Question

The answers are shown, but how do you get the answer to part IV ONLY. Please show all work and explain

NAME: 13. Consider a species that is divided into two subpopuiations that are each individually in Hardy- Weinberg equilibrium with the following genotype frequencies FIAA) FIAa) Flaa) 049 0.42 0.09 0.04 0.32 0.64 i. Assuming that the two subpopulations are of equa! size, calculate the overall genotype frequencies in the global population. Use your answers to complete the blanks in the table above. 3 pts] ii. Calculate the expected frequency for cach genotype in the global population to show that they are not the same as the observed values above and, thus, that the global population is not in Hardy Weinberg equilibrium. 3 pts] Expected F(AA)-045-0.2025 Expected F(Aa)-2 * 0.45 * 0.55-0.495 .Expected Faa)-0ss-0.302s ii. Given that there appears to be random mating (and Hardy-Weinberg equilibrium) within each subpopulation, what causes the global population to not be in Hardy-Weinberg equilibrium?[4 pts] Population subdivisione, the Wahlund Effect) iv. In Subpopulation, researchers have also measure allele frequencies at a second locus on a different chromosome and found that the frequencies ofthe ..В" and " h" alleles are 0.9 and 0.1, respectively. Assuming that this subpopulation is in linkage equilibrium, calculate the expected frequencies for each of the following haploid genotypes. 4 pts] Expected F(AB)-0.7 0.9-0.63 Expected F(Ab)-0.7 0.1 0.07 Expected F(aB)- 0.30.9-027 Expected F(ab)-0.30.1-0.03

Answer 1

In part IV, the frequencies of B and b are given 0.9 and 0.1.

In order to find out the expexted frequencies of AB,Ab and ab; we have to find out theindividual frequency of A and a.

AA represents homozygous genotype, having 2 copies of the gene.

To find out frequency of individual allele A for sub population 1 = √AA

A= √0.49 ( given)

= 0.7

Similarly for homozygous genotype aa ,

individual frequency of a =√aa

=√0.9

=0.3

Now, the frequencies of the haploid genotype can be calculated using the individual genotypes.

F(AB) = 0.7 X 0.9   

= 0.63

F(Ab) =0.7 x 0.1

= 0.07

F(aB) = 0.3 X 0.9

=0.27

F(ab) = 0.3 x 0.1

=0.03