## Transcribed Text

X^2=∑(observed-expected)^2/expected
Using chi squared requires that first you have a genetic hypothesis because the expected numbers are the number of organisms you expect based on the size of the data set and the hypothesis. The observed are what was recorded in a particular experiment. To illustrate this I will calculate the chi squared for the midterm question 3 assuming the transmission is through an autosomal recessive trait and then I will do this treated the sex of the child as part of the data. Finally I will compare these to the value I calculated on the KEY for the hypothesis that this is a sex-linked trait.
Based on this hypothesis we would expect 25% of the children in generations I and II to be affected and 75% to be unaffected. There are 9 children total hence the expect # of affected is 9x0.25=2.25 and the unaffected in 9X0.75=6.75. Assuming 1 degree of freedom (2data classes-1=1) this is an excellent fit well above the 0.05 threshold.
X^2=(2.25-3)^2/2.25+(6-6.75)^2/6.75=0.33
What happens if we consider the sex of the child and divide the data into 4 data classes?
Data class expected observed Observed-expected (o-e)2/e
Affected males in generation II and II 1 3 2 4
Unaffected males 3 1 2 4
Affected females 1.25 0 1.25 1.25
Unaffect females 3.75 5 1.25 0.41
X^2=(3-1)^2/1+(1-3)^2/3+(0-1.25)^2/1.25+(5-3.75)^2/3.75=9.66
Given 3 degrees of freedom (4 data classes-1=3) we still cannot reject this hypothesis since P falls between 0.05 and 0.01, but clearly it is not fitting the data as well as the sex-linked hypothesis does. Note the sex linked hypothesis had a chi squared of 1 which had a P value of 0.85 fairly close to the perfect fit of 1.00.
Hence lets go over some pointers:
How should you divide the data set? Consider the potential hypothesis and make sure you set up the chi squared to give you the most rigorous test of the hypothesis.
Only consider the data for the individuals you can assigned explicitly to a data class. You cannot without more information divide these progeny into homozygous dominate, and heterozygous.
Calculate the expected by multiplying the total number of individuals in your data set (as you defined in parts 1 and 2) by the fraction expected based on the hypothesis.
I will do one more example. I will calculate chi squared for the data given to you in question 2 on the midterm. Some of you noted this was not following the expected ratios hence lets investigate this.
Table 2 below shows the banding pattern for a cross between parents P1 and P2 producing the 6 progeny I1-6. The parents are double heterozygotes for two independently assorting genes.
Table 2
RFLP data
Size (kb) P1 P2 I1 I2 I3 I4 I5 I6
5.5 X X X X X X
3.5 X X X X X X X
2.5 X X X X X X
2 X X X X X X
Based on the question we would expect 1/4 of the progeny to be homozygous at both loci, ½ to homozygous at one loci and heterozygous at the other, and ¼ to be heterozygous at both loci. Now lets calculate the expected numbers.
Data class expected observed o-e (o-e)2/e
Homo at both 1.5 3 1.5 1.5
Hetero at both 1.5 2 0.5 0.167
Homo at 1 and hetero at the other 3 1 2 1.33
X2=4.5 and 2 degrees of freedom this is still a believable data set although clearly not perfect since the P value is about 0.125 which is above the 0.05 threshold
Problems:
Use the data from midterm question section 2.1 and calculate chi squared for the breeders original hypothesis.
Use the data in problem 5.18 from the text to calculate a chi squared for the linkage between the F and the G genes assuming that these are F1 progeny numbers from a cross to a triply homozygous recessive fly (fgh/fgh).
You are doing routine breeding experiments on Drosophila melanogaster strains collected from around the world. You have isolate a female fly, from a stock of flies capture near Chernobyl, Russia, that has a dominate mutation in a gene we will call W which makes the wings very small (tiny wings). Normally the wt version of this gene (+) found on one end of chromosome 2. You assume she is heterozygous for the mutation (W/+). You cross her with a normal male (+/+). You isolate the males with tiny wings and cross these with normal females (+/+). You note the crosses do not result in as many offspring as seen in other experiments. You get the following data. Explain what happen to the original fly’s chromosomes? Draw a picture of the chromosomes of the original fly.
Data class Number of progeny
Tiny wing males 0
Tiny wing females 205
Normal wing males 198
Normal wing females 0

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