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ANSWERS
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
Red eyes
31
33
2
4
(o-e) 2
e
0.1212
Sepia eyes
13
11
2
4
0.3636
0.4848
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.
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Copyright © 2010 Pearson Education, Inc., publishing as Pearson Benjamin Cummings
The cross is autosomal. 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 mode of inheritance is sex–linked. Notice that 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+ 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 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. If the chi–square value calculated for Case 1 is less than the value of p = 0.05,
the null hypothesis is valid. Since x2 = .4848, your results support the hypothesis.
Although your observed results are not precisely 3:1, there is little variation from the
expected results. The probability of obtaining results so close to the expected merely by
chance is less than 0.05 or 5%. The deviation in your experiment is not regarded as
statistically significant.
____________________________________________________________________________________
Copyright © 2010 Pearson Education, Inc., publishing as Pearson Benjamin Cummings