Question

1.) In a cross involving Drosophilia melanogaster, an F₂ population included 272 flies with long (normal) wings and 60 flies with dumpy wings. Calculate X² and fill in the blanks below. Do these results approximate a 3:1 ratio?

Phenotype	O	E	(O-E)	(O-E)2	(O-E)2/E
Normal
Dumpy
Totalts					X2=8.4980

a. In interpreting this X² value, you have ___________ degrees of freedom.

b. In this case, do you accept/reject the hypothesis that these data approximates a 3:1 ratio?

c. What is the probability that the deviations are due to chance alone.

Could you please give an explaination also. Thanks

Answer 1

Answer 1)

Expected individuals:

Normal: (3/4) ×332=249

Dumpy=(1/4) ×332=83

Now look in the table for chi square test:

The chi square experimental value is 8.498

A) Degree of freedom=n-1 =2-1 =1

Here n is number of phenotype= 2

so degree of freedom is 1.

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b) Now the chi square value from table at degree of freedom 1 and at p=0.005 is 3.841.

As our chi square experimental value is more than chi square value from table. That's why null hypothesis is rejected.

c) The probability of deviation is the difference between experimental and value from table.

= 8.498 - 3.841= 4.651