Population Genetics and Evolution

Hardy-Weinberg Equilibrium

Initial Class Frequencies:

AA: 0  Aa: 14 aa: 0

My Initial Genotype: Aa

F1 Genotype: Aa

F2 Genotype: aa

F3 Genotype: aa

F4 Genotype: Aa

F5 Genotype: AA

Final Class Frequencies:

AA: 2 Aa: 10 aa: 2

p: 0.5 q: 0.5

1. What does the Hardy-Weinberg equation predict for the new p and q?

Hardy-Weinberg predicts that with no selection, p will equal .5 and q will equal .5. This means there is a 50/50 chance of p and q being present in the genotype, which implies that there is a 50% chance of the offspring being heterozygous. This means there is a 25% chance of having a homozygous recessive offspring and a 25% chance of having a heterozygous offspring.

2. Do the results obtained in this simulation agree? If not, why?

The results in this simulation are very similar to the prediction brought about by the Hardy-Weinberg equation. This is because the Hardy-Weinberg equation predicted the offspring would be a heterozygote 50% of the time. Instead, the offspring was a heterozygote approximately 71.4% of this time. Because the sample size was small, this deviance from the Hardy-Weinberg prediction can be accounted for. If the percentage of offspring that are heterozygotes is less than 50% then the Hardy-Weinberg prediction would not agree with the simulation.

3. What major assumption(s) were not strictly followed in this experiment?

The major assumptions not strictly followed in this experiment were the large breeding population and random mating. There were only fourteen people in the room in order to carry out the experiment, which is very different from the natural environment, where there are thousands if not millions of potential mates to choose from. Additionally, even though the class tried to ensure that the mating was random, people were still more likely to choose their friends over other, unfamiliar peers. This was simply because people in general, especially teenagers, are more comfortable with interacting with people they know rather than people they don’t.

Selection

Initial Class Frequencies:

AA: 0 Aa: 14 aa: 0

My Initial Genotype: Aa

F1 Genotype: Aa

F2 Genotype: AA

F3 Genotype: Aa

F4 Genotype: AA

F5 Genotype: AA

Final Class Frequencies:

AA: 9 Aa: 5 aa: 0

p: 1 q: 0

1. How do the new frequencies of p and q compare to the initial frequencies in Case I?

The new frequencies of Case II were very different from the initial frequencies in Case I. In Case II, the homozygous recesive offspring didn’t survive, and the parents of the homozygous recessive organism would have to mate again. Therefore, the q value would have to be zero while the p value would have to be 1.  Because of this, the only possible genotypes viable in this gene pool were AA and Aa, which left only one recessive allele for every three dominant alleles. Thus, the odds were greater for an AA offspring as opposed to an Aa offspring. On the other hand, in the case with no selection, the odds were greatest for the production of a heterozygous offspring.

2. What major assumption(s) were not strictly followed in this simulation?

As mentioned in Case I, in which there was no alteration or selection in the gene pool, there was no true random mating and the breeding population wasn’t large. However, in the second case, the homozygous recessive offspring didn’t live to reproduce.  This violated the assumption that there was no selection. If there was no selection, then all of the genotypes would have an equal chance of surviving and reproducing. The homozygous recessive genotype had no chance of surviving, and because there were more dominant alleles than recessive ones in the gene pool . Thus the number of homozygous dominant results were greater than the heterozygous results in the case where the homozygous recessive genotypes were completely removed.

3.Predict what would happen to the frequencies of p and q if you simulated another five generations.

If another five generations were simulated, the amount of p

4. In a large population would it be possible to completely eliminate a deleterious recessive allele? Explain.

It would be impossible to completely eliminate a deleterious recessive allele. This is because even though some organisms will express the trait in its homozygous recessive form and die before they are able to reproduce, others will be carriers for the trait. The carriers for this trait are heterozygous, so they do not express this trait as a phenotype. They are able to live their lives normally. If two heterozygous organisms mate, they have the potential to produce a homozygous recessive offspring, which would outwardly express the allele.

