Answer to Question #343142 in Genetics for sahil

Question

A cross was made between pure breeding plants differing in two characters (colour and shape of seed) and the progenies were intercrossed to produce F2 population. Four types of phenotypes; yellow round (1200), yellow wrinkled (386), green round (372) and green wrinkled (144) were observed in F2. Use the chi-square test to determine if these results are due to independent assortment 

Answer 1

In the case of a dihybrid cross, the ratio of phenotypes in F2 will be 9:3:3:1.

If we count all the offspring in F2, there are 2102 in total. Theoretically, in each phenotypic class there will be such a number of offspring:

• (2102/16)*9 = 1182.375 (against 1200);
• (2102/16)*3 = 394.125 (against 386 and 372);
• (2102/16)*1 = 131.375 (against 144).

Using the following formula, we perform a chi-square test:

$χ^2 = Σ(E−T)^2/T$

(1200-1182.375)^2/1182.375 + (386-394.125)^2/394.125 + (372-394.125)^2/394.125 + (144-131.375)^2/131.375 = 0.26 + 0.17 + 1.24 + 1.21 = 2.88.

If we look at the chi-square table, with a probability of 0.05 and a degree of freedom of 3, the chi-square value is 7.815. For the null hypothesis to be accepted, the calculated value must be less than 7.815.

The calculated chi-square value actually turned out to be less than 7.815, which means that the null hypothesis is accepted in this case. Crossing results showed independent assortment of genes in gametes.