PERFECT SCORE BIOLOGY 2011
BAHAGIAN SEKOLAH BERASRAMA PENUH DAN
SEKOLAH KECEMERLANGAN
KEMENTERIAN PELAJARAN MALAYSIA
PERFECT
SCORE
BIOLOGY
2011
Teacher’s Module
PAPER 3
QUESTION 1
1
PERFECT SCORE BIOLOGY 2011
Question 1 :
No.
1
Questions
Marks
Student
notes
A group of students carried out an experiment to study the effect of the concentration of
glucose on the activity of yeast . Diagram 1.1 shows the method used by the students.
The initial height of the coloured liquid in the manometer is shown in Diagram 1.2.
The experiment was repeated using different concentrations of glucose. Table 1.1 shows the
results of the experiment after 10 minutes.
rubber tubing
clip
Manometer with
coloured liquid
Rubber stopper
Glass tube
Initial height of
coloured liquid
Boiling tube containing yeast
suspension
Diagram 1.1
Initial height of
coloured liquid :
1 cm
DIAGRAM 1.2
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Percentage concentration of
glucose / %
Final height of coloured liquid in the
manometer after 10 minutes /cm
10
3
15
5
20
8
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No.
Questions
Complete Table 1.2 by recording the height of coloured
liquid in the manometer after 10 minutes
(a)
(b)
(i)
Based on Table 1.1, state two observations .
1.
At 10% concentration of glucose ,the final
height of coloured liquid after 10 min is 3 cm
2.
(c)
Student
notes
3
3
At 20% concentration of glucose , the final
height of coloured liquid after 10 min is 8 cm
(ii) State the inference which corresponds to the observation in
1(b(i).
1. Low activity of yeast in lower concentration of
glucose, less carbon dioxide is released
2.
Marks
3
High activity of yeast in high concentration of
glucose, more carbon dioxide is released
Complete Table 1.2 for the three variables based on the
experiment.
Variable
Method to handle the variable
Manipulated variable:
The concentration of
glucose
Responding variable:
Height of coloured liquid//
The rate of yeast activity
Use different concentration of
nutrients/glucose
Record the height of coloured
liquid by using a metre rule //
Calculate rate of yeast
respiration using formula:
= height of coloured liquid
time
Controlled variable :
Volume of yeast
suspension /mass of
yeast/volume of
glucose/pH/light
intensity/temperature/time
taken
Fix the volume of 100cm3 of
yeast suspension /the mass of
4 g of yeast /pH5 /light
intensity at distance of 50cm
/temperature at room
temperature/time taken for 10
minutes
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PERFECT SCORE BIOLOGY 2011
(d)
(e)
State the hypothesis for the experiment.
The higher/ lower the concentration of glucose, the
higher / lower the rate of yeast activity
(i)
Based on Table 1.1, construct a table and record the results
of the experiment which includes the following aspects:



3
3
Percentage concentration of glucose
Height of coloured liquid
The rate of the activity of yeast
Percentage
concentration
of glucose (%)
Height of
coloured liquid
(cm)
The rate of the
activity of yeast
(cm/min)
10
3
0.3
15
5
0.5
20
8
0.8
Table 1.1
(e)
(ii) Draw a graph of the rate of the activity of yeast against the
concentration of glucose
(iii
)
Based on the graph in 1(e)(ii), state the relationship between
the rate of the activity of yeast and the concentration of
glucose. Explain your answer.
3
3
When the concentration of glucose increases/decreases,
the rate of yeast activity increases/decreases, more
substrate for yeast to use for energy production, more
yeast reproduced.
(f)
Based on the experiment, define anaerobic respiration in
yeast operationally.
3
An anaerobic respiration is when yeast using glucose to
produce gas that causes the rising of liquid in
manometer tube and the process is affected by
concentration of glucose
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(g)
The experiment is repeated by using 1 ml of 0.1 mol dm-3 of
sodium hydroxide solution is added into the boiling tube.
Predict the manometer reading after 10 minutes. Explain
your prediction.
3
1 cm, not increase, sodium hydroxide is alkali, the
medium is not suitable for yeast.
(h)
The following list is part of the apparatus and material used
in this experiment.
3
Yeast, metre rule, coloured liquid, electronic
balance, glucose solution, measuring cylinder
Complete Table 1.3 by matching each variable with the
apparatus and material used in the experiment.
Variables
Apparatus
Material
Manipulated
Measuring
cylinder
Glucose
Responding
Coloured liquid
Metre ruler
electronic balance
Yeast
Controlled
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Question 2 :
No.
