BI 255 · Bethune-Cookman University · Fall 2026
Correlation tells you that two variables are related. Regression goes further: it describes the relationship with a mathematical equation, allowing you to predict one variable from another and to test whether the predictor explains a significant portion of the variation.
Simple linear regression fits a straight line: y = β₀ + β₁x + ε
By the end of this lab you will be able to:
lm() in Rsummary(lm())predict()The Bradford assay estimates protein concentration by measuring absorbance at 595 nm. A standard curve is produced by measuring known concentrations of BSA (bovine serum albumin) — this is a textbook linear regression problem in biochemistry.
# BSA standard concentrations (ug/mL) and measured absorbance (A595)
bsa_conc <- c(0, 50, 100, 150, 200, 250, 300, 350, 400)
absorbance <- c(0.041, 0.189, 0.352, 0.508, 0.667, 0.821, 0.979, 1.128, 1.276) +
rnorm(9, 0, 0.018)
# Always visualise first
plot(bsa_conc, absorbance,
pch = 16,
col = "#2C5F8A",
cex = 1.4,
xlab = "BSA Concentration (ug/mL)",
ylab = "Absorbance at 595 nm",
main = "Bradford Assay Standard Curve")
# Fit the regression model
bradford_model <- lm(absorbance ~ bsa_conc)
# Examine the full summary
summary(bradford_model)
Call:
lm(formula = absorbance ~ bsa_conc)
Residuals:
Min 1Q Median 3Q Max
-0.0140223 -0.0117579 -0.0006154 0.0096677 0.0188824
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 2.534e-02 8.184e-03 3.096 0.0174 *
bsa_conc 3.161e-03 3.438e-05 91.936 4.75e-12 ***
---
Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
Residual standard error: 0.01332 on 7 degrees of freedom
Multiple R-squared: 0.9992, Adjusted R-squared: 0.9991
F-statistic: 8452 on 1 and 7 DF, p-value: 4.745e-12summary(lm()) Output
Coefficients table:
| Term | Estimate | Meaning |
|---|---|---|
(Intercept) |
β₀ | Expected absorbance when [BSA] = 0 (should be near 0 for blank) |
bsa_conc |
β₁ | Increase in absorbance per 1 ug/mL increase in BSA |
Overall model fit:
| Statistic | Meaning |
|---|---|
Multiple R-squared |
Proportion of variation in y explained by x |
F-statistic, p-value |
Overall test: does x significantly predict y? (H₀: β₁ = 0) |
Residual standard error |
Average distance of data points from the fitted line |
An R² of 0.998 means 99.8% of the variation in absorbance is explained by BSA concentration — ideal for a calibration curve.
library(ggplot2)
# Create a data frame for ggplot
bradford_df <- data.frame(conc = bsa_conc, abs = absorbance)
# New x values for smooth confidence band
new_conc <- data.frame(bsa_conc = seq(0, 400, by = 5))
# Confidence band (uncertainty in the mean prediction)
ci_band <- predict(bradford_model, newdata = new_conc,
interval = "confidence", level = 0.95)
# Prediction band (uncertainty for a new single measurement)
pi_band <- predict(bradford_model, newdata = new_conc,
interval = "prediction", level = 0.95)
band_df <- data.frame(new_conc,
fit = ci_band[, "fit"],
ci_lo = ci_band[, "lwr"],
ci_hi = ci_band[, "upr"],
pi_lo = pi_band[, "lwr"],
pi_hi = pi_band[, "upr"])
ggplot(bradford_df, aes(x = conc, y = abs)) +
geom_ribbon(data = band_df, aes(x = bsa_conc, ymin = pi_lo, ymax = pi_hi),
fill = "#A8C4D9", alpha = 0.4, inherit.aes = FALSE) +
geom_ribbon(data = band_df, aes(x = bsa_conc, ymin = ci_lo, ymax = ci_hi),
fill = "#5B8DB8", alpha = 0.5, inherit.aes = FALSE) +
geom_line(data = band_df, aes(x = bsa_conc, y = fit),
colour = "#2C5F8A", linewidth = 1.2, inherit.aes = FALSE) +
geom_point(colour = "#2C5F8A", size = 3) +
annotate("text", x = 50, y = 1.2,
label = paste0("R² = ", round(summary(bradford_model)$r.squared, 4)),
hjust = 0, size = 4) +
labs(title = "Bradford Assay Standard Curve",
subtitle = "Dark band = 95% CI for mean; light band = 95% PI for new observation",
x = "BSA Concentration (ug/mL)",
y = "Absorbance (595 nm)") +
theme_classic(base_size = 13)
Confidence interval (CI): the plausible range for the mean absorbance at a given concentration — gets narrower with more data.
Prediction interval (PI): the plausible range for the absorbance of a single new observation at that concentration — always wider than the CI because it incorporates measurement-to-measurement variability.
When using a standard curve to estimate an unknown protein concentration, the prediction interval is more honest — you are predicting a single new measurement, not the average of many.
