The Method

Example I

Example II

F2 Statistical Summary

Overall Summary

Homework Assignment #6 Questions

Homework Assignment #6 Answers

F Statistical Summary

We can now summarize those statistics for each type of F family with p = 0.2967. We can calculate the X = c/I.

Family n i c I c2/I df
F2 repulsion 1415 1.235 -76.26 1747.5 3.33 1
F2 coupling 1764 3.158 +123.22 5570.6 2.73 1
Sum   46.96 7318.1 6.06 2
Total X2   (46.96)2/7318.1 = 0.30 1
Heterogeneity X2   5.76 1

We have a pooled X2 = 0.30 w/1df which is not significant. We accept = 0.2967 as a good fit for the pooled data. The significant heterogeneity X2 tells us that we cannot find one value of that uniformly fits both types of data.

Correction for = 49.96/7318.1 = 0.007

The new estimate of p is 0.2967 + 0.007 = 0.303. There is little change in p and this new estimate is close enough. Now to find the standard error of this combined estimate of p.

Repulsion (p = 0.303) Coupling (p = 1 - 0.303)
i =        2(1+2(0.303))     
     (2+(0.303))(1-0.303))
i =       2(1+2(0.697))            (2+(0.697))(1-0.697))
= 2.3670 = 1.2459
= 1.8999
= 3.9436 = 3.0858
= 1.2780
I = Ni = 1415(1.2459)
I = Nlll= 1762.8
I = Ni = 1764(3.0858)
I = Nlll= 5443.9



= 0.0118

= + s.e. = 0.303 + 0.0118

First we estimate p seperately for F2 repulsion and coupling families using the product method.

Source of data i n I pI
F2 repulsion 0.244 1.1555 1415 1635.0 398.94
F2 coupling 0.318 2.9279 1764 5164.8 1642.21

Combined value of p by weighting =

= 398.94 + 1642.41 = 2041.35
= 1635.0 + 5164.81 =  6800

= 0.300

This method was much easier than Fisher's Scoring Method. However, Fisher's scoring method did not permit us to test the overall X2 for the combined estimate of p or to test for heterogeneity.

For a discussion of progeny testing of F2 data see:

Kramer, H.H., and C.R. Burnham. 1947. Genetics 32:379-390

Immer, F.R. 1934. Genetics 19:119-136.

Copyright 2000©, Ted Helms

Back | Home | Top | Next
Home Forward Back