00:01
This question regarding the red -green colorblind is a recessive x -linked.
00:11
So look at the pedigree and figure out the chance of the couple in second generation produce a first child with a colorblind if it's a boy.
00:21
So let's say if it's x -linked recessive, this means that the first generation male must be a recessive x lower or y.
00:28
This is the recessive, show recessive phenotype.
00:31
Male has only one x chromosome.
00:33
Amazon.
00:33
Female x, at least one copy of dominant allele shows normal phenotype.
00:39
We have no idea about the second.
00:41
Now, if you looked at the second generation female, she is phenotypically normal, so she inherits a x capital r from the mother.
00:50
And she has no choice but to inherit the x lower r from the father, because the other x of this woman comes from her father, which has x lower r only.
01:03
So the woman is a heterozygous.
01:05
Now the second generation male show phenotype, so recessive phenotype.
01:11
That means this male must have x lower r y genotype.
01:16
Male only has one x chromosome, the allele decide its phenotype.
01:20
So from there we know that we have a cross between the heterozygous female and x lower r y male.
01:28
So let's do the pendant square.
01:30
The parents, the two alleles separate from each other.
01:42
X capital r and x lower r separate from each other.
01:46
And x lower r and y from father.
01:50
So then in f1 generation, you put the two alleles back.
02:00
So you can see the top one has two x chromosomes, female.
02:03
The bottom, male, one y chromosome.
02:06
So the first question says that what is the probability that the first child will be colorblind if it's a boy? so let's look at the bottom.
02:14
So if it's a boy, you have two possible genomes...