Lab 8 Population Genetics 
Introduction:
G. H. Harding and W. Weinberg both came up with the idea that evolution could be viewed as changes in the frequency of alleles in a population. They used the letter “p” to represent and “A” allele and the letter “q” to represent the “a” allele. So, in a population of 100 individuals and 40% of the alleles are “A”, then “p” is .40, “q” would equal .60. The frequency at which the alleles show up is known as their allele frequency. The frequency of the possible combinations of the alleles can be figured using this equation:
p2 + 2pq + q2 = 1.0
Hardy and Weinberg also came up with 5 conditions in which a population’s frequencies would remain constant from generation to generation. One of the conditions is that the population has to be large. Another condition is the mating has to be random. Another condition that is very important is that there isn’t any mutation of the alleles. Another condition is there isn’t any migration going on. The last condition is there can’t be any natural selection.
Using these methods, allele frequency and evolution can be measured in a population. Evolution is difficult to work with in a natural population, so the class will serve as a model to represent a population under different conditions.
Hypothesis:
The purpose of this lab is to learn how to measure allele frequencies in a population under certain conditions and observe evolution and natural selection take its role.
Materials:
Materials used in this lab involved: PTC tasting paper, calculators, and note cards.
Methods:
Exercise 8A: Everyone in the class needs to get a piece of PTC paper and put it into their mouths. If you can taste a bitter flavor, then you are considered a taster, if you can’t taste a bitter flavor, then you are considered a nontaster. In order to calculate the percent of tasters in the class, divide the number of tasters by the total number of students in the class. Calculate the percent of the nontasters the same way and record the numbers into the table. To figure out the allele frequency of q, calculate the square root of the percent of nontasters. To calculate the frequency of p, subtract the frequency of q from 1. Record your answers in the graph.
Exercise 8B CASE I: This part of the lab simulates a population breeding under the conditions of HardyWeinberg. The entire class will participate in this case. Each student needs to randomly pick a partner genotype or sex doesn’t matter. When the students have a partner, then they must all get 4 cards. Each card either has an “A” or an “a” written on it, the students need to be sure to get 2 cards with matching letters. When that’s all sorted out, the students will randomly pick a card out of their hand and lay it down. This simulates the alleles of the first offspring. Note whether or not the cards show a heterozygote, homozygous dominant, or homozygous recessive in a table. The student pair will repeat the procedure once more. Once each pair of students has drawn cards twice and noted the genotypes in a table, they must then assume the genotypes of their offspring. Say one genotype was Aa and another was AA, then one student will have all 4 cards being A, a, A, a and the other student will have all his/her cards be A, A, A, A. Once the exchanging has been done, the students then must rotate, randomly, to another partner. All in all, each student will rotate 5 times with different partners each time representing 5 generations of offspring. All the genotypes must be counted after then ends of each generation and put into a graph.
After the generations have been “born” and the genotypes recorded, frequencies can be calculated. To calculate “p”, multiply the total number of offspring with the genotype AA by 2. Then multiply the total number of offspring with the genotype Aa by 1. Add up the 2 values and divide that number by the total number of genotypes multiplied by 2, representing the total number of alleles in the population. To calculate “q”, multiply the total number of offspring with genotypes aa by 2 and add the total number of offspring with the genotypes Aa. Next, divide the sum of the 2 values by the total number of alleles in the population, in this case, the total number of genotypes multiplied by 2. There you have it!
CASE II: This case is exactly like CASE I except that the aa genotypes won’t be allowed. This simulates that the environment will favor some genotypes over others, in this case, aa. If the 2 students draw aa, then they must try again until they get a surviving offspring. Once the students have reached 5 generations, calculate the frequencies for p and q. Remember to record the genotypes after each generation.
CASE III: This case simulates sicklecell anemia in human genomes. Just like in CASE II, the homozygous recessive, aa, alleles never survive. Individuals that are heterozygous to the disease are slightly more resistant than individuals who are homozygous dominant. With this in mind, if the pair draws up AA, then they must flip a coin to see whether or not the offspring will survive or not, heads survives, tails dies. The student pair must always try to get 2 offspring, no matter how long it takes. Remember to record the genotypes after each generation. Once everything is finished, calculate the frequencies for p and q.
CASE IV: The case is just like CASE I except after each generation is produced, there will be no random selection of new mates. The purpose of this case is to simulate isolated populations and genetic drift. Each group must no interact in any way in order to represent isolated populations. Record the genotypes after each generation and calculate the frequencies for p and q as usual.
Results:
Exercise 8A:

