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a_sh-v [17]
2 years ago
7

Consider a population of 1099 apple fly maggots. In this population, you observe that the allele frequency of the dominant allel

e at the ADH locus, where are are two alleles, is 0.554. What is the expected number of heterozygotes if this population is at HWE (please give your answer as an integer)?
Biology
1 answer:
Tatiana [17]2 years ago
5 0

Answer:

543

Explanation:

Given is the frequency of the dominant allele in the population (p)= 0.554

Since the locus has two alleles and the population is in Hardy Weinberg equilibrium, the frequency of the recessive allele (q) of the same locus would be= 1-0.554= 0.446

Frequency of heterozygous genotype in the given population = 2pq= 2 x 0.554 x 0.446= 0.4941

Now, the number of heterozygous genotype in the given population= 0.4941 x 1099= 543.0159 = 543

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The recombination frequency between three genes A, B, and C located on the same chromosome is obtained as follows:
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Based on the given recombination frequency, Genes B and C are closest to each other. This is because they bear the recombination frequency of 10%.

<h3>What is Recombination frequency?</h3>

Recombination frequency may be defined as a proportion of recombinant offspring delivered in a genetic cross between two organisms. It is always expressed in percentages.

The complete question is: The recombination frequency between three genes A, B, and C located on the same chromosome is obtained as follows:

  • A and C - 28%
  • A and B - 18%
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Based on the result, which two genes are closest to each other?

Hence, the answer to this question is, that Genes B and C are closest to each other. This is because they bear the recombination frequency of 10%.

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Therefore, it is well described above.

To learn more about Recombination frequency, refer to the link:

brainly.com/question/7299933

#SPJ1

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