GENETICMAPPING

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+ = yellowFruit 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