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Gene Mapping with a Three-Point Cross

The page Genetic Linkage and Linkage Maps shows how carrying out three different dihybrid test crosses in the corn plant reveals:

Here we shall see how a single test cross of a trihybrid corn plant ; that is,

reveals

Hypothetical breeding data are shown in this table.

Group Expressed Alleles
(Phenotype)
Crossovers Number Totals
1 CShBz None; the parentals 479 952
2 cshbz 473
3 C|shbz Single; between C and others 15 28
4 c|ShBz 13
5 CSh|bz Single, between Bz and others 9 18
6 csh|Bz 9
7 C|sh|Bz Double recombinants 1 2
8 c|Sh|bz 1
    Totals 1000  


Eight different phenotypes — representing the 8 possible genotypes (23 = 8) are produced. Scoring them reveals

But adding the distances between C and Sh and Sh and Bz gives a map distance between C and Bz of 5.0 cM not the 4.6 cM revealed by the data (and the same number that a C,c,Bz,bz dihybrid cross would have produced).

Why the discrepancy?

Because the double recombinants restored the parental configuration, they were missed in the scoring. So the two rare classes of double recombinants need to be added (twice) to the data.

28 + 18 + 2 + 2 = 50/1000 = 5%

to get the true value.

So the map of this region of the chromosome is:

This exercise underscores the rule that the closer the intervals examined, the more accurate the map.

A three-point cross also gives the gene order immediately.

The procedure is:

  1. Determine the rarest classes (here, C,sh,Bz and c,Sh,bz) because two crossovers between a pair of loci will be rarer than one.
  2. In these two groups, the alleles that specify the trait that was not seen in the parents (sh and Sh) occupy the middle locus.
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19 February 2011