2
Questions
Marks
Student notes
Lemna minor is a species of free-floating aquatic plants from the duckweed family
Lemnaceae. The plants grow mainly by vegetative reproduction: two daughter plants
bud off from the adult plant.
An experiment is carried out to investigate the effect of abiotic factor such as pH on
Lemna sp. growth. Experiment is done under controlled conditions: 12 hours a day
light exposure and using the same Knop’s solution.
Petri dish is filled with 20 ml Knop’s solution with different pH value and 5 Lemna sp.
each.
The Knop’s solution is treated by adding acid or alkali to achieve the pH value needed.
** Knop’s solution is a solution which contains essential nutrient for plants growth.
Petri dish
Knop’s solution
Lemna minor
Figure 1
After 7 days, the observation is made and the result shown in Table 1.1
.
pH
value
Petri dish
Number of Lemna sp.
3
4
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5
5
7
8
9
11
11
5
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13
1
Table 1.1
No.
Questions
(a)
Marks
Student notes
State the number of Lemna sp. in the spaces
provided in Table 1.1
3
(b)
(i)
Based on Table 1, state two different observations .
Able to state any two observations correctly according to 2
criteria:
 pH ( Manipulated Variable)
 Number of Lemna sp (Responding Variable)
3
Sample answers:
1. At pH 2 (Knop solution), the number of Lemna sp is 4
2. At pH 8 (Knop solution), the number of Lemna sp is 11
3. At pH 12 ( Knop solution), the number of Lemna sp is
1
4. At pH 12 (Knop solution), the number of Lemna sp
grow is less than at pH 2/4/6/8/10
5. At pH 8 (Knop solution), the number of Lemna sp is
more than at pH2/4/6/10/12
(ii)
*1,2 &3 is a horizontal observation
*4 & 5 is a vertical observation
State the inferences which corresponds to the observations
in 1(b)(i).
Able to make one logical inference for each observation
based on the criteria
3
 suitable abiotic factor
 Favourable for Lemna sp growth
Sample answers:
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PERFECT SCORE BIOLOGY 2011
1. Strong acidic condition is not favorable for Lemna
growth.
2. Weak/slight alkaline // neutral condition is most
favorable for Lemna growth.
3. Strong alkaline is not favorable for Lemna growth.
4. Strong alkaline condition is the least favorable for
Lemna growth compare with other conditions.
5. Neutral/Slight alkaline condition is the best/moss
favorable condition for Lemna growth.
*1,2 &3 is a horizontal inference
*4 & 5 is a vertical inference
(c)
Complete Table 1.4 to show the variables involved in
the experiment and how the variables are operated.
Variables
3
How the variables are operated
Manipulated:
pH
Add/Use acid or alkali to the
Knop solution to get different
pH
condition//
Use
pH
solution: pH2, pH4, pH6, pH8,
pH10,pH12 // change/alter the
medium condition
Responding:
Count and record the number
of Lemna sp. plants after 7
days.
Number of Lemna sp
Fixed:
Light exposure /
Volume of Knop
solution
(d)
Fix 12 hours light exposure
every day /
Maintain the volume at 20ml
State the hypothesis for this experiment.
Able to state a hypothesis to show a relationship between the
manipulated variable and responding variable and the
hypothesis can be validated, based on 3 criteria:
 manipulated variable
 responding variable
 relationship
Sample answer :
1. In low pH, number of Lemna sp is less than in a
higher pH.
2. The higher pH the higher number of Lemna sp.
3. In a neutral condition the number of Lemna sp.
3
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plants is the highest /the most.
4. The more alkali the medium is the less number of
Lemna sp.
(e)
(i)
Construct a table and record the results of the
experiment.
Your table should contain the following title.

pH of water

Number of Lemna sp.
Able to draw and fill a table
labeled with complete unit
Sample answers
with all columns and rows
pH of water
2
4
6
8
10
12
(e)
(ii)
3
Number of Lemna sp
4
5
8
11
5
1
Plot a graph showing the number of Lemna sp against the pH
in the graph below
3
Able to plot a graph with 3 criteria:

A(axis): correct title with unit and uniform scale

P (point) : transferred correctly

S (Shape): able to joint all points, smooth graph, bell
shape.
(iii) Referring to the graph in (e) (ii), describe the relationship
between the Lemna sp growth and the condition of the
medium.
