AP Biology · Official Lab Investigations · All 13 Labs
Laboratory
Investigations
AP Biology requires 25% of class time devoted to inquiry-based labs. The 13 official investigations span all four Big Ideas and directly test the six science practices. Exam FRQs are frequently framed in experimental contexts — knowing these labs deeply is essential for full scores.
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INV 1
Artificial Selection
Unit 7 — Natural Selection · Big Idea 1: Evolution
🎯 Big Question
Can humans alter the phenotypic distribution of a population by selecting for specific traits over multiple generations?
🧪 Common Model Organisms
Wisconsin Fast Plants (Brassica rapa) — trichome density or leaf morphology. Brine shrimp hatching viability at various salinities. Radishes or other rapid-generation plants.
📊 Independent Variable
The trait being selected for (e.g., high vs. low trichome density); the selection criterion applied by the experimenter each generation
📈 Dependent Variable
Mean phenotype of the trait in each subsequent generation (e.g., average trichome number); frequency distribution of the trait across generations
🔬 Key Measurement
Count and record trait measurements in each generation; compare mean ± SE across generations using bar graphs with error bars
📋 Key Procedures
1. Measure the trait in the parental (P) generation — establish baseline distribution.
2. Select only individuals with the most extreme phenotype (highest or lowest) as parents.
3. Allow selected individuals to reproduce; measure offspring (F1 generation) — is the mean shifting?
4. Repeat selection for multiple generations (F2, F3...) and graph mean ± standard error each generation.
5. Compare your selection line to a randomly-mated control line (no selection pressure).
2. Select only individuals with the most extreme phenotype (highest or lowest) as parents.
3. Allow selected individuals to reproduce; measure offspring (F1 generation) — is the mean shifting?
4. Repeat selection for multiple generations (F2, F3...) and graph mean ± standard error each generation.
5. Compare your selection line to a randomly-mated control line (no selection pressure).
🔑 Key Concepts Tested
• Heritability: If the trait responds to selection, it has a heritable component — it is encoded in alleles that can be passed to offspring.
• Mechanism parallel to natural selection: Artificial selection demonstrates all four of Darwin's postulates in action — variation exists, it is heritable, not all individuals reproduce (selected by humans), and there is differential reproduction based on phenotype.
• Selection differential: The greater the gap between selected parents and population mean, the greater the response to selection each generation.
• Limits to selection: After many generations, response may slow as genetic variation is exhausted (most favorable alleles go to fixation) or as counterselective effects emerge.
• Mechanism parallel to natural selection: Artificial selection demonstrates all four of Darwin's postulates in action — variation exists, it is heritable, not all individuals reproduce (selected by humans), and there is differential reproduction based on phenotype.
• Selection differential: The greater the gap between selected parents and population mean, the greater the response to selection each generation.
• Limits to selection: After many generations, response may slow as genetic variation is exhausted (most favorable alleles go to fixation) or as counterselective effects emerge.
High-Frequency Exam Points
Control group requirement: The control is a randomly mated group of the same species with NO selection applied — same conditions, same number of generations, but parents chosen at random. Without this control, you cannot attribute trait change to selection (it could be environmental change, random drift, etc.).
Data analysis: The primary analysis for artificial selection experiments is tracking the shift in mean trait value across generations in the selected line vs. the randomly mated control. Bar graphs with error bars (mean ± SE) are plotted generation by generation — a directional shift in the selected line confirms a heritable response to selection. Chi-square is not the standard analysis for continuous trait data in this context; it is more appropriate for categorical count data (e.g., Mendelian ratios, behavior choices).
Connect to evolution: Exam FRQs may ask you to compare results from artificial selection to natural selection — the mechanism is identical; only who/what applies the selective pressure differs (humans vs. environment).
INV 2
Mathematical Modeling: Hardy-Weinberg
Unit 7 — Natural Selection · Big Idea 1: Evolution
🎯 Big Question
How do allele frequencies change (or not change) in populations under different conditions? What factors disrupt Hardy-Weinberg equilibrium?
