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Modern Genetic Analysis Chapter 8. Chromosome Mutations
Solved Problem

1.

A corn plant is obtained that is heterozygous for a reciprocal translocation and therefore is semisterile. This plant is backcrossed to a chromosomally normal strain that is homozygous for the recessive allele brachytic (b), located on chromosome 2. A semisterile F1 plant is then backcrossed to the homozygous brachytic strain. The progeny obtained show the following phenotypes:

a. What ratio would you expect if the chromosome carrying the brachytic allele is not involved in the translocation?
b. Do you think that chromosome 2 is involved in the translocation? Explain your answer, showing the conformation of the relevant chromosomes of the semisterile F1 and the reason for the specific numbers obtained.
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Solution
a. We should start with the methodical approach and simply restate the data in the form of a diagram, where

To simplify the diagram, we do not show the chromosomes divided into chromatids (although they would be at this stage of meiosis):

All the progeny from this cross will be heterozygous for the chromosome carrying the brachytic allele, but what about the chromosomes involved in the translocation? In this chapter, we have seen that only alternate-segregation products survive and that half these survivors will be chromosomally normal and half will carry the two rearranged chromosomes. The rearranged combination will regenerate a translocation heterozygote when it combines with the chromosomally normal complement from the normal parent. These latter typesthe semisterile F1sare diagrammed as part of the backcross to the parental brachytic strain:

In calculating the expected ratio of phenotypes from this cross, we can treat the behavior of the translocated chromosomes independently of the behavior of chromosome 2. Hence, we can predict that the progeny will be

This predicted 1:1:1:1 ratio is quite different from that obtained in the actual cross.

b. Because we observe a departure from the expected ratio based on the independence of the brachytic phenotype and semisterility, it seems likely that chromosome 2 is involved in the translocation. Let's assume that the brachytic locus (b) is on the orange chromosome. But where? For the purpose of the diagram, it doesn't matter where we put it, but it does matter genetically because the position of the b locus affects the ratios in the progeny. If we assume that the b locus is near the tip of the piece that is translocated, we can redraw the pedigree:

If the chromosomes of the semisterile F1 segregate as diagrammed here, we could then predict

Most progeny are certainly of this type, so we must be on the right track. How are the two less-frequent types produced? Somehow we have to get the b+ allele onto the normal orange chromosome and the b allele onto the translocated chromosome. This must be achieved by crossing-over between the translocation breakpoint (the center of the cross-shaped structure) and the brachytic locus. To represent this crossover, we must show chromatids, because crossing-over occurs at the chromatid stage:

The recombinant chromosomes produce some progeny that are fertile and nonbrachytic and some that are semisterile and brachytic (these two classes together constitute 69 progeny of a total of 682, or a frequency of about 10 percent). We can see that this frequency is really a measure of the map distance (10 m.u.) of the brachytic locus from the breakpoint. (The same basic result would have been obtained if we had drawn the brachytic locus in the part of the chromosome on the other side of the breakpoint.)

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