# bio 101 Fruit Flies Lab Week2

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What Can Fruit Flies Reveal about Inheritance?

Lab Notebook

Chi-Square test for Case 1

Phenotype

Observed No. (o)

Expected No. (e)

(o-e)

(o-e) 2 (o-e) 2 e Red eyes

3

Sepia eyes

1

2 (to the nearest ten-thousandth)

Questions

1. Why is it important to remove the adults in the parental generation? It is important to keep the generations separate so that you know you are crossing only F1 flies.

2. What generation will their offspring be? The new offspring are the F2 generation.

3. Based on the data obtained, is the cross in Case 1 monohybrid or dihybrid? Explain. The cross is monohybrid because only one trait –eye color– is involved. Further, the data show a 3 : 1 ratio. A dihybrid cross would have resulted in a 9 : 3 : 3 : 1 ratio.

4. Is the cross in Case 1 sex–linked or autosomal? Explain. Since the pattern seems to be the same for males and females, the inheritance is probably autosomal (the gene is not on a sex chromosome).

5. Based on the data obtained, is the most likely mode of inheritance in Case 2 autosomal or sex–linked? Explain. The pattern of inheritance is not the same for males and females (there are no females with white eyes). This is strong evidence for sex linkage.

6. From the data presented, what is the genotype of the parental (before the F1 generation; not shown here) generation? X+X+ x XwY

7. Determine the degrees of freedom. This is the number of categories (red eyes or sepia eyes) minus one. For the data in Case 1, what is the number of degrees of freedom? There are 2 categories minus 1. The degrees of freedom is therefore 1.

8. Find the probability (p) value for 1 degree of freedom in the 0.05 row. Compare this with the chi–square value you calculated in your Lab Notebook. What can you say about the null hypothesis? It is 3.84. Compare this to the chi-square value you calculated earlier. If the chi-square value calculated for your experiment is less than the value of p =

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