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Three Point Mapping Experiment: Backward Analysis
In a three-point cross, the following are essential:
1. Knowledge of whether genes/loci are autosomal or X-linked,
dominance relationship.
2. Heterozygous individual (heterozygous at all loci) to produce
gametes.
3. Accurate determination of genotypes of gametes that are
produced by the heterozygous individual.
4. Sufficient number of progeny from the linkage testcross so
that all crossover classes are represented.
Determine: gene order, map distances between genes,
coefficient of coincidence and interference
Three Point Mapping: Backward Analysis
wing morphology
cu
cu+ (+)
curled: wings curled upward
straight wings
cu tx gro
+
+
+
cu tx gro
+
+
+
(P)
wing angle
tx
tx+ (+)
Linkage
Testcross
taxi: wings 75° from axis
wings not held out
cu tx gro
cu tx gro
+
cu tx gro
(F1)
bristles
+
+
gro
groucho: clumps above eyes
gro+ (+) no extra bristles
Three autosomal loci are linked, but
suppose gene order is unknown.
Conduct linkage crosses as shown on
the right.
Generate progeny.
Suppose n = 1000
Three Point Mapping: Backward Analysis
Progeny Phenotypes/Gametes from Heterozygote
Phenotype
Counts
Class
Phenotype
cu tx gro
269
NCO
cu +
+
NCO
+ tx gro
196
?
cu + gro
3
?
+ tx
4
+
+
324
cu tx
+
1
+
+ gro
2
593
3
+
+
Counts
Class
201
?
397
?
?
7
?
GENE ORDER: The locus in the middle is the one that differs from the parentals.
A
B
C
A
b
a
b
c
a
B
C
c
AbC
aBc
DCO class is expected
to have the fewest
number of progeny.
Three Point Mapping: Backward Analysis
Phenotype
Counts
Class
Phenotype
Counts
Class
cu gro tx
269
NCO
cu
+
201
SCO I
+
+
+
324
593
+
gro tx
196
397
cu
+
tx
1
DCO
cu gro +
3
SCO II
+
gro +
2
3
+
4
7
+
+
tx
Region
Recombination Frequency
Map Distance
cu - gro
gro - tx
cu - tx
(397 + 3)/1000 = 0.40 or 40%
(7 + 3)/1000 = 0.01 or 1%
40 m.u.
1 m.u.
41 m.u.
Interference (I) = 1 - C
Coefficient of Coincidence (C)
# observed DCO =
# expected DCO
3
=
0.75
I = 1 - 0.75 = 0.25
(0.4)(0.01)(1000)
I > 0 means fewer obs DCO than expected.
I < 0 means more obs DCO than expected.