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Computational Biology

(BIOSC 1540)

Sep 26, 2024

Lecture 10:
Differential gene expression

Announcements

A04 is due tomorrow by 11:59 pm
- Will post A04 solutions Monday morning
All I am doing this weekend is grading
Exam is one week from today (October 3rd)
- The review guide will be posted tomorrow

After today, you should be able to

Why are we learning about differential gene expression?

What is Differential Gene Expression?

Differential Gene Expression (DGE): The process of identifying and quantifying changes in gene expression levels between different sample groups or conditions

Sample Collection: Gather samples from different conditions (e.g., healthy vs. diseased).
RNA Sequencing (RNA-seq): Quantify gene expression levels using high-throughput sequencing technologies.
Read Mapping and Quantification: Align RNA-seq reads to a reference genome and quantify expression (e.g., using Salmon).
Statistical Analysis: Identify genes with significant expression differences between conditions.

Case study: Breast cancer

Objective: Identify genes differentially expressed between triple-negative breast cancer (TNBC) and hormone receptor-positive breast cancer
Findings:
- TNBC shows upregulation of genes involved in cell proliferation and metastasis.
Implications:
- Targets for specific therapies.
- Improved classification and prognosis of breast cancer subtypes.

Differential gene expression provides statistical tools to identify changes between samples

After today, you should be able to

Explain the purpose of statistical models and hypothesis testing

What is a statistical model?

A statistical model is a mathematical tool that describes how data are generated

Gene expression

Normal

Cancerous

Is there an apparent difference in gene expression between the two conditions?
If so, is it real, or could it have happened by random chance or experimental flaws?

It helps us answer:

Statistical models help us make sense of complex data by identifying patterns and determining whether differences are meaningful or just due to chance

Hypothesis testing in RNA-seq data

After fitting a statistical model, we need to perform hypothesis testing to see if the difference in expression between conditions is statistically significant

Null Hypothesis (H₀): There is no difference in gene expression between the two conditions

Gene expression

Normal

Cancerous

Alternative Hypothesis (H₁): There is a significant difference in gene expression between the conditions

We have two hypotheses:

We reject the null hypothesis when our statistical test demonstrates that the observed difference, if any, is unlikely to have happened by random chance

The P-value is the probability of the null hypothesis being true

What is the probability that any difference is either (1) nonexistent or (2) due to random chance (i.e., "getting lucky")

Probability value (p-value):

The higher the p-value, the more our model supports the null hypothesis

The lower the p-value, the more our model supports the alternative hypothesis

Gene expression

Normal

Cancerous

p \approx 0.9999

Gene expression

Normal

Cancerous

p \approx 0.00001

Differential gene expression uses statistical models for hypothesis testing

Ensures that we are not biasing our data or our interpretation

After today, you should be able to

Discuss the challenges of working with gene expression data

The nature of count data

RNA-seq generates count data – the number of RNA fragments that map to each gene

Gene expression

Normal

Cancerous

Example: 573,282 TPM

Discrete data requires us to use special statistical tools

Data that can only take specific values (like whole numbers)
In RNA-seq, we measure the number of reads mapped to a gene, so the data are count-based
You can't have "half a read" or a decimal number of reads.

What is discrete data:

For example, you cannot use a normal distribution because it requires continuous data

Binomial: A Simple Model for Discrete Counts

The Binomial distribution models the number of successes in a fixed number of independent trials, where each trial has the same probability of success

n

k

P

p

P \left( X = k \right) = \frac{n!}{k! \left( n - k \right)!} p^k \left( 1 - p \right)^{n - k}

Number of trials

Number of successes

Probability

Probability of success

RNA-seq analogy: Each read can be considered a "trial," and the probability that a read maps to a specific gene is the "probability of success."

Limitations of Binomial distributions for RNA-Seq

Main limitation: Assumes that the probability of success is constant between samples
Smaller limitation 1: The number of possible trials can be very large, especially when sequencing at a high depth
Smaller limitation 2: The probability of expression is very small for many genes because they are either lowly expressed or not at all

The Poisson distribution simplifies computation and allows for varying probabilities

P \left( X = k \right) = \frac{n!}{k! \left( n - k \right)!} p^k \left( 1 - p \right)^{n - k}

Computations with low p and high n are computationally demanding

Poisson distribution: A baseline for modeling discrete counts

The Poisson distribution is a statistical tool used to model the number of events (or counts) that happen in a fixed period of time or space, where:

