Baylee Roy
AP Biology, 4th Period
AP Lab 8: Population Genetics and Evolution
(Adapted from the 2001 Student Lab Manual)
Purpose: In this lab, you will:
1. learn about the Hardy-Weinberg law of genetic equilibrium
2. study the relationship between evolution and changes in the allele frequency by
using your class to represent a sample population
Prelab questions:
1. What is the Hardy-Weinberg equation?
a. The Hardy-Weinberg equation is “p2+2pq+q2=1” and it is used to
calculate the genetic variation of a population at equilibrium by
calculating the frequency of the alleles in the population.
2. Which part of the equation represents: the homozygous dominant, the
heterozygotes, and the homozygous recessive?
a. The heterozygotes is represented by “2pq”
b. The homozygous recessive is represented by “q2”
c. The homozygous dominant is represented by “p2”
3. How does the equation p+q=1 relate to Hardy-Weinberg?
a. This relates to the Hardy-Weinberg equation because the sum of the
alleles (dominant and recessive) must combine to form the genotype in 1
population, making the sum 1 (100%).
4. What are the 5 conditions for Hardy-Weinberg equilibrium to exist?
a. No mutation
b. Random mating
c. No natural selection
d. Large population size
e. Must be isolated; no migration or emigration
Exercise A: Estimating Allele Frequencies for a Specific Trait Within a Sample
Population
For this exercise, you will use the class as a sample population to find the allele
frequencies for PTC tasting. For a person expressing the dominant allele, T, PTC has a
bitter taste. If a person is homozygous recessive, s/he cannot taste PTC.
1. Using the PTC taste-test papers, tear off a short strip and press it to your tongue
tip. Tasters will sense a bitter taste.
2. Calculate the frequency of the taster population by dividing the total number of
tasters by the total number of students in the class.
3. Calculate the frequency of the non-taster population by dividing the number of
non-tasters by the number of students in the class.
4. Use the Hardy-Weinberg equation to determine the frequencies (p and q) of the
two alleles.
Data for Exercise A:
Baylee Roy
AP Biology, 4th Period
Phenotypes
Class
Population
N. American
Population
# and % Tasters # and % Non(p2+2pq)
Tasters (q2)
20
71%
8
29%
55%
45%
Allele
Frequency
Based on the HW equation
P
q
.54
.46
.33
.67
Exercise B: Case Studies
Case 1-A Test of an Ideal Hardy-Weinberg Population
The class will represent a breeding population. In order to simulate random mating,
choose another student at random. For this simulation, we will assume that gender an
genotype are irrelevant to mate selection. The class is a population of randomly mating
heterozygous individuals with an initial gene frequency of p=.5 and q=.5. For the
phenotypes, AA=.25, Aa=.50, and aa=.25. Your initial genotype is therefore Aa.
Record this on your data chart.
You will be given 2 cards. These are your gametes. One card will have A and the other
will have a. Each parent will contribute 1 gamete to each child.
1. Turn the cards over so that the letters don’t show and shuffle them. The card on
top will contribute to the production of your 1st offspring. Your partner will do
the same.
2. The product of your 2 cards will be the genotype of your 1st offspring. Record
this on your data chart and produce another offspring with your partner. Be sure
to reshuffle your cards as in step 1 before making your offspring. Your partner
can record the 2nd child on their data chart.
3. Each of you assumes the genotype of your child. You should obtain the necessary
cards from the “gene pool” pile at the front of the room.
4. You then need to find a new partner and repeat 1-3.
5. You’ll do this for 6 Generations, recording the genotypes each time
Data for Case 1
Initial Class Frequencies:
AA= .25
Aa= .50
aa= .25
My initial Genotype= Aa
F1= aa
F2= aa
F3= Aa
F4= Aa
F5= Aa
Using the following equations, calculate p and q:
Baylee Roy
AP Biology, 4th Period
Number of A Alleles Present at the 5th generation=Total # of A alleles__________
Total number of alleles in class
Number of a Alleles Present at the 5th generation=Total # of a alleles___________
Total number of alleles in class
p=
.5
q=
.5
Final Class Frequencies:
(# of people who had it in class)
AA= 6
AA= .25
Aa=
Aa=
16
.50
aa= 6
aa= .25
Questions for Case 1
1. What does the H-W equation predict for p and q?
a. It predicts that the new p and q are the same as the initial p and q.
2. Do the results you obtained agree? Why or why not?
a. My specific results do not agree to the simulation because the population
size was too small to obtain true results in relation to the H-W rules of
equilibrium.
3. What major assumption(s) were not strictly followed in this simulation?
a. In this lab the assumption of a large population size was not followed (one
of the major rules for H-W equilibrium).
Case 2: Selection
Now, we’re going to make this a little more realistic. In the natural environment, not all
genotypes have the same rate of survival. For this simulation, you will assume that the
homozygous recessive individuals never survive, while the others survive most of the
time.
1. Repeat the procedure as you did in Case 1, except that any aa offspring will die.
Since we need a constant population size, you and your partner need to reproduce
until you make 2 surviving offspring.
