AP Biology
Lab Review
Based on the PPT from Ms. Foglia
Big Idea 1: Evolution
Lab 1: Artificial Selection
• Concepts:
• Natural selection = differential reproduction in a population
• Populations change over time  evolution
• Natural Selection vs. Artificial Selection
Lab 1: Artificial Selection
• Description:
•
•
•
•
Use Wisconsin Fast Plants to perform artificial selection
Identify traits and variations in traits
Cross-pollinate (top 10%) for selected trait
Collect data for 2 generations (P and F1)
Sample Histogram of a Population
Lab 1: Artificial Selection
Analysis & Results:
• Calculate mean, median, standard deviation, range
• Are the 2 populations before and after selection (P and F1)
actually different?
• Are the 2 sub-populations of F1 (hairy vs. non-hairy) different?
• Are the means statistically different?
• A T-test could be used to determine if 2 sets of data are
statistically different from each other; these are often used in
examples where the sample size is small
Lab 2: Mathematical Modeling:
Hardy-Weinberg
• Concepts:
• Evolution = change in frequency of alleles in a population
from generation to generation
• Hardy-Weinberg Equilibrium
• Allele Frequencies (p + q = 1)
• Genotypic Frequencies (p2+2pq+q2 = 1)
• Conditions:
1.
2.
3.
4.
5.
large population
random mating
no mutations
no natural selection
no migration
Lab 2: Mathematical Modeling:
Hardy-Weinberg
Analysis & Results:
• Null model: in the absence of random events that
affect populations, allele frequencies (p,q) should
be the same from generation to generation (H-W
equilibrium)
• Analyze genetic drift and the effect of selection on
a given population
• Manipulate parameters in model:
• Population size, selection (fitness), mutation, migration,
genetic drift
Lab 2: Mathematical Modeling:
Hardy-Weinberg
• Real-life applications:
• Cystic fibrosis, polydactyly
• Heterozygote advantage (Sickle-Cell Anemia)
Lab 2: Mathematical Modeling:
Hardy-Weinberg
ESSAY 1989
Do the following with reference to the Hardy-Weinberg model.
a. Indicate the conditions under which allele frequencies (p and q) remain constant
from one generation to the next.
b. Calculate, showing all work, the frequencies of the alleles and frequencies of the
genotypes in a population of 100,000 rabbits of which 25,000 are white and
75,000 are agouti (brownish).
(In rabbits the white color is due to a recessive allele, w, and agouti is due to a
dominant allele, W.)
c. If the homozygous dominant condition were to become lethal, what would
happen to the allelic and genotypic frequencies in the rabbit population after
two generations?
a. Indicate the conditions under which allele frequencies (p and q) remain
constant from one generation to the next.
1. large population
2. random mating
3. no mutations
4. no natural selection
5. no migration
b. Calculate, showing all work, the frequencies of the alleles and frequencies of
the genotypes in a population of 100,000 rabbits of which 25,000 are white
and 75,000 are agouti (brownish). (In rabbits the white color is due to a
recessive allele, w, and agouti is due to a dominant allele, W.)
25,000/ 100,000 = .25 so .25 = q2
(a) q = .5 and (A) p = .5
Homozygous Rec = .25 = 25,000
Homozygous Dom (agouti) = p2 = .25 = 25,000
Heterozygous (agouti) = 2(p)(q) = 2(.5)(.5) = .5 = 50,000
c. If the homozygous dominant condition were to become lethal, what would
happen to the allelic and genotypic frequencies in the rabbit population after
two generations? The frequency of the ‘A’ allele would decrease and the
frequency of the ‘a’ allele would increase.
