00:01
So, this is a cross about the small fish found in japan.
00:05
So, you know that you start out with a homozygous brown fish, which is controlled by two different alleles, or two different genes, b and r.
00:20
They're homozygous brown.
00:22
So, the parent, if you look at the genotype, is going to be homozygous capital b and capital r.
00:32
So, this is going to be brown.
00:38
And this homozygous brown fish was crossed with a homozygous white.
00:43
So, we know that the genotype is homozygous lower b and lower r.
00:52
So, then you'll have a f1.
00:55
So, this means that the brown parents will produce one type of allele or gametes, capital b and capital r.
01:01
So, the two alleles are the same.
01:04
So, you can tell that there's only one type of gamete produced by the brown.
01:10
The same with white parents.
01:12
The two alleles are the same.
01:13
So, lower b and lower r.
01:16
So, now you put the two gametes together.
01:18
The f1 pick up one allele from each of the parents.
01:20
So, the genotype of f1 is heterozygous b and heterozygous r.
01:26
It's still going to show brown.
01:30
The next step, you will back cross f1 with a homozygous white parent.
01:38
And you produce 260 brown, 253 blue, 249 red, and 238 white.
01:48
So, first one, you want to give the genotype of the back cross progeny.
01:52
So, we're talking about the f2.
01:55
So, this answer the first question.
01:58
Now, you have f1 parents, heterozygous for both and cross with the white.
02:05
Again, the genotype is homozygous lower b and lower r.
02:10
So, this is the back cross parent.
02:13
So, from this case, you can see that the f1 parents produce four different possible gametes because according to law of independent assortment, the two, or let's say law of segregation and independent assortment, the two allele of the same gene separate, then assort with another gene allele independently.
02:35
So, this means the capital b and lower b will separate from each other.
02:41
And same with capital r and lower r.
02:44
And b and r will assort independently.
02:46
So, which means you can have capital b and capital r, capital b, lower r, lower b, capital r, and lower b, lower r.
02:58
So, you have four different gametes.
03:00
They each have one fourth of the chance.
03:02
Now, for the white parents, again, it's a homozygous recessive.
03:06
So, there are only one type of gametes produced from the white parent.
03:11
So, it's a lower b and lower r.
03:12
So, these are the gametes production.
03:16
Now, once you have the gametes, you put the two gametes back together.
03:20
So, you have f2 first type, you have capital b, r with lower b, r.
03:26
So, you have capital b, lower b, capital r, lower r, the first kind.
03:31
Second, you have the second type of gametes.
03:33
So, you have capital b, lower b, lower r, lower r.
03:39
Or you may have a third gamete, lower b, lower b, capital r, and lower r.
03:45
Lastly, both recessive, lower b, lower b, lower r, and lower r.
03:50
So, again, these are all four genotype of the back cross, which is f2.
03:57
So, these are the genotype of the back cross progeny.
04:02
Now, let's take a look at the phenotype of each one.
04:05
You have both capital b and capital r.
04:08
This is going to be a brown.
04:11
The second one, you have a capital b, but homozygous lower r.
04:16
So, this is blue.
04:20
The third one has homozygous lower b, but with a capital r.
04:23
This is red.
04:25
The last one is going to be white.
04:27
Now, since the gamete is one to one to one to one ratio, these four f2 is also going to be the same one to one to one to one ratio.
04:38
So, each one fourth.
04:40
So, now we know theoretically you should see these four fish, they are in a one to one to one to one ratio.
04:47
So, let's do our punnett square, which is second question.
04:51
So, it says they use punnett square to, or use a chi -square test to compare the observed number of back cross was number expected.
05:00
So, in this case, let's say we have a total fish of 260 plus 253 plus 249 plus 238.
05:21
So, you get a total of 1000.
05:24
So, since each of the category should be one fourth of a total, so you would expect to see brown one fourth of 1000, which is 250.
05:41
And then blue also 250, red 250, and white 250.
05:48
So, from there, we calculate expected number.
05:50
So, let's calculate our chi -square chart.
05:54
The first one called the phenotype.
06:00
So, we have brown, blue, red, and white.
06:12
The second column we call observed.
06:15
This is what you actually observed.
06:18
So, we get 260 brown, and 253 blue, and 249 red, and lastly 238 white.
06:35
The next column called expected.
06:39
So, this is what we just calculated here.
06:42
So, we're going to have 250 each, because each has one fourth of the total.
06:50
The next column is called deviation.
06:53
So, this is going to be the difference between o and e...