Answer to Question #341240 in Genetics for Thanay kumar

Question

A cross was made between pure breeding plants differing in

two characters (colour and shape of seed) and the progenies

were inter-crossed to produce F2 population. Four types of

phenotypes; yellow round (1108), yellow wrinkled (382), green

round (366) and green wrinkled (140) were observed in F2.

Answer 1

In order to test for independent inheritance of genes, we perform a chi-square test. In the case described above, in the F2 population there will be such a distribution of phenotypes:

9:3:3:1

The total number of offsprings is 1996, so each phenotypic class will have the following number of offsprings:

• (1996/16)*9 = 1122.75 (against 1108);
• (1996/16)*3 = 374.25 (against 382 and 366);
• (1996/16)*1 = 124.75 (against 140).

Using the following formula, we calculate the chi-square test:

$χ^2 = Σ(E−T)^2/T$

(1108-1122.75)^2/1122.75 + (382-374.25)^2/374.25 + (366-374.25)^2/374.25 + (140-124.75)^2/124.75 = 0.19 + 0.16 + 0.18 + 1.86 = 2.39.

Based on the chi-square table, with a probability of 0.05 and a degree of freedom of 3, the chi-test value is 7.815, and if the calculated value is less than this value, then the null hypothesis is accepted.

The calculated value is actually less than the theoretical value, which means that the genes are inherited independently.