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?

so chance that their daughter is colorblind= 0%

Step-by-step 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%

the answer is c just got it on ed

step-by-step explanation:

the number of possible outcomes that do not show a 1 on the first spinner and show the number 4 on the second spinner is 2.

spinner 1 spinner 2

1                     1

1                     2

1                     3

1                     4

2                   1

2                   2

2                   3

2                   4   = not 1, and 4

3                   1

3                   2

3                   3

3                   4 = not 1, and 4

only 2 outcomes have passed the given condition.

step-by-step explanation: