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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

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  • Category:- Biology
  • Reference No.:- M9367334

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