Question

Both Martha and her mother are normal (not color-blind or anemic) and Martha's mother is not a carrier for either disorder. Martha's father is color-blind and anemic. George and Martha marry and have a daughter with normal phenotype named Kate. Question A. If George and Martha have another child, what is the probability the child will be a son with normal phenotype?

Answer 1

1/4

Step-by-step explanation:

The probability of having a son with normal phenotype would be 1/4.

Both color blindness and anemia are X-linked and are linked together. The genes for X-linked disorders are found on the sex chromosomes with the male being XY and female being XX. The male child gets his X chromosome from the mother and Y from the father while the female child gets one of her X chromosome form the father and the other from the mother.

For recessive disorders, two recessive alleles are needed by the female on the two X chromosome to become affected while only one allele is needed on one X chromosome of the male to become affected.

Assuming the allele for color blindness is a and that of anemia is b.

George is color blind but not anemic, the genotype would be
`X^(aB)Y`

Martha is normal but her dad is color blind and anemic, this means that Martha is a carrier for both disorders with genotype
`X^(AB)X^(ab)`

Crossing the two genotypes

`X^(aB)Y` x
`X^(AB)X^(ab)`

offspring:
`X^(AB)X^(aB), X^(aB)X^(ab), X^(AB)Y, X^(ab)Y`

`X^(AB)X^(aB)` = normal phenotype female

`X^(aB)X^(ab)` = color blind, non anemic female

`X^(AB)Y` = normal phenotype male

`X^(ab)Y` = color blind, anemic male

Probability of having a son = 1/2

Probability of producing a normal phenotype child = 1/2

Hence,

Probability of a son with normal phenotype = 1/2 x 1/2 = 1/4.