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Epigenomic and Regulatory Data · Tutorial 784 of 795

Peak Calls and Continuous Chromatin Signal

You will compare peak-based and track-based summaries, calculate each from genomic examples, and identify when the two representations lead to different conclusions.

Intermediate 9 min read

What You'll Learn

  • Distinguish binary peak calls from continuous-valued chromatin signal tracks.
  • Calculate called-interval coverage, mean track value, and integrated signal with appropriate units.
  • Decompose an average track value into contributions from peak and nonpeak bases.
  • Compare samples when peak coverage and signal magnitude rank them differently.
  • Weight unequal-width genomic bins correctly and recognize common interpretation pitfalls.

Two Representations of Chromatin Signal

Chromatin sequencing data can be summarized as peak intervals or as a signal track. A peak caller reports genomic intervals judged to show enrichment under its model and settings. A signal track assigns a numerical value to genomic positions or bins, retaining variation in signal magnitude. These are related representations, but they are not interchangeable: peak calls reduce signal to an interval label, while a track preserves more of its quantitative detail.

The distinction matters when summarizing a region or comparing samples. A region may have a high fraction of bases inside called peaks but modest signal there; another may have no called interval yet show moderate signal across much of its length. A peak-coverage summary and a track-based average answer different questions.

Definition: For sample \(s\), let \(R\) be a genomic region of length \(L\) base pairs. Let \(P_s\) be the union of the sample's called peak intervals, and define the peak indicator \(I_s(x)=1\) if base \(x\) lies in \(P_s\), and \(I_s(x)=0\) otherwise. Let \(Y_s(x)\) be the signal-track value at base \(x\). The called-interval coverage fraction is the fraction of bases in \(R\) that lie in \(P_s\); the mean track value averages \(Y_s(x)\) over \(R\); the integrated signal sums those values over \(R\).

Here, “continuous” means that track values can vary in magnitude rather than being just peak/no-peak labels; it does not imply that a computational track must vary smoothly at every position. A track may be defined at single bases or at bins. For clarity, the examples first use base-level values. When a track is binned, summaries must account for the number of base pairs represented by each bin.

Three Summaries, Three Questions

For a base-level track, the called-interval coverage fraction, mean track value, and integrated signal over \(R\) are:

$$ C_s(R)=\frac{\sum_{x\in R} I_s(x)}{L}, \qquad \bar{Y}_s(R)=\frac{1}{L}\sum_{x\in R}Y_s(x), \qquad A_s(R)=\sum_{x\in R}Y_s(x). $$

The fraction \(C_s(R)\) is dimensionless and can be reported as a percentage. The mean \(\bar{Y}_s(R)\) has the same units as the track values, here called signal units. The integrated signal \(A_s(R)\) has units of signal-unit base pairs. If signal is defined as a normalized or transformed score, the units and interpretation should follow that score's definition; it should not automatically be called a read count or a physical concentration.

Coverage summarizes where peak intervals were called, not how strong the track is. The mean summarizes signal magnitude across the whole region, including positions outside called intervals. The integrated signal combines magnitude and region length. For a fixed region, the mean is the integrated signal divided by its length; across regions of different lengths, integrated signals can differ partly because one region is longer.

Worked Example: Peak Coverage and Signal in One Region

Consider a hypothetical 10-base-pair region with track values, in signal units, of \(0,0,1,2,5,6,5,2,1,0\) at positions 1 through 10. For this illustration only, call a peak at positions with signal at least 4. The called interval is therefore positions 5 through 7, covering 3 of the 10 bases.

The called-interval coverage fraction is \(3/10=0.300\), or 30%. The mean track value uses all 10 positions:

$$ \bar{Y}(R)=\frac{0+0+1+2+5+6+5+2+1+0}{10} =\frac{22}{10}=2.2\text{ signal units}. $$

The integrated signal is \(22\) signal-unit base pairs. The mean within the three called bases is \((5+6+5)/3=5.33\) signal units. Thus, peak coverage is 30%, but the region-wide mean is 2.2 signal units; neither value can be substituted for the other. The nonzero values outside the called interval contribute to the track mean and integrated signal even though those bases do not count toward peak coverage.

Why the Summaries Can Disagree

A peak call is a thresholded and model-dependent classification, not a direct measurement of all signal in a region. Peak-calling procedures may use local background, smoothing, or other decision rules; the simple numerical threshold in the example above is only an illustration. Changing the caller or its settings can change the intervals. A track summary instead uses the values in the chosen track and does not, by itself, decide which positions constitute a peak.

The relationship between the mean and peak coverage can be written exactly. Let \(\bar{Y}_{P}\) be the mean track value on bases in \(R\) that are called peaks, and let \(\bar{Y}_{N}\) be the mean on the remaining bases. If both sets are nonempty, the total signal is the sum over those two disjoint sets. Dividing that sum by \(L\) gives:

$$ \bar{Y}(R) =\frac{L C(R)\bar{Y}_{P}+L[1-C(R)]\bar{Y}_{N}}{L} =C(R)\bar{Y}_{P}+[1-C(R)]\bar{Y}_{N}. $$

This identity shows why peak coverage alone cannot determine the mean: the average signal inside and outside called intervals also matters. If there are no called bases or no nonpeak bases, the corresponding subgroup mean is undefined, and the summary should instead be calculated directly over the bases that exist.

Worked Example: A Ranking Reversal Between Two Samples

Suppose a hypothetical 5-base-pair region has the following track values in two samples. For this illustration, a base is called a peak when its value is at least 4.

