Asked Jan 10, 2019

I asked this question before but the solution was not correct and I'm really trying to figure out how to do this to get the correct answer. Please show work!
DNA typing is used to compare evidence DNA (E) left at a crime scene to two suspects
(S1 and S2). Suspect 1 is excluded by the evidence, but suspect 2 remains included. What is
the frequency of suspect 2's genotype if the allelic frequencies in the population are f(A1) =
0.1, f(A2) = 0.2, and f(A3) = 0.7, and the population is at Hardy–Weinberg equilibrium?

The answer is 0.49.


Image Transcriptionclose

S1 S2 A2


Expert Answer

1 Rating
Step 1

According to the question it is required to determine the frequency of suspect 2’s genotype and the allelic frequencies of the population are given. It is given that the population is in Hardy–Weinberg equilibrium.

The frequency of the allele is given as

f(A1) =0.1,

f(A2) = 0.2

f(A3) = 0.7.

Step 2

According to the Hardy-Weinberg’s principle, the genotypic frequency for the heterozygote is determined by 2pq.


p represents the frequency of  allele 1, and q represents the frequency of allele 2.

Step 3

For the suspect 2, the allele 1 is A1 and the allele 2 is A2. The genotype of the Suspect 2 is A1A2.



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