Eight Classes, One Chromosome: Reading a Three-Point Testcross
A genetics student crosses a triple heterozygote (AaBbCc, all three dominant alleles inherited from one parent) to a tester (aabbcc) and scores 1,000 progeny without molecular markers. The counts are:
- ABC 390, abc 380
- aBC 52, Abc 48
- ABc 40, abC 45
- AbC 22, aBc 23
All three genes lie on the same chromosome. The student's lab partner argues that because all eight phenotypic classes appeared in the progeny, the genes must be assorting independently, and that the parental classes are simply "a bit overrepresented by chance."
1. Determine the gene order and construct a map with distances between adjacent genes. Show the reasoning that fixes the order.
Read the worked answer
Step 1 - identify the parental classes. The two largest classes are ABC (390) and abc (380), totaling 770/1000 = 77%. These are the nonrecombinant (parental) haplotypes.
Step 2 - identify the double crossovers. The two rarest classes are AbC (22) and aBc (23), 45/1000 = 4.5%. DCOs are always the smallest classes.
Step 3 - find the middle gene. Compare a DCO haplotype with a parental haplotype and see which single gene has switched:
- AbC versus A B C: A matches, C matches, B has flipped. So B is in the middle; the order is A–B–C.
Step 4 - distances (recombinants = SCOs for that interval + all DCOs).
- A–B: aBC (52) + Abc (48) + DCO 45 = 145 → 14.5 cM
- B–C: ABc (40) + abC (45) + DCO 45 = 130 → 13.0 cM
Map: A -14.5- B -13.0- C. Expected DCO frequency from the product rule is 0.145 × 0.130 ≈ 1.9%, but 4.5% was observed, so there is positive interference (coefficient of coincidence ≈ 0.045/0.019 ≈ 2.4, i.e. negative interference - worth flagging as unusual and possibly indicating scoring error in the rare classes).
2. Evaluate the partner's claim that eight classes mean independent assortment, and state what a chi-square test would actually be testing here.
Read the worked answer
The claim confuses two different things: which classes can appear versus at what frequencies they appear.
- Independent assortment predicts equal frequencies across all eight classes: 125 each.
- The observed data are nowhere near that: ABC and abc alone are 77% of the progeny, and the rarest class is 22.
- All eight classes appear precisely because the genes are linked and crossing over occurs - recombination between linked loci generates the four nonparental haplotypes, including the DCOs. Eight classes are the expected outcome of linkage in a triple heterozygote, not evidence against it.
Chi-square: testing against a 1:1:1:1:1:1:1:1 expectation would produce an enormous statistic (deviations of 265 and 255 in the parental classes) and reject independence decisively. What is being tested is whether the frequencies fit 1:1:1:1:1:1:1:1 - a hypothesis about ratios, not about the existence of classes.
A fair follow-up: the partner's phrase "overrepresented by chance" is not a testable statement; the appropriate chance model is the equal-ratio null, and it fails badly. The correct interpretation is linkage with recombination, and the parental excess reflects the physical distance between the loci.
3. The direct A-to-C recombination frequency is 18.5%, but the A–B–C map totals 27.5 cM. Explain the discrepancy and state what it implies about the data.
Read the worked answer
Compute the A–C frequency from the parental combinations. Parental coupling for A and C is A C / a c. A–C recombinants are the haplotypes where A and C are in the other combination:
- A c: Abc (48) + ABc (40) = 88
- a C: aBC (52) + abC (45) = 97
- Total 185/1000 = 18.5%
Why it is an underestimate: the 45 double crossovers (AbC, aBc) change the middle gene B but restore the parental A–C combination, so they are scored as nonrecombinants for A and C - exactly the same reason DCOs make two-point distances short. The discrepancy 27.5 − 18.5 = 9.0 cM is roughly twice the DCO frequency (2 × 4.5 = 9.0), which is the arithmetic signature of undetected double crossovers.
Implications:
- Recombination frequency is not additive beyond short distances; map distances are additive, recombination fractions are not.
- A two-point cross between A and C alone would have reported 18.5 cM and badly underestimated the true distance - the three-point cross is what allowed the correction.
- Recombination frequencies above ~50% are uninformative; for larger distances, mapping functions or molecular markers are needed.