🧪 Method
Simulation (card/bead model or computer): randomly draw alleles to simulate random mating, then add selection, genetic drift, mutation, or migration to observe effects on allele frequencies
📋 Key Equations
p + q = 1; p² + 2pq + q² = 1. Calculate allele frequencies from phenotype data (starting from q² for recessive homozygotes)
🔑 Key Concepts Tested
• Five HWE conditions: Large population, no migration, no mutation, random mating, no natural selection. Testing each by violating one at a time and observing allele frequency changes.
• Genetic drift is greater in small populations: In small simulations, allele frequencies drift randomly even without selection. In large simulations, they stay near the predicted HWE values.
• Selection changes allele frequencies directionally: When phenotypes with recessive alleles have lower fitness, q decreases over generations — but elimination is very slow once q becomes small (many recessive alleles are hidden in heterozygotes).
• Gene flow homogenizes populations: When simulating migration between two populations with different allele frequencies, frequencies converge toward each other.
• Genetic drift is greater in small populations: In small simulations, allele frequencies drift randomly even without selection. In large simulations, they stay near the predicted HWE values.
• Selection changes allele frequencies directionally: When phenotypes with recessive alleles have lower fitness, q decreases over generations — but elimination is very slow once q becomes small (many recessive alleles are hidden in heterozygotes).
• Gene flow homogenizes populations: When simulating migration between two populations with different allele frequencies, frequencies converge toward each other.
High-Frequency Exam Points
Null hypothesis for HWE: H₀ = The population is in Hardy-Weinberg equilibrium (allele frequencies are not changing due to evolution). Rejection of H₀ means at least one evolutionary force is acting.
Calculation chain — always start from q²: If 9 out of 100 individuals show the recessive phenotype → q² = 0.09 → q = 0.30 → p = 0.70 → expected heterozygotes = 2pq × 100 = 2(0.70)(0.30) × 100 = 42. Always show all steps — AP graders award partial credit for each correct calculation.
INV 3
Comparing DNA Sequences to Investigate Evolutionary Relationships
Unit 7 — Natural Selection · Big Idea 1: Evolution
🎯 Big Question
How can similarities and differences in DNA (or protein) sequences be used to determine evolutionary relationships among organisms?
🧪 Method
Bioinformatics analysis (computer-based): align homologous gene sequences (e.g., cytochrome c, FOXP2, rRNA) across species; count sequence differences; construct a phylogenetic tree based on similarity
📊 Key Data Type
Pairwise percent sequence identity between species; number of nucleotide or amino acid differences; these are used as a proxy for evolutionary distance
🔑 Key Concepts Tested
• Molecular clock: The rate of neutral mutations accumulates relatively steadily over time. Greater sequence divergence → more time since divergence from common ancestor → more distantly related.
• Shared sequences = common ancestry: Highly conserved sequences (e.g., histone genes, rRNA) are nearly identical across all eukaryotes — strong evidence for universal common ancestry.
• Molecular data vs. morphological data: Molecular phylogenies often reveal evolutionary relationships not apparent from morphology (e.g., whales are more closely related to hippos than to fish despite similar body shape).
• Outgroup selection: The outgroup is the least related species, used to root the tree and distinguish ancestral from derived character states.
• Limitations: Convergent sequence evolution can mislead (two unrelated lineages evolve the same mutation). Gene transfer (especially in prokaryotes) complicates tree building.
• Shared sequences = common ancestry: Highly conserved sequences (e.g., histone genes, rRNA) are nearly identical across all eukaryotes — strong evidence for universal common ancestry.
• Molecular data vs. morphological data: Molecular phylogenies often reveal evolutionary relationships not apparent from morphology (e.g., whales are more closely related to hippos than to fish despite similar body shape).
• Outgroup selection: The outgroup is the least related species, used to root the tree and distinguish ancestral from derived character states.
• Limitations: Convergent sequence evolution can mislead (two unrelated lineages evolve the same mutation). Gene transfer (especially in prokaryotes) complicates tree building.
High-Frequency Exam Points
Reading a sequence alignment: If given a table of percent sequence identity, the pair with the HIGHEST percent identity is most closely related (least diverged). To find the most recent common ancestor of any two species, find the node where their branches connect — the deeper (further from tips) the node, the more ancient and distant the relationship.
