The data given for each lettered set below represent initial

The data given for each lettered set below represent initial population values for D, H, and R, respectively. For each set of values, determine the equilibrium values of p and q and the genotypic frequency of AA, Aa, and aa at Hardy-Weinberg equilibrium. (Keep in mind that D, H, and R are the relative genotypic frequencies for homozygous dominants (D), heterozygotes (H), and homozygous recessives (R).) (0.00, 1.00, 0.00) (1/3, 1/3, 1/3) (0.70., 0 30, 0.00) (0.00, 0.70, 0.30) (0.40, 0.30, 0.30) (0.10, 0.30, 0.60) (0.10, 0.70, 0.20) (0.20, 0.20, 0.60)

Solution

(a)GENOTYPE FREQUENCIES:

AA (p²) = 0

Aa (2pq) = 1

aa (q²) = 0

ALLELE FREQUENCIES:

Freq of A = p = p2 + 1/2 (2pq) = 0+ 1/2 (1) = 0.5

Freq of a = q = 1-p = 1 - 0. 5= 0.5

EXPECTED GENOTYPE FREQUENCIES (assuming Hardy-Weinberg):

AA (p²) = (0.5) ² = 0.25

Aa   (2pq) = 2 (0.5)(0.5) = 0.5

aa (q²) = (0.5)² = 0.25

p²+2pq + q²=1;    0.25+0.5+0.25=1

(b)GENOTYPE FREQUENCIES:

AA (p²) = 0.333 (1/3)

Aa (2pq) = 0.333(1/3)

aa (q²) = 0.333(1/3)

ALLELE FREQUENCIES:

Freq of A = p = p2 + 1/2 (2pq) = 0.333+ 1/2 (0.333) = 0.5

Freq of a = q = 1-p = 1 - 0. 5= 0.5

EXPECTED GENOTYPE FREQUENCIES (assuming Hardy-Weinberg):

AA (p²) = (0.5) ² = 0.25

Aa   (2pq) = 2 (0. 5)(0.5) = 0.5

aa (q²) = (0.5)² = 0.25

p²+2pq + q²=1;    0.25+0.5+0.25=1

(c)GENOTYPE FREQUENCIES:

AA (p²) = 0.70

Aa (2pq) = 0.30

aa (q²) = 0

ALLELE FREQUENCIES:

Freq of A = p = p2 + 1/2 (2pq) = 0.70+ 1/2 (0.30) = 0.85

Freq of a = q = 1-p = 1 - 0. 85= 0.15

EXPECTED GENOTYPE FREQUENCIES (assuming Hardy-Weinberg):

AA (p²) = (0.85) ² = 0.7225

Aa   (2pq) = 2 (0. 85)(0.15) = 0.255

aa (q²) = (0.15)² = 0.0225

p²+2pq + q²=1;    0.7225+0.255+0.0225=1

In the same way for all other options, we can calculate the both allelic and genotypic frequencies.