE CALLED ELECTROMAGNETIC WAVES
Properties of EM Waves

EM waves are transverse in nature.

They do not require any medium to travel.

They all travel with the speed of 3 x 10 8 m/sec
For details please refer to the PPT – EM waves.
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Worksheet 1
Reflection of Light
1.
The diagrams, given below, are incorrect. Point out the reason that makes them
incorrect and redraw the correct diagram.
a.
________________________________________________________________________
2.
State the law of reflection and use it to find the value of angle of reflection for a ray
incident normally on a plane reflecting surface..
Observe the diagram at the right.
Which angle corresponds to the
(a)
angle of incidence ____.
(b)
angle of reflection ____.
(c)
When an incident ray of light makes an angle of 23° with the mirror surface
what would be the angle of reflection ?
6.
We usually observe the reflection of sun, from the windows of distant houses only
around dawn or dusk times. Why? Explain your answer by drawing appropriate
light rays on the given diagram.
70
7.
The diagrams, given below show a protractor (having markings at 10° intervals)
embedded in the setup. Draw the reflected ray in each case and state the
corresponding values of the angle of incidence and the angle of reflection.
Worksheet 2
1.
What is the difference between diffuse reflection and regular reflection? State the
conditions under which each is likely to occur?
2.
Do light rays continue to obey the laws of reflection when they undergo diffuse
reflection?
3.
Only one of the following diagrams corresponds to diffuse reflection? Identify the
diagram.
71
4.
When a beam of light undergoes diffuse reflection, would it be correct to say that
the individual rays within the beam do NOT follow the law of reflection. Justify
your answer.
5.
Five rays of light are approaching a microscopically rough surface as shown
below. For each incident ray, draw the approximate position of the normal line
and the corresponding reflected ray of light.
Worksheet 3
1.
The diagram, given below, shows an object (denoted by a dark circle) kept in front
of a plane mirror. For the different light rays incident on the mirror, use
appropriate
geometrical
devices
to
draw,
accurately,
the
reflected
ray
corresponding to each of these incident rays.
2.
Using dotted lines, extend each of the reflected rays in the above diagram and
check whether or not do all of these extended dotted lines intersect at the same
location? What is the name given to the common intersection location point? How
does its perpendicular distance, from the mirror, compare with the perpendicular
distance of the object from the mirror?
72
3.
For the images formed by a plane mirror, identify the statements that are logically
valid as per the laws of reflection.
a. The location of an image is different for different observers.
b. Observers at different locations will sight along different lines at the same
image.
c. Every image is located on the mirror surface and at the same location for
different observers.
d. Every image is located on the mirror surface, but at a different location for
different observers.
e. All observers (regardless of their location) will sight at the same image
location.
4.
The path of four incident rays, originating from an object, incident on a mirror, are
as shown below. Identify the letter which corresponds to the location of the image.
5.
The path of four reflected rays, corresponding to some incident rays emerging
from an object and getting reflected from a plane mirror, are shown below.
Identify the letter on the right which corresponds to the location of the image.
Justify your answer.
73
Worksheet 4
1.
The optical density is the property of a medium that provides a relative measure
of the speed at which light travels in that medium. Does the speed of propagation
of light increase or decrease with an increase in the optical density of the medium?
2.
The index of refraction value of a transparent material is a numerical value that
provides a relative measure of the speed of light in that particular material. How
does the speed of light, in a medium, depend on its index of refraction value?
3.
4.
Calculate the speed of light in the following materials. (Given c = 3X 108m/s)
a. water (n = 1.33):
b. glass (n = 1.50):
c. ice (n = 1.31):
d. diamond (n = 2.42):
How does the path of a ray of light change when it goes from a medium having a
greater speed of light to a medium having lower speed of light?
5.
Consider the refraction of light in the five diagrams below. Compare the optical
densities of the two media for each of the five diagrams.
6.
Consider the diagram at the right in answering the following
(a) How many boundaries have been shown in the diagram?
(b) Does light propagate faster in medium 1 as compared to
medium 2?
(c) Does light propagate slower in medium 2 as compared to medium 3.
74
7.
In each diagram, draw the "missing" ray (either incident or refracted).
8.
A ray of light is shown passing through three consecutive
layered materials. Observe the diagram carefully and arrange
the three materials in order of increasing index of refraction.
9.
The apparent location of a fish is shown in the
diagram. Is the real position of the fish different
from the apparent position? What is the reason
for this difference? To successfully catch the fish
where should the boy aim below, or above the
position, where the fish appears to be? Explain.
75
Worksheet 5
1. A convex lens, of power +5D, has an object kept at a distance of 40 cm from it.
Find the position magnification and nature of the image formed by the lens.
2. A convex lens, of focal length 25cm, forms an erect image of an object whose size is 5
times that of the object.
Find the position of the object. Check your calculated result with a ray diagram
drawn to scale.
3. A lens, with a focal length of magnitude 40cm, is to be used to get an image half the
size of the object.
What can be the type of the lens used? Where would the object be located with
respect to the lens?
(Hint: The image formed being a diminished image, we can use either a convex lens
(+ve focal length) or a concave lens (- ve focal length). The convex lens would forms
a real, inverted (m is –ve) and diminished image. The concave lens would form a
virtual, erect (m is +ve) and diminished image)
4. Find the position, magnification and nature of the image of an object formed by a (i)
convex lens (ii) Concave lens. When the focal length has a magnitude of 30cm and
the object is at a distance of 45 cm from the lens in both the cases.
5. A lens, of power -4D, has an object kept at a distance of 25 cm from it.
Find the position, magnification and nature of the image formed by this lens.
Draw a ray diagram to support your calculated results.
76
Revision Worksheet
Reflection
1. Three cards, with holes, are arranged as shown.
a. Would a person be able to see the lamp, if the
three cards were set up as shown?
b. When one of the cards is moved sideways;
what would the observer see?
c. Does this experiment reveal some important
property of light?
2. Suggest an experiment to show that light travels in straight
lines.
Describe its essential details using an appropriate labelled
diagram.
3.
A ray of light strikes a plane mirror as shown
(i) Draw the path taken by the ray in the diagram and state
the underlying law.
(ii) Name an instrument that works on the basis of reflection
of light from mirrors.
4.
A ray of light is incident onto a plane mirror in a periscope.
(i) Complete the ray diagram.
(ii) Give one use of a periscope.
5. A simple periscope was designed by
using two plane (flat) mirrors as
shown If an observer looks through
the periscope at the word ‘Science’
77
written on a card pinned to the laboratory wall. Would the pupil see first image or
the second image through the periscope?
Justify your answer.
6. Why is the word AMBULANCE painted in reverse, on the front of the
ambulances?
7. Look at the photograph of the wader, a bird that feeds in shallow
water.
(i) Is the image of the bird shown here, being produced by
reflection or by refraction?
(ii) Give a reason for your answer.
Refraction
8. Explain, in brief the meaning of the term refraction of light?
9. Give an application of the phenomenon of light.
10. Suggest another way in which the direction of a light ray can be
changed apart from refraction.
11. Three narrow beams of light (rays) hit a lens, as shown
Which of these rays that passes through the lens
without bending,
Which rays are bent by the lens in the diagram.
Dispersion
78
12. A narrow beam (ray) of white light is incident on the equipment A. It emerges out as
shown.
The light passes through A to form a band of colours.
(i) Name the piece of equipment labelled A.
(ii) Name the colour labelled B.
13. When a ray of white light enters a triangular glass prism,
and passes through it and emerges as a band of coloured
light.
What does the experiment illustrated here, tells us about
the composition of white light?
(i)
Name the phenomenon associated with this separation of white light into
different colours?
(ii)
What name is given to the band of coloured light produced?
(iii) What is the colour of the light labelled X and the colour of the light labelled Y at the
two extreme ends of the band of light emerging out of the prism.
14. When Thunder and lightning occur at a place,
(i) Which is detected first, the flash of lightning or the clap of thunder?
(ii) What does this tell us about the speed of light with respect to the speed of
sound?
15. If one is viewing a fireworks display from a distance, he/she sees the fireworks
explode before hearing the sound of their explosions, why?
16. A light ray, incident on one side of a glass block ,passes from
air to glass on this side and emerges out from glass to air , on
the other end.Which of the rays P, Q, R or S, represents the
path taken by light ray on leaving the glass? State the reason,
if any, for your choice.
79
General Questions
1. What is meant by a luminous object? Give examples.
2. Give two observations that suggest that light travels in straight lines.
3. Explain, using a diagram, how a shadow of a person is formed on a sunny day.
4. What is the name given to the change in direction of light as it enters a prism?
5. Describe, with the aid of a diagram, an experiment to show the refraction of light as
it passes through a lens.
6. Draw diagrams to show how(i) a Concave Lens, and (ii) a Convex Lens works.
7. Why do rainbows often appear after a heavy shower of rain?
8. Explain: the only part of your body that you can kiss in the mirror is your mouth?
80
Rubrics of Assessment for Learning
Unit 4 - Light

Parameter
Beginning
Approaching
Meeting
Exceeding
Learner is able to
(1)
(2)
(3)
(4)
Understand and describe the
nature of Light

Differentiate between regular
and diffused reflection

State and explain the laws of
reflection

Understand and illustrate
Image formation by plane
mirror

Understand and explain the
phenomenon of refraction.

Explain refractive Index

Explain, understand and
illustrate the image
formation by lenses

Practically apply the laws of
reflection and refraction.

Explain and understand the
phenomenon of dispersion

Explain scattering of light

Explain the structure and
function of human eye

Understand the causes and
81
correction of common eye
defects
Resources
http://www.enchantedlearning.com/subjects/anatomy/eye/label/labeleye.shtml
www.youtube.com/watch?v=kc2o73FyN3I
http://hyperphysics-astro.gsu.edu
http://www.good-science-fair-projects.com
www.physicsclassroom.com
en.wikipedia.org/wiki/light
82
Chemistry
Carbon and its Compounds
Unit-4
SYLLABUS COVERAGE
Core
S
Y

A
B
U
S
Names and structure of hydrocarbons,

Structural isomerism.
alcohols ,carbonyl compounds ,carboxylic

Natural macromolecules-
acids upto ‗carbon atoms per molecule
Structure of Proteins, Fats

Homologous series and functional group
and Carbohydrates.

Fractional distillation of petroleum and
L
L
Extension

Hydrolysis of Fats,
uses of fractions
Proteins and

Alkanes-Properties and Bonding
Carbohydrates.

Alkenes- Preparation and properties

Difference between saturated and
unsaturated hydrocarbons.

Addition polymerisation

Alcohols-Preparation by fermentation

Use of alcohol as a fuel and solvent

Acids-Preparation by oxidation of alcohols

Esterification

Condensation Polymerization- Structure of
Nylon and Terylene

Pollution caused by non biodegradable
plastics.
83
Matrix
CONTENT
INTENDED LEARNING
SKILL
Communicating
1.Warm up-
Students appreciate that the number
Whole class
of compounds formed by carbon out Inferring
approach, for
numbers the compounds formed by
brain storming,
all of the elements known put
ask initiating
together.
questions
suggested in
teacher notes
2.Pre content
Students participate in a group
Comparing
activity in which the goal is to
Communicating
discover the common thread
Inferring
between a series of illustrations.
Content-1
 Appreciate that some elements Predicting
Bonding in
form compounds by sharing of Hypothesizing
Carbon
electrons to form covalent bonds.
Compounds

Understand reasons for tetra
valence of carbon.

Correlate the bonds formed as
single, double or triple to the
number of pairs of electrons
shared between the atoms.

Learn
writing
structure
for
electronic
dot
formation
of
covalent compounds.

Worksheet-1
84
Classifying

Content -2
Learning by doing,
to IUPAC system of
Observation
nomenclature and also derive
their structures from the given
Nomenclature of
Organic
Name the compounds according
names.

Worksheet-2

learn that a series of compounds
Experiment/ Laboratory
Homologous
with same functional group is
Manage, record and
series
called Homologous series.
communicate
discover that in a homologous
information
Compounds
Content-3

series the difference in the
formula of two adjacent
compounds is of ‗-CH2‘ and
difference in molecular mass is ‗14 u‘.

appreciate that chemical
properties of members of a
Homologous series are same.

formulate the general formula
for a homologous series.

discover the next members of a
Homologous series.
ACTIVITY: Build models of
organic molecules in which
carbon atoms may bond with
each other as well as with atoms
of other elements.
85
Worksheet-3
Content-4
Understanding of the principle of
Resourcefulness, careful
fractional distillation and its
observation, application,
Fractional
application.
distillation of
Activity: Analysing the graph for
petroleum
components of a sample obtained
after fractional distillation.
Worksheet-4
Content-5
Classification of haydrocarbons.
Resourcefulness, careful
The Physical and
Properties of hydrocarbons.
observation, application,
Chemical
Uses of hydrocarbons in our day-to-
Making connections,
Properties of
day life.
Classification
Hydrocarbons
Activity: properties of
Observation and
hydrocarbons
conclusions
Worksheet-5
Setting up an experiment.
Preparation and properties of
Reasoning,
alcohols.
Making connections,
Uses of alcohols in our day-to-day
Classification
Content-6
Alcohols
life.
Activity: Properties of alcohols
Content-7
Preparation and properties of
Reasoning,
Carboxylic Acid
carboxylic acids.
Making connections,
Uses of carboxylic acids in our day-
Classification
to-day life.
Observation and
Activity- properties of carboxylic
conclusions
acid.
Setting up an experiment.
86
WORKSHEET-6 for content 5 to 7.
Content-8

Polymes-Nylon
and Terylene

Describe macromolecules in
Learning by observation
terms of large molecules built up
and reasoning.
from small units (monomers).
Application
Deduce structures of different
knowledge
of
macromolecules having different
units and/or different linkages.

Describe the pollution problems
caused by non-biodegradable
plastics
Post content
Name some typical uses of plastics
Learning
and of man-made fibres
technological
Extension activities
advancements
about
Revision worksheet
Assessment Worksheet.
Learning outcomes - Core
At the end of this unit students would be able to –

Appreciate that some elements form compounds by sharing of electrons to form
covalent bonds.

Understand that carbon with four valence electrons forms only covalent bonds.

Understand reasons for tetravalence of carbon.

Correlate the bonds formed as single, double or triple to the number of pairs of
electrons shared between the atoms.

learn writing electronic dot structure for formation of covalent compounds.

calculate and identify the number and types of bonds in a compound formed by
sharing of electrons.
87

write structures of organic molecules in various ways and classify the organic
compounds

name the compounds according to IUPAC system of nomenclature and also
derive their structures from the given names.

learn that a series of compounds with same functional group is called
Homologous series.

discover that in a homologous series the difference in the formula of two
adjacent compounds is of ‗-CH2‘ and difference in molecular mass is ‗-14 u‘.

appreciate that chemical properties of members of a Homologous series are
same.

formulate the general formula for a homologous series.

discover the next members of a Homologous series.

Describe petroleum as a mixture of hydrocarbons and its separation into useful
fractions by fractional distillation.

Recall chemical properties of hydrocarbons, alcohols and carboxylic acids.

Write chemical equations for the organic reactions with main products formed
during the reaction.

Draw correlations between the type of functional group and the chemical
properties of organic compounds.

Recognise various reagents like -a reducing or an oxidizing agent used in organic
reactions.

Describe macromolecules in terms of large molecules built up from small units
(monomers).

Deduce the structure of the polymer product from a given alkene and vice versa

Deduce structures of different macromolecules having different units and/or
different linkages.

Name some typical uses of plastics and of man-made fibres

Describe the pollution problems caused by non-biodegradable plastics
88
Learning outcomes – Extension
At the end of this unit students would be able to –

Describe and identify structural isomerism

Describe proteins as possessing the same (amide) linkages as nylon but with
different units

Describe the hydrolysis of proteins to amino acids (Structures and names are not
required.)

Describe fats as esters possessing the same linkage as Terylene but with different
units

Describe soap as a product of hydrolysis of fats

Describe the acid hydrolysis of complex carbohydrates (e.g.starch) to give simple
sugars

Describe the fermentation of simple sugars to produce ethanol(and carbon
dioxide) (Candidates will not be expected to give the molecular formulae of
sugars.)
89
Cross Curricular Links
SUBJECT
VISUAL ART
ACTIVITY
Making models of isomers of carbon compounds.
Depicting the 3-D, structure of simple hydrocarbons.
Writing dot and cross structures.
English
Writing a project report on properties of carbon which
make it different from other elements of the periodic
table.
Mathematics
Counting the number of possible isomers for a given
molecular formula.
Analysis of the graph for fractional distillation of crude
oil.
ICT
Power point presentations.
Research work done by the students.
Video:
http://www.youtube.com/watch?v=Kjn5Ht0Vn30
Social sciences
Need for bonding to be stable related to need of a
family and society.
90
Teachers’ Notes
For Teacher Student Support Material (TSSM)
For a warm up to this unitTeacher may ask the students to make a list of things we use or consume in our day to
day life, and classify them on the basis of material of which they are made, we find that
many products are made up of carbon compounds.
Things made of metal
Things made of
Others
glass/clay
Chasis of cars and other
Crockery Tumblers,
Toothpaste, comb,
vehicles Utensils,
Show pieces etc.
Toothbrush(plastic),
furniture etc.
Food, soap, hair oil,
Cooking oil,
clothes, shoes (leather),
books etc.
Most articles in the ‘Others‘ column are carbon compounds, thus carbon forms most of
the things around us and not to last all the living things. A simple method to
distinguish the carbon compounds from others is to burn them, when most of the
carbon compounds on burning produce carbon dioxide.
Pre-content activityAfter the initial warm up the teacher may ask the students to fill the KWL sheets in
groups of two and let the students discuss their need to learn from this unit.
Let the students perform the following group activityAim - Discover a common thread between the given series of illustrations.
Classroom dynamics- Each student is given a different illustration and only allowed to
describe the illustration verbally to members of the group. The activity presents an
91
opportunity to observe how group dynamics evolve and how the participatory process
is enhanced by a common goal and a nonthreatening environment of collaboration. This
provides the impetus for the activity described herein.
The Activity
Create a series of 8.5 in. × 11 in. illustrations from a commercially available clip art
library (Figure 1), in which each successive picture focuses on one detail from the
previous picture. Along with scenes from everyday life, each illustration contains some
snippet of organic chemistry such as a molecule from an object in the scene, a reaction
from the course, or a key concept. Divide the class into equal groups of 10–15 students
and give each student a page from the set of illustrations. Tell the students that, aside
from organic chemistry, the series of illustrations have another thing in common and
that the contest is to see which group can discover this common thread first. The one
stipulation is that students cannot visually reveal their assigned illustration to a fellow
group member. Rather, they are only allowed to describe the illustration verbally to one
another. Care is taken to randomize the illustrations prior to distribution.
Discussion
The activity takes approximately 15 minutes after each student has been assigned an
illustration. In general the group dynamics develop slowly, with each student turning
to his or her neighbor and inquiring about the contents of the other‘s
illustration.Eventually, small subgroups of 3–4 students form. Within these subgroups
of the main group, one student describes his or her illustration to the entire group.
Eventually one student will recognize that an item from the orator‘s description also
appears in his or her illustration. As students in the group begin to realize that each
illustration contains an object from another illustration, the subgroups become larger
until the common thread is solved. The ultimate ―winner‖ is the group that correctly
arranges itself in the order of the illustrations as shown in Figure 1. Instructors will
want to take note of any leaders that emerge, as this can aid in the assigning of small
92
groups for future collaborative classroom activities. To solve the puzzle, it is almost
better to not know any organic chemistry (i.e., to do the activity at the beginning of the
semester), so that students do not dwell too much on the organic structures and
concepts of each illustration. In any event, the activity creates a situation in which every
group member is required to contribute to the conversation.
Summary
This activity can serve many purposes. It can serve as an icebreaker at the beginning of
a course. It can be used to reveal the prominence of organic chemistry in everyday life.
It can be used as a jumping point for group work or other collaborative activities in a
class. Finally, it can be used as a fun activity in celebration of a year of organic
chemistry.
http://ochem.jsd.claremont.edu/pubs.dir/JCE_2004_p513(Poon).pdf
93
Content-1
Bonding in Carbon Compounds
• Teacher may give the worksheet 1 to the students after teaching bonding in carbon
taking care to cover the following points-
Carbon has 4 electrons in outermost shell so it can neither give electrons nor
accept electrons. So it form bond by sharing its valence electrons with atoms
of other elements. The shared electrons belong to the outermost shells of both
the atoms and lead to both atoms attaining the noble as configuration.
-
The bond formed by sharing electrons is called covalent bonds.
-
Not only carbon, but many other elements form molecules by sharing
electrons.
-
When covalent bond is formed by sharing one pair of electron is called single
covalent bond and is represented by a line between the two atoms.
-
When covalent bond is formed by sharing two pair of electrons is called
double bond and is represented by two lines between the two atoms.
-
When covalent bond is formed by sharing 3 pairs of electrons between two
atoms is called Triple bond and is represented by three lines between the two
Suggestive remediation
Some students may not know the molecular formula of methane, ethene and
propyne, teacher may help them.

Some students may not be able to recollect the number of valence electrons in
carbon and hydrogen atom and their placement, they may be helped.

Some students may miss the bond while counting different type of bonds.
94
Content-2
Nomenclature of Organic Compounds

The teacher may give the worksheet-2 to the students after explaining the
concept of nomenclature of hydrocarbons.

A few students may not be able to write the correct name or structure of the
simple hydrocarbons, they may be helped by the teachers.

Some students may confuse while putting prefix for alcohol and aldehyde
teacher should take care to explain the difference to the students.

Some students may forget to count the carbon of carboxyl group, in the base
chain.

