4. [20 points total] In a sample of a population, 100 individuals were genotyped

Question

4. [20 points total] In a sample of a population, 100 individuals were genotyped using a molecular technique that produced the following banding patterns on an electrophoretic gel: Type of banding pattern: "D"H" "R" Band "S" Band "F' Of the 100 total individuals, 18 have banding pattern "D", 4 have the type "H" banding pattern, and all the rest (78) have the type "R" pattern. Assume that the two different bands (F and S) represent two different alleles of the same gene a. [5 points] What are the Hardy-Weinberg expected genotype frequencies? Is the population in Hardy-Weinberg equilibrium? Explain how you can tell. b. [5 points] Estimate the inbreeding coefficient, F, of the sampled population (i.e., the difference between expected and observed htergzyggsities as a fraction of the expected heterozygosity). What does the value of F tell you about this population? c. [5 points] Provide the best explanation for these results. d. [5 points] If the "F" allele were dominant and advantageous how would its frequency change over time in a population in which it is initially rare? (Hint: drawing a graph may help your explanation. Make sure you include not only what happens initially, but what the dynamics are after several generations.)

Explanation / Answer

4a. Here we have following data-

Genotypes-

SS = 18

FS = 78

FF= 4

Under hardy- Weinberg expectation, the allelic frequencies must remain constant from generations to generations with an aggregation to 1.

Means, p+q = 1

Here p and q are allelic frequencies of two alleles p and q.

Therefore, the ideal allelic frequency for an allele is 0.5.

Here the allelic frequency of the allele S = Numbers of S allele / Total alleles

= [(2 × Homozygous population of the allele)+ Heterozygous population] ÷ [2 × Population]

S = [(18 x 2) + 78] /200

S = 0.57

The allelic frequency of the allele F =[(4 x 2) + 78] /200

F = 0.43

F+S = 0.57+ 0.43

F+S =1.

Therefore, it follows the hardy- Weinberg law.

4b-

For an ideal cross there would be 1:2:1 ratio amongst the individuals. According to this amongst 100 individuals the number of heterozygotes will be-

2 x 100 /4

=50; this is the expected one.

The observed values are 78

The inbreeding coefficient F = [Observed ~ Expected] / Expected

F= [78 ~ 50] / 50

F =0.56

4c-

In this situation, the dominant allele is S, the inbreeding coefficient is 0.56. Therefore, here are quick chances that dominant allele will reach to the maximum frequency level 1.

4d-

Present time heterozygocity = F x S =0.47 x0.53

The next generation heterozygocity would be-

H x F = 0.47 x0.53 x 0.56

=0.14

Therefore, the dominant allele will increase at the rate of 0.14.

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