00:01
Okay, so here we're still working on mendel's experiments and heredity.
00:04
We're going to use the probability method to find our genotypes.
00:09
And in this case, we are doing a cross from a heterozygous and a homozygous recessive.
00:23
So if we actually go ahead and do this cross, we have four possibilities for genes from the heterozygous.
00:36
But just the one from the homozygous.
00:41
So we're having a smaller pool of possibilities here.
01:00
Okay, and so our total number is going to be out of four, whereas if we were doing the two, across between two of the homozygous, i guess, we would, of course, have the 16.
01:13
And so when we want to look at the probability method, we're looking for a recessive in a.
01:24
Okay, so how many possibilities of combinations are there for b? well, there's just two.
01:29
There's the heterozygous or the recessive.
01:36
Okay, so there's two of these, and we would multiply that by the one possibility for a, which is what we're looking for.
01:45
Okay, so that's going to give us two out of four, because the total number is four here, and that would be one -half.
01:52
Our homozygous, i guess, recessive in a.
01:54
So now we want to look for dominant in both genes.
02:08
Okay, so in this instance, how many ways can the ab gene be dominant? and that only occurs in one way.
02:24
Multiply that by the number of ways the b gene can be dominant.
02:27
And again, they can only be dominant in this heterosigous form.
02:31
So that means one -fourth are going to be dominant in both.
02:37
If we want recessive in both, we're basically going to do the same thing, right? there's one way that a can be recessive in this combination.
02:45
There's one way that b can be recessive in this combination.
02:49
So we get one fourth again.
02:53
And then, just to be a little bit more complicated, we're looking for recessive in either a or b.
03:00
Okay, so for this one, if we start with recessive in a, there'd be one way it can be recessive in a, and we multiply that by both ways that be.
03:17
Can be, which is going to be either headers, i guess, or recessive.
03:22
We're going to add that to the two ways that, basically all the ways that a can be multiplied by the recessive b allele.
03:46
Okay, then we have to be careful here because we're double counting when both are recessive.
03:57
So we also need to subtract one to make sure we don't double count.
04:03
So we're subtracting one time.
04:05
Basically, we're subtracting the number of times that we get aab, which in this crosses 1.
04:11
Okay, so that's 2 plus 2 minus 1 is going to be 3 out of 4.
04:20
Should have recessive features.
04:24
And either a or b.
04:25
If we go back to our little grid, we can see that that's true.
04:31
Okay, and so then moving on, b in some ways it's trickier, and in some ways it's easier.
04:37
We're doing the standard cross, where it's heterozygous in every gene.
04:49
Okay, and in this case, we have three genes, so it's going to be a total of 64 that we're working with.
04:59
And the first thing we want to know is recessive in all traits.
05:02
So we want to find the probability of a, a, b, b, c.
05:09
Okay, so there's one way we can get a, a, multiply that by the.
05:16
So one way we can get recessive b times the one way we can get recessive c, and that will be one out of 64.
05:26
So next we want recessive in c and dominant in the other two.
05:35
Okay, so we're looking for a, anything, dominant in b, and then recessive in c.
05:44
So there's three ways we can have dominance in a.
05:48
Homozygous or kind of both flavors of heterazegis...