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NAME: ________________________________________ DATE: _________ BLOCK: _____
In order to study how gene frequencies can change over time (evolve), a benchmark is needed
to compare evolving populations with non-evolving populations. Gene pools that are not
evolving were described mathematically by the Hardy Weinberg Theorem in 1908.
p2 + 2pq + q2 = 1
where p is the dominant allele frequency and q is the recessive allele frequency
This theorem states that the allele frequencies of a population remain constant over time as
long as certain criteria are met. The following conditions must be true for the Hardy Weinberg
Theorem for non-evolving populations.
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



Large population size
No gene flow
No mutations
Random mating
No natural selection
In this lab you will use the Hardy Weinberg Theorem to calculate gene frequencies.
Remember, a population has evolved if the gene frequencies of a population change over
generations. Note that the Hardy Weinberg Theorem can only work for genes with two alleles.
For example, this would not work for blood type because there are multiple alleles.
There are several parts to this lab. Each part represents a real-life scenario so that you can
see the possible effects of each on gene frequencies.
Exercise A: Estimating allele frequencies for a specific trait
1. Use PTC test papers to determine if you are a taster (dominant trait) or a nontaster
(recessive).
2. Use Hardy-Weinberg to determine the frequency of p and q. Complete the table below.
Populations
Class
Population
North
American
Population
Allele Frequency
Based on Hardy
Weinberg
Phenotypes
Tasters (p2 + 2pq)
#
%
Nontasters (q2)
#
%
0.55
0.45
1
p
q
PART ONE: Hardy Weinberg Equilibrium
Assume that “A” represents a dominant allele for brown eyes and “a” represents the recessive
allele for blue eyes.
1. Pick up two index cards: an “A” card and an “a” card. This means that everyone starts
out heterozygous (Aa).
2. Find your first partner and mate to make your first offspring.
3. The offspring are made by randomly selecting one card from each person. The
resulting baby will have a genotype based on the cards chosen. For example, the baby
may be aa. Each person should take back the card that they contributed to the baby.
4. Next, repeat step 3 so that you are making a second baby for the other person.
5. Now, in nature this does not happen, but in order to simulate generations, you will
become your offspring. In the example above, you would become aa. There will be
times when you have to get different cards because your genotype changes.
6. Next, you find a new partner randomly and mate again.
7. Each person will do this for 5 generations (make five babies).
8. Record the genotypes of each generation (each baby that you make) in Table 1.
9. Return to your desk and report your individual data to Mrs. Flick as instructed.
10. Record the class data in the calculation chart.
PART TWO: Natural Selection
You will repeat mating steps as in part one with one difference. There has been a change in
the environment. A new emergent virus from the rainforest has jumped from birds to humans.
Scientists have found that people who are homozygous for blue eyes (aa) are more
susceptible to the virus and more likely to die because the virus can only enter the brain
through the blue-eyed pigment. In other words the environment is selecting for those people
who are AA and Aa. To represent this in our simulation, babies who are aa die.
1. If you make an aa baby, it dies. You must keep trying to make a baby until you make
one that survives.
2. Just as in Case One you must make 5 babies and become the baby you made for each
round.
3. Record the genotypes of each generation (each baby that you make) in Table 1.
4. Return to your desk and report your individual data to Mrs. Flick as instructed.
5. Record the class data in the calculation chart.
PART THREE: Homozygous Advantage
You will repeat mating steps as in part one with the following difference. This time the “a”
represents the recessive allele for sickle cell anemia and the “A” represents the dominant
normal allele. It turns out that people who are heterozygous (Aa) for sickle cell are protected
against the severe symptoms malaria. This means that people who are AA have a higher
change of dying from malaria and people who are aa have higher chance of dying from sickle
cell.
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1. If you make an aa or AA baby, pick up a penny and flip it. If the penny lands heads up
the baby dies. If the penny lands tails up, the baby survives.
2. If the baby survives you will become that baby (either aa or AA) in the next generation.
If the baby dies, you must keep trying to make a baby that will survive.
3. Record the genotypes of each generation (each baby that you make) in Table 1.
4. Return to your desk and report your individual data to Mrs. Flick as instructed.
5. Record the class data in the calculation chart.
PART FOUR: Genetic Drift
You will repeat mating steps as in part one with the following difference. You must mate in
small groups of 4 or 5 people. You may not mate with anybody outside of your group.
Assume that each group represents a small founder population on separate islands.
1. Record the genotypes of each generation (each baby that you make) in Table 1.
2. Return to your desk and report your individual data to Mrs. Flick as instructed.
3. Record the class data in the calculation chart.
Table 1: Baby Data
Part One: H-W Equilibrium
My initial genotype
Aa
Part Two: Natural Selection
My initial genotype
Aa
F1 genotype
F1 genotype
F2 genotype
F2 genotype
F3 genotype
F3 genotype
F4 genotype
F4 genotype
F5 genotype
F5 genotype
Part Three: Heterozygote
Advantage
My initial genotype
Aa
Part Four: Genetic Drift
My initial genotype
F1 genotype
F1 genotype
F2 genotype
F2 genotype
F3 genotype
F3 genotype
F4 genotype
F4 genotype
F5 genotype
F5 genotype
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Aa
QUESTIONS
Show your calculations for all answers involving math!
