Question

A mole has two alleles for fur color, B = brown fur and b = white fur. In a population of 100 moles, 12 have white fur.  a) What is the brown fur allele frequency?  b) The mole population is at Hardy-Weinberg Equilibrium. How does the gene pool of the population change over time? 

Answer 1

I'll assume $\mathrm B$ is completely dominant over $\mathrm b$. Recall the Hardy-Weinberg equations:

$\begin{cases}p^2+2pq+q^2=1\\p+q=1\end{cases}$

where $p$ represents the allele frequency for brown fur, or the number of copies of the allele $\mathrm B$ within the population; and $q$ represents the allele frequency for white fur, or the number of copies of $\mathrm b$. In the first equation, the squared terms refer to the frequencies of the corresponding homozygous individuals, while $2pq$ is the frequency of the heterozygotes.

We're told that 12 individual moles have white fur, so we know for sure that there are 12 homozygous recessive individuals, which means

$q^2=\dfrac{12}{100}=\dfrac3{25}\implies q=\dfrac{\sqrt3}5\approx0.346$


from which it follows that

$p=1-\dfrac{\sqrt3}5\approx0.654$

Over time, H-W equilibrium guarantees that the allele frequencies $(p,q)$ do not change within the population. For example, suppose we denote the frequency of the $\mathrm B$ allele in generation $n$ by $p_n$. Then


$p_n={p_{n-1}}^2+\dfrac{2p_{n-1}q_{n-1}}2$

That is, the frequency $\mathrm B$ in the $n$-th generation has to match the frequency of $\mathrm B$ attributed to the $\mathrm{BB}$ and half the $\mathrm{Bb}$ individuals of the previous generation.

$p_n={p_{n-1}}^2+p_{n-1}q_{n-1}=p_{n-1}(p_{n-1}+q_{n-1})=p_{n-1}$

The same goes for $q_n$.