Jack Gerlitz
Population Genetics and Evolution Lab
Objectives:
Before doing the lab, students should know the Hardy-Weinberg equation and how it
determines the frequency of alleles in the population. As well as how natural selection is able to
alter allele frequencies, and its effect on a particular genotype.
After the lab, students will be able to calculate allele frequencies in the gene pool of a
population using the Hardy-Weinberg equation. Also, be able to talk about how natural
selection or other causes of microevolution are factors that can affect the maintaining of HardyWeinberg equilibrium.
Introduction:
G.H. Hardy and W. Weinberg came up with an idea that would predict the frequency of an
individual allele in a population. In this idea, p relates to the dominant allele, and q relates to
the recessive allele. When p and q are added together, they must equal to 1. Diploid
combinations can be predicted as well with the Hardy Weinberg equation. However there are 5
conditions that must be met in order for equilibrium to occur. The population must be large,
mating is random, mutation of alleles is nonexistent, migration does not occur, and there is no
natural selection.
In this experiment, the class is used as a simulated, controlled population to maintain the five
conditions of Hardy Weinberg equilibrium. This way, it will be possible to show the equation in
effect.
Materials:
Part A:



PTC Paper Strip
Lemon Head Candies
Calculator
Part B:
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4 allele cards per person,( Two labeled a, Two labeled A.)
Coin
Calculator
Procedure:
Part A:
1. Take a short strip of PTC paper, and press it against your tongue. If the individual
can taste it, it will be a bitter taste.
2. Calculate the decimal for the frequency of each allele. Take the amount of
tasters divided by the amount of people in the class. And the same for non
tasters.
3. Now use the Hardy-Weinberg Equation to determine the frequencies of each
allele. Once found, record in data table.
Part B: Case I
1. Turn your four cards over so the letters are face down. With your partner, flip one card
over from each pile. These two cards will represent the alleles of the first offspring. One
partner record in the data sheet.
2. Repeat for second offspring. The other partner should record these offspring alleles on
the data sheet.
3. Now, take the cards that represent your off springs alleles. If your offspring has AA, then
your should have cards that have A,A,A,A.
4. Go to another partner, and repeat steps 1 and 2.
5. Repeat 3 and 4 until you have done all 5 generations.
6. Calculate the q and p frequencies.
Case II
1. Start with the same initial genotype as you did in the previous case.
2. Repeat the same steps in case one to determine the generations. But whenever a
homozygous recessive offspring is reproduced, the child does not survive. You must
reproduce again until there is a surviving offspring.
3. Calculate the p and q frequencies.
Case III
1. Start with the same initial genotype as in the previous cases.
2. Do the same steps, homozygous recessive never survives, and if it is homozygous
dominant, it only survives if the coin toss is tails.
3. Do this for 5 generations. Calculate the p and q frequencies.
4. Repeat for another 5 generations.
Case IV
1. Start with the same initial genotype as in the previous cases.
2. In this case, the class will be broken up into small populations. Breeding will occur only
between the small populations. This way no one will have contact with other individuals from
another population.
3. Reproduce five generations as before.
4. Calculate the frequencies of p and q.
Data
Part A
Phenotypes
Tasters
Class
#
Population
8
N.A. Population
Allele Frequency of H-W Equation
Non tasters
p
q
%
#
%
.34
.66
57
6
43
.33
.67
.55
.45
Calculations: Percents:
Allele Frequencies
8/14= .57
q²=.43 ------ q= .66
6/14=.43
1-.66= p ----- p= .34
q²=.45 ------- q=.67
1-.67=p------- p= .33
Part B:
Case I
Initial Genotype
First Generation Genotype
Second Generation Genotype
Third Generation Genotype
Fourth Generation Genotype
Aa
Aa
Aa
Aa
AA
Fifth Generation Genotype
AA
Case II
Initial Genotype
First Generation Genotype
Second Generation Genotype
Third Generation Genotype
Fourth Generation Genotype
Fifth Generation Genotype
Aa
AA
AA
AA
Aa
Aa
Case III
Initial Genotype
First Generation Genotype
Second Generation Genotype
Third Generation Genotype
Fourth Generation Genotype
Fifth Generation Genotype
Aa
Aa
AA
Aa
Aa
AA
Sixth Generation Genotype
Seventh Generation Genotype
Eighth Generation Genotype
Ninth Generation Genotype
Tenth Generation Genotype
Aa
Aa
Aa
Aa
Aa
Case IV
Initial Genotype
First Generation Genotype
Second Generation Genotype
Third Generation Genotype
Fourth Generation Genotype
Fifth Generation Genotype
Aa
aa
Aa
AA
Aa
AA
Final Class Frequency Tables
Case I
AA
Aa
aa
p
q
Calculations:
Number of Offspring with Genotype AA: 3 x 2 = 6 A alleles
