Question

100 POINTS! A hypothetical population of 10,000 humans has 6840 individuals with the blood type AA, 2860 individuals with blood type AB, and 300 individuals with a blood type BB.

What is the frequency of each genotype in this population?

What does the frequency of the A allele?
What is the frequency of the B allele?
If the next generation contain 25,000 individuals, how many individuals would have blood type BB, assuming the population is in Hardy-Weinberg equilibrium?

Answer 1

The Hardy - Weinberg equations are meant to connect a statistical model to primarily Biological characteristics. This is a classic example of how these equations should be used.

This equilibrium has 5 conditions that must be obeyed. Two of the most important are that your sample must be a closed sample which means that no new members are let into the tested group, and none of the group can leave. The second condition is that in order to declare equilibrium the second sample taken must confirm the first. So this is very theoretical.

Equations
A + b = 1
A^2 + 2Ab + b^2 = 1

Method
The examples given always start with the recessive characteristics. In this case, there were 300 out of 10000 that were bb. So you have to figure out that if bb = 300/10000 what does b equal.

b^2 = 300/10000 = 0.03
b = sqrt(0.03)
b = 0.1732

From this result, using A + b = 1 you get that
A = 1 - b
A = 1 - 0.1732
A = 0.8268

Finding AA and Ab
A = 0.8268
A*A = 0.8268^2
AA = 0.6836

Ab = 0.8268 * 0.1732
Ab = 0.1432
2Ab = 2*0.1432
2Ab = 0.2864

Confirming Hardy Weinberg
If the sample has 10000 members total, then
AA = 0.6836 * 10000 = 6836 members.
2Ab = 0.2868 * 10000 = 2868
bb = 0.03 * 10000 = 300
The total is 10000 as expected.

Do these results follow the second Hardy Weinberg Equation
AA + 2Ab + bb = 0.6836 + 0.2868 + 0.03
AA + 2Ab + bb = 1.0004 which can be explained as a rounding error.

If the sample next year is 25000, what should the results be?
The sample states that in the second year, the sample size is 25000. Equilibrium will be established if the 25000 is calculated with the same ratios.

AA = 0.6836 * 25000 = 17090
2Ab = 0.2868 * 25000 = 7170
bb = 0.03 * 25000 = 750

Confirmation
17090 + 7170 + 750 = 25010 which again can be a rounding error.
Note: you can get a confirmation by just multiplying the numbers you got for 10000 by 2.5
For example bb = 2.5 * 300 = 750
For 2Ab = 2868 * 2.5 = 7170
I'll leave AA to you

Answer 2

Here's what we know:
10,000 individuals
6,840 individuals have blood type AA
2,860 individuals have blood type AB
300 individuals have blood type BB

AA genotype frequency: 68.4%
AB genotype frequency: 28.6%
BB genotype frequency: 3%

The A allele occurs 6,840 * 2 + 2,860 * 1 = 16,540 times, which is a frequency of 82.7%, meaning the B allele occurs 3,460 times, which is a frequency of 17.3%. In the next generation, 3%, or 750, individuals would have blood type BB.