1. Three-point testcross 2. Double cross-overs 3....

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




Female’s gametes determine eight phenotypic classes:Parental, no COv cv+ ct+ 580 v+ cv ct 592




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



Select any other two loci, for example v and ct.

Parental: v ct+ Recombinant: v ct

v+ ct v+ ct+




(191 / 1448) X 100% = 13.2 %

v ct

13.2 m.u.

Step 3: map any other two lociThree-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

?


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




(93 / 1448) X 100% = 6.4 %

ct cv

6.4 m.u.

Step 3: map the remaining two lociThree-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




Female’s gametes determine eight phenotypic classes The loci in REAL order:Parental, no COv cv+ ct+ 580 v+ cv ct 592




Total 1448


v ct+ cv+

v+ ct cv

v ct cv

Step 5: draw the loci in REAL order


13.2 6.4





SCO (ct, cv)v cv ct+ 45 v+ cv+ ct 40



Total 1448


v ct+ cv+

v+ ct cv

v ct+ cv+

v+ ct cv

v ct cv

Step 6: single crossover classes


13.2 6.4






SCO (v, ct)v cv ct 89 v+ cv+ ct+ 94


Total 1448


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


13.2 6.4







DCOv cv+ ct 3 v+ cv ct+ 5

Total 1448


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


13.2 6.419.6

v+/ v+ . cv / cv . ct / ct X v / v . cv+/ cv+ . ct+ / ct+ (in the arbitrarbey order!)






DCOv cv+ ct 3 v+ cv ct+ 5

Total 1448


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


13.2 6.419.6

v+/ v+ . cv / cv . ct / ct X v / v . cv+/ cv+ . ct+ / ct+ (in the arbitrary order!)



Female’s gametes determine eight phenotypic classes DCO flips alleles in the center Parental: MOST ABUNDANTv cv+ ct+ 580 v+ cv ct 592



DCO : LEAST ABUNDANTv cv+ ct 3 v+ cv ct+ 5


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

Drosophila linkage map was a prerequisite to a detailed analysis of its genome

Calculating RF for two X-linked human genes

A linkage map of the human X chromosome, one of the first human linkage maps

1. Three-point testcross 2. Double cross-overs 3....

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