00:03
In the plant arabidopsis, a geneticist is interested in development of trichome, a small projections on the leaf.
00:10
A large screen turn up two mutant, you have a and b that have no trichomes.
00:16
The mutant seem to be potentially useful in studying the trichome development.
00:20
Now if they are determined by a single gene, we know that.
00:28
So let's name this single gene a, and then finding the normal and abnormal function of these gene will be instructive.
00:39
Now each plant was crossed with wild type.
00:41
In both cases, the next generation f1 had normal trichomes.
00:46
Then f1 plant were self -assumed and resulting f2 were followed.
00:52
So from mutant a, you observe 602 normal, then you also see 198 no trichome.
01:16
And in mutant b, you observe a lot less, 267 normal and 93 no trichome.
01:34
So the question says that select the genotype of the f2 plant that are consistent with the pattern inheritance.
01:40
So first of all, we know that it's a single gene.
01:43
So a single gene have two different forms.
01:46
So according to our f1, since f1, all of them have a wild type, so that normal trichome.
02:02
So this tells us that the wild type must be dominant.
02:12
So we named this gene capital a.
02:15
Now no trichome must be recessive because when you have a heterozygous, it doesn't show up.
02:30
So this is the conclusion we draw from f1.
02:33
The normal is going to be dominant allele wild type and no trichome is recessive, lower a.
02:40
So from there, we know it's a single gene.
02:43
And since f1 are all normal, we know that it's an autosomal recessive inheritance pattern.
03:01
So from there, we should be able to figure out the cross.
03:07
So first of all, the mutant is going to be homozygous, lower a cross with normal or wild type homozygous dominant a.
03:21
Now f1 are all heterozygous capital a, lower a...