NEET BIOLOGY

GENETICS AND EVOLUTION

GENETIC BASIS OF INHERITANCE

Question [CLICK ON ANY CHOICE TO KNOW THE RIGHT ANSWER]
In a randomly breeding population, 16% of them carry recessive trait. Assuming C is dominant over c, what percent are the carrier individuals?
A
36%
B
48%
C
63%
D
84%
Explanation: 

Detailed explanation-1: -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.

Detailed explanation-2: -Solution: 16% population means q2 is 0.16; In a population 0.16 = 400, total population is 0.16/400 = 2500.

Detailed explanation-3: -If the homozygous recessive genotype makes of 16% of the population the fraction of the recessive allele is the square root of the q (recessive allele fraction )= √¯ 0.16 = 0.4. The dominant allele is 1-p = 1-0.4 = 0.6.

Detailed explanation-4: -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.

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