Question

07.900 Thi 3. Le 1. You have sampled a population in which you know that the percentage of the homove recessive genotype(a) i
gous 3. Lets say that brown fur coloring is dominant to gray fur coloring in mice. If you have 168 brown mice in a populatio
5. If 51% of the population carries at least one copy of the recessive allele. What is the predicted frequency of individuals
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Answer #1

Hardy-Weinberg equilibrium:

  • Hardy-Weinberg equation describes: The most possible distribution of the population genotype of the population, if the frequencies of the alleles are known.
  • If allele frequencies are p and q, then the genotype frequencies are p2, q2, 2pq.
  • A population is considered to be in Hardy-Weinberg equilibrium when: In a randomly mating population, allele frequency and genotype frequency remain constant.

Thus, under Hardy-Weinberg equilibrium:

p+ q=1 and

p2+2pq + q2 = 1.

1.

Total = 100

  • Number of genotypes for individuals= aa = recessive homozygous condition =36%
  • Genotype frequency for aa = q2= 36/100 =0.36

A. Answer: genotype frequency for aa = 0.36

Allele frequency for a = q= 0.6

B. Answer: frequency of a allele = 0.6

As per Hardy-Weinberg equilibrium p+q = 1

Thus, p = 1- 0.6 = 0.4 = frequency of A allele

C. Answer: 0.4

D. Genotype frequency for AA = p2= (0.4)2= 0.16 (Answer)

E. Heterozygous genotype frequency (Aa) = 2 x p x q =2 x 0.6x 0.4 =0.48 (Answer)

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