Three-Point Test Cross


A three-point test cross is a technique used in genetic mapping when considering the inheritance of three alleles. It can be used to order three loci on a chromosome, and map the distance between these loci in centimorgans (cM). The centimorgan is equivalent to the frequency of recombination; a map distance of 1cM across a chromosome is equivalent to a recombination frequency of 1% for that chromosomal region.

The three-point test cross involves a mating between a triple heterozygote and a triply recessive homozygote. Once offspring phenotypes have been identified and counted in classes, the procedure is as follows:

  1. Sum the offspring to get a total number of progeny
  2. Identify the offspring with parental genotypes (these will have highest frequency classes)
  3. Identify the offspring with double recombinant genotypes (these will have the lowest frequency classes)
  4. Identify which locus has swapped position, relative to the other two, in the double recombinant genotypes compared to the parental genotypes - this locus is the middle locus on the chromosome, and the other two loci may be placed either side (in any order)
  5. Draw a map of the chromosome, and divide the spaces between loci into two 'regions'
  6. For the first region identify the two classes of single recombinant offspring where a recombination event has occured in this area. Sum the offspring of these classes and the offspring of the double recombinant classes to give you a recombinant value for that region
  7. The map distance for this region is then (recombinant value)/(total number of progeny) x 100
  8. Repeat steps 6 & 7 for the second 'region' to give this region a map distance


The following is a fictitious example to illustrate the process:

In Neptunian aliens:

  • red skin (R) is dominant to yellow skin (r)
  • big antennae (B) are dominant to small antennae (b)
  • green polkadots (G) are dominant to having no polkadots (g)

These characteristics are linked on the same chromosome.

A male alien with red skin, big antennae and green polkadots is crossed with a female alien with yellow skin, small antennae and no polkadots. The following offspring arise from this mating:

Red skin, big antennae, green polkadots = 479
Yellow skin, small antennae, no polkadots = 473
Red skin, small antennae, no polkadots = 15
Yellow skin, big antennae, green polkadots = 13
Red skin, big antennae, no polkadots = 9
Yellow skin, small antennae, green polkadots = 9
Red skin, small antennae, green polkadots = 1
Yellow skin, big antennae, no polkadots = 1

You can use the above data to order the loci on their chromosome, and deduce the map distance between them.

From the above phenotypes, you can deduce genotypes. When you see the genotypes, it is easier to identify where recombination events have occured. The table below is in the same order as the phenotypic list above.


Group
Genotype
Crossovers
Number
Totals
1
RBG
None; the parental genotypes
479
952
2
rbg
473
3
R|bg
Single, between locus R and the others
15
28
4
r|BG
13
5
RB|g
Single, between locus G and the others
9
18
6
rb|G
9
7
R|b|G
Double recombinants
1
2
8
r|B|g
1


Totals
1000


The B locus has swapped in the double recombinants, compared to the parental genotypes, so the B locus is in the middle of the chromosome and the other two loci may be put either side (in any order):

R----------------------------B-----------------------------G
<---------region 1--------><--------- region 2--------->

The chromosome can be divided into region 1 and region 2; we will deal with these regions separately in calculating map distance.

For region 1, we are looking for recombination events between locus R and locus B. There were 28 offspring with single recombination events in this region, and 2 with double recombination events, so the map distance can be calculated as follows:

[(28 + 2)/1000] * 100 = 3cM


For region 2, we are looking for recombination events between locus B and locus G. There were 18 offspring with single recombination events in this region, and 2 with double recombination events, so the map distance can be calculated as follows:

[(18 + 2)/1000] * 100 = 2cM

So the map of the chromosome carrying these characteristics is:

R----------------------------B------------------------------G
<-----------3cM-----------> <----------- 2cM------------>