Respuesta :
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
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
Hardy Weinberg principle in genetics asserts that in the lack of evolutionary influence the alleles or the recurrence of genotypes will be unchanged.
It is calculated by using the formula:
p²+2pq+q²=1
Where,
- frequency of dominant purebred = p²
- frequency of heterozygous= 2pq
- frequency of recessive purebred = q²
The correct answer is:
option D. 0.42
The explanation for this is:
p+q=1
Where,
- p: frequency of a1 allele
- q: frequency of a2 allele
- p²: frequency of a1a1 genotype
- 2pq: frequency of a1a2 genotype
- q²: frequency of a2a2 genotype.
70% of the given population contains a1 allele so,
p= 70% = 0.7
If p=0.7 and p+q=1
Then,
q = 1 - p
= 1 - 0.7
= 0.3
Hence the population of the flies that carries a1a2 genotype will be:
= 2pq
= 2 × 0.7 × 0.3
= 0.42
Therefore, the population of flies containing both a1 and a2 alleles is 0.42.
To learn more about Hardy-Weinberg equilibrium follow the given link:
https://brainly.com/question/419732