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    Genetic 342 A&B Kenneth G Wilson Professory of Botany  | 
  
Trihybrid cross Assuming Independent Assortment
There are two possible solutions
Classical Solution
In Drosophila the following wild type alleles are used for analysis v (vermilion eyes), cv (cross-veinless) and ct (cut or snipped wing edges). Flies doubly recessive and homozygous (v+v+,cvcv,ctct) are crossed with flies singly recessive and homozygous (vv, cv+cv+, ct+ct+). Female progeny that are vv+, cvcv+, ctct+ are then test crossed with the tripley recessive males (vv, cvcv, ctct)
P1=v+v+,cvcv,ctct x vv, cv+cv+, ct+ct+
F1 =Wild Type - V+V, cv+cv, ct+ct x vv, cvcv, ctct (gametes v, cv, ct)
| F1 gamete types | Obs. | Description | ||
v  | 
      cv+  | 
      ct+  | 
      580  | 
      Parental  | 
    
v+  | 
      cv  | 
      ct  | 
      592  | 
      Parental  | 
    
v  | 
      cv  | 
      ct+  | 
      45  | 
      Crossover Region I  | 
    
v+  | 
      cv+  | 
      ct  | 
      40  | 
      Crossover Region I  | 
    
v  | 
      cv  | 
      ct  | 
      89  | 
      Crossover Region II  | 
    
v+  | 
      cv+  | 
      ct+  | 
      94  | 
      Crossover Region II  | 
    
v  | 
      cv+  | 
      ct  | 
      3  | 
      Double Crossover  | 
    
v+  | 
      cv  | 
      ct+  | 
      5  | 
      Double Crossover  | 
    
1448  | 
      ||||
Finding the center gene by comparing parental to double crossover.
580+59 2  | 
      Parental  | 
    
45+40  | 
      Crossover Region I  | 
    
89+94  | 
      Crossover Region II  | 
    
3+5  | 
      Double Crossover  | 
    
1448  | 
      
C/O  | 
      Chromosome region  | 
      % Recombination  | 
    
45+40  | 
      Crossover Region I  | 
      [(45+40)+(3+5)]x100/1448=6.4%  | 
    
89+94  | 
      Crossover Region II  | 
      [(89+94)+(3+5)]x100/1448=13.2%  | 
    
3+5  | 
      Double Crossover  | 
      |
Map data
  v<---13.2--->ct<--6.4-->cv 
Can we calculate the Interference?
     The double cross overs are always the smallest in number. In this     problem that is 5+3.   The expected frequency of double crossovers     can be calculated by multiplying the   frequency of a crossover between     v and ct by the frequency of a crossover between ct  and cv. 0.132 x 0.064     = 0.0084 
The expected number of double crossovers is equal to the expected frequence     of  double crossovers times the total number of individuals . 0.0084 x     1448 = 12
    
     Interference (I) = 1 -([# of dble c/o]/[exp # of dble c/o]) = 1-     (8/12) = 0.33 }
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