3
Able to state clearly and accurately the relationship between
the condition of medium and Lemna growth based on the
criteria:
(f)
 P1- Alkali, acidic or neutral (abiotic factor)
 P2- Lemna sp. growth
Sample answer:
(Associates each of the condition with the Lemna growth)
1. In the acidic medium the Lemna sp. growth is
less, and increase when the medium become
neutral but decrease when in alkali condition.
2 Lemna sp. grow very well in neutral medium and
less growth rate in alkali or acidic medium
Based on the experiment, define operationally the abiotic
factor in an ecosystem.
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PERFECT SCORE BIOLOGY 2011
Able to explain the abiotic factor operationally base on 3
criteria:
 Lemna sp (organism)
 affected (growth)
 pH of medium (abiotic factor in ecosystem)
Sample answer:
1. Abiotic factor is pH of the medium that affect the
Lemna sp growth in an ecosystem.
(g)
The effluent from laundry shop flows into a pond nearby,
predict the population of Lemna sp in the pond. Explain your
answer.
3
3
Able to predict the result accurately base on 2 criteria.
 Expected population of Lemna sp
 The reason of the answer
 Not suitable for growth
(h)
Sample answer:
P1- No Lemna sp found/ very small population of Lemna
sp,
P2- Because water is contaminated with soap/detergent
contain alkali,
P3- Which is not suitable/favourable for Lemna to grow
Classify the biotic and abiotic factors from the list
provided below.
Humidity, light intensity, decomposer,
parasites,
symbiotic
organism,
soil
texture, invertebrates, topography
Able to classify all 4 pairs of the abiotic and biotic factors in
ecosystem
Sample answer
Abiotic factors
Biotic factors
Humidity
Decomposer
Light intensity
Parasite
Soil texture
Symbiotic organism
Topography
invertebrates
3
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Question 3:
No.
1.
Questions
Marks Student
notes
A group of students conducted an experiment to study the effect of light intensity on the
population distribution of Lichen on the tree trunk. He placed a 10 cm x 10 cm transparent
quadrat on the East-facing surface of the tree trunk. He counted the number of squares that
contained half or more than half of the areas covered by the Lichen. Square with less than
half of the covered areas were not included.
The procedures were repeated for the surfaces that face the direction of North (N), south (S)
and west (W).
Figure 1 shows how a quadrat is placed on the tree trunk. Each small square represent 1
cm2.
Figure 1
Table 1 shows the areas covered by the Lichen on the different surface of the tree trunk.
Direction/position of
Total surface area covered by
surface
Lichen
East
60 cm2
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South
10 cm
35 cm2
North
10 cm
45 cm2
West
10 cm
52 cm2
Table 1
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a)
b)
Count the total surface area of Lichen for each quadrat and record
the answer in the spaces provided in Table 1.
(i)
3
State two different observation based on the diagram in Table 1.
Observation 1:
At the surface facing east (MV), the total surface area of Lichen
is 60 cm2 (RV).
Observation 2:
At the surface facing south (MV), the total surface area of Lichen
is 35 cm2 (RV).
3
(ii) State the inferences from the observation in 1 (b) (i).
Inference from observation 1:
At the east aspect is most suitable for the growth of Lichen
because it receives more light intensity, so higher rate of
photosynthesis.
Inference from observation 2:
At the south aspect is least suitable for the growth of Lichen
because it receives less light intensity, so lower rate of
photosynthesis.
(c)
3
Complete Table 2 based on this experiment.
Variable
Manipulated
variable
Method to handle the variable
Direction
facing on the
tree trunk //
Responding
variable
Use different direction on the tree trunk such as
east, north, south and west.
Total surface
area coverage
by Lichen
Count and record the total surface area
coverage by lichen by using the quadrat.
Constant
variable
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PERFECT SCORE BIOLOGY 2011
Quadrat size
Fix the size of quadrat at 10 cm X 10 cm.
Type of
organism
Fix the organism use in the experiment that
is Lichen
Sampling time Sampling experiment is carried out at same
time
Table 2
d)
3
State the hypothesis for this experiment:
1. The total surface area of Lichen on the tree trunk (RV) is
higher (R) when the light intensity is high (MV).
2. When the Lichen is facing east (MV), the total surface area
covered by Lichen/population of Lichen (RV) is increase (R).
3. The higher the light intensity (MV), the higher (R) the total
surface area covered by Lichen / the higher the population of
Lichen (RV).
e)
(i)
3
Construct a table and record all data collected in this experiment.
