This is the prefered consensus algorithm for reading data with Phred decibel scale confidence values. As will become clear from the follwing description, it is more complicated than the other algorithms, but produces a much more useful result.
A difficulty in designing an algorithm to calculate the confidence for a consensus derived from several readings, possibly using different chemistries, and hopefully from both strands of the DNA, is knowing the level of independence of the results from different experiments - namely the readings. Given that sequencing traces are sequence dependent, we do not regard readings as wholly independent, but at the same time, repeated readings which confirm base calls may give us more confidence in their accuracy. In addition, if we get a particularly good sequencing run, with consequently high base call confidence values, we are more likely to believe its base call and confidence value assignments. The final point in this preamble is that the Phred confidence values refer only to the probability for the called base, and they tell us nothing about the relative likelihood of each of the other 3 base types appearing at the same position. These difficulties are taken into account by our algorithm, which is described below.
In what follows, a particular position in an alignment of readings is referred to as a "column". The base calls in a column are classified by their chemistry and strand. We currently group them into "top strand dye primer", "top strand dye terminator", "bottom strand dye primer" and "bottom strand dye terminator" classes.
Within each class there may be zero or many base calls. For each class we check for multiple occurrences of the same base type. For each base type we find the highest confidence value, and then increase it by an amount dependent on the number of confirming reads. Then Bayes formula is used to derive the probabilities and hence the confidence values for each base type.
To further describe the method it is easiest to work through an example. Suppose we have 5 readings with the following characteristics covering a particular column.
Dye primer, top strand, 'A', confidence 20 Dye primer, top strand, 'A', confidence 10 Dye primer, top strand, 'T', confidence 20 Dye terminator, top strand, 'T', confidence 10 Dye primer, bottom strand, 'A', confidence 5
Hence there are three possible classes.
Examining the "dye primer top strand" class we see there are three readings (A, A and T). The highest A is 20. We add to this a fixed quantity to indicate one other occurence of an A in this set. For this example we add 5. Now we have an adjusted confidence of 25 for A and 20 for T. This is equivalent to a .997 probability of A being correct and .99 probability of T being correct. To use Bayes we split the remaining probabilies evenly. A has a probability of .997 and so the remaining .003 is spread amongst the other base types. Similarly for the .01 of the T. The result is shown in the table below.
| A C G T --+----------------------- A | .997 .001 .001 .001 T | .0033 .0033 .0033 .990
Bayesian calculations on this table then give us probabilities of approximately .766 for A, .00154 for C, .00154 for G and .231 for T.
The other classes give probalities of .033 for A, C, G and .9 for T, and .316 for A, and .228 for C, G and T.
To combine the values for each class we produce a table for a further Bayesian calculation. Once again we fill in the probabilities and spread the remainder evenly amongst the other base types.
| A C G T -----------+-------------------------- Primer Top | .766 .00154 .00154 .231 Term Top | .0333 .0333 .0333 .9 Primer Bot | .316 .228 .228 .228
From this Bayes gives the final probabilities of .135 for A, .0002 for C, .0002 for G and .854 for T. This is what would be expected intuitively: the T signal was present in both dye primer and dye terminator experiments with 1/100 and 1/10 error rates whilst the A signal was present on both strands with 1/100 and 1/3 error rates. Hence the consensus base is T with confidence 8.4 (-10*log10(1-.854)).
If a padding character is present in a column we consider the pad as a separate base type and then evenly divide the remaining probabilities by 4 instead of 3.