The students may be asked to prepare a written report of the project undertaken,
using the format given below:
*
Aim of the Project
*
Introduction
*
Theory
*
Procedure
*
Result
Content-3
Homologous series
Through this Activity, students will explore and build models for molecules in which
carbon atoms may bond with each other as well as with atoms of other elements. Take
care that students keep in mind that each carbon atom, can make only four covalent
bonds. When carbon atoms bond with each other, they form a carbon chain.
Compounds containing carbon chains and same general formula form a homologous
series.
A homologous series is a set of compounds that conforms to a general mathematical
ratio and general structure. As students build models of the members of some
95
homologous series, they will discover some series contain functional groups. A
functional group is responsible for characteristic properties and reactions. Students can
recognize a functional group by the collection and arrangement of specific atoms.
Content-4
Fractional distillation of petroleum
Explain the formation of petroleum to the students and through the given activity
explain the process of distillation to separate various fractions of petroleum. Explain to
the students the difference between the process of fractional distillation and simple
distillation. They should understand that the components of a mixture with a difference
in boiling points as more than 250C may be separated using simple distillation whereas
for components having their boiling points quite close fractional distillation process is
to be used.
Explaining the distillationThe teacher may use the following similySurely, you must be fond of fettucini and spaghetti. While eating these, you find that
short strands are easy to pull out than longer ones. This helps to explain the distillation
of crude oil. Instead of spaghetti think of molecules.
As the oil is heated, the small molecules boil of first. These gas molecules are then
condensed (turned back to liquid) in the cold receiving tube.
Small hydrocarbons have lower boiling points than larger ones. They are easier to
separate from the mixture of molecules in crude oil.
As heating is carried on, the temperature of the crude oil rises. The large molecules are
then boiled off and are collected.
96
Performing the given activity would help the students to appreciate the process. The
teacher may ask the following questions 1. Note the colour and viscosity (how thick it is) of each of the fraction collected.
2. Light each fraction with a lighted splint and ask the students to observe what
happens also ask them to check for the amount of soot produced.
3. Why is the receiving tube kept in cold water?
4. You need a fuel for your car. Why would you use the first fraction collected?
5. Why would you use the last fraction for lubricating a car engine?
6. The graph of various fractions should be given to the students to analyse.
Content-5
The Physical and Chemical Properties of Hydrocarbons
The teacher may ask the students to answer the following questions before doing the
experiment
Q1. What is the difference between satuarated and unsaturated hydrocarbon?
Q2. Give the name and structure of a saturated hydrocarbon and an unsaturated
hydrocarbon each having 5 carbons.
Q3. What would you infer about the physical property of an unknown compound
if it dissolves in non polar solvent and does not dissolve in polar solvent?
After performing the experiments A, B, C and D under content 5 teacher may ask the
following questions to reinforce the concept.
Q1. What characteristic features differentiate an alkane, an alkene, and an alkyne?
97
Q2. Of the following compounds which will react with bromine- Ethane, Ethene
and Propane.
Q3. What observation tells us that a compound reacts with bromine?
Q4. How does KMnO4 test help us to distinguish between an alkane and an alkyne?
Q5. What observation tells us that a compound reacts with KMnO4?
Content-6
Preparation and Properties of Alcohols
Preparation of Ethanol by Fermentation
This experiment would require approximately half an hour for fermentation to take
place. Teacher may describe the fermentation of simple sugars to produce ethanol (and
carbon dioxide) using this experiment. It may be tried that different groups of students
get different sugars- sucrose (common sugar), maltose, glucose etc., so as to bring forth
the concept that simple sugars undergo fermentation.
Fermentation of glucose using yeast
Beer and wine are produced by fermenting glucose with yeast. Yeast contains enzymes
that catalyse the breakdown of glucose to ethanol and carbon dioxide. In this
experiment, a glucose solution is left to ferment. Students then test for fermentation
products.
Yeast has an enzyme called zymase and this catalyses the fermentation process.
Glucose zymase → Ethanol + carbon dioxide
C6H12O6 (aq) → 2C2H5OH(aq) + 2CO2(g)
98
An alternative to delivery tube to transfer carbon-dioxide is to use a big size balloon
and put it on the mouth of the flask. Once the balloon gets inflated it may be cautiously
taken out and the gas may be transferred to the test tube containing lime-water.
Brainstorming questions
Some possible questions to ask students:
1. How do you know fermentation is taking place?
2. Which gas does limewater test for?
3. Suggest methods for measuring the speed of this reaction.
Properties of alcohols:
This experiment can be done completely by extension students if the use of sodium is
closely supervised. With core students, the sodium reaction and possibly the acid
dichromate reaction should be done by the teacher. The experiments will take about 45
minutes.
Technical Notes
1. Prepare the small pieces of sodium in advance of the lesson. Using forceps,
remove a large piece of sodium (Highly flammable, Corrosive) from the oil, and
place on a tile. Ensure that conditions are dry. Using a scalpel or sharp knife, cut
some of the sodium into a few small pieces no larger than 2 x 2 x 2 mm. Place
these small pieces in a separate bottle of oil. Return the larger piece to its bottle.
Dispose of any small pieces of unused sodium by dissolving them in ethanol until all
trace has disappeared and the fizzing has stopped. Then pour the solution down the
sink with plenty of water.
The use of sodium by students must be supervised by the teacher.
2. The beakers must be absolutely dry.
99
Teaching Notes
Lower homologues of alcohols are fully miscible with water. This is because the –OH
groups hydrogen bond with the water. Higher alcohols are less soluble since the
hydrocarbon chain is hydrophobic in nature (repels water) and plays hindrance in
formation of hydrogen bonds in water.
The pH of alcohols will show that they are neutral.
(Note that, if indicator solution is used, ethanol at least will give an acid colour. This is because
ethanol is the solvent used to prepare the indicator solution, and diluting the dyes puts the
mixture out of balance.)
1. Alcohols burn with a fairly clean, blue flame.
C2H5OH + 3O2 → 2CO2 + 3H2O
2. Alcohols react with sodium to form hydrogen gas.
C2H5OH + Na → C2H5ONa + ½H2
(sodium ethoxide)
3. Alcohols get oxidised to aldehydes, which have a sour but fruity smell.
C2H5OH + [O] → CH3CHO + H2O
(ethanal)
4. Alcohols react with carboxylic acids in presence of conc. H2SO4 to form esters
which have a fruity odour.
C2H5OH + CH3COOH → CH3COOC2H5 + H2O
(ethylethanoate)
5. Teacher may ask the student to repeat the same set of experiment with any other
alcohol and observe.
100
These experiments show that alcohols react similarly in all these reactions. They make
clear the concept of functional group in organic chemistry. The reactions of all alcohols
are same as the reactions are due to –OH functional group. Complete oxidation turns
primary alcohols into acids, while secondary alcohols are only oxidised to ketones
under these conditions.
Content-7
Carboxylic Acid
The acidic reactions of ethanoic acid
Students test ethanoic acid with full-range universal indicator solution, magnesium,
sodium hydroxide solution and sodium carbonate solution. (Extension: copper
carbonate). They compare these reactions with those of hydrochloric acid of the same
concentration. This shows that ethanoic acid is a weak acid
Lesson organisation
This experiment can be carried out by the students in about 30 min, There are a lot of
colourless solutions involved, so students will have to be organised. Ask the students to
put labels on the test-tubes with their contents.
Students could be asked to write equations for the reactions that occur. The writing of
formulae of salts of divalent ions with ethanoate can be taken up as brainstorming
activity.
Teaching Notes
The discussion of weak and strong acids is probably more suited to extension students
rather than core students:
Ethanoic acid is a weak acid which means it does not fully dissociate into ions in water.
CH3COOH ⇌ H+ + CH3COOHydrochloric acid is a strong acid and dissociates fully.
HCl → H+ + Cl101
This means that the concentration of H+ ions in 0.4 mol dm–3 HCl is higher than that in
0.4 mol dm–3 ethanoic acid, so its pH is lower and its acid reactions are faster. When
alkalis are added, however, the ethanoic acid equilibrium position moves to the right,
so eventually all the acid reacts, as shown in the equations below:
1. Reaction with a strong base.
NaOH + HCl → NaCl + H2O
NaOH + CH3COOH → CH3COONa + H2O
2. Reaction with a metal carbonate
Na2CO3 + 2HCl → 2NaCl + CO2 + H2O
Na2CO3 + 2CH3COOH → 2CH3COONa + CO2 + H2O
3. Reaction with active metals (metals more reactive than hydrogen).
Mg + 2HCl → MgCl2 + H2
Mg + 2CH3COOH → (CH3COO)2Mg + H2
Technical Notes
The magnesium should be scraped with a spatula, or rubbed with sandpaper, to
remove any oxide coating. Magnesium ribbon is attractive to students who might try to
remove it from the laboratory and light it. Any subsequent fires will be very hot and
hard to extinguish. It is advisable therefore to keep it under teacher control. Cutting is
best done with scissors, as attempts to tear with the fingers can result in cuts.
Worksheet-6
The formative assessment work-sheet 6 in TSSM may be given to the students to assess
their understanding of the chemical properties shown by hydrocarbons, alcohols and
carboxylic acids.
A revision of the following concepts may be done before the assessment.
-
Hydrocarbons burn in oxygen to give carbon dioxide, water and heat.
102
-
Alcohol when heated with Alkaline KMnO4 or acidified potassium dichromate,
are converted to carboxylic acid.
-
Unsaturated hydrocarbons add hydrogen in the presence of catalysts such as
Palladium or nickel to give saturated hydrocarbons.
-
In the presence of sunlight, saturated hydrocarbons undergo substitution
reacting with chlorine.
-
Alcohol react with sodium metal to form sodium alkoxide and hydrogen gas.
-
On heating ethanol with excess conc. H2SO4 at 443K ethene and water is formed.
-
When carboxylic acid and alcohol react in presence of acid, sweet smelling ester
is formed and ester react in presence of an acid or a base to give back the alcohol
and carboxylic acid.
-
Carboxylic acid reacts with a base like NaOH to give sodium salt of carboxylic
acid and water.
-
Carboxylic acid reacts with carbonates and hydrogen carbonates to give salts,
water and carbon dioxide.
Suggestive Remediation:
Some students may find it difficult to write the correct formula of an organic
compound. Teacher may encourage the students to write the general formulae and the
functional groups of organic compounds in their syllabus on a chart and revisit the
naming of organic compounds.
-
Some students may find it difficult to recall the products of the chemical reaction
or balance the equation. The teacher may ask them to tabulate all the reactions of a
particular functional group together and draw correlations in the reactions.
Correlations serve as a memory tool.
-
The revision exercise may be repeated with the students followed by another
formative assessment worksheet.
103
Content-8
Polymers
Demonstration experiment to be done by the teacher -Formation of Condensation
Polymer by carrying out polymerisation reaction to create Nylon 6-10
Aim: To show the formation of a condensation polymer.
Scientific Principle Behind Condensation Polymerisation:
Nylon is a synthetic polyamide polymer which falls in the category of fibres.
An amide (CONH-) linkage is formed by condensation of an amino group (-NH2) with
a carboxylic group (-COOH) with the loss of a water molecule The numbers written
after nylon indicate the number of carbon atoms in the monomer chains. The first
number that follows Nylon represents the number of carbon atoms in diamine and the
second that in dicarboxylic acid. In fact to form a polymer we need functional group on
both the ends of the carbon chain. This may be achieved either by polymerisation of a
diamine with a dicarboxylic acid or one compound having both the groups, i.e.an
aminoacid, eg. Aminocaproic acid- H2N(CH2)5COOH, this is a monomer of Nylon6.
This is an example of a homopolymer in which the single monomer containing the two
required functional groups for formation of an amide linkage is used for the formation
of the complete polymer.
Nylon 6-10 is an example of a co-polymer as it is formed by the copolymerisation
reaction of hexamethylenediamine and sebacic acid. In this demonstration the acid
chloride, sebacyl (or Sebacoyl) chloride, is used instead of sebacic acid. The equation is:
Many diamines and diacids (or diacid chlorides) can be reacted to make other
condensation products that are described by the generic name "nylon."
One such
product is an important commercial polyamide, nylon 6-6, which can be prepared by
substituting adipoyl chloride for Sebacoyl chloride in the procedure described here.
The equation is:
104
Materials Required:

50 ml 0.50 M hexamethylenediamine (1,6-diaminohexane), H2N(CH2)6NH2, in 0.5
M sodium hydroxide, NaOH (This may be prepared by dissolving 3.0 g of
H2N(CH2)6NH2 and 1.0 g NaOH in 50 ml distilled water. The reagent bottle of
Hexamethylenediamine should be by placed in hot water to melt the solidified
diamine (melting point: 39-40o C.) so that it may be decanted.

50 ml 0.2 M Sebacoyl chloride, ClCO(CH2)8COCl, in hexane (This may be
prepared by dissolving 1.5 ml to 2.0 ml Sebacoyl chloride in 50 ml hexane.)

gloves, plastic or rubber (check for the ones that will not dissolve in hexane)

250 ml beaker or crystallizing dish

forceps

2 stirring rods

food-coloring dye (optional)

phenolphthalein (optional)
Technical Notes:
• Hexamethylenediamine is irritating to the skin, eyes, and respiratory system.
• Sodium hydroxide is extremely caustic and can cause severe burns. Contact with
the skin and eyes must be prevented.
• Sebacoyl chloride is corrosive and irritating to the skin, eyes, and respiratory
system.
• Hexane is extremely flammable. Hexane vapor can irritate the respiratory tract
and, in high concentrations, be narcotic.
105
Procedure:
1.
Place the hexamethylenediamine solution in a 250-ml beaker.
2.
Slowly pour the Sebacoyl chloride solution through the side walls of the beaker so
as it forms a second layer on top of the diamine solution. Care should be taken to
minimize agitation at the interface of the two solutions.
3.
With the help of forceps, grasp the polymer film that forms at the interface of the
two solutions and pull it carefully from the center of the beaker.
4.
Wind the polymer thread carefully on a stirring rod.
5.
Wash the polymer thoroughly with water or ethanol.
Food coloring dyes or phenolphthalein can be added to the lower (aqueous) phase
to enhance the visibility of the liquid interface. Some of the dye will be taken up
with the polymer, but can be removed by washing with water.
Disposal:
1.
The remaining reactants should be mixed thoroughly to produce nylon. The solid
nylon should be washed before being discarded in a solid waste container.
2.
The remaining liquid should be discarded in a solvent waste container or should
be neutralized with either sodium bisulphate (if basic) or sodium carbonate (if
acidic) and flushed down the drain with water.
Resource: http://matse1.matse.illinois.edu/polymers
Chemical demonstrations: a handbook for teachers of chemistry, By Bassam Z.
Shakhashiri
106
Extension
Structural Isomerism

Compounds having the same molecular formula but different structures are
called structural isomers and the phenomenon of existence of two or more
structures having the same molecular formula is called structural isomerism, for
eg.Butane having molecular formula C4H10 has two structural isomers n-butane
and Iso-butane
CH3 -- CH2 – CH2 – CH3
CH3 – CH – CH3
|
CH3
n-butane

Iso-butane
Carbon compounds in which no carbon atom of the chain is linked to more than
two other carbon atoms are called straight chain compounds for eg. When all the
five carbon atoms of pentane having molecular formula C5H12 are arranged in a
straight line then it is a straight chain compound.
CH3 – CH2 -- CH2 -- CH2 – CH3
n-pentane

Carbon compounds in which atleast one carbon of the chain is linked to three or
four other carbon atoms are called branched chain compounds for eg. Pentane
having molecular formula C5H12 can also have the following structures having
branches.
CH3 – CH – CH2 – CH3
CH3
|
|
CH3
CH3 -– C -– CH3
|
CH3
Iso-pentane
neo-pentane
107
Polymers: Biological Macromolecules
Learning Objectives:

Recognise proteins, fats and carbohydrates as the main bio-molecules which
constituent organic matter and are essential part of our food.

Describe proteins as possessing the amide linkages as nylon but with different
monomer units.

Describe the structure of proteins as:

Describe the hydrolysis of proteins to amino acids (Structures and names are not
required.)

Describe fats as esters possessing the same linkage as Terylene but with different
monomer units.

Describe soap as a product of hydrolysis of fats

Describe complex carbohydrates in terms of a large number of sugar units,
considered as

, joined together by condensation polymerisation.
Describe the acid hydrolysis of complex carbohydrates (e.g.starch) to give simple
sugars
There are four basic biological macromolecules also called as bio-molecules which
comparise all living things or organisms. These are carbohydrates, lipids, proteins and
nucleic acids. These polymers are composed of different monomers and serve different
functions.

Carbohydrates - composed of monomers which are polyhydroxy aldehydes or
ketones. Carbohydrates are necessary for energy storage.

Proteins - composed of amino acid monomers and have a wide variety of
functions

Lipids- include fats, phospholipids and steroids. Lipids help to store energy,
cushion and protect body organs insulate the body and form cell membranes.
108

Nucleic Acids- include DNA and RNA. Nucleic acids contain instructions for
protein synthesis and allow organisms to transfer genetic information from one
generation to the next.
Carbohydrates:
Carbohydrates are primarily produced by plants and form a very large group of
naturally occurring organic compounds. Some common examples are cane sugar,
glucose, starch, etc. Most of them have a general formula, Cx(H2O)y, and were
considered as hydrates of carbon from where the name carbohydrate was derived. For
example, the molecular formula of glucose (C6H12O6) fits into this general formula,
C6(H2O)6. But all the compounds which fit into this formula may not be classified as
carbohydrates. Acetic acid (CH3COOH) fits into this general formula, C2(H2O)2 but is
not a carbohydrate. Similarly, rhamnose, C6H12O5 is a carbohydrate but does not fit in
this definition. A large number of their reactions have shown that they contain specific
functional groups. Chemically, the carbohydrates may be defined as optically active
polyhydroxy aldehydes or ketones or the compoundswhich produce such units on hydrolysis.
Some of the carbohydrates, which are sweet in taste, are also called sugars. The most
common sugar, used in our homes is named as sucrose whereas the sugar present in
milk is known as lactose. Carbohydrates are also called saccharides (Greek: sakcharon
means sugar).
Carbohydrates are classified on the basis of their behaviour on hydrolysis. They have
been broadly divided into following three groups.
(i)
Monosaccharides: A carbohydrate that cannot be hydrolysed further to give
simpler unit of polyhydroxy aldehyde or ketone is called a monosaccharide.
About 20 monosaccharides are known to occur in nature. Some common
examples are glucose, fructose, ribose, etc.
(ii)
Oligosaccharides: Carbohydrates that yield two to ten monosaccharide units, on
hydrolysis, are
called oligosaccharides.They
are
further
classified
as
disaccharides, trisaccharides, tetrasaccharides, etc., depending upon the
109
number of monosaccharides, they provide on hydrolysis. Amongst these the
most common are disaccharides. The two monosaccharide units obtained on
hydrolysis of a disaccharide may be same or different. For example, sucrose on
hydrolysis gives one molecule each of glucose and fructose
whereas maltose gives two molecules of glucose only.
(iii) Polysaccharides: Carbohydrates which yield a large number of monosaccharide
units on hydrolysis are called polysaccharides.Some common examples are
starch, cellulose, glycogen, gums, etc. Polysaccharides are not sweet in taste,
hence they are also called non-sugars.
Examples of Carbohydrates
Carbohydrate
Monosaccharides
(simple sugars)
Disaccharides
(double sugars)
Polysaccharides
Name
Formula
Source
Glucose
C6H12O6
fruits
Fructose
C6H12O6
fruits, honey
galactose
C6H12O6
not naturally occurring
sucrose (glucose + fructose) C12H22O11
sugar cane, sugar beet
lactose (glucose + galactose) C12H22O11
milk
maltose (glucose + glucose)
C12H22O11
germinating grain
Starch
-(C6H10O5)-n energy store in plants
Glycogen
-(C6H10O5)-n
Cellulose
-(C6H10O5)-n plant fibre
110
energy store in animals
(liver and muscles)
Lipids:
A lipid is a fat, oil or wax. A lipid contains carbon, hydrogen and some oxygen atoms.
Lipids are mixtures of glycerides, mostly triglycerides. Triglyceride means tri-esters of
glycerol, and so lipids contain ester functional groups
O
||
- C -O- , or, -CO-OTriglycerides are formed in a condensation reaction between glycerol and fatty acids. A
fatty acid is a long chain carboxylic acid. Fats are classified as saturated or unsaturated.

A saturated fat contains only single bonds between the carbon atoms in the fatty
acid componenet.

An unsaturated fat contains 1 or more double bonds between the carbon atoms
in the fatty acid component. An unsaturated fat can be classified as either
monounsaturated or polyunsaturated. A monounsaturated fat contains only 1
double bond between the carbon atoms in the fatty acid component. A
polyunsaturated fat contains more than 1 double bond between the carbon atoms
in the fatty acid component.
Fatty Acids:
Fatty acids are long chain carboxylic acids. Fatty acids can be classified as:
Saturated: only single bonds are present between the carbon atoms.
Unsaturated: more than one double bond in the chain.
Unsaturated fatty acids can be classified as either:
Monounsaturated: only 1 double bond between carbon atoms in the chain.
Polyunsaturated: more than 1 double bond between carbon atoms in the chain.
111
Type
Saturated
(no double
bonds)
General
Formula
Example
CH3(CH2)14COOH
CnH2n+1COO palmitic acid
H
CH3(CH2)16COOH
stearic acid
Monounsaturate
CH3(CH2)7CH=CH(CH2)7COOH
d
CnH2n-1COOH
oleic acid
(1 double bond)
Properties
solids
unreactive
relatively
non-polar
insoluble in
water
more in
animal fats
softer than
saturated
fats
reactive
relatively
non-polar
insoluble in
water
likely to be
oils
CnH2n-3COOH
reactive
(2 double
Polyunsaturated
CH3(CH2)4CH=CHCH2CH=CH(CH2)7CO relatively
bonds)
(> 1 double
OH
non-polar
CnH2n-5COOH
bond)
linoleic acid
insoluble in
(3 double
water
bonds)
more in
vegetable
oils
Formation of Trigylcerides (Lipids)
A triglyceride is formed by the condensation reaction between glycerol and 3 fatty acid
molecules. The products are a triglyceride (containing 3 ester functional groups) and
water.
112
glycerol
+ fatty acids
H
O
||
|
||
H- C
O
|
|
O
HO -C- (CH2)nCH3
H
H- C -OH
-----> triglyceride
+
||
----->
H- C
H
||
H- C
HO -C- (CH2)nCH3
-(CH2)nCH3
+ H2O
||
O-
C
-(CH2)nCH3
O
|
O
C
H2O
O
|
|
|
O-
HO -C- (CH2)nCH3
H- C -OH
H- C -OH
-
+ water
||
O-
C
-(CH2)n-
H2O
CH3
|
H
Proteins:
Proteins are the most abundant biomolecules of the living system. Chief sources of
proteins are milk, cheese, pulses, peanuts, fish, meat,etc. They occur in every part of the
body and form the fundamental basis of structure and functions of life. They are also
required for growth and maintenance of body. The word protein is derived from Greek
word, ―proteios‖ which means primary or of prime importance. All proteins are
polymers of α-amino acids.
Amino acids contain amino (–NH2) and carboxyl (–COOH) functional groups.
Depending upon the relative position of amino group with respect to carboxyl group,
the amino acids can be classified as α, β, γ, δ and so on. Only α-amino acids are
obtained on hydrolysis of proteins. They may contain other functional groups also.
113
proteins are the polymers of α-amino acids and they are connected to each other by
peptide bond or peptide linkage. Chemically, peptide linkage is an amide formed
between –COOH group and –NH2 group. The reaction between two molecules of
similar or different amino acids, proceeds through the combination of the amino group
of one molecule with the carboxyl group of the other. This results in the elimination of a
water molecule and formation of a peptide bond –CO–NH–. The product of the reaction
is called a dipeptide because it is made up of two amino acids. For example, when
carboxyl group of glycine combines with the amino group of alanine we get a
dipeptide, glycylalanine.
If a third amino acid combines to a dipeptide, the product is called a tripeptide. A
tripeptide contains three amino acids linked by two peptide linkages. Similarly when
four, five or six amino acids are linked, the respective products are known as
tetrapeptide, pentapeptide or hexapeptide, respectively. When the number of such
amino acids is more than ten, then
the products are called polypeptides. Water is eliminated when the amino acids react to
form a protein. This is known as a condensation reaction, or a condensation
polymerisation reaction.
114
TEACHER
STUDENT
SUPPORT
MATERIAL
115
AN OVERVIEW
“There would have been no life on this universe,
If carbon didn’t have its properties diverse.”
This unit provides an overview of the organic compounds, i.e. the compounds of carbon.
Through this topic the students would be able to appreciate the huge number of
compounds, natural or man-made having carbon as the major component of their
backbone.
The unit starts with the special properties of carbon which makes it possible for it to
form a large number of carbon compounds so much so that the life on this universe is
there only because of the carbon compounds.Followed by the different types of
compounds formed by carbon starting from the simplest- hydrocarbons to the complex
life forming polymers.
Pre content: Creating the context –completion of the ‗Know‘ and ‗Want to know‘
sections of the KWL sheets.
I Know
I Want to Know
(about this topic)
(about this topic)
I have learnt
Content-1
Bonding in Carbon Compounds
Learning Objective: The students will be able to-
appreciate that carbon forms compounds by sharing electrons, bond formed is
called covalent bond.
-
understand the bonds formed by sharing one pair of electron is single covalent
bond, by sharing two pairs of electrons is double covalent bond and by sharing
three pairs of elections is triple covalent bond.
116
-
learn writing electronic dot structure of compounds.
-
identify the number and types of bonds in a compound.
Bonding in Carbon Compounds - Covalent bond

Carbon is a versatile element. All living structures, soaps, detergents, clothes,
medicines, furniture made up of wood, newspapers, books and various food
items are made up of carbon compounds.

Atoms combine to impart stability to the participating atoms. They do so either
by loosing, gaining or by sharing electrons.

Ionic bonds are formed by complete transfer of electrons from one atom to the
other. For example,
Sodium Chloride
Thus ionic compounds consist of ions.

Covalent bonds are formed by mutual sharing of electrons between the two
atoms. For example,
H.
+
H. →
H ..
H
Thus, covalent compounds consist of molecules.

Ionic bonds are formed between atoms of metals and non metals while covalent
bond result when atoms of non metals combine.

Covalently bonded molecules have strong bonds within the molecules but their
intermolecular forces of attraction are weak. Therefore, melting points and
boiling points of covalent compounds are quite low as compared to those of ionic
compounds.

Carbon can neither loose nor gain electrons due to understated reasons-
117
(i)
If it loses four electrons present in the valence shell to some atom/ atoms
and become C4+ ion which has the configuration of helium (2), a noble gas,
it would require a large amount of energy to remove four electrons
leaving behind a highly charged cation with six protons in the nucleus
holding on to just two electrons.
(ii)
If it gains four electrons to form C4- ion which has the configuration of
Neon (2, 8), a noble gas, it would result in a highly unstable anion because
it would be difficult for a nucleus with six protons to hold on 10 electrons.

Covalency of an element is the number of electrons which an atom of the element
shares with other atoms in the formation of covalent molecule.

Types of covalent bonds- The covalent bonds are of three types. A single covalent
bond is formed when each of the two combining atoms share one electron each, a
double bond is formed when each atom contributes two electrons and a triple
bond is formed when each atom contributes three electrons.

Covalent molecules in which bonding atoms are same, are known as
homoatomic molecules while the molecules in which these are different are
called heteroatomic molecules.

Catenation and tetravalency are the two factors noticed in carbon due to which
there exist such a large number of carbon compounds.The property of self
linking of carbon atoms through covalent bonds to form long, straight or
branched chains and rings of different sizes is called catenation.No other element
exhibits the property of catenation to the extent seen in carbon compounds.
Silicon also shows catenation but to a lesser extent. Silicon forms compound with
hydrogen which contains chain only upto seven or eight silicon atoms, but these
compounds are very reactive. The carbon-carbon bond is strong and hence
stable.

Tetravalency of Carbon-Carbon has a valency of four. Therefore it is capable of
bonding four other atoms of carbon or some other elements. Due to its small size
118
carbon forms very strong bonds with many elements such as hydrogen, oxygen,
nitrogen etc. This further increases the number of carbon compounds.

Organic compounds Compounds of carbon containing usually hydrogen and one
and more other elements such as oxygen, nitrogen, sulphur, halogens,
phosphorous,etc. are called Organic compounds and the branch of chemistry
which deals with the study of organic compounds is called Organic chemistry.

Carbon forms a large number of compounds because due to its poperty of
catenation it can form- .

straight chain compounds
CH3 – CH2 – CH2 – CH3
Carbon compounds in which no carbon atom of the chain is linked to more than
two other carbon atoms are called straight chain compounds for eg. When all the
five carbon atoms of pentane having molecular formula C5H12 are arranged in a
straight line then it is a straight chain compound.
CH3 – CH2 -- CH2 -- CH2 – CH3

branched chain compounds

Carbon compounds in which at least one carbon of the chain is linked to three or
four other carbon atoms are called branched chain compounds for eg. Pentane
having molecular formula C5H12 can also have the following structures having
branches.
CH3 – CH – CH2 – CH3
CH3
|
|
CH3
CH3 -– C -– CH3
|
CH3
119

cyclic compounds
Compounds of carbon in which carbon atoms are arranged in a ring are called
cyclic compounds.

