equation for chi aka (x^2) x^2= sum of [((observed- expected)^2 )/ expected] I k

Question

equation for chi aka (x^2)

x^2= sum of [((observed- expected)^2 )/ expected]

I know how to solve of chi, but how do you know the expected. The ratio of the genotypes are not 9:3:3:1, so how do you find the expected number of offspring?

3832 E01-Homework week #3 1. You've performed a dihybrid testcross for the unlinked A and B genes. Below are the numbers of offspring in each genotypic class. A/a B/b A/a b/b a/a B/b a/a b/b 97 102 112 89 400 a) What is the X2 value for this data? (As with all math-based problems... you need to show your work. Shown work is a non-insignificant part of the points you'll earn on math-based questions ALWAYS.) b) OK. So do these results significantly differ from the expected results? (yes or no) 2. The loci for the D and G genes of a plant are 20 map units apart on chromosome #3. If a cis heterozygote were testcrossed, roughly what percentage of the offspring would express the dominant D phenotype but the recessive G phenotype? Make sure you explain yourself/show whatever work there is to show.

Explanation / Answer

Please note that the expected ratio will not be 9:3:3:1. Since it is a test cross, so the expected ratio will be 1:1:1:1.

The original cross is: AaBb X aabb

This will give AaBb, Aabb, aaBb and aabb in the ratio 1:1:1:1.

So, the table will be:

Chi square value = 2.78 (Answer to part 'a')

Degrees of freedom = (no. of rows -1) (no. of columns -1) = (4-1) (4-1) = 9

Tabulated value (at 0.05 level of significance) = 16.919 (from chi-square table).

Null hypothesis is accepted when calculated value is less than tabulated value.

Here calculated value (2.78) < Tabulated value (16.9)

So, null hypothesis is accepted.

b) No, these results do not differ significantly from expected results. And this data is statistically significant.

observed expected (O-E)^2 (O-E)^2/ E AaBb 97 100 9 0.09 Aabb 102 100 4 0.04 aaBb 112 100 144 1.44 aabb 89 100 121 1.21

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