Can a colorblind father and normal sighted homozygous mother produce a daughter that is a carrier?

Answer 1

Color blindness is an X-linked trait. That means it is carried in the X chromosome, which differentiates whether a baby will be a girl or a boy. Women have two X chromosomes (XX), and men have an XY combination. If a woman is a carrier for color blindness, only one of her chromosomes will be affected (we'll call it a little "x"), and for that reason she will not be colorblind. Men, on the other hand, only have one X chromosome, so any time they carry the colorblindness gene, they will be colorblind.

A child inherits one chromosome from each parent. He/She will get an X chromosome from his/her mother, and an X from her father (if a girl) or a Y from his father (if a boy).

So, If a woman is a carrier, Xx, and a man is normal, XY, they have several different chances for different offspring:

XX (a normal girl)

XY (a normal boy)

Xx (a normal girl who carries the colorblindness gene)

xY (a colorblind boy)

The short answer is, that if a woman has a boy, he has a 50% chance of being colorblind.

Answer 2

Color blindness is an X-linked trait, so if a mother is a carrier of the trait on one of her two X chromosomes, her female children have a 50% chance of carrying the trait but a 0% chance of being color blind. Her male children will have a 50% chance of being color blind and carrying the trait. The only way a female can display the trait of color-blindness is if her father is color-blind and her mother is a carrier of the trait.

Answer 3

vary likely