Cyclic compounds are of the following two types :
(a) Cyclic compounds having single bond between the carbon atoms are called
saturated cyclic compounds.Cyclohexane with the molecular formula C6H12
is an example of saturated cyclic compound.
(b) Cyclic compounds having one or more double or triple bonds between the
carbon atoms are called unsaturated cyclic compounds. Benzene with the
molecular formula C6H6 is an unsaturated cyclic compound.
120
Worksheet-1
Instruction:
Look at the following diagram showing electronic dot structure carefully and answer
their questions that follow:
I
Q.1
II
III
Name the molecule whose electronic dot structure is depicted in diagram I and
III
_______________________________________________________________________
Q.2
Identify the type of bond being formed in the molecules shown above.
_______________________________________________________________________
Q.3
How many pairs of electrons are being shared between the two atoms of oxygen?
_______________________________________________________________________
Q.4
How many bonds are formed between two atoms of Oxygen?
_______________________________________________________________________
Q.5
Write down the structural formula of the compound being shown in diagram III.
____________________________________________________________________
121
Q.6
Write the electronic dot structure of MethaneH
|
H-– C -– H
|
H
_______________________________________________________________________
Q.7
Write the electronic dot structure of ethene.
H
|
H
|
H—C = C -– H
_______________________________________________________________________
Q.8
Write the electronic dot structure of propane.
CH3 – CH2 -– CH3
_______________________________________________________________________
Q.9
Give the number of different type of bonds present in one molecule of propane.
_______________________________________________________________________
122
Content-2
Nomenclature of Organic Compounds
Learning objectives:
The students will be able to-
Gain expertise in the skill of naming hydrocarbons
-
Write the structural formula of hydrocarbon
The Hydrocarbons
Compounds of carbon and hydrogen are called hydrocarbons. These are of the following
two types :
(a)
Saturated hydrocarbons containing open chain structures are called alkanes.
Their general formula is CnH2n+2
Where n= 1, 2, 3, 4 …etc. Some examples of alkanes are methane (CH4), ethane
(CH3CH3), propane (CH3CH2CH3), etc.
(b)
Unsaturated hydrocarbons containing open chain structures are called alkenes
and alkynes.
Those which contain one or more double bonds are called alkenes.
General formula is CnH2n where n= 1, 2, 3, 4 …etc. Some examples of alkenes
are ethene (C2H4), propene (C3H6), etc.
Organic compounds, i.e. carbon compounds containing one or more triple
bonds are called alkynes.
General formula is CnH2n-2 where n= 1, 2, 3, 4 …etc. Some examples of alkynes
are ethyne (C2H 2), propyne (C3H 4), etc.

The Isomers
Compounds having the same molecular formula but different structures are
called structural isomers and the phenomenon of existence of two or more
structures having the same molecular formula is called structural isomerism, for
123
eg.Butane having molecular formula C4H10 has two structural isomers n-butane
and Iso-butane
CH3 -- CH2 – CH2 – CH3
CH3 – CH – CH3
|
CH3
n-butane

Iso-butane
A functional group may be defined as an atom or a group of atoms present in a
molecule which largely determines its chemical properties for eg.
H
|
H
|
H – C – C – OH
|
H
→ Functional group
|
H
Ethanol
Nomenclature of Organic Compounds
Many organic compounds have two names:
(a)
Trivial or common names: These names were given after the source from which
the organic compounds were first isolated. For example acetic acid got its name
from acetum ( Latin: acetum means vinegar ) since it is present in vinegar.
(b) IUPAC names: International Union of Pure and Applied Chemistry (IUPAC)
have given certain rules to systemize the nomenclature of organic compounds.
The names based upon these rules are called IUPAC names. The IUPAC name
of the compound is built from four parts. These are arranged in the sequence
below: Prefix….Word root…..Primary suffix…. Secondary suffix.
Prefix: It denotes the substituent group (branches) if present in the organic compound.
The prefixes for a few substituent groups are
124
Substituent group
Prefix
--F
Fluoro
--Cl
Chloro
--Br
Bromo
--I
Iodo
--NO2
Nitro
--CH3
Methyl
--C2H5
Ethyl
--C3H7
Propyl
Word Root: Denotes the number of carbon atoms in the longest possible chain. The
word roots used for different number of carbon atoms are:
Number of carbon atoms
Word Root
1 one
Meth
2 two
Eth
3 three
Prop
4 four
But
5
five
Pent
6
six
Hex
7
seven
Hept
8
eight
Oct
9
nine
Non
10
ten
dec
Primary Suffix: This denotes the nature of carbon to carbon bond in the organic
compounds. For example,
ane:
Primary suffix for C—C bond
ene:
Primary suffix for C=C bond
yne:
Primary suffix for C=C bond
125
Secondary Suffix: Secondary suffix is used to represent the functional group if present
in an organic molecule and is attached to the primary suffix while writing the IUPAC
name. If the secondary suffix starts with a vowel or vowel sounding word then terminal
‗e‘ of the primary suffix is dropped.

The IUPAC name of any organic compound can be derived by using the
following rules:
(a)
Select the longest chain of carbon atoms known as parent chain which
may or may not be straight.
(b)
The parent chain must include the carbon atoms involved in the double
bond, triple bond, and also those involved in terminal functional groups (-CHO , --COOH etc.)
(c)
The carbon atoms of the parent chain are numbered starting from one end
in such a way that the carbon atoms involved in the multiple bond get the
least possible number, For e.g.
5
4
3
2
1
1
CH3—CH2—CH=CH—CH3
2
3
4
5
CH3—CH2—CH=CH—CH3
(Correct numbering)
(Wrong numbering)
(d) In case a terminal functional group is present, the numbering of the chain
must begin from the carbon atom of the functional group. For example,
2
3
4
1
2
3
4
5
CH3—CH2—CH—CH2—CH3
CH3—CH2—CH—CH2—
CH
1
|
|
C=O
C=O
|
|
H
H
(Correct numbering)
(Wrong numbering)
126
(e)
The positions of the prefixes or suffixes are indicated by numerical figures
which are mentioned before their names and are separated by hyphens (-).
For example;
O
4
3
1│|
2
CH3—CH—CH2—C—OH
Prefix: bromo
|
Word root: but
Br
Primary suffix: ane
Secondary suffix: oic acid
IUPAC name: 3-Bromobutanoic acid

Summary of functional groups
Functional
Structure
group
carboxylic
- COOH
acids
Prefix
Suffix
none
-oic acid
aldehydes
-CHO
none
-al
ketones
-CO -
none
-one
alcohols
-OH
hydroxy- -ol
fluorine
-F
fluoro-
none
chlorine
-Cl
chloro-
none
bromine
-Br
bromo-
none
iodine
-I
iodo-
none
127
Worksheet-2
Q1
Classify the following as Alkane ,Alkene and Alkyne and write their IUPAC
name .
S.NO
Molecular Formula
Alkane /Alkene
IUPAC Name
/Alkyne
Q2
1
C4H10
2
C2H2
3
C3H4
4
C6H12
5
C5H8
Write the IUPAC and Common names for the following :-
S.NO
MOLECULAR
IUPAC
FORMULA
NAME
1
CH3CHO
2
HCHO
3
CH3Br
4
HCOOH
128
STRUCTURAL COMMON NAME
FORMULA
S.NO
MOLECULAR
IUPAC
FORMULA
NAME
5
C2H5COOH
6
CH3COCH3
7
C4H9OH
8
C2H4
9
C2H2
10
C5H12
11
C4H8
12
C3H4
13
C2H5OH
14
CH3Cl
15
CH3COOH
129
STRUCTURAL COMMON NAME
FORMULA
Q2
Write the Structural Formula for the following :S.NO
IUPAC NAME
1
Propanol
2
Pentanal
3
Propanone
4
Chlorobutane
5
Ethanol
6
Methanoicacid
S.NO
STRUCTURAL FORMULA
COMMON NAME
1
Acetone
2
Acetaldehyde
3
Formaldehyde
4
Formicacid
STRUCTURAL FORMULA
130
Q3
Write the IUPAC name of the
1. Alcohol derived from Pentane.
2. Aldehyde derived from Butane.
3
Q4
Q5.
Carboxylic acid derived from Methane.
Write the structural formula for each of the following:
(a) propane (b) ethyne
(c) propene (d) pentyne
(e) butyne
(f) butene
(g) ethane
(i) pentane
(j) ethane
(k) propyne (l) pentene
(h) methane
Name the compound represented by:
H
H H
│ │ │
(a) H--C ≡ C--H
H H H
(b) C = C--C―H
H
H
H
│
H―C―C―C―C―H
(d)
H―C―H
│ │ │ │
│
H H H
H
H
H H H
H
│ │ │ │
(e)
│
H
│ │ │ │
(c)
│
H―C―C―C―C═ C ― H
│ │ │
H H H
│
H
131
Content-3
Homologous Series
Learning Objectives:
The students will be able to:
-
learn that a series of compounds with same functional group is called
Homologous series.
-
discovers that in a homologous series the difference in the formula of two
adjacent compounds is of ‗-CH2‘ and difference in molecular mass is ‗-14 u‘.
-
appreciate that chemical properties of members of a Homologous series are
same.
-
formulate the general formula for a homologous series.
-
discover the next members of a Homologous series.
Homologous Series

A homologous series may be defined as a family of organic compounds having the
same functional group, similar chemical properties and the successive members
of which differ by a –CH2 unit or 14 mass units.

For example, the compounds given below belong to the alcohol family.
CH3OH methyl alcohol (methanol)
CH3CH2CH2OH propyl alcohol (1-propanol)
CH3CH2.CH2.CH2OH butyl alcohol (1-butanol)

Each Homologous series can be given a General formula in which number of
carbon atom is denoted by n and then number of hydrogen is related to n. If any
other element like ‗O‘ is present in the compound, its number is also written.
For example, alkanes can be represented by the formula CnH2n+2.
CH4
Methane

C2H5
C3H8
Ethane
C4H10
Propane
Butane
If one knows the general formula of a Homologous series, the molecular formula
of other members can be written.
132

Each member of a homologous series has a common difference of -CH2 from the
next higher or lower member.

Common general methods of preparation exist for all members of the series.

All members exhibit similar chemical behavior.

An increase in molecular mass of members within a homologous series show a
similar regular gradation of the physical properties, such as, physical state,
melting and boiling points etc.
Aim: To build models of organic molecules in which carbon atoms may bond with each
other as well as with atoms of other elements.
Materials Required : per Lab Station ball and stick molecular model kit
Procedure:
1. Obtain a ball and stick model kit. The instruction sheet in the box lists the colour
code for the balls. Let one ball represent an atom. A stick is used to show one
covalent bond. Springs are used to represent two or more bonds between the
same two atoms.
2. Make a model of methane, CH4. Record the structural formula in Data Chart I.
3. Make models of successive members of the series in which the hydrocarbon,
methane, is the first member. (A hydrocarbon is a compound composed of
hydrogen and carbon only.) Record on data table.
DATA CHART - I
Alkane Series CnH2n+2
Member
Molecular
Structural
Formula
Formula
Name
First
Methane
Second
Ethane
Third
Propane
Fourth
Butane
133
4. Study Data Chart I and answer the following questions:
(a) For every additional carbon atom, how many hydrogen atoms are needed?
___________________________________________________________________
(b) The consecutive members of the homologous series differ by CH2. Does ethane
belong to a homologous series?
___________________________________________________________________
5. Select two carbon atoms and connect them by two springs. Complete the model by
adding as many hydrogen atoms as possible. The model represents a molecule of
ethene, also called ethylene, the first member of the Alkene Series. Record the
structural formula in Data Chart II.
DATA CHART - II
Alkene Series CnH2n
Member
First
Molecular
Formula
Name
Structural Formula
C2H4
H
|
H
ethene(ethylene)
|
H—C = C – H
Second
propene
(propylene)
Third
butene
(butylene)
134
6.
Continue to add carbon atoms until all of the information for Data Chart II has
been found.Keep the pair of double-bonded carbon atoms at one end of the
molecular model. The other carbon bonds are single bonds
7.
Study Data Chart II and answer the questions.
Are the members of the Alkene Series members of a homologous series?
___________________________________________________________________
How do you know?
_____________________________________________________________________
8.
Make a model for a methanol molecule. Methanol is an alcohol whose structural
formula is
H
│
H ― C― O ―H
│
H
The O-H is known as a functional group; it is found in all alcohols.
9.
Continue to add carbon atoms until all of the information for the Data Chart below
has been found.Keep the O-H group in all of your models. Write the structural
formulas for the models you make.
135
DATA CHART III
Member
Molecular
Structural
Formula
Formula
First
Name
Methanol
(methyl alcohol)
Second
Ethanol
(ethyl alcohol)
Third
Propanol
(propyl alcohol)
Fourth
Butanol
(butyl alcohol)
10.
Study the models of the alcohols you made. Are they members of a homologous
series?
Explain.
_____________________________________________________________________
Worksheet-3
1.
How is the presence of only single-bonded carbon atoms in a homologous series
indicated in the name of
(a)
the series?
______________________________________________________________
(b)
the name of a member of the series?
______________________________________________________________
136
2.
How is the presence of one double bond in the carbon chain indicated in the
name of a homologous series?
_________________________________________________________________
3.
Study the Data Chart for the Alkene Series.
(a)
Is there any member of this series that does not have twice as many
hydrogen atoms as carbon atoms?
_________________________________________________________________
(b)
The general formula for the Alkene Series is CnH2n. What does this
formula tell you?
_________________________________________________________________
Is the general formula a formula for a single compound or does it apply to
every member of the series?Explain.
__________________________________________________________________
4.
Study the Data Chart for the Alkane Series.
(a)
In the formula for ethane, are there exactly twice as many hydrogen atoms
as carbon atoms?
__________________________________________________________________
Are there more than twice as many carbon atoms?
__________________________________________________________________
If so, how many?
__________________________________________________________________
(b)
How many carbon atoms are in a molecule of propane?
__________________________________________________________________
137
How many hydrogen atoms?
__________________________________________________________________
Do you think that the general formula for the Alkane Series might be
CnH2n ? Explain.
__________________________________________________________________
5.
The general formula for the Alkyne Series is CnH2n-2. What does this formula tell
you about the formula for any member of this series?
________________________________________________________________________
6.
The name ending of the Alkyne Series indicates a triple bond in the carbon chain.
Write the formulas for the first four members of this series
_______________________________________________________________________
7.
Molecular Formula
Structural Formula
Ethyne
Ethyne
Propyne
Butyne
Butyne
Propyne
Pentyne
Pentyne
Complete the Table.
Name of Series
Name endings of
General Formula
Bonding in the
Members
for Series
Carbon Chain
Alkane
Alkene
Alkyne
138
8.
Do you think that there can be a homologous series of alcohols containing two
OH groups? Explain.
________________________________________________________________________
________________________________________________________________________
2.
What would be the general formula for a homologous series in which a bromine
atom has been substituted for one hydrogen atom in Alkane compounds?
________________________________________________________________________
Resource:
http://mhvpschool.com/science1/web_documents/organicmodelbuilding.pdf
Content-4
Fractional distillation of petroleum
Learning objectives:
The students will be able to –

Understand that petroleum is a mixture of various hydrocarbons

Appreciate that hydrocarbons can be separated by fractional distillation of
petroleum.
Fractional Distillation of Petroleum (CRUDE OIL)
Petroleum, or crude oil, is a complex mixture of substances. Even long before the
dinosaurs lived on this earth, most of the earth was just oceans with tiny sea organisms
called the phytoplanktons. With passage of time these died and sank to the bottom of
the sea and slowly got covered with silt and sand. The sand and sediments formed
139
impervious, sedimentary rocks. Pressure due to the weight of these rocks and heat of
the earth converted the dead plants and animals into petroleum and natural gas over a
million of years. Wells are dug as deep as 5 to10 km to reach the deposits of oil and gas.
Down the memory lane:
In 1806, in West Virginia, USA, people were digging wells to get salt water (brine) for getting
salt. But they were really upset to see a black oily substance also came out floating on the brine.
They didn’t know at that time the same dirty liquid will run the whole world one day!! The first
oil well was drilled in 1859 in Pennsylvania, USA.
Petroleum was first discovered where it seeped to the surface in shallow pools. People
gradually learned that this raw material could be separated into substances that were
more useful than the original crude mixture. Refined petroleum is used to make items
such as asphalt, gasoline, kerosene, lubricants, waxes and plastics.
Different components of crude oil are separated by fractional distillation. The process of
fractional distillation is based on the difference in physical property of boiling point of
the components. This separation is done by heating, evaporating, cooling and
condensing the original crude mixture until you have separated out the different
materials that make it up. Recall that boiling point is the temperature at which a liquid
changes to a gas (vapour) at normal atmospheric pressure. Different substances have
different boiling point
Fuels like coal and petroleum also contain elements like nitrogen and sulphur. Their
combustion results in the formation of oxides of nitrogen (i.e., nitrogen dioxide, nitric
oxide, etc.) and sulphur (i.e., sulphur dioxide, sulphur trioxide, etc.) which are the
major pollutants in the environment.
Distillation of Crude Oil
At the refinery
140
Crude oil is a thick, black, smelly liquid containing a mixture of hydrocarbons. These
hydrocarbons are separated into groups of useful substances (called fractions) at the
refinery by the process of Fractional Distillation.
We can distil crude oil in the lab to enable a better understanding of what happens in
the refinery.
Demonstration
Set up the apparatus in a fume cupboard as shown in figure

Heat the crude oil gently at first. Collect a few drops of the first liquid that
collects. Keep heating and change the receiving tube. Now collect the liquid that
distills up to 150 C. Again change the tube and collect the liquid when you heat
strongly.

Pour each fraction collected onto a watch glass.
Fractions
Low b.pt
(up to 800C)
Low b.pt
(800-1500C)
Low b.pt
(up to 1500C)
Size of
molecules
Small
Color
Colourless
Thickness
(viscosity)
Runny
How it burns
Medium
Yellow
Thicker
lights easily(highly
flammable),clean flame
harder to light some smoke
Large
Dark
Orange
Thick
(syrupy)
difficult to light,
smoky flame
As the molecules get larger in size, the fractions get less volatile.
141
Worksheet-4
1.
Explain the characteristic properties of matter which govern that the process of
fractional distillation. Why and how does this process work?
2.
What step in the Fossil Fuel Resource Cycle does fractional distillation
demonstrate? Why is this step important?
3.
Petroleum is a complex mixture of substances that can be separated into useful
substances using the process of fractional distillation. Identify 3 products formed
from the distillation of petroleum and explain their uses.
4.
Why does the temperature rise as the distillation proceeds?
5.
What do the plateaus on the graph of temperature versus time represent?
6.
Why is there a relatively rapid rise in temperature between the plateaus on the
graph of temperature versus time?
7.
Graphs given is a typical graph of temperature versus time obtained from the
distillation of a 10 ml sample of the crude oil. Each plateau on the graph above 25
ºC represents the temperature at which a given substance separates from the crude
oil.
Use information presented in the graph to answer the following questions:
142
Graph: Temperature vs. Time – for distillation of 10ml sample of crude oil
a.
How many fractions (different substances) are there in the crude oil?
_________________________________________________________________________
b.
What are the temperatures at which those substances should boil/distil off?
_________________________________________________________________________
c.
When is it optimal to change collecting tubes during the distillation process?
_________________________________________________________________________
143
Content-5
The Physical and Chemical Properties of Hydrocarbons
Learning Objectives:
The students will be able to –

To examine the physical properties of hydrocarbons

To observe reactions characteristic of each of the three main classes of
hydrocarbons
Physical and Chemical Properties of Hydrocarbons
Chemical Properties of Carbon Compounds
(a)
Combustion: Combustion means heating a substance strongly in excess of oxygen
or air as a result, it gets oxidized. Both carbon and hydrogen present in organic
compounds get oxidized during combustion to form carbon dioxide and water
respectively along with the release of large amount of heat and light. For example
C
Carbon
+
O2
→
CO2
+
heat and light
Oxygen
Carbon dioxide
144
*
Saturated hydrocarbons burn in excess of air with a blue flame but
unsaturated hydrocarbons burn with yellow flame with lots of black smoke.
The reason being that the carbon content of unsaturated compounds is more
than the hydrogen content and hence carbon is not completely burnt and the
unburnt carbon deposits as a soot.
*
Even saturated hydrocarbons burn with a sooty flame due to incomplete
combustion if the supply of air is limited. This is the reason for the presence
of air holes in the burners of gas/kerosene stoves. When the holes are open,
complete combustion of gas occurs and a blue flame is obtained. If on the
other hand, holes are closed, enough supply of air is not available and the gas
burns incompletely to produce a yellow sooty flame.
*
When a fuel is ignited it may burn with a flame or without a flame. The
reason being that a flame is produced only when gaseous substances burn.
When wood or charcoal is ignited, the volatile substances present vapourise
and burn with a flame in the beginning. A luminous flame is produced when
the atoms of gaseous substance are heated and begin to glow. The colour
produced by each element is a characteristic property of that element. For
example, a greenish blue flame is produced when copper wire is heated in the
flame of a gas stove.
(b)
Oxidation: Addition of oxygen to any substance is called oxidation and the
substance which are capable of adding oxygen to other substances are called
oxidising agents. Combustion, in fact, means complete oxidation. In addition to
combustion, there are reactions in which partial oxidation occurs which is carried
in the presence of certain oxidising agents like alkaline potassium permanganate,
acidified potassium dichromate etc. For example
145
KMnO4/KOH, Heat
CH3CH2OH +
2[O]
→
CH3COOH
K2Cr2O7/ H2SO4, Heat
Ethanol
(c)
Ethanoic acid
Addition reaction: In these reactions, the attacking species adds to the molecule
of unsaturated hydrocarbon which gets converted to saturated hydrocarbon. For
example, both ethyne (HC=CH) and ethene (H2C=CH2) are unsaturated
hydrocarbons add hydrogen in the presence of a catalyst such as nickel, platinum
or palladium to form saturated hydrocarbon ethane (H3C—CH3). This process is
called catalytic hydrogenation.
H H
|
|
H—C=C—H
Ni/Pt
+
H2
→
H
H
|
|
H—C—C--H
|
473 K
H
Ethene (Unsaturated)
|
H
Ethane (saturated)
This addition reaction is commonly used in the hydrogenation of vegetable oils
in presence of nickel as catalyst to form fats (vegetable ghee). Vegetable oils
generally have long unsaturated carbon chains while animal fats have saturated
carbon chains.
Ni/Pt
Vegetable oil
(Liquid)
(d)
+
H2
→
473 K
Vegetable ghee
(solid)
Substitution reaction: Reactions which involve the direct replacement of an atom
or a group of atoms in an organic molecule by another atom or group of atoms
146
without any change in the rest of the molecule are called substitution reactions.
These reactions are very common in saturated hydrocarbons. For example
chlorine reacts with methane in the presence of heat or light to form substitution
products
Sunlight
CH4
+
Methane
Cl2
→
CH3Cl
Chlorine
+
HCl
Chloromethane
Hydrogen chloride
(Substitution product)
However with excess of chlorine all the hydrogen atoms of methane are replaced
one by one by one to form a number of products
Sunlight
CH3Cl
+
Cl2
Chloromethane
→
CH2Cl2
Chlorine
+
HCl
Dichloromethane
Hydrogen chloride
Sunlight
CH2Cl2
+
Cl2
Dichloromethane
→
CHCl3
Chlorine
+
HCl
Trichloromethane
Hydrogen chloride
Sunlight
CHCl3
+
Cl2
Trichloromethane
→
Chlorine
Hydrogen chloride
147
CCl4
+
HCl
Carbon tetrachloride
Aim:

To examine the physical properties of hydrocarbons.

To observe reactions characteristic of each of saturated and unsaturated
hydrocarbons.
Materials required:
Heptane,
Cyclohexene,
and
dichloromethane(methylene
an
unknown
hydrocarbon,
chloride),ethanol,dichloromethane,
1%
0.05%
bromine
in
KMnO4
in
water,5% sodium chloride solution, eye dropper,small test tubes,large test tubes,
evaporating dish, wire guage, tongs,wood splints and matchstick box.
Procedure:
Perform the tests described in A,B,C and D on each of these compounds-Heptane,
cyclohexene and unknown hydrocarbon
A. Solubility:

Place 10 drops of each hydrocarbon (heptanes,cyclohexene and unknown) in
small separate clean, dry test tubes. Cautiously compare their odours.

Add about 15 drops of methylene chloride to each tube, Shake the tubes and let it
stand for 2 minutes and note in the data table whether the compound is soluble,
insoluble or partially soluble. If the compound is soluble, the two layers will
blend together to form a single homogenous layer. If the compound is insoluble,
two distinct layers will form. If the compound is slightly soluble, the mixture
may appear cloudy.

Repeat the solubility test using ethanol.

Repeat the solubility test a third time using water as a solvent.
148
SOLUBILITY
Solubility in
Hydrocarbon
Odour
Methylene
Chloride
Solubility in
Solubility in
Ethanol
water
Heptane
Cyclohexene
Unknown
Hydrocarbon
B.
Combustion:

Place about 8 drops of heptanes in a clean, dry white evaporating dish on a
wire guage and ignite it with a burning splint. (Keep another wire guage at
hand to put over the evaporating dish to distinguish the flame if necessary).

Observe the colour of the flame, the smoke given off and whether a residue
is left. Record your observations and repeat the test for each of the other
hydrocarbons.
COMBUSTION TEST RESULT
Hydrocarbon
Flame
Smoke
Heptane
Cyclohexene
Unknown
hydrocarbon
149
Residue
C.
Reaction with Bromine:

Place 10 drops of each hydrocarbon in separate clean, small, dry test tubes.

