1. Sepia& Ebony have a published distance of 44.7 M.U. For that data, run a Chi-square test to see if the data obtained fits with that distance. The following questions will walk you through that process
a. First, determine expected frequencies of each individual progeny possibility based on the 44.7 M.U. distance. Do this by showing a punnet square for frequencies. Think about what M.U.s mean and how that affects gamete probabilities. Assume that crossing over occurs equally between males and females.
b. From your punnet square, calculate the expected frequencies of each phenotype based on a 44.7 M.U. distance.
c. Determine the expected number of flies of each category (round to the nearest hundredth of a fly).
|
WT
|
Ebony (body)
|
Sepia (eye)
|
Sepia&Ebony
|
Total (contributing)
|
Observed
|
2313
|
317
|
274
|
454
|
3358
|
Expected
|
|
|
|
|
|
|
Female and male combined |
|
sepia (se) & ebony(e) |
wild type |
just dark bodies |
just darker eyes |
darker eyes and bodies |
Total |
|
2313 |
317 |
274 |
454 |
3358 |