Question

You discover a population of guppies and find it to be at Hardy–Weinberg equilibrium with respect to the S locus. Suppose there are two alleles at this locus and that the frequency of the s allele is 0.4. Of the individuals that carry at least one s allele, what fraction are homozygotes?

Answer 1

Answer:

A quarter (25%) of the individuals that carry at least one s allele are homozygotes.

Explanation:

Knowing the frequency of s, we can find the frequency of both alleles, because:

$p+q= 1\\p= frequencyS\\q= frecuencys\\\\So, if q=0.4 \\1-q=p\\1-0.4= 0.6= p\\\\p= 0.6 and q=0.4$

Now, if the population of guppies is at Hardy-Weinberg equilibrium, we can find how many heterozygotes and homozygotes are, like this:

$p^{2}+2pq+q^{2}=1$

We know that:

$p^{2}= frequencySS\\ 2pq= frequencySs\\q^{2}= frequencyss$

So, we need to know what is the frequency of guppies Ss (2pq) and ss (s2):

$2pq= 2 (0.6)*(0.4)= 0.48\\q^{2}= (0.4)^{2}= 0.16$

Finally, to find the fraction of homozygotes (ss) of the individuals that carry at least one s allele, we need to divide the frecuency of homozygotes ss into the frequency of heterozygotes Ss:$\frac{2pq}{q^{2} }=\frac{0.16}{0.64}= 0.25$

0.25 or 25% of the guppies carring the allele s are homozygotes.