Please explain the question, show your work. Thanks! ANSWERS: 11. 0.367, 0.46, 0

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

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