Man has 2/3 chance of being a carrier. (He does not have the disease). Woman carrier risk must be calculated with p^2 + 2pq + q^2. q^2 = 1/40,000 q = 1/200, p is roughly = 1 2pq = 1/100 = Carrier frequency .
Risk of having a child thus equals 2/3 x 1/100 x 1/4 = 1/600 Because: 2/3 = Man Carrier risk 1/100 = Female Carrier Risk 1/4 = Chance they each pass on the recessive gene to their offspring.
This the same as @keycompany's explanation with more explanation those that need it...
The patient is unaffected by the particular autosomal recessive disease and his brother has the autosomal recessive disease --> this means each of their parents carries the AR allele. Now, for the patient to be unaffected given that both his parents have the AR allele, it means that the patient has a 2/3 chance that he is simultaneously unaffected and a carrier.
The patient's partner is unaffected and normal Hardy-Weinberg genetics (as stated in the problem)..when this is specifically mentioned, you are to assume that the partner has a carrier frequency for the AR allele, which equal to 2pq.
The disease has a frequency in the population of 1/40000. This is q^2 --> q^2 = 1/40000. Solving for q, you get q = 1/200.
Carrier frequency is 2pq. However, for a rare disease, 2pq ≈ 2q. So, the carrier frequency for the partner = 2q = 2 * 1/200 = 1/100.
Now, the Q asks about an offspring of the patient and his partner being affected, so to have an affected child, there is a 1/4 chance of having an affected child (becasuse the patient and the partner both have a 1/2 chance of passing the allele --> you multiple these together 1/2 * 1/2).
So, multiply 2/3 * 1/100 * 1/4 = 1/600. This is P(patient being a carrier) * P(partner being a carrier) * P(having an affected offspring together).
The probability of the father being a carrier is 2/3 since it is known that he doesn’t have the disease. Then the probability of him passing it on to his kid is:
1/2 * 2/3 = 1/3
With the Hardy-Weinberg Principle, you can figure out the probability of the mother being a carrier:
q = sqrt(1/40,000) = 1/200
2pq = 2 * 1/200 * 199/200, which approx is 1/100, and the probability of the child getting this allele is
1/100 * 1/2 = 1/200
1/200 * 1/3 = 1/600
why there is probability of father being a carrier is 2/3
We assume this is a recessive disease. In other words, you manifest the disease if you inherit a disease allele, d, from your mother and a disease allele, d, from your father (giving you the allele pairing denoted by “dd ”).
dd = you have the disease.
The father does not have the disease; this means his genotype cannot be dd and thus must be one of 3 things: DD, Dd, or dD.
Notice that in only two of these 3 possibilities (⅔) does the father (potentially) have a disease allele, d, that he might pass on to his progeny.
submitted by ∗drdoom(1203)
Calculations for dad. The probability of the father being a carrier is 2/3 since it is known that he doesn’t have the disease. Then the probability of him passing it on to his kid is 1/2, thus:
Calculations for mom. With the Hardy-Weinberg Principle, you can figure out the probability of the mother being a carrier:
q = sqrt(1/40,000) = 1/200
2pq = 2 * 1/200 * 199/200, which is approx 1/100.
For the child to get the allele from mom, two things need to happen: (1) mom must be a carrier [“heterozygote”] and (2) mom must pass the allele to child:
Puting it all together. Now, combine all together:
= (probability of dad being carrier) * (probability of dad passing on disease allele) * (probability of mom being carrier) * (probability of mom passing on disease allele)
2/3 * 1/2 * 1/100 * 1/2
1 in 600