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Question

conline.broward.edu/d21/ms/dropbox/user/folder submit files.d21?db 408379&grpid-0aspv-Faise;&bp-08ou; BSC2011 Assignment 7 -Hardy Weinberg Equilibrium Theorem, CCRS Deletion Mutation The Hardy-Weinberg Equilibrium Theorem, which states that polymorphism frequencies in a population remain constant under specific conditions used in population genetic studies. This equation is applied consisting of only two possible alleles (dominant and recessi traits p2 + 2pq + q2 = 1 and p + q = 1 p- frequency of the dominant allele in the population q frequency of the recessive allele in the population p2- percentage of homozygous dominant individuals q2 percent of homozygous recessive individuals 2pqn entage of heterozygous individuals The equation allows biologists to determine whether evolution has occurred in a given population. Any changes in the gene frequencies in the population over time can be detected. The theorem essentially states that if no evolution is occurring, thern an equilibrium of allele frequencies will remain in effect in each succeeding generation of sexually reproducing individuals. In order for equilibrium to effect (i.e. that no evolution is occurring) then the following five co met 1. No mutations must occur so that new alleles do not enter the population. 2. No gene flow can occur (i.e. no migration of individuals into, or out of, the population) 3. Random mating must occur (i.e. individuals must pair by chance 4. The population must be large so that no genetic drift (random chance) carn cause the allele freauencies to change Submit Cancel oe@#3 rch up

Explanation / Answer

Total population = 1318

Number of alleles = 2636

Homozygous WT = 1102

Number of WT alleles = 2204

Frequency of homozygous WT allele, p2 = 2204/2636 = 0.836 = 83.6%

p = 0.914

p+q = 1

q = 1 - 0.914 = 0.086

Frequency of homozygous mutants for deletion allele, q2 = 0.007396

Frequency of heterozygotes, 2pq = 2 X 0.086 X 0.914 = 0.157

p2 + 2pq + q2 = 0.836 + 0.157 + 0.00734 = 1

1. Susan has 0.007396 (=0.74%) chance that she can be homozygous for the deletion. So, there is a greater chance that she can be infected with HIV if she participates in multi-partner sex without protection.

2. Carriers for deletion, 2pq = 0.157 = 15.7%

3. Number of heterozygotes, 2pq = 15.7%

4. Before HIV appeared, the population may not be at HW equilibrium. The infection exerts selection pressure to maintain the deletion allele in the population.

5. In the absence of HIV treatment, the frequency of deletion mutant alleles would increase as it confers resistance.

6. If the HIV treatment is available, there won't be any change in the frequency of alleles or mutant alleles will be eliminated.

7. The mutant allele frequency would increase.