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 +0  (nbme22#34)

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

So, 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

Thus:

1/200 * 1/3 = 1/600

brill45  You did this right but I think made it a bit more complicated than needed to be. To figure out the likelihood of a random person in the population passing on a recessive allele, you just need to do squareroot(q2) to get to q. And just use that q to multiply with the 1/3 you got earlier. The q value tells you the allele frequency in the population, whether the person is a carrier or homozygous recessive both. You still got it, but just letting you know the easier route in case you see it again!




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