The three common genotypes at the hemoglobin locus have very different phenotypes: SS individuals suffer from severe anemia, AS individuals have a relatively mild form of anemia but are resistant to malaria, and AA individuals have no anemia but are susceptible to malaria. The frequency of the S allele among the gametes produced by the first generation of a central African population is 0.2.

(a) Assuming that mating occurs at random, what are the frequencies of the three genotypes among zygotes produced by this population?

I know that this is 0.2 for the S allele (q in the hardy weinberg equation) and 0.8 for the A allele (p in the hardy weinberg equation).

So for the frequencies I got AA = 0.64 AS = 0.32 SS = 0.04

(b) In this area, no SS individuals survive to adulthood, 70% of the AA individuals survive, and all of the AS individuals survive. What is the frequency of each of the three genotypes among the second generation of adults?

I don't know how to do this! I don't understand what to do! I tried multiplying

0.64 by 0.7 (because only 70% survive), but I know that is wrong answer. I know you have to divide 0.64 x 0.7 by the total that survived, but I don't understand why you have to do that. You didn't have to divide for part a, why do you have to divide for part b?

Also for some reason the frequency of the AS is not 0.32 for the answer. I don't understand why! Please Please help me.

(c) What is the frequency of the S allele among gametes produced by these adults? I dont know this either!



1 Answer 1


Your answer to a) is correct (standard Hardy-Weinberg equation).

For b) you have to consider that you are calculating proportions of the population, which should sum to 1. Since 30% of the AA individuals and all SS individuals die, this is not the case after selection, which is why you should divide the frequencies by the total proportion that survive (also for the AS genotype), such as:

$$ \begin{array}{c|cccc} \text{genotype} & \text{freq before} & w_{survival} & \text{freq after selection} & \text{corrected freq}\\ &\text{selection}&&&\text{after selection}\\ \hline AA & 0.64 & 0.7 & f(AA)\cdot w_{AA}=0.45 & 0.58\\ AS & 0.32 & 1 & f(AS)\cdot w_{AS}=0.32 & 0.42 \\ SS & 0.04 & 0 & f(SS)\cdot w_{SS}=0.0 & 0.0\\ \text{sum} & 1 & & 0.768 & 1 \end{array} $$

here $w$ is the fitness of genotypes. And for your comment on why you didn't have to divide by the sum of freqencies in a): you could do this, but since the sum of genotype frequencies is 1 it doesn't make a difference.

For c) you use the standard formula for calculating allele frequencies from genotypes, so that $f(A)=f(AA)+f(AS)/2= 0.79$ and $f(S)=f(SS)+f(AS)/2= 0.21$.


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