All answers must start with the equation that was used to answer the question, a

Question

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4. In a population of diploids, 1 individual out of 100 exhibits the homozygous recessive phenotype. What is the expected frequency of heterozygous genotypes.         

4b. In the above population, 25 heterozygous individuals are observed. Is the population in HWE?

5. If a genome is 4.2 Mb in length, with an average mutation rate of 1.8 x 108 mutations/site/generation, how many mutations should occur in the average gamete in a single generation?

Explanation / Answer

By HWE

p^2 + 2pq + q^2 = 1

p^2 = frequency of homozygous dominant individuals

q^2 = frequecy of homozygous recessive individuals

2pq = frequency of heterozygous individuals

Given

q^2 = 1 / 100 = 0.01

q = 0.1

p + q = 1

p = 1 - 0.1 = 0.9

p^2 = 0.81

p^2 + 2pq + q^2 = 1

0.81 + 2pq + 0.01 = 1

2pq = 1 - 0.82 = 0.18

Frequency of heterozygous genotypes = 0.18 i.e 18 are heterozygotes in a population of 100.

4b . The number of heterozygotes = 18 , So if 25 heterozygotes are observed then population would not be in HWE.

5.

No of sites = 56 x 4.2

mutations that occur in the average gamete in a single generation = 1.8 x 10^-8 * 56 * 4.2 = 4.24 x 10^-6 mutations

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