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Lecture 6: Gene mapping by 3-point testcross
1. Three-point testcross2. Double cross-overs3. Interference4. Mapping of X chromosome in
humans
With the increase of map distances, an increasing fraction of cross-overs becomes “invisible”
A double crossover between linked genes gives parental gametes
A single crossover between linked genes generates recombinant gametes
Frequency of double crossovers increases with the increase of distance between linked genes. This results in underestimation of map distance
A triple heterozygote is used to visualize these events. The three genes are simultaneously mapped by three-point testcross
Mutant alleles: v, determines vermilion eyes (bright red)cv, crossveinless wingsct, cut wing edges
P: v+/ v+ . cv / cv . ct / ct X v / v . cv+/ cv+ . ct+ / ct+ (in the arbitrary order!)
Gametes: v+ cv ct v cv+ ct+
F1: ♀ v+ / v . cv / cv+ . ct / ct+ is test-crossed with triple recessive homozygous maleX ♂ v / v . cv / cv . ct / ct
Female’s gametes determine eight phenotypic classes:
v cv+ ct+ 580 v+ cv ct 592
v cv ct+ 45 v+ cv+ ct 40
v cv ct 89 v+ cv+ ct+ 94
v cv+ ct 3 v+ cv ct+ 5
Total 1448
Three-point testcross
Mutant alleles v, determines vermilion eyes (bright red)cv, crossveinless wingsct, cut wing edges
P: v+/ v+ . cv / cv . ct / ct X v / v . cv+/ cv+ . ct+ / ct+ (in the arbitrary order!)
Gametes: v+ cv ct v cv+ ct+
F1: ♀ v+ / v . cv / cv+ . ct / ct+ is test-crossed with triple recessive homozygous maleX ♂ v / v . cv / cv . ct / ct
Female’s gametes determine eight phenotypic classes:Parental, no COv cv+ ct+ 580 v+ cv ct 592
v cv ct+ 45 v+ cv+ ct 40
v cv ct 89 v+ cv+ ct+ 94
v cv+ ct 3 v+ cv ct+ 5
Total 1448
Step 1: identify parental classesThree-point testcross
Select any two loci, for example v and cv.
Parental: v cv+ Recombinant: v cv
v+ cv v+ cv+
The progeny with recombinant phenotypes : 45 + 40 + 89 + 94 = 268
Map distance (in map units) is determined by Recombination Frequency
RF = recombinant progeny / total progeny %
(268 / 1448) X 100% = 18.5 %
v cv
18.5 m.u.
Step 2: map any two lociThree-point testcross
v cv ct+ 45 v+ cv+ ct 40
v cv ct 89 v+ cv+ ct+ 94
Select any other two loci, for example v and ct.
Parental: v ct+ Recombinant: v ct
v+ ct v+ ct+
The progeny with recombinant phenotypes : 89 + 94 + 3 + 5 = 191
Map distance (in map units) is determined by Recombination Frequency
RF = recombinant progeny / total progeny %
(191 / 1448) X 100% = 13.2 %
v ct
13.2 m.u.
Step 3: map any other two lociThree-point testcross
v cv ct 89 v+ cv+ ct+ 94
v cv+ ct 3 v+ cv ct+ 5
Three-point testcross
we are now trying to put the two maps together:
v ct cv13.2
18.5
ct v cv13.2 18.5
OR
but we do not know yet which one is correct
?
Three-point testcross
the solution depends on the ct to cv distance:
v ct cv13.2 5.3
18.5
ct v cv13.2 18.5
OR
and we need to determine this distance by finding the RF(ct, cv)
31.7
Parental: cv+ ct+ Recombinant: cv ct+
cv ct cv+ ct
The progeny with recombinant phenotypes : 45 + 40 + 3 + 5 = 93
Map distance (in map units) is determined by Recombination Frequency
RF = recombinant progeny / total progeny %
(93 / 1448) X 100% = 6.4 %
ct cv
6.4 m.u.
Step 3: map the remaining two lociThree-point testcross
v cv ct+ 45 v+ cv+ ct 40
v cv+ ct 3 v+ cv ct+ 5
Three-point testcross
the distance fits better with the fist map
v ct cv13.2 6.4
19.6
However, the v to cv distance must be 19.6 m.u. instead of 18.5 m.u. that we have determined during Step 2
?
...still puzzled
ct v cv13.2 18.5
31.7
Step 4: Choosing the right map
Linkage map is known:
13.2 6.4
P: v+/ v+ . cv / cv . ct / ct X v / v . cv+/ cv+ . ct+ / ct+ (in the arbitrary order!)
Gametes: v+ cv ct v cv+ ct+
F1: ♀ v+ / v . cv / cv+ . ct / ct+ is test-crossed with triple recessive homozygous maleX ♂ v / v . cv / cv . ct / ct
Female’s gametes determine eight phenotypic classes The loci in REAL order:Parental, no COv cv+ ct+ 580 v+ cv ct 592
v cv ct+ 45 v+ cv+ ct 40
v cv ct 89 v+ cv+ ct+ 94
v cv+ ct 3 v+ cv ct+ 5
Total 1448
Three-point testcross
v ct+ cv+
v+ ct cv
v ct cv
Step 5: draw the loci in REAL order
Linkage map is known:
13.2 6.4
P: v+/ v+ . cv / cv . ct / ct X v / v . cv+/ cv+ . ct+ / ct+ (in the arbitrary order!)