Why is molecular data preferred over morphology? Morphology can be misleading due to convergent evolution (dolphins and sharks look similar but are not closely related). DNA sequences mutate at a relatively consistent rate and are less subject to convergence at the molecular level — especially for highly conserved genes like cytochrome c or rRNA.
INV 4
Diffusion and Osmosis
Unit 2 — Cells · Big Idea 2: Energetics
🎯 Big Question
How does solute concentration affect the movement of water across a selectively permeable membrane? How can we determine the osmotic concentration of living cells?
🧪 Parts of the Lab
Part A: Dialysis tubing bags with sucrose solutions placed in water — measure mass change. Part B: Potato cores / celery / eggs in sucrose solutions of varying concentrations — measure % mass change. Water Potential calculations using Ψ = Ψs + Ψp
📊 Independent Variable
Concentration of sucrose solution (external) — typically 0.0, 0.2, 0.4, 0.6, 0.8, 1.0 M
📈 Dependent Variable
Percent change in mass of the potato core (or dialysis bag): % Δmass = [(final − initial)/initial] × 100
📋 Key Procedures & Calculations
1. Weigh each potato core (or dialysis bag) before placing in solution — record initial mass.
2. Submerge in different sucrose concentrations for ~30–45 minutes; blot dry and reweigh.
3. Calculate % mass change = [(final mass − initial mass) / initial mass] × 100.
4. Graph % mass change (y-axis) vs. sucrose concentration (x-axis). The x-intercept (0% mass change) = isotonic point = the molarity equal to the internal solute concentration of the potato cells.
5. For water potential: Ψs = −iCRT (i = ionization constant, C = molar concentration, R = 0.0831 L·bar/mol·K, T = temp in Kelvin). Ψp = 0 for cut potato (no cell wall pressure possible). Therefore Ψ = Ψs.
2. Submerge in different sucrose concentrations for ~30–45 minutes; blot dry and reweigh.
3. Calculate % mass change = [(final mass − initial mass) / initial mass] × 100.
4. Graph % mass change (y-axis) vs. sucrose concentration (x-axis). The x-intercept (0% mass change) = isotonic point = the molarity equal to the internal solute concentration of the potato cells.
5. For water potential: Ψs = −iCRT (i = ionization constant, C = molar concentration, R = 0.0831 L·bar/mol·K, T = temp in Kelvin). Ψp = 0 for cut potato (no cell wall pressure possible). Therefore Ψ = Ψs.
🔑 Key Concepts Tested
• Mass gain = water entering = hypotonic external solution (external [solute] < potato [solute]).
• Mass loss = water leaving = hypertonic external solution (external [solute] > potato [solute]).
• No mass change = isotonic — this solution concentration equals the internal osmotic concentration of potato cells.
• Water potential gradient drives osmosis: Water moves from higher Ψ (less concentrated) to lower Ψ (more concentrated).
• Mass loss = water leaving = hypertonic external solution (external [solute] > potato [solute]).
• No mass change = isotonic — this solution concentration equals the internal osmotic concentration of potato cells.
• Water potential gradient drives osmosis: Water moves from higher Ψ (less concentrated) to lower Ψ (more concentrated).
High-Frequency Exam Points
Percent change, not absolute change: Always use percent change (not raw mass change) to control for initial mass differences between samples. This is a frequent point source on FRQs.
X-intercept on the graph = isotonic point: Drawing a best-fit line through the data and identifying where it crosses x = 0 gives the osmotic concentration of the potato cells. This is the defining result of Part B.
Water potential calculation steps: Ψs = −iCRT. For a 0.4 M sucrose solution (non-ionizing, i=1) at 25°C (298 K): Ψs = −(1)(0.4)(0.0831)(298) = −9.9 bars. Since Ψp = 0, Ψ = −9.9 bars. Water moves from higher Ψ to lower Ψ.
A student places 5 potato cores in sucrose solutions of 0.0, 0.2, 0.4, 0.6, and 0.8 M. After 30 minutes, percent mass changes are +8%, +3%, 0%, −6%, and −12%, respectively. What is the estimated osmotic concentration of the potato cells, and what would happen to a potato cell placed in 0.8 M sucrose?