To each sample add 2 drops of a 1% solution of bromine solution. Shake
each test tube and record results.

Observe whether the colour of the bromine solution is lost or retained when
it is mixed with the hydrocarbon. This test will tell whether a compound is
unsaturated or not.
BROMINE TEST RESULTS
OBSERVATION
HYDROCARBON
Heptane
Cyclohexene
Unknown Hydrocarbon
D.
Reaction with Potassium Permanganate:

Place 20 drops of each hydrocarbon in separate large test tubes.

Add 10 drops of a 0.05% potassium permanganate solution and two drops
of sodium carbonate solution to each test tube and shake.

Observe the colour of the coloured layer that separates out. Make your
observation within 30 seconds of mixing and record your observation in the
data sheet.

This test will tell whether a compound is unsaturated or not.
150
POTASSIUM PERMANGANATE TEST RESULTS
Hydrocarbon
Observation
Heptane
Cyclohexene
Unknown Hydrocarbon

Use your test result to classify the unknown hydrocarbon as saturated or
unsaturated hydrocarbon.
Resource: httpchemistry.olivet.educlasseschem100pdfLabsHydrocarbons%20Lab.pdf
Worksheet-5
1.
Write the complete structural formulas for the two known hydrocarbons used in
this experiment.
a. Heptane:
b. Cyclohexene:
151
2.
Write the complete structural formulas of the three solvents used and label each
as polar, non polar or slightly polar.
a. Methylene Chloride:
b. Ethanol:
c. Water:
3.
Write and balanced chemical equation for combustion of nonane and pentene .
________________________________________________________________________
________________________________________________________________________
4.
Write a balanced chemical equation for each of the known hydrocarbons that
reacted with bromine solution.
________________________________________________________________________
________________________________________________________________________
5.
When the KMnO4 reacted to give a brown solid, what was the identity of the
brown solid?
Give both the formula and the name of the compound.
________________________________________________________________________
152
6.
Based upon your solubility tests do you consider the hydrocarbons to be mostly
polar or non polar? Explain you reasoning.
________________________________________________________________________
________________________________________________________________________
7.
When hydrocarbons burn in such a manner that a lot of soot is produced, what is
the chemical name and formula of the soot?
_______________________________________________________________________
8.
An unknown hydrocarbon dissolved in methylene chloride, but was not soluble
in water. It reacted with Br2 to decolourize the bromine solution. It turned the
KMnO4 solution from purple solution to brown solid. It burned to produce a
small amount of smoke and some black residue. What can you conclude about
the nature of the hydrocarbon from these results?
________________________________________________________________________
9.
Write the chemical equations for the substitution of chlorine in Methane in the
presence of sunlight. Also give the IUPAC names of the products formed in each
step.
________________________________________________________________________
________________________________________________________________________
153
Content-6
Preparation of Alcohols
Aim: To prepare ethanol from glucose by fermentation in the presence of yeast.
Materials Required:
Conical flask (100 cm3), Measuring cylinder (50 cm3), Delivery tube, Boiling tube, Cotton
wool, Warm water 30–40 °C, Glucose, 5 g, Yeast,1 g, Limewater
Procedure:
1. Put 2 g of glucose in the conical flask and add 25 ml of warm water. Stir the
content of the flask to dissolve the glucose.
2. Add about 1 g of yeast to the solution and loosely close the top of the flask with
cotton wool.
3. Wait while fermentation takes place.
4. Remove the cotton wool and pass the invisible gas using the delivery tube into
another tube containing limewater.
5. Gently shake the limewater in the boiling tube and observe what happens.
6. Replace the cotton wool in the top of the flask.
Properties of Alcohols
Learning Objectives:
The students will be able to –
154

Write chemical equations for the organic reactions with main products formed
during the reaction.

Draw correlations between the type of functional group and the chemical
properties of organic compounds.

Recognise various reagents like -a reducing or an oxidizing agent used in organic
reactions.
Ethanol or Ethyl Alcohol
Physical properties:
(a)
Ethanol is a colourless liquid with a characteristic smell called alcoholic
smell and a burning taste.
(b)
Its freezing point is 156 K while its boiling point is 351 K.
(c)
It is soluble in water in all proportions.
(d)
It is neutral towards litmus.
*
A solution containing 95% alcohol and 5% water is called rectified spirit.
Completely pure or 100% alcohol is known as absolute alcohol. However
intake of even a small quantity of pure alcohol can be lethal.
Chemical properties:
(a)
Reaction with sodium metal: Ethanol reacts with sodium metal to form
sodium ethoxide with the evolution of hydrogen gas. It is accompanied by
brisk effervescence.
Cold
(b)
2CH3CH2OH +
2Na
Ethanol
Sodium
→
2CH3CH2ONa
+
H2
Sodium ethoxide
Reaction with concentrated sulphuric acid: When ethanol is heated
strongly with concentrated sulphuric acid to about 443 K, it loses a
molecule of water and forms ethene.Concentrated sulphuric acid is a
dehydrating agent as it has strong affinity for water.
Conc H2SO4
155
CH3CH2OH
---------------»
CH2=CH2
+
H2O
443 K
Ethanol
Ethene
Water
This reaction is called dehydration because it involves the removal of
molecule of water from ethanol.
Uses:
Ethanol is a very useful substance. It is used
(a)
as an important alcoholic beverage in the form of beer, wine, whisky,
brandy etc.
(b)
As a fuel
Ethanol readily burns to give carbon dioxide and water without causing
any pollution and can be used as a fuel in its own right, or in mixtures
with petrol (gasoline). "Gasohol" is a petrol / ethanol mixture containing
about 10 - 20% ethanol.
Also used as an additive in petrol. Ethanol is mixed with petrol in some
proportions.
(c)
as a solvent- in many perfumes, paint, cosmetics,lacquers,varnishes, etc.
and for polishing wooden furniture
(d)
as a starting material for the manufacture of ether, chloroform, acetic acid
and acetaldehyde etc.
(e)
as an antifreeze in the radiators of automobiles in cold countries.
(f)
as an antiseptic for wounds in the form of rectified spirit.
(g)
in the preparation of dyes, cosmetics and transparent soaps.
(h)
in spirit levels and low temperature thermometers..
156
Aim: To study the physical and chemical properties of alcohols.
Materials required: 2 Test tubes, 2 Boiling tubes, 2 Beakers (100 cm3), Tin lid, Wooden
splint, Bunsen burner Test tube holder, Pipette, Universal indicator paper (full range,
pH 1-14) Forceps for picking sodium piece, Small pieces of sodium, Ethanol and
Propan-1-ol, Potassium dichromate solution, (0.1 mol dm–3); Sulphuric acid, (1 mol dm–
3),
Anti-bumping granules
Procedure
1.
Solubility- Place a few drops of the ethanol in a test-tube and add an equal number
of drops of water. Observe and record in the data table.
2.
Nature- Put a drop of the ethanol on a piece of full-range indicator paper. Note the
pH and record in the data table.
3.
Combustion- Place a few drops of ethanol on a tin lid on a heat resistant mat.
Ignite the alcohol with a lit matchstick and observe the flame.
4.
Reaction with sodium metal
Using forceps place two small pieces of sodium on a piece of filter paper and
dab the pieces of sodium with the filter paper to remove any excess oil.

take ethanol in a two separate, dry, 100 ml beakers and to each, add a small
piece of sodium. Observe the reaction and record in the data table.
5.
Oxidation Reaction-Take about 2ml of dilute sulphuric acid in a boiling tube and
add five drops of potassium dichromate solution, two drops of ethanol and a few
anti- bumping granules. Heat the mixture until it just boils.
Observe the changes and record in the data table.
6.
Repeat all the above steps with Propan-1-ol and note observation in the following
data table.
157
Observation :
Data Table
Property
Ethanol
Propan-1-ol
Solubility
Nature
Combustion
Reaction with sodium
metal
Oxidation(Reaction with
acidified K2Cr2O7)
Conclusion:
__________________________________________________________________________
__________________________________________________________________________
158
Content-7
Carboxylic Acid
Learning Objectives:
The students will be able to
Write chemical equations for the organic reactions with main products formed
during the reaction.

Draw correlations between the type of functional group and the chemical
properties of organic compounds.

Compare the acidic strength of Ethanoic acid with any other mineral acid

Recognise various reagents like -a reducing or an oxidizing agent used in organic
reactions.
Ethanoic Acid or Acetic Acid
It is generally present in many fruits and their sour taste is because of the acid. A dilute
aqueous solution of acid containing only 5 to 8 percent acid is called vinegar, which is
used for preservation of sausage, pickles etc.
Physical properties:
(a)
It is colourless, pungent smelling liquid.
(b)
It is soluble in water in all proportions.
(c)
The acid boils at 391 K
(d)
On cooling, pure Ethanoic acid freezes to form ice like flakes which
resemble a glacier. Due to this property, pure Ethanoic acid is often called
glacial acetic acid or glacial Ethanoic acid.
Chemical properties: The chemical characteristics of the acid are mainly of the carboxyl
group.
(a)
Acidic nature: On adding a few drops of blue litmus solution to the
aqueous solution of the acid, the solution acquires a red colour.But
ethanoic acid is a weaker acid than mineral acids like H2SO4, HCl and
159
HNO3 since mineral acids are completely ionized while carboxylic acids
are partially ionized.
(b)
Reaction with alkalis or bases: Like mineral acids, Ethanoic acid reacts
with a base like sodium hydroxide to give a salt and water.
CH3COOH
+
Ethanoic acid
NaOH
→
CH3COONa +
Sodium hydroxide
H2O
Sodium ethanoate
Water
(Acid)
(Base)
(Salt)
This reaction between an acid and a base to form a salt and water is called
neutralization reaction.
(c)
Reaction with sodium carbonate and sodium hydrogen carbonate:
Ethanoic acid reacts with Sodium carbonate or sodium hydrogen
carbonate to form a salt and water with the evolution of carbon dioxide
gas.
2CH3COOH + Na2CO3 →
2CH3COONa +
Ethanoic acid Sodium
H2 O
Sodium ethanoate
+ CO2
Water
Carbon dioxide
Carbonate
CH3COOH + NaHCO3 →
CH3COONa +
H2 O
Ethanoic acid Sodium
Sodium ethanoate
Carbon dioxide
hydrogen
+ CO2
Water
carbonate
(d)
Esterification: Reaction between a carboxylic acids and an alcohol to form
an ester is called the esterification reaction. For example, Ethanoic acid
160
reacts with ethanol in presence of concentrated sulphuric acid (catalyst) to
form ethyl ethanoate an ester.
CH3COOH
+
Ethanoic acid
C2H5OH
→
Ethanol
(a carboxylic acid)
CH3COOC2H5
+
H2O
Ethyl ethanoate
(An alcohol)
(An ester)
Esters are sweet smelling substances. So they are widely used in making
perfumes and flavouring agents.
Esters on heating with an aqueous acid or a base give back the original
alcohol and the original carboxylic acid. For example,
Heat
CH3COOC2H5
+
Ethyl ethanoate
NaOH →
Sodium
CH3COONa +
C2H5OH
Sodium ethanoate
Ethanol
Hydroxide
This reaction is known as saponification reaction because it is used in the
preparation of soap.
Aim: To study the properties of Ethanoic acid and compare its acidic strength with
hydrochloric acid.
Materials Required:
Each group of students will need:
Test-tube rack
Test-tubes, 6
Beakers (100 ml), 2
Glass rod
Dropping pipettes, 2
Universal indicator pH colour chart
161
Magnesium ribbon two-1 cm strips
Ethanoic acid solution, (0.05 mol dm–3 )15 ml
Hydrochloric acid solution, (0.05 mol dm–3 ) 15 ml
Sodium hydroxide solution,( 0.4 mol dm–3 ) 5 ml
Sodium carbonate solution, (0.4 mol dm–3 ) 5 ml
Universal indicator solution (full range, pH 1-14)
Procedure:
1.
Place six test-tubes in a test tube rack.
2.
In three of the test-tubes take about 1 ml of ethanoic acid solution.
3.
In the other three tubes take about 1 ml of hydrochloric acid.
4.
pHAdd two- three drops of full-range Universal indicator solution to one of the
ethanoic acid tubes and note the pH.
Add two-three drops of full-range Universal indicator solution to one of the
hydrochloric acid tubes and note the pH.
5.
Reaction with sodium carbonate
Take 1 ml of sodium carbonate solution in a small beaker.
Use a dropping pipette to add one drop of sodium carbonate solution to the
ethanoic acid tube which contains the indicator. Stir the tube with a glass rod
and note observations if any. Continue to add drops until the pH is neutral.
Count the number of drops you have used.

Repeat the above procedure of adding the sodium carbonate to the tube
containing hydrochloric acid and indicator.
6.
Reaction with sodium hydroxide
Take 1 ml of sodium hydroxide solution in a small beaker.
Use a dropping pipette to add one drop of sodium hydroxide solution to the
ethanoic acid tube which contains the indicator. Stir the tube with a glass rod
162
and note observations if any. Continue to add drops until the pH is neutral.
Count the number of drops you have used.

Repeat the above procedure of adding the sodium hydroxide to the tube
containing hydrochloric acid and indicator.
7.
Reaction with magnesium
Add a small piece of magnesium ribbon to the remaining hydrochloric acid
tube. Note the observation and try to identify the gas given off.

Repeat the above procedure using the third ethanoic acid tube and compare
the rate of reaction with that of hydrochloric acid by noting the rate at which
the gas is given off.
Observations:
Properties
Ethanoic acid
Hydrochloric acid
pH
Reaction with Sodium
Carbonate
(No. of drops used)
Reaction with Sodium
hydroxide
(No. of drops used)
Reaction with
Magnesium ribbon
Conclusion:
______________________________________________________________________________
___________________________________________________________________________
163
Worksheet-6
Write balanced chemical equation for the following chemical reaction:
a)
Combustion of ethane.
________________________________________________________________________
b)
Reaction of ethanol with hot alkaline potassium permagnate
_______________________________________________________________________
c)
Reaction of ethene with Hydrogen in presence of palladium or nickle
________________________________________________________________________
d)
Reaction of Methane with chlorine in presence of sunlight.
________________________________________________________________________
e)
Reaction of ethanol with Sodium Metal
________________________________________________________________________
f)
When ethanol is reacted with hot conc. Sulphuric Acid.
________________________________________________________________________
g)
Reaction of ethanoic acid with ethanol in presence conc. H2SO4.
________________________________________________________________________
h)
Reaction of ethylethanoate (CH3COOC2H5) with Sodium hydroxide.
________________________________________________________________________
i)
Reaction of acetic acid with sodium hydroxide.
________________________________________________________________________
j)
Reaction of acetic acid with Sodium bicarbonate
________________________________________________________________________
164
Content-8
Polymers-Nylon and Terylene
Learning Objectives:
The students will be able to
explain the terms - monomer,polymer and polymerisation and appreciate their
importance

describe the preparation and properties of nylon and terylene

appreciate the importance of polymers in daily life.
Polymers
The word ‗polymer‘ is coined from two Greek words:poly means many and mer means
unit or part. The term polymer is defined as very large molecules having high
molecular mass (103-107u). These are also referred to as macromolecules, which are
formed by joining of repeating structural units on a large scale. The repeating structural
units are derived from some simple and reactive molecules known as monomers and
are linked to each other by covalent bonds. This process of formation of polymers from
respective monomers is called polymerisation. Polymers can be classified on the basis of
mode of polymerisation into two sub groups.
1.
Addition polymers
The addition polymers are formed by the repeated addition of monomer molecules
possessing double or triple bonds, e.g., the formation of polythene from ethene and
polypropene from propene. However, the addition polymers formed by the
polymerisation of a single monomeric species are known as homopolymers,
e.g.,polythene.
The polymers made by addition polymerisation from two different monomers are
termed as copolymers, e.g., Buna-S, Buna-N, etc.
165
2.
Condensation polymers
The condensation polymers are formed by repeated condensation reaction
between two different bi-functional or tri-functional monomeric units. In these
polymerisation reactions, the elimination of small molecules such as water,
alcohol, hydrogen chloride, etc. take place. The examples are terylene (dacron),
nylon 6, 6, nylon 6, etc. For example, nylon 6, 6 is formed by the condensation of
hexamethylene diamine with adipic acid.
(a)
Preparation of nylon
(i)
Nylon 6,6: It is prepared by the condensation polymerisation of
hexamethylenediamine with adipic acid under high pressure and at
high temperature.
Nylon 6, 6 is used in making sheets, bristles for brushes and in textile
industry.
(b)
Preparation of Terylene/Dacron
Terylene is manufactured by heating a mixture of ethylene glycol and
terephthalic acid at 420 to 460 K in the presence of zinc acetateantimony trioxide
catalyst. Dacron fibre (terylene) is crease resistant and is used in blending with
cotton and wool fibres and also as glass reinforcing materials in safety helmets,
etc.
166
Polymers already have a range of applications that far exceeds that of any other class of
material available to man. Current applications extend from adhesives, coatings, foams,
and packaging materials to textile and industrial fibres, electronic devices, biomedical
devices, optical devices, and precursors for many newly developed high-tech ceramics.
167
1500’s
British explorers discover
the ancient Mayan civilizationin
Central America. The Mayans are assumed to be among the first to find an
application for polymers; as their children
were fond of playing with balls made
from local
rubber trees.
HISTORY
OF POLYMERS
1917
X-ray crystallography is invented as a method of analyzing
crystal structures. Eight years later,
this method is used by M. Polanyi to
discover the chemical structure ofcellulose. This establishes the fact that polymer unit cells contain sections of long
chain molecules rather than small
molecular species.
1839
Charles Goodyear discovers
vulcanization, by combining natural rubber with sulfur and heating it
to 270 degrees Fahrenheit. Vulcanized
rubber is a polymeric substances that is
much more durable than its natural counter part. Its most common use today is in
automobile tires.
1907
The oldest recorded synthetic plastic is fabricated by Leo
Bakeland. Bakelite’s hardness and
high heat resistivity made it an excellent
choice as an electrical insulator.
1500PRESENT
1920
Staudinger published his
classic paper entitled “Uber
Polymerization.” Publication of this
paper heralded a decade of intense research and presented to the world the
development of modern polymer theory.
1927
Large scale production of
vinyl-chloride resins begins. This
polymeric compound continues to be
widely used today to make plumbing
(PVC) pipe, Euttery tile, and bottles.
3
168
Applications of Polymers

Polymeric materials are used to improve soil aeration, supply mulch, and
support plant growth and health.

Many biomaterials, in particularly for heart valve replacements and blood
vessels, are prepared from polymers like Dacron, Teflon and polyurethane
169

Polymers are used for the preparation of plastic containers ,clothing, floor
coverings, garbage disposal bags, and packaging. These are economically less
expensive than the more traditional materials.

In industries polymers are used for the manufacture of automobile parts,
windshields for fighter planes, pipes, tanks, packing materials, insulation, wood
substitutes, adhesives, matrix for composites, elastomers etc.

Playground equipment, various balls, golf clubs, swimming pools, and
protective helmets are often produced from polymers.
Quick Facts on Plastic Pollution

A plastic milk jug takes 1 million years to decompose.

Recycled plastic can be used to make things like trash cans, park benches,
playground equipment, decks etc.

Special fleece-like fabrics used in clothes and blankets can be made out of
recycled plastic bottles.

Plastic bags and other plastic garbage thrown into the ocean kill as many as 1
million sea creatures every year.

A plastic cup can take 50 - 80 years to decompose.

Recycling plastic saves twice as much energy as burning it in an incinerator.

The worldwide fishing industry dumps an estimated 150,000 tons of plastic into
the ocean each year, including packaging, plastic nets, lines, and buoys.

Nearly all of the plastic in any form, ever made still exists today.
Major Plastic Resins and Their Uses
Resin Name
Common Uses
Examples of Recycled Products
Polyethylene
Soft drink bottles, peanut
Liquid soap bottles, strapping,
Terephthalate (PET butter jars, salad dressing
170
fiberfill for winter coats,
or PETE)
bottles, mouth wash jars
surfboards, paint brushes, fuzz on
tennis balls, soft drink bottles, film
Soft drink based cups, flower pots,
High density
Milk, water, and juice
Polyethylene
containers, grocery bags, toys,
(HDPE)
liquid detergent bottles
Polyvinyl Chloride Clear food packaging,
or Vinyl (PVC-V)
Low density
Polyethylene
(LDPE)
drain pipes, signs, stadium seats,
trash cans, re-cycling bins, traffic
barrier cones, golf bag liners, toys
Floor mats, pipes, hoses, mud flaps
shampoo bottles
Bread bags, frozen food bags,
Garbage can liners, grocery bags,
grocery bags
multi purpose bags
Manhole steps, paint buckets,
Polypropylene
(PP)
Ketchup bottles, yogurt
videocassette storage cases, ice
containers, margarine, tubs,
scrapers, fast food trays, lawn
medicine bottles
mower wheels, automobile battery
parts.
Polystyrene (PS)
Video cassette cases, compact
License plate holders, golf course
disk jackets, coffee cups,
and septic tank drainage systems,
cutlery, cafeteria trays, grocery desk top accessories, hanging files,
store meat trays, fast-food
food service trays, flower pots,
sandwich container
trash cans
Resource: http://matse1.matse.illinois.edu/polymers/polymers.html
171
Worksheet - 7
1.
Define the following terms: monomer, polymer, and addition polymerization.
2.
Define polymerisation and give three examples of common polymers that you
come across in your daily life.
3.
Write the names of monomers and structure used for getting the following
polymers ?
(i)
Polyvinylchloride
(ii)
polyethene
(iii) Teflon
(iv) Nylon
4.
Using the given monomer for polymerization
Cl
Cl
│
│
C == C
│
│
Cl
Cl
Write the chemical reaction for the polymerization of n molecules of
tetrachloroethylene to form a polymer.
5.
Identify the monomer in the following polymeric structures.
( ii)
172
6.
Which artificial polymer is present in bubble gum or chewing gum ?
7.
Name the polymer used for making i) medicinal Capsule ii) Electrical goods.
8.
Is (CH2 — CH — C6H5)n a homo polymer or a copolymer?
9.
Which colligative property is used to determine the molecular masses of the
polymers ?
RUBRICS OF ASSESSMENT FOR LEARNING
Unit 4, Carbon and its Compounds
Parameter

Beginning
Approaching
Meeting
Exceeding
(1)
(2)
(3)
(4)
Is able to assign IUPAC
names and structure to
hydrocarbons ,alcohols,
carbonyl compounds,
carboxylic acids upto 6
carbon atoms per
molecule

Homologous series and
functional group.

Fractional distillation of
petroleum and uses of
fractions

Alkanes-Properties and
Bonding
173

Alkenes- Preparation and
properties

Difference between
saturated and
unsaturated
hydrocarbons.

Addition polymerisation

Alcohols-Preparation by
fermentation

Use of alcohol as a fuel
and solvent

Acids-Preparation by
oxidation of alcohols

Esterification

Polymerization
Structure of Nylon and
Terylene

Pollution caused by non
biodegradable plastics.

Natural macromoleculesStructure of Proteins, Fats
and Carbohydrates.

