00:02
So let's assume the f2 generation, the following result obtained from drosophila that hybrid cross.
00:09
So we get a wild type of 1 ,750.
00:17
A vestigial has 535 and ebony has 500 and both ebony and vestigial 150.
00:37
So the question asks you to write down the cross setup to obtain the result and make sure you deal with one cross at a time.
00:52
You can write down p, f1, f2 and also set up chi -square table with the result book and calculate the chi -square and p value for the result.
01:03
Now state whether you accept or reject your hypothesis.
01:07
So although i do not see the firm part of the question, i can deduce that you're talking about the cross of two genes.
01:15
So most likely you have a wild type fly crossed with a vestigial ebony double recessive mutant fly.
01:24
So this is going to be apparent.
01:26
So let's give each of the mutant a key.
01:32
Say the vestigial, we call this v and the wild type is a v plus because the vestigial is recessive allele.
01:46
And for ebony, same thing.
01:47
Ebony is going to be e and the wild type phenotype when it comes to ebony body is going to be e plus.
01:56
So with the keys, let's go ahead and write down the parent.
02:01
So the parent must be a homozygous dominant cross with homozygous recessive.
02:06
So you have a v plus, v plus, e plus, e plus.
02:11
This is going to be the wild type cross with vestigial wing and ebony body.
02:18
So v, v, e, e, both recessive.
02:28
Now for f1, each parents produce one type of allele, v plus, e plus.
02:37
This is again, come from the wild type parents and the vestigial ebony parent produce only one type of allele, v and e, because they're homozygous.
02:46
The two alleles are the same.
02:48
Now, once you put the two alleles back together, all f1 will have heterozygous v plus v, e plus e genotype.
02:58
And the phenotype is again, going to be wild type because you have a v plus and e plus dominant allele.
03:04
The next step, you are crossing f1 heterozygous.
03:10
This is the hybrid cross.
03:11
Now i'm not going to expand the entire punnett square because it's a 16 punnett square, which is exactly like the genetic book about the mendel's dihybrid cross with the pe plan, green and yellow, and round and wrinkled example.
03:33
So we know that when you have a heterozygous parent dihybrid cross, you will have a nine to three to three to one ratio at the end.
03:43
So i'm going to write down this ratio.
03:45
So for f2, you can tell that for each parents, it produced four different alleles or gametes, capital v, capital e, or capital v plus v plus, let's call it this, an e or v e plus and v e.
04:21
So you can see that for each parent, you can have four different gametes and then the other parent produced the same four.
04:35
So from there, you can actually do your 16 punnett square.
04:38
So you can fill in this chart.
04:53
You can see that the other parents produce the same type of gametes, v plus e plus v plus e, v e plus, and v and e.
05:04
So i'm just going to, i'm not going to do the entire thing, but you can see that, or you can see i put v and e and e back.
05:13
I'm going to only do one of it.
05:14
So i'm not going to finish the entire thing.
05:16
But as you can see that you can have a 16 square punnett square, but at the end, you'll have four different phenotype.
05:24
One is v plus e plus, at least one dominant allele.
05:28
This is going to be the wild type.
05:30
It has nine out of 16.
05:33
The second category was a v plus, but homozygous lower e, abney.
05:37
It will be three out of 16.
05:41
The next one will be v homozygous with e plus, which means it's going to be dominant, which will have a wild type abney, when it comes abney, allele gene, but vestigial only.
05:57
This is also three out of 16.
06:04
Lastly, v and e homozygous recessive.
06:07
This is going to be abney and vestigial.
06:14
It should be one out of 16 ratio.
06:19
Now, once we have that, we are able to predict what we have.
06:26
So we have a total f2, according to the question, 1750 plus 535 plus 500 plus 150.
06:48
So you get 2935 f2.
06:56
So out of all the offspring, the first one wild type, according to our ratio is nine out of 16 times the total 2935.
07:06
But you have about 1650.
07:14
The next category abney, it should be nine, sorry, three out of 16 times total 2935.
07:26
Then you should expect to see 550.
07:31
The next one vestigial.
07:37
So three out of 16 as well, we also expect to see about 550.
07:44
And lastly, you have e and b, this is going to be one out of 16.
07:59
So 2935 out of 16 is 183, 184.
08:13
All right, now, let's actually go ahead and do the chi square analysis.
08:21
The first is phenotype category...