Osmotic concentration: The 0% mass change point occurs at 0.4 M sucrose — this is the isotonic point. The potato cells have an internal solute concentration approximately equivalent to 0.4 M sucrose. At this concentration, there is no net movement of water in either direction.
In 0.8 M sucrose (hypertonic): The external solution (0.8 M) has a higher solute concentration than the potato cells (~0.4 M). The external solution therefore has a lower water potential than the cells. Water moves out of the potato cells by osmosis → cells lose water → cells shrink → the potato tissue becomes flaccid and limp. In plant cells, severe water loss causes plasmolysis — the plasma membrane pulls away from the cell wall.
In 0.8 M sucrose (hypertonic): The external solution (0.8 M) has a higher solute concentration than the potato cells (~0.4 M). The external solution therefore has a lower water potential than the cells. Water moves out of the potato cells by osmosis → cells lose water → cells shrink → the potato tissue becomes flaccid and limp. In plant cells, severe water loss causes plasmolysis — the plasma membrane pulls away from the cell wall.
INV 5
Photosynthesis (Floating Leaf Disk / Spectrophotometry)
Unit 3 — Cellular Energetics · Big Idea 2: Energetics
🎯 Big Question
What environmental factors (light intensity, CO₂ concentration, wavelength) affect the rate of photosynthesis? How is photosynthesis rate measured?
🧪 Floating Leaf Disk Method
Leaf disks are vacuum-infiltrated with bicarbonate solution (CO₂ source) — they sink because spaces are filled with liquid. Under light, photosynthesis produces O₂, which fills intercellular spaces → disks float. Rate measured as ET₅₀ (time for 50% of disks to float) or number of disks floating at set intervals.
📊 Independent Variable
Light intensity (distance from lamp), wavelength of light (colored filters), CO₂ concentration (bicarbonate concentration), temperature
📈 Dependent Variable
Rate of photosynthesis — measured as: number of disks floating per minute OR time for 50% of disks to float (shorter time = faster rate)
⚠️ Controls
Dark control (disks in bicarbonate, NO light) — demonstrates that disk flotation depends on light-driven photosynthetic O₂ production; without light, no net O₂ is produced and disks remain sunken or re-sink. Water control (no bicarbonate, no CO₂ source) — shows baseline without photosynthesis substrate, confirming CO₂ is required
🔑 Key Concepts Tested
• O₂ produced by photosynthesis is from WATER, not CO₂. The O₂ that makes disks float comes from photolysis of water in Photosystem II.
• Bicarbonate provides CO₂ for the Calvin cycle: HCO₃⁻ + H⁺ → H₂O + CO₂.
• Compensation point: Light intensity at which photosynthesis rate = cellular respiration rate (no net O₂ gain or loss). At compensation point, disks stop sinking but also stop rising — net change = 0.
• Light saturation point: Light intensity above which further increase has no effect on photosynthesis rate (other factors become limiting — CO₂ or temperature).
• Limiting factors: At low light intensity, light is limiting. At high light with low CO₂, CO₂ is limiting. At optimal light and CO₂, temperature may limit enzyme-catalyzed Calvin cycle reactions.
• Bicarbonate provides CO₂ for the Calvin cycle: HCO₃⁻ + H⁺ → H₂O + CO₂.
• Compensation point: Light intensity at which photosynthesis rate = cellular respiration rate (no net O₂ gain or loss). At compensation point, disks stop sinking but also stop rising — net change = 0.
• Light saturation point: Light intensity above which further increase has no effect on photosynthesis rate (other factors become limiting — CO₂ or temperature).
• Limiting factors: At low light intensity, light is limiting. At high light with low CO₂, CO₂ is limiting. At optimal light and CO₂, temperature may limit enzyme-catalyzed Calvin cycle reactions.
High-Frequency Exam Points
Why do disks sink when infiltrated with bicarbonate? The air spaces in the leaf mesophyll are replaced with liquid (bicarbonate solution), increasing density → sinks. When O₂ is produced by photosynthesis, it gradually replaces the liquid in intercellular spaces → disk density decreases → floats. This is the key mechanism — always explain at the cellular level.