Lab 3: Comparing DNA Sequences using BLAST 
Evolutionary Relationships
• Concepts:
• Bioinformatics: combines statistics, math modeling,
computer science to analyze biological data
• Genomes can be compared to detect genetic
similarities and differences
• BLAST = Basic Local Alignment Search Tool
• Input gene sequence of interest
• Search genomic libraries for identical or similar sequences
Lab 3: Comparing DNA Sequences using BLAST 
Evolutionary Relationships
• Description:
• Use BLAST to compare several genes
• Use information to construct a cladogram (phylogenetic tree)
• Cladogram = visualization of evolutionary relatedness of species
Lab 3: Comparing DNA Sequences using BLAST 
Evolutionary Relationships
Lab 3: Comparing DNA Sequences using BLAST 
Evolutionary Relationships
• Use this data to construct a cladogram of the major plant groups
Lab 3: Comparing DNA Sequences using BLAST 
Evolutionary Relationships
• Fossil specimen in China
• DNA was extracted from preserved tissue
• Sequences from 4 genes were analyzed using BLAST
Lab 3: Comparing DNA Sequences using BLAST 
Evolutionary Relationships
• Analysis & Results:
• BLAST results: the higher the score, the closer the alignment
• The more similar the genes, the more recent their common ancestor 
located closer on the cladogram
Lab 3: Comparing DNA Sequences using BLAST 
Evolutionary Relationships
Answer is C
Big Idea 2: cellular
processes: energy and
communication
Lab 4: Diffusion & Osmosis
• Concepts:
•
•
•
•
Selectively permeable membrane
Diffusion (high  low concentration)
Osmosis (aquaporins)
Water potential ()
•  = pressure potential (P) + solute potential (S)
• Solutions:
• Hypertonic
• hypotonic
• isotonic
Lab 4: Diffusion & Osmosis
Lab 4: Diffusion & Osmosis
• Description:
• Surface area and cell size vs. rate of diffusion
• Cell modeling: dialysis tubing + various solutions
(distilled water, sucrose, salt, glucose, protein)
• Identify concentrations of sucrose solution and solute
concentration of potato cores
• Observe osmosis in onion cells (effect of salt water)
Lab 4: Diffusion & Osmosis
Potato Cores in Different Concentrations of Sucrose
Lab 4: Diffusion & Osmosis
• Conclusions
• Water moves from high water potential ( )
(hypotonic=low solute) to low water potential ()
(hypertonic=high solute)
• Solute concentration & size of molecule
affect movement across selectively permeable membrane
Answer is A
Answer is B
Lab 4: Diffusion & Osmosis
ESSAY 1992
A laboratory assistant prepared solutions of 0.8 M, 0.6 M, 0.4 M, and
0.2 M sucrose, but forgot to label them. After realizing the error, the assistant
randomly labeled the flasks containing these four unknown solutions as flask A, flask
B, flask C, and flask D.
Design an experiment, based on the principles of diffusion and osmosis, that the
assistant could use to determine which of the flasks contains each of the four
unknown solutions.
Include in your answer:
a. a description of how you would set up and perform the experiment;
b. the results you would expect from your experiment; and
c. an explanation of those results based on the principles involved.
Be sure to clearly state the principles addressed in your discussion.
A laboratory assistant prepared solutions of 0.8 M, 0.6 M, 0.4 M, and
0.2 M sucrose, but forgot to label them. After realizing the error, the assistant randomly labeled the
flasks containing these four unknown solutions as flask A, flask B, flask C, and flask D.
Design an experiment, based on the principles of diffusion and osmosis, that the assistant could use to
determine which of the flasks contains each of the four unknown solutions.
Include in your answer:
a. a description of how you would set up and perform the experiment;
b. the results you would expect from your experiment; and
c. an explanation of those results based on the principles involved.
Be sure to clearly state the principles addressed in your discussion.
Think about our lab that we did…. Talk about putting a cell (potato) in each of the solutions – mass
before and then mass after and see in which solution did it gain/lose the most weight. The principle
involved is simple osmosis. The potato will lose weight in hypertonic solutions and gain weight in the
hypotonic solutions. Based on how much weight is gained/lost in each of the solutions, you would be
able to put them in order.
