Question

For Drosophila Melanogaster how to determine whether the loci are assorting independently (lets say we are talking about two traits, ebony body and curved wing)

Would you have to perform a Chi-square testing for it?

Answer 1

Answer:

First we need to know the observed numbers.

Then we have to do testcross. The expected Dihybrid testcross ratio 1:1:1:1.

Based on the values, the table is prepared as below:

Phenotype	Observed(O)	Expected (E)	O-E	(O-E)2	(O-E)2/E
WT wing and body	44	45	3	9	0.15
Scalloped wing and WT body	42	45	-3	9	0.15
WT wing and ebony body	48	45	-2	4	0.2
Scalloped wing and ebony body	46	45	2	4	0.2
Total	180	180			0.7

Chi-square value = 0.70

Degrees of freedom = Number of catergories – 1

Df = 4 -1= 3

Critical value at probability of 0.05 = 7.81 (standard value)

The chi-square value of 0.70 is less than critical value of 7.81. We can accept the null hypothesis and the two genes are assorted independently.

----------------------------------------

Answer:

First we need to know the observed numbers.

Then we have to do testcross. The expected Dihybrid testcross ratio 1:1:1:1.

Based on the values, the table is prepared as below:

Phenotype	Observed(O)	Expected (E)	O-E	(O-E)2	(O-E)2/E
WT wing and bristles	44	45	3	9	0.15
apterous wing and WT body	42	45	-3	9	0.15
WT wing and spinless bristles	48	45	-2	4	0.2
apterous wing and pinless bristles	46	45	2	4	0.2
Total	180	180			0.7

Chi-square value = 0.70

Degrees of freedom = Number of catergories – 1

Df = 4 -1= 3

Critical value at probability of 0.05 = 7.81 (standard value)

The chi-square value of 0.70 is less than critical value of 7.81. We can accept the null hypothesis and the two genes are assorted independently.