# Genetic risk calculations Click on the image to see a larger version.

Solution:

a ) The frequency of homozygotes for the morbid allele is 0.0025
Let p be the frequency of the normal allele (N), 0<p<1
Let q be the frequency of the morbid allele (h), 0<q<1
p+q=1.

The frequency of the genotype NN is  p x p
the frequency of the genotype hh is q x q
and the frequency of the genotype Nh is pq+qp=2pq (since Nh and hN are the same genotype)

Now, we already have the frequency of genotype hh = qxq=0.0025
hence q is the square root of 0.0025 which is 0.05

Therefore, the frequency of heterozygotes in the population is 2pq = 2 (1-q)q= 2(1-0.05)0.05=0.095
this is the proba for the mother to be heterozygote, we should multiply this by 1/2 to get the proba of the child being homozygote for the morbid allele i.e. affected.

so the answer is 0.095 x 1/2 = 0.0475

b) If the couple already have an affected child, then the mother is surely heterozygote, no doubts about that or probabilities, so the risk for the other child to be affected is 1/2.

Let me know if you have any questions. (I know you’re not used to this kind of questions, but hey, that’s what the official exams are for )

c) The frequency at birth 1/40000 is actually q xq, therefore q is the square root of 1/40000 which is 1/200, p =1-q; hence p = 199/200,

The frequency of heterozygotes is 2pq, this is the proba for the mother to be heterozygote 2(1/200)(1/199)

The proba for the father to be heterozygote is 2/3.

Therefore, the risk for III-1 to be heterozygous = 2/3 x 2(1/200)(1/199) x 1/4

d) II-2 and II-3 are both heterozygotes as their father is affected. III-1 and III-2 can be heterozygotes if they received the morbid allele from II-2 and II-3 respectively OR if they received it from II-1 and II-4.  The proba for II-1 and II-4 to be heterozygotes is 2pq = 2(1/200)(1/199)

To sum up, there are 3 cases for each of the parents (III-1 and III-2) that would lead to them being heterozygote, we’ll consider III-1’s case since the calculation is the same for III-2:

1st case: II-1 is homozygote NN (proba of which is pxp = (1/199)x(1/199) and the morbid allele is transmitted from II-2

2nd case : II-1 is heterozygous Nm (proba of which is 2pq) and the morbid allele is transmitted from II-2

3rd case: II-1 is heterozygous and the morbid allele is transmitted by her

Add them up and you’ll get the proba of III-1 beign hetero

the same value applies to III-2

the child IV-1 has a risk of 1/4 x risk of III-1 being hetero x risk of III-2 being hetero

In conclusion, this problem is way beyond the scope of the official exams, it’s more suitable for students studying population genetics in college, so just disregard the possibility that you might face such a question in the Lebanese Bacc.