Q23-5TYU
Question
A fruit fly population has a gene with two alleles, A1 and A2. Tests show that 70% of the gametes produced in the population contain the A1 allele. If the population is in Hardy-Weinberg equilibrium, what proportion of the flies carry both A1 and A2?
(A) 0.7
(B) 0.49
(C) 0.42
(D) 0.21
Step-by-Step Solution
VerifiedThe option ‘0.7’ is false.
The option ‘0.49’ is false.
The option ‘0.42’ is true.
The option ‘0.21’ is false.
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.
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.
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.
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.
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.