Backcrosses And Cosegregation

As noted above, plants from mutagenized populations frequently carry multiple mutations. Some of these may affect the phenotype of interest. It js advisable to backcross mutant plants to their wild-type parents several times to remove extraneous mutations. Each time a mutant is crossed to wild type, one half of the mutant genome is replaced with wild-type genes. Subsequent analysis of the F2 population allows the plants that are homozygous for the recessive mutation of interest to be recovered, but does not increase the liklihood of removing unlinked secondary mutations because only one quarter of the plants will have lost such a mutation (an "improvement"), whereas one half of the plants will be heterozygous and one quarter of the plants will be homozygous for the secondary mutation. Consequently, each backcross removes half of the unlinked secondary mutations, and the probability that a particular unlinked mutation remains after n rounds of backcrossing is (l/2)n. The chance of such a mutation remaining after four rounds of back-crossing is 0.062 (i.e., 0.54). Of course, if a secondary mutation is linked to the mutation of interest, the probability of removing it in a backcross is much less than one half and approaches zero as the distance between the primary and secondary mutations decreases.

Confirmation that a phenotype of interest is due to a particular mutation can be obtained by identifying the corresponding wild-type gene, introducing that gene into mutant plants, and demonstrating that the mutant phenotype is rescued (complementation). In the case of dominant mutations, it is necessary to demonstrate that a transgene construct carrying the dominant allele transforms wild-type plants to the mutant phenotype.

If a mutant exhibits two or more phenotypes, it is usually necessary to determine whether both phenotypes result from the same mutation (pleiotropy) or from two or more different mutations. These possibilities can usually be distinguished by multiple rounds of backcrossing or by testing for cosegregation of the phenotypes, which is generally much faster.

Consider the probability (p) of phenotypes caused by two unlinked recessive mutations cosegregating. The probability that they are not caused by two unlinked, recessive mutations is 1 - p. Among the F2 progeny of a backcross to wild-type, if the two mutations are recessive and unlinked, the probability of a plant that is homozygous for one mutation (genotype aa showing phenotype "a") also being homozygous for the second mutation (genotype bb, hypothetical cause of phenotype "b") is one quarter. The probability that all of n plants that display phenotype "a" will also display phenotype "b" is l/4n. For example, the probability that two recessive phenotypes, which cosegregate among four plants, are caused by two unlinked mutations is 0.004 (i.e., 0.254); the probability that this is not true is 0.996 (i.e., 1-0.004). Thus, the phenotypes are almost certainly caused by the same mutation, or by linked mutations.

Of course, the possibility that the two phenotypes are caused by two linked mutations, rather than one single mutation, is much more difficult to exclude, with the difficulty increasing with decreasing distance between the mutations. (The probability that aa and bb will cosegregate is (1 - r)2", where r is the recombination frequency between the two genes.) Consider two genes with 1% recombination between them that are presumed to account for two phenotypes. To test this hypothesis, enough plants must be examined to be 95% confident that a recombinant has been recovered. If the recombinant has both phenotypes, then these phenotypes must be caused by a mutation in only one gene (aa); on the other hand, if one recombinant is found that lacks one of the phenotypes, then the phenotypes are caused by mutations in two genes (aa and bb).

How many plants are enough? The probability of finding a recombinant (i.e., that aa and bb will not cosegregate) is 1 - (1 - r)2n, and a 95% probability of finding a recombinant is reached when 0.95 = 1 -(0.99)2°. Solving for n yields:

In the case of insertion mutations, it is desirable to determine whether the insertion caused the mutant phenotype. The situation is complicated by the fact that insertion mutants often have insertions at more than one site. If the insertion is tracked by a dominant drug resistance phenotype, and the mutant phenotype is recessive, many plants that display the drug resistance will not display the mutant phenotype, even if one of the insertions causes the phenotype. However all plants that display the mutant phenotype must carry this insertion. If the mutant phenotype is recessive, and there is one unlinked insertion, then the probability that n plants displaying the mutant phenotype will also display drug resistance is (3/4)n, or 0.056 for 10 plants. If there are two unlinked insertions, the probability is (15/16)" or 0.040 for 50 plants.

Obviously, the possibility that a linked insertion does not cause the phenotype is much more difficult to exclude. Now that isolation of sequences adjacent to insertion sites has become straightforward, many investigators prefer to test whether a particular insertion caused the phenotype by introducing the corresponding wild-type sequence into the mutant plants and testing for complementation of the mutant defect (see Chapter 6).

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