Interference in a tri-hybrid test cross

Interference occurs where one cross-over decreases the probability of others nearby:
        Compare the observed vs expected number (or frequency) of double-recombinants

For the data set
Genotype
 Phenotype
 Count
 Class
E F G efg
 'EFG'
 370
 P
E F g efg 'EFg'
 8
 D
E f G efg 'EfG'
 37
 II
E f g efg 'Efg'
 95
 I
e F G efg 'eFG'
 85
 I
e F g efg 'eFg'
 43
 II
e f G efg 'efG'
 12
 D
e f g efg 'efg'
 350
 P


we calculated he gene map

    E--------20cM--------G---10cM---F

where # EG recombinants = 95 + 85 + (8 + 12) = 200 , and 200/1000 = 0.20 cM
and    # GF recombinants = 47 + 37 + (8 + 12) = 100 ,  and 100/1000 = 0.10 cM
so that # double recombinants expected = (0.20)(0.10)(1000) = 20
                                                                    and f(Dexp) = 20/1000 = 0.020
and    # double recombinants observed = (8 + 12) = 20
                                                                    and f(Dobs) = 20/1000 = 0.020
 
Interference (I) = 1 - (observed # D) / (expected # D) = 1 - [(8 + 12) / (0.20)(0.10)(1000)]
                          = 1 - f(Dobs) / f(Dexp)                            = 1 - [(0.008 + 0.012) / [(0.20))(0.10)]
               
          In these data, I = 1 - (8+12)/(20) =  0 :       # expected = # observed     no interference

If instead
Genotype
 Phenotype
 Count
 Class
E F G efg
 'EFG'
 360
 P
E F g efg 'EFg'
 3
 D
E f G efg 'EfG'
 42
 II
E f g efg 'Efg'
 100
 I
e F G efg 'eFG'
 90
 I
e F g efg 'eFg'
 48
 II
e f G efg 'efG'
 7
 D
e f g efg 'efg'
 360
 P


where # EG recombinants = 100 + 90 + (3 + 7) = 200 , and 200/1000 = 0.20 cM
and    # GF recombinants  = 42 + 48 + (3 + 7)   = 100 , and 100/1000 = 0.10 cM

so and f(Dexp) = (0.20)(0.10) as before
but now
         # double recombinants = (3 + 7) = 10 and f(Dobs) = 10/1000 = 0.010
 
       Interference (I) = 1 - (observed # D) / (expected # D) = 1 - [(3 + 7) / (0.20)(0.10)(1000)]
                                 = 1 - f(Dobs) / f(Dexp)                            = 1 - [(0.003 + 0.007) / [(0.20))(0.10)]
               
          In these data, I = 1 - (3+7)/(20) =  0.5
                                 I = 1 - (0.003 + 0.007)/(0.020) = 0.5      

Half as many observed as expected     0.50 interference
All text material © 2016 by Steven M. Carr