Question

Anthropologists examined the gene for Rh blood type in two villages on the northern coast of Nunavut, Canada. This blood type has two alleles, "plus" and "minus". The "plus" allele had a frequency of 60% in Taloyoak, and a frequency of 50% in Kugaaruk. For the remaining part of the question, we will assume that the two villages have the same population size.

Step 1a: For Taloyoak, what is the frequency of the "minus" allele?

Step 1b: For Kugaaruk, what is the frequency of the "minus" allele?

Step 2a: What is the average allele frequency for "plus" in the total population including both villages?

Step 2b: What is the average allele frequency of the "minus" allele in the total population including both villages?

Step 3a: What is the heterozygosity in Taloyoak?

Step 3b: What is the heterozygosity in Kugaaruk?

Step 4: What is the average of the heterozygosities in Taloyoak and Kugaaruk (HS)?

Now go back to the average allele frequencies you calculated in step 2a and 2b.

Step 5: What is the heterozygosity based on the total population allele frequencies (HT)?

Step 6: Calculate FST = (HT-HS)/HT.

Answer 1

Step 1(a):

Since a person can either have Rh plus or minus allele, the frequency of minus allele in Taloyoak is

(100 - 60)% = 40%

Step 1(b):

Since a person can either have Rh plus or minus allele, the frequency of minus allele in Kugaaruk is

(100 - 50)% = 50%

Step 2(a):

Average allele frequency for "plus" for both the villages is (60 + 50)/2 % = 55%.

Step 2(b):

Average allele frequency for "minus" for both the villages is (40 + 50)/2 % = 45%.

Step 3(a):

The expected heterozygosity is given by 2(freuency of plus allele)(fequency of minus allele).

So, expected hetrozygosity in Taloyoak is = 2(0.40)(0.60) = 0.48

Step 3(b):

The expected heterozygosity is given by 2(freuency of plus allele)(fequency of minus allele).

So, expected hetrozygosity in Kugaaruk is = 2(0.50)(0.50) = 0.50

Step - 4

Now, the average of the heterozygosities in Taloyoak and Kugaaruk = (0.48+0.50)/2 = 0.49

So, HS = 0.49

Step - 5

Heterogygosity based on total population = 2(0.55)(0.45) = 0.495

So, HT = 0.495

Step 6:

FST = (0.495 - 0.49)/0.495 = 0.1010...