SampleTrack values, positions 1–5 (signal units)Called bases
A5, 5, 0, 0, 01 and 2
B3, 3, 3, 3, 3None

Sample A has peak coverage \(2/5=0.400\), or 40%, and mean signal \((5+5+0+0+0)/5=2.0\) signal units. Its integrated signal is \(10\) signal-unit base pairs. Sample B has peak coverage \(0/5=0\), mean signal \((3+3+3+3+3)/5=3.0\) signal units, and integrated signal \(15\) signal-unit base pairs.

By called-interval coverage, A is higher; by mean and integrated signal, B is higher. These results are consistent: A has stronger signal at fewer positions, while B has moderate signal at every position. Neither ranking is universally correct without specifying whether the question concerns called-peak coverage or the magnitude of the track across the region.

Comparing Samples and Regions Fairly

A numerical comparison is meaningful only when the quantities being compared are defined consistently. For peak coverage, use the same genomic assembly, region boundaries, and a clearly specified peak-calling approach and settings. If called intervals overlap, count their union within the region rather than counting overlapping bases more than once. For a track-based comparison, use the same region and ensure that signal tracks have compatible definitions and scales. If tracks were normalized or transformed differently, a difference in track values may reflect the processing as well as the underlying chromatin signal.

For tracks represented in unequal-width bins, an unweighted average of bin values gives each bin equal influence, not each base pair. If bin \(j\) has width \(w_j\) and value \(y_j\), the region-wide mean and integrated signal are:

$$ \bar{Y}(R)=\frac{\sum_j w_j y_j}{\sum_j w_j}, \qquad A(R)=\sum_j w_j y_j. $$

These expressions treat the value in each bin as applying across that bin. If the track uses a different bin-level definition, its interpretation must be checked before applying these formulas.

Worked Example: Unequal-Width Bins

A hypothetical 10-base-pair region is represented by three bins: a 2-base-pair bin with value 4 signal units, a 3-base-pair bin with value 2 signal units, and a 5-base-pair bin with value 0 signal units. Suppose the first bin is called a peak and the other two are not.

Peak coverage is \(2/10=0.200\), or 20%. The correct region-wide mean weights each bin by its width:

$$ \bar{Y}(R)=\frac{(2)(4)+(3)(2)+(5)(0)}{2+3+5} =\frac{14}{10}=1.4\text{ signal units}. $$

The integrated signal is \((2)(4)+(3)(2)+(5)(0)=14\) signal-unit base pairs. By contrast, the unweighted average of the three bin values is \((4+2+0)/3=2.0\) signal units. That calculation gives the 2-base-pair bin the same weight as the 5-base-pair bin, so it is not the mean across the 10 base pairs.

Diagnostics, Alternatives, and Limitations

Useful checks begin with the data representation. Confirm the genome assembly and coordinate convention, verify that region lengths agree with the intervals or bins being summarized, and inspect whether overlapping intervals were merged for coverage calculations. For binned tracks, check bin widths and any missing values. A surprising mean may result from a few very large track values, while a peak fraction may change sharply with calling settings; examining both the underlying track and the interval calls helps identify such cases.

The main alternatives are complementary rather than competing measurements. Peak-based summaries are useful when the question is how much of a region is covered by intervals called under a chosen procedure. Track-based summaries are useful when the question concerns signal magnitude across the full region. A combined description can report both coverage and mean or integrated signal, making clear that the summaries measure different features. No single summary captures location, width, and magnitude in full detail.

These summaries are descriptive. A difference in peak coverage or track mean does not, by itself, establish a statistically reliable sample difference or a causal effect. The interpretation also depends on how the track was produced, how comparable its scale is across samples, and which regions were selected for summary. Treating a called peak as if every base within it had equal signal, or treating an uncalled base as if its signal were zero, discards information that may be present in the track.

  • Common mistake: “Sample A has more peak coverage, so it must have more chromatin signal overall.” Coverage counts called bases, not signal magnitude. Compare the track mean or integrated signal as well if the question concerns magnitude.
  • Common mistake: “The track mean is signal units per base.” Averaging track values over bases leaves the mean in signal units. The integrated sum, not the mean, has units of signal-unit base pairs.
  • Common mistake: Averaging unequal-width bins without weights. Use bin widths so that each base pair contributes appropriately.
Key takeaway: Peak intervals summarize a thresholded location decision; continuous-valued tracks summarize signal magnitude. State which quantity is being reported, its units, and how the genomic region and signal representation were defined.

Check Your Understanding

Answer each question, then compare your reasoning with the solutions.

  1. A 20-base-pair region contains 8 bases within called peaks. What is its called-interval coverage fraction and percentage?
  2. Across a 4-base-pair region, track values are \(1,2,3,4\) signal units. Find the mean track value and integrated signal, including units.
  3. Explain how a sample can have zero called-interval coverage but a positive mean track value.
  4. A region has two bins: width 2 with value 5, and width 6 with value 1. Find its width-weighted mean and explain why the unweighted bin average differs.

Solutions. (1) The fraction is \(8/20=0.400\), or 40%. (2) The mean is \((1+2+3+4)/4=2.5\) signal units; the integrated signal is \(10\) signal-unit base pairs. (3) A peak caller can classify every base as outside called intervals while the track still has positive subthreshold values; peak coverage is then zero, but the average of those values can be positive. (4) The weighted mean is \([(2)(5)+(6)(1)]/(2+6)=16/8=2\) signal units. The unweighted average is \((5+1)/2=3\) signal units because it gives each bin equal weight despite their different widths.

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