the best value of the coupling-ratio from a given set of data, 487 
normal distribution of errors is assumed in the calculation of the 
tables, but even from the empirical point of view the method 
suggested seems much better than any now in use. 
Suppose the two factors to be A and B, and let the gametes 
be produced by the heterozygote in the following proportions : 
AB Ab aB ab 
aie O— BY AOD eB). iB 
then, assuming random mating, zygotic forms will be produced in 
the proportions : 
AB Ab aB ab 
(p? + 0°5) (0:25 — p?) (0°25 — p?) {DP 
Let the observed proportions of zygotes be fifrfsfi, where 
fit fot+fs+ fs=1. Then we have to make a minimum the 
quantity 
(p+05-fi? , (025 —p'— fry (Wo ek 4, GES 
p +05 0-25 — p? 0-25 —p? je 
Differentiating with respect to p and equating to zero, we 
find 
(f2+ fe—fe—fe) p+ (O5fe+ fe + fe— 05 fe) p' 
a + 0°25 f2+ 01875 f2—0:0625 f2) p' 
+ 00625 f2p? — 0015625 fP=0 22. ceeces.--s es (i). 
This is an equation of the fourth degree for p?. A first approxi- 
mation to the root required may be obtained by Collins’ method 
or from the formula 
(P= OPE GAAS I= Jb) sossesnssooon occ (2) 
(which gives the value of p that makes the sum of the squares of 
differences least), or by comparison with various calculated series, 
and the solution is then readily obtained by Newton’s method. 
To take the data of the preceding note as an illustration, the 
values of the proportions f are 0°5634, 0:2207, 0:2019, and 0:0141; 
writing w for p?, this gives the equation 
— 0:22814721 + 0:248083.2' + 0:0025662? 
+ 0000012442 — 0:0000031094 = 0. 
The data suggested a gametic ratio 1:3:38:1, which gives 
| p=0125, p?=0:015625. Trial shewed that 0°0156 was not a 
very close approximation to a root; 002 proved nearer to a 
: 
| 
solution, and Newton’s method gave by two approximations 
0: 019715... Hence p=0'1404 and this gives a ratio 1 : 2°56, 
The observed frequencies were then compared with the frequencies 