Hydrolysis of Fats,
Proteins and
Carbohydrates.
174
Videos:
Carbon Compounds: http://www.youtube.com/watch?v=Kjn5Ht0Vn30
Polymerization: http://www.youtube.com/watch?v=uCXBu8j6dlk&feature=related
Resources:
Web Links
http://en.wikipedia.org/wiki/Fractional_distillation Fractional Distillation
www.chemguide.co.uk/organicprops/alcohols/oxidation.html covers the theory well.
www.mp-docker.demon.co.uk/as_a2/topics/oxidation_of_alcohols/index.html
provides useful ―quizzes‖ on the work
www.gcsechemistry.com/rc16.htm General information on glucose fermentation.
http://ochem.jsd.claremont.edu/pubs.dir/JCE_2004_p513(Poon).pdf
Class room activity- Journal of Chemical Education
www.chemguide.co.uk/organicprops/acids/acidity.html Gives details of these and
other properties of ethanoic acid.
www.scool.co.uk/topic_quicklearn.asp?loc=ql&topic_id=1&quicklearn_id=1&subject_i
d=21&ebt=&ebn=&ebs=&ebl=&elc Gives simple acid-alkali facts
Read more: Plastic - water, environmental, United States, types, impact, industrial,
liquid, toxic, power, sources, disposal, use, oil, Pollution Problems
http://www.pollutionissues.com/Pl-Re/Plastic.html#ixzz1PUexYU00
175
Biology
Heredity and Evolution
Unit-4
Syllabus Coverage
Extension
Core
S

Accumulation of variations
during reproduction
Y
L

Rules for the inheritance

Structure of DNA
determination in human beings
 Evidences that help in tracing
evolutionary relationships
 The steps in species formation
(Deoxyribonucleic acid)
L

How do traits get expressed?
A

Origin of life

How the original simple
B
forms gave rise to the present
day wide variety of complex
U
S
 The chromosomal basis of sex
organisms

our solar system

Evolution and Classification
176
and human evolution
Matrix
Content/
Intended learning
Concept
Skill
Structure of
Understand the ladder model of DNA and the 
Comprehend,
DNA
base pairing rule
Construct
Traits
Understand that genetic basis of traits
Identify, relate
Know that the traits are passed from one
and analyze
generation to next and are donated by both
parents
Gene,
Learn and understand various terms related to
Compare,
genotype,
Genetics
Contrast,
phenotype,
Examine
homologous
chromosomes
Rules for
Deduce the rules of Genetic inheritance
Comprehend,
Inheritance
Justify the role of genes in Inheritance
Understand
Genetic
Identify the genetic variation and justify them
Grasp and
Variation
Connect
Chromosomal
Understand and relate with the chromosomal
Comprehend,
basis of sex
inheritance
compare
 Relate that elements in the solar system can form
Relate to real life
determination
in human
beings
our solar
system and
organic molecules in appropriate suitable
situation, Decide,
life
physical conditions.
Describe,
 select a planet most likely to support life as
we know it, in a fictional solar system
177
Identify,
Understand,
 Describe what makes a planet habitable
solve problem
 Identify that appropriate suitable conditions
were present on earth when it was
 Cooling for organic molecules to be formed
origin of life
A- Abiogenesis
 Describe the process of synthesis of organic
Relate to real life
molecules from inorganic components of
situation, Decide,
earth
Describe,
 Undestand that organic molecules have the
Identify
B-“Inorganic
capacity to combine and behave like living
Understand,
and Organic
molecules.
handle Lab,
Compounds‖
1. recognize coacervates
instruments,
Creativity,,
2. recognize the life-like properties of
coacervates.
Show,
recognize,define
3. Show that under suitable conditions,
complex life-like cell-like structures can
be produced naturally from simple
materials with simple changes
 Define Abiogenesis
 Identify the scientific theory Abiogenesis as
the most natural explanation and acceptable
theory to explain how life originated .
1. Explain the origin of life on earth need
not have required supernatural forces
2. investigate the chemical compounds in
living things and contrast these with the
major elements of the Earth‘s crust
3. compare inorganic compounds to organic
178
Investigate,
compare,
Discover,
determine
compounds and examine examples of
each group of organic compounds
4. discover why carbon is referred to as the
unparalleled element.
5. determine the rationale behind the
scientific classification system.
Evolution





Relate environmental change to changes in
Relate to real life
organisms
situation, Decide,
Describe the importance of adaptation in
Describe,
avoiding predation
Identify
Explain how natural selection causes
Understand,
populations to change
solve problem,
Discover that fittest survive in the given
discover,explain,
situation
demonstrate
Demonstrate the mechanism of Natural
selection
A Long Time- a

time line
Explain that Life has been on Earth for a long
Relate to real life
time.
situation, Decide,
Describe,
Identify
Understand,
explain
evidences that

Describe fossils as evidence of past life.
elate to real life
help in tracing

prove that life has gradually evolved with
situation, Decide,
the help of Study of fossils
Describe,
evolutionary
relationships
Identify
Understand,
prove
179
Evolution and
Classification
 Construct a cladogram with the help of given Relate to real life
groups of organisms and some of their
situation, Decide,
distinguishing characteristics.
Describe,
 interpret and analyze the cladogram in terms
Identify
of how it shows common ancestry and
Understand,
degrees of evolutionary relationship,
create, interpret
 describe that modern classification is based
on evolution theory.
the steps in
 Analyse that while natural selection explains
Relate to real life
species
evolutionary modifications within lineages,
situation, Decide,
formation
speciation explains evolutionary branching
Describe,
and diversification.
Identify
 explainthat speciation involves genetic
Understand,
differentiation, ecological differentiation
explain, analyse,
(niche separation) and reproductive
Discover,
isolation.
provide
 Give evidence that isolation of members of a
evidences
species in different environments may result
in the formation of a number of subspecies.
human
 handle and read the measuring instruments.
Relate to real life
evolution
 identify the appropriate skeletal and dental
situation,
features and landmarks
Decide,Describe,I
required for the measurements and
dentify
descriptions.
Understand,use
 describe features of a given specimen as
of lab.
either similar to, different from or the same
Instruments,reco
as those present in another specimen.
gnize,construct,a
 recognize the sequence pattern in which
180
nalyse
several human skull features appeared over
time.
 construct and justify a taxonomic
classification of the specimens.
 Explain that transitional forms in an
evolutionary sequence are generally mosaic;
some traits evolve more rapidly than others.
 analyse that modern humans have not
evolved from modern apes: both have
evolved from a common ancestor
 recognize some of the patterns revealed and
see how hominines have changed over time
181
Scope Document
UNIT 4: Heredity and Evolution
Reproduction with variation is a major characteristic of life. The earliest single-celled
organisms reproduced by duplicating their genetic material and then dividing in two.
The two resulting daughter cells were identical to each other and to the parent cell,
except for mutations that occurred during the process of gene duplication. Such errors,
although rare, provided the raw material for biological evolution. The combination of
reproduction and errors in the duplication of genetic material results in biological
evolution, a change in the genetic composition of a population of organisms over time.
The diversification of life has always been driven by variation in the physical
environment. There are cold places and warm places, as well as places that are cold
during some parts of the year and warm during other parts. Some places (oceans, lakes,
rivers) are wet; others (deserts) are usually very dry. No single kind of living thing can
perform well in all these environments. In addition, living things generate their own
diversity. During the process of evolution, as the plants evolved, they became the main
source of food for other living things. The plant eaters were, in turn, potential food for
other organisms. And when most of the living things die, they become food for still
other organisms.
For a long period of time, there was no life on Earth. Then there was an extended period
of only unicellular life, fol-lowed by a proliferation of multicellular life. In other words,
the nature and diversity of life has changed over time. Identification of the processes
that result in biological evolution was one of the great scientific advances of the
nineteenth century.
182
Learning outcomes - Foundation
At the end of this unit students should be able to :

Recognize how variations are accumulated during reproduction

Understand various Rules for inheritance

Explain the Structure of DNA (Deoxyribonucleic acid) and highlight its
important ascpects

Summarize how the traits get expressed in individuals?

Conceptualize Origin of life in detail

Understand how did the original simple forms gave rise to the present day wide
variety of complex organisms

Learn, in detail, about our solar system

Identification of processes that result in Evolution

Learn why there was a need for Classification
Learning outcomes – Extension
At the end of this unit students should be able to :

Explain the chromosomal basis of sex determination in human beings

Give evidences that help in tracing evolutionary relationships

Discuss the steps in species formation and human evolution
183
Teachers’ Notes
Reproductive processes give rise to new individuals that are similar but with some
variations. Thus the similarities between parents and their offsprings are due to
‗heredity‘ and the dissimilarities are due to ‗variations‘. Continuity of life is maintained
through the heredity and evolution. Heredity explains that the organisms resemble
each other because they arise form common ancestor. Evolution is the orderly changes
of various forms through a slow but continuous process. It gives rise to more complex
body designs even while the simpler body designs continue to flourish.
Teacher Note for Various Activities
Activity 1
Structure of DNA
Teacher Activity:
Discuss with students what they know about how genetic information is transmitted
from one generation to the next. Ask them why scientists would be interested in
knowing more about this topic and why they as students are interested.
The teacher will then explain the structure of DNA and the base pairing rule. The
difference in nucleotide and nucleoside can be shown using cartoon figures and also
videos on you tube.
The structure of a nucleotide can be drawn and cut out in a puzzle form. Photocopy the
two sets of puzzle pieces. Cut out the pieces and place a set of both correct and incorrect
pieces in an envelope. Enough sets should be made so that students can work in teams
of two.
Student activity:
184
The students will solve the puzzle of the nucleotide.
The students can also role model as reporters for Newspaper Science section. They
have been assigned to write a series of one-page articles informing readers about why
DNA has been heralded as the most important molecule ever discovered. The first piece
should explain the basic structure of the DNA molecule. These articles should be
written for the general public.
The students can also be given various nitrogen bases and they can pair them according
to the Chargaff rule
Activity 2
Genetic Traits
The teacher may begin the class with the following questions:
What traits do you share in common with others in your family? What traits are unique
to you?
Invite the students to sit in a circle. Hold up one game card at a time to show the picture
of a trait. All students that have this trait should stand or raise their hand. Continue
with each game card to find similarities and differences among family members. The
students would then be required to do the same activity at home with his /her family
members and record the observations.
Students mark their traits on tree leaf cut-outs
Student activity:
185
Students review and become familiar with basic genetic concepts and terms, such as
DNA, chromosome, gene, allele, homozygous, heterozygous, recessive and dominant
genes, genotype, phenotype. They will understand
complex traits, Mendelian
inheritance, and Punnett Square. Students apply these to identify and examine several
examples of simple and complex genetic traits in several characters in Harry Potter.
Students also examine inheritance patterns of magical ability in Harry Potter, and use
the concepts they have learned to identify possible genotypes of the magical ability
demonstrated by several characters in the series.
Activity 3
Genes as Decision Makers
The purpose of this activity is to extend the knowledge that genes are the units of
inheritance for individual characteristics and decide the phenotype of an organism. The
teacher may give the following concept using a chart or an aid like spoon and fork/
socks/ribbons.
Most genes have two or more variations, called alleles. For example, the gene for
hairline shape has two alleles – widow‘s peak or straight. An individual may inherit
two identical or two different alleles from their parents. When two different alleles are
present they interact in specific ways. For the traits included in this activity, the alleles
interact in what is called a dominant or a recessive manner. The traits due to dominant
alleles are always observed, even when a recessive allele is present. Traits due to
recessive alleles are only observed when two recessive alleles are present.
Review with the students Mendel's experiments with the pea plant.
Student activity:
186
The students can make posters of animals and their offspring. Students will recognize
their likenesses and differences.
Students will turn toward their partners and observe their physical traits such
as eyes, hair, cheekbones etc.
Students will complete the activity sheet using the mirrors in which they will
check their own physical traits for their own recessive and dominant traits.
The students can also make a list of their traits and compare it with their family
members. They can also depict the same in a tree.
Activity 4
Rules of Inheritance
Introduce the use of Punnett squares to predict the inheritance of traits in living
things.The teacher may prepare blank punette squares where the students will enter the
probability.
Additional activityFill two bags with peas. The first bag will represent the original cross between a yellowpea parent and a green-pea parent. This bag should be filled with only yellow peas. As
you are describing the cross between the two parents, ask the students what they think
the offspring (F1) generation will look like. After students have offered some
suggestions, have one student pull a handful of peas out of this first bag. The students
will be able to see that only yellow peas come out of the bag, just as only offspring
producing yellow peas result from this cross.
The second bag will represent the cross between two yellow F1 offspring plants. In this
bag, put a combination of yellow and green peas: ¾ yellow and ¼ green. As you
187
describe the cross between the two F1 plants, ask students what they think the offspring
of this cross will look like. Then, have a student take a handful of peas out of the second
bag. The peas in the second bag represent the F2 offspring of this cross. Thus, students
will be able to see that this time, some of the offspring are green. If the students count
up the number of yellow peas and the number of green peas in the handful drawn from
the bag, they will be able to see that the F2 plants produce peas at an approximate ratio
of 3 yellow: 1 green.
Activity 5
Genetic Variation
Human Genetic Variation has two central objectives. The first is to introduce students to
major concepts related to human genetic variation. The second objective is to convey to
students the relationship between basic biomedical research and the improvement of
personal and public health.
Students conduct a class wide inventory of human traits, construct histograms of the
data they collect, and play a brief game that introduces students to major concepts
related to human genetic variation and the notion of each individual's uniqueness.
The teacher may discuss various aspects which will lead to genetic variation.
Student activity:
The students will do a comparative study of the traits inherited by the offspring as
compared to their parents. They may do a research on the nature of alleles influencing
the phenotype.
Activity 6
Sex Determination
188
The teacher may pose following questions at the beginning of the class.
What‘s the evolutionary value of mixing genetic information (recombination)?
Why don‘t brothers and sisters look more alike? Why don‘t children look more like
their parents? What sex chromosomes do YOU have?
Discuss the normal sex chromosome pairing and mention some unusual genetic
conditions that result from extra sex chromosomes or the absence of a sex chromosome.
Why doesn‘t nature favor the abnormal sex complements?
A man who carries a defective (mutated) Y chromosome will pass the defective
chromosome on to
which of his children?
-all his children
-half of his children
-only his sons
-only his daughters
Student activity: the students will draw a concept map to show sex linked inheritance.
189
Activity 7
In the previous year ,students learned how the presence or absence of a carbon cycle
explains very existence of life .Here students will learn important differences among
Mars, Earth, and Venus based on carbon cycle operating there. This activity helps to
identify the conditions necessary to support on-going life on a planet. After considering
the conditions needed for life as we know it, students select the most suitable planet for
survival
in a fictitious planetary system, on which to crash land. This leads to a
discussion about what conditions makes a planet habitable. By explaining why Earth is
able to support life, this activity serves as a lead-in for the next step, Origin of Life.
Activity 8
In any introduction of lesson dealing with the possible scenarios of a natural origin of
life (as opposed to a supernatural cause) it is important to make clear the distinction
between evolution and the origin of life. These are not parts of the same process.
Mechanisms operating and environmental conditions, when life came into existence
may be quite different than the mechanisms and conditions which cause and control
evolution. In addition, any indication that life began billions of years ago, left no
evidence of the process (no "molecular fossils"), and has not taken place since. In
contrast, evolution (which provides an explanation for the diversity of life) has left
fossil indicators of past lives, and shows evidences of going on today. Therefore,
evolution is much easier to study, with an abundance of various types of evidence. With
the continuation of research going on today, we may have enough circumstantial
evidence to describe a plausible scenario for how life began.
The concept of "special creation" combines both processes ,evolution and the origin of
life and means: "life in all its forms was created by divine power all at once." But if we
can give evidences that evolution has produced the diversity of life, we are left with
190
only the question about how life began. In science, we cannot use supernatural forces to
explain natural phenomena.
Upon finishing the evolution introduction unit, the point may be raised that: "If
evolution shows us how life has changed over millions of years and how different
species began. The next question may occur is 'how did life begin?'. Teacher may look at
some of the ideas scientists have explored, and some of their evidence. At this stage a
brief introduction or review of some basic terminology and chemical concepts related
to basics of the chemistry of life and molecular biology ,(elements and compounds, atoms
and molecules, ionic and covalent bonds, acids, basis, and pH, organic and inorganic,
simple inorganic molecules H2, H2O, NH3, CH4 and CO2, organic molecules like
amino acids, proteins, sugars and carbohydrates) may be given to students .
1.
The study of major organic groups (proteins, carbohydrates, lipids, nucleic acids)
and the pH concept, may lead to an introduction to ideas and studies bearing on
the origin of life. It would be very helpful if students have
seen some live
amoebas under microscope.
2.
Teacher may inform that "we are going to create molecules needed to create life!
(well, almost). At least, we will take a look at some of the structural and behavioral
aspects of molecules of a living cell, using very simple materials, something
analogous to the likely conditions on the early pre-life earth."
3.
Teacher may want to precede the activity with some discussion of the Urey and
Miller experiment (simulating one pre-life scenario of simple gases exposed to
electrical discharges, producing amino acids and other organic compounds).
4.
Teacher may prepare the mixtures and lab set-ups the day before the activity and
might want to assign reading material dealing with origin of life ideas and studies
It's also helpful for students to read over the purposes, preparation and procedure
for the activity before doing it.
191
5.
Students can work in pairs, sharing the equipment or prepares his/her own tube,
and tries to find coacervates under the microscope. Not all coacervates produced
will have the desired "amoeba-like" appearance, so as the teacher moves during
the activity, announce over any preparations which look good, and encourage
others to take a peep..
6.
It may be useful to do a quick demo of the procedure to entire class just before
students do the activity.
7.
If possible, teacher may cover this with a video camera - microscope set up to
display (and videotape) the better-looking coacervates found. When such tapes are
played back/ fast-forwarded, they can more clearly reveal the changes in shape,
size, and movements (which are usually fairly slow).
8.
Teacher may point out to the class that cell membranes today have evolved as
double layers of phospholipids (fat), with proteins inserted here and there serving
various functions which control transport of materials through the membrane,
and/or provide identity markers for the cells. The carbohydrate + protein --->
coacervate model may not be the most accurate portrayal of a likely pre-life
scenario. An activity which does show a more plausible scenario is described in
the section given below. It uses lecithin to produce little vesicles. Teacher may
want to try this, and if it works, he/she may have students try it.
ANOTHER ACTIVITY ON
THE ORIGIN OF LIFE
PHOSPHOLIPID VESICLE
FORMATION
Materials: microscopes, slides, cover slips, Lecithin (from local pharmacy) food
coloring, water dropper, blotting paper
192
Procedure: Students can work in teams of4-6. One small drop of lecithin solution to
be placed on a slide, and one drop of colored water is placed next to the lecithin
solution. Cover slip is placed so it covers both the lecithin and water. Students should
carefully examine the boundary between the water and lecithin at the start of the
period and after regular intervals during the rest of the period, drawing a picture of
what they see each time. They should be allowed to switch between low, medium and
high power, but should label each sketch made with the magnification power used.
Twenty five minutes into the period, students should exchange the colored water for
clear water by placing a piece of blotting paper along the water end of the cover slip
to draw out the colored water. A drop of clear water should be added at the lecithin
end and be sucked through. Repeat as needed to get fairly clear water around the
vesicles that have formed. They should be visible as colored spots containing
undiluted water and food coloring. Sketch a few vesicles.
Activity 9A
Teacher Note:

This activity demonstrates an important part of evolution: adaptation. Adaptation is "an
alteration or adjustment in structure or habits, often which are inheritable, by which a
species or individual improves its condition in relationship to its environment." In order to
survive in a changing environment, an animal needs to adapt, to develop traits that fit it to
where it lives. For example, mountain goats have evolved to have padded hooves to grip
the rocks on which they climb, giraffes have gained height as compare to their ancestors to
reach the leaves of tall trees, and frogs have long, sticky tongues to catch insects to eat
without moving, and are colored brown or green to match their environments so that they
can camouflage. In fact, nearly everything about an animal is an adaptation of one kind or
another. Think of a trait that an animal has, and then find out how that trait helps it to
survive successfully or reproduce in its environment.
193

Plants also adapt to their environments in order to survive successfully. For example, cacti
are able to store water because they live in dry places, and they have leaves modified in to
spines to discourage animals from eating them. The colour and fragrance of flowers of
flowering plants are an adaptation for attracting bees and butterflies, which enable the plant
to pollinate for reproduce.
Activity 9B
Teacher Note

Teacher may reinforce the concept of Natural selection by motivating students to
analyze how color affects the organism's ability to survive in certain environments

Teacher may tell students about the Industrial Melanism in England.
Industrial Melanism is a term used to describe the adaptation of a population in
response to pollution. One example of rapid industrial melanism occurred in
populations of peppered moths in the area of Manchester, England from 1845 to 1890.
Before the industrial revolution, the trunks of the trees in the forest around Manchester
were light grayish-green due to the presence of lichens. Most of the peppered moths in
the area were light colored with dark spots. As the industrial revolution progressed, the
tree trunks became covered with soot and turned dark. Over a period of 45 years, the
dark variety of the peppered moth became more common.
This will help students to understand the concept of Natural Selection.
194
Activity 10A
Teacher Note:
The earliest fossils date to about 3.7 billion years ago. This is the beginning of the
Precambrian Era. Life for most of the Precambrian consisted of unicellular bacteria, but
late Precambrian fossils include jelly fish like and wormlike life. The Paleozoic Era
began about 550 million years ago and was marked by the rise of arthropods (e.g.,
crustaceans and insects) as well as fish and amphibians. The Mesozoic Era (beginning at
250 mya) featured, dinosaurs, swimming reptiles, flying reptiles and early mammals. In
the Cenozoic Era (from 65 mya to present), mammals rose to dominate the large fauna
and birds colonized every continent.
It is interesting to note that all life today descended from Precambrian life and that
there are several examples of present-day life that are very similar to ancient forms.
Examples include jelly fish, corals, sea stars, and dragonflies. Remarkably, opossums
and shrews closely resemble mammals that shared the Earth in Mesozoic period.
Teacher may suggest high-interest books or videos about prehistoric life available in
library or in bookstores. Examples:
Maia: A Dinosaur Grows Up
by
John
R.
Horner
tells
an
engaging
tale.
Dinosaurs: The Biggest, Baddest, Strangest, Fastest by Zimmerman and Olshevsky is an
excellent picture book with interesting text.
Explore these links for additional information on the topics covered in this lesson:

Fossils

Geologic Time

History of Life

www.youtube.com/watch?v=ucSVdtd3rRc
Teaching Tips:
195
The numbers on the timeline will be impressive for students but teacher must ensure to
get across the idea that life has been on Earth for an extremely long time and it hasn‘t
always stayed the same.
Activity 10B
Teacher Note:
Fossils are formed when some living organism dies and instead of being destroyed by
scavengers, weather or other forces, it is preserved by being covered by sediment,
which over a long period of time, becomes rock. This most often occurs in lake or bay
environments where there is not a great deal of motion in the water. Sediment slowly
covers the deceased organism and covers it with protective layers preventing it to
decay, that later becomes rock. There are many other ways organisms or their traces can
be preserved (as an insect in amber or dinosaurs covered by a landslide or flood).
Teacher may suggest high-interest books or videos about fossils available in library or
in bookstores. Examples:

The UCMP website has many photos of fossils.

Real fossils can often be found in ―nature‖ stores or at rock shops.

Dinosaurs Walked Here and Other Stories Fossils Tell by Patricia Lauber Fossil Book
(The)

The Fossil book
by Gary Parke
Explore this link for additional information on the topics covered in this lesson:


Fossils
www.youtube.com/watch?v=SEDfRy6DQns
Teaching Tips:
196
Inform throughout the activity that the class is making impressions of living thing and
not fossils. We can learn about past looking at such impressions. This is similar to how
we can learn from fossil imprints that we find in ancient sediments.
Activity 11
Teacher Note
Teacher may revise the major groups of organisms (kingdoms, and at least the major
phyla, and classes of animals), their defining features, and common examples of each
which was done in an earlier grade.
Teacher in the casual discussion make students familiar of degrees of similarity of
structure being associated with degrees of biological relationship (analogous to family
traits in people indicating relationships) and may use it as a tool to introduce the
concept of descent with modification.
Teacher may explain about comparative anatomy, and also that different numbers of
shared derived characters exist between different groups. Based on this students can
draw a diagram of branching lines which connect those groups, showing their different
degrees of relationship. These diagrams look like trees and are called "phylogenetic
trees" or "cladogram" (CLAY-doe-grams); Teacher may provided examples The
organisms are at the tips of the stems. The shared ,derived features of the homologous
structures are shown on the cladogram by solid square boxes along the branches, and
common ancestors are shown by open circles. The more derived structures two
organisms share, the closer is their evolutionary relationship -- that is, the more recently
their common ancestor lived. On the cladogram, close relationships are shown by a
recent fork from the supporting branch. The closer the fork in the branch between two
organisms, the closer is their relationship.
197
For extension and further practice, teacher may provide with another set of organisms
and their distinguishing characteristics. On a separate sheet, student may prepare a table
for analyzing the data, draw a Venn diagram, and draw a cladogram for those
organisms, similar to the work done in activity
Activity 12
Teacher Note:
The Class activity explains quickly how species evolve. This is just a demonstration to
help students internalize concept of change occurring in a given population.
Teacher has to emphasize that the speciation evolves over a very long duration, Teacher
may take help of Time Line(-Activity-10A) to make students realize how long it has
taken for a cell to evolve and then period of Organic Evolution.
Activity 13
Teacher Note:
The teacher may emphasize on how well humans can be used as evidence to support
the idea that modern species are evolutionarily related to one another and descended
from now-extinct forms.
These casts which are required for the activity, are commercially available, Check
SKULLS: PRICE COMPARISON chart for online addresses and prices for
recommended casts of skulls.
Drawings of specimens may also be used for study but are not nearly as good as the
actual skull replicas. A set of 7 drawings of hominid profiles can be downloaded from
the site hominid drawings.
198
In order to accommodate a greater number of measurements, please find added a
collection of 28 HOMINID PHOTOS: 4 different views each of 7 skulls: front, top,
right side, and an under-view of each skull. In fact, teachers may refer for an excellent
craniometry- in an online article in the NABT journal The American Biology Teacher for
March 2007:
199
Teacher
Student
Study
Material
200
Activity 1
Learning outcome: The students will be able to understand the structure of DNA. The
students will be divided into groups and the teacher will explain the structure of DNA
highlighting the nucleotide.
Background Information: DNA is a polymer. The monomer
units of DNA are nucleotides, and the polymer is known as
a "polynucleotide." Each nucleotide consists of a 5-carbon
sugar (deoxyribose), a nitrogen containing base attached to
the sugar, and a phosphate group. There are four different
types of nucleotides found in DNA, differing only in the
nitrogenous base. The four nucleotides are given one letter
abbreviations as shorthand for the four bases.

