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

Question

All answers must start with the equation that was used to answer the question, and all work must be shown.

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

Ans 1:

(i) To explore the Hardy-Weinberg equation, we can examine a simple genetic locus at which there are two alleles, A and a. The Hardy-Weinberg equation is expressed as:

p2 + 2pq + q2 = 1

q2 = 1/100 = 0.01

Therefore, q= 0.1

We know p+ q =1

we found out q=0.1

Therefore, p=1-0.1=0.9

Now the frequency of heterozygous individuals is = 2pq = 2 x 0.1 x 0.9 = 0.18

(ii) Assuming Hardy Weinberg Equilibrium, number of heterozygous individuals = 2pq x N = 2 x 0.9 x0.1 x 100= 18 (N= population size)

Since 25 heterozygous individuals are observed thus it shows that population is not in HWE as the frequencies seems to have changed.

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