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R function for a hacky approximation of the F_Tt factors in the Bryant et al SNAPP method as specialized for use by a homework assignment in the Kelly and Holder "Likelihood methods in Biology" class
calc.F_Tt <- function (r_Tb, n_Tt, r_Tt, param.vec) {
if (r_Tt > n_Tt) {
return (0.0);
}
# unnumbered eqn on the bottom of page 5
Ne <- param.vec[2];
mu <- param.vec[3];
tau <- param.vec[4];
# The probability of coalescence along the Turkish branch goes to 1 as tau gets big or Ne gets small
prob.coalescence = 1 - exp(-tau/(2.0*Ne))
prob.no.coalescence = 1 - prob.coalescence
# Given that there is no coalescence, each branch will have length tau along the Turkish population history
# the following formula would apply to each branch (and are the same as the probabilities for the Spanish population)
expNeg2TauMu <- exp(-2*tau*mu)
prob.no.mut.given.no.coal <- (1 + expNeg2TauMu)/2.0;
prob.mut.given.no.coal <- 1 - prob.no.mut.given.no.coal;
# Here is some very crude averaging over branch lengths that Bryant et al don't do...
# we'll assume that if there is coalescence it averages out to occurring 1/3 of the way from the present to the
# time of divergence. No real justificaion for this...
hacky.total.branch.length.given.coal = (4/3) * tau # hack mentioned above
expNegHackyBranchLen <- exp(-2*hacky.total.branch.length.given.coal*mu)
prob.no.mut.given.coal <- (1 + expNegHackyBranchLen)/2.0;
prob.mut.given.coal <- 1 - prob.no.mut.given.no.coal;
# 2/3 of tau is contributed by the common ancestral branch,
# 1/3 of tau is contributed by each terminal (for a total of 4/3)
# so, given that there was coalescence, and only one mutation in the Turkish branch, and
# our new hack. 50% of mutations will be on the ancestral branch of the gene tree,
# and 25% will be on each of the terminal's
prob.mut.common.anc.branch = .5*prob.mut.given.coal; # hack mentioned above, again
prob.mut.terminal.branch = .25*prob.mut.given.coal; # hack mentioned above, again
if (r_Tb == 1 ) {
#eqn 10 - eqn 14
if (n_Tt == 2) {
#eqn 10 - eqn 12
prob.tree.shape = prob.no.coalescence
if (r_Tt == 1) {
# eqn 11 would be the correct form to use
#
# here we'll use the hacky way...
# same # of lineages in state 0 at the beginning and end. This probability will be dominated
# by the scenario of no mutations...
prob.two.mutations = prob.mut.given.no.coal * prob.mut.given.no.coal
prob.zero.mutations = prob.no.mut.given.no.coal * prob.no.mut.given.no.coal
prob.mutation.scenario = prob.two.mutations + prob.zero.mutations
} else {
# eqn 10 and 12 are identical
# hacky way -> one lineage mutated, and the other did not...
prob.mutation.scenario = prob.mut.given.no.coal * prob.no.mut.given.no.coal
}
} else {
#eqn 13 and eqn 14 are the same, so only one branch of code here
# the hacky way is to assume there was only one mutation, and it must have been on a terminal...
prob.tree.shape = prob.coalescence
prob.mutation.scenario = prob.no.mut.given.coal*prob.mut.terminal.branch
}
} else {
#eqn 10 - eqn 14
if (n_Tt == 2) {
#eqn 5 - eqn 5
prob.tree.shape = prob.no.coalescence
if (r_Tt == r_Tb) {
# eqn 5 would be the correct way...
# here is the hack...
prob.mutation.scenario = prob.no.mut.given.no.coal * prob.no.mut.given.no.coal
} else if (r_Tt == 1) {
# eqn 6 is the correct way...
# hacky way -> one lineage mutated, and the other did not...
prob.mutation.scenario = prob.mut.given.no.coal * prob.no.mut.given.no.coal
} else {
#eqn 7 is the correct way.
# if r_Tb is 0 or 2 and r_Tt is 0 or 2, but r_Tt != r_Tb, then both lineages had mutations
prob.mutation.scenario = prob.mut.given.no.coal * prob.mut.given.no.coal
}
} else {
#eqn 8 or eqn 9
prob.tree.shape = prob.coalescence
if (r_Tt == 1) {
if (r_Tb == 2) {
# eqn 8
# no mutations required when a state-0 lineage splits into two state-0 lineages
prob.mutation.scenario = prob.no.mut.given.coal
} else {
# eqn 9
# a mutations before splitting is the easiest way to turn a state-0 lineage splits into two state-1 lineages
prob.mutation.scenario = prob.mut.common.anc.branch
}
} else {
# eqn 9
if (r_Tb == 0) {
# eqn 8
# no mutations required when a state-1 lineage splits into two state-1 lineages
prob.mutation.scenario = prob.no.mut.given.coal
} else {
# eqn 9
# a mutations before splitting is the easiest way to turn a state-1 lineage splits into two state-0 lineages
prob.mutation.scenario = prob.mut.common.anc.branch
}
}
}
}
return (prob.tree.shape*prob.mutation.scenario)
}
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