SME
Genetics · Problem type
Three-point cross: interference
Use this for “Genes 12 cM and 20 cM apart give 15 double crossovers in 1000. What is the interference?”
Diagram
A–C = 12 + 20 = 32 cM
Expected doubles = 0.12 × 0.2 × 1,000 = 24
c.o.c.
=
15
24
=
0.625
I = 1 − 0.625 = 0.375
This is an example. Type your own values to start a new problem.
d₁
Distance A–B
example · cM
d₂
Distance B–C
example · cM
dAC
Distance A–C
worked out (not typed) · cM
N
Offspring
example
E
Expected double crossovers
worked out (not typed)
O
Observed double crossovers
example
c.o.c.
Coefficient of coincidence
calculated
I
Interference
calculated
Formulas
dAC
=
d₁
+
d₂
32 = 12 + 20
E
=
d₁
100
×
d₂
100
×
N
24
=
12
100
×
20
100
×
1,000
c.o.c.
=
O
E
0.625
=
15
24
I
=
1
−
c.o.c.
0.375 = 1 − 0.625
Assumptions
•
The middle gene B is the one that switches in the double crossovers.
•
With no interference, crossovers in the two intervals are independent: their chances multiply.
•
Interference I = 1 − c.o.c.: one crossover makes a second one nearby less likely.
Step-by-step
WE KNOW
d₁ = 12 cM, d₂ = 20 cM, N = 1,000, O = 15
FIND
distance A–C (dAC), expected double crossovers (E), coefficient of coincidence (c.o.c.), interference (I)
Step 1 · Find distance A–C (dAC)
Use dAC = d₁ + d₂
Map distances add along the chromosome.
dAC = 12 + 20
dAC = 32 cM
Step 2 · Find expected double crossovers (E)
Use E = d₁ ÷ 100 × d₂ ÷ 100 × N
Two independent crossovers: multiply the two chances, then the offspring.
E
=
12
100
×
20
100
×
1,000
E
=
0.12
×
0.2
×
1,000
E
=
0.024
×
1,000
E
=
24
Step 3 · Find coefficient of coincidence (c.o.c.)
Use c.o.c. = O ÷ E
How many of the expected double crossovers turned up.
c.o.c.
=
15
24
c.o.c. = 0.625
Step 4 · Find interference (I)
Use I = 1 − c.o.c.
The share of expected doubles that a first crossover prevented.
I
=
1
−
0.625
I
=
0.375
Check
Put the numbers back in. Both sides should match.
32 = 12 + 20 ✓
24
=
12
100
×
20
100
×
1,000
✓
0.625
=
15
24
✓
0.375 = 1 − 0.625 ✓
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