Lecture 6: Gene mapping by 3-point testcross
1.
2.
3.
4.
Three-point testcross
Double cross-overs
Interference
Mapping of X chromosome in
humans
With the increase of map distances, an increasing
fraction of cross-overs becomes “invisible”
A single crossover
between linked genes
generates recombinant
gametes
A double crossover between linked
genes gives parental 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 threepoint testcross
Three-point testcross
Mutant alleles:
v, determines vermilion eyes (bright red)
cv, crossveinless wings
ct, cut wing edges
P: v+/ v+ . cv / cv . ct / ct
Gametes:
F1:
X
v / v . cv+/ cv+ . ct+ / ct+ (in the arbitrary order!)
v+ cv ct
v cv+ ct+
♀ v+ / v . cv / cv+ . ct / ct+
is test-crossed with triple recessive homozygous male
X
♂ v / v . cv / cv . ct / ct
Female’s gametes determine eight phenotypic classes:
v cv+ ct+
v+ cv ct
580
592
v cv ct+
v+ cv+ ct
45
40
v cv ct
v+ cv+ ct+
89
94
v cv+ ct
v+ cv ct+
3
5
Total 1448
Three-point testcross
Mutant alleles
v, determines vermilion eyes (bright red)
cv, crossveinless wings
ct, cut wing edges
P: v+/ v+ . cv / cv . ct / ct
Gametes:
F1:
Step 1: identify parental classes
X
v / v . cv+/ cv+ . ct+ / ct+ (in the arbitrary order!)
v+ cv ct
v cv+ ct+
♀ v+ / v . cv / cv+ . ct / ct+
is test-crossed with triple recessive homozygous male
X
♂ v / v . cv / cv . ct / ct
Female’s gametes determine eight phenotypic classes:
Parental, no CO
v cv+ ct+
580
v+ cv ct
592
v cv ct+
v+ cv+ ct
45
40
v cv ct
v+ cv+ ct+
89
94
v cv+ ct
v+ cv ct+
3
5
Total 1448
Three-point testcross
Step 2: map any two loci
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 ct+
v+ cv+ ct
45
40
v cv ct
v+ cv+ ct+
89
94
v
cv
18.5 m.u.
Three-point testcross
Step 3: map any other two loci
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 cv ct
v+ cv+ ct+
89
94
v cv+ ct
v+ cv ct+
3
5
v
ct
13.2 m.u.
Three-point testcross
we are now trying to put the two maps together:
v
ct
cv
13.2
18.5
?
OR
ct
v
13.2
cv
18.5
but we do not know yet which one is correct
Three-point testcross
the solution depends on the ct to cv distance:
v
ct
13.2
cv
5.3
18.5
OR
ct
v
13.2
31.7
cv
18.5
and we need to determine this distance by finding the RF(ct, cv)
Three-point testcross
Parental: cv+
cv
Step 3: map the remaining two loci
ct+
Recombinant:
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 %
v cv ct+
v+ cv+ ct
45
40
v cv+ ct
v+ cv ct+
3
5
ct
cv
6.4 m.u.
Three-point testcross
Step 4: Choosing the right map
the distance fits better with the fist map
v
ct
13.2
cv
6.4
19.6
ct
v
13.2
31.7
cv
18.5
?
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
Three-point testcross
Step 5: draw the loci in REAL order
Linkage map is known:
v
ct
13.2
P: v+/ v+ . cv / cv . ct / ct
Gametes:
F1:
X
cv
6.4
v / v . cv+/ cv+ . ct+ / ct+ (in the arbitrary order!)
v+ cv ct
v cv+ ct+
♀ v+ / v . cv / cv+ . ct / ct+
is test-crossed with triple recessive homozygous male
X
♂ v / v . cv / cv . ct / ct
Female’s gametes determine eight phenotypic classes
Parental, no CO
v cv+ ct+
580
v+ cv ct
592
v cv ct+
v+ cv+ ct
45
40
v cv ct
v+ cv+ ct+
89
94
v cv+ ct
v+ cv ct+
3
5
Total 1448
The loci in REAL order:
v
ct+
cv+
v+
ct
cv
Three-point testcross
Step 6: single crossover classes
Linkage map is known:
v
ct
13.2
P: v+/ v+ . cv / cv . ct / ct
Gametes:
F1:
X
cv
6.4
v / v . cv+/ cv+ . ct+ / ct+ (in the arbitrary order!)
v+ cv ct
v cv+ ct+
♀ v+ / v . cv / cv+ . ct / ct+
is test-crossed with triple recessive homozygous male
X
♂ v / v . cv / cv . ct / ct
Female’s gametes determine eight phenotypic classes
Parental, no CO
v cv+ ct+
580
v+ cv ct
592
SCO (ct, cv)
v cv ct+
v+ cv+ ct
45
40
v cv ct
v+ cv+ ct+
89
94
v cv+ ct
v+ cv ct+
3
5
Total 1448
The loci in REAL order:
v
ct+
cv+
v+
ct
cv
v
ct+
cv+
v+
ct
cv
Three-point testcross
Step 6: single crossover classes
Linkage map is known:
v
ct
13.2
P: v+/ v+ . cv / cv . ct / ct
Gametes:
F1:
X
cv
6.4
v / v . cv+/ cv+ . ct+ / ct+ (in the arbitrary order!)
