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Three-Point Cross: Gene Order and Double Crossovers
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A three-point cross follows three linked genes at once. It gives the gene order and the map distances in one experiment, and it catches double crossovers that a two-point cross undercounts.
The three rules
- The two most common classes are the parentals (no crossover).
- The two rarest classes are the double crossovers.
- Compare a double crossover class to a parental class. The one gene that switched is the middle gene.
Worked example
A heterozygote A B C / a b c is test-crossed to a b c / a b c. The 1080 offspring:
| Phenotype | Count | Class |
|---|---|---|
| A B C | 430 | Parental |
| a b c | 440 | Parental |
| a B C | 62 | Single crossover |
| A b c | 58 | Single crossover |
| A B c | 38 | Single crossover |
| a b C | 42 | Single crossover |
| A b C | 5 | Double crossover |
| a B c | 5 | Double crossover |
Gene order. A b C differs from the parental A B C only at B, so B is in the middle. The order is A, B, C. (If the order is not given in the problem, the order you started with may be wrong; the double crossover test fixes it.)
Distance A to B. Count every offspring with a crossover between A and B, including double crossovers: 62 + 58 + 5 + 5 = 130. 130/1080 = 12.0%, so 12.0 map units.
Distance B to C. 38 + 42 + 5 + 5 = 90. 90/1080 = 8.3 map units.
Distance A to C. 12.0 + 8.3 = 20.4 map units. Counting only the offspring that look recombinant for A and C gives 200/1080 = 18.5%, which is too low, because a double crossover swaps the middle gene and leaves A and C in their parental combination. Add each double crossover twice (2 x 10 = 20) to get 220/1080 = 20.4%.
Coefficient of coincidence and interference
- Expected double crossovers = 0.120 x 0.083 x 1080, about 10.8.
- Observed = 10.
- Coefficient of coincidence = observed/expected, about 0.92.
- Interference = 1 - coincidence, about 0.08.
Positive interference means one crossover makes a nearby second crossover less likely, which is the usual case.
Why double crossovers are the rarest
A double crossover needs two independent events, so its frequency is roughly the product of the two single-crossover frequencies. Here that is about 1 percent, versus 8 to 12 percent for the single classes.
Common traps
- Picking the middle gene as the one that is most different from the parentals. It is the one that switched in the double crossover class.
- Leaving out the double crossovers when computing each interval. Each interval must include them.
- Using the three-point order you were given without checking. The data decide.
- Expecting distances to add exactly after a rough count. They add once double crossovers are counted in every interval they span.
One-minute recap
Parentals are most common, double crossovers are rarest, and the gene that flips in a double crossover is in the middle. Each interval counts its single crossovers plus all double crossovers. Distances then add, and coincidence compares observed to expected doubles. For the two-point basics, see the recombination frequency note.
Check yourself: 5-question mini-quiz
1. In a three-point test cross, which classes are the double crossovers?
The two rarest phenotype classes, since two independent crossovers are needed.
2. A double crossover class is A b C and the parental class is A B C. Which gene is in the middle?
B, because it is the only gene that switched.
3. Why must double crossovers be counted in both intervals?
A double crossover has a crossover in each interval, so it is a recombinant for each pair of neighbors, and leaving it out underestimates the distances.
4. Distances of 12 and 8 map units are found for A to B and B to C. What is A to C?
20 map units, if double crossovers are counted correctly.
5. Observed doubles are 10 and expected are 10.8. What is interference?
Coincidence is 10/10.8, about 0.92, so interference is about 0.08.
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Keep studying
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- All free notes
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Download your free sample: 3-page Amino Acids, Proteins and Enzymes cheat sheet (PDF)
Meanwhile, start with the free notes.