2. Colorblindness is a sex linked trait on the X chromosome. Its frequency s 5%

Question

2. Colorblindness is a sex linked trait on the X chromosome. Its frequency s 5% in the population. (a) A male is color blind if he received the allele for colorblindness. What fraction of the male population is colorblind? (b) A female is color blind if she received two alleles for colorblindness. Assuming that the probability of receiving the allele is 5% independently from each of her parents, what fraction of the female population is colorblind? (c) Assuming that the population is made up 1/2 men and 1/2 women, find the fraction of the population that is color blind. (d) If a person chosen at random is color blind, find the probability that the person is female. (e) If six randomly chosen individuals are color blind, find the probability one or fewer are female.

Explanation / Answer

Given p(X=X0) (Probability of X chromosome carrying Color blind allelle) = 5% = 0.05,

So q = 1-p = 0.95

a) Males are XY, So they can be color blind if they receive colorblind X from mother. ie

p(X0Y) = p(X0) = 0.05 = 5%

b) Females are XX so

p(X0X0) = 0.05 * 0.05 = 0.0025 = 0.25%

c) 5% males and 0.25% females are colorblind, So fraction of colorblind = (5+0.25)/2 = 2.625%

d) Let there be x males and x females.

Number of colorblind females = 0.0025x

Number of colorblind males = 0.05x

Total color blind = 0.0525x

P( female given color blind) = 0.0025x/( 0.0525x ) = 0.0476 = 4.76%

e) P( female given colorblind ) = 4.76%

So P( male given colorblind ) = 1-4.76% = 95.24%

P( one or fewer males in a group of 6 colorblind ) = P( 1 colorblind female ) + P( No colorblind female )

=6C1 * (0.0476) * (0.9524)^5 + 6C0 * (0.0476)^0 * (0.9524)^6 = 0.9701 = 97.01%

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