Genetics (Hardy-Weinberg)

[regeneration]

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Hi,

In BR Bio, Chapter 9, Passage 5, you are given the following:

Red-green colored blindness is a sex-linked recessive trait:

8% of males have it, <1% females have it, and 15% females are carriers.

The question given is what % of females would be expected to exhibit color blindness. My solution to this was (0.08*0.15*0.5), i.e. The chance that a male with the sex-linked trait would mate with a female who is a carrier (multiplied by 1/2 because a female could get the wildtype or the mutant allele from the mom); i'm ignoring females with the disease mating with males with the disease because the chance of that is an order of magnitude smaller.

BR's solution is simply (0.08)^2, and they are saying q = 0.08 (it seems they are just ignoring the female carriers).

Edit: Do we only count dominants for p and q, and count carriers as 2pq? In that case, it would make sense not to include the female carriers for p^2. What's wrong with my solution though?

Why do we ignore female carriers for X-linked recessive traits in females?

 

[chem4ever]

Hi,

In BR Bio, Chapter 9, Passage 5, you are given the following:

Red-green colored blindness is a sex-linked recessive trait:

8% of males have it, <1% females have it, and 15% females are carriers.

The question given is what % of females would be expected to exhibit color blindness. My solution to this was (0.08*0.15*0.5), i.e. The chance that a male with the sex-linked trait would mate with a female who is a carrier (multiplied by 1/2 because a female could get the wildtype or the mutant allele from the mom); i'm ignoring females with the disease mating with males with the disease because the chance of that is an order of magnitude smaller.

BR's solution is simply (0.08)^2, and they are saying q = 0.08 (it seems they are just ignoring the female carriers).

Edit: Do we only count dominants for p and q, and count carriers as 2pq? In that case, it would make sense not to include the female carriers for p^2. What's wrong with my solution though?

Why do we ignore female carriers for X-linked recessive traits in females?

Does the question tell you that the population is in Hardy-Weinberg equilibrium? If yes, then if 8% of the males have it, that means that 8% of the alleles are mutant (because it only takes one mutant allele for the males so if there are 100 males in a population and 8% have the trait, then 8 of the alleles must be mutated. I wouldn't be able to split up the males and females like that but if you're in HWE, then everything is distributed statistically perfect.)

So if 8% of the alleles are mutant, and females need both alleles to have the trait, then the probability of a female having the trait is 0.08 * 0.08 = (0.08)^2

Hope I didn't confuse you too much.