Question

1.)If the population frequencies of two alleles at a locus are B = 0.5 and b = 0.5, what is onepossible set of frequencies for the three resulting genotypes that would NOT reflect Hardy- Weinberg equilibrium?

2.)In a population that is in Hardy-Weinberg equilibrium, the frequency of the homozygous recessive genotype is 0.09. What is the frequency of individuals that are homozygous for the dominant allele?

3.)In humans, Rh-positive individuals have the Rh antigen on their red blood cells, while Rh- negative individuals do not. If the Rh-positive phenotype is produced by a dominant gene (A), and the Rh-negative phenotype is due to its recessive allele (a), what is the frequency of the Rh- positive allele if 84% of a population is Rh-positive and the population is in Hardy-Weinberg equilibrium?

Answer 1

1).

Hardy-Weinberg equilibrium:
p+q = 1
p2+2pq+q2 = 1

The frequency of dominant allele B, p = 0.5
Frequency of recessive allele b, q = 0.5

If the population is in H-W equilibrium,
The frequency of dominant homozygotes BB, p2 = 0.25
The frequency of recessive homozygotes bb, q2 = 0.25
The frequency of heterozygotes Bb, 2pq = 0.5 However, it is given that the population is not in H-W equilibrium. So, other than 0.25.0.25, and 0.5 frequency values would indicate deviation from H-W equilibrium.
Examples:
1. p2 = 0.3, q2 = 0.4, 2pq = 0.3

2. p2 = 0.7, q2 = 0.1, 2pq = 0.2

3. p2 = 0.5, q2 = 0.4, 2pq = 0.1

2). Given frequency of the homozygous recessive

.09=q2

then q= root over .09=0.30

since p+q=1

then p= 1-.3=0.70

homozygous dominant are represented as p2 = .70*.70=0.49

so, the frequency of individuals that are homozygous for the dominant allele is 0.49