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

Verified step by step guidance

1. First, understand that the Hardy-Weinberg equilibrium principle states that allele and genotype frequencies in a population will remain constant from generation to generation in the absence of other evolutionary influences.
2. Given that 70% of the gametes contain the A1 allele, the frequency of A1 (p) is 0.7. Consequently, the frequency of the A2 allele (q) can be calculated as 1 - p, which is 1 - 0.7.
3. Calculate the frequency of the A2 allele: q = 1 - 0.7 = 0.3.
4. In Hardy-Weinberg equilibrium, the proportion of heterozygous individuals (carrying both A1 and A2 alleles) is given by the formula 2pq, where p is the frequency of A1 and q is the frequency of A2.
5. Substitute the values of p and q into the formula: 2pq = 2 * 0.7 * 0.3. This will give you the proportion of flies that are heterozygous (A1A2).

Key Concepts

Hardy-Weinberg Equilibrium

The Hardy-Weinberg equilibrium is a principle that states that allele and genotype frequencies in a population will remain constant from generation to generation in the absence of evolutionary influences. These influences include mutation, selection, gene flow, genetic drift, and non-random mating. It provides a mathematical model to study genetic variation in populations.

Allele Frequency

Allele frequency refers to how common an allele is in a population. It is calculated as the number of copies of a specific allele divided by the total number of all alleles for that gene in the population. In this question, the allele frequency of A1 is given as 0.7, meaning 70% of the gametes carry the A1 allele.

Genotype Frequency Calculation

In a population in Hardy-Weinberg equilibrium, the genotype frequencies can be calculated using the allele frequencies. For two alleles, A1 and A2, the frequency of the heterozygous genotype (A1A2) is given by 2pq, where p is the frequency of A1 and q is the frequency of A2. Given p = 0.7, q can be calculated as 1 - p = 0.3, leading to a heterozygous frequency of 2(0.7)(0.3) = 0.42.