In a small tribal population, the frequencies of two alleles A and a at a partic
ID: 35835 • Letter: I
Question
In a small tribal population, the frequencies of two alleles A and a at a particular locus were 0.3 and 0.7, respectively. However, not all the individuals with genotype aa could live up to the reproductive age and the relative fitness of this genotype was found to be 0.5. The remaining genotypes had a relative fitness of 1. What is the expected percentage of heterozygotes among newborns in the next generation?
My answer is : 37 % but the answer given 43.52 %
I did the following : First I assumed the population to be of 100 individuals and then I counted the no of individuals with respective genotypes. Then I assumed that each person leaves 1 offspring except the aa ones which leave 0.5 offsprings and then I calculated the new genotype frequency from this.
Explanation / Answer
Your reasoning is incorrect, because once you have the number of individuals surviving after the 50% mortality you can't just give them all one identical offspring, you have to mate them with one another.
However, I can't get the answer given, instead I get 47.9%
Here is my reasoning:
f(A)=0.3
f(a)=0.7
calculate genotype frequencies:
AA = 0.32 = 0.09
aa = 0.72 = 0.49
Aa = 2 * 0.3 * 0.7 = 0.42
(these add to 1 as expected)
use 1000 individuals for nice round numbers
number AA = 90
number aa = 490
number Aa = 420
Now apply 50% mortality in aa individuals:
number AA => 90
number aa => 245
number Aa => 420
now calculate new allele frequencies in mating population:
f(A) = (90 + 90 + 420)/1510 = 600/1510
f(a) = (245 + 245 + 420)/1510 = 910/1510
calculate proportion of heterozygote offspring
= 2 * f(A) * f(a)
= (1200 * 910)/(1510 * 1510)
= 0.479
I hope someone can see where I'm going wrong.