Question

In a population of 200 individuals, frequency of alleles responsible for sickle-cell anaemia in the population is 0.4. Find the number of heterozygous individuals in the population.

• A. 80
• B. 40
• C. 32
• D. 96

Answer 1

Answer: (A)

80

It is given that the frequency of allele responsible for sickle cells anemia (HB${}^S$) is 0.4 which means the frequency of a normal allele ( HB ) is 1 - 0.4 = 0.6. This is in accordance with the Hardy-Weinberg principle.

Also, according to the Hardy Weinberg principle frequency of heterozygous individual is given by 2Aa, where 'A' and 'a represent frequencies of two alleles in a population.

Hence, the frequency of the heterozygous combination of alleles is 2×0.4×0.6 = 0.48 and the number of individuals is equal to 0.48× 200 = 96.

So, the correct option is '96'.