A is for adenine

G is for guanine

C is for cytosine

T is for thymine
Purine Bases
Adenine and guanine are purines. Purines are the larger of the two types of bases found
in DNA.
Pyrimidine Bases
Cytosine and thymine are pyrimidines. The 6 stoms (4 carbon, 2 nitrogen) are
numbered 1-6. Like purines, all pyrimidine ring atoms lie in the same plane.
These are called "bases" because that is exactly what they are in chemical terms. They
have lone pairs on nitrogens and so can act as electron pair donors (or accept hydrogen
201
ions, if you prefer the simpler definition). This isn't particularly relevant to their
function in DNA, but they are always referred to as bases anyway.
Nucleosides
A nucleoside is one of the four DNA bases covalently attached to the C1' position of a
sugar. The sugar in deoxynucleosides is 2'-deoxyribose. The sugar in ribonucleosides is
ribose. Nucleosides differ from nucleotides in that they lack phosphate groups.
The teacher may conduct a role play with children playing the role of

Pentose sugar

Phosphate

Nitrogenous base
DNA model
A group of students will be required to build a large DNA demo model in which the
base sequence codes for the name,. Flat rectangular sheets of cardboard could serve as
the base-pair steps; flat pentagonal pieces as deoxyribose sugars, and flat round pieces
as the phosphate groups
1. Hand out magnetic whiteboards and bags with DNA pieces to each pair of
students.
2. Students assemble their DNA double strands.
3. Point out the ladder structure and the complimentary base pairing.
4. Have students count how many A's, T', G's and C's their DNA molecule contains.
5. Direct students to "unzip" their DNA molecule by sliding the magnetic pieces of
each strand apart.
6. Students then assemble the new, complimentary DNA strands.
7. The students will identify the original DNA strands and the new ones.
202
Worksheet 1
1.
Inside what structure are the cell's chromosomes found?
2.
How many chromosomes are there in nearly every one of your body's cells?
3.
What is the shape of the DNA molecule?
4.
How many different types of bases are there in a molecule of DNA?
5.
What is DNA?
6.
Where is DNA found?
7.
In what way does DNA provide the code of life?
8.
What are the four letters in a DNA molecule, and what do they represent?
9.
What types of things do genes control in a developing organism?
10. What are master control genes and how do they function?
11. What does the master control gene called the "eyeless gene" control? What
happens when it is removed or damaged?
12. What happened when the eyeless gene was transferred from a mouse to a fly?
13. Where does an individual organism get the genes it has?
14. Can an individual organism change the sequence of its genes during its lifetime?
15. How are the traits of an organism related to its DNA sequence?
16. What would happen if there was a "mistake" in the DNA sequence?
17. Where is the genome (complete genetic code) of an organism found?
18. Describe the basic structure of DNA.
19. What are genes? How are they related to chromosomes? How are genes related to
strands of DNA that make up an organism's genetic code?
20. What is the relationship between an organism's genes and its traits?
21. Is there a one-to-one relationship between genes and traits? In other words, does
one gene code for one trait?
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Activity 2
Learning outcome: The students will be able to follow
the concepts of genetic traits and understand its role in
inheritance.
Background Information: Genes are the portions of an
organism's DNA that carry the code responsible for
building that organism in a very specific way. Genes -and, thus, the traits they code for -- are passed from
parent to offspring. From generation to generation, well-understood molecular
mechanisms reshuffle, duplicate, and alter genes in a way that produces genetic
variation. This variation is the raw material for evolution.
Evolution is not a random process. The genetic variation on which natural selection acts
may occur randomly, but natural selection itself is not random at all. The survival and
reproductive success of an individual is directly related to the ways its inherited traits
function in the context of its local environment. Whether or not an individual survives
and reproduces depends on whether it has genes that produce traits that are well
adapted to its environment.
Sexual reproduction allows an organism to combine half of its genes with half of
another individual's genes, which means new combinations of genes are produced
every generation. In addition, when eggs and sperm are produced, genetic material is
shuffled and recombined in ways that produce new combinations of genes. Sexual
reproduction thus increases genetic variation, which increases the raw material on
which natural selection operates. Genetic variation within a species -- also known as
genetic diversity -- increases a species' opportunity for change over successive
generations.
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Evolution is not a random process. The genetic variation on which natural selection acts
may occur randomly, but natural selection itself is not random at all. The survival and
reproductive success of an individual is directly related to the ways its inherited traits
function in the context of its local environment. Whether or not an individual survives
and reproduces depends on whether it has genes that produce traits that are well
adapted to its environment.
The teacher may discuss the meaning of ‗traits‘ by doing the following exercise with the
students1.
Present the following scenario to your students: ―An exchange student from
Australia is coming to stay with your family for a month. You go to the airport to
pick her up and need to describe yourself to her so that she can find you in the
crowd at the airport. In 2-3 sentences, how would you describe yourself?‖
2.
After 5 or 6 students have shared, draw attention to some of the general categories
of responses. Note how some descriptors are biologically based (eye color,
ethnicity, hair color, height, etc.), whereas others are environmental (clothing,
accessories, dyed hair, etc.). In this class, we will focus on the biological
descriptors.
3.
Go over the vocabulary. ―Characteristics‖ are the general category of descriptions
(height, eye color, etc.) whereas ―traits‖ are the precise description of an
individual‘s characteristics (5‘2‖, blue eyes, etc.)
The teacher may then discuss how traits are inherited and are distinct for each
individual.
Each parent passes on half of their DNA to their offspring. DNA is the code that
determines how every single thing in the body is built. So each offspring gets the
instructions for half of their body building plan from the mother and half from the
father.
205
The teacher may explain how traits and proteins are linked and how exactly a trait gets
exhibited.
The red color seen in these flowers, is ...
... an example of a genetic trait. Traits are small parts of the phenotype of an
organism, such as the red color seen in the flower petals.
Under the magnifying power of a microscope, the flower petal cells can be
seen to contain red granular material that is absorbing the rays of white light
and only reflecting the "red" wavelengths. That is why the flower petals look
red to our eyes.
It takes the tools of molecular and cellular biology to see that, in the cytoplasm
of the cell, a chemical reaction is taking place in which a colorless molecule is
being converted into a red pigment molecule As with all chemical reactions
within cells, this conversion reaction is catalyzed by an enzyme catalyst.
Without this enzyme catalyst, the reaction would proceed so slowly that little
or no pigment would be produced.
The enzyme catalyst makes the chemical reaction proceed at "life speed".
When the enzyme is present in the cell it can produce a lot of pigment and
pigment granules. As these granules accumulate in the cells the flower petals
turn red - the genetic trait.
The "red" trait, therefore is the product of a chemical reaction catalyzed by the
enzyme.
Almost all enzymes are proteins.
Almost all traits are produced by the action of proteins.
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Worksheet 2
Yes or no/Multiple choice traits:
Trait
Self
1:
2:
Eye color – what color are your eyes?
brown
brown
brown
blue
blue
blue
green
green
green
hazel
hazel
hazel
yes
yes
yes
no
no
no
yes
yes
yes
no
no
no
Dimples – do you have dimples on
yes
yes
yes
your cheeks?
no
no
no
Earlobe attachment – are your
yes
yes
yes
no
no
no
yes
yes
yes
no
no
no
Freckles – do you have freckles? Say ―yes‖ only if
you have LOTS of freckles all over your nose and
cheeks. Sun freckles do not count.
Tongue rolling - can you roll your tongue into a
tube?
earlobes attached to the side
of your face?
Widow‘s peak – do you have a
widow‘s peak (hairline has a
V)?
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Cleft chin – do you have a
cleft chin?
Hair color – what color is your hair?
Hair texture – what is the texture of your hair?
yes
yes
yes
no
no
no
black
black
black
brown
brown
brown
blond
blond
blond
red
red
red
curly
curly
curly
wavy
wavy
wavy
straight straight straight
Activity 3
Learning outcome: The students will be able to
relate that genes are the units of inheritance for
individual
characteristics
and
decide
the
phenotype of an organism.
Many
thousands
of
genes
have
been
discovered, including many that have roles in
disease. These genes are scattered throughout
the human genome. But what is the human genome? The physical appearance of the
bulk of the human genome is 46 long, thin structures known as chromosomes. The
genome of all individuals (with a few exceptions) looks like that in figure on the right.
Genes are located on chromosomes. Genes are simply regions on chromosomes that
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code for polypeptides. Genes are sections or segments of DNA that are carried on the
chromosomes and determine specific human characteristics, such as height or hair
color. Because each parent gives you one chromosome in each pair, you have two of
every gene (except for some of the genes on the X and Y chromosomes in boys because
boys have only one of each).
Some characteristics come from a single gene, whereas others come from gene
combinations. Because every person has from 25,000 to 35,000 different genes, there is
an almost endless number of possible combinations!
Genes hold the instructions for making protein products (like the enzymes
to digest food or the pigment that gives your eyes their color). As your cells duplicate,
they pass this genetic information to the new cells. Genes can be dominant or recessive.
Dominant genes show their effect even if there is only one copy of that gene in the pair.
For a person to have a recessive disease or characteristic, the person must have the gene
on both chromosomes of the pair.
Student activity: The student will make a chromosome with a variety of things like
straws, socks, spoons and any other material to show the chromosomes and the concept
of homologous chromosome following.
The teacher would reinforce the fact that homologous chromosomes would carry the
gene for the same trait at the same location.
Genotype and Phenotype: To clear the concept of
genotype and phenotype the teacher may show a
picture of eye showing the color
to depict its
genotype and external appearance and may ask
questions on the same highlighting the terms.
209
Worksheet 3
1.
What do you understand by homologous chromosome? A hint is given in the
picture below.
2.
Complete the following using appropriate words.
Homologous
chromosomes are
a
matching
pair
of chromosomes containing
the
__________in the same order. One of the homologous chromosomes is received from the
father, and the other is received from the mother. Because there are two chromosomes in
the pair, the pair contains two different ______at every loci, one on each chromosome. This
allows for a great deal of variation in individuals, by means of dominant traits, recessive
traits, and polygenic traits, and masking genes. This type of variation, which accounts for
the bulk of population diversity does not come from _________ but the process known
as genetic recombination.
3.
On the basis of the
characteristics shown
below in the picture
give
the
dominant
and recessive alleles:
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Activity 4
Learning outcome: The students will be able to understand the rules for inheritance
Background Information: Mendel determined that an organism inherits two copies of
the genetic material that determines an individual‘s physical traits, one copy coming
from each the male and female parent. Mendel observed that for each trait, sometimes
what is inherited from one parent masks what is inherited from the other. He called the
hidden trait recessive and the expressed trait dominant.
The rules for inheritance of such traits in human beings are related to the fact that both
the father and the mother contribute practically equal amounts of genetic material to the
child. This means that each trait can be influenced by both paternal and maternal DNA,
thus the trait seen in the child be? Mendel worked out the main rules of such
inheritance and it is interesting to look at some of his experiments from more than a
century ago. Mendel used number of contrasting visible characters of garden peasround/wrinkled seeds, tall/tall/short plant, and white/violet flowers and so on. He
took pea plants with different characteristics-a tall plant and a short plant, produced
progeny from them and calculated the percentages of tall or short progeny. Mendel
chose seven physical traits (now referred to as phenotypes) to study: flower color and
placement, pod color and shape, pea color and shape, and plant size. These were all
easily observable properties of the plants and so could be quickly counted. Mendel‘s
goal was to reveal the genetic makeup (now called genotype) underlying each variety of
pea plant and to understand how each trait was inherited.
Activity: The teacher may have a role play of the students to show the inheritance rule.
The child may become alleles. The taller children may represent dominant alleles and
the three laws of Mendel can be depicted.
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Activity: In the Harry Potter series, characters
are born with or without magical ability. Those
with magical ability also show very strong,
normal or weak ability.
Assuming that the magical ability is inherited,
identify the possible phenotypes and genotypes
of the following characters: Harry, Hermione,
Ron, Dumbledore, Aunt Petunia, and Mr. Filch
Hints: Start by identifying phenotypes which will provide possible genotypes. Also
consider whether simple Mendelian or complex traits apply to the magical ability traits.
Worksheet 4
Use your knowledge of genetics to complete this worksheet.
1.
Use the information for SpongeBob‘s traits to write the phenotype (physical
appearance) for each item.
(a) LL-______________
(e) Rr-_______________
(b) yy-_______________
(f) ll- _______________
(c) Ss-_______________
(g) ss- _______________
(d) RR - _____________
(h) Yy -______________
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2.
Use the information in the chart above to write the genotype (or genotypes) for
each trait below.
3.
4.
(a) Yellow body - ___________
(e) Stubby nose - ___________
(b) Roundpants - ___________
(f) Round eyes - ____________
(c) Oval eyes - ______________
(g) Squarepants - ___________
(d) Long nose - _____________
(h) Blue body - ____________
Determine the genotypes for each using the information in the chart
(a) Heterozygous round eyes -_____
(c) Homozygous long nose - ______
(b) Purebred squarepants - ______
(d) Hybrid yellow body - ______
One of SpongeBob‘s cousins, SpongeBillyBob, recently met a cute squarepants gal,
SpongeGerdy, at a local dance and fell in love. Use your knowledge of genetics to
answer the questions below.
(a)
If SpongeGerdy‘s father is a heterozygous squarepants and her mother is a
roundpants, what is her genotype? Complete the Punnett square to show the
possible genotypes that would result to help you determine Gerdy‘s
genotype. What is Gerdy‘s genotype? _________
(b)
SpongeBillyBob is heterozygous for his squarepants shape. What is his
genotype?
(c)
Complete the Punnett square to show the possibilities that would result if
Billy Bob & Gerdy had children.
(d) List the possible genotypes and phenotypes for the kids.
(e)
What is the probability of kids with squarepants? _____ %
213
Activity 5
Learning outcome: The students will be able to identify the source of genetic variation
in a population and explain how genetic variations are not always harmful.
Background Information: Genetic variation describes naturally occurring genetic
differences among individuals of the same species. This variation permits flexibility and
survival of a population in the face of changing environmental circumstances.
Consequently, genetic variation is often considered an advantage, as it is a form of
preparation for the unexpected. The ultimate source of genetic variation is differences in
DNA sequences. Most genetic differences result in variations that do not adversely
affect us. Some differences however result in disease. Some variations in fact provide
some organisms with a survival advantage.
Sometimes, there can be random fluctuations in the numbers of alleles in a population.
These changes in alleles assortment, called genetic drift, can either increase or decrease
by chance over time.
Typically, genetic drift occurs in small populations, where alleles which are n less
proportion face a greater chance of being lost. Once it begins, genetic drift will continue
until the involved allele is either lost by a population or is the only allele present at a
particular gene locus within a population. Both possibilities decrease the genetic
diversity of a population.
Genetic variation in a population is derived from a wide assortment of genes and
alleles. The persistence of populations over time through changing environments
depends on their capacity to adapt to shifting external conditions. Sometimes the
addition of a new allele to a population makes it more able to survive; sometimes the
addition of a new allele to a population makes it less able. Still other times, the addition
of a new allele to a population has no effect at all, yet the new allele will persist over
generations because its contribution to survival is neutral.
214
There are three primary sources of genetic variation, which we will learn more about:
1. Mutations are changes in the DNA. A single mutation can have a large effect, but
in many cases, evolutionary change is based on the accumulation of many
mutations.
2. Gene flow is any movement of genes from one population to another and is an
important source of genetic variation.
3. Sex can introduce new gene combinations into a population. This genetic
shuffling is another important source of genetic variation.
Activity:
1. Show the students a variety of pictures of wolves or cats and pictures of a
number of herding animals in their herds. Ask them if the animals of one type
look exactly alike. If students see zebras or giraffes as identical to their relatives,
ask them to look more closely.
2. Help the class to see that the patterns on the faces of zebras or on the necks of
giraffes vary from one individual to another.
3. Repeat this process with a number of different species of animals.
4. Go outside and select two trees or shrubs of the same species. Ask students how
they are alike and how they are different. Leaf shape, color, and limb
arrangement all vary between individuals.
5. Close the lesson with this question: ―If an animal or plant is the same species
does that mean it is exactly the same as all others of its type?‖ Students should be
able to communicate, in their language, that individuals vary.
215
Worksheet 5
1.
_____________ is the father of genetics.
a. Gregor Mendel
b. Benjamin Franklin
c. Thomas Jefferson
d. Albert Einstein
2.
In a monohybrid cross between a homozygous dominant parent and
homozygous recessive parent, one would predict the offspring to be.....
a. all heterozygous
b all homozygous dominant
c. all homozygous recessive
d. all heterozygous recessive
3.
An organism with two identical alleles for a trait is...
a. heterozygous
b. homozygous
c. recessive
d. dominant
216
4.
In Mendel's experiments, a trait that dissappeared in the F1 generation but
reappeared in the F2 generation was always a ____________________.
a. dominant trait
b. heterozygous trait
c. recessive trait
d. homozygous trait
5.
This law states a pair of factors are separated during formation of the gametes.
a. Law of Independent Assortment
b. Law of Segregation
c. Law of Individualism
d. Law of Separate Alleles
6.
An individual heterozygous for a trait and an individual homozygous recessive
for the trait are crossed and produce many offspring that are....
a. all the same genotype
b. of two different phenotypes
c. of three different phenotype
d. all the same phenotype
217
Activity 6
Chromosomal Basis of Sex Determination
Learning outcome: The students will be able to understand the chromosomal basis of
sex determination in human beings.
Background Information: It is not so long ago that women were blamed if they failed to
produce a son for their husband and clearly it was thought that the power of sex
determination lay within the body of the woman. During this century the chromosomal
basis of human sex determination has been demonstrated and in the last few years
some of the genes responsible have been identified. The sexual identity of an individual
is determined at several levels, chromosomal sex, gonadal sex, somatic sex and sexual
orientation.
1 All chromosomes besides X and Y are autosomes.
2 Sex chromosomes of the normal male are X and Y; the normal female has two
X‘s.
3 The
X
and
Y
chromosomes
share
similarities and differences:

They share genes in common. Half
of the genes found on the Y are also
found on the X and these genes
have nothing to do with a particular
sex.
Their
gene
commonality
suggests the sex chromosomes probably evolved from a pair of identical
autosomes.

The Y chromosome is 1/3 the size of the X and contains approximately 30
genes compared to the 1000 genes of the X.
218

The Y chromosome has genes that are male specific in that their
expression determines maleness (sex determination), external organs and
internal accessory organs, and sperm production (fertility).
4. All genes on a given chromosome (autosome or sex chromosome) belong to the
same linkage group (and are often called linked genes). Genes of the linkage
group do not obey Mendel‘s Laws (review).
5. Crossing over between chromosomes (homologous chromosomes) creates
recombinant chromosomes by rearranging genes that were previously linked.
CLASS DISCUSSION on the following concepts (This could be done as a class
discussion or group assignment)
1. What‘s the evolutionary value of mixing genetic information (recombination)?
2. Why don‘t brothers and sisters look more alike?
Why don‘t children look more like their parents?
3. What sex chromosomes do YOU have?
Discuss the normal sex chromosome pairing and mention some unusual
genetic conditions that result from extra sex chromosomes or the absence of a
sex chromosome. Why doesn‘t nature favor the abnormal sex complements?
4. A man who carries a defective (mutated) Y chromosome will pass the defective
chromosome on to which of his children?

all his children

half of his children

only his sons **

only his daughters
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5. Historically, women have been held responsible (blamed!) for not producing
the male heir. Are women responsible for the sex of the child? Which parent
actually determines the gender of the child?
Activity 7
Our Solar System
Learning Objectives: After completing this activity, students will be able to:

Relate that elements in the solar system can form organic molecules in appropriate suitable
physical conditions.

select a planet most likely to support life as we know it, in a fictional solar system

