Use the Hardy-Weinberg equation

1. Take the square root of 1/1600, and that will give you the frequency of the recessive allele = 1/40.
2. Calculate the frequency of the dominant allele with p+q=1, which is p= 0.975.
3. They are telling you to calculate the frequency of the disease carriers, which is with the equation 2pq.
4. They want only the disease carriers in which deletion is present. To calculate this, use the q value (1/40) and multiply by 80% in this should give you 0.02.
5. Finally, calculate for 2Pq 2 (0.975)(0.02)= 0.04 = 1/25.

hate this whole scramble thing: In one line: 2 * q * 80%.

This is for diseased individuals (two q alleles).I = 1/1600 = q^2 The frequency of q = 1/40

Now carriers is 2 p q. P is close to one assuming HW eqm (p+q=1). There's an additional step in this question due to the two different mutations.

so 2(q) = 1/20. 80% of these carriers are deletions so multiply 1/20 * 0.8 = 1/25

p + q = 1 p^2 + 2pq + q^2 = 1 if q^2 = 1/1600 = 0.00063 then q = sqrt(q^2) = 0.025 solve for p to get p = 1 - r = 1 - 0.025 = 0.975 the heterozygous carriers = 2pq = 1 - p^2 = 1 - 0.95 = 0.5 q^2 can be dropped b/c it's much smaller than p^2. The deletion is responsible for 80% of the mutations. 0.8 x 0.5 = 0.04 = 4/100 = 1/25

There might be an easier way to do this, but it worked for me.