# Question eeaeb

May 24, 2017

36% of the population

#### Explanation:

Let's first establish a few things.

The $\text{Hardy-Weinberg equation}$ can be seen in two forms:

color(white)(aaaaaaa)color(magenta)(p^2+2pq+q^2=1)color(white)(aaa)andcolor(white)(aaa)color(orange)(p+q = 1

Where:
color(magenta)(p^2 = "frequency of homozygous dominant genotype"
color(magenta)(2pq = "frequency of heterozygous genotype"
color(magenta)(q^2 = "frequency of homozygous recessive genotype"

and

color(orange)(p = "frequency of dominant allele"
color(orange)(q = "frequency of recessive allele"

$- - - - - - - - - - - - - - - - - - - - -$

You are told that 16 % of the population expresses the recessive trait. How can you express a recessive trait? Well, if both alleles of a gene are recessive (homozygous recessive), then the recessive trait is expressed. This is usually indicated with a lowercase letter $\left(\text{Ex. aa}\right)$

Since we know that 16 % of the population expresses the recessive trait and $\textcolor{m a \ge n t a}{{q}^{2}}$ represents the frequency of the homozygous recessive genotype, then we know the following

$\textcolor{w h i t e}{a a a a a a a a a a a a a a a} \textcolor{m a \ge n t a}{{q}^{2}} = 0.16$

Now we want to find the percentage of the population that would be homozygous for the domnaint trait so somehow we need to get from

$\textcolor{w h i t e}{a a a a a a a a a a a a a a a} \textcolor{w h i t e}{a} \textcolor{m a \ge n t a}{{q}^{2}} \to \textcolor{m a \ge n t a}{{p}^{2}}$

Steps

Take the square root of $\textcolor{m a \ge n t a}{{q}^{2}}$
- $\sqrt{\textcolor{m a \ge n t a}{{q}^{2}}} \to \sqrt{\textcolor{m a \ge n t a}{0.16}} \to q = 0.4$

Find p from q
- color(orange)(p+q = 1
- $p = 1 - 0.4$
- $\textcolor{\mathmr{and} a n \ge}{p} = 0.6$

Square $\textcolor{\mathmr{and} a n \ge}{p}$ t o get $\textcolor{m a \ge n t a}{{p}^{2}}$
${p}^{2} \to {\left(0.6\right)}^{2} \to 0.36$

So, $\textcolor{m a \ge n t a}{{p}^{2}} = 0.36$ meaning 36%# of the population wouuld be homozygous for the dominant genotype.