Bio101r4 Fruit Flies Lab Week2 1 
David Khal
Process of Science
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
31
33
2
4
0.1212
Sepia eyes
13
11
2
4
0.3636
2 (to the nearest ten-thousandth)
0.4848
Questions
1. Why is it important to remove the adults in the parental generation?
To keep the tests accurate, it is important to separate the adults from the parental generation so you know you are only crossing the F-1 flies. 2. What generation will their offspring be? F2 generation.
3. Based on the data obtained, is the cross in Case 1 monohybrid or dihybrid? Explain.
The cross in case 1 has to be monohybrid since there was only one trait passed. Instead of a dihybrid cross which would have involved two observed traits. 4. Is the cross in Case 1 sex–linked or autosomal? Explain.
Autosomal, since the patterns are the same for both the males and the females.
5. Based on the data obtained, is the most likely mode of inheritance in Case 2 autosomal or sex–linked? Explain.
Sex linkage, because the pattern of inheritance is not the same for males and females.
6. From the data presented, what is the genotype of the parental (before the F1 generation; not shown here) generation?
X+X and X+Y
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 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?
The probability value for 1 degree of freedom in the 0.05 row is 3.84 and the chi-squared value is 0.4848. There is little statistically significant difference between observed results of the experiment and the expected results.
Process of Science
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
31
33
2
4
0.1212
Sepia eyes
13
11
2
4
0.3636
2 (to the nearest ten-thousandth)
0.4848
Questions
1. Why is it important to remove the adults in the parental generation?
To keep the tests accurate, it is important to separate the adults from the parental generation so you know you are only crossing the F-1 flies. 2. What generation will their offspring be? F2 generation.
3. Based on the data obtained, is the cross in Case 1 monohybrid or dihybrid? Explain.
The cross in case 1 has to be monohybrid since there was only one trait passed. Instead of a dihybrid cross which would have involved two observed traits. 4. Is the cross in Case 1 sex–linked or autosomal? Explain.
Autosomal, since the patterns are the same for both the males and the females.
5. Based on the data obtained, is the most likely mode of inheritance in Case 2 autosomal or sex–linked? Explain.
Sex linkage, because the pattern of inheritance is not the same for males and females.
6. From the data presented, what is the genotype of the parental (before the F1 generation; not shown here) generation?
X+X and X+Y
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 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?
The probability value for 1 degree of freedom in the 0.05 row is 3.84 and the chi-squared value is 0.4848. There is little statistically significant difference between observed results of the experiment and the expected results.