0.8 M – potato will lose the most mass
0.6 M – potato will lose the second most mass
0.4 M – potato will remain mostly the same with losing just a bit
0.2 M – potato will gain mass
*potatos are about 0.3 molarity
Lab 5: Photosynthesis
• Concepts:
• Photosynthesis
• 6H2O + 6CO2 + Light  C6H12O6 + 6O2
• Ways to measure the rate of photosynthesis:
• Production of oxygen (O2)
• Consumption of carbon dioxide (CO2)
Lab 5: Photosynthesis
• Description:
• Paper chromatography to identify pigments
• Floating disk technique
• Leaf disks float in water
• Gases can be drawn from out from leaf using syringe
 leaf sinks
• Photosynthesis  O2 produced  bubbles form on
leaf  leaf disk rises
• Measure rate of photosynthesis by O2 production
• Factors tested: types of plants, light intensity, colors of
leaves, pH of solutions
Plant Pigments & Chromatography
Floating Disk Technique
To make comparisons between
experiments, a standard point of
reference is needed. Repeated testing
of this procedure has shown that the
point at which 50% of the disks are
floating (the median or ET50) is a
reliable and repeatable point of
reference. In this case, the disks
floating are counted at the end of each
time interval. The median is chosen
over the mean as the summary
statistic. The median will generally
provide a better estimate of the central
tendency of the data because, on
occasion, a disk fails to rise or takes a
very long time to do so. A term coined
by G. L Steucek and R. J Hill (1985)
for this relationship is ET50, the
estimated time for 50% of the disks to
rise. That is, rate is a change in a
variable over time. The time required
for 50% of the leaf disks to float is
represented as Effective Time = ET50.
Lab 5: Photosynthesis
• Concepts:
• photosynthesis
• Photosystems II, I
• H2O split, ATP, NADPH
• chlorophylls & other
plant pigments
•
•
•
•
chlorophyll a
chlorophyll b
xanthophylls
carotenoids
• experimental design
• control vs. experimental
Answer is A
Answer is D
Lab 6: Cellular Respiration
• Concepts:
• Respiration
• Measure rate of respiration by:
• O2 consumption
• CO2 production
Lab 6: Cellular Respiration
• Description:
• Use respirometer
• Measure rate of respiration (O2 consumption) in various seeds
• Factors tested:
•
•
•
•
•
•
Non-germinating seeds
Germinating seeds
Effect of temperature
Surface area of seeds
Types of seeds
Plants vs. animals
Lab 6: Cellular Respiration
Lab 6: Cellular Respiration
Lab 6:• Cellular
Respiration
Conclusions:
•
•
•
•
temp = respiration
germination = respiration
Animal respiration > plant respiration
 surface area =  respiration
Calculate Rate
Lab 6: Cellular Respiration
Answer is A
Answer is A
Answer is C
Lab 6: Cellular Respiration
ESSAY 1990
The results below are measurements of cumulative oxygen consumption by germinating and dry
seeds. Gas volume measurements were corrected for changes in temperature and pressure.
Cumulative Oxygen Consumed (mL)
Time (minutes)
0
10
20
30
40
Germinating seeds 22°C
0.0
8.8
16.0
23.7
32.0
Dry Seeds (non-germinating) 22°C
0.0
0.2
0.1
0.0
0.1
Germinating Seeds 10°C
0.0
2.9
6.2
9.4
12.5
Dry Seeds (non-germinating) 10°C
0.0
0.0
0.2
0.1
0.2
a. Plot the results for the germinating seeds at 22°C and 10°C.
b. Calculate the rate of oxygen consumption for the germinating seeds at 22°C, using the time
interval between 10 and 20 minutes.
c. Account for the differences in oxygen consumption observed between:
1. germinating seeds at 22°C and at 10°C
2. germinating seeds and dry seeds.
d. Describe the essential features of an experimental apparatus that could be used to measure
oxygen consumption by a small organism. Explain why each of these features is necessary.
Lab 6: Cellular Respiration
Cumulative Oxygen Consumed (mL)
Time (minutes)
0
10
20
30
40
Germinating seeds 22°C
0.0
8.8
16.0
23.7
32.0
Dry Seeds (non-germinating) 22°C
0.0
0.2
0.1
0.0
0.1
Germinating Seeds 10°C
0.0
2.9
6.2
9.4
12.5
Dry Seeds (non-germinating) 10°C
0.0
0.0
0.2
0.1
0.2
a. Plot the results for the
germinating seeds at
22°C and 10°C.
I would say mL of O2 consumed instead of
“corrected difference” and this graph shows
all 4 lines when we only needed 2 (this was
all I could find online)
**make sure graph has a title, axis are
labeled, lines labeled
Lab 6: Cellular Respiration
Cumulative Oxygen Consumed (mL)
Time (minutes)
0
10
20
30
40
Germinating seeds 22°C
0.0
8.8
16.0
23.7
32.0
Dry Seeds (non-germinating) 22°C
0.0
0.2
0.1
0.0
0.1
Germinating Seeds 10°C
0.0
2.9
6.2
9.4
12.5
Dry Seeds (non-germinating) 10°C
0.0
0.0
0.2
0.1
0.2
b.