Phenotypes 
Allele Frequency Based on the HW Equation 


Tasters (p2 + 2pq) 
Nontasters (q2) 
P 
q 





Class Population 

# 
% 
# 
% 
.53 
.47 


7 
77.78 
2 
22.22 


North American Population 
0.55 
0.45 
.329 
.671 

1. What is the percentage of heterozygous tasters (2pq) in your class? 49.82%
2. What percentage of the North American population is heterozygous for the taster trait? 44.15%
Exercise 8B CASE I:

AA 
Aa 
aa 
F1 
1 
5 
2 
F2 
2 
4 
2 
F3 
1 
6 
1 
F4 
1 
5 
2 
F5 
1 
5 
2 
Total 
6 
25 
9 
Frequencies:
P= .46
Q= .54
AA=.21 Aa=.50 aa=.29
1. What does the HardyWeinberg equation predict for the new p and q? That the frequency of AA alleles is 46% while the frequency of the aa alleles is 54%.
2. Do the results you obtained in this simulation agree? If not, why not? The results that I obtained agree because it’s normal for the heterzygotes to be large in number because they can carry on both alleles.
3. What major assumption(s) were not strictly followed in this simulation? All of the factors in the environment that could change the results.
CASE II:

AA 
Aa 
aa 
F1 
2 
6 
0 
F2 
6 
2 
0 
F3 
5 
3 
0 
F4 
5 
3 
0 
F5 
4 
4 
0 
Total 
22 
18 
0 
Frequencies:
P= .78
Q= .23
AA=.61 Aa=.36 aa=.05
1. How do the new frequencies of p and q compare to the initial frequencies in Case I? Well, p has gone up a lot and q has gone down a lot.
2. How has the allelic frequency of the population changed? It has gotten smaller.
3. Predict what would happen to the frequencies of p and q if you simulated another five generations. The p value would continue to increase, and the q value would decrease.
4. In a large population, would it be possible to completely eliminate a deleterious recessive allele? Explain. No because if the recessive alleles are being carried on by the heterozygous alleles, then they can’t be eliminated.
CASE III:

AA 
Aa 
aa 
F1 
3 
5 
0 
F2 
1 
7 
0 
F3 
2 
6 
0 
F4 
2 
6 
0 
F5 
4 
4 
0 
Total 
12 
28 
0 
Frequencies:
P= .65
Q= .35
AA=.42 Aa=.46 aa=.12
1. Explain how the changes in p and q frequencies in Case II compare with Case I and Case III. The changes in Case II astounded the results in Case I, but in Case III, the results are similar because the aa genotype is wiped out.
2. Do you think the recessive allele will be completely eliminated in either Case II or Case III? No because if the recessive alleles are being carried on by the heterozygous alleles, then they can’t be eliminated.
3. What is the importance of heterozygotes in maintaining genetic variation in populations? They contain both the dominant and the recessive allele, which make a variation right off the bat.
CASE IV:

AA 
Aa 
aa 
F5 
7 
18 
15 
Frequencies:
P=.40
Q=.60
AA=.16 Aa= .48 aa=.36
1. What do your results indicate about the importance of population size as an evolutionary force? When a population is very large, there is more diversity. When there is a small population with only a couple of offspring reproducing at each generation, and then eventually, there will be no diversity at all.
Error Analysis:
Not many things could have affected the results to this lab unless someone didn’t compute the correct information into their calculators.
Conclusions:
Based on this lab, I can conclude that allelic frequencies appear normal in a perfect environment like Case I. In the rest of the cases however, as we related more to the real world, changes occurred in the frequencies. In 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 are half recessive. In Case III, the heterozygotes being more resistant to sicklecell had an affect on the results, they were dominant. In Case IV, evidence of the populations becoming fixed is showing as the dominant AA alleles slowly disappear.