On the Analysis of Optical Mapping Data - University of Wisconsin ...

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unknown probability p, say, of success. It is possible to argue that the probability should
depend on proximity to other cuts, but the idea is difficult to formalize. Instead, the issue
is dealt with using a desorption model for small fragments (see below). An optical map can
also contain false cuts, i.e. apparent restriction sites that correspond to no restriction site
in the true map, perhaps due to random breakage of DNA or image errors. The locations
of such spurious cut sites in optical maps may be modeled as the realizations of a homogeneous
Poisson process, with rate ζ, say, per Kb of DNA. These models have been used by
Anantharaman et al. (1999) and Valouev et al. (2006).
Length measurement errors: Consider an optical map with n fragments of measured
lengths X 1 , . . .,X n . Assuming no cut errors, each fragment has a corresponding true but
unobserved length, which we denote by µ i , i = 1, . . .,n. Recall that each X i is calculated as
the product Y i R i , where Y i is the total fluorescent intensity of the pixels that constitute the
fragment, and R i is a scale factor to convert fluorescent intensities to base pairs, estimated
using standards. Restricting our attention to the marginal distribution of X i , we may treat
Y i and R i as independent latent variables within a given image. We can assume without loss
of generality that the true scale factor is 1. It is natural to assume that the distribution of
Y i depends only on µ i . Valouev et al. (2006) note that Y i is the sum of intensities of several
pixels. Assuming these terms to be i.i.d., the expected number of terms is proportional to
µ i . Invoking the Central Limit Theorem, they postulate that for some σ,
Y i ∼ N ( )
µ i , σ 2 µ i
This model additionally has the following desirable property: denoting the N (µ, σ 2 µ) density
by f µ , if Y i ∼ f µi , Y j ∼ f µj and Y i and Y j are independent, then Y i + Y j ∼ f µi +µ j
. This
is relevant when adjacent fragments are reported as one due to a missing cut. Valouev et al.
(2006) ignore scaling and postulate that the observed lengths X i = Y i . If we instead assume
that the mean E(R i ) = 1 and variance V (R i ) = τ 2 > 0 without making any further
assumptions about the distribution of R i , we have
E(X i ) = E(E(Y i R i |R i )) = µ i