Question

A fruit fly population has a gene with two alleles,A1andA2. Tests show that 70% of the gametes produced in the population contain theA1allele. If the population is in Hardy-Weinberg equilibrium, what proportion of the flies carry bothA1andA2?

(A) 0.7

(B) 0.49

(C) 0.42

(D) 0.21

Short answer

1. The option ‘0.7’ isfalse.

2. The option ‘0.49’ isfalse.

3. The option ‘0.42’ is true.

4. The option ‘0.21’ is false.

Step-by-step solution

Fruit fly

In entomology, the tiny two-winged insects that are mainly attracted to and feed on ripened fruits and vegetables and are found everywhere in the world are called fruit flies.

Explanation of option (A)

The principle of population genetics regarding the genetic variation and allele frequency within the population that will remain constant across generations is called the Hardy-Weinberg equilibrium.

Thus, the value 0.7 is only of p, based on the Hardy-Weinberg equilibrium.

Therefore, the given option is false.

Explanation of option (B)

The different conditions involved in the Hardy-Weinberg equilibrium are no mutation, random mating, no natural selection, and a large population size.

Thus, the result of 2pq is 0.42 instead of 0.49.

Therefore, the given option is false.

Explanation of option (C)

When there is a frequent occurrence of a gene in the population of various generations, this frequency is called allele frequency.

Thus, based on the Hardy-Weinberg equation, the remaining recessive allele frequency is 30%, the p will be equal to 0.7, and q is 0.3. The heterozygous allele is obtained by multiplying 2 with p and q. The result will be 0.42.

Therefore, the given option is true.

Explanation of option (D)

The two distinct kinds of alleles based on their traits are recessive and dominant alleles.

Thus, 0.21 is not the correct result obtained from the Hardy-Weinberg equation.

Therefore, the given option is false.