Question

You test-cross plants you suspect to be AaBb dihybrids and look at 100 offspring, with the following results: Phenotype AB 22 Ab 28 ab 29 ab 21100 total (for the purposes of this quiz, do not combine parental and recombinant classes) You wish to test the hypothesis that the data are best explained as a dihybrid test cross AaBb x aabb. 1)For a chi-square test, what expected number would you use for the aB class? 2)Calculate the chi-square value for these data given the model that the results are from a dihybrid test cross of two genes on separate chromosomes. X2 value=?? 3)Using the chi-square table provided, determine the probability that the model would give the observed data. Enter the answer as your best estimate of a single probability value between 0 and 1.0 4)Given this probability, would you accept or reject your model? Select one: a. Accept b. Reject You collect more plants, so you've now looked at 1000 offspring, with the following results: Phenotype AB 222 Ab 280 aB 280 ab 218 1000 total (for the purposes of this quiz, do not combine parental and recombinant classes) Calculate the chi-square value for this larger dataset given the model that the results are from a dihybrid test cross of two genes on separate chromosomes. X^2 value: determine the probability that the model would give the larger observed dataset. Enter the answer as your best estimate of a single probability value between 0 and 1.0.

Answer 1

1) For a dihybrid test cross, the expected ratio is 1:1:1:1. So, the ratio expected for aB will be (100/4) = 25

2)

Null hypothesis: Ho: The population is assorting independently.

Alternate hypothesis: H1: The population is not assorting independently.

Please note that test cross expected phenotypic ratio should be 1:1:1:1 in case of independent assortment (part ‘b’. Here, this is not the case.

	Observed	Expected	(O-E)^2	(O-E)^2/E= chi square value
AB	22	25	9	0.36
Ab	28	25	9	0.36
aB	29	25	16	0.64
ab	21	25	16	0.64
Total	100			2

Calculated value = 2 (chi-square value)= Answer 2

Degree of freedom = (rows-1) (columns-1) = (4-1)* (4-1) = 9

The value for 9 degree of freedom is not given in the table. So, cannot be calculated from your table.

From internet resources,

Tabulated value = taken from chi square table = 16.91

Level of significance (given) = 0.05

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

Here, tabulated value>calculated value, so, null hypothesis should be accepted.

So, the population is assorting independently.