Why inspect normalized expression?
Normalization aims to make expression measurements more comparable by reducing technical differences while retaining meaningful biological variation. A normalization method can produce values on a convenient scale without making every relevant feature of the data comparable. Diagnostics help reveal patterns that merit investigation: shifted sample distributions, intensity-dependent differences between samples, or variability that still changes with average expression.
Let \(y_{gj}\) denote the normalized, transformed expression value for gene \(g\) in sample \(j\). For sequencing data, \(y_{gj}\) might be log-CPM; for array data, it might be a normalized log-scale intensity. The plots below assume that the values being compared are on the same defined expression scale. The samples should also have undergone the intended preprocessing, and the gene set used for comparisons should be stated. Different filtering, missingness, or transformations can change a plot’s appearance.
No single plot establishes that normalization is correct. Distributional alignment is a useful check, not a universal target: a real biological difference in overall expression composition can make two properly processed samples look different. Conversely, a normalization procedure may force distributions to align while obscuring a genuine global shift. Interpret visual patterns alongside the study question and the assumptions of the normalization method.
Sample-wise boxplots and density curves
A boxplot summarizes the distribution of expression values across genes within each sample. Compare sample medians, interquartile ranges (IQRs), whiskers, and outlying values. Large differences can flag a sample-scale discrepancy, unusual spread, or a small group of extreme measurements. A density curve displays the distribution in more detail and can reveal differences in skewness, shoulders, or multiple modes that a boxplot compresses.
These comparisons require a common scale and a comparable set of genes. A difference in medians or curve shapes is a prompt to ask why, not an automatic instruction to force the distributions to match. In particular, if many genes truly shift together, global alignment can be biologically inappropriate. Density curves also depend on smoothing choices, so small wiggles should not be treated as distinct biological populations without further evidence.
Worked Example: Reading Sample Boxplots
Suppose a hypothetical normalized log-expression matrix is summarized across the same 1,000 genes in three samples. The five-number summaries (minimum, first quartile, median, third quartile, maximum) are: sample A, \(2.0, 5.1, 6.0, 7.1, 10.0\); sample B, \(2.2, 6.1, 7.0, 8.1, 10.2\); and sample C, \(1.8, 5.0, 6.1, 7.2, 12.0\). All values are on the same log-expression scale.
For sample A, \(IQR=7.1-5.1=2.0\); for B, \(IQR=8.1-6.1=2.0\); and for C, \(IQR=7.2-5.0=2.2\). Samples A and B have similar central spread, but B’s median is \(7.0-6.0=1.0\) log-expression unit higher. C has a median close to A’s, a slightly wider IQR, and a larger maximum.
The pattern suggests checking why B’s distribution is shifted and whether C’s upper tail is driven by a few genes. It does not prove that B is technically misnormalized or that C is a failed sample. For example, a genuine sample-wide biological shift or a difference in cell composition could also affect the distribution. Density curves and gene-level comparisons can clarify which parts of the distributions differ.
MA plots: differences across expression intensity
An MA plot compares two samples gene by gene. For samples \(j\) and \(k\), define the average expression \(A_g\) and difference \(M_g\) for gene \(g\) by
Thus, \(A_g\) is the average log-expression and \(M_g\) is the log-expression difference, in the same log-scale units. If comparable samples have been normalized successfully and most genes are not expected to shift systematically, the center of \(M_g\) should be near zero across much of the intensity range, without a persistent trend in \(A_g\). This expectation depends on the biological comparison and the normalization assumptions; it is not a requirement that every gene have \(M_g=0\).
A global offset and an intensity-dependent bias are different patterns. To see why, suppose the difference between two samples can be represented approximately as
where \(c\) is a constant sample offset, \(f(\theta_g)\) varies with the gene’s underlying expression intensity, \(d\) is the size of an intensity-dependent component, and \(\varepsilon_g\) is gene-specific variation. A nonzero \(c\) shifts the plot vertically. If \(d\ne 0\), the expected difference can change with intensity, producing a trend or curvature. Centering the overall differences can remove a constant offset from a summary, but it cannot by itself remove a trend across intensity.
Worked Example: Detecting an Intensity-Dependent Pattern
Consider a hypothetical pair of normalized samples, with \(M_g=y_{gj}-y_{gk}\) and \(A_g=(y_{gj}+y_{gk})/2\). For one low-intensity gene, the values are \(y_{gj}=3.1\) and \(y_{gk}=3.0\), so \(M_g=3.1-3.0=0.1\) and \(A_g=(3.1+3.0)/2=3.05\). For one high-intensity gene, the values are \(10.0\) and \(10.5\), so \(M_g=-0.5\) and \(A_g=10.25\).
