Likelihood analysis of multiple loci

Section author: Gavin Huttley

We want to know whether an exchangeability parameter is different between alignments. We will specify a null model, under which each alignment get’s it’s own motif probabilities and all alignments share branch lengths and the exchangeability parameter kappa (the transition / transversion ratio). We’ll split the example alignment into two-pieces.

from cogent3 import load_aligned_seqs, make_tree, make_table
from cogent3.evolve.models import HKY85
from cogent3.recalculation.scope import EACH, ALL
from cogent3.maths.stats import chisqprob

aln = load_aligned_seqs("data/long_testseqs.fasta")
half = len(aln) // 2
aln1 = aln[:half]
aln2 = aln[half:]

We provide names for those alignments, then construct the tree, model instances.

loci_names = ["1st-half", "2nd-half"]
loci = [aln1, aln2]
tree = make_tree(tip_names=aln.names)
mod = HKY85()

To make a likelihood function with multiple alignments we provide the list of loci names. We can then specify a parameter (other than length) to be the same across the loci (using the imported ALL) or different for each locus (using EACH). We conduct a LR test as before.

lf = mod.make_likelihood_function(tree, loci=loci_names, digits=2, space=3)
lf.set_param_rule("length", is_independent=False)
lf.set_param_rule("kappa", loci=ALL)
lf.set_alignment(loci)
lf.optimise(show_progress=False)
lf

HKY85

log-likelihood = -9168.3331

number of free parameters = 2

Global params
kappalength
3.980.13
Locus motif params
locusACGT
2nd-half0.350.190.220.24
1st-half0.380.180.210.22
all_lnL = lf.lnL
all_nfp = lf.nfp
lf.set_param_rule("kappa", loci=EACH)
lf.optimise(show_progress=False)
lf

HKY85

log-likelihood = -9167.5373

number of free parameters = 3

Global params
length
0.13
Locus params
locuskappa
1st-half4.33
2nd-half3.74
Locus motif params
locusACGT
2nd-half0.350.190.220.24
1st-half0.380.180.210.22
each_lnL = lf.lnL
each_nfp = lf.nfp
LR = 2 * (each_lnL - all_lnL)
df = each_nfp - all_nfp

Just to pretty up the result display, I’ll print(a table consisting of the test statistics created on the fly.)

make_table(
    header=["LR", "df", "p"], rows=[[LR, df, chisqprob(LR, df)]], digits=2, space=3,
)
LRdfp
1.5910.21

1 rows x 3 columns