Hardy Weinberg and X-linked conditions

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

What if the condition was x-linked?

Lets use hemophilia as an example This is an X-linked recessive condition Possible male genotypes are:

XHY or XhY In the H-W system, male genotypes are no

different than just “p” and “q” p2 and q2 are not possible in males!

Females and X-linkage

Females can be

XH XH or XH Xh or Xh Xh

Females therefore fit all three HW genotypic frequencies (p2 & 2pq & q2) since they are diploid for X chromosomes

The frequency of hemophilia in woman is 16%. Calculate all other percentages

16% is basically q2 of HW√.16 = 0.4 q = 0.4 p = 0.6Female Carriers:

2pq = 2(0.6) (0.4) = .48 48% Female Normal

p2 = (0.6)2 = 0.36 36%

What about the boys?

q = 0.4 p = 0.6Male Afflicted:

Same as q 0.4 40% Males Normal

Same as p 0.6 60%

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