Calculate the rate of oxygen consumption for the germinating seeds at 22°C,
using the time interval between 10 and 20 minutes.
Rate = ∆Y/ ∆X
(16.0 – 8.8)
= 7.2 = .72 mL/ min
(20-10)
10
**make sure to show work and make sure to include units!
Lab 6: Cellular Respiration
Cumulative Oxygen Consumed (mL)
Time (minutes)
0
10
20
30
40
Germinating seeds 22°C
0.0
8.8
16.0
23.7
32.0
Dry Seeds (non-germinating) 22°C
0.0
0.2
0.1
0.0
0.1
Germinating Seeds 10°C
0.0
2.9
6.2
9.4
12.5
Dry Seeds (non-germinating) 10°C
0.0
0.0
0.2
0.1
0.2
c. Account for the differences in oxygen consumption observed between:
1. germinating seeds at 22°C and at 10°C  increased temp causes increased respiration/
metabolism so they use more O2 at a higher temp
2. germinating seeds and dry seeds  germinating seeds are active and thus will use more O2,
dry seeds are dormant and will respire very little
d. Describe the essential features of an experimental apparatus that could be used to measure
oxygen consumption by a small organism. Explain why each of these features is necessary. Needs
to be a closed system that separates oxygen consumption and carbon dioxide release (like KOH
that turns CO2 into a solid). Must be able to control temp (water bath), control volume (glass
beads), have a timer, equal numbers of organisms in each setup, etc.
Big Idea 3: genetics and
information transfer
Lab 7: Mitosis & Meiosis
• Concepts:
• Cell Cycle (G1  S  G2  M)
• Control of cell cycle (checkpoints)
• Cyclins & cyclin-dependent kinases (CDKs)
• Mitosis vs. Meiosis
• Crossing over  genetic diversity
Lab 7: Mitosis & Meiosis
Lab 7: Mitosis & Meiosis
Lab 7: Mitosis & Meiosis
• Description:
• Model mitosis & meiosis (pipecleaners, beads)
• How environment affects mitosis of plant roots
• Lectin - proteins secreted by fungus
• Root stimulating powder
• Count # cells in interphase, mitosis
• Observe karyotypes (cancer, mutations)
• Meiosis & crossing over in Sordaria (fungus)
Lab 7: Mitosis & Meiosis
Lab 7: Mitosis & Meiosis
Abnormal karyotype = Cancer
Meiosis: Crossing over in Prophase I
Lab 7: Mitosis & Meiosis
• Observed crossing over in fungus (Sordaria)
• Arrangement of ascospores
Sordaria Analysis
total crossover
% crossover =
total offspring
distance from
=
centromere
% crossover
2
Lab 7: Mitosis & Meiosis
ESSAY 1987
Discuss the process of cell division in animals. Include a description of mitosis and
cytokinesis, and of the other phases of the cell cycle. Do not include meiosis.
ESSAY 2004
Meiosis reduces chromosome number and rearranges genetic information.
a. Explain how the reduction and rearrangement are accomplished in meiosis.
b. Several human disorders occur as a result of defects in the meiotic process. Identify
ONE such chromosomal abnormality; what effects does it have on the phenotype of
people with the disorder? Describe how this abnormality could result from a defect in
meiosis.
c. Production of offspring by parthenogenesis or cloning bypasses the typical meiotic
process. Describe either parthenogenesis or cloning and compare the genomes of the
offspring with those of the parents.
