Question

In garden peas, inflated pod shape (I) is dominant to pinched pod shape (i) and yellow pea colour (Y) is dominant to green (y) pea colour. Plants heterozygous for these characteristics were selfed (note that this is not a test cross), and the following numbers of phenotypes were seen in the progeny: 556 inflated, yellow; 193 inflated green; 184 pinched, yellow; 61 pinched green Test the data for the involvement of two independently assorting genes using the Chi-square test (don’t forget to state your Null Hypothesis).

Answer 1

Answer:

Total number of progeny = 994

IiYy x IiYy---Parents

Ii x Ii = I_ (3/4) & ii (1/4)

Yy x Yy = Y_(3/4) & yy (1/4)

Expected number of each phenotype.

I_Y_ = inflated pod and yellow pea = ¾ * ¾ = 9/16 *994 = 559.125

I_yy = inflate pod and green pea = ¾ * ¼ = 3/16 * 994 = 186.375

iiY_ = pinched pod and yellow pea = ¼ * ¾ = 3/16 * 994 = 186.375

iiyy = pinched pod and green pea = ¼ * ¼ = 1/16 * 994 = 62.125

Phenotype	Observed(O)	Expected (E)	O-E	(O-E)2	(O-E)2/E
inflated pod and yellow pea	556	559.13	-3.13	9.77	0.02
inflate pod and green pea	193	186.38	6.63	43.89	0.24
pinched pod and yellow pea	184	186.38	-2.38	5.64	0.03
pinched pod and green pea	61	62.13	-1.13	1.27	0.02
Total	994	994.00			0.30

Chi-square value = 0.30

Degrees of freedom = Number of phenotypes – 1

Df = 4-1=3

Critical value =7.81

The chi-square value of 0.30 is less than the critical value of 7.81. We can accept the null hypothesis and the two genes are assorted independently.