# A single locus has alleles A and a. The frequency of allele A is 0.6. If the homozygous recessive genotype has a relative fitness level of 50% to both the homozygous dominant genotype and the...

A single locus has alleles A and a. The frequency of allele A is 0.6. If the homozygous recessive genotype has a relative fitness level of 50% to both the homozygous dominant genotype and the heterozygous genotype in the current generation, what would be the frequency of allele A in the next generation?

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### 1 Answer

In the current generation, the frequency of the dominant allele A is given as p = 0.6. Therefore the frequency of the recessive allele a is q = 0.4, since p + q = 1. The population is assumed to not yet be in Hardy-Weinberg equilibrium (ie the frequencies are still changing from generation to generation).

The three possible obtainable genotypes in the population are AA Aa and aa. The homozygous ("identical pair") recessive genotype is aa, and we are given that its *relative *fitness level w in the current generation is 50%, meaning that compared to AA and Aa there is a relative reduction of 50% in survival rate. That is, the recessive phenotype has relatively 50% the survival seen in the non-recessive (dominant) phenotype (assuming there is complete dominance of trait A, ie genotypes AA and Aa appear the same and hence have the same phenotype)/

We can write this more clearly as a table (hopefully the alignment of this will come out clearly in the text, but if not the entries are comma separated)

AA Aa aa sum

relative freq, 1, 1, 1-s = 0.5,

freq, p^2, 2pq, q^2, 1,

new freq (almost), 1.p^2, 1. 2pq, (1-s)q^2, p^2 + 2pq + (1-s)q^2

= 1 - sq^2,

new freq (sum=1), p^2/(1-sq^2), 2pq/(1-sq^2), q^2(1-s)/(1-sq^2), 1,

= p'p', = 2p'q', = q'q',

**Therefore the new frequency of the dominant A allele in the next generation on is**

`p' = sqrt(p^2/(1-sq^2)) = p/sqrt(1-sq^2) = 0.6/sqrt(1-0.5(0.4^2)) = 0.6/(1-0.08) = 0.6/0.92 = 0.652 `

**to 3sf**

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