Gametes: v+ cv ct v cv+ ct+
F1: ♀ v+ / v . cv / cv+ . ct / ct+ is test-crossed with triple recessive homozygous maleX ♂ v / v . cv / cv . ct / ct
Female’s gametes determine eight phenotypic classes The loci in REAL order:Parental, no COv cv+ ct+ 580 v+ cv ct 592
SCO (ct, cv)v cv ct+ 45 v+ cv+ ct 40
v cv ct 89 v+ cv+ ct+ 94
v cv+ ct 3 v+ cv ct+ 5
Total 1448
Three-point testcross
v ct+ cv+
v+ ct cv
v ct+ cv+
v+ ct cv
v ct cv
Step 6: single crossover classes
Linkage map is known:
13.2 6.4
P: v+/ v+ . cv / cv . ct / ct X v / v . cv+/ cv+ . ct+ / ct+ (in the arbitrary order!)
Gametes: v+ cv ct v cv+ ct+
F1: ♀ v+ / v . cv / cv+ . ct / ct+ is test-crossed with triple recessive homozygous maleX ♂ v / v . cv / cv . ct / ct
Female’s gametes determine eight phenotypic classes The loci in REAL order:Parental, no COv cv+ ct+ 580 v+ cv ct 592
SCO (ct, cv)v cv ct+ 45 v+ cv+ ct 40
SCO (v, ct)v cv ct 89 v+ cv+ ct+ 94
v cv+ ct 3 v+ cv ct+ 5
Total 1448
Three-point testcross
v ct+ cv+
v+ ct cv
v ct+ cv+
v+ ct cv
v ct+ cv+
v+ ct cv
v ct cv
Step 6: single crossover classes
Linkage map is known:
13.2 6.4
P: v+/ v+ . cv / cv . ct / ct X v / v . cv+/ cv+ . ct+ / ct+ (in the arbitrary order!)
Gametes: v+ cv ct v cv+ ct+
F1: ♀ v+ / v . cv / cv+ . ct / ct+ is test-crossed with triple recessive homozygous maleX ♂ v / v . cv / cv . ct / ct
Female’s gametes determine eight phenotypic classes The loci in REAL order:Parental, no COv cv+ ct+ 580 v+ cv ct 592
SCO (ct, cv)v cv ct+ 45 v+ cv+ ct 40
SCO (v, ct)v cv ct 89 v+ cv+ ct+ 94
DCOv cv+ ct 3 v+ cv ct+ 5
Total 1448
Three-point testcross
v ct+ cv+
v+ ct cv
v ct+ cv+
v+ ct cv
v ct+ cv+
v+ ct cv
v ct+ cv+
v+ ct cv
v ct cv
Step 7: double crossover classes
Linkage map is known:
13.2 6.419.6
v+/ v+ . cv / cv . ct / ct X v / v . cv+/ cv+ . ct+ / ct+ (in the arbitrarbey order!)
Gametes: v+ cv ct v cv+ ct+
F1: ♀ v+ / v . cv / cv+ . ct / ct+ is test-crossed with triple recessive homozygous maleX ♂ v / v . cv / cv . ct / ct
Female’s gametes determine eight phenotypic classes The loci in REAL order:Parental, no COv cv+ ct+ 580 v+ cv ct 592
SCO (ct, cv)v cv ct+ 45 v+ cv+ ct 40
SCO (v, ct)v cv ct 89 v+ cv+ ct+ 94
DCOv cv+ ct 3 v+ cv ct+ 5
Total 1448
Three-point testcross
v ct+ cv+
v+ ct cv
v ct+ cv+
v+ ct cv
v ct+ cv+
v+ ct cv
v ct+ cv+
v+ ct cv
v ct cv
Step 8: refining the map distance
RF (v,cv) = 45+40+89+94+ 2*3 + 2*5 / 1448 = 0.196 or 19.6 m.u.we need to account for DCO recombinations
The 18.5 m.u. distance between the loci v and cv (Step 2) was underestimate because it did not account for the DCO progeny
Linkage map is known:
13.2 6.419.6
v+/ v+ . cv / cv . ct / ct X v / v . cv+/ cv+ . ct+ / ct+ (in the arbitrary order!)
Gametes: v+ cv ct v cv+ ct+
F1: ♀ v+ / v . cv / cv+ . ct / ct+ is test-crossed with triple recessive homozygous maleX ♂ v / v . cv / cv . ct / ct
Female’s gametes determine eight phenotypic classes DCO flips alleles in the center Parental: MOST ABUNDANTv cv+ ct+ 580 v+ cv ct 592
SCO (ct, cv)v cv ct+ 45 v+ cv+ ct 40
SCO (v, ct)v cv ct 89 v+ cv+ ct+ 94
DCO : LEAST ABUNDANTv cv+ ct 3 v+ cv ct+ 5
Three-point testcross
v ct+ cv+
v+ ct cv
v ct+ cv+
v+ ct cv
v ct+ cv+
v+ ct cv
v ct cv+
v+ ct+ cv
v ct cv
Step 9: learning some tricks
this pair flipped with respect to the others, thus it is in the center!
1. Determine Parentals and DCO2. Find the pair of alleles that flips3. The respective locus is in the center
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