Your table should have the following aspect:
 Title with correct unit
 Position of direction
 Total surface area covered by Lichen
Position of direction
East
South
West
North
Total surface area covered by
Lichen (cm2)
60
35
52
45
3
(ii) Use the graph paper provided to answer this question.
Using the data in 1 (e) (i), draw a bar chart graph to show the
relationship between the population of Lichen against the directions
facing on the tree.
The population of Lichen is represented by the total surface area
covered in the quadrat.
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3
(f)
Based on the graph in 1 (e)(ii), explain the relationship between the
population distribution of Lichen and the light intensity.
P1 – Population of Lichen / Total surface area covered by Lichen
P2 – Position direction of quadrat
P3 – Degree of light intensity
Sample answer:
1. Population of Lichen / The total surface area covered by
Lichen is higher at east direction which receives high
light intensity.
2. Population of Lichen / The total surface area covered by
Lichen is low at south direction which receives low light
intensity.
3. Population of Lichen / The total surface area covered by
Lichen is higher at east direction than at the south
direction because Lichen at east direction receives high
light intensity so rate of photosynthesis is higher.
(g)
3
State the operational definition for population distribution of Lichen.
P1 – Total surface area covered by Lichen
P2 – Size of quadrat
P3 – Abiotic factor that influence the population distribution
Sample answer:
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1. Population distribution is defined as total surface area
covered by Lichen (P1) within the quadrat size of 10 cm x 10
cm at different direction of compass (P2) which influence by
the light intensity (P3).
(h)
3
Lightning strike the tree and cause the tree to fall. The Lichen under
study is then exposed to direct sunlight from 7.00 a.m. to 6.00 p.m.
daily.
Predict what will happen to the total surface area covered by Lichen
after a month.
Explain your prediction.
P1: Prediction of total surface area of Lichen
P2: Effect of light intensity
P3: Effect on the Lichen
Sample answer:
Size of total surface area covered by lichen is increase / more
than 60 cm2 because Lichen receive more sunlight / light
intensity, so more photosynthesis by Lichen and more growth to
Lichen.
(i)
3
The following is a list of biotic and abiotic factors.
pH of water, pigeon orchid, humidity, bird,
temperature, Elodea sp.
3
Classify these factors in the Table 3.
Abiotic factors
pH of water
Humidity
Temperature
Biotic factors
Pigeon orchid
Bird
Elodea sp
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Question 4 :
No.
4
Questions
Marks
Student notes
An experiment was carried out to investigate the water pollution level or BOD in three different
locations from a suspected polluted Rivers. Three water samples are collected from these
three locations and labelled as P, Q and R as in Diagram 1.
200 ml of each sample is put in a reagent bottle and added with 1 ml of 0.1% methylene blue
solution. All the bottles are kept in dark cupboard.
Observations are made every minute to see the changes in the methylene blue colour.
Sample P
Sample Q
Sample R
Each sample is added
with methylene blue
solution
Diagram 1
Table 1 shows the results of this experiment.
Water sample
P
Q
R
Time taken for
methylene
blue solution
become
colourless
10 minutes
23 minutes
42 minutes
Table 1
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No.
Questions
(a)
Marks
Student notes
Record the time taken for methylene blue solution become
colourless in the boxes provided in Table 1.
3
(b)
(i)
Based on Table 1, state two different observations .
Able to state any two observations correctly according to the
criteria:
o Sample
o Time taken
o Become colourless
Sample answers:
1. Time taken for methylene blue to become colourless
for sample P is 10 minutes.
2. Time taken for methylene blue to become colourless
for sample R is 42 minutes
3. Time taken for methylene blue to become colourless
for sample Q is 23 minutes
4. Time taken for sample P is 10 minutes that is shorter
than time taken for sample R that is 42 minutes to
become colourless
3
(ii) State the inferences which corresponds to the observations in
1(b)(i).
Able to make one logical inference for each observation
based on the criteria
o
o
o
Sample
Oxygen concentration
Duration of time for methylene blue to become
colourless
Sample answers:
1. In sample P, oxygen concentration is low, the
methylene blue become colourless very fast/ less
time taken
2. Oxygen concentration in sample R is high, the
methylene blue become colourless slow/ longer
time taken
3. Oxygen concentration in sample P is lower than
oxygen concentration in sample R, the time taken for
methylene blue to become colourless is shorter.
3
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Complete Table 2 based on this experiment.
(c)
Variables
How the variables are
operated
Manipulated:
Water sample
Water sample is collected
from
three
different
locations.
