Question

Information for question # 21,22, and 23. Allele frequencies at a wing length locus are measured for a natural population of a migratory cricket species as P(A1) = 0.4 and Q (A2) = 0.6. Observed genotype frequencies in the field population are shown below. Compute the expected frequency of the three genotypes if the population is in Hardy-Weinberg Equilibrium for this locus?

Observed frequency Expected frequency Show your work

A1A1 = 0.28

A1A2 = 0.22

A2A2 = 0.50

21. The expected frequency of the genotypes are:

a) A1A1 =0.25 A1A2 = 0.50 A2A2 = 0.25

b) A1A1 =0.28 A1A2 = 0.22 A2A2 = 0.50

c) A1A1 =0.16 A1A2 = 0.48 A2A2 = 0.36

d) A1A1 =0.36 A1A2 = 0.48 A2A2 = 0.16

e) None of the above

22. Compute the value of the inbreeding coefficient (F) for this population. The inbreeding coefficient is

A) 0 B) 0.043 C) 0.43 D) 0.54 E) 1

SHOW YOUR WORK IN THIS SPACE

23. Based on your calculation is there evidence for assortative mating in this population?

a) Yes b) No c) Not enough information

EXPLAIN YOUR ANSWER IN THIS SPACE

Answer 1

So we have been given allele frequency in the question. Value for A1 is 0.4 and for A2 it us 0.6.

Now to calculate expected frequency according to hardy weinberg law, the formula is :-

p2 + 2pq + q2​​​​​​

In the question p = 0.4 and q = 0.6

So it becomes :

(0.4)2 + 2 (0.4)(0.6) + (0.6)2

0.16 + 0.48 + 0.36

So the expected frequency are :-

A1A1 = 0.16 A1A2 = 0.48, A2A2 = 0.36

So the right option is Option (C)

Question 22 :-

To compute the value of inbreeding coefficient, the formula is :-

1 - Observed frequency of heterozygote / Expected frequency of heterozygotes

So putting the values from the question, we get

1 - 0.22 / 0.48

= 0.54

So 0.54 is the inbreeding coefficient.

The right option is Option (D) 0.54

Question 23 :-

Assortative mating means that there is a preference in the population for a particular phenotype. Let say A1 prefer A1 and A2 prefer A2. In assortative mating, this result in accumulation of homozygous alleles and the variations which usually occur due to random mating and heterozygosity it is reduced.

So the right option is Option (a) Yes

Reason :- In assortative mating, there's a preference for similar phenotype and similar traits among mates. This result in high frequency of similar alleles (Homozygous) while heterozygosity declines. In our question, Observed Heterozygous frequency is 0.22 which is very low than expected frequency of 0.48. Similarly there's a very steep rise in A2 homozygous. Thus A2 individuals are involved in assortative mating and it is directly affecting the frequency of Homozygous A1 and Heterozygous individuals.