MAPPING: THREE POINT TEST CROSSES

rvsd 1/26/94, 1/23/95, 1/26/96, 1/27/97, 21 Jan 00, 23 Jan 02, 23 Jan 04, 23Jan06, 23 Jan 08

[SGML P 103-] gmslg: p. 128-, 7^th: p 150-151, 9^th: pp129-165

Three point test cross allows the ordering of three linked genes:

recombination of the middle gene with retention of the two outer parental genes requires double cross over, much less likely.

HOW TO DETERMINE GENE ORDER FROM THREE POINT TEST CROSS:

1)         identify the two parental types from the eight phenotypes possible (largest class)

2)         identify the two double cross overs (smallest class)

3)         pick one double cross over, which parent does it most closely resemble? The recombined gene is in the middle.

Example: Cross AA/BB/CC x aa/bb/cc = F1: Aa/Bb/Cc                         ABC    parental

                                       Test cross F1 times aa/bb/cc                                 ABc

                          Get 2 x 2 x 2 = 8 classes of progeny phenotypes:              AbC     double cross over

   Can usually pick out middle gene on inspection of data                        Abc

   without any calculations.                                                                            aBC

             Parental                                     =          two largest classes        aBc      double cross over

             double recombinants                =          two smallest classes      abC

                                                                                                                     abc       parental

If order is a-b-c, then double crossovers required for which phenotypes?

Drosophila cross:          v           vermillion eyes

                                       cv         crossveinless

                                       ct:         cut wing edges

Cross: +/+ cv/cv  ct/ct, x v/v  +/+/  +/+ (nB: Parental = +, ct, cv and v, +, +)

             F₁ = v^+/v  cv^+/cv  ct⁺/ct is test crossed, progeny counted:

test cross progeny_:

             v + +               580       (parental)            

             + cv ct               592       (parental)           

             v cv ct                89

             + + +                94 

             v cv +                45 

             + + ct                40  

             v + ct                 3         These show double crossovers

             + cv +              5           should be counted twice in RF

             total:                  1448

To map: count the number of cross overs between two markers and divide by total progeny.

recombinant types:                                                                        %

      between markers          recom types number        recombination frequency

      for v and cv:                    v cv and + +           268      RF = 18.5^* (Apparent, important, see below)

      for v and ct:                     v ct and + +            191      RF = 13.2

      for cv and ct:                   cv + and + ct            93      RF = 6.4

* For outside markers (v&cv): must add double cross overs (v + ct and + cv +) times two:

      89 + 94 +45 + 40 (3x2) + (5x2) = 284

      284/1448 = 19.6 map units