A female that is a carrier for colorblindness has children with a male that has normal vision.?

Question

What is the genotype of the mother?
What is the genotype of the fatther?
Using a Punnett square determine the population of the F1 generation.
What percentage of their boys will be colorblind?
What percentage of their girls will be colorblind?

Please simplify as best as you can on your answer, so I can understand it, I'm just learning Biology.
Thank you

Answer 1

First off, all females have two X chromosomes, so their genotype is XX. All males have an X chromosome and a Y chromosome, so their genotype is XY. Any female who is a carrier of the trait has only one allele for colorblindness, which is represented by XX+ (where the X+ represents the color blind allele). Any female who is colorblind has the allele for colorblindness on both chromosomes, represented by X+X+. Since color blind is an X-linked trait, males who have the allele for colorblindness express the trait no matter what; that is, there is no such thing as male carriers. Males can be X+Y (color blind) or XY (normal vision).

Genotype of mom= XX+
Genotype of dad= XY

For a Punnett Square, draw 4 boxes. The mom will be the row parent and the dad will be the column parent. Put the dad's genotype above the top 2 boxes, and the mom's genotype to the left of the 2 leftmost boxes. Then, for each box, look at the allele of the dad and the allele of the mom in that particular column/row and cross them to get one of the F1 generation's possible genotypes.

[If you already understand Punnetts, just ignore that long explanation]

After you make the Punnett Square, you will see that you get 4 different genotypes: XX (normal female), XX+ (carrier female), X+Y (color blind male) and XY (normal male).

1 of the 2 boys will be colorblind, so 50% are colorblind.
0 of the 2 girls will be colorblind, so 0% are colorblind.

Answer 2

You know that colorblindness is only on x chromosomes, right?

so C= color
and c= colorblind

I'm assuming that the mother can see well, but is only a carrier, right?

so mother's genotype: XCXc (the dominant trait rules over the recessive trait)

and the father's genotype: XCY (so not a carrier and not colorbind)

you put that into a punnet square (mother is XC and Xc, father is XC and Y)

children's genotypes are : XCXC, XcXC, XCY, and XcY

so none of the girls are colorblind. and 50 percent of boys will be colorblind and 50 percent will not.

Answer 3

'C' represents the allele for normal vision
'c' represents the allele for colour blindness

*an allele is different forms of the same gene
* "X" and "Y" are sex chromosomes

father's genotype: XcY

* the y chromosome does not carry any alleles

mother's genotype: XCXc

*only the normal vision is expressed because the alleles are in there heterozygous form and therefore only the dominant allele is expressed which in this case is "C" which is the allele for normal vision but because she has the "c" allele this means that she is a carrier for colour blindness and therefore can be inherited by the next generation offsprings (her children).

XcY XCXc

F1 XCXc----- XcXc----XCY------XcY

therefore:
XCXc--- female with normal vision but is a carrier

XcXc----female with colour blindness (recessive ['c'] homologus form

XCY----- male with normal vision

XcY------ male with colour blindness

Hence:

1/2 of the boys will be colour blind (50%)

1/2 of the girls will be colour blind

*

Answer 4

female: C c
male: C y since the male carries only 1 "x" chromosome

C y cross C c

1 CC has color vision, female
1 Cc has color vision, female
1 Cy has color vision, male
1 cy colorblind, male

25% will be colorblind, male

Answer 5

The gene responsible for colour blindness has not expressed, in that boy. But its presence is sure. It will certainly express, in future generation.