Lab 7: Mitosis & Meiosis
ESSAY 1987
Discuss the process of cell division in animals. Include a description of mitosis and
cytokinesis, and of the other phases of the cell cycle. Do not include meiosis. Simply
explain cell cycle (G1, S, G2, Mitosis, cytokinesis)  see notes on topic
ESSAY 2004
Meiosis reduces chromosome number and rearranges genetic information.
a. Explain how the reduction and rearrangement are accomplished in meiosis. Chromosomes
are paired in tetrads first, they get split in half so the cells after the first division of meiosis
have ½ the number of chromosomes as the parent cell. Rearrangement is due to
independent assortment, where maternal and paternal chromosomes line up randomly in
the tetrads, and crossing over which occurs in Prophase I. Crossing over is when a piece of
one chromosomes breaks off and switches places with a piece on the homologous
chromosome.
b. Several human disorders occur as a result of defects in the meiotic process. Identify ONE
such chromosomal abnormality; what effects does it have on the phenotype of people with
the disorder? Describe how this abnormality could result from a defect in meiosis.
c. Production of offspring by parthenogenesis or cloning bypasses the typical meiotic process.
Describe either parthenogenesis or cloning and compare the genomes of the offspring with
those of the parents.
Lab 7: Mitosis & Meiosis
b. Several human disorders occur as a result of defects in the meiotic process. Identify
ONE such chromosomal abnormality; what effects does it have on the phenotype of
people with the disorder? Describe how this abnormality could result from a defect in
meiosis. Down Syndrome is trisomy 21 (an extra copy of chromosome #21). This is
characterized by low muscle tone, small stature, an upward slant to the eyes, a single
deep crease across the center of the palm, and some mental deficiencies. Down
Syndrome is caused by nondisjunction – chromosomes not separating correctly
during meiosis. This can happen in either meiosis I or meiosis II.
c. Production of offspring by parthenogenesis or cloning bypasses the typical meiotic
process. Describe either parthenogenesis or cloning and compare the genomes of the
offspring with those of the parents. Parthenogenesis is an example of asexual
reproduction where growth and development of embryos occur without fertilization.
In animals, parthenogenesis means development of an embryo from an unfertilized
egg cell. This would produce clones and all offspring would be genetically identical to
the parent cell.
Lab 8: Bacterial Transformation
Concepts:
• Transformation: uptake of foreign DNA from
surroundings
• Plasmid = small ring of DNA with a few genes
• Replicates separately from bacteria DNA
• Can carry genes for antibiotic resistance
• Genetic engineering: recombinant DNA = pGLO
plasmid
Lab 8: Bacterial Transformation
Lab 8: Bacterial Transformation
Lab 8: Bacterial Transformation
• Conclusions:
• Foreign DNA inserted using vector (plasmid)
• Ampicillin = Selecting agent
• No transformation = no growth on amp+ plate
• Regulate genes by transcription factors (araC protein)
Plates that have only ampicillinresistant bacteria growing include
which of the following?
A. I only
B. III only
C. IV only
D. I and II
Plates that have only ampicillinresistant bacteria growing include
which of the following?
A. I only
B. III only
C. IV only
D. I and II
Answer is C
Answer is A
Answer is A
Answer is B
Answer is C
Lab 9: Restriction Enzyme Analysis of DNA
• Concepts:
• Restriction Enzymes
• Cut DNA at specific locations
• Gel Electrophoresis
• DNA is negatively charged
• Smaller fragments travel faster
Lab 9: Restriction Enzyme Analysis of DNA
• Description
Lab 9: Restriction Enzyme Analysis of DNA
• Determine DNA fragment sizes
Lab 9: Restriction Enzyme Analysis of DNA
• Conclusions:
•
•
•
•
Restriction enzymes cut at specific locations (restriction sites)
DNA is negatively charged
Smaller DNA fragments travel faster than larger fragments
Relative size of DNA fragments can be determined by distance travelled
• Use standard curve to calculate size (remember we made this from the marker when we
KNEW how many bases were in each band)
Lab 8-9: Biotechnology
ESSAY 1995
The diagram below shows a segment of DNA with a total length of 4,900 base pairs.
The arrows indicate reaction sites for two restriction enzymes (enzyme X and enzyme Y).
En zyme
X
En zymeEn zyme
Y
En zy me
X
X
D NA Seg men t
Len g th (b ase p airs)
4 00
500
1,20 0
1,30 0
1 ,5 0 0
a. Explain how the principles of gel electrophoresis allow for the separation of DNA fragments
b. Describe the results you would expect from electrophoretic separation of fragments from the following
treatments of the DNA segment above. Assume that the digestion occurred under appropriate
conditions and went to completion.
I. DNA digested with only enzyme X
II. DNA digested with only enzyme Y
III. DNA digested with enzyme X and enzyme Y combined
IV. Undigested DNA
c. Explain both of the following:
1. The mechanism of action of restriction enzymes
2. The different results you would expect if a mutation occurred at the recognition site for
enzyme Y.
Lab 8-9: Biotechnology
ESSAY 1995
The diagram below shows a segment of DNA with a total length of 4,900 base pairs.
The arrows indicate reaction sites for two restriction enzymes (enzyme X and enzyme Y).
En zyme
X
En zymeEn zyme
Y
En zy me
X
X
D NA Seg men t
Len g th (b ase p airs)
4 00
500
1,20 0
1,30 0
1 ,5 0 0
a. Explain how the principles of gel electrophoresis allow for the separation of DNA fragments. Gel
electrophoresis separates segments of DNA based on their size. The DNA is put into the gel at the
negative end and it runs to the positive. The electric current that moves through the buffer and the gel
allow the negatively charged DNA to be attracted to the positive end of the chamber. The smaller
pieces move through the gel faster than the larger pieces.
Lab 8-9: Biotechnology
ESSAY 1995
The diagram below shows a segment of DNA with a total length of 4,900 base pairs.
The arrows indicate reaction sites for two restriction enzymes (enzyme X and enzyme Y).
En zyme
X
En zymeEn zyme
Y
En zy me
X
X
D NA Seg men t
Len g th (b ase p airs)
4 00
500
1,20 0
1,30 0
1 ,5 0 0
b. Describe the results you would expect from electrophoretic separation of fragments from the following
treatments of the DNA segment above. Assume that the digestion occurred under appropriate conditions
and went to completion.
I. DNA digested with only enzyme X  4 pieces; sizes in base pairs are: 400, 1700, 1300, 1500
II. DNA digested with only enzyme Y  2 pieces; sizes in base pairs are: 900, 4000
III. DNA digested with enzyme X and enzyme Y combined  5 bands; sizes in base pairs are: 400,
500, 1200, 1300, 1500
IV. Undigested DNA  one huge band
Lab 8-9: Biotechnology
ESSAY 1995
The diagram below shows a segment of DNA with a total length of 4,900 base pairs.
The arrows indicate reaction sites for two restriction enzymes (enzyme X and enzyme Y).
En zyme
X
En zymeEn zyme
Y
En zy me
X
X
D NA Seg men t
Len g th (b ase p airs)
4 00
500
1,20 0
1,30 0
1 ,5 0 0
c. Explain both of the following:
1. The mechanism of action of restriction enzymes  Restriction enzymes identify specific
splice sites in DNA and cut there. Restriction sites are palindromes which read the same
forward as backward on the opposite strands of the DNA molecule. Restriction enzymes can
either create blunt ends or sticky ends.
2. The different results you would expect if a mutation occurred at the recognition site for
enzyme Y. I would expect the Y enzyme not to cut the piece of DNA, so if you digested the
whole piece of DNA with both the X and Y enzymes, it would only be cut at the X restriction
sites.
