# Hardy Weinberg Equilibrium problem

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.

• Are you saying that f(A)=0.3 and f(a)=0.7? – fileunderwater Nov 15 '13 at 14:55
• @fileunderwater yes. – biogirl Nov 15 '13 at 16:27

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.

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.

• I got 47.89% when only working with frequencies, so your answer seems right. My feeling was also that both answers were incorrect, therefore my comment to the question. – fileunderwater Nov 15 '13 at 16:16
• thanks @fileunderwater, I've been tearing my hair out over this! You should post your calculation - I expect it is more elegant than mine. – Alan Boyd Nov 15 '13 at 16:18
• This is pretty much what I was getting as well, and I couldn't figure out how to get the 43.52%. – MCM Nov 15 '13 at 16:22

This is basically the same solution as @AlanBoyd's answer, but since he asked me to, I will post my solution as well (also a reason to try out $\mathcal{MathJax}$ a bit more).

Assuming:

$f(A)=p=0.3, \\f(a)=q=0.7$

and the standard formulas:

$f(AA)=p^2\\ f(Aa)=2pq\\ f(aa)=q^2$

gives these genotypes in the parental generation (G1), before and after selection (using the fitness vector $w$):

$$\begin{array}{c|cccc} \text{genotype} & \text{freq G1} & w & \text{freq breed.pop} & \text{freq breed. rescaled}\\ \hline AA & 0.09 & 1 & f(AA)\cdot w_{AA}=0.09 & 0.119\\ Aa & 0.42 & 1 & f(Aa)\cdot w_{Aa}=0.42 & 0.556 \\ aa & 0.49 & 0.5 & f(aa)\cdot w_{aa}=0.245 & 0.325\\ \text{sum} & 1 & & 0.755 & 1 \end{array}$$

The genotype frequencies in the breeding population (rescaled so that they sum to one) are then used to calculate allele frequencies as:

$$f(A)=f(AA)+\frac{f(Aa)}{2} = 0.397\\ f(a)=f(aa)+\frac{f(Aa)}{2} = 0.603$$

These are then used to calculate the genotype frequencies in the next generation (G2), using the same HW-equations as in the beginning:

\begin{align} f(AA)&=0.397^2=0.158 \\ f(Aa)&=2\cdot0.397\cdot0.603=0.479 \\ f(aa)&=0.603^2=0.363 \\ \end{align}

I do not understand where the given answer $f(Aa)=0.435$ is coming from.