Responding variable
Time taken for methylene
blue
to
become
colourless is recorded by
using a stopwatch.
Time taken to
decolourise methylene
blue
Fixed variable
Metlhylene blue
concentration / volume/
volume of water
sample
0.1% of Methylene blue is
used for all experiments/
1 ml volume/ 200 ml of
water sample.
3
Table 2
(d)
State the hypothesis for this experiment.
Able to state a hypothesis to show a relationship between the
manipulated variable and responding variable and the
hypothesis can be validated, base on 3 criteria:
 manipulated variable
 responding variable
 relationship
Sample answer :
1. The most polluted water has shortest time for
methylene blue to become colourless.
2. Sample water P is the most polluted has shortest time
for methylene blue to become colourless.
3. Sample water R is less polluted compare to water
samples P and Q, has longest time for methylene blue
to become colourless,
(e)
(i)
3
Construct a table and record all the data collected in this
experiment based on the following criteria:


Water sample
Time taken
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Able to tabulate a table and fill in data accurately base on
three criteria:
o Table draw with labeled column.
o Sample
o Time taken with unit.
Sample answers :
Water Sample
Time taken ( minutes)
P
10
Q
23
R
42
3
(f)
Based on the data in 1(e) draw a bar chart of time taken for
methylene blue solution become colourless against water
samples.
Able to draw a bar chart base on criteria:
o
o
o
(g)
Correct chart
Axis with correct scale
Correct value
What is the relationship between time taken, oxygen
concentration and BOD value of water in this experiment?
Able to state clearly and accurately the relationship between:
o time taken
o oxygen content
o BOD value
Sample answer:
1. The shorter time taken for methylene blue to
become colourless, less oxygen in the water and
BOD value is high.
(h)
3
3
Based on the result of this experiment, state the operational
definition for BOD
Able to explain BOD base on experiment correctly according
to the criteria:
o Amount of oxygen in the water sample
o used by microorganisms
o shown by time taken
Sample answer:
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PERFECT SCORE BIOLOGY 2011
2. BOD is amount of oxygen in the water sample that
used by microorganisms and can be shown by
time taken of methylene blue to become
colourless.
(i)
3
This experiment is repeated by using water sample from
chicken farm areas. Predict the time taken for methelyne blue
to become colourless.
Able to predict the result accurately.
o Expected time
o Compare to which
o Reason
Sample answer:
The time taken for methylene blue to become colourless
is 5 minutes, less than water sample P, because chicken
farm water can be contaminated with chicken faeces/ or
any other answer.
(j)
3
Arrange the water samples from the most polluted to the least
polluted.
Able to arrange the 3 level of polluted water
Sample answer:
Types of water
Polluted
P
Most
Q
Moderate
R
Least
3
Most polluted  least polluted
P  Q

R
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Question 5 :
No.
5
Questions
Marks
Student notes
Transpiration is the evaporation of water from a plant to the surroundings. The rate of transpiration
is affected by environmental factors such as temperature.
A group of students carried out an experiment to study the effect of temperature on the rate of
transpiration. Diagram 1 shows the set up of the apparatus. An air bubble was trapped in the
capillary tube. The apparatus was placed in an air-conditioned room at 20oC.
The time taken for the air bubble to move a distance of 10 cm was recorded. The experiment was
repeated for a second time to get average readings.
The experiment is repeated by placing the apparatus at three more different temperatures: an airconditioned room at 25oC , an air-conditioned room at 30oC and in a non air-conditioned room at
35oC.
Diagram 1
Table 1 shows the reading of stopwatch for air bubble to move a distance of 10 cm at different
temperature
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Temperature
o
C
Time taken for air bubble to move a distance of 10 cm (min)
First reading
Second reading
Average
Reading
20
39
41
40.0
25
28
32
30.0
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Temperature
Suhu oC
Time taken for air bubble to move a distance of 10 cm (min)
First reading
Second reading
Average
Reading
30
20
20
20.0
11
9
10.0
35
No.
Questions
(a)
Record the time taken for the air bubbles to move a distance
of 10 cm and average reading in Table 1.
(b)
(i)
Marks
Student notes
3
Based on Table 1, state two different observations .
1. When temperature is 20oC, the average time taken for
air bubble to move a distance of 10 cm is 40 minutes
2. When temperature is 35oC , the average time taken for
air bubble to move a distance of 10 cm is 10 minutes.
