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Modern Genetic Analysis Chapter 12. Genomics
Solved Problem

1.

A Neurospora geneticist has just isolated a new mutation that causes aluminum insensitivity (al) in a strain of Oak Ridge background (see page 344). She wishes to clone the gene by positional cloning and therefore needs to map it. For reasons that we do not need to go into, she suspects that it is located near the tip of the right arm of chromosome 4. Luckily, there are three RFLP markers (1, 2, and 3) available in that vicinity, so the following cross is made:

One hundred progeny are isolated and tested for al and the six RFLP alleles 1O, 2O, 3O, 1M, 2M, and 3M. The results were as follows, where O and M represent the RFLP alleles, and al and + represent al and al+:

a. Is the al locus in fact in this vicinity?
b. If so, to which RFLP is it closest?
c. How many map units separate the three RFLP loci?
See answer

Solution

This is a mapping problem, but with the twist that some of the markers are traditional types (which we have encountered in preceding chapters) but others are molecular markers (in this case, RFLPs). Nevertheless, the principle of mapping is the same as we used before; in other words, it is based on recombinant frequency. In any recombination analysis, we must be clear about the genotype of the parents before we can classify progeny into recombinant classes. In this case, we know that the Oak Ridge parent must contain all O alleles, and Mauriceville all M alleles; therefore, the parents were

and knowing this makes determining recombinant classes easy. We see from the data that the parental classes are the two most common (34 and 36). We first of all notice that the al alleles are tightly linked to RFLP 1 (all progeny are al 1O or +1M). Therefore, the al locus is definitely on this part of chromosome 4. There are 6 + 4 = 10 recombinants between RFLP 1 and 2, so these loci must be 10 map units apart. There are 12 + 8 = 20 recombinants between RFLP 2 and 3; that is, they are 20 map units apart. There are 6 + 4 + 12 + 8 = 30 recombinants between RFLP 1 and 3, showing that these loci must flank RFLP 2. Therefore, the map is

There are evidently no double recombinants, which would have been of the type M O M and O M O.

Notice that there is really no new principle at work in the solution of this problem; the real challenge is to understand the nature of RFLPs and to translate this understanding into genotypes from which to study recombination. If you still don't understand RFLPs, you might ask yourself how exactly the RFLP alleles are tested experimentally.

See question

© 1999 by W. H. Freeman and Company. All rights reserved.