THE TEXTBOOK APPROACH TO 3-FACTOR DISTANCES

The traditional way of teaching three-factor crosses cannot actually be used to map a new gene relative to other genes. This method also sets a bad example of how one does science.

Let’s start with a typical example, using wild type and a strain homozygous for recessive mutations in three genes in maize or Drosophila. The mutant strain is written as abc/abc, but the order of the genes is not known. The relative map order could be abc, acb, or cab

1. The strains are mated together and the F1 progeny are mated to the triply homozygous strain. The second cross is a test cross that allows the genotypes to be determined.

abc/abc X +++/+++ —> abc/+++

abc/+++ X abc/abc —> several different genotypes.

The phenotypes of the progeny from the second cross are

ABC    169

AB      18

AC       9

BC       0

A        1

B       10

C       21

WT     172

2. The progeny are rearranged so that those with reciprocal chromosomes are grouped together and so that the number of progeny decreases.

ABC    169

WT     172

AB      18

C       21

AC       9

B       10

BC       0

A        1

3. The rarest group represents individuals with chromosomes that underwent two recombination events. This group is used to determine the map order of the genes. (Sometimes this group is missing because none of these events took place.) In the example, this is the group with BC and A. The only way an a or a bc chromosome can be produced is if a is between b and c, so the order is cab. In general the single gene is in the middle. Note: that these data give a relative map position NOT a direction; in the example the order could be given as bac.

4. Map distance is calculated between c and a (p) and between a and b (q).

(a) For double crossovers to occur, recombination must occur in both intervals, so the probability is the product pq= (BC + A)/Total = 1/400

(b) A single recombination in the first interval (between c and a producing the C and AB progeny) must be accompanied by the lack of recombination in the second, i.e., p(1-q) = (C +AB)/Total = 39/400

(c) Similarly for recombination only in the second interval: q(1-p) = (CA + B)/Total = 19/400

(d) We can solve these equations to get that p=10 & q=5

5. One reason people like to use examples like this one is that they allow one to point out that numbers do not also work out, and this is evidence for interference.

(a) Interference (I) is defined as I = 1 - obtained doubles/expected doubles

This ratio is called the coefficient of coincidence. In the example I = 0.5.

(b)Interference can be positive or negative, perhaps by preventing or allowing subsequent pairing. Unfortunately the numbers are so small here that the value for I is meaningless.

This method of solving for 3-factor distances, however, has several problems.

1. The main purpose of a mapping experiment is to position a new gene c relative to two previously existing genes a and b. The textbook method relies on having the abc/abc strain, but for a new mutation c, such a strain would not exist. Moreover, generating the abc/abc strain would give the answer that was needed (the relative position of the genes). I believe that the method was not really begun to obtain map positions, but rather to try to determine chromosome size (in map units). These experiments, which were done in the 1910s, used genes whose relative positions were known. Because the data were so good, they were used in textbooks starting in the 1920s. They have been passed on by one textbook writer to another to the present.

2. The desire to show the double recombinant class leads to the use of large map distances, but most often we need to determine map distances in small intervals. Thus, these problems give the incorrect impression that mapping is done over large intervals.

3. The procedure requires that all of the individuals be counted; this requires an enormous amount of unnecessary effort.

4. As indicated above the values to be used for interference are of very small numbers, so their significance is not great.