Drummond:Akashi's Test
Akashi's test on a single gene
Akashi's test is very simple. Suppose you have two aligned codon sequences (a target sequence and an orthologous sequence) and a list of preferred codons. The question we wish to answer: Is there an association between preferred codons and conserved amino acids, controlling for differences between amino acids?
From the aligned codon sequences, build a 2x2 table with entries a, b, c, and d like this:
| AA=Ser | Conserved | Variable |
| Preferred | [math]\displaystyle{ a }[/math] | [math]\displaystyle{ b }[/math] |
| Unpreferred | [math]\displaystyle{ c }[/math] | [math]\displaystyle{ d }[/math] |
for each amino acid. You'll usually have 18 tables; W and M have no synonymous codon alternatives and therefore don't contribute to Akashi's test.
- [math]\displaystyle{ a }[/math] = the number of codons in your target sequence that encode amino acid AA, are PREFERRED, and encode an AA which is unchanged (CONSERVED) in the orthologous sequence
- [math]\displaystyle{ b }[/math] = the number of codons in your target sequence that encode amino acid AA, are PREFERRED and encode an AAwhich is different (VARIABLE) in the orthologous sequence
- [math]\displaystyle{ c }[/math] = the number of codons in your target sequence that encode amino acid AA, are UNPREFERRED and encode an AA which is unchanged (CONSERVED) in the orthologous sequence
- [math]\displaystyle{ d }[/math] = the number of codons in your target sequence that encode amino acid AA, are UNPREFERRED and encode an AA which is different (VARIABLE) in the orthologous sequence
Now the statistics. Assuming no association -- that is, assuming that the probability of a codon being preferred (which we designate [math]\displaystyle{ p }[/math]) is independent of the probability that it encodes a conserved amino acid (which we designate [math]\displaystyle{ q }[/math]) -- we can write down the expected value and variance of [math]\displaystyle{ a }[/math], [math]\displaystyle{ E(a) }[/math] and [math]\displaystyle{ V(a) }[/math]:
[math]\displaystyle{ n = a + b + c + d }[/math]
[math]\displaystyle{ \hat{p} = \frac{a + b}{n} }[/math]
[math]\displaystyle{ E(a) = }[/math]
