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MCAT Biology : Population Genetics and Hardy-Weinberg

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Example Question #31 : Population Genetics And Hardy Weinberg

A rare recessive mutation causes rabbits that are normally white to be pink. If one out of every 625 rabbits is pink, what percentage of the population is heterozygous?

Possible Answers:

Correct answer:

Explanation:

We can use the Hardy-Weinberg equilibrium formulas to calculate the allele frequencies.

$\displaystyle p + q = 1$

$\displaystyle p^2+2pq+q^2=1$

We know that the frequency of homozygous recessive (pink) rabbits is $\frac{1}{625}$ $\displaystyle \frac{1}{625}$. This is equal to q^2 $\displaystyle q^2$ in the Hardy-Weinberg calculation. We can use this information to solve for $\displaystyle q$, the recessive allele frequency.

$q^2=\frac{1}{625}$ $\displaystyle q^2=\frac{1}{625}$

$q = \sqrt{\frac{1}{625}} = 0.04$ $\displaystyle q = \sqrt{\frac{1}{625}} = 0.04$

Now that we know the value of $\displaystyle q$, we can solve for the value of $\displaystyle p$.

p+q=1 $\displaystyle p+q=1$

$\displaystyle p = 1- 0.04 = 0.96$

The frequency of heterozygotes is equal to 2pq $\displaystyle 2pq$ in the Hardy-Weinberg calculation. Now that we know the frequency of each allele, we can complete this calculation.

$\displaystyle 2pq = 2(0.96)(0.04) = 0.077$

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MCAT Biology : Population Genetics and Hardy-Weinberg

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