00:01
So this question tells us that color blindness is an x -linked inherited disease, and it says that a woman who carries the color blindness gene on one of her x chromosomes, but not on the other, will have normal vision, and a man who carries a gene on his only x chromosome is colorblind.
00:17
It says that if the woman with normal vision who carries the color blindness gene on one of her x chromosomes has a child with a man who has normal vision, what is the probability that if they have a child it will be colorblind? so first of all, we'll solve this by using a punnet square, which i have on the screen here.
00:35
First we need to recognize that for women, their chromosomes are xx, and for men, they have one x and a y.
00:46
So it says that this woman has normal vision, the woman that we're dealing with, with the xx chromosomes on the top here, but she does carry the color blindness gene on one of her x chromosomes.
01:00
So let's just let us see here denote color blindness.
01:04
So there it is on one of her x chromosomes.
01:07
And it says that the man, for the x y chromosomes, on the left side of the square here, has normal vision.
01:14
It tells us in the first part of the problem that a man will have normal vision if he does not have color blindness gene on his x chromosome.
01:25
So that means that his x chromosome must not have that little seed, that color blindness trait.
01:31
So now we're ready to put this information into our punnett square.
01:38
So in our top left square, we'll have x with the color blindness trait from the mother and then we'll have another x.
01:46
So this would be a female that is not colorblind because as the problem said, for a female to be colorblind, both of the x chromosomes must carry the colorblind gene...