Once a three-point test cross has revealed gene order, the same offspring counts can be used to estimate the genetic distance of each adjacent interval.

Suppose the inferred order is

$$A-B-C.$$

Using the test-cross counts

the total offspring number is

$$N=1000.$$

The parental classes are $ABC$ and $abc$, and $AbC$ and $aBc$ are double-crossover classes.

Count recombinants separately for each interval

For interval $A-B$, a gamete is recombinant when the $A/B$ association differs from the parental phase. The relevant classes are

$$Abc,\quad aBC,\quad AbC,\quad aBc.$$

The double-crossover classes must be included because each double crossover contains one crossover in the $A-B$ interval.

Thus

$$r_{AB}=\frac{76+72+12+10}{1000}=0.170,$$

so

$$A-B\approx17.0\ \mathrm{cM}.$$

For interval $B-C$, the recombinant classes are

$$ABc,\quad abC,\quad AbC,\quad aBc,$$

so

$$r_{BC}=\frac{68+64+12+10}{1000}=0.154,$$

and

$$B-C\approx15.4\ \mathrm{cM}.$$

The resulting map is

A --------17.0-------- B -------15.4------- C

and the map distance along the chromosome from $A$ to $C$ is

$$17.0+15.4=32.4\ \mathrm{cM}.$$

The outer-marker recombinant fraction misses double crossovers

If we ignore the middle locus $B$ and classify offspring only by $A$ and $C$, the classes recombinant for the outer markers are

$$Abc,\quad aBC,\quad ABc,\quad abC.$$

Their total is

$$76+72+68+64=280,$$

so the direct outer-marker recombination frequency is only

$$r_{AC}=\frac{280}{1000}=0.28.$$

A naive two-point conversion would therefore suggest only

$$28\ \mathrm{cM},$$

which is smaller than the interval-sum map distance

$$32.4\ \mathrm{cM}.$$

The missing information is carried by the double-crossover classes $AbC$ and $aBc$: each contains a crossover in both adjacent intervals, yet the two outer alleles have returned to their parental association. They look nonrecombinant if $B$ is not observed.

This is why double-crossover classes contribute to the recombinant count for both adjacent intervals, while a direct endpoint comparison can miss them.

Three-point mapping therefore solves two problems at once: it identifies gene order and exposes crossover events that a pairwise outer-marker comparison hides.