Describe what makes a planet habitable

Identify that appropriate suitable conditions were present on earth when it was cooling
Content: Compared to other cosmic bodies, the Earth itself has a very hospitable
climate to support life. Our neighboring planets, Venus and Mars, range the extremes in
temperatures: the surface of Venus, at a temperature of 900º F, is hot enough to melt
lead, while in Martian nights the temperature drops to 220º F below zero.
Earth’s favorable climate: Earth‘s temperature is comfortably ―just right,‖for life to
survive. Is Venus just hot because it‘s closer to the Sun, and Mars cold because it‘s
farther away.
There are three reasons to the answer to this problem. The first reason is like insulation,
which keeps planets warm, ( d u e t o infamous greenhouse effect ) The greenhouse
effect makes Earth cozy. Without the greenhouse effect the Earth would be frozen,
more than 60º F colder.
We receive energy from our Sun in the form of light. The Sun is a ball of very hot gas,
220
heated by thermonuclear reactions at its core. All hot objects emit electromagnetic
radiation, which includes radio waves, visible light, and X-rays, as well as ultraviolet
and
infrared
radiation. The
kind
of
radiation
emitted
de pends on the
temperature (the higher the temperature, the shorter the wavelength)..
So the greenhouse effect works like this: the Sun, being very hot, emits visible light.
The light from the Sun passes through the Earth‘s atmosphere, which is transparent to
visible
light
(that‘s
why
our
eyes evolved to be sensitive to this kind of
electromagnetic radiation), and warms the surface of the Earth, which in turn
reradiates(reflects) the energy, now
as
infrared
radiation,
because the
Earth‘s
surface isn‘t as hot as the Sun.
But the Earth‘s atmosphere while transparent to visible light, is not transparent to
infrared light. So heat energy, in the form of infrared radiation, stays trapped in
Earth‘s atmosphere, and the Earth becomes much warmer than it would be, without
the greenhouse effect.
While most of the Earth‘s CO2 is locked in t h e E a r t h ‘ s crust, it doesn‘t stay there
forever. The action of plate tectonics, the motion of the Earth‘s surface, that is, as
chunks of the Earth‘s crust gets pushed together, some of the rocks get pushed
deeper into the interior, where they are subjected to heat and pressure. Such heat and
pressure initially changes limestone to marble. But under even greater heat and
pressure, the CO2 is released from the rock, and makes it way back to the surface
where it is emitted into the atmosphere through volcanic action. Hence volcanoes are a
source of CO2.
This is the complete carbon cycle: rainwater removes CO2
from the atmosphere
and puts it in the crust, and volcanic action releases CO2 from the crust and puts it
back in the atmosphere.
221
Venus has no water! Early in its history Venus may have had water, but it is too
close to the Sun to retain it. When water
molecules rise high in an atmosphere,
ultraviolet radiation splits the
water
molecules
into
their component gases,
oxygen and hydrogen, and
the lighter hydrogen molecules escape into space.
While Earth‘s lower atmosphere is about one percent water vapor
the upper
atmosphere, where ultraviolet radiation can penetrate, is very dry: a cold trap, a
combination
of
pressure
and temperature, prevents water vapor from rising
high in the
Earth‘s atmosphere. Venus has a cold trap, too, but because Venus is
closer to the Sun its cold trap is much higher in the atmosphere and Venusian
water molecules rise high enough to be broken apart by ultraviolet radiation.
Therefore the carbon cycle is incomplete on Venus: without water, CO2 cannot be
removed from the atmosphere. Venus does have volcanoes, however. Radar mappings
of Venus by interplanetary probes indicate volcano-like mountains, and there is other
evidence for volcanoes as well. The atmosphere of Venus is full
and
of sulfur dioxide
sulfur particulates. Sulfur and sulfur dioxide is highly reactive and cannot
remain
regularly
long
an
replenishing
interplanetary
content
in
of
atmosphere; therefore something (volcanoes) must be
the sulfur.
This
theory
is
bolstered
by data
from
probes, which have detected large fluctuations in the sulfur
the Venus
atmosphere,
as
well
as radio signals reminiscent of
lightning—and lightning is often found in volcanic plumes.
And Mars? The carbon cycle is also broken on Mars, but opposite to Venus. Mars has
no active volcanoes to replenish the CO2
in its atmosphere. We know Mars once
had running water—we can still see billion-year-old river beds where water once
ran—and the water may still be there, locked up in the ice caps and in permafrost
beneath the surface. And it seems likely that Mars has CO2
still locked up in its
crust, deposited there billions of years ago by the action of water. If you could
release that CO2 you could warm up Mars again.
222
A fully active carbon cycle acts as a thermostat, regulating a planet‘s climate. The
carbon cycle has similar negative feedback. Suppose the Earth gets too warm. Then
more water will evaporate from the oceans, and the additional precipitation will
remove CO2 from the atmosphere, moderating the greenhouse effect and cooling the
planet. If the planet cools too much, less water will evaporate and there will be less
precipitation to remove CO2; the CO2 will build up, warming the planet.
This carbon cycle thermostat helps to explain a mystery about Earth‘s long term
climate. Computer models of our Sun show that it has gotten progressively brighter
over its five-billion-year lifetime, by about twenty-five percent. Since a mere two
percent change in the Sun‘s luminosity would (all other things being equal) plunge
the Earth into a deep ice age, one might expect the surface to have only
recently
defrosted But two hundred million years ago the Earth was in fact warmer than it
is today. So all things were not equal. The carbon cycle explains how the Earth‘s climate
can compensate for changes in the Sun‘s luminosity. It seems likely that the Earth‘s
atmosphere had somewhat more CO2
half a billion years ago than today; as the
Sun slowly grew brighter, the carbon cycle deposited more CO2 in the crust, keeping
the temperature ―just right.‖
The carbon cycle is a crude and slow thermostat, however. It takes millions of years
to compensate, enough for slow changes in the Sun, but not enough to immediately
correct for humans dumping tons of extra CO2
every year. So don‘t be complacent
about global warming— the carbon cycle won‘t protect us!
The story is not quite finished. We‘ve learned that Venus is too hot because it has a
runaway greenhouse effect, caused by a broken carbon cycle, from too little water;
Mars is too cold because its carbon cycle is also broken, lacking active volcanoes,
and therefore it has too small a
greenhouse effect. Earth is lucky, with a fully
functioning carbon cycle and therefore a moderate and moderated greenhouse effect.
223
But why do Earth and Venus have active volcanoes, and Mars none? We know
that Mars once had volcanoes. In fact Olympus Mons on Mars is the largest volcano,
albeit extinct for billions of years, in the solar system. You could fit the entire state of
Rhode Island in the caldera of Olympus Mons. But no Martian volcanoes are active
today. To understand why, we must delve deep into the interior of the planets.
The interior of the Earth is hot. Part of this heat is generated by the natural decay of
radioactive elements in the rocks, and part of the heat is left over from the formation
of the Earth five billion years ago—when gravity pulled together bits of gas and
dust to form our planet. And there are three ways a planet can lose heat from its
interior. The first is simple conduction. If you stick a poker into a fire and hang
onto it, heat will slowly travel up the poker to your hand and burn you. This is the
main way small planets and moons lose heat, for the smaller the object the faster
it cools. Imagine baking potatoes of different sizes. When you take them out, the
bigger one cools much slower than the small one. And so this is what happened to
Mars. With only about one-eighth the mass of the Earth, Mars is a small potato
and cooled so rapidly it lost its heat to power
volcanoes. Jupiter, on the other
hand, as the largest planet in the solar system, is still immensely hot in its gaseous
interior: in fact Jupiter radiates 1.7 times as much thermal energy as it receives from
the Sun.
There are two other ways to transport heat from the interior of a planet to its surface,
both of which can produce volcanoes. One is convection. You can, with care,
produce convection on your stove top. Take some soup—thick tomato or some sort
of cream soup might work well—and heat it slowly. Don‘t bring it to a boil; but if
you get the temperature right, you will see soup upwelling from the bottom. Hot soup
expands and becomes lighter, and floats to the top, while cold material on the top
is denser and sinks, forming a cyclic pattern.
Convection
drives
plate
tectonics: the source of continental drift, building of
mountain ranges, earth- quakes, and some kinds of volcanoes.
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Not all volcanoes arise from plate tectonics. Sometimes a little ―worm‖ of hot rock
will force its way up through the crust, forming what is known as a shield volcano, with
long, gentle slopes. The Hawaiian Islands are formed from shield volcanoes. This
process is called advection. The volcanoes on Venus, and the ex- volcanoes on Mars,
were all formed by advection.
Venus is the same size as Earth, and so its interior stays hot enough to support
volcanism. Exactly why Venus lacks plate tectonics is not clear. Perhaps its crust
is too thick and brittle; or perhaps water is necessary to ―lubricate‖ the action of
plate tectonics. But in contrast, Mars, being so much smaller than Venus and Earth,
lost its interior heat too quickly to support volcanoes for long.
So now we can conclude that Venus is too close to the Sun to retain water,
which breaks its carbon cycle, leading to an overabundance of greenhouse gases
and a well-cooked surface. Mars is too small to have a hot interior, and thus no
volcanoes to pump greenhouse gases into its atmosphere, and so is cold due to
lack of insulation. We Earthlings are lucky: our planet has water and volcanoes and a
regulated climate.
Student activity:
Engage (10 minutes)
1. Ask students to recap what they have learned about the differences in the
environments of Mars, Venus, and Earth, and why the three planets are different.
2. Ask students the following questions
a. What makes a planet habitable?
b. What conditions need to exist in a planet in order for life to survive?
c. What temperature range is suitable for life to exist?
d. What kind of atmosphere does life need?
e. Does life need liquid water?
f. What do organisms need in order to eat?
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g. Do organisms need light?
h. Does it matter how massive the planet is?
3. Once students have described a good environment for life, show the video: Evolution
Caves: Extreme Conditions for Life-You Tube .
4. Direct students to identify two types of extreme environments on Earth as they view the
video. Expand students' definition of a good environment for life by discussing extreme
environments, including black smokers (deep sea vents), algae in Antarctica, and
bacteria in the Yellowstone hot springs. These organisms all live in places that would be
uninhabitable for humans, yet they are able to thrive. Explain that students will now
examine what makes a world habitable.
Explore (20 minutes)
1. Divide students into small teams (five students each). Have them create a list of
items they would need to bring with them to survive an extended trip on a
spaceship into space (the supplies that are necessary for life as we know it). Each
student should record on a piece of paper the list their team develops. (These will
be collected at the end of class).
2. Once the teams have developed lists of what they would need, inform them that
something has gone terribly wrong on their well-packed space ship. They will need
to crash land in the nearest planetary system. Give each team the Crash Landing!
Student Work Sheet (a profile of the planetary system-worksheet-1B) and instruct
the teams to decide on which planet or moon to crash. Teams should discuss the
planetary system for no more than 10 minutes. Tell students to record their
selections and the reasons why they chose that planet or moon on the back of their
student activity sheets.
Explain (10 minutes)
1. Bring the whole class back together. Have each team share which planet or moon
they thought was best for a crash landing. Remind presenters to elaborate on their
group's reasoning.
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Elaborate (10 minutes)
1. Ask students if they think there is any one factor that is essential for life. Remind
students about the extremophiles they saw in the video at the beginning of class.
2. Introduce the concept of the "habitable zone" as the area around a star where water
can be liquid. Show the video: What is the Habitable Zone? While watching the
video, ask students to listen for the planets in our solar system that are in the
"habitable zone." Discuss where in our own solar system the habitable zone is
located. (Just inside Earth's orbit to just outside Mars' orbit.)
3. Discuss the possibility of a second habitable zone in any other solar system(See
Science & Resources for help in leading this discussion.) Close the discussion by
reexamining the list of planets and moons in the Borong System. Review which of
them have liquid water. Point out that some of the life forms that students have
examined today could live on some liquid water worlds, but not others. Liquid
water is the most basic requirement for all life as we know it, but other factors are
important for the particular organisms that students examined as well.
4. Evaluate
1. Collect the Crash Landing! Student Work Sheet and check to see if students'
reasoning for choosing a habitable planet makes sense. Also collect students' lists
of what they would need for an extended space trip.
Worksheet 7A
1.
Explain why is Venus too hot, Mars too cold, and the Earth just right for liquid water?
2.
Define the greenhouse effect and explain its impact on the survival of life.
227
3.
Describe the carbon cycle and explain its impact on living organisms.
4.
Explain why Venus is too hot, Mars too cold, and the Earth just right to support
life.
Worksheet-7 B
Crash Landing!
Your spaceship has developed a snag. Luckily, you are passing through the Borong
System, which consists of a sun-like star surrounded by six planets, some of which
have moons. Your ship has barely enough fuel and guidance ability to allow you to
select a nearby place to crash-land. Below are profiles of each of the planets and moons
in the Nonog system. The information is sketchy, but it's all your sensors had time to
collect before going off-line due to the damage caused by the meteoroid. Good luck.
Planet 1 (closest to the star)
Planet 4
Mass: 1.5 (Earth = 1)
Mass: 1.5
Tectonics: Active volcanoes and seismic
Tectonics: Active volcanoes and
activity detected.
seismicactivity detected.
Atmosphere: CO2, N, and H20
Atmosphere: N, O2, and ozone layer
Average Temperature: 651 degrees C
Average Temperature: 2 degrees C
Description: Thick clouds surround the
Description: Cold oceans, covered with
planet. No surface is visible through the
ice along much of the globe. Some liquid
clouds.
water around equator.
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Planet 2
Planet 5
Mass: 0.5
Gas Giant with one large moon.
Tectonics: No activity detected.
Moon: Sulfur dioxide (SO2) atmosphere.
Atmosphere: Thin CO2 atmosphere
Many volcanoes and hot springs on
detected.
surface. Temperatures in hot spots can be
Average Temperature: 10 degrees C
up to 600 degrees C. Other spots away
Description: Polar ice caps, dry riverbeds,
from volcanic heat can get as low in
and many craters can be seen from orbit.
temperature as 145 degrees C.
Planet 3
Planet 6
Mass: 1
Gas giant with four large, rocky satellites
Tectonics: Active volcanoes and seismic
(moons). Moons have no appreciable
activity detected.
atmosphere. Ice detectable on one.
Atmosphere: CO2, H20,O2
Temperature: 30 degrees C
Description: Liquid water oceans cover
much of the surface. Volcanic island chains
make up most of the dry land.
Answer the following
1.
Which planet do you select for landing ?
2.
Give reasons for your decision.
http://www.voyagesthroughtime.org/planetary/sample/lesson5/z_act3.htm
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Activity 8
Origin of Life- Abiogenesis
Learning Objectives :
1.
After completing this activity, students will be able to:

Understand the process of synthesis of organic molecules from inorganic
components of earth.

Explain
that organic molecules have the capacity to combine and behave like
living molecules.
1. recognize coacervates
2. recognize the life-like properties of coacervates.
3. Show that under suitable conditions, complex life-like cell-like
structures can be produced naturally from simple materials with
simple changes

Define Abiogenesis

Identify the scientific theory Abiogenesis as the most natural explanations
and acceptable theory to explain how life originated .
2. Explain the origin of life on earth need not have required supernatural forces.
Content:
Audio video activity
You Tube--The Origin of Life - Abiogenesis - Dr. Jack Szostak
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Student Activity 8A
Creating Coacervates
Material:
1 safety goggles & lab aprons
2 compound microscope, slides, cover slips
3 test tube rack with clean small culture tubes (13x100 mm works well)... one tube
per student.
4 one medicine dropper per tube
5 one dropping bottle with 0.1M HCl solution
6 pH paper in dispenser with color code
7 one 50-ml beaker with coacervate mix (5 parts of 1% gelatin solution + 3 parts 1%
gum arabic soln.). You can make the 1% solutions day before lab, (adding a pinch
of mold inhibitor seems to help). Mix the two solutions (5:3 ratio) day of lab, then
dispense into little beakers. For 5 classes, 500 ml of gelatin solution and 300 ml of
gum arabic solution should be ample. Gelatin is a protein; gum arabic is a
carbohydrate.
For current sources of Gum Arabic, see Resources, below.
Gelatin can be
purchased in grocery stores, or from most school chemical supply catalogs.
8. a sample student worksheet is provided at the end of lesson.
Procedure:
1 FOR SAFETY, do not forget to wear goggles and aprons. You will be working with
dilute acid in dropping bottles, and accidental splattering can occur. BE CAUTIOUS!
2 Set up your microscope and clean a slide and cover slip.
3 Half fill your culture tube with some of the mix, a 5:3 ratio of protein (gelatin) :
carbohydrate (gum-arabic). With the help of a dropper, put a drop of the mixture on pH
paper. Read and record the pH of the mix.
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4 Add a drop of acid to the tube; cover it with a stopper and invert once to mix. It should
turn cloudy. If the cloudiness disappears, add one more little drop of acid and invertmix. Do this until it remains cloudy. (This should take no more than 1-3 drops total; if
you require to add more, there's something wrong; start again). When it stays cloudy,
put another sample drop on pH paper. Record the pH.
5 Place a drop of the cloudy suspension on a slide, add cover slip carefully, and search
under low power (40-100x) for something which looks like mucus , clear irregular and
some what transparent bodies with rounded shape, often with tiny bubbles inside.
Starting with a small diaphragm opening at first might help, as coacervates often have
very thin delicate outlines, and may be hard to see if light is too bright, and the small
aperture causes these fine lines to appear thicker and darker. Movement will not be
notice d immediately; you might see some after viewing for awhile. You can go to higher
power (400x) for closer look. Draw a typical coacervate or two. If you have a chance,
take a look at some of the coacervates found by other students.
6 Add drops of acid (invert-mixing after each drop), count the drops until extreme
cloudiness disappears (2-3 additional drops). Record the pH.
7 Before doing the activity , try predicting the approximate pH with each added drop;
record these in the "predict. pH" column of your data table. Also, if your text has a
picture of a coacervate, draw it on your worksheet, as a "predicted coacervate"
8 When finished with the lab, clean up and begin the discussion questions.
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Worksheet-8A
A. Coacervates expected: [copy here the picture of coacervates as shown below]
B. Expected pH levels: (see "pH of Mix Predict." column in the "Results" table below).
[Assume pH of the original mix will be about 5; indicate (in that column) the pH
expected as acid is added...]
1. Coacervate formation expected at pH level:
2. Coacervates expected to disappear, as more acid is added, at pH level:
Example of Coacervate
C.
Predicted Coacervate
RESULTS:
Drops HCl pH of Mix pH of Mix Appearance of Mix: (clear,
Added
Predict.
Actual
cloudy)
0
5
Sketches of
coacervates
formed
at pH
1
2
seen at
x in
microscope
3
4
5
6
233
D.
DISCUSSION: [Read ALL questions BEFORE the activity; After the activity, list
their numbers, and answer each briefly, based on your results and results of
others. Answer #1 before doing the activity; this one does not require activity
data to answer.]
1
Compare the materials you have used to make coacervates in laboratory
with those that might have been present in the ancient oceans?
2
At what pH did the coacervate droplets form?
3
Did the pH change as expected (up or down, as a result of adding more
acid to the solution between coacervate formation and clearing)?
4
When dilute hydrochloric acid was added beyond a certain point, the
coacervates disappeared. What might you add to the test tube to make the
coacervates reappear?
5
How might the coacervate droplets be made more visible under the
microscope?
http://www.indiana.edu/~ensiweb/lessons/coac.ws.pdf
Student Activity 8B
“Inorganic and Organic Compounds”
Learning Outcomes: Students will be able to:
1 investigate the chemical compounds in living things and contrast these with the
major elements of the Earth‘s crust
2 compare inorganic compounds to organic compounds and examine examples of
each group of organic compounds
3 discover why carbon is referred to as the unparalleled element.
4 determine the rationale behind the scientific classification system..
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Materials:

a pack of playing cards,

a set of organic compound pictures that are classified into 4 categories , per
group of students (Each organic compound set should include an assortment of
pictures from the four groups: carbohydrates, lipids, nucleic acids, and proteins,
and each molecule should be labeled as to which group they are in along with its
molecular formula ).
Pre-Activity:

Ask each student to write numbers from 1 to 5 on a sheet, two times. Label one
set of numbers ―Human Body‖ and the second set of numbers ―Earth‘s crust‖.

Give the students 5 minutes to write down in descending order which elements
they think are most common in both the human body and the earth‘s crust.

Ask the students if they are surprised, and if they know
o why some elements in the human body are not the same as the elements
in the Earth‘s crust (theory of evolution).
o Where does all the carbon come from in the human body?
o Where does all the carbon come from in living things?
o How do organic and inorganic molecules differ from each other?
Procedure:
1.
Arrange the class into four small groups using a pack of playing cards. Ask that
all the hearts meet together, all the spades, all the clubs, and all the diamonds. (do
not use all 52 cards of the pack; use the same number of cards as you have
students).
2.
Distribute one set of organic compound pictures to each small group.
Each
organic compound set should include an assortment of pictures from the four
groups: carbohydrates, lipids, nucleic acids, and proteins, and each molecule
235
should be labeled as to which group they are in along with its molecular formula.
3.
groups may be guided:

to spend some time carefully observing the materials in each organic
compound group.

to pay particular attention to the chemical formula beside each picture

to find reasons why some elements in one group are the same as the other
elements in the other group, and why some elements in the other group are
different
4.

to prepare report to explain their reasoning

to set a time limit of 25 minutes
As students study the materials, the teacher may move
through the classroom,
and amongst each group to ensure that students are on the right path and are
taking initiative. 5. After 25 minutes has past, or when the students appear to
have classified the material, regroup the small groups into one large group.
6.
Debrief by asking key questions ―Carbon: The Unparalleled Element‖ and
―Variations in Carbon Skeletons‖
7.
Assign homework.
236
Workheet 8B
1
What do you understand by organic molecules?
2
What do you understand by inorganic molecules?
3
Which element/elements do all carbohydrates, lipids, nucleic acids, proteins have
in common?
4
Write one unique characteristic of each organic group?
5
In which category would you like to place water?
6
―Carbon: The Unparalleled Element‖ to point out the differences between inorganic
and organic compounds, and to illustrate how special the element carbon is.
http://www.usask.ca/education/coursework/mcvittiej/bio30unit1/lessons/lesson05.
htm
237
Activity 9
Evolution
Learning Outcome: After completing this activity, students will be able to:

Describe the importance of adaptation in avoiding predation

Relate environmental change to changes in organisms

Explain how natural selection causes populations to change

Give evidences that fittest in the given situation survive

Demonstrate the mechanism of Natural selection
Content:
Evolution
Charles Darwin is among one of the first scientists to
predict the origin of species. His most famous book, On the
Origin of Species by Means of Natural Selection (1859), is a
landmark in human understanding of evolution of life
forms in nature.
According to the theory of evolution, life originated more
than 3.4 billion years ago, when the earth's environment
Darwin, the founder of
Evolution.
was much different than that of today, e.g. lack of
significant amounts of free oxygen in the atmosphere. Experiments have shown that
rather complicated organic molecules, including amino acids, can arise spontaneously
under conditions that are believed to be present in the earth's primitive environment.
Concentration of such molecules evidently led to the synthesis of active chemical
groupings of molecules, such as proteins, and eventually interactions among chemical
compounds. A rudimentary genetic system eventually evolved and was elaborated by
natural selection into the complicated mechanisms of inheritance known today. The
earliest organisms must have fed on nonliving organic compounds/molecules, but
238
chemical and solar energy sources were soon tapped. Photosynthesis freed organisms
from their dependence on organic compounds and also released oxygen so the
atmosphere and oceans gradually became more hospitable to advanced life forms.
The earliest organisms of which fossilized provide evidence that they were already
cells, resembling modern bacteria (see Cell). These simple unicellular forms
(prokaryotes) were at first anaerobic (living without oxygen), but they diversified into
an array of adaptive types from which cyanobacteria (formerly known as blue-green
algae) descended, and later aerobic photo synthesizers. Advanced cells (eukaryotes)
may have evolved through the amalgamation of a number of distinct simple cell types
over a long period of time.. Single-celled eukaryotes then developed complex modes of
living and advanced types of reproduction that led to the appearance of multicellular
plants and animals of diverse kind.
Student Activity:
Material:
a. one Dice
b. Paper and pen
Procedure:

This activity is about creating animals (imaginary) with useful adaptations.

Select the animal's environment by answering the MCQ given with the help of
the dice, because that is how Darwin's theory of evolution and adaptation work:
the individual organisms that are best adapted to their particular environments
survive, so the adaptations gradually appear in more and more of the
population. Different adaptations are complementary to different environments:
For a rabbit living in the Arctic, white fur would be helpful to avoid being seen
by predators. For a rabbit living in the woods, being white would make it more
conspicuous, but being brown would be helpful.

To "build a beast,"
239
o first roll a dice for each category given below to determine the conditions
under which the animal lives. Make a list of the features identified for
each category
Then prepare a detailed write up giving and explaining the adaptive
o
features you have identified .Draw an animal which has
these
adaptations (with adaptive features) to be able to live well in that kind of
environment you have decided with the help of dice..
A)
WHERE DOES IT LIVE?
1 - mountains
2 - flatlands
3 - rocky, harsh
4 - small island
5 – desert
6 - in ice filled cave
B)
HOW MUCH WATER IS THERE?
1 - almost none; dry and barren
2 – water is present in the part of the year, the rest part of the year is drought
3 - lots of precipitation all year
4 - near a coastline
5 - in a river
6 - in the ocean
C)
WHAT IS THE CLIMATE/WEATHER LIKE?
1 - hot and humid
2 - hot and dry
3 - moderate
4 - cold, rainy, and windy
5 - seasons change from hot to cold
6 - sub-zero temperatures
240
D)
WHAT DOES IT EAT?
1 - leaves from tall plants
2 - small animals
3 - berries, plants, and
4 - water animals
5 – Grass
6 - flying insects
E)
WHAT EATS IT?
1 - Carnivores chase it,kill and eat it
2 - vampire butterflies land on it and suck it dry
3 - buzzbugs lay eggs the burrow into its skin
4 - parasite suck nutrition from it
5 - wolf-like animals
6 – Snake like reptiles leap out of the sand and swallow it
241
Worksheet 9A
Instruction-Read the information given in the Concept Map at different places and with
the help of this information fill up the gaps.
Activity 9B
Heredity and Evolution
Material required-Circles cut from newspaper and white sheets. Record sheet,
Worksheet, Forceps.
Procedure-- In lab, teacher may simulate how predators locate prey in different
environments.
1
Take out 30 circles from newspaper and 30 circles from white sheet provided in the
poly bag. (can be made with hole punch). These are symbolic to two types of Moth.
242
2.
Place a sheet of white paper on the table and have one student spread 30 white
circles and 30 newspaper circles over the surface (ensure that the other student)
isn't looking.
3.
The other student is a "predator", who will now use forceps to pick up as many of
the circles as he can in 15 seconds. Record the data in chart below.
4.
This trial will be repeated with a newspaper background. Record the data in chart
below.
5.
Reverse the role and repeat the exercise.
Starting Population
Trial
Background
Newspaper
White
1
White
20
20
2
White
20
20
3
Newspaper
20
20
4
Newspaper
20
20
Number Picked up
Newspaper
White
Worksheet- 9B
Instruction-Observe the data you just
filled in the table. . Answer the following
questions on the basis of pattern you have observed,
1 How are prey selected by predators?
2 Which moth (circle) coloration is the best adaptation for a dark (newspaper)
background? (This is our Trial 1)
3 Give one evidence for your answer (for Q. 2).
243
4 4. Which colored moth would be found in larger number in the next generation
after trial ‗1‘?
5 Which colored moth would be dominating the population after three
generations?
6 What is this phenomenon called as?
Observe the table given below and answer the questions.
Year
No. of
No. of
Light
Dark
Moths
Moths
Captured Captured
2
547
110
3
474
192
4
392
210
5
254
279
6
225
337
7
180
412
8
147
510
9
84
550
10
54
601
7. What happens to the population of Light Moths after nine years? Why?
8. What happens to the population of Dark Moths after nine years? Why?
244
Activity 10A
A Long Time- a time line
Learning Outcome: After completing this activity, students will be able to:

Explain that Life has been on Earth for a long time.
Content: A timeline is a visual way of displaying a list of events in chronological order.
It is typically a graphic design showing a long bar labeled with dates alongside itself on
one side and (usually) events labeled on points where they would have happened on
the other side.
Time scale: Timelines can use any time scale, depending on the subject and data. Most
timelines use a linear scale, where a unit of distance is equal to a set amount of time.
This time scale is dependent on the events in the timeline. A timeline of evolution can
cover millions of years, whereas a timeline about the September 11, 2001 can take place
over minutes. While most timelines use a linear timescale, for very large or small time
spans, Refer to the following sources for the timeline of the evolution of life which
outlines the major events in the development of life on the plan Earth. .

Source- http://en.wikipedia.org/wiki/Timeline

http://departments.oxy.edu/biology/bbraker/courses/bio105/images/15.1%2
0The%20Geological%20Time%20Sca.JPG
Materials:

38 meters of measuring tape (used for athletic purposes) in meters.

