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A recessive disorder affects 1 person in 1600 in a large population that is in Hardy-Weinberg…

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Try it on paper first. A hint, the key idea and the worked solution are here when you need them.

A recessive disorder affects $1$ person in $1600$ in a large population that is in Hardy-Weinberg equilibrium.

What is the frequency $q$ of the recessive allele?

What per cent of the people are carriers?

How many carriers are expected in a town of $20000$ people?

Need a reminder? The key ideaShow

Chi-square is $\sum \dfrac{(O - E)^2}{E}$ , with the classes minus 1 degrees of freedom. A value below the critical value (3.84, 5.99, 7.81 for 1, 2, 3 degrees of freedom) means that the ratio fits. For Hardy-Weinberg, $p + q = 1$ and $p^2 + 2pq + q^2 = 1$ . The affected share is $q^2$ and the carriers are $2pq$ .

The worked solution

Have you tried it?

Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.

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