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3. When temperature is 20oC ,the average time taken
for air bubble to move a distance of 10 cm is
longer than the average time taken when
temperature is 35oC
3
(ii) State the inferences which corresponds to the observations
in 1(b)(i).
(c)
1. (When temperature is low) , the amount of water lost
from the leaf is low(P1). So the rate of transpiration is
low (P2)
2. (When temperature is high) , the amount of water lost
from the leaf is high(p1). So the rate of transpiration is
high (P2)
3. When the temperature is higher/lower, the amount of
water lost from the leaf is higher/lower. So the rate of
transpiration is higher/lower when the temperature is
higher/lower
Complete Table 2 based on this experiment.
Variable
Method to handle the variable
Manipulated Variable
Temperature
3
Place the
apparatus/potometer at
different temperature / 20 oC,
25 oC, 30 oC and 35 oC
Responding Variable
1.Rate of transpiration
1.Calculate and record the
rate of transpiration by using
formula : Distance / time
2. Time taken for air
bubble to move a
distance of 10 cm
2. Record the time taken for
air bubble to move a distance
of 10 cm by using stopwatch
Constant Variable
1.Type of plant
1.Use the same plant
2.Distance travelled by
air bubble
2.Fix the distance travelled by
air bubble at 10cm
3
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(d)
State the hypothesis for this experiment.
Able to make a hypothesis based on the following aspects
P1 : MV- Temperature
P2 : RV – rate of transpiration
H : Relationship (Higher…. Higher)
Sample Answer:
1. The higher the temperature, the higher the rate of
transpiration//vice versa
(e)
(i)
Construct a table and record all the data collected in this
experiment.
Your table should have the following aspects:
- Temperature
- Average time taken for air bubbles to move a
distance of 10 cm .
Rate of transpiration
Rate of transpiration = Distance
Time
Able to construct a table based on the following aspects
1. Title with correct unit - 1 mark
2. Data
- 1 mark
3. Rate of transpiration - 1 mark
Sample Answer
Average time taken for air by
Temperature
air bubble to move a
Suhu oC
distance of 10 cm (min)
20
25
30
35
(e)
3
Rate of
transpiration
cm/min
40.0
30.0
20.0
10.0
0.25
0.33
0.5
1.0
(ii) Using the data in 1(e)(i), draw the graph of the rate of
transpiration against the temperature
Able to draw the graph correctly
Axes : Uniform scales on both horizontal and vertical axis
with correct unit – 1 mark
Points : All points plotted correctly - 1 mark
Curve : smooth without touching the axes - 1 mark
3
3
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PERFECT SCORE BIOLOGY 2011
(f)
Based on the graph in 1(e)(ii), explain the relationship
between the rate of transpiration and temperature.
Able to explain the relationship between the rate of
transpiration and temperature based on the following
aspects.
P1 – State the relationship
P2 – kinetic energy of water
P3 – evaporation
Sample Answer
When the temperature increases, the rate of
transpiration increases. When the temperature
increases, kinetic energy of water molecules (in the leaf)
increases, causes the rate of evaporation increase.
(g)
(h)
Based on the result of this experiment, state the operational
definition for process of transpiration.
Able to define operationally the process of transpiration
based on the following aspects:
P1 – water loss from plant at different places
P2 – Air bubble in capillary tube move at 10 cm
P3 – The rate of transpiration is influenced by temperature
Sample Answer
Transpiration is a process where water is lost from the
plant when it is placed at different temperature which
causes the air bubble in capillary tube move a distance
of 10 cm. The rate of transpiration is influenced by the
temperature.
If the surface of the leaves of a plant at temperature of 35 oC
are covered with vaselin, predict the time taken for air
bubble to move a distance of 10 cm. Explain your prediction.
3
3
Able to predict the outcome of the experiment based on the
following aspects
P1 : Correct prediction
P2 : Effect
P3 : Reason
Sample Answer
Time taken for air to move a distance of 10 cm is more
than 10 minutes. Rate of transpiration decreases
because vaselin covered the stomata/stomata closed
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PERFECT SCORE BIOLOGY 2011
(i)
The following list is a factor that affecting transpiration.
Relative humidity
cuticle
Kelembapan relatif
3
kutikel
air movement
pergerakan angin
stomata
stomata
light intensity
keamatan cahaya
Classify the factors into two group in Table 3.
Environmental factor
Morphology factors
1. Relative humidity
1. Cuticle
2. Air movement
2. Stomata
3. Light intensity
Table 3
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PERFECT SCORE BIOLOGY 2011
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