Hardy Weinberg principle

Question

The principle is that sum total of all allelic frequencies is 1.

Individual frequencies for example can be named p,q.In a diploid cell , p and q represent the frequency of allele A and a respectively . The frequency of AA individuals in a population is p² as the probability of an allele A with a frequency of p appear on both the chromosomes of a diploid individual is simply the product of the probabilities. Similarly for aa it will be q².Can anybody explain that how it will be 2pq for Aa?

I am getting it as pq.

Answer 1

You almost said it all! Here are two ways to think of this problem

Finding the missing probability

Let's denote the fraction yz genotype in the population with $f(yz)$. The sum of of fraction of all genotypes must be equal to 1. In equation it means

$$f(aa) + f(aA) + f(AA) = 1$$

Here, of course I don't make a difference between $f(aA)$ and $f(Aa)$ (whether A is inherited from the mother or from the father). Knowing as you said

$$f(aa) = p^2$$

and

$$f(AA) = q^2$$

and feeding this value back to the first equation result into

$$p^2 + f(aA) + q^2 = 1$$

Solving for $f(aA)$ yields to

$$f(aA) = 2pq$$

Directly calculate the probability

Another to reach this result is to ask the following questions

What is the probability that the maternal allele is A?

The answer is $q$

What is the probability that the paternal allele is a?

The answer is $p$

What is the probability that the maternal allele is a?

The answer is $p$

What is the probability that the paternal allele is A?

The answer is $p$

Therefore, $$f(aA) = pq + pq = 2pq$$