next up previous
Next: Summary Up: Analyzing the genetic structure Previous: Introduction

An example from Isotoma petraea

To make the differences in implementation and calculation clear, I'm going to use data from 12 populations of Isotoma petraea in southwestern Australia surveyed for genotype at GOT-1 [2] as an example throughout these discussions (Table [*]).

Table 1: Genotype counts at the $GOT-1$ locus in Isotoma petraea (from [2]).
  Genotype  
Population $A_{1}A_{1}$ $A_{1}A_{2}$ $A_{2}A_{2}$ $\hat p$
Yackeyackine Soak 29 0 0 1.0000
Gnarlbine Rock 14 3 3 0.7750
Boorabbin 15 2 3 0.8000
Bullabulling 9 0 0 1.0000
Mt. Caudan 9 0 0 1.0000
Victoria Rock 23 5 2 0.8500
Yellowdine 23 3 4 0.8167
Wargangering 29 3 1 0.9242
Wagga Rock 5 0 0 1.0000
``Iron Knob Major'' 1 0 0 1.0000
Rainy Rocks 0 1 0 0.5000
``Rainy Rocks Major'' 1 0 0 1.0000


Let's ignore the sampling problem for a moment and calculate the $F$-statistics as if we had observed the population allele frequencies without error. They'll serve as our baseline for comparison.

\begin{eqnarray*}
\bar p &=& 0.8888 \\
\hbox{Var}(p) &=& 0.02118 \\
F_{st} &=&...
...st}} \\
&=& \frac{0.3221 - 0.2143}{1 - 0.2143} \\
&=& 0.1372
\end{eqnarray*}



Subsections
next up previous
Next: Summary Up: Analyzing the genetic structure Previous: Introduction
Kent Holsinger 2008-08-18