The events are independent of each other
Each event has a constant average rate

P \left( X = k \right) = \frac{\lambda^k e^{-\lambda}}{k!}

\lambda

Expected average of X

k

Number of events or counts

P

Probability

Provides an accurate distribution of counts if your mean and variance are approximately equal

Poisson distribution becomes inaccurate when variance > mean

RNA-seq data are noisy (i.e., high variance) and incompatible with Poisson distribution

Parity plots with mean and variance show deviations with Poisson distributions

Higher counts typically have a larger variance

Count mean

Count variance

Mean = variance line

Overdispersion in RNA-Seq

Overdispersion: It happens when the variance in the data is larger than what is predicted by simpler models (e.g., Poisson distribution)

Expected variance for Poisson-distributed data equals the mean: $Variance=μ\text{Variance} = \mu$
Variance is often larger than the mean for RNA-Seq: $Variance>μ\text{Variance} > \mu$

Overdispersion may reflect biological variability between samples not captured by the experimental conditions

Differences in RNA quality
sequencing depth,
biological factors like different cell types within the same tissue

Negative Binomial distribution accounts for high dispersion

P(X = k) = \frac{\Gamma(k + \frac{1}{\alpha})}{k! \, \Gamma(\frac{1}{\alpha})} \left(\frac{1}{1 + \alpha \mu}\right)^{\frac{1}{\alpha}} \left(\frac{\alpha \mu}{1 + \alpha \mu}\right)^k

k

Observed number of counts

\mu

Mean or expected value of counts

\alpha

Dispersion parameter, controlling how much the variance exceeds the mean

\Gamma \left( \cdot \right)

Gamma function, which generalizes the factorial to floats

\text{Var}(X) = \mu + \alpha \mu^2

If $α=0\alpha = 0$ , the Negative Binomial distribution reduces to the Poisson distribution

The Challenge of zeros in RNA-seq data

RNA-seq data frequently contains zero counts for some genes because not all genes are expressed under all conditions

Most statistical models account for variance, but not that zeros can dominate counts

For example, if we have a high expected mean with Poisson distribution we can still have zeros or very low counts

In these circumstances, we have to use zero-inflated models

We will ignore these for now

After today, you should be able to

Discuss fitting of statistical models

Why are statistical models important in RNA-seq?

RNA-seq data is messy: counts vary, there are lots of zeros, and data doesn’t follow simple patterns

Step 1: The model looks at the data from both groups
Step 2: It considers how much variation there is within each group
Step 3: The model calculates how likely it is to see the average difference if there is no real difference between the groups (just by chance)

We need models to account for this complexity and figure out which genes are differentially expressed in a meaningful way

Define the model for each gene

A statistical model predicts each sample's count data
(number of reads mapping to each gene)

It accounts for the mean (average expression for a gene) and the dispersion (how much the expression varies across samples)

Fit parameters using optimization algorithms

MLE tries to find the model parameters that make the observed counts most likely

We use maximum likelihood estimation (MLE) to estimate the parameters $μ\mu$ (mean) and $α\alpha$ (dispersion) for each gene.
MLE finds the values of the parameters that maximize the likelihood of observing the data given the model.

It does this by adjusting the model until the predicted counts match the actual counts as closely as possible (i.e., minimize the error)

After today, you should be able to

Understand statistical tests used for gene expression data

Wald’s Test for Gene Expression Differences

Wald’s Test: A statistical test that helps us determine whether the estimated log fold change between two conditions is significantly different from zero.
Null Hypothesis (H₀): The log fold change between conditions is zero (no difference in expression between the conditions).
- Log Fold Change (β₁) = 0 means that the gene is expressed at the same level in both conditions.
Alternative Hypothesis (H₁): The log fold change between conditions is not zero (there is a difference in expression).

Log Fold Change

Positive Log Fold Change: Indicates higher expression in the condition of interest (e.g., diseased).
Negative Log Fold Change: Indicates lower expression in the condition of interest.
Log Fold Change of Zero: Means no difference between conditions.

