Mendelian Genetics

Question

In humans, color blindness and hemophilia are two sex linked recessive characters. A normal (not color blind or hemophiliac) woman whose mother was colorblind and whose father has hemophilia marries a color blind, normal clotting man. They produce a son and a daughter. Assuming that crossing over has not occurred during meiosis, what is the probability that:

a. the son is color blind but not hemophiliac
b. the daughter is phenotypically normal
c. the son is color blind and has hemophilia
d. the son is phenotypically normal
e. the daughter is color blind but not hemophiliac

Answer 1

The "c" superscript stands for color blindness, and the "h" stands for hemophilia. It's recessive X-linked so the mom needs two X^c to be color blind whereas the father only needs one to have hemophilia.

 

(X^cX^c ) ×  [X^h Y]

 

       (X^cX^h )

(phenotypically normal woman)

Carrier of both color blindness and hemophilia

 

-- > marries a color blind man

 

           (X^cX^h) × [X^cY]

 

(X^cX^c) ; (X^hX^c ) ; [X^cY] ; [X^hY]

 

50% daughters have color blindness

50% daughters have normal phenotype but are carriers of both disorders.

50% of sons have color blindness

50% of sons have hemophilia

 

that means c. and d. are 0% chance