00:01
So in the 4 o 'clock plant, the wild have a little encode of red flower, whereas the recessive encode of white flower.
00:08
A true -breeding red flower plant is crossed with the true -riding white flower plants.
00:14
The f -1 generation were then self -fertilized to produce f2.
00:19
The following result were obtained.
00:22
So f -1 generation, or all pink flower plant as expected.
00:27
So this is an incomplete dominance.
00:32
And f2 generation generate 55 red flowers, 95 pink flower, and 50 white flowered.
00:41
With this data, answer the following question.
00:44
Propose a hypothesis that allows us to calculate the expected value based on medallion inheritance pattern.
00:50
And then test the goodness of fit between the data and your hypothesis using a kynx4 test and interpret what your calculated kai square result means.
01:01
So in this case, let's use two letters to represent the two alleles.
01:06
Let's say capital r is dominant, it stands for red flower, and lower r stands for white.
01:14
So since it's an incomplete dominance, when you have a heterozygars, it's going to show a paint.
01:20
So we have three different genotype.
01:23
Homozegas, capital r, red flower, heterozygous.
01:27
Zyghurs, kaptal r, lower r, pink, and homo zegars llor are white.
01:36
So in this case, we have this cross that have a true -breeding red flower crossed with the true -breeding white flower.
01:44
So the cross, the parental generation is going to be homo -zegers -kapt -r, cross with homo -zegars lower -r.
01:50
So that the two alleles are the same, so it will only produce one type of gamete, which is cap -to -r.
01:58
And for the white flower, again, it produced one type of gammy, which is lower art.
02:04
So all f1 is going to be capital r and lower r, one allele from each of the parent, and 100 % pink.
02:15
Now for f2, the f1 self -fertilized.
02:19
So the idea is that we have capital r, lower r cross with another capital r lower r.
02:25
So the two allele separate, and then you have punny square, capital r, lower r on top, cap to r on the side.
02:50
So according to panis square, one fourth of the progeny is going to be homozygars, capital r, this is red, and half heterozygars pink, and one fourth homozygars recessive, lower r, white.
03:09
So if we look at the ratio of the three phenotypes, it's one to two to one ratio.
03:22
Now since we have this expected ratio, let's take a look at our questions.
03:27
So in our case, we have 55 red flower versus 95 pink flowered and 50 white flowered in f2 generation.
03:36
So a total f2 can be calculated by 55 plus.
03:46
Plus 95 plus 50 equals 200.
03:55
Now according to our, the ratio, one of two to one ratio, let's calculate expected value.
04:08
So the red flower is gonna be one fourth of 250.
04:19
Heterozygars pink, half of 200.
04:27
So we expect to see 100 pink and then homozygars recessive white again is one fourth of 200, which is again 50.
04:39
So these are the expected value.
04:42
Let's do the kite square value to verify our hypothesis.
04:47
So our null hypothesis obviously is red versus pink versus white is 1 to 2 to 1 ratio.
05:07
So now we're going to verify this no hypothesis.
05:11
So let's complete this chi -square chart.
05:17
So the very first column we call it phenotype.
05:28
So you have three phenotype, red, pink, and then you can have a total underneath.
05:45
The next column, we called it observed value.
05:49
So this comes from the question.
05:52
What are the three observed value? we have 55, red, 95, pink, and 50...
