# Population Genetics: Hardy-Weinberg Question?

If the percentage of a homozygous recessive genotype (rr) is 47%, what is the frequency of:

1. The rr genotype

2. The r allele

3. The R allele

4. The Rr genotype

5. The R phenotype

What I sort of understand is the formula to use is p² + 2pq + q² =1 and p+q=1

p² = homozygous dominant 2pq = heterozygous q² = homozygous recessive

and p= dominant allele q= recessive allele.

Detailed explanations are greatly appreciated!!

### 1 Answer

- Paul JacksonLv 76 years agoFavorite Answer
1. 47%, r is recessive so only rr has the rr phenotype.

2. The frequency of the r allele is q.

q^2 = 0.47

q = √0.47 = 0.69 (rounded)

3. The frequency of the R allele is p = 1 - q = 0.31

4. Rr is the homozygote

2pq = 2( 0.686)(0.314) = 0.43

5. both the dominant hetrozygote and the homozygote have the R phenotype.

p^2 + 2pq = (0.31)^2 + 0.43 = 0.53

Note that

p^2 + 2pq + q^2 = 0.53 + 0.47 = 1

From a biological perspective we describe the situation this formula models as follows. In an isolated population with a gene with a single dominate allele and a recessive allele where neither of phenotypes cause any mating preference or change in reproduction rate, the population will eventually reach the equilibrium state described by the Hardy-Weinberg equation.

This means that if 100 RR's and 100 rr's land on a big isolated island and all start to mate, and we leave them alone for several generations, when we come back can expect to find that 25% percent of the population will be RR. 50% of the population will be Rr, and the other 25% will be rr.

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