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 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? 0.7 0.09 0.21 0.42 0.49

Question

Melanie Murillo

Biology

7 May, 03:56

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 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? 0.7 0.09 0.21 0.42 0.49

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Answers (1)

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Kennedy Poole · Answer 1

The answer is 0.42.

According to the Hardy-Weinberg equilibrium:

p + q = 1

p² + 2pq + q² = 1

where p is a frequency of a1 allele, q is a frequency of a2 allele, p² is a frequency of a1a1 genotype, 2pq is a frequency of a1a2 genotype, q² is a frequency of a2a2 genotype.

If 70% of the gametes produced in the population contain the a1 allele, then:

p = 70% = 0.7.

If p = 0.7 and p + q = 1, then q = 1 - p = 1 - 0.7 = 0.3

So, the proportion of the flies that carry both a1 and a2 (a1a2 genotype) is:

2pq = 2 * 0.7 * 0.3 = 0.42