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!!
- 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
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.