The data given for each lettered set below represent initial population values f
ID: 314758 • Letter: T
Question
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)Explanation / Answer
(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.
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