The easiest cases are where we can count recombinants. For example, in a
double backcross ++/pt 
pt/pt involving two recessive petal color mutants p
claret colored) and t (crimson-magenta) of the poppy Papaver Rhoeas (+ =
wild-type petals are scarlet), Philp (1934) observed the following:
The recombination fraction is (28+30)/(123+28+30+124) = 0.19.
In another experiment with the same mutants, ++/pt 
++/pt, Philp observed:
What estimate do these give of the recombination fraction?
This problem, the estimation of recombination fractions in phase-known
intercrosses with both traits exhibiting dominance, had been solved some
years earlier by Fisher and Balmukand (1928). Let us consider the cross
AB/ab x AB/ab, denoting the recombination fraction between the loci in males
by r and that in females by 
(We do this because in general there are large
differences between male and female recombination fractions.)
Male gametes with genetic constitution AB, Ab, aB and ab are produced with
frequencies 
and 
respectively, and similarly for
female gametes (replace r with 
). Forming all 16 possible combinations,
taking into account the dominance of A over a and B over b, we find that
the four distinguishable phenotypes arise with the following frequencies:
where 
Clearly only t is identifiable from the phenotypic
frequencies, but if we suppose (as is often done) that 
then r can
be recovered from the relation 
In practice, we would estimate
t (or r) by maximum likelihood, and the paper by Fisher and Balmukand gives
a clear discussion of the relative efficiencies of other estimators.
Estimation of recombination fractions with human pedigree data is less straightforward, and we will discuss this later.
Exercise 4. Check that the 16 cells collapse into 4 with the probabilities indicated.