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Answer:
See the answer below
Explanation:
<u>From the problem, color is X-linked. Males have just one (XY) chromosome while females have two (XX). Hence, for a recessive trait, males will need only one allele to be affected while females will need two alleles.</u>
Dominant form allele for normal vision = C
Recessive form allele for color deficiency = c
Genotype of a man with normal color vision =
Genotype of a woman with color deficiency =
Crossing the two:
x
<em>normal female normal female affected male affected male</em>
(a) All their sons would be affected for color deficiency. Hence, the probability of them having a color-deficient son is 100% or 1.
(b) None of their daughters is supposed to be affected for color deficiency. Hence the probability of them having a color-deficient daughter is 0.
<em>If the woman gives birth to a color-deficient but otherwise normal daughter and husband sues for a divorce on the ground of adultery, the case will definitely stand up in the court because none of their daughters can be affected for color deficiency based on genes of the man and the woman. </em>
<em>In order to produce an affected daughter, the father will need to supply an affected X chromosome in conjunction with the one that will be supplied by the woman. </em>
<em>In this case, the man does not have an affected chromosome, hence it is not possible for them to have an affected daughter except that the woman slept with another man who happens to have an affected X chromosome.</em>
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