Three Point Test Cross: Multiple Point Gene Mapping

Gene mappers are motivated to map all of the tens of thousands of genes found on the chromosomes of plant or animals. Analyzing data from crosses to determine map distances for two genes at a time makes the process time consuming and tedious. Therefore geneticists will often attempt to map as many genes as possible from a single set of progeny. We will go through the simplest multiple point mapping example, a three point testcross, to demonstrate the process.

We will use our three corn seed trait loci again and use data from one cross to map these three loci. The first step would be to obtain a trihybrid individual that is heterozygous at all three loci and then perform a testcross with this trihybrid.

           Parents: CCssWW (CsW/CsW)       X       ccSSww (cSw/cSw)

           Trihybrid: CcSsWs (CsW/cSw)         X       ccssww (csw/csw)

                                                                     Test cross offspring

We observe eight phenotypes of seeds in the testcross progeny because the trihybrid can make eight kinds of gametes. As we knew, these three genes are linked and so the uneven ratio of phenotypes reflects the combinations of two parental and six recombinant gametes. The parental gametes are still the most frequently produced type and the six recombinant gametes are made when crossing over occurs between the loci. We can systematically account for these crossovers if we follow the following four steps. First we will explain the steps and then we will show how to use them to map the three genes.

Step 1: Identify the parental gametes. 

These are CsW and cSw, that combination which came from the parent generation and the combination made by the trihybrid at the highest frequency.

Step 2: Classify the recombinants.

If we observe the gene combination in each of the six recombinant gametes, we can ask ourselves if the gamete has a new combination for each pair of genes. For example CSw has a new combination for the CS and the Cw compared to the parental gametes but the same combination for Sw as the cSw parental gamete. Therefore, these 116 gametes represent crossovers between C,c and S,s as well as C,c and W,s. This information is recorded (see below).

Step 3: Determine recombinant gamete frequency. 

Once all six recombinant gametes have been classified, the total number of crossovers between the three loci can be added up and crossover percentage determined between each pair of loci. This number will reflect gene distance but one more step is needed to complete the process.

Step 4: Add in the double crossover gametes.

Two of the six recombinant gametes were made as a result of double crossovers between the two loci that are furthest apart. These crossovers have been added to the map distances between the middle locus and the two outside loci. Now we need to add these double crossovers to the outside loci distance.

Applying these fours steps to our three point test cross data would work as follows:

Fig. 2, 3. Double crossovers need to occur to produce all dominant allele (CSW) and all recessive allele (csw) gametes. (Image by D. Lee)

When two crossovers occur between the C,c and W,w loci the parental combinations of CW and cw are restored. Therefore we did not count the CSW and csw gametes as representing crossovers when they actually represent two crossovers each (Fig 2). Thus we should add up the double crossovers (3 + 4 = 7), multiply times two and add these 14 crossovers to the C,c to W,w distance (1508 + 14 = 1522 / 7000 = 21.7 map units). Now we have double checked our map distances and have our three point map complete.

It can be intimidating to see all the data generated from a three point test cross. This information, however, can be used systematically to save the geneticists time and map three genes in one experiment. The three point data also provides a more accurate measure of map distances compared to two point data when genes are farther apart on a chromosome. This is because a third gene in between the more distant loci can account for double crossovers that would not be detectable in a two point analysis. For this reason, map distances tend to be underestimated when genes are further apart as we saw to a modest extent in this example (21.5 two-point vs. 21.7 three-point).

How would a geneticist work with a four point test cross? There would be sixteen phenotypes in the progeny as a result of the hybrid parent making two parental gametes and fourteen recombinant gametes. These recombinant gametes would be a result of single, double and even triple crossovers that occurred in prophase I. While the data would be tedious to work with, one data set could be analyzed to reveal the map of four genes. It is not surprising that geneticists now have written computer programs which will perform these types of calculations, save time and reduce the chance of calculator error. Geneticists mapping genes in economically important plants and animals will also generate mapping populations that are a result of crossing parents that have different alleles at hundreds or thousands of loci. Even though the development of computers and programs has become a part of modern gene mapping the process of indirectly measuring crossover frequency by observing the inheritance of trait combinations is the basis of generating these gene maps.