v+ cv ct
v cv+ ct+
♀ v+ / v . cv / cv+ . ct / ct+
is test-crossed with triple recessive homozygous male
X
♂ v / v . cv / cv . ct / ct
Female’s gametes determine eight phenotypic classes
Parental, no CO
v cv+ ct+
580
v+ cv ct
592
SCO (ct, cv)
v cv ct+
v+ cv+ ct
45
40
SCO (v, ct)
v cv ct
v+ cv+ ct+
89
94
v cv+ ct
v+ cv ct+
3
5
Total 1448
The loci in REAL order:
v
ct+
cv+
v+
ct
cv
v
ct+
cv+
v+
ct
cv
v
ct+
cv+
v+
ct
cv
Three-point testcross
Step 7: double crossover classes
Linkage map is known:
v
ct
13.2
P: v+/ v+ . cv / cv . ct / ct
Gametes:
F1:
X
cv
6.4
v / v . cv+/ cv+ . ct+ / ct+ (in the arbitrary order!)
v+ cv ct
v cv+ ct+
♀ v+ / v . cv / cv+ . ct / ct+
is test-crossed with triple recessive homozygous male
X
♂ v / v . cv / cv . ct / ct
Female’s gametes determine eight phenotypic classes
Parental, no CO
v cv+ ct+
580
v+ cv ct
592
SCO (ct, cv)
v cv ct+
v+ cv+ ct
45
40
SCO (v, ct)
v cv ct
v+ cv+ ct+
89
94
DCO
v cv+ ct
v+ cv ct+
3
5
Total 1448
The loci in REAL order:
v
ct+
cv+
v+
ct
cv
v
ct+
cv+
v+
ct
cv
v
ct+
cv+
v+
ct
cv
v
ct+
cv+
v+
ct
cv
Three-point testcross
Step 8: refining the map distance
Linkage map is known:
v
ct
13.2
v+/ v+ . cv / cv . ct / ct
Gametes:
X
cv
6.4
19.6
v / v . cv+/ cv+ . ct+ / ct+
v+ cv ct
The 18.5 m.u. distance between
the loci v and cv (Step 2) was
underestimate because it did not
(in the arbitrarbeyaccount
order!)for the DCO progeny
v cv+ ct+
RF (v,cv)
2*3 + 2*5
/ 1448
= 0.196
or 19.6
♀ v /=
v 45+40+89+94+
cv / cv ct / ct
is test-crossed
with triple
recessive
homozygous
male m.u.
X
♂ v / v cv / cv ct / ct
we need to account for DCO recombinations
F1:
+
.
+ .
+
.
Female’s gametes determine eight phenotypic classes
Parental, no CO
v cv+ ct+
580
v+ cv ct
592
SCO (ct, cv)
v cv ct+
v+ cv+ ct
45
40
SCO (v, ct)
v cv ct
v+ cv+ ct+
89
94
DCO
v cv+ ct
v+ cv ct+
3
5
Total 1448
The loci in REAL order:
v
ct+
cv+
v+
ct
cv
v
ct+
cv+
v+
ct
cv
v
ct+
cv+
v+
ct
cv
v
ct+
cv+
v+
ct
cv
.
Three-point testcross
Step 9: learning some tricks
Linkage map is known:
v
ct
13.2
v+/ v+ . cv / cv . ct / ct
Gametes:
F1:
X
cv
6.4
19.6
v / v . cv+/ cv+ . ct+ / ct+ (in the arbitrary order!)
1.v+ cvDetermine
Parentals
and DCO
ct
v cv+ ct+
2. Find the pair
of alleles that flips
♀ v / v cv / cv ct / ct
is test-crossed with triple recessive homozygous male
X
♂ v / v cv / cv
3. The respective locus is in the center
+
.
+ .
+
.
Female’s gametes determine eight phenotypic classes
Parental: MOST ABUNDANT
v
ct+
cv+
v cv+ ct+
v+ cv ct
v+
ct
cv
v
ct+
cv+
v+
ct
cv
v
ct+
cv+
v+
ct
cv
v
ct
cv+
v+
ct+
cv
SCO (ct, cv)
v cv ct+
v+ cv+ ct
SCO (v, ct)
v cv ct
v+ cv+ ct+
580
592
this
pair flipped
45
40
with
respect to
the
others, thus it
89
is94in the center!
DCO : LEAST ABUNDANT
v cv+ ct
3
v+ cv ct+
5
DCO flips alleles in the center
.
ct / ct
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