1. Three Point Mapping: Forward Analysis
Single Crossovers Region I (SCO I)
jv se+ h+
jv+ se h
6.8 m.u.
0.5 m.u. jv se h
jv+
se+ h+
Region: I II
Single Crossovers Region II (SCO II)
jv se h+
jv+ se+ h
Double Crossovers (DCO)
jv se+ h
jv+ se h+
Prob(XO in Region I) = 0.068
Prob(XO in Region II) = 0.005
Prob(XOs in Region I and II) =
No Crossovers, Parental (NCO)
jv se h
jv+ se+ h+
(0.068) x (0.005) =

2. Three Point Mapping: Forward Analysis
Prob(DCO) = (0.068) x (0.005) =
Prob(SCO I) = =
Prob(SCO II) = =
Prob(NCO) =
1 - Prob(SCO I) - Prob(SCO II) - Prob(DCO) =
6.8 m.u.
0.5 m.u. jv se h
jv+
se+ h+
Region: I II
Double Crossovers (DCO)
jv se+ h
jv+ se h+
Single Crossovers Region I (SCO I)
jv se+ h+
jv+ se h
Single Crossovers Region II (SCO II)
jv se h+
jv+ se+ h
No Crossovers, Parentals (NCO)
jv se h
jv+ se+ h+

3. Three Point Mapping Experiment: Backward Analysis
In a three-point cross, the following are essential:
Knowledge of whether genes/loci are autosomal or X-linked, dominance relationship.
Heterozygous individual (heterozygous at all loci) to produce gametes.
Accurate determination of genotypes of gametes that are produced by the heterozygous individual.
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

4. Three Point Mapping: Backward Analysis
wing morphology
cu curled: wings curled upward
cu+ (+) straight wings
wing angle
tx taxi: wings 75° from axis
tx+ (+) wings not held out
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.
cu tx gro
(P)
cu tx gro
Linkage
Testcross
cu tx gro
cu tx gro
(F1)
cu tx gro
Generate progeny.
Suppose n = 1000

5. Three Point Mapping: Backward Analysis
Progeny Phenotypes/Gametes from Heterozygote
Phenotype
Counts
Class
Phenotype
Counts
Class
cu tx gro
269
NCO cu
201 ?
593
397
324
NCO
+ tx gro
196 ?
cu tx + 1 ?
cu + gro 3 ? 3 7
gro 2 ?
+ tx + 4 ?
GENE ORDER: The locus in the middle is the one that differs from the parentals.
A B C
A b C
A b C
DCO class is expected to have the fewest number of progeny.
a b c
a B c
a B c

6. I > 0 means fewer obs DCO than expected.
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 tx 4 7
Region
cu - gro
gro - tx
cu - tx
Recombination Frequency
( )/1000 = 0.40 or 40%
(7 + 3)/1000 = 0.01 or 1%
Map Distance
40 m.u.
1 m.u.
41 m.u.
Coefficient of Coincidence (C)
# observed DCO = = 0.75
# expected DCO (0.4)(0.01)(1000)
I > 0 means fewer obs DCO than expected.
I < 0 means more obs DCO than expected.
Interference (I) = 1 - C
I = = 0.25