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=(12)/(100)=\frac3{25}\implies q=\frac{\sqrt3}5\approx0.346`


from which it follows that


`p=1-\frac{\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+\frac{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`.