EVOLUTION
HARDY WEINBERG EQUILIBRIUM
Question
[CLICK ON ANY CHOICE TO KNOW THE RIGHT ANSWER]
|
|
.81
|
|
.9
|
|
.1
|
|
none of the above
|
Detailed explanation-1: -We know that the frequency of the recessive homozygote genotype is q2 and equal to 0.09.
Detailed explanation-2: -The frequency of homozygous dominant plants (p2) is (0.6)2 = 0.36. Out of 100 individuals, there are 36 homozygous dominant (YY) plants. The frequency of heterozygous plants (2pq) is 2(0.6)(0.4) = 0.48. Therefore, 48 out of 100 plants are heterozygous yellow (Yy).
Detailed explanation-3: -In the question, the genotype frequency of recessive population is (q2) = (16% or 0.16). Hence, frequency of recessive allele = q = 0.16 = 0.4. Since p+q = 1; thus the frequency of dominant allele = p = 1-q = 1-0.4 = 0.60.
Detailed explanation-4: -On sampling a population, the frequency of homozygous recessive genotype (aa) is 36%.