# Three unknown samples were measured at these absorbances:
unknown_abs <- c(0.55, 0.88, 1.10)
# Invert the regression: concentration = (absorbance - intercept) / slope
intercept <- coef(bradford_model)[1]
slope <- coef(bradford_model)[2]
predicted_conc <- (unknown_abs - intercept) / slope
cat("Predicted concentrations (ug/mL):\n")Predicted concentrations (ug/mL):print(round(predicted_conc, 1))[1] 166.0 270.4 340.0# Or use predict() directly (predict absorbance from concentration)
# For inverse prediction, you can also use the equation above
# Let's confirm: what absorbance do we expect at 175 ug/mL?
predict(bradford_model,
newdata = data.frame(bsa_conc = 175),
interval = "prediction") fit lwr upr
1 0.5784624 0.5452118 0.611713Linear regression assumes:
par(mfrow = c(2, 2))
plot(bradford_model)
par(mfrow = c(1, 1))| Plot | What to look for |
|---|---|
| Residuals vs Fitted | Points scattered randomly around 0 — no pattern. Curved pattern = nonlinearity. Funnel = heteroscedasticity. |
| Normal QQ of residuals | Points on the diagonal line — confirms normality of residuals. |
| Scale-Location | Square root of standardised residuals vs fitted — flat line = constant variance. |
| Residuals vs Leverage | Identifies influential points. Points outside Cook’s distance lines have disproportionate influence on the slope. |
Allometric scaling describes how physiological variables scale with body mass. Basal metabolic rate (BMR) scales approximately with body mass raised to the power 0.75 — on a log-log scale, this becomes a straight line.
set.seed(88)
# Mammal body mass (kg) and BMR (kcal/day) — realistic scale range
body_mass <- c(0.02, 0.05, 0.1, 0.5, 1, 3, 10, 20, 50, 100, 200, 500, 1000, 3000, 5000)
bmr <- 70 * (body_mass^0.75) * exp(rnorm(15, 0, 0.15))
# Linear regression on log-log scale
log_mass <- log10(body_mass)
log_bmr <- log10(bmr)
allometry_model <- lm(log_bmr ~ log_mass)
summary(allometry_model)
Call:
lm(formula = log_bmr ~ log_mass)
Residuals:
Min 1Q Median 3Q Max
-0.134950 -0.037428 0.006082 0.038943 0.146347
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 1.84964 0.02494 74.18 <2e-16 ***
log_mass 0.74805 0.01218 61.40 <2e-16 ***
---
Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
Residual standard error: 0.07967 on 13 degrees of freedom
Multiple R-squared: 0.9966, Adjusted R-squared: 0.9963
F-statistic: 3769 on 1 and 13 DF, p-value: < 2.2e-16# The slope on log-log scale = the allometric exponent
cat("Allometric exponent (slope on log-log scale):",
round(coef(allometry_model)[2], 3), "\n")Allometric exponent (slope on log-log scale): 0.748 cat("Expected from Kleiber's law: ~0.75\n")Expected from Kleiber's law: ~0.75plot(log_mass, log_bmr,
pch = 16, col = "#2C5F8A", cex = 1.4,
xlab = "log₁₀ Body Mass (kg)",
ylab = "log₁₀ BMR (kcal/day)",
main = "Allometric Scaling of Basal Metabolic Rate")
abline(allometry_model, col = "#C8102E", lwd = 2)
legend("topleft",
legend = c(paste0("slope = ", round(coef(allometry_model)[2], 3)),
paste0("R² = ", round(summary(allometry_model)$r.squared, 3))),
bty = "n")
Many biological relationships follow power laws: y = a × x^b. On a log-log scale, this becomes log(y) = log(a) + b × log(x) — a straight line. The slope b is the allometric exponent, and the intercept log(a) gives the proportionality constant.
This is why biologists often plot mass-related data on log-log axes — it linearises power-law relationships and makes regression analysis valid.
Close your notes. Answer these on your own — you have 3 minutes. We’ll go through the answers together after.
1. In the output of summary(lm()), the coefficient for the predictor has estimate = 0.0032 and p-value = 0.0001. What does this tell you?
2. A regression model gives R² = 0.45. Which statement is correct?
a) 45% of observations are correctly predicted b) The model explains 45% of the variance in y c) The slope is 0.45 d) 45% of data points lie on the regression line3. You run a Bradford assay and get an absorbance of 0.72 for an unknown sample. Explain step by step how you would use a linear regression model to estimate the protein concentration.
4. In a Residuals vs Fitted plot, the points form a U-shaped curve. What does this indicate and what should you do?
5. True or False: A prediction interval for a new observation is always narrower than the confidence interval for the mean at the same x value.
1. The slope is 0.0032, meaning that for each one-unit increase in x, y increases by 0.0032 on average. The p-value of 0.0001 means this slope is highly significantly different from zero — we reject H₀: β₁ = 0. The predictor x significantly explains variation in y. However, the biological importance depends on the units and context, not just statistical significance.
2. b) The model explains 45% of the variance in y — R² (the coefficient of determination) is the proportion of variance in the response variable accounted for by the predictor. It does not mean 45% of predictions are “correct” (which has no clear meaning in regression) or that 45% of points lie on the line.
3. Step 1: Identify the model intercept (β₀) and slope (β₁) from coef(model). Step 2: Set up the equation: absorbance = β₀ + β₁ × concentration. Step 3: Solve for concentration: concentration = (absorbance − β₀) / β₁ = (0.72 − β₀) / β₁. Step 4: Check that 0.72 falls within the range of your standard curve (do not extrapolate beyond it).
4. A U-shaped (curved) pattern in Residuals vs Fitted indicates non-linearity — the true relationship between x and y is not linear, even though you fitted a straight line. You should consider: (a) adding a quadratic term (x²) to the model, (b) log-transforming x or y, or (c) fitting a curve instead of a line. A linear model is not appropriate for these data.
5. False — The prediction interval is always wider than the confidence interval. The CI quantifies uncertainty in the mean prediction; the PI additionally accounts for individual observation variability (the residual variance). At the same x value, PI = CI width plus the residual standard error contribution. A PI will always be broader.
Before you leave, make sure you can:
The R built-in dataset trees contains girth, height, and volume for 31 black cherry trees. Fit a linear regression of Volume on Girth. Check all four diagnostic plots. Then try regressing log(Volume) on log(Girth) and compare R² and the diagnostic plots. Which model fits better? Write a two-sentence biological interpretation of the slope in each model.