Each of the two isolated populations is in Hardy-Weinberg equilibrium (HWE) and

Question

Each of the two isolated populations is in Hardy-Weinberg equilibrium (HWE) and the number of individuals in different genotype categories in the two populations is shown below:

A. If the populations join together to form a single population, calculate the allele and genotype frequencies in the population immediately after merger.

B. Use the chi-square test to determine if the merged population is in HWE.

C. In class we found that a population of 50 Aa and 50 aa individuals is not in HWE. Would the population be in HWE after one generation of random mating?

Explanation / Answer

If the two populations join together, the new numbers for the different genotypes would be:

AA - 68

Aa - 64

aa - 68.

Genotype frequencies would be - 34% for AA and aa(68/200 x 100). 32% for Aa(64/200 x 100).

Allele frequencies - total number of alleles = 400.

Frequency of A = 68 x 2 + 64/400 = 200/400 = 0.5

Frequency of a will be the same.

--------------

Since p + q = 1, the population is in HWE. So the expected and observed frequencies would be the same.

---------

The genotype frequencies would be 50/100 * 100 = 50% for both.

Alelic frequencies would be 100 + 50 /200 = 0.75(a). The A allele is 0.25. (50/200)

This is in HWE. After one generaiton of mating also the equilibrium will continue

Related Questions

Navigate

• Browse (All)
• Browse E
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.