The following 3 questions are linked^^^^^^^^^^^^^^^^^ Whip-poor-wills. ( Camprim
ID: 62189 • Letter: T
Question
The following 3 questions are linked^^^^^^^^^^^^^^^^^
Whip-poor-wills. (Camprimulgus vociferus! The population has found in two adult phenotypes: rufous(red) and gray. An ornithologist investigates the genetics of this color variation and discovers that rufous individuals are one of two genotypes (RR or Rr), while gray individuals are rr. The frequency of these genotypes at Generation 1 in northern New Jersey are: 60% RR, 20%Rr, and 20% rr.
1. If the Hardy-Weinberg Law holds for these genetic variants, what is the frequency of the R allele in this population?
(A) 0.3
(B) 0.4
(C) 0.5
(D) 0.6
(E) 0.7
2. What will the phenotype frequency be in the next generation (Generation+1)?
(A) 60% rufous, 40% gray
(B) 80% rufous, 20% gray
(C) 51% rufous, 49% gray
(D) 91% rufous, 9% gray
(E) 49% rufous, 51% gray
3 What will the dominant allele frequency be in the generation after Generationt+2 ?
(A) 0.3
(B) 0.4
(C) 0.5
(D) 0.6
(E) 0.7
Explanation / Answer
1. The frequency of RR or p2 = 60% or 0.6,
p = 0.77 or 0.7 (approximately) and the frequency of q = 1-0.7 = 0.3
Thus, the correct option is (E) 0.7.
2. In the next generation, the frequency of AA is (p2) = 0.7* 0.7 = 0.49 or 49%
The frequency of aa allele (q2) = 0.3* 0.3 = 0.09 or 9%
The frequency of Aa is = 2*pq = 2* 0.7* 0.3 = 0.42 or 42%
Thus, the number of rufous individuals = 49+42 = 91%
Number of grey individuals = 9%.
Thus, the correct option is (D) 91% rufous, 9% gray