Analysis of a tri-hybrid test cross


Theoretical:
Given three loci, linked on a single chromosome in the following order
 
  -----A----------B----------C-----

Then, a cross between ABC//ABC x abc//abc tester involves

  -----A----------B----------C-----
  -----A----------B----------C-----

  -----a----------b----------c-----

  -----a----------b----------c-----


Recombination may not occur at all (parental type),
    or may occur between A & B, or B & C (
single crossovers)
    or between both: a double crossover

  During Meiosis I, single cross-over between the A & B loci looks like:

  -----A----------B--------C-----  => ABC Parental
  -----A---\  /---B--------C----- 
=> aBC Single recombinant
            X        
  -----a---/  \---b--------c----- 
=> Abc Single recombinant
  -----a----------b--------c-----  => abc Parental

A single cross-over between the B & C loci looks like:

  -----A--------B----------C-----  => ABC Parental
  -----A--------B---\  /---C-----
  => abC Single recombinant
                     X
  -----a--------b---/  \---c-----
  => ABc Single recombinant
  -----a--------b----------c-----  => abc Parental

Each crossover changes the phase of one locus with respect to (wrt) its nearest neighbor.
The frequency of either single crossover is proportional to the distance between loci,
        and increases with distance

 The frequency of a double crossover is the product of these frequencies:
          therefore a double-crossover is the rarest type:

  -----A----------B----------C-----
  -----A---\  /---B---\  /---C-----

            X         X
  -----a---/  \---b---/  \---c-----
  -----a----------b----------c-----

A double crossover will reverse the phase of the middle locus

  -----A----------B----------C----- => ABC
  -----A----------b----------C----- => AbC

  -----a----------B----------c-----
=> aBc
  -----a----------b----------c----- => abc

 Thus, when you compare the most common types ( the parental ABC & abc )
            with the rarest ( the double-recombinants AbC & aBc )
           
the cis / trans phase of the middle locus is reversed wrt both outside loci
       Here: B is reversed wrt A & C (and b wrt a & c): B is the middle locus



Experimental
:
Consider three loci E, F & G, linked on a single chromosome, but in an unknown order


The following cross is constructed:
     EEFFGG    x   eeffgg  (P) => EeFfGg   x    eeffgg  (tester)  (F1)
which produces the following counts among 1,000 F2 offspring.

       Use of a tester means the genotype of the offspring can be inferred directly from the phenotype

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

(1) Identify the four pairs of phenotype classes:
    one parental (P) class, which is most frequent, and
                                       which shows the same phase relationships as the parents

    two  single recombinant (I & II) classes, between the middle locus and each outside locus
         
    one double recombinant (D) class, between the middle locus and both outside loci
                                       As this requires two events, this is the rarest class

    The two members of each recombinant pair should occur in approximately equal numbers,
         because each crossover produces two products

(2) To determine gene locus order:
    Most common phenotypes EFG  efg are parental: no recombinants
    Rarest phenotype EFg & efG will be the double-recombinants:
          Compare phases of [EFG efg] vs [EFg efG]
        * The G locus changes phase ("flips") wrt E & F, therefore G is in the middle *

   -----E-----G-----F-----

(3) To determine the
frequency of recombination between locus pairs,
       count all  classes in which the phase changes from the parental types
     (i) for E & G

          EG & eg are parental types
          Eg & eG are recombinant types between E & G: (Class I):
          count all classes with this type
            # (EFg + efG + eFG + Efg)
               [(8    + 12)  + (95   + 85)] / 1000 = 0.20 = 20% = 20 m.u. = 20 cM

     (ii) for G & F
          GF & gf are parental types
          Gf & gF are recombinant types between F & G (Class II):
          # (EFg + efG + EfG + eFg)
             [(8   +  12)  + (37   + 43)] / 1000 = 0.10 = 10% = 10 m.u. = 10 cM

    (iii) for E & F
          EF & ef are parental types
          Ef & eF are recombinant types between E & F
          # (EfG + eFg + eFG + Efg)
             [(37   + 43)  + (95   + 85)] / 1000 = 0.26 = 26% = 26m.u. = 26 cM

(4)    Map data then look like this:
              The distance between E & F is greatest, so these are on the outside

    E--------20cM--------G---10cM---F
    E-----------26cM-----------F


          but 0.26 < (0.20 + 0.10)     WHY?           


In the double recombinants,

  --A----B----C--   ABC
  --A----b----C-- 
  AbC

  --a----B----c-- 
  aBc
  --a----b----c--   abc

so double recombinants resemble parental type
       but each represents 2 cross-overs between outside ( E & F ) loci

(3.iii') Then the corrected count of crossovers between E & F
           [(37 + 43) + (95 + 85) + (2)(8 + 12)] = 0.3 = 30 m.u. = 30 cM

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

       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


All text material 2013 by Steven M. Carr