Exercise A: Tasters
1. What is the percentage of heterozygous tasters (2pq) in the class?
2. What percentage of the North American Population is heterozygous for the taster trait?
PART ONE: HARDY-WEINBERG EQULIBRIUM
3. What does H-W predict for the new p and q (for the 5th generation)?
4. Do the results you obtained in this simulation agree with the above prediction? ____
Why or why not?
5. What major conditions were not strictly followed in this simulation?
PART TWO: NATURAL SELECTION
6. How do the initial frequencies of p and q compare to the final frequencies?
7. Has evolution occurred in this simulation? Explain your answer.
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8. Predict what would happen to the p and q frequencies if you simulated another 5
generations?
9. In a large population, would it be possible to completely eliminate a recessive allele –
even one that is homozygous lethal? Explain your answer.
PART THREE: HETEROZYGOUS ADVANTAGE
10. How do the initial frequencies of p and q compare to the final allele frequencies?
11. In a large population, do you think the recessive allele will be eliminated in this
example? Explain your answer.
CASE FOUR: GENETIC DRIFT
12. How do the initial frequencies of p and q compare to the final allele frequencies?
13. What do your results indicate about the importance of population size as an
evolutionary force?
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HARDY-WEINBERG PROBLEMS
14. In Drosophila, the allele for normal length wings is dominant over the allele for vestigial
wings. In a population of 1,000 individuals, 360 flies have the recessive phenotype.
How many individuals would you expect to be homozygous dominant and heterozygous
for this trait?
15. The allele for the ability to roll one’s tongue is dominant over the allele for the lack of
this ability. In a population of 500 individuals, 25% are nonrollers. How many
individuals would you expect to be homozygous dominant and heterozygous for this
trait?
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16. The allele for the hair pattern called widow’s peak is dominant over the allele for no
widow’s peak. In a population of 1,000 individuals, 510 have widow’s peaks. How
many individuals would you expect of each of the three possible genotypes?
17. In the US, about 16% of the population is Rh negative. The Rh negative allele is
recessive to the Rh positive allele. If a student population of high school in the US is
2000, how many students would you expect for each of the three possible genotypes?
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18. In certain African countries, 4% of the newborn babies have sickle-cell anemia, which is
a recessive trait. Out of a random population of 1,000 newborn babies in these
countries, how many would you expect for each of the three possible genotypes?
19. In a certain population, the dominant phenotype of a certain trait occurs 91% of the
time. What is the frequency of the dominant allele?
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CALCULATION CHARTS
Part One: Initial Allele Frequencies
Genotypes
AA
Aa
Number of
each
genotype
Multiply
by…
Part One: Final Allele Frequencies
Number
of A's
Genotypes
AA
Aa
Total number of A's
Genotypes
Aa
aa
Number of
each
genotype
Multiply
by…
Number of
each
genotype
Multiply
by…
Number
of A's
Total number of A's
Number
of a's
Genotypes
Aa
aa
Total number of a's
Number of
each
genotype
Multiply
by…
Number
of a's
Total number of a's
Total Number of genes
p
q
Total Number of genes
p
q
Part Two: Initial Allele Frequencies
Part Two: Final Allele Frequencies
Genotypes
AA
Aa
Number of
each
genotype
Multiply
by…
Number
of A's
Genotypes
AA
Aa
Total number of A's
Genotypes
Aa
aa
Number of
each
genotype
Multiply
by…
Number of
each
genotype
Multiply
by…
Number
of A's
Total number of A's
Number
of a's
Genotypes
Aa
aa
Total number of a's
Number of
each
genotype
Multiply
by…
Total number of a's
Total Number of genes
p
q
Total Number of genes
p
q
9
Number
of a's
Part Three: Initial Allele Frequencies
Genotypes
AA
Aa
Number of
each
genotype
Multiply
by…
Part Three: Final Allele Frequencies
Number
of A's
Genotypes
AA
Aa
Total number of A's
Genotypes
Aa
aa
Number of
each
genotype
Multiply
by…
Number
of a's
Genotypes
Aa
aa
Genotypes
Aa
aa
Multiply
by…
Multiply
by…
Number
of a's
Part Four: Final Allele Frequencies
Number
of A's
Genotypes
AA
Aa
Total number of A's
Number of
each
genotype
Number of
each
genotype
Total Number of genes
p
q
Part Four: Initial Allele Frequencies
Multiply
by…
Number
of A's
Total number of a's
Total Number of genes
p
q
Number of
each
genotype
Multiply
by…
Total number of A's
Total number of a's
Genotypes
AA
Aa
Number of
each
genotype
Number of
each
genotype
Multiply
by…
Number
of A's
Total number of A's
Number
of a's
Genotypes
Aa
aa
Total number of a's
Number of
each
genotype
Multiply
by…
Total number of a's
Total Number of genes
p
q
Total Number of genes
p
q
10
Number
of a's