Question

At an autosomal gene locus in humans, the allele for brown eyes is dominant over the allele for blue eyes. At another gene locus, located on the X chromosome, a recessive allele produces colorblindness while the dominant allele produces normal color vision. A heterozygous brown-eyed woman who is a carrier of colorblindness has a child with a blue-eyed man who is not colorblind. An ultrasound test shows that the child is a girl. What is the probability that she will be colorblind?

Answer 1

Answer:There is a 0% chance since the man is not a carrier of this gene

Step-by-step explanation:For this girl to be colour blind she will have to inherit two recessive genes (cc) for this disorder because this is a recessive allele, it requires two recessive affected genes to express itself , meaning one from her mother and one from her father but since her father doesn't carry such gene it is unlikely to express itself in the child.

Answer 2

so chance that their daughter is colorblind= 0%

Explanation:

colorblindness is a sex-linked recessive trait,

let the alleles be XC- normal, Xc- colour-blind

the woman is heterozygous for this gene, so her genotype is XCXc

man is normal so his genotype is XCY

( man is XY and woman is XX)

XCY * XCXc

see table at the attachment corner below

probability of getting a colorblind progeny given that progeny is a girl= number of colorblind girl/total number of girls=0/2=0

so chance that their daughter is colorblind= 0%