Question

A student makes a monohybrid eross with Drosophila(fruit flies). She crosses two heterozygotes for

Answer 1

Assuming a significance level of 0.05, X² < P₀.₀₅, so there is no enought evidence to reject the null hypothesis. We fail to reject it. She got the predicted ratios.

Available data

• cross between two heter0zyg0us flies for eye color, Ww

• expected ratio 3:1

• Observed numbers 210 white-eyed and 680 red-eyed flies.

The total number of individuals is 210 + 680 = 890

Let us first take the expected numbers from the expected ratio,

4 -------------890

3 ------------X = 667.5

1 -------------X = 222.5

                        Red-eyed individuals      White-eyed individuals

Observed                     680                                   210

Expected                      667.5                                222.5

(Obs-Exp)²/Exp             0.234                               0.702

Now we need to get the chi-square value

X² = Σ(Obs-Exp)²/Exp

X² = 0.936

Degrees of freedom = n - 1 = 2 - 1 = 1

Significance level = 0.05

Critical value = 3.841

X² < P₀.₀₅

0.936 < 3.841

p value is greater than 0.05, so there is not enough evidence to reject the null hypothesis. The genotypes might be in equilibrium, and there might be independent assortment.

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