00:01
Huntington's disease is autosomal dominant.
00:04
And we're looking at this.
00:06
If individual 4 has three children, the normal man, what is the chance that two of them will have the disorder? so it's dominant.
00:14
You only need one copy to have the condition.
00:18
So let's look, first of all, at our individual.
00:24
They have the condition.
00:25
So we know they have at least one dominant allele.
00:28
So i'm going to have capital a as being for huntington's, little a as being normal.
00:36
Okay, but if we look at this person's parents, one of them had huntington's, the other didn't.
00:44
So the other must have given them a normal allele.
00:47
This person has to be a heterozygote.
00:50
The only way they could be homozygous is if both are their parents at huntington's.
00:57
Okay, so they are heterozygous.
00:59
They are having three children with a normal man, so this is the cross we're getting.
01:05
How do we work out the probability that two of their three children will have the disorder? first, we need to calculate the chance that one of them will have it.
01:14
So if we perform this cross, we can see that there are two possible outcomes...
