AP BIOLOGY

EVOLUTION

HARDY WEINBERG EQUILIBRIUM

Question [CLICK ON ANY CHOICE TO KNOW THE RIGHT ANSWER]
A population of 150 individuals has an allele frequency of 0.3 for the dominant allele (B) and a frequency of 0.7 for the recessive allele (b). Use the Hardy-Weinberg equation to determine the frequency of homozygous recessive (bb) individuals in the population.
A
0.09
B
0.42
C
0.49
D
0.21
Explanation: 

Detailed explanation-1: -Since 30% will have the recessive allele, then in a population of 200 organisms, there will be 200⋅30%=60 who will have the recessive allele.

Detailed explanation-2: -To determine q, which is the frequency of the recessive allele in the population, simply take the square root of q2 which works out to be 0.632 (i.e. 0.632 x 0.632 = 0.4). So, q = 0.63. Since p + q = 1, then p must be 1-0.63 = 0.37.

Detailed explanation-3: -Another aspect of alleles and segregation is the concept of allele frequencies within a population. On an individual level, a person has a 0.5 chance of passing on one of two alleles for a given gene.

Detailed explanation-4: -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.

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