00:01
So everyone, in this question we have been said that there are true breeding parents bearing yellow seeds that were crossed with plants bearing the green seeds.
00:14
So here in the f1 generation, the heterozygous condition will be formed and all the seeds will be yellow because yellow is the dominant allele here.
00:27
And now when this f1 generation was self -fertilized, the f2 progeny were produced.
00:38
And the f2 progeny produced will have the genotype, capital y, capital y, capital y, capital y, small y, small y and small y.
00:51
So here we can see that the phenotypic ratio is 3 is to 1.
00:58
That is yellow to green and the genotypic ratio here is 1 is 2 is to 1 so we have been given that in the f2 progeny 6400 are bearing the yellow seeds that means the phenotype diacus 3 is to 1 out of these 3 air are 6400 and 22 2 and 2 to 4 are bearing the green seeds.
01:34
So we need to perform a kai square test.
01:39
So here let us write the phenotypes that are yellow and green and let us here write the total number of the offsprings that are produced.
01:53
So the observed number of this progeny.
01:58
Here in the yellow seeds are 6 ,400 and in the green seeds are 2 to 4.
02:05
So the total will be 8 ,624.
02:12
Now we will find the expected.
02:15
So the expected will be as we obtain 3 ratio in the yellow.
02:21
So this here will be 3 divided by 4 because here 4 progenies are produced.
02:28
So it will be 3 divided by 4 multiplied by total, that is 8 ,624.
02:36
So this here will be equals to 6 ,468.
02:43
And this will be 1 divided by 4 multiplied by 8 ,624.
02:50
This here will be equals to 2 ,156.
02:56
Now we will find the observed minus expected.
03:02
So that here will be 6400 minus 6468...