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Hardy Weinberg and X-linked conditions

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

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Hardy Weinberg and X-linked conditions

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  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: 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!

  4. 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

  5. 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 = 0.6 • Female Carriers: 2pq = 2(0.6) (0.4) = .48 48% • Female Normal p2 = (0.6)2 = 0.36 36%

  6. 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%

  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

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