Probability applied to genetics

Please see the attached Word document for a detailed solution for this exercise. The solution shows how the frequencies of the overall sample is a function of the population stratification.

You are reviewing a scientific study where the epidemiologists sampled 1000 people from Charleston, South Carolina and measured variation at two loci, A and B. The haplotype frequencies in the sample population are:

AB = 0.406
Ab = 0.214
aB = 0.214
ab = 0.166

Toward the end of the article you are reviewing, you read that the 1000 people sampled were actually 600 non-hispanic whites [p(A)=.5, p(B) = .5] and 400 African Americans [p(A) = 0.8 and p(B) = 0.8]. Assuming linkage equilibrium within each ethnic group, you check to see if the haplotype frequencies (and thus linkage disequilibrium) found in the overall sample of 1000 is really a function of population stratification. What do you conclude? p="the probability."

Solution:

The key to understanding this problem is to calculate the probabilities separately for each of the two groups, non-Hispanic whites and African Americans. Then weight each of these two distributions by their proportions in the sample. Lastly, combine the weighted values into a single value. This should result in the relative frequencies given in your table. Complete details follow:

First create a table similar to the one above for non-Hispanic Whites only. You are given that P(A) = 0.5 and P(B) = 0.5.

This means that P(a) = 1 - P(A) = 0.5 and P(b) = 1 - P(B) = 0.5 since:

A and a are complementary events and their probabilities must sum to 1.
B and b are complementary events and their probabilities must sum to 1.

That means if you consider non-Hispanic whites only the table of relative frequencies should be:

P(AB) = P(A)P(B) = (0.5)(0.5) = 0.25
P(Ab) = P(A)P(b) = (0.5)(0.5) = 0.25
P(aB) = P(a)P(B) = (0.5)(0.5) = 0.25
P(ab) = P(a)P(b) = (0.5)(0.5) = 0.25

Next, create a table similar to the one above for African Americans only. You are given that P(A) = 0.8 and P(B) = 0.8.

This means that P(a) = 1 - P(A) = 0.2 and P(b) = 1 - P(B) = 0.2 since:

A and a are complementary events and their probabilities must sum to 1.
B and b are complementary events and their probabilities must sum to 1.

That means if you consider African Americans only the table of relative frequencies should be:

P(AB) = P(A)P(B) = (0.8)(0.8) = 0.64
P(Ab) = P(A)P(b) = (0.8)(0.2) = 0.16
P(aB) = P(a)P(B) = (0.2)(0.8) = 0.16
P(ab) = P(a)P(b) = (0.2)(0.2) = 0.04

Then you should combine these two distributions into a single distribution by weighting them by their proportions in the sample size. If we let the "Wh" represent non-Hispanic whites and "Af" represent African Americans, we know that a randomly selected individual from the group of 1000 has the following probabilities. P(Wh) = 0.6 and P(Af) = 0.4. When we combine the two distributions with the weight we get:

P(Wh)P(AB) = P(Wh)P(A)P(B) = (0.6)(0.5)(0.5) = 0.15
P(Wh)P(Ab) = P(Wh)P(A)P(b) = (0.6)(0.5)(0.5) = 0.15
P(Wh)P(aB) = P(Wh)P(a)P(B) = (0.6) (0.5)(0.5) = 0.15
P(Wh)P(ab) = P(Wh)P(a)P(b) = (0.6) (0.5)(0.5) = 0.15
P(Af)P(AB) = P(Af)P(A)P(B) = (0.4)(0.8)(0.8) = 0.256
P(Af)P(Ab) = P(Af)P(A)P(b) = (0.4)(0.8)(0.2) = 0.064
P(Af)P(aB) = P(Af)P(a)P(B) = (0.4)(0.2)(0.8) = 0.064
P(Af)P(ab) = P(Af)P(a)P(b) = (0.4)(0.2)(0.2) = 0.016

Finally combine the contributions of each of the ethnic groups from the table to get the relative frequencies shown in the original table.

P(AB) = P(Wh)P(AB) + P(Af)P(AB) = 0.15 + 0.256 = 0.406
P(Ab) = P(Wh)P(Ab) + P(Af)P(Ab) = 0.15 + 0.064 = 0.214
P(aB) = P(Wh)P(aB) + P(Af)P(aB) = 0.15 + 0.064 = 0.214
P(ab) = P(Wh)P(ab) + P(Af)P(ab) = 0.15 + 0.016 = 0.166.

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