Transcript gene mapping

GENETIC
MAPPING
III
The problem of double crossovers in
genetic mapping experiments
Consider a cross to map 2 genes, a and b
They are some distance apart, but
mappable
The heterozygote is in tetrad stage:
a+ b+/ab
A single crossover generates recombinant
chromosomes
Which give recombinant gametes and
eventually recombinant progeny
A 2-strand, double crossover restores the
original arrangement of the marker genes
So all progeny are scored as parental, with
no recombinants
It looks exactly as if there has been no
crossing over
There have been two crossover events
which will be uncounted
And since recombination frequency is a
measure of map distance, this means that
the distance between the genes will be
underestimated
How can we avoid these errors?
Two general ways:
1. Map closely linked genes
Double crossovers rarely occur within
map distances < 10 cM
2. Do three-point testcrosses, rather than
two-point
These involve 3 genes within a relatively
short section of chromosome
The rationale for using these is illustrated
in the next slide.
As before, a 2-strand double crossover gives
gametes that are nonrecombinant for genes a
and b
BUT, notice that the resulting gametes are
recombinant with respect to c
The gene in the middle reveals the occurrence
of a double crossover
3-point crossovers are routinely used for
mapping, because they allow us to correct for
double crossovers, and determine the gene
order
Suppose we want to map 3 genes in a plant
Fruit color: p = purple; p+ = yellow
Fruit shape: r = round; r+ = elongated
Juiciness: j = juicy; j+ = dry
What is the order, and map distances, of
these 3 genes?
We set up our testcross with a triply
heterozygous parent, in coupling phase (in
this case) and count the offspring
We know that is the genes were unlinked,
we would expect eight phenotypic classes
of progeny
For this kind of trihybrid cross, we expect
the same classes, but not in the same
proportions
Because of linkage, some phenotypic
classes may have 0 individuals; if so, that’s
important to note
Here are the
eight
phenotypic
classes of
progeny
These are parentals. Note that they are in
approximately equal numbers
These recombinants both involve the p
gene
Notice that they are in about equal
numbers, and are rarer than the parentals
These recombinants involve the r gene
They are rarer still
These recombinants are the rarest.
Gene j is involved
We expect double crossovers to be rarer
than single crossovers
So it follows that recombinants due to
double crossovers will be the rarest class
We can use this fact to help us order the
genes.
How?
Recall our earlier example:
Notice how the double crossover restored the
outside genes to the parental arrangement,
but the middle gene has its orientation
changed
So the gene which is in a recombinant
arrangement in the rarest, double crossover
class of progeny, must be the middle gene.
We can see that p and r are in their
parental configuration, but j is in a new
arrangement
So, j must be the gene in the middle
The order must be p , j , r
Now that we know the correct gene order,
we can interpret the data to generate map
distances:
For the p - j distance, we need to add
together all the recombinant progeny
resulting from crossovers in Region I
This includes both single crossovers and
the double crossovers (since they also
involve this region of the chromosome)
So, the percentage of recombinants =
[(52+46) + (4+2)]/500 x 100% =
104/500 x 100% = 20.8%
So, p and j are 20.8 cM apart
We do the same sort of calculations to find
the distance between j and r
We again add together the single
crossovers (this time from Region II) and
the double crossovers
[(22+22) + (4+2)]/500 x 100% =
50/500 x 100% = 10.0%
So, j and r are 10.0 cM apart
Our linkage map now looks like this.
To get the distance between p and r, we
simply add the inner distances
= 30.8 cM