Question

Use the chi-square test for goodness of fit to determine if the data shown below are consistent with an inheritance pattern of simple dominance. Use p = 0.05 for determining significance.

Cross: Aa x Aa

Offspring:

70 dominant phenotype

30 recessive phenotype

A. The data are consistent with a simple dominance pattern of inheritance.

B. The data are not consistent with a simple dominance pattern of inheritance.

Answer 1

The chi-square test is a test for determining whether variation in two sets of data (maybe under differing conditions) holds any significance or not. Regarding this particular instance, there is two hypotheses that can be postulated: (a) the null hypothesis that data here does not deviate enough from the normal pattern of inheritance, and (b) the alternate hypothesis: the dta observed is signifying an alternated mode of inherotance differnt from the simple dominance pattern of inheritance. . To do this with this question, we need two sets of data: one is the expected number of progenies for each phenotype and the other is the observed number of progenies for each phenotype (given in the question). To find out what to expect let's draw a Punnett square of the cross: Aa X Aa:

Gametes	A	a
A	AA (dominant)	Aa (dominant)
a	Aa (dominant)	aa (recessive)

So, we see that a cross between Aa and Aa should result in 3/4 offsprings with dominant phenotypes, and 1/4 offsprings with a recessive phenotype. Hence, Out of 100 ( the total number of samples that were observed), 3/4 of 100=75 offsprings is expected to have the dominant phenotype, and 1/4 of 100= 25 offsprings are expected of having recessive phenotype. So, now that we have two sets of data we can finally tabulate in the chi-square table:

Phenotype	Observed (O)	Expected (E)	(O-E)	(O-E)2	(O-E)2 / E
Dominant	70	75	-5	25	0.333
Recessive	30	25	5	25	1
	1.333

So, the X²value get is the calculated X²value =1.333. Now, to be able to determine the significance, we need to compare it with chi-square value from the chi-square table corresponding to degrees of freedom (df) governing this test. The df is calculated as number of categories minus 1. Since there are two categories here (dominant and recessive), df=2-1=1. The X² value for df=1 for p=0.05 is 3.84 which is greater than the calculated X²value. Since calculated X²did not exceed the tabulated X²value, the change is not significant, and the null hypothesis stays (it could not be rejected).

Thus, (A) the data are consistent with a simple dominance pattern of inheritance.