Genetics Q - Hardy Weinberg and Calculating Allele Frequencies

[mineralwater]

Hey everyone,

 

This is the question:

 

"Two spaceships land on Planet Zork. One holds 50 people from an entirely brown-eyed population. The other carries 50 from a Hardy-Weinberg population in which there is eye colour variation. 32 of the passengers have brown eyes and 8 blue. The blue eyed allele, b, is obviously recessive to brown, B.

What is the frequency of the alleles B and b on the planet immediately after the landing?"

 

That's the exam question. First of all, I'm going to assume that there is an error in the question, because 8 + 32 =/= 50. I'm going to change it to 18 blue eyed people. And I correct in assuming this is a typo?

 

Calculating the frequency of the b allele is simple; because b is recessive to B, the 18 individuals who have blue eyes must be homozygous for b at the eye colour locus. Because each of the blue-eyed individuals has 2 copies of the b allele, and because the total number of people on the planet is 50 + 50 = 100, the frequency of b homozygotes must be 2(18) / 100 = 36 / 100 = 0.36.

 

The 2 HW equations:

p + q = 1

p^2 + 2pq + q^2 = 1

 

We have just calculated the frequency of bb homozygotes in the HW population. This is represented in the equations as q^2. So q^2 is 0.36.

We can use this to calculate q = [sqrt]q^2 = [sqrt]0.36 = 0.6.

So q = 0.6, which represents the frequency of b.

 

If we know the frequency of b, we can calculate the frequency of B in the HW population by using p + q = 1.

p = 1 - q

= 1 - 0.6

= 0.4.

So the frequency of B in the HW population is 0.4.

 

There is no eye colour variation in the entirely brown-eyed population. So I am going to assume that they are all BB homozygotes. Am I correct in making this assumption?

The frequency of B in the entirely brown-eyed population must then come to 1.

The overall frequency of B must therefore be (1 + 0.4) / 2 = 0.7

The overall frequency of b must be (0 + 0.6) / 2 = 0.3.

 

So I have calculated the overall frequencies of alleles b and B to be 0.3 and 0.7 respectively.

 

Are there any mistakes in my methodology/final answer?

Is there a simpler method to calculator this?

 

Many thanks in advance.

Edited April 29, 2012 by mineralwater

[John]

For your first assumption, a distribution of 32 and 18 in the second population leads to much nicer numbers than (32 or 42) and 8, so you're probably correct there. Given the wording of the question, your other assumptions are probably fair too. I also don't see any errors in your calculations or reasoning, except that

 

Because each of the blue-eyed individuals has 2 copies of the b allele' date=' and because the total number of people on the planet is 50 + 50 = 100, the frequency of b homozygotes must be 2(18) / 100 = 36 / 100 = 0.36.[/quote']

is a bit unclear to my eyes. If you look at the frequency of homozygous recessive individuals with respect to the total number of people, you get .18, not .36. However, I assume this isn't what you meant, since it doesn't mesh with your later addition of 100% B frequency on the first ship with the lower B frequency on the second ship.

Unless I'm blind and/or crazy, your final answer is correct.