# Why is it $q^2$ for the individual count in hardy weinberg?

My understanding:

In Hardy-Weinberg problems the frequency of a homozygous recessive genetic occurrence in a population is $q^2$. So if 1 in 100 people in a population have albinism (homozygous recessive disorder) then we say the frequency is $q^2=1/100$.

We then say, to find the frequency of the allele count $q$ that $q=\sqrt{1/100}= 1/10$.

I don't understand why we say this. Why would the allele count be the square root of the population frequency? There's 2 alleles per person. Why isn't it x2 or /2 instead? I suppose my problem is understand what exactly is p and q.

• p and q are allele frequencies. Did you search anything about Hardy-Weinberg equilibrium. – WYSIWYG May 30 '15 at 14:21
• Yes but I'm having problems seeing why why square them. – Paze May 30 '15 at 14:22
• Squaring is for genotypes. – WYSIWYG May 30 '15 at 14:24
• @paze If you think so,you are dramatically overestimating the difficulty of this question. – fileunderwater May 30 '15 at 15:15
• Okay . If p is probability of allele-A then what is the probability of A mating with another A; p×p=p² – WYSIWYG May 30 '15 at 15:28

First of, let me correct your equation: $q = \sqrt{\frac{1}{100}} = \frac{1}{10} = 0.1 ≠ 10$.

From allele frequency to genotype frequency

Imagine you were to randomly sample an allele from a population of allele where the allele A is present at frequency $q$. What is the probability that you draw allele A? Answer: $P(A) = q$. Now, put this allele back in the pool and imagine you have to draw two alleles. What is the probability that the two alleles are A. Well it is the probability that the first allele is A times the probability that the second allele is A, it is $P(A) \cdot P(A) = q \cdot q = q^2$.

From genotype frequency to allele frequency

Therefore, under Hardy-Weinberg conditions, if the allele A is at frequency $q$, then the homozygote genotype AA is at frequency $q \cdot q = q^2$. Now let's denote the frequency of the genotype AA by $f_{AA}$. You know that $f_{AA} = q^2$ from the above though experiment. If you take the square root of both sides, you get $\sqrt{f_{AA}} = \sqrt{q^2} = q$. In words, the frequency of the allele A is the square root of the frequency of the genotype AA.

Recessivity and dominance

Btw, you'll note that those calculations tell nothing about the patterns of dominance/recessivity of alleles (Hardy-Weinberg principle assumes no selection anyway).

• also 1/100 != 1/10 – shigeta Jun 1 '15 at 13:43
• Ha ha oops. I forgot the square root. Thank you! – Remi.b Jun 1 '15 at 15:35

To understand population genetics and the Hardy-Weinberg equilibrium you first need to clearly separate genotype frequencies (the frequency of individuals in a population having a genotype) and allele frequencies (the frequency of an allele in the population gene pool). In your question p and q represents allele frequencies, while p2, q2 and 2pq represents genotype frequencies (if $p=P(A)$, $p^2$ is $P(AA)$). Genotype frequencies can be thought of as the probability that two particular alleles are located in the same individual, and the calculation of these probabilities is a simple application of the multiplication rule of probabilites for independent events.

Therefore, if the allele frequency p is 0.2 the probability of having two copies of this allele is $p \times p = p^2 = 0.2^2 = 0.04$. If there are only two alleles in the population, q has to be 0.8, which means that the other two genotype frequencies are $2pq = 2\times0.2\times0.8 = 0.32$ and $q^2 = 0.8^2 = 0.64$. As you can see, these genotype frequencies also sum to 1, since they represent the genotypes of all individuals in the population,

In Hardy-Weinberg problems the frequency of a homozygous recessive genetic occurrence in a population is q2.

Sort of but not quite. If the locus has two possible alleles, $A$ and $a$, the frequency of each allele in the population is $p$ and $q$ respectively. If the population size ($N$) is 500 diploid individuals, there are 1000 loci ($2*N$, because of diploidy) which each could have either $A$ or $a$, and each of the 500 individuals could have the genotypes $AA$, $Aa$, or $aa$. If we know that the population of loci has 100 copies of $A$ then $p = 100/1000 = 0.1$, and $q$, the frequency of $a$, must then be $1-p = q = 0.9$.

The value of $q^2$ tells us how many individuals in the population, under the assumptions of HWE, will be of the genotype $aa$ - it says nothing about the recessivity (or dominance) of the allele, nor the phenotype. In the example above, $q^2 = 0.9^2 = 0.81$ - under assumptions of HWE 81% of the population will be genotype $aa$.

In terms of math it is simple probability theory. If we have a bag of coloured marbles, 5 red and 5 blue, the frequency of red ($p$) is 0.5, and blue ($q$) is 0.5. If we draw one marble from the bag, the probability of the marble being blue, $P(blue)$, is equal to the frequency of blue, $q$, which is 0.5. If we then return the marble to the bag and make a second draw, the probability of drawing a blue marble is again $b$, 0.5. The probability of both of the drawn marbles being blue $P(blue) * P(blue) = 0.5 * 0.5$, or $q^2$. The probability of drawing one red and one blue is $2 * p * q$, and probability of two red marbles is $p^2$.