Question

Assume you have a population of stinkbug larvae scored for their genotypes at the

PGI-2 allozyme locus:

PGI-2a/PGI-2a: 125 individuals

PGI-2a/PGI-2b: 250 individuals

PGI-2b/PGI-2b: 125 individuals

Now assume that only 70% of the heterozygotes survive to adulthood and become stinkbugs.

a) Calculate the observed genotype and allele frequencies for the initial larvae.

b) Calculate the expected Hardy-Weinberg genotypic frequencies for the larvae. Are the

larvae in Hardy-Weinberg equilibrium?

c) What are the genotypic and allele frequencies among adults in the population?

d) Calculate the expected Hardy-Weinberg genotypic frequencies and determine if the

adults are in Hardy-Weinberg equilibrium.

e) Do you find your answer surprising? Explain why or why not?

Answer 1

Answer a :)

At larvae stage:

Observed genotype frequency of PGI-2a/PGI-2a = 125/500

Observed genotype frequency of PGI-2a/PGI-2a = 0.25

Observed genotype frequency of PGI-2a/PGI-2b= 250/500

Observed genotype frequency of PGI-2a/PGI-2b= 0.5

Observed genotype frequency of PGI-2b/PGI-2b= 125/500

Observed genotype frequency of PGI-2b/PGI-2b= 0.25

Observed allele frequency for allele PGI-2a =

Observed allele frequency for allele PGI-2a = 0.5

Observed allele frequency for allele PGI-2b = 1- 0.5

Observed allele frequency for allele PGI-2b = 0.5

Answer b :)

The expected Hardy-Weinberg genotypic frequencies are the same as the observed.

It is because the law of segregation suggests 1:2:1 genotype frequency and we are already getting the 1:2:1 ratio in the observed population. Therefore, the larvae are in Hardy-Weinberg equilibrium.

For alleles p and q, the Hardy-Weinberg equilibrium is p+q =1.

For 1:2:1 ratio

Genotype frequency pp = ¼

Genotype frequency pp = 0.25

Genotype frequency qq = ¼

Genotype frequency qq = 0.25

Genotype frequency pq = 2/4

Genotype frequency pq = 0.5

Allele frequency p = √pp

Allele frequency p = √0.25

Allele frequency p = 0.5

Allele frequency q = √qq

Allele frequency q = √0.25

Allele frequency q = 0.5

Answer c:)

At adults stage:

Genotype of allele Frequencies = Frequency x Survival rate

Therefore, genotype frequency of PGI-2a/PGI-2a = 0.25 x 70%

Genotype frequency of PGI-2a/PGI-2a = 0.175

Genotype frequency of PGI-2a/PGI-2b= 0.5 x 70%

Genotype frequency of PGI-2a/PGI-2b= 0.35

Genotype frequency of PGI-2b/PGI-2b= 0.25 x 70%

Genotype frequency of PGI-2b/PGI-2b= 0.175

Allele frequency for allele PGI-2a = 0.5 x 70%

Allele frequency for allele PGI-2a = 0.35

Allele frequency for allele PGI-2b = 0.5 x 70%

Allele frequency for allele PGI-2b = 0.35

Answer d:)

Here the expected Hardy-Weinberg genotypic frequencies are also same as answer b.

Answer e :)

Yes, we find our answer surprising because the observer frequencies in adults are showing that these frequencies are not in Hardy-Weinberg equilibrium.

In observed adult’s frequencies the addition of the two alleles’ frequencies PGI-2a and PGI-2b is:

0.35 +0.35 = 0.7

However, the Hardy-Weinberg equilibrium is 1 [p+q= 1]. The observed adults’ frequencies are 0.3 less than the Hardy-Weinberg equilibrium.

""""""""""""""Please rate the answer"""""""""""""""""