A guided workflow for turning eight offspring classes from a three-point testcross into gene order, map distances, coefficient of coincidence, and interference.
The big idea
Eight classes tell one chromosome story
In a three-gene testcross, each offspring phenotype reports the gamete produced by the heterozygous parent. With three genes there are eight possible gamete classes: two parental (nonrecombinant), four single-crossover classes, and two double-crossover classes.
① Find extremesThe two most frequent classes are usually parental; the two least frequent are usually double crossovers.
② Find the middle geneCompare a parental class with its corresponding DCO. The allele that switches identifies the middle gene.
③ Map each intervalFor each interval, count its two SCO classes plus both DCO classes.
④ Test crossover independenceCompare observed DCOs with the number expected from the two recombination frequencies.
Step 1
Determine the parental and double-crossover classes
Start with the observed counts—not the phenotype names. Under the standard assumptions of a mapping cross, parental gametes are recovered most often, while DCO gametes are recovered least often because two crossover events are less common than zero or one.
Class
Gamete
Observed
Interpretation
A
+++
350
Parental
B
abc
350
Parental
C
a++
70
SCO
D
+bc
70
SCO
E
++c
60
SCO
F
ab+
60
SCO
G
+b+
20
DCO
H
a+c
20
DCO
Important: “largest = parental” and “smallest = DCO” is a mapping rule based on the expected frequencies in a standard three-point testcross. Biological effects such as viability differences can distort observed class sizes.
Step 2
Use the DCO classes to identify the middle gene
Do not guess gene order from the labels. Align each DCO with the parental class it most closely resembles. In a DCO, the outside alleles return to the parental arrangement while the middle allele is the one that differs.
Parental: + + + ↔ DCO: + b +
↑
only b changes
Therefore: b is the middle gene
a
interval 1
b
interval 2
c
The reverse order, c—b—a, represents the same map. What matters is that b lies between a and c.
Step 3
Choose the correct recombinants for each interval
Once the order is known, classify the four remaining classes by asking which interval must have crossed over. Then calculate each interval separately.
Interval
Relevant SCO classes
Add DCO classes?
Recombinants
a—b
a++ (70) + +bc (70)
Yes: 20 + 20
180
b—c
++c (60) + ab+ (60)
Yes: 20 + 20
160
Recombination frequency (RF) = recombinants in the interval / total offspring
Map distance (cM) ≈ RF × 100
a—b: (70 + 70 + 20 + 20) / 1000 × 100 = 18 cM
b—c: (60 + 60 + 20 + 20) / 1000 × 100 = 16 cM
a
18 cM
b
16 cM
c
The classic mistake: forgetting to include the DCO offspring in both adjacent intervals. Each DCO contains one crossover in interval a—b and another in interval b—c.
Step 4
Calculate coefficient of coincidence (C) and interference (I)
If crossovers in the two intervals occurred independently, the expected frequency of double crossovers would be the product of the two interval recombination frequencies.
Expected DCO frequency = RF₁ × RF₂
Expected DCO number = RF₁ × RF₂ × N
= 0.18 × 0.16 × 1000
= 28.8 expected DCO offspring
Observed DCO = 20 + 20 = 40
Coefficient of coincidence (C) = observed DCO / expected DCO
C = 40 / 28.8 = 1.389
Interference (I) = 1 − C
I = 1 − 1.389 = −0.389
In this teaching dataset, C > 1 and therefore I < 0: more DCOs were observed than expected under independence. Positive interference would instead produce fewer DCOs than expected, giving C < 1 and I > 0.
Try it
Interactive mapping calculator
After you have identified the correct classes, enter the pooled counts below. The calculator handles the arithmetic; you still have to decide which offspring belong in each category.
Quick concept check
Why are DCO offspring included when calculating both interval distances?
Exactly. A double crossover contributes one recombination event to each adjacent interval.
Which comparison reveals the middle gene?
Yes. The allele that changes between a parental chromosome and the matching DCO identifies the middle locus.
Take-home workflow
Four moves. Every three-point map.
1. Highest counts → identify the two parental classes.
Lowest counts → identify the two DCO classes.
2. Compare parental ↔ DCO → the allele that switches is the middle gene.
3. For EACH interval:
RF = (SCOs for that interval + ALL DCOs) / N
distance ≈ RF × 100 cM
4. Expected DCO = RF₁ × RF₂ × N
C = observed DCO / expected DCO
I = 1 − C