Example I

For F families

L =       n!      (m)(m)(m)(m)
       a!b!c!d!

class	A_B_	A_bb	aaB_	aabb	Total
observed	a	b	c	d	n
expected	m	m	m	m

2p = 0.5(2) = 1, so we can leave out the 2p above. Now we substitute p=1/2 in the above equation.

Now the amount of information per individual () for F family is:

We set p = 0.5, then:

Now the total amount of information for a specific F family when p=1/2 is ni = I = n16/9.

Fisher showed that c/I is distributed as X² w/1df for a test of significance for deviation from 50% recombination. Each c²/I can be calculated for each type of family which are combined to obtain the combined estimate of p.