Adhesive tape

Pictures of Cenozoic, Mesozoic, Paleozoic, and Precambrian living organisms.
The websites below should help with this:
245
- www.amnh.org/exhibitions/Fossil_Halls/Timelines/index.html/
- www.rth.org/tarpits/

http://csls-text2.c.u-tokyo.ac.jp/images/fig/fig01_02.gif
- www.humboldt.edu/~natmus/Exhibits/Life_time/Mural.web/MurIndex.htm
- www.peabody.yale.edu/mural/

Advance Preparation:
- Put up the timeline with appropriate numbers using a scale of 10 meters
equals to a billion years (1 centimeter equals one million years) on the wall of
the class room. Mark Cenozoic: 0 to 65 cm, Mesozoic: 65 cm to 2.5 m,
Paleozoic: 2.5 m to 5.5 m, Precambrian: 5.5 m to 38 m on the time line.
-
Copy and cut out appropriate and commonly known organisms ‘pictures
belonging to different eras.
-
Cut out a very small picture of a girl and a boy.
Procedure:
1. Ask students if they notice anything different on the classroom wall. What is on it?
Tell them that this is called a timeline. A timeline is a way to show the events which
happened in the past.
2. Tell the students that they are going to use the timeline to take an imaginary trip
back in time. ―We‘ll put pictures of things that lived a very long time ago on our
timeline.‖
3. Starting with the illustration of a girl and a boy on the very edge of the timeline, ask
students to go backwards in time, placing appropriate illustrations on the timeline
as they find out what lived in each era and mentioning how many years ago each
lived.
246
Worksheet-10A
1. What does the timeline tell us?
2. ―What do we know from looking at the timeline?‖
3.
What is your general observation about the kinds of living organismsyou see on the
timeline?
4.
―How are theyi similar? How are they different?
5. How long the timeline has been?
6. Where are human Being on the Time Line?
http://www.ucmp.berkeley.edu/education/lessons/a_long_time.html
247
Activity 10B
Evidences of Evoution
Learning Outcome: After completing this activity, students will be able to

Describe fossils as evidence of past life.

prove that life has gradually evolved with the help of Study of fossils
Materials:

Two packets of modeling clay for each group (two colors)

An assortment of leaves or small plastic animals

Examples of actual leaf or animal fossils, if possible

Time line from lesson ―A Long Time‖
Instructions
—
Flatten the clay to the approximate size of the leaf you have selected.
—
Prepare a sample impression by pressing two colors of clay together with a leaf in
between. Separate the clay layers and discard the leaf.
—
Display both the clay impressions.
Procedure:
1. Teacher may show the entire class the imprint she has made and ask them what
they think created the impression.
2. Teacher may explain students the instructions to make imprint so that they can
get to make their own imprints.
3. Distribute two colors of clay to each pair of students and ask them to flatten each
piece of clay until it is about 1/2" thick. Have them select a leaf or a small plastic
animal and press it between the layers of clay.
248
4. Students may be told to separate their layers of clay and give each pair a chance
to share the results with the class. As each pair shares its imprint, have the class
guess what the original object looked like for each of the sets of imprints. Ask
students to make a note of feature identified which will help them to identify
each of the imprints.
5. Display a number of actual fossils if available. Ask students to tell what they
know about fossils. Ask the class what they think the animals or plants looked
like before they were fossilized. An explanation may be given that fossils are
formed when some living organism dies and then is covered for a very long time
without being destroyed.
6. Teacher may explain that the real fossils were not pressed between clay by
people, but were formed by natural forces. Tell as much of this story as students
are ready for.
7. Teacher may ask students to collect information from net about fossils.
8. Ask students if they think they made real fossils in the lab.. ―What makes you
think that?‖ Ask them how their imprints and fossils are different. Teacher may
use the time line to talk about fossils and how long ago the life forms lived.
http://evolution.berkeley.edu/evolibrary/search/lessonsummary.php?&thisaudience
=K-2&resource_id=7
249
Worksheet-10B
Instruction- Observe the diagram Set-1 and Set-2. Set-1 shows forelimbs of Bird and
Human while Set-2 has wings of Bat and Bird. Read the questions carefully and write
answers.
Q.1
In the given image, which pair represents an analogy ?
Q.2
In the given image, which pair represents a homology?
Q.3
Define homology:
Q.4.
Define analogy.
Q.5.
What is one similarity between homologous and analogous organs?
Q.6. ‎ What is one difference between homologous and analogous organs ?
Q.7.
Are limbs of a frog and limbs of a Lizard homologous or analogous ? why?
Q.8
Birds and bats have wings, but squirrels and lizards do not. Are birds and bats
more closely related to each other than to squirrels or lizards? Why?
250
Activity 11
EVOLUTION- Classification
MAKING CLADOGRAMS
Learning Outcome: After completing this activity, students will be able to:

Construct a cladogram with the help of given groups of organisms and some of
their distinguishing characteristics.

interpret and analyze the cladogram in terms of how it shows common ancestry
and degrees of evolutionary relationship,

Explain that modern classification is based on evolution theory.
Content: One way to discover how groups of organisms are related to each other
(phylogeny) is to compare the anatomical structures (body organs and parts) of
many different organisms. Corresponding organs and other body parts of these
organisms, that are alike in basic structure and origin (although appear different
apparently) are said to be homologous structures (for example, the front legs of a
horse, wings of a bird, flippers of a whale, and the arms of a person are all
homologous to each other). When different organisms share a large number of
homologous structures, it is considered as strong evidence that they are related to
each other( although they may not appear so) . When organisms are related to each
other, it means they must have had a common ancestor at some time in the past.
251
Student Activity 11A
Procedure: Using your textbook and the explanations below, determine which of the
characteristics each animal has. In the Data Table provided (on your Cladogram
Worksheet-11A), place an "Y" in the box if the animal has the characteristic.
Read the explanations of Characteristics and accordingly fill up the Data Table:
Set -1: Dorsal nerve cord (running along the back or "dorsal" body surface) Notochord
(a flexible but supporting cartilage-like rod running along the back or "dorsal"
surface)
Set -2: Paired appendages (paired legs, paired arms, wings, fins, flippers, antennae)
Vertebral column ("backbone")
Set-3: Paired legs
Set -4: Amnion (a membrane that holds in the amniotic fluid surrounding the embryo;
may or may not be inside an egg shell)
Set -5: Mammary glands (milk-secreting glands that nourish the young in females)
Set -6: Placenta (structure attached to inside of uterus of mother, and joined to the
embryo by the umbilical cord; provides nourishment and oxygen to the growing
embryo)
Set -7: Canine teeth short (same length as other teeth) Foramen magnum forward
(spinal cord opening, located forward in occipital bones, under skull)
252
Student Activity 11B
M ake a Venn diagram on worksheet-11B, placing your seven animals in groups to
illustrate those characteristics which different animals have in common on the basis of
Data Table. See example below:
Human: Foramen magnum forward
Horse: Placenta
Tuna: Backbone
Student Activity 11C
Using the Venn diagram of the groupings just completed (as a guide), draw a cladogram
on the Worksheet-11 C to illustrate the ancestry of these animals. The diagram should
reflect shared characteristics as time proceeds. An example is shown below. Notice how
the different animals are all placed at the same time level (across the top) since they all
live today.
Example of Cladogram
253
TIME
TUNA HORSE HUMAN
Foramen magnum
forward
Placenta
Backbone
SHARED CHARACTERISTICS CLADOGRAM
WORKSHEET-11A
Step 1: DATA TABLE
Animals
SETS
TRAITS
Kangaroo Lamprey
Rhesus Bullfrog Human Snapping Tuna
Monkey
SET 1 Dorsal Nerve
Cord
Notochord
SET 2 Paired
Appendages
Vertebral
column
SET 3 Paired legs
SET 4 Amnion
(Amniotic
254
Turtle
sac)
SET 5 Mammary
Glands
SET 6 Placenta
SET 7 Canine teeth
shortForame
n magnum
fwd
TOTALS of
Ys-------->
Note: Put ‗ Y‘ to show the presence of given feature
Worksheet-11B
Venn Diagram:
Worksheet-11 C
Cladogram:
255
Worksheet-11 D
Answer the following
1.
Explain at least three
types of information which can be obtained from a
cladogram.
Three types of new information shown by a cladogram:
2.
1.
Shows ...
2.
Shows ...
3.
Shows ...
Three previously unknown vertebrates have been found in a rain forest in North
east India..
One animal is very similar to a Garden lizard. The second animal resembles a
large rat. The third is similar to a fish. Place these animals on your cladogram and
give reason why you placed them where you did Reason for placing each branch
where you did:
" Lizard-like" animal:
"Rat-like" animal: fish-like" animal
Resource: http://www.indiana.edu/~ensiweb/lessons/mclad.ws.pdf
256
Activity 12A
Speciation
Learning Outcome: After completing this activity, students will be able to:

Analyse that while natural selection explains evolutionary modifications within
lineages, speciation explains evolutionary branching and diversification.

Describe
that
speciation
involves
genetic
differentiation,
ecological
differentiation (niche separation) and reproductive isolation.

Give evidence that isolation of members of a species in different environments
may result in the formation of a number of subspecies.
Content:
Audio-video Activity
Speciation – the formation of species
http://www.yteach.com/page.php/resources/view_all?id=species_behavioural_iso
lation_geographical_isolation_mechanical_isolation_anatomical_isolation_morpholo
gical_isolation_post_zygotic_isolation_after_fertilization_isolation_reproductive_isol
ation_temporal&from=search/
Speciation – the formation of species.url
257
Student Activity-12A
This is an activity to effectively get across to students to help them understand
different ways new populations can emerge in nature.The organisms can
be
reproductively separated from the parent population, and eventually evolve into a
new species. It is the basis for genetic drift and "bottleneck" speciation!
Material:
Get two large bowls: one pink( "mamma"bowl) and oneblue ("papa" bowl pink
beads = all the mommas in the population, blue beads = all the pappas.
In each bowl place equal numbers of blue and pink beads, say 50 blue + 50 pink (or
any two colors).
Small white bowls, one per student in the class (these are "baby" cups).
A big plastic (1-liter) bottle of water (with nossel on) for dramatic "realism" and
sound effects of running water, or rushing river,
Procedure: (about 10-15 minutes)
Hand out little "baby" bowls - one per student.
Take your two cups (a "mamma" cup and a "papa" cup) around to each student, ask
each to quickly reach in and pick (one hand into one cup, the other into the other
cup, at the same time) a bead from each bowl (without trying to select any color) and
place those two beads into his/her little bowl. Quickly go around the room until each
student has a pair of beads in his/her little bowl. Each "pick" means a pair of alleles
received in a random "mating" from the population's "gene pool", and the bowls are
their "babies".
Ask everyone with two blue beads (alleles) to raise a hand. Note where raised hands
are concentrated. If no concentration, ask all with two pink beads to do the same.
258
VARIATION: Have students raise one hand for each blue bead they have; that way
you can get a quick sense of where all blue beads are concentrated - or pink beads if
you ask the same for pink beads.
Wherever you see a pathway to go between ,consciously separate that cluster of
students with an unusually high concentration of one color or the other from the
other students, start moving through that pathway.
As you go, splash water from your bottle , making "slosh-slosh" sounds, and convey
the message that"I am a river - growing larger and larger" You could also say "I am a
crevice zone, pushing apart your continent, and making the Indian Ocean," or "I am
the rising sea level,because of greenhouse effect , making lowland hills into islands."
Use your creative imagination and lots of energy to dramatize whatever "barrier" you
decide to build.
When done, announce that the population (students) with a high proportion of one
color is now permanently separated from the other population, and they can no
longer mate with the other population because of the physical barrier. Due to a
higher proportion of (blue) genes, this little population has a specific character e.g.
mostly greenish in colour,some what different from the parent population, so they
can camouflage and escape predators, so they survive, and over many generations
evolve into a new species with typically matching colour like surroundings and
accompanying features.. - Their new gene frequencies should be distinctively
different from the parent population. You can take tally of total number of blue beads
in each population, then the number of pink beads in each population, and note the
different proportions in each.
259
Worksheet 12A
Biological Species Concept: ―A species is a group or groups of organisms that can
actually or potentially interbreed in nature and are reproductively isolated from any
other such groups.‖
1. Do you find any problem with this definition? If yes, identify 2 problems with
this definition.
2. Giveone other definition of species. What is one advantage and what is one
disadvantage of this alternative definition?
teacherweb.com/NM/BosqueSchool/.../SpeciationWorksheet.doc - Si
Activity 12B
Learning Outcome: After completing this activity, students will be able to:

Analyse that while natural selection explains evolutionary modifications within
lineages, speciation explains evolutionary branching and diversification.

Describe
that
speciation
involves
genetic
differentiation,
ecological
differentiation (niche separation) and reproductive isolation.

Give evidence that isolation of members of a species in different environments
may result in the formation of a number of subspecies.
Content:
Mechanisms: the processes of evolution
260
Evolution is the process involving change by which modern organisms have descended
from ancient ancestors. Evolution is responsible for both the remarkable similarities we
see across vast variety of life forms and the amazing diversity of that life —
Change at genetic level is the fundamental mechanism to the process of evolution is
change. which is also called as genetic variation. The mechanisms of evolution focuses
on:

Descent means the genetic differences which are found in a generation and that
are heritable and passed on to the next generation;

Mutation, migration (gene flow), genetic drift, and natural selection act as
mechanisms of change in a population.;

The importance of genetic variation in a population ;

The random nature of genetic drift operating in nature and the effects is
reduction in genetic variation in a population ;

How variation, differential reproduction, and heredity result in evolution by
natural selection;
261
Descent with Modification
Evolution only occurs when there is a change in gene frequency within a population
over time. These genetic differences are heritable and can be passed on to the next
generation — which is what really matters in evolution: long term change.
Compare these two examples of change in beetle populations found in nature. Study
and identify which one is an example of evolution?
Example 1. Beetles on a diet Imagine a year or two of drought in
which there are very few plants available that these beetles can
eat.
All the beetles have the same chances of survival and
reproduction, but because of food
shortage, the beetles in the
population are a little smaller in size than the preceding generation
of beetles.
Example 2. Beetles of a different color Most of the beetles in the
population (say 80%) have the genes for bright green coloration
and a few of them (20%) have a gene that makes them more
brown.
After some generations later, things have changed: brown beetles
are more common than they used to be and make up 70% of the
population and green may constitute 30% of the population..
Which example illustrates descent with modification — a change
in gene frequency over time?
The difference in weight in example 1 came about because of environmental influences
— the low food supply and this generation of small-bodied beetles will produce beetles
that will grow to normal size if they have a normal food supply.— Therefore, example 1
262
is not evolution. Because the small body size in this population was not genetically
determined,
The changing color in example 2 is definitely evolution: these two generations of the
same population are genetically different. But how did it happen?
Mechanisms of change
Each of these four processes is a basic mechanism of evolutionary change.
Possibility -1 Mutation
A
mutation
(Change
in
the
structure
or
number
of
chromosome/DNA) could cause parents with genes for bright
red coloration to have offspring with a gene for 0range
coloration. That would make genes for orange coloration more
frequent in the population than they were before the mutation.
Possibility-2 Migration
Some individuals from a population of orange beetles might have joined a population
of red beetles. That would make genes for orange coloration more frequent in the red
beetle population than they were before the orange beetles migrated into it.
Possibility -3 Genetic Drift
Imagine that in one generation, three orange beetles happened to have six offspring
survive to reproduce. Several red beetles were killed when someone stepped on them
and had no offspring. The next generation would have a few more orange beetles than
263
the previous generation — but just by chance. These chance changes from one
generation to the other generation are known as genetic drift.
Possibility -4
Natural selection Imagine that red beetles are easier for birds to spot (and hence, eat).
Orange beetles get camouflaged in the presence of orange flowers and hence are a little
more likely to survive to produce offspring. They pass their genes for orange coloration
on to their offspring. So in the next generation, orange beetles are more common than in
the previous generation.
All of these mechanisms can cause changes in the frequencies of genes in populations,
and so all of them are mechanisms of evolutionary change. It is important to note,
natural selection and genetic drift cannot operate unless there is genetic variation in the
given population. — . If the population of beetles were 100% red, selection and drift
would not have had any effect because their genetic make-up could not change.
So, what are the sources of genetic variation?
264
Genetic Variation
Without genetic variation, some of the basic mechanisms of evolutionary change cannot
operate.
There are three primary sources of genetic variation, which we will learn more about:
1. Mutations are changes in the DNA structure or number of chromosomes. A
single mutation can have a large effect, but in many cases, evolutionary change is
based on the accumulation of many mutations. Mutations are random. Not all
mutations matter to evolution.
2. Gene flow is any movement of genes from one population to another and is an
important source of genetic variation.
3. Sexual reproduction is responsible
for new gene combinations into a
population. This genetic shuffling is another important source of genetic
variation.
Three main reasons which can bring about Genetic variation during sexual
reproduction.

Independent assortment of chromosome during meiosis

Reciprocal recombination of linked genes on chromosomes by crossing over in
meiosis

Random fertilization of gametes
265
Gene Flow

Gene flow — also called migration — is any movement of genes from one
population to another. Gene flow includes lots of different kinds of events, such as
pollen being blown to a new destination or people moving to new cities or countries.
If genes are carried to a population where those genes previously did not exist, gene
flow can be a very important source of genetic variation. In the graphic below, the
gene for brown coloration moves from one population to another.
http://evolution.berkeley.edu/evolibrary/article/0_0_0/evo_24.
Speciation Worksheet – 12B
Part 1 = Multiple Choice
(Select the best answer for each question.)
Use the following information to answer the next question.
There are two important pathways that can lead to new species. In one pathway,
divergence, members of a parent species are separated from each other and
prevented from interbreeding. In the other pathway, the old species is gradually
replaced by a new one.
266
1.
2.
What is the name of the other speciation pathway mentioned above?
A.
transformation
B.
isolation
C.
gradualism
D.
adaptation
Which of the following statements about speciation is not true?
A.
Speciation increases biological diversity.
B.
Speciation is the development of one or more species from existing one
species.
C.
Two populations need not be prevented from interbreeding for speciation to
occur.
D.
Speciation occurs due to accumulating differences from mutations and
natural selection.
3.
Which of the following cases is not an example of a biological barrier?
A.
Two species of snakes occupying different habits in the same area rarely
encountering each other.
B.
Populations of cichlid fishes dived into small pools due to a decrease in lake
levels.
4.
C.
The distinct courtship songs of birds living in the same region.
D.
Pheromones produced by mosquitoes and bees to attract mates.
When reproductive isolation occurs between groups in a population, new species
have evolved. Differences in which of the following characteristics would most
likely result in reproductive isolation?
267
5.
A.
Coloration of skin
B.
Mating behaviors
C.
Blood types
D.
Body size
For speciation to occur, two populations must be prevented from interbreeding.
Interbreeding can be prevented by a mountain range which is a/an ___i____
barrier, or by behavior that is a/an ___ii___ barrier. The statement given above is
completed by the information in row: The correct option is--------Row
6.
I
ii
A.
Geographical
ecological
B.
Biological
geographical
C.
Ecological
biological
D.
Geographical
biological
A new species is formed from pre-existing species. What is this process called?
A. mutation
B. speciation
C. adaptation
D. natural selection
7.
Female spiders use pheromones to attract males of their own species. This is an
example of a(n)
A. natural selection
B. mechanical barrier
C.
ecological barrier
D. biological barrier
268
8.
Two gametes are prevented from uniting to form a zygote. Which process has
occurred?
A. speciation
B. natural selection
C. prezygotic isolation
D. postzygotic isolation
Part 2 = Match the following
Match each of the examples above to the type of prezygotic or postzygotic barrier below.
A. offspring die young when two species of salamanders mate – they do not make it
to maturity
B. two different species of birds have different mating calls so they cannot
recognize each other as mates
C. one type of spotted skunk breeds in the spring while another breeds in the fall
D. garter snakes that live in water and one on land cannot mate although they live
in the same geographic area
E. mules are produced by mating a horse and a donkey, but mules are sterile and
cannot breed
F. if gametes of red and purple sea urchins do not fuse, there will be no zygote
produced
G. although some strains of cultivated rice can be cross-mated, after a few
generations, sterile offspring are created
_____
1. reduced hybrid fertility
_____
2. habitat isolation
_____
3. temporal isolation
269
_____
4. reduced hybrid viability
_____
5. gametic isolation
_____
6. behavioral isolation
_____
7. hybrid breakdown
www.strathmorehighschool.com/docs/.../Speciation%20Worksheet.doc - Similar
Activity-13
Human Evolution
Learning Outcome: After completing this activity, students will be able to:

handle and read the measuring instruments.

identify the appropriate cranial and dental features and landmarks required for
the measurements and descriptions.

describe features of a given specimen as either similar to, different from or the
same as those present in another specimen.

recognize the sequence pattern in which several human skull features appeared
over time.

construct and justify a taxonomic classification of the specimens.

understand that transitional forms in an evolutionary sequence are generally
mosaic; some traits evolve more rapidly than others.

analyse that modern humans have not evolved from modern apes: both have
evolved from a common ancestor

recognize some of the patterns revealed and see how hominines have changed
over time
270
Student Activity-13
Material:
1.
Plastic/POP cranium casts of modern apes i.e., chimpanzee or gorilla specimens,
male and female, humans and fossil hominines. . A modern human skull may be
made available from the skeleton standing in the corner of the lab,
2.
Plastic/POP casts of a Neanderthal cranium, 450,000 year old Homo erectus
("Peking"
is
the
most
widely
available
and
least
expensive)
and
an
australopithecine cranium(preferably the "robust" 1.8 million year old Olduvai
number 5 or "Zinjanthropus") as an example of an early hominine 3. Sliding
calipers/ hinge calipers and rulers with metric scales. Commonly available plastic
sliding calipers may be purchased from
hardware or arts & crafts stores..
Human Evolution Patterns
Students describe, measure and compare cranial casts from contemporary apes modern
humans and fossil "hominins" (The purpose of the activity is for students to discover
for themselves some of the similarities and differences that exist between these forms,
and to see the pattern of the gradual accumulation of traits over time, leading to
modern humans in order to analyze the patterns Human Evolution .
Instruction:
1. Work in groups of 5 students so that everyone can be actively involved in the
activity.
2. Plan well so that each one in the group gets a turn to observe and measure
different parameters as mentioned in the check list..
3. When taking a measurement, use the sliding calipers (except for #11 & #12 which
may require the hinge calipers) and remember to...
271
4. Always measure in millimeters [mm] and round off to whole numbers.
5. Please do not spoil by marking
with pencil or pen on these crania.
Procedures:
1. Have students work in groups of -5 since normally there arey fewer cranial casts
than students available.
2. Students may work in groups (that move from one "Skull Station" to another.
3. Each student should have a copy of the Hominid Cranium Comparison Checklist
as the details of each measurement and observation are given on it. Each student
should also have her/his own observation table for recording descriptions and
measurements.
4. Have all students label the columns on their data worksheets with the names of
each specimen. 5. Have students take turns to observe, measure/describe and
record the 18 items on the Checklist and have hands on experience individually ..
6. Remind students to record all measurements in millimeters 7. Ask students to
support each specimen in the palms of their hands carefully and not mishandle
like bowling balls with their fingers stuck into the eye orbits and nasal cavity!
8. After the students have measured and described the specimens in the observations
sheet, have them analyse and describe the patterns represented by their findings.
This can be done in a variety of ways:
- a) list those features that all of the specimens have in common;
- b) identify those features which help to distinguish between the specimens;
- c) describe the changes that occur in only the hominin crania over time;
272
Obsevation
Table- Write the description /measurement , as per what
you have
observed with the help of check list.
Sr.
Feature
No.
of the
Human
Chimpanzee-
Chimpanzee-
Male
Female
Homoerectus
robust
cranium
1
2
17
18
Worksheet-13A
I.
BRAINCASE: (items #1-7)
1. Does the FOREHEAD (frontal bone) look more flat when the skull is held in
normal anatomical position [NAP] (i.e., with the eyes oriented forward)?
2. Is a SUPRAORBITAL BROWRIDGE present?
3. Is the BROWRIDGE DIVIDED present? If yes, is it
in the middle, or
CONTINUOUS?
4. What is the SHAPE OF THE BRAINCASE (front to back) when viewed from
top?
5. Is a SAGITTAL CREST present?
273
6. In NAP, is the orientation of FORAMEN MAGNUM more downward OR more
to the rear?
7. Is the MASTOID process relatively flat OR does it noticeably project?
----------------------------------------------------------II. FACE: (5 items: #8-12)
8. Are the NASAL BONES raised (arched) OR flat?
9. Measure the MAXIMUM width of the NASAL OPENING [mm].
10. Measure the maximum length of the NASAL OPENING [mm].
11. Measure the length of the MAXILLA (the upper jaw) [mm].
12. Measure the BIZYGOMATIC BREADTH using the hinge caliper if necessary
[mm]. (This is the width of the face from the widest part of one zygomatic arch
to the widest part of the other zygomatic arch.)
---------------------------------------------------------III. DENTITION: (6 items #13-18)
13. Do the tooth rows diverge towards the back OR are they more straight-sided
and parallel to one another?
14. When viewed from the side, are the INCISORS angled out OR are they vertical?
15. Measure and calculate the COMBINED WIDTH or BREADTH of the 4
INCISORS together.
16. Does the CANINE tooth protrude above the chewing surfaces of the other
teeth?
274
17. Is a CANINE DIASTEMA present?
18. Measure the LENGTH of the LEFT 2 PREMOLARS and 3 MOLARS together by
measuring from the back of the last molar to the front of the first premolar. This
helps us to determine the length of the chewing surface of the "cheek teeth".
[mm].
Worksheet-13B
Answer the following
a.
What could be the reason for the canine tooth to be reduced in size so much from
earlier to later hominines? –
b.
Why the face flattens over time in hominines?
c.
How is the position of the foramen magnum related to the body posture and
locomotors pattern of the animal?
d.
What parts of the braincase enlarge first and which ones enlarge later in the
hominines?
e.
What behavioral and cognitive functions are associated with these cerebral areas?
f.
Do you think we have really lost the brow ridge?
http://www.indiana.edu/~ensiweb/lessons/hom.cran.html
275
UNIT 4
Heredity and Evolution (Class X)
Rubrics of Assessment
1
2
3
4
5
6
Parameter
Student is able to
understand the
importance of variation
and how these
accumulate during
reproduction
Student can learn the
various rules of
inheritance
Students can be given an
insight into the structure
of DNA
Students may also be
explained about the
expression of traits
Students may next move
on to origin of life
Beginning Approaching
(1)
(2)
Students may be
enlightened with the
fascinating world of
evolution and further
classification of
organisms
http://www.indiana.edu/~ensiweb/lessons/hom.cran.html
276
Meeting Exceeding
(3)
(4)
Resources
Activity-7
Our Solar System
http://www.voyagesthroughtime.org/planetary/sample/lesson5/z_act3.htm
Activity-8A
Abiogenesis
Audeo video activity
You Tube--The Origin of Life - Abiogenesis - Dr. Jack Szostak
http://www.indiana.edu/~ensiweb/lessons/coac.ws.pdf
Activity: 8B
“Inorganic and Organic Compounds”
http://www.usask.ca/education/coursework/mcvittiej/bio30unit1/lessons/lesson05.
htm
Activity-9
Evolution
Activity-10A
Time Line
http://www.ucmp.berkeley.edu/education/lessons/a_long_time.html
Activity-10B
Evidences of evolution
http://evolution.berkeley.edu/evolibrary/search/lessonsummary.php?&thisaudience
=K-2&resource_id=7
Activity-11
EVOLUTION- Classification
http://www.indiana.edu/~ensiweb/lessons/mclad.ws.pdf
Activity-12 Speciation
Audio-video Activity
Speciation – the formation of species
http://www.yteach.com/page.php/resources/view_all?id=species_behavioural_isolat
ion_geographical_isolation_mechanical_isolation_anatomical_isolation_morphological_
277
isolation_post_zygotic_isolation_after_fertilization_isolation_reproductive_isolation_te
mporal&from=search/
Speciation – the formation of species.url
Content: http://evolution.berkeley.edu/evolibrary/article/0_0_0/evo_24.
Worksheet: www.strathmorehighschool.com/docs/.../Speciation%20Worksheet.doc Similar
Activity-13
Human Evolution
http://www.indiana.edu/~ensiweb/lessons/hom.cran.html
Resource: http://www.indiana.edu/~ensiweb/lessons/mclad.ws.pdf
278
CENTRAL BOARD OF SECONDARY EDUCATION
Shiksha Kendra, 2, Community Centre, Preet Vihar,
Delhi-110 092 India