Transcript Hardy Weinberg and X

Hardy Weinberg and
X-linked conditions
Thus far…
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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 xlinked?
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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
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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
is basically q2 of HW
 √.16 = 0.4
q = 0.4 p = 0.6
 Female Carriers:
2pq = 2(0.6) (0.4) = .48 48%
 Female Normal
p2 = (0.6)2 = 0.36
36%
 16%
What about the boys?
q
= 0.4
p = 0.6
 Male Afflicted:
Same as q
0.4 40%
 Males Normal
Same as p
0.6 60%
Tips…
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In X linked recessive conditions
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The value for q is the same as the frequency in
males
In X linked dominant conditions
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The value for p is the same as the male frequency