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Bio,pls ans as soon as possible

dear student 
according to hardy weinberg principle,
if 70% of the gametes have A1 allele than the rest of the gametes will have allele A2.
so using p+q=1
we can say A1+A2=1
and A1 = 70% = 0.7
so A2 = 1-0.7 = 0.3

now coming to the frequency of alleles, we use 
A12 + 2A1A2 + A22 = 1, where A1A2 represents the heterozygous character and using values of A1 and A2 from above we can conclude that A1A2 = 2 * 0.7 * 0.3 =0.42

so correct answer will be option 1

Regards

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View Full Answer
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......
HOPE IT WILL HELP YOU............
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