Heterozygote Advantage

Initial Class Frequencies:

AA: 0 Aa: 14 aa: 0

My Initial Genotype: Aa

F1 Genotype: Aa

F2 Genotype: Aa

F3 Genotype: Aa

F4 Genotype: Aa

F5 Genotype: Aa

Final Class Frequencies:

AA: 3 Aa: 11 aa: 0

p: 1 q: 0

Next Round:

F6 Genotype: Aa

F7 Genotype: Aa

F8 Genotype: Aa

F9 Genotype: AA

F10 Genotype: Aa

Final Class Frequencies:

AA: 4 Aa: 10 aa: 0

p: 1 q: 0

1. Explain how the changes in p and q frequencies in Case II compare with Case I and Case III.

Case II, the case in which there was selection, had extremely different frequencies than the simulation in which there was no selection. Case II demonstrated a heterozygous advantage. In the heterzygote advantage, the homozygous recessive genotype didn’t survive. This meant there would be zero chance of the offspring being homozygous recessive. The p value was 1 and the q value was zero. When there was no selection the p value was .5 and the q value was .5 because there was an equal chance of the offspring containing a dominant or recessive allele. The heterozygote advantage was more similar to the selection case, but in the heterozygous advantage there was a slim chance of the offspring being anything other than heterozygous because if the offspring was homozygous dominant, a coin was flipped to decide if the offspring would survive. If the offspring was heterozygous then it automatically survived. While this yielded more heterozygous offspring, the final class frequency was 3 AA and 11 Aa in this case compared to the selection one, which was 9 AA and 5 Aa, the p and q values were exactly the same. Because there was no chance of the offspring being homozygous recessive, both of the p values were one and both of the q values were 0.

2. Do you think the recessive allele will be completely eliminated in either Case II or Case III?

The recessive allele will be eliminated in neither of the cases. However, there is a greater possibility that the recessive allele will become less and less common in the case of selection because there is a possibility of either having a homozygous dominant or a heterozygote, which means that out of those possibilities three of the alleles are dominant and one is recessive. Therefore, there are simply more dominant alleles present, which means that the dominant alleles are more likely to be passed on. In case three where the heterozygous genotype has the advantage, there was a very slim chance that the zygote would be anything but heterozygous, thus it would be impossible for the allele to be completely eliminated in Case III.

3. What is the importance of heterozygotes (the heterozygous advantage) in maintaining genetic variation in populations?

It is very important to have heterozygotes in a population. If there were no heterozygotes, then there would only be homozygous genotypes, and there would be no chance of parents passing on a trait to their child that did not show up in their phenotype; the child would have the exact same genetic makeup as their parents. Heterozygous genotypes are absolutely necessary in order to have variation in a population.

Genetic Drift

Initial Class Frequencies

AA: 0 Aa: 14 aa: 0

p: 0.5 q: 0.5

My Initial Genotype: Aa

F1 Genotype: aa

F2 Genotype: Aa

F3 Genotype: aa

F4 Genotype: Aa

F5 Genotype: AA

Final Class Frequencies:

AA: 3 Aa: 1 aa: 0

p: .866 q: .134

1. Explain how the initial genotypic frequencies of the populations compare.

The initial genotypic frequencies and the final genotypic frequencies were significantly different. The initial genotypic frequencies were 0.5 for p and 0.5 for q. There was equilibrium between the p and q values, meaning there was equilibrium between the number of dominant and number of recessive alleles. Because the initial class frequency was set at equilibrium and there was no selection or advantage in any genotype, the final class frequency should have been at equilibrium as well. However, the p value was at .866 and the q value was at .134. This indicated that there were significantly more dominant alleles than recessive ones. The gene pool was almost flooded with dominant alleles, with the final genotypes being three homozygous dominant offspring, one heterozygous offspring, and no homozygous recessive offspring. If our group was to remain mating, there would be no possibility of having a homozygous recessive genotype. Additionally, because the gene pool was so small and there was only one recessive gene left, the recessive gene could have been lost forever thanks to one generation where only dominant alleles were donated to the offspring.;p

2. What do your results indicate about the importance of population size as an evolutionary force?

The larger the population is, the more diversity there will be. The initial p and q values were .5, meaning that there was a 50/50 chance of either allele being in an organism. By the end of the simulation, the p value was .65 and the q value was .35, meaning that there was more likelihood of an allele being dominant versus and allele being recessive. Therefore, the population was becoming less and less diverse because the gene pool was small. If there was one round where several people received dominant alleles, then the rest of the population from there on out would be more and more likely to have dominant alleles. When there is a large population, however, and several organisms donate dominant alleles by coincidence, it doesn’t matter because the rest of the population balances them out. The larger the population and the more samples available, the closer the final class frequencies will be to the initial class frequencies.