2. Again, proceed through 5 generations.
Data for Case 2
Initial Class Frequencies:
AA= .25
Aa= .50
My initial Genotype= Aa
F1= Aa
F2= AA
F3= AA
aa= .25
Baylee Roy
AP Biology, 4th Period
F4=
F5=
AA
AA
p=
q=
.81
.19
Final Class Frequencies:
(# of people who had it in class)
AA= 17
Aa=
AA= .6561 Aa=
10
aa= 0
.3078 aa= .0361
Questions for Case 2
1. How do the new frequencies of p and q compare to the initial frequencies in Case
1?
a. The frequency of p has increased whereas the frequency of q decreased.
This is due to the fact that in this simulation natural selection has favored
the dominant allele (p) for the fittest survival in Case 2, contributing to its
increase in frequency over the 5 generations of reproduction
2. What major assumptions were not strictly followed in this simulation?
a. The major assumption of a large population was not followed in this
simulation as we only had a population of 27, skewing the data we
received.
3. Predict what would happen to the frequencies of p and q if you simulated another
five generations?
a. If we simulated another 5 generations in case 2, the q frequency would
continue to drop (inversely causing the p frequency to increase). However,
the q frequency will never decrease to 0, as the heterozygous (the carriers
of the recessive q allele) will continue to pass the q allele into the gene
pool.
4. In a large population, would it be possible to completely eliminate a deleterious
(fatal) recessive allele? Explain
a. It is not possible to completely eliminate a fatal recessive allele in a large
population because the heterozygotes in the population can carry the “fatal
allele” without dying themselves (as you only die from the allele if
homozygous dominant). So they will continue to contribute the allele in
the gene pool of the population; never actually “eliminating” the q allele.
Case 3: Heterozygote Advantage
It’s easy to see what happens to the lethal recessive allele in the population. However,
the data for many human populations shows an unexpectantly high frequency of the
sickle-cell allele in certain populations. Thus, our simulation does not accurately reflect
the real situation because the heterozygous individuals are slightly more resistant to
malaria than the homozygous dominant individuals.
Baylee Roy
AP Biology, 4th Period
1. In this round, you’ll repeat your procedure as in Case 2. However, if your
offspring is AA, flip a coin. If the result is heads, the individual does not survive.
If the result is tails, the individual survives.
2. Again, simulate 5 generations as before, keeping the population size the same by
mating with your partner until you produce 2 surviving offspring.
3. If time permits, start with the F5 genotype and simulate another 5 generations.
Data for Case 3
Initial Class Frequencies:
AA= .25
Aa= .50
aa= .25
My initial Genotype= Aa
F1= Aa
F2= Aa
F3= Aa
F4= Aa
F5= Aa
p=
q=
.72
.28
Final Class Frequencies (after F5):
(# of people who had it in class)
AA= 12
Aa=
AA= .5184 Aa=
15
aa= 0
.4032 aa= .0784
Final Class Frequencies (after F10): ***(we never went up to F10 in class)***
AA= -Aa= -aa=
-Questions:
1. Explain how the changes in p and q frequencies in Case 2 compare with Case 1
and Case 3.
a. The frequency of the p allele was higher in Case 2 than in Case 1 and 3
because the homozygous recessive in Case 2 was selected against; causing
a major increase in the dominant allele (as only those with a dominant
allele could live to contribute their genes in the gene pool). In turn, the
natural selection working against the q allele in Case 2 causes the
frequency of q to be lower in Case 2 than in Case 1 and 3.
2. Do you think the recessive allele will be completely eliminated in either case 2 or
Case 3?
a. The recessive allele will never by completely eliminated in either case,
because the heterozygous allele will remain in the population, contributing
the recessive alleles in the gene pool.
Baylee Roy
AP Biology, 4th Period
3. What is the importance of heterozygotes (the heterozygote advantage) in
maintaining genetic variation in populations?
a. The heterozygotes allow for there to be a greater genetic diversity, by
carrying both of the alleles. This way, neither allele can be completely
eliminated in a population.
Case 4: Genetic Drift
Recall that small populations have problems maintaining the genetic diversity of their
gene pool
1. For this, you’ll be divided into a small population. You can only mate within
your population.
2. Repeat the mating procedure as you did for Case 1.
Data for Case 4
Initial Class Frequencies:
AA= .25
Aa= .50
aa=
.25
My initial Genotype= Aa
***I died in a forest fire, disabling me from reproducing***
p=
q=
.46
.54
Final Class Frequencies:
(# of people who had it in class)
AA= 2
Aa=
AA= .2116 Aa=
9
aa= 3
.4968 aa= .2916
Questions for Case 4
1. Explain how the initial genotypic frequencies of the population compare.
a. The genotypic frequencies of the population in Case 4 have stayed
generally around the same range. There was a slight decrease in
homozygous dominant (.25 to .21) and a slight increase in homozygous
recessive (.25 to .29); but no major changes in the heterozygous (.50 to
.49).
2. What do your results indicate about the importance of population size as an
evolutionary force?
a. My results indicate that having a smaller population size, contributes to an
increase in genetic drift. As due to the few numbers of organisms, the
population will be more heavily impacted than a population with a large
number of organisms. Therefore evolutionary forces will have a heavier
and faster impact when it is applied to a smaller population when
compared to a larger population.