Question

A dihybrid cross is performed in Drosophila - one parent is heterozygous for both genes (B/b ; F/f) and is test-crossed with a homozygous recessive type (b/b ; f/f). (where B = black body; b = brown body; F = forked bristles; f = unforked bristles.)

The results are:

black, forked 230

black, unforked 210

brown, forked 240

brown, unforked 250

How many degrees of freedom would you use in this case?

What would be the expected number for each phenotype?

Calculate the chi-square value for this cross.

Assuming p=0.05, what critical value would you use to compare the calculated chi-square value against?

Do you reject or not reject the expected ratio in this case

Answer 1

Answer:

Expected testcross progeny would be 1:1:1:1 ratio according to Mendel experiments.

Total progeny = 930

Expected values of each phenotype = ¼*930 = 232.5

Phenotype	Observed(O)	Expected (E)	O-E	(O-E)2	(O-E)2/E
black, forked	230	232.50	-2.500	6.250	0.027
black, unforked	210	232.50	-22.500	506.250	2.177
brown, forked	240	232.50	7.500	56.250	0.242
brown, unforked	250	232.50	17.500	306.250	1.317
Total	930	930			3.763

Chi-square value = 3.763

Degrees of freedom = number of categories – 1

Df = 4-1 = 3

Critical value at 0.05 = 7.815

The chi-square value of 3.763 is less than the critical value of 7.815. So we can accept the null hypothesis. Hence the population is not in Hardy-Weinberg equilibrium