Suppose the plotted genes are grouped into three intensity ranges. The mean \(M_g\) is \(0.30\) for the low range, \(0.02\) for the middle range, and \(-0.25\) for the high range. The pattern moves from positive differences at low intensity to negative differences at high intensity. The middle range is close to zero, but the full plot is not well described by a single overall center: the intensity pattern matters.
This is evidence of an intensity-dependent pairwise discrepancy worth investigating. It does not show which sample is responsible or whether the difference is technical. The comparison is most informative when the samples are biologically comparable for the diagnostic purpose and the expected pattern under the normalization assumptions is understood.
Mean–variance plots
A mean–variance plot asks whether variability across samples depends on a gene’s average expression. For each gene \(g\), calculate the sample mean and sample variance of its normalized expression values across \(n\) samples:
Plot \(s_g^2\) against \(\bar{y}_g\), often using a logarithmic vertical axis if variances span a wide range. A systematic increase or decrease in variance with mean indicates that the transformed values may retain a mean-dependent scale pattern. A transformation such as log-CPM or a variance-stabilizing transformation can reduce such dependence, but the plot checks what remains in the data actually analyzed.
The variance here is across the samples included in the calculation. It can combine technical variation with genuine differences among biological samples. If samples belong to distinct biological groups, their mean differences can contribute to a gene’s across-sample variance. Consequently, an upward trend is not automatically a normalization failure, and a flat trend is not proof that all sources of variation have been handled. State which samples and genes went into the calculation, and interpret the plot in that context.
Worked Example: Calculating Gene-Wise Variance
Four hypothetical samples have transformed expression values for two genes. Gene L has values \(1, 2, 3, 4\), and gene H has values \(8, 12, 16, 20\). For gene L, \(\bar{y}_L=(1+2+3+4)/4=2.5\). Its squared deviations sum to \((1-2.5)^2+(2-2.5)^2+(3-2.5)^2+(4-2.5)^2=5\), so \(s_L^2=5/(4-1)=1.667\).
For gene H, \(\bar{y}_H=(8+12+16+20)/4=14\). Its squared deviations sum to \((8-14)^2+(12-14)^2+(16-14)^2+(20-14)^2=80\), so \(s_H^2=80/3=26.67\). In this small hypothetical example, the higher-mean gene also has substantially higher variance.
Two genes are far too few to establish a general trend, but the calculation shows what each plotted point represents: one gene’s mean and its variance across samples. In a real diagnostic plot, a broad upward pattern would prompt assessment of residual mean–variance dependence on this scale. The numbers alone do not distinguish technical variability from biological variation.
Combining plots and checking failure modes
The plots answer related but distinct questions. Boxplots and density curves compare whole-sample distributions. An MA plot examines pairwise differences as a function of average intensity. A mean–variance plot examines how across-sample variability relates to gene-level average expression. Agreement among diagnostics can strengthen a concern, while disagreement can help localize it. For example, similar sample medians do not rule out an MA trend: positive differences at low intensity and negative differences at high intensity can partly cancel in an overall summary.
A practical review should first confirm that plots use the intended normalized scale and a clearly defined, comparable gene set. Then inspect whether patterns are broad or driven by a few genes, whether they recur across relevant sample pairs, and whether they are plausible given the biology and preprocessing. Plot labels, axes, sample identities, and units matter: an MA plot with reversed sample order changes the sign of every \(M_g\), though not the presence of an intensity trend.
Common failures include treating identical distribution shapes as proof of valid normalization, treating any sample-wise difference as a technical defect, and interpreting a mean–variance trend without considering group composition. Very small gene sets, extensive missingness, and a few extreme values can also make summaries unstable or misleading. These plots are diagnostics, not formal tests and not substitutes for understanding how the normalization method behaves.
Check Your Understanding
Try each question, then compare your reasoning with the solutions below.
- Two samples have equal medians but different IQRs. What does this suggest, and what does it not establish?
- For a gene, sample \(j\) has \(y_{gj}=7.2\) and sample \(k\) has \(y_{gk}=6.8\). Calculate \(M_g\) and \(A_g\).
- An MA plot has mean differences near \(0.2\) at low intensity and \(-0.2\) at high intensity. Why can an overall average near zero be misleading?
- Why might a mean–variance plot show an upward trend even if the transformed expression values have no technical defect?
- What does a density curve add to a boxplot, and what is one limitation of density curves?
Solutions. (1) The samples have similar medians but different central spreads; investigate the distributional difference. This alone does not establish its cause or prove normalization failed. (2) \(M_g=7.2-6.8=0.4\) log-expression units and \(A_g=(7.2+6.8)/2=7.0\) log-expression units. (3) Positive and negative differences at different intensities can cancel in an overall average, hiding systematic intensity dependence. (4) Across-sample variance can include genuine biological differences, including differences between biological groups, as well as technical variation. (5) A density curve can reveal skewness or multiple modes that a boxplot compresses; its appearance depends on smoothing and can overemphasize small fluctuations.