Download Problem13.19

Problem 13.19 addresses the use of semisterility in corn to localize a gene onto a specific
chromosome. This sort of localization is an essential first step to many more detailed
genetic studies and the procedures referred to in this question are excellent examples of
how this work is done. Full understanding of this question requires a knowledge both of
translocations and of the basic genetic principle of recombination.
There are two phenotypes involved. The first is the fact of semisterility itself. The
characteristic appearance of an ear of corn that is missing about half of the kernels
(ovules) shows that the kernels on that ear are carrying a translocation in the
heterozygous configuration. The second phenotype is leaf color. Plants homozygous for
the yg allele have yellow-green leaves, while those with the yg+ allele (homozygous or
heterozygous have green leaves, which are the wild type.
In order to make the analysis easier, we will assume that alternate and adjacent-1
segregation occur with equal frequency and that adjacent-2 segregation does not occur.
It is important to see that the question actually involves three generations of plants. It is
best to look at each cross.
________________________
Cross 1: Semisterile green-leafed plant X fertile yellow-green-leafed plant
The semisterile parent is homozygous for yg+
The F1 from this cross will have the green phenotype and all will be heterozygous for the
yg and yg+ alleles. Half of the these F1 will be completely fertile and half will be
semisterile.
All of these viable offspring arose from alternate segregation.
Cross 2: semisterile F1 progeny from cross 1 X fertile yellow-green
This is a test cross and is also referred to as a back cross, since one of the original parents
was fertile and yellow-green.
At this point, we must consider two possible options.
Option 1 is that the yg gene is NOT on the translocated chromosome. In this case, the
two phenotypes, leaf color and semisterility, will assort independently so that half are
fertile, half are semisterile, half are yg+ / yg , half are yg / yg. Therefore, there will be
four phenotypes in equal proportions:
semisterile, green
semisterile, yellow-green
fertile, green
fertile, yellow-green
Option 2 is that the yg gene IS on the translocated chromosome. In this case, the color
and semisterility are linked, although this is a new use of the term linked. This is shown
in the following diagram:
This figure is the appearance of the chromosomes at the start of meiosis, with the four
chromosomes marked as usual. There is, of course a yg allele on both of the N2 sister
chromatids, but only one is drawn in so that the diagram is easier to read. Similarly, there
is a yg+ allele on both of the T1 sister chromatids, but only one is shown here.
Balanced gametes produced by this individual will be of 2 types:
N1 and N2
T1 and T2
Since the dominant allele is on the translocated chromosome in this example, then the
offspring will display two different phenotypes:
fertile, yellow-green (from the N1 and N2 segregation)
semisterile, green (from the T1 and T2 segregation)
Again we need to look at the next generation for these final phenotypes.
The final part (c) of the question notes that there were a few fertile, green and semisterile,
yellow-green progeny from this second cross. How is this possible, and what genetic
distance can be determined?
The only way these rare phenotypes can appear is if there were a crossover between the
breakpoint and the yg gene. If this happened, then after the crossover, the chromosomes
will be arranged as shown in the following figure.
The recessive allele is now with the translocated chromosome, so the semisterile
offspring will have yellow-green leaves and the fertile offspring will have green leaves.
The genetic distance that can be calculated is the number of map units from the break
point to the yg gene. In this way, we have a new kind of genetic marker, namely the
location of the break point in the translocation.