Estimate Parameters from the Negative Binomial Model

For each gene, the Negative Binomial model gives us an estimated log fold change $β^1\hat{\beta}_1$

\hat{\beta}_1

It also gives us a standard error (SE) for this estimate, which tells us how uncertain we are about the estimate of log fold change $β^1\hat{\beta}_$

SE \left( \hat{\beta}_1 \right)

The Wald statistic is calculated as

\text{Wald statistic} = \frac{\hat{\beta}_1}{SE \left( \hat{\beta}_1 \right)}

This statistic tells us how many standard deviations the estimated log fold change is away from zero (no difference)

Likelihood Ratio Test

To compute a p-value, a likelihood ratio test (LRT) can be used $β^1\hat{\beta}_1$

The idea is to compare the likelihood of the data under

the null model (same expression in both conditions)
the alternative model (different expression levels in each condition) $β^1\hat{\beta}_1$

Log-Likelihood of Negative Binomial

\log \mathcal{L}(r, \mu | X) = X \log \left( \frac{\mu}{\mu + r} \right) + r \log \left( \frac{r}{\mu + r} \right)

For each condition, you compute the log-likelihoods:

\mathcal{L}_{\text{A}} = \sum_{i} \log \mathcal{L}(r_A, \mu_A | X_i)

\mathcal{L}_{\text{B}} = \sum_{i} \log \mathcal{L}(r_B, \mu_B | X_i)

LRT Statistic

The LRT statistic is: $β^1\hat{\beta}_1$

\text{LRT} = -2 \left( \log \mathcal{L}_{\text{null}} - \left( \log \mathcal{L}_{\text{A}} + \log \mathcal{L}_{\text{B}} \right) \right)

\log \mathcal{L}_{\text{null}} = \sum_{i} \log \mathcal{L}(r_0, \mu_0 | X_i)

The log-likelihood under the null hypothesis (assuming a common mean $μ0\mu_0$ for both conditions)

The LRT statistic approximately follows a chi-squared distribution with 1 degree of freedom under the null hypothesis

The p-value is computed as:

p = 1 - F_{\chi^2_1}(\text{LRT})

k would be 1

After today, you should be able to

Interpret common visualizations used in differential expression analysis

Visualizing Significance vs. Magnitude of Expression Changes

Interpretation:

Top Corners: Genes with high significance and large fold changes (both upregulated and downregulated)
Center: Genes with little to no change or low significance

A volcano plot displays the relationship between each gene's statistical significance (p-value) and the magnitude of change (fold change).

MA Plots

An MA plot visualizes the relationship between the average expression (A) and the log fold change (M) for each gene.

Interpretation:

Center Line (M=0): No change in expression
Spread: Indicates variability in fold changes across different expression levels

Usage: Identifying trends or biases in expression data, such as mean-dependent variance.

Heatmaps

Components:

Rows: Genes
Columns: Samples
Color Intensity: Represents expression level (e.g., red for upregulation, blue for downregulation)

A heatmap displays the expression levels of multiple genes across different samples using color gradients

Interpretation: Identifying clusters of co-expressed genes and sample groupings based on expression profiles.

Principal Component Analysis (PCA) Plots

PCA transforms high-dimensional gene expression data into principal components that capture the most variance

Axes: Principal components representing the most significant sources of variation

Interpretation:

Sample Clustering: Samples from similar conditions cluster together.
Outliers: Samples that do not group with others may indicate technical or biological variability.

Usage: Assessing batch effects, overall data structure, and sample quality

Announcements

After today, you should be able to

Why are we learning about differential gene expression?

What is Differential Gene Expression?

Case study: Breast cancer

After today, you should be able to

Explain the purpose of statistical models and hypothesis testing

What is a statistical model?

Hypothesis testing in RNA-seq data

The P-value is the probability of the null hypothesis being true

Differential gene expression uses statistical models for hypothesis testing

After today, you should be able to

Discuss the challenges of working with gene expression data

The nature of count data

Binomial: A Simple Model for Discrete Counts

Limitations of Binomial distributions for RNA-Seq

Poisson distribution: A baseline for modeling discrete counts

Poisson distribution becomes inaccurate when variance > mean

Parity plots with mean and variance show deviations with Poisson distributions

Overdispersion in RNA-Seq

Negative Binomial distribution accounts for high dispersion

The Challenge of zeros in RNA-seq data

After today, you should be able to

Discuss fitting of statistical models

Why are statistical models important in RNA-seq?

Define the model for each gene

Fit parameters using optimization algorithms

After today, you should be able to

Understand statistical tests used for gene expression data

Wald’s Test for Gene Expression Differences

Log Fold Change

Estimate Parameters from the Negative Binomial Model

Likelihood Ratio Test

LRT Statistic

After today, you should be able to

Interpret common visualizations used in differential expression analysis

Visualizing Significance vs. Magnitude of Expression Changes

MA Plots

Heatmaps

Principal Component Analysis (PCA) Plots

Before the next class, you should