3
3
8
32%
68%
Number of Offspring with Genotype Aa:
3 x 1 = 3 A alleles
Total = 9 A alleles
P= 9/28= 32%
𝑝 = 𝑡𝑜𝑡𝑎𝑙 # 𝑜𝑓 𝐴 𝑎𝑙𝑙𝑒𝑙𝑒𝑠 ÷ 𝑡𝑜𝑡𝑎𝑙 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑎𝑙𝑙𝑒𝑙𝑒𝑠 𝑖𝑛 𝑝𝑜𝑝𝑢𝑙𝑎𝑡𝑖𝑜𝑛
Number of Offspring with Genotype aa:
8 x 2 = 16 a alleles
Number of Offspring with Genotype Aa:
3x1=3
Total= 19 a alleles
Q= 19/28=68%
𝑞 = 𝑡𝑜𝑡𝑎𝑙 # 𝑜𝑓 𝑎 𝑎𝑙𝑙𝑒𝑙𝑒𝑠 ÷ 𝑡𝑜𝑡𝑎𝑙 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑎𝑙𝑙𝑒𝑙𝑒𝑠 𝑖𝑛 𝑝𝑜𝑝𝑢𝑙𝑎𝑡𝑖𝑜𝑛
Case II
AA
Aa
aa
p
q
5
9
0
68%
32%
5 x 2 = 10 A alleles
0 x 2 = 0 a alleles
9 x 1 = 9 Alleles
9 x 1 = 9 a alleles
Total = 19 A alleles
Total= 9 a alleles
P= 19/28=68%
Q= 9/28=32%
1 x 2 = 2 A alleles
0 x 2 = 0 a alleles
13 x 1 = 13 A alleles
13 x 1 = 13 a alleles
Total = 15 A alleles
Total= 13 a alleles
P= 15/28= 54%
Q= 13/28= 46%
Case III
Table 1
AA
Aa
aa
1
13
0
p
q
54%
46%
Table 2
AA
Aa
aa
p
q
4
10
0
64%
36%
4 x 2 = 8 A alleles
0 x 2 = 0 a alleles
10 x 1 = 10 A alleles
10 x 1 = 10 a alleles
Total = 18 A alleles
Total = 10 a alleles
P = 18/28= 64%
Q= 10/28= 36%
7
2
5
57%
43%
7 x 2 = 14 A alleles
5 x 2 = 10 a alleles
2 x 1 = 2 A alleles
2 x 1 = 2 a alleles
Total= 16 A alleles
Total= 12 a alleles
P = 16/28 = 57%
Q= 16/28 = 43%
Case IV
AA
Aa
aa
p
q
Conclusion:
Case 1
- The Hardy-Weinberg equation predicts that the new p will be 32% and the new q will be 68%.
- The results I have obtained in this simulation do not agree with the Hardy-Weinberg
equilibrium because the experiment was of a small population, where in order to maintain
equilibrium, the population must be larger.
- The major assumptions that were not strictly followed was the condition of a larger
population, our population was only 14 people.
Case II
- The new frequencies of case II are slightly altered to those of the initial frequency of case
number 1.
- The major assumption that was not followed in this simulation was that within the population,
there was no selection. But in this case, the genotype aa was a selected against because
someone with that genotype did not reproduce.
- If I simulated another 5 generations, the frequency of p would increase, as the frequency of q
would decrease.
- If there was a large population, it would be impossible to eliminate a deleterious recessive
allele because in this case, heterozygous is not selected for, so the recessive allele is still
carrying on.
Case III
- The changes in p and q frequencies in Case II were the exact opposite of the frequencies in
case I and were very similar to the frequencies after the second half of generations in case
number 3.
- The recessive allele was no eliminated in case II or case III, however I think after a few more
generations in case II, it will be eliminated.
- The importance of heterozygote advantage is that it keeps the recessive allele alive in the
population.
Case IV
- The initial genotypic frequencies are close to the final ones, but only altered a tiny bit. This is
due to the lack of migration.
- My results say that the size of the population must be large in order to increase the genetic
variation, which would increase the force of evolution.
Hardy Weinberg Problems
- In a population of 1,000 drosophila, if 360 show the recessive trait of vestigial wings, I would
expect 160 to have a homozygous dominant 480 heterozygous.
- If a population of 500 people, 25% show the recessive phenotype for attached earlobes, I
would expect to see 125 homozygous dominant, and 250 heterozygous.
- If a population of 1,000, 510 showed the dominant phenotype for widows peak, I would
expect to see 90 homozygous dominant, 420 heterozygous, and 490 homozygous recessive for
this trait.
- If a population of a high school is 2,000, and 16% of the population have Rh negative, I would
expect to see 720 dominant homozygotes, 960 heterozygotes, and 320 recessive homozygotes
for this trait.
- If a population of new born babies in Africa is 1,000, and 4% have sicle cell anemia, which is
recessive, I would expect 640 homozygous dominant, 320 heterozygotes, and 40 recessive
homozygotes.
- If a certain population has a dominant phenotype of a certain trait which occurs 91% of the
time, the frequency of the dominant allele is .7.
In conclusion, Case I allelic frequencies appear normal in a perfect environment that has been
simulated. However, the rest of the cases are related more to the world we live in. As expected,
changes occurred in the frequencies of the alleles. According to Case II the homozygous dominant alleles
became the dominant genotype when the aa genotype died because as the recessive alleles became
scarcer, the heterozygous alleles also suffered because they have therecessive allele. In Case III, the
heterozygotes being more resistant to sickle-cell anemia had an effect on the results, they were the
most frequent and the recessive allele was able to survive. In Case IV, evidence of the populations
becoming controlled is present as the dominant homozygous alleles slowly disappear.
.