Please explain the question, show your work. Thanks! ANSWERS: 11. 0.367, 0.46, 0
ID: 316721 • Letter: P
Question
Please explain the question, show your work. Thanks!
ANSWERS: 11. 0.367, 0.46, 0.17; .60, .40; 0.36 0.48 0.16; close
11. A researcher recorded genotypes at a gene (A) in ater fleas" (Daphnia) collected from a pond. The gene is involved in protecting individuals from the damaging effects of UV light. A1A A1A A2A 69 26 55 What are the observed genotype frequencies? A1A A1A A2A2 What are the observed allele frequencies? fr(A1) E fr (A2) What are the expected frequencies, according to the A1A1 A1A A2A2 Hardy-Weinberg principle? Are the genotypes close to H-W proportions, or different (at least 1 comparison >5%)?Explanation / Answer
The observed genotype frequencies can be calculated as
A1A1(p2)= 55/(55+69+26) = 55/150 = 0.3666
A1A2 (2pq) = 69/(55+69+26) = 69/150 = 0.466
A2A2 (q2) = 26/(55+69+26) = 26/150 = 0.17
The Observed allele frequencies
fr(A1) = p2 + 1/2(2pq) = 0.366 + 1/2(0.466) = 0.59999 = 0.60
fr(A2) = q = 1-p = 1-0.6 = 0.4
The expected frequencies as per Hardy-weinberg
A1A1 = (p)2 = (0.599)2= 0.36
A1A2 = 2pq = 2*0.597*0.399 = 0.48
A2A2= (q)2 = (0.3999)2 = 0.159 = 0.16