1. Hardy Weinberg and X-linked conditions

2. Thus far… Hardy Weinberg Problems we have completed implied diploidy The traits that we analyzed were autosomal traits There is no variation in frequencies of p and q or genotypic frequencies for males vs.. females for autosomal conditions I.E. If tongue rolling has a frequency of 0.8 in males, it is also 0.8 in females

3. What if the condition was x- linked? Lets use hemophilia as an example This is an X-linked recessive condition Possible male genotypes are: X H Y or X h Y In the H-W system, male genotypes are no different than just “p” and “q” p 2 and q 2 are not possible in males!

4. Females and X-linkage Females can be X H X H or X H X h or X h X h Females therefore fit all three HW genotypic frequencies (p 2 & 2pq & q 2 ) since they are diploid for X chromosomes

5. The frequency of hemophilia in woman is 16%. Calculate all other percentages 16% is basically q 2 of HW √.16 = 0.4q = 0.4p = 0.6 Female Carriers: 2pq = 2(0.6) (0.4) =.4848% Female Normal p 2 = (0.6) 2 = 0.3636%

6. What about the boys? q = 0.4p = 0.6 Male Afflicted: Same as q0.4 40% Males Normal Same as p0.660%

7. Tips… In X linked recessive conditions The value for q is the same as the frequency in males In X linked dominant conditions The value for p is the same as the male frequency