Question
This test requires constructing a contingency table whose values are the number of synonymous versus nonsynonymous substitutions and polymorphisms in the gene of interest. For 10 points each:
[10h] Name this doubly-eponymous statistical test used to distinguish between positive and negative forms of natural selection by comparing variation within a species and between species.
ANSWER: McDonald–Kreitman test [prompt on MKT]
[10m] Departure from the null hypothesis of the McDonald–Kreitman test is quantified by an index named for this adjective. A theory of molecular evolution described by this adjective was developed by Motoo (“mo-TOH”) Kimura.
ANSWER: neutral [accept neutrality index; accept neutral theory of molecular evolution]
[10e] McDonald and Kreitman developed their namesake test while studying the gene that codes for alcohol dehydrogenase, which oxidizes this two-carbon alcohol produced during fermentation into acetaldehyde.
ANSWER: ethanol [or ethyl alcohol; or EtOH; or C2H6O]
<MS, Biology>
Summary
| Great Lakes | 2025-02-01 | Y | 6 | 6.67 | 67% | 0% | 0% |
| Lower Mid-Atlantic | 2025-02-01 | Y | 1 | 10.00 | 100% | 0% | 0% |
| Midwest | 2025-02-01 | Y | 1 | 10.00 | 100% | 0% | 0% |
| Northeast | 2025-02-01 | Y | 3 | 13.33 | 100% | 33% | 0% |
| Overflow | 2025-02-01 | Y | 3 | 13.33 | 100% | 33% | 0% |
| Pacific Northwest | 2025-02-01 | Y | 2 | 10.00 | 100% | 0% | 0% |
| South Central | 2025-02-01 | Y | 2 | 20.00 | 100% | 100% | 0% |
| UK | 2025-02-01 | Y | 1 | 20.00 | 100% | 100% | 0% |
Data
| Michigan A | Kenyon | 0 | 0 | 10 | 10 |
| Michigan B | Case Western B | 0 | 0 | 0 | 0 |
| Michigan D | Ohio State A | 0 | 0 | 0 | 0 |
| Michigan State | Case Western A | 0 | 0 | 10 | 10 |
| Ohio State B | Carnegie Mellon B | 0 | 0 | 10 | 10 |
| Carnegie Mellon A | Michigan C | 0 | 0 | 10 | 10 |