Big Idea 4: interactions
Lab 10: Energy Dynamics
• Concepts:
• Energy from sunlight  drives photosynthesis (store
E in organic compounds)
• Gross Productivity (GPP) = energy captured
• But some energy is used for respiration (R)
• Net primary productivity (NPP) = GPP – R
• Energy flows! (but matter cycles)
• Producers  consumers
• Biomass = mass of dry weight
Lab 10: Energy Dynamics
Pyramid of Energy
Pyramid of Biomass
Pyramid of Numbers
Lab 10: Energy Dynamics
• Description:
• Brassica (cabbage)  cabbage white butterfly larvae
(caterpillars)
Lab 10: Energy Dynamics
• Measuring Biomass:
• Cabbage  mass lost
• Caterpillar  mass gained
• Caterpillar frass (poop)  dry mass
Lab 10: Energy Dynamics
Lab 10: Energy Dynamics
• Conclusions:
Lab 10: Energy Dynamics
• Conclusions:
• Energy is lost (respiration, waste)
• Conservation of Mass
• Input = Output
Answer:
125 – (60 + 5) = 60 g/m2
Lab 11: Transpiration
• Concepts:
• Transpiration
• Xylem
• Water potential
• Cohesion-tension hypothesis
• Stomata & Guard cells
• Leaf surface area & # stomata vs. rate of transpiration
Lab 11: Transpiration
Lab 11: Transpiration
• Description:
• Determine relationship between leaf surface area, #
stomata, rate of transpiration
• Nail polish  stomatal peels
• Effects of environmental factors on rate of transpiration
• Temperature, humidity, air flow (wind), light
intensity
Analysis of Stomata
Rates of Transpiration
Lab 11: Transpiration
• Conclusions:
•
•
•
•
transpiration:  wind,  light
transpiration:  humidity
Density of stomata vs. transpiration
Leaf surface area vs. transpiration
Answer:
Answer:
Answer:
Answer:
Lab 12: Animal Behavior
• Concepts:
• Experimental design
• IV, DV, control, constants
• Control vs. Experimental
• Hypothesis
• innate vs. learned behavior
• choice chambers
•
•
•
•
•
temperature
humidity
light intensity
salinity
other factors
Lab 12: Animal Behavior
• Description:
• Investigate relationship between environmental factors
vs. behavior
• Betta fish agonistic behavior
• Drosophila (fruit fly) behavior
• Pillbug kinesis
Lab 12: Animal Behavior
Lab 12: Animal Behavior
• Hypothesis Development
Keep this in mind if
you have to write a
hypothesis on the AP
exam… what makes a
good hypothesis?
• Poor:
I think pillbugs will move toward the wet side of a choice
chamber.
• Better:
If pillbugs are randomly placed on two sides of a wet/dry
choice chamber and allowed to move about freely for
10 minutes, then more pillbugs will be found on the wet
side because they prefer moist environments.
Lab 12: Animal Behavior
• Experimental Design
sample size
Lab 12: Animal Behavior
• Data Analysis:
•
•
•
•
Chi-Square Test
Null hypothesis: there is no difference between the conditions
Degrees of Freedom = n-1
At p=0.05, if X2 < critical value  accept null hypothesis (any differences
between observed and expected due to CHANCE)
Lab 12: Animal (Fruit Fly or Pillbug) Behavior
ESSAY 1997
A scientist working with Bursatella leachii, a sea slug that lives in an intertidal habitat in the
coastal waters of Puerto Rico, gathered the following information about the distribution of the
sea slugs within a ten-meter square plot over a 10-day period.
time of day
average distance
between individuals
12 mid
4am
8am
12 noon
4pm
8pm
12 mid
8.0
8.9
44.8
174.0
350.5
60.5
8.0
a. For the data above, provide information on each of the following:
• Summarize the pattern.
• Identify three physiological or environmental variables that could cause the slugs to
vary their distance from each other.
• Explain how each variable could bring about the observed pattern of distribution.
b. Choose one of the variables that you identified and design a controlled experiment to test
your hypothetical explanation. Describe results that would support or refute your
hypothesis.
Lab 12: Animal (Fruit Fly or Pillbug) Behavior
ESSAY 2002
The activities of organisms change at regular time intervals. These changes are called biological rhythms. The
graph depicts the activity cycle over a 48-hour period for a fictional group of mammals called pointy-eared
bombats, found on an isolated island in the temperate zone.
a. Describe the cycle of activity
for the bombats. Discuss how
three of the following factors
might affect the physiology and/or
behavior of the bombats to result in
this pattern of activity.
• temperature
• food availability
• presence of predators
• social behavior
b. Propose a hypothesis regarding the effect of light on the cycle of activity in bombats. Describe a
controlled experiment that could be performed to test this hypothesis, and the results you would
expect.
Lab 13: Enzyme Activity
• Concepts:
• Enzyme
• Structure (active site, allosteric site)
• Lower activation energy
• Substrate  product
• Proteins denature (structure/binding site changes)
Lab 13: Enzyme Activity
• Description:
• Determine which factors affecting rate of enzyme reaction
• H2O2  H2O + O2
catalase
• Measure rate of O2 production
Turnip peroxidase  Color change (O2 produced)
Lab 13: Enzyme Activity
• Conclusions:
• Enzyme reaction rate affected by:
•
•
•
•
pH (acids, bases)
Temperature
Substrate concentration
Enzyme concentration
Calculate Rate of Reaction
Answer is B