1) Using what you know from observing the F2 progeny, go back to your observations of the phenotypes of the F1 generation. What patterns of inheritance can you discern?
· Is the trait sex-linked?
· What is the dominant version of the trait, and what is the recessive version?
2) Formulate two hypotheses, H0 and HA (the null hypothesis and the alternate hypothesis). These should have the following characteristics:
· They are mutually exclusive (they cannot both be true)
· Together, they are all-encompassing (one or the other MUST be true)
· The null hypothesis makes a specific, quantitative, Mendelian prediction for each phenotypic category
Example:
H0: seed pod color is determined by a single gene that exhibits simple dominance, and green is dominant to yellow. The offspring of a monohybrid cross will be in the phenotypic ratio of 3:1 (3 green: 1 yellow)
HA: seed pod color is NOT determined by a single gene that exhibits simple dominance. The offspring of a monohybrid cross will NOT be in the phenotypic ratio of 3:1 (3 green: 1 yellow)
3) Build a table of observed and expected values:
|
Phenotype |
Observed # |
Expected proportion |
Expected # |
Chi squared |
|
|
|
|
|
|
|
|
|
|
|
|
|
Sum: |
|
1 |
|
|
4) Calculate the chi-squared value for each category, and then sum for all categories.
Chi2 = (observed # - expected #)2/expected #
5) Compare your calculated chi-squared value (for the sum of all categories) to the critical value from a chi-squared table (which will be provided).
· Use a cutoff of alpha = 5%
· Degrees of freedom = # of categories minus 1
6) Make a decision about your null hypothesis.
· If your value is less than or equal to the critical value, accept H0
· If your value is greater than the critical value, reject H0 and accept HA
7) Draw a biological conclusion.
· If you accepted H0, what does this mean? ("The data support the conclusion that..." note that you haven't proven anything)
· If you rejected H0, what might be a reasonable, alternate biological explanation? CHECK to be sure that this explanation is consistent with the data.
|
Degrees of freedom |
Alpha = 0.10 |
Alpha = 0.05 |
Alpha = 0.025 |
Alpha = 0.01 |
|
1 |
2.71 |
3.84 |
5.02 |
6.63 |
|
2 |
4.61 |
5.99 |
7.38 |
9.21 |
|
3 |
6.25 |
7.81 |
9.35 |
11.3 |
|
4 |
7.78 |
9.49 |
11.1 |
13.3 |
|
5 |
9.24 |
11.1 |
12.8 |
15.1 |
|
6 |
10.6 |
12.6 |
14.4 |
16.8 |
|
7 |
12.0 |
14.1 |
16.0 |
18.5 |
|
8 |
13.4 |
15.5 |
17.5 |
20.1 |
|
9 |
14.7 |
16.9 |
19.0 |
21.7 |
|
10 |
16.0 |
18.3 |
20.5 |
23.2 |
Values are from An Introduction to Biostatistics, by T. Glover and K. Mitchell. McGraw-Hill publishers. New York, 2002.
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