More genetic frequencies stuff, need help (ASAP)(2 questions) please show work also. Thank you very much.?

Question

1. Assume a population is in Hardy-Weinberg equilibrium for a given genetic autosomal trait. What proportion of individuals in the population are heterozyous for the gene if the frequency of the recessive allele is 1%?

2. Assume that the probability of a sex-linked gene for color blindness is 0.09=q and the probability of the normal allele is 0.91=p. This means tha tthe probability of X vhromosomes carrying the color blindness allele is 0.09 and the probability of X vhromosomes carrying the normal allele is 0.91.
a. What is the probability of having a color-blind male in the population?

b. What is the probability of a color-blind female?

Answer 1

Best Answer - Chosen by Voters

1. If Fr(a)=0.01=q, then Fr(A)=0.99=p, according to HW equation #1. Then, HW equation #2 says that the frequency of heterozygotes is given by the expression 2pq, which is 2(0.01)(0.99)=0.0198 or almost 2%.

2a. The probability of having a "bad" X is 0.09, and the probability of being a male is 0.5, so the probability of a male with a bad X is (0.5)(0.09)=0.0450

2b. The probabliity of hAveing a bad X is 0.09, and the probability of being a female is 0.5, so the probability of being a female with 2 bad X chromosomes is (0.5)(0.09)(0.09)=0.0041

iF WE ALREADY KNOW WE ARE DEALING WITH A MALE OR FEMALE, REMOVE THE 0.5 FACTOR...