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