Negative control with no bicarbonate: Without a CO₂ source, the Calvin cycle cannot run even if light reactions proceed → no net O₂ production → disks remain sunken. This confirms that CO₂ is required for photosynthesis, not just light.
Graph shape: Rate vs. light intensity shows saturation kinetics — initially linear (light is limiting), then plateau (something else is limiting). This is analogous to enzyme saturation (Km and Vmax).
INV 6
Cellular Respiration
Unit 3 — Cellular Energetics · Big Idea 2: Energetics
🎯 Big Question
How do factors like temperature, germination state, and organism type affect the rate of cellular respiration?
🧪 Method — Respirometer
Closed chambers (vials) containing germinating seeds or small organisms + KOH (absorbs CO₂). As O₂ is consumed and CO₂ is absorbed by KOH, gas volume decreases. A colored dye in a capillary tube shows the volume change. Rate of respiration = rate of O₂ consumption = rate of dye movement per unit time.
📊 Independent Variable
Temperature (cold water vs. room temp); germinating seeds vs. dry (non-germinating) seeds; type of organism (germinating peas, crickets, mealworms)
📈 Dependent Variable
Rate of O₂ consumption (mL O₂/min) — calculated from volume change in respirometer over time
⚠️ Control
Vial with glass beads (no living organisms) of equivalent volume — corrects for pressure/temperature changes due to physical factors; should show no volume change if conditions are stable
📋 Key Calculations
Rate of respiration = (volume change in experimental vial) − (volume change in control vial) / time
The control correction is critical because temperature changes in the water bath cause volume changes in all vials regardless of respiration — subtracting the control removes this artifact.
To correct for organism mass: express rate as mL O₂ / g · min to allow comparison between organisms of different sizes.
The control correction is critical because temperature changes in the water bath cause volume changes in all vials regardless of respiration — subtracting the control removes this artifact.
To correct for organism mass: express rate as mL O₂ / g · min to allow comparison between organisms of different sizes.
🔑 Key Concepts Tested
• Germinating seeds respire faster than dry seeds — active metabolic processes (cell division, protein synthesis) require ATP → more cellular respiration.
• Temperature effects: Higher temperature → faster enzyme-catalyzed respiration (up to the optimal temperature). Cold temperatures slow respiration significantly. Ectotherms show a more direct dependence on ambient temperature than endotherms; endotherms can generate metabolic heat to maintain body temperature, but respiration rates in isolated tissues or small organisms can still be affected by experimental temperature.
• KOH absorbs CO₂: Without KOH, CO₂ released by respiration would replace O₂ consumed, and no net volume change would be detected. KOH ensures only O₂ consumption is measured.
• Relationship to cellular respiration equation: C₆H₁₂O₆ + 6O₂ → 6CO₂ + 6H₂O. For every mole of O₂ consumed, one mole of CO₂ is produced (respiratory quotient = 1 for carbohydrates).
• Temperature effects: Higher temperature → faster enzyme-catalyzed respiration (up to the optimal temperature). Cold temperatures slow respiration significantly. Ectotherms show a more direct dependence on ambient temperature than endotherms; endotherms can generate metabolic heat to maintain body temperature, but respiration rates in isolated tissues or small organisms can still be affected by experimental temperature.
• KOH absorbs CO₂: Without KOH, CO₂ released by respiration would replace O₂ consumed, and no net volume change would be detected. KOH ensures only O₂ consumption is measured.
• Relationship to cellular respiration equation: C₆H₁₂O₆ + 6O₂ → 6CO₂ + 6H₂O. For every mole of O₂ consumed, one mole of CO₂ is produced (respiratory quotient = 1 for carbohydrates).
High-Frequency Exam Points
Why is a non-living (bead) control essential? Temperature changes, barometric pressure changes, and any leak in the system will affect all vials equally. The bead control accounts for these physical changes. The corrected rate = (experimental change) − (control change) gives the true biological rate of O₂ consumption.
Fermentation extension: This lab is often extended to test fermentation using yeast + different substrates (glucose, sucrose, starch, fructose) — measure CO₂ production with a CO₂ sensor or inverted-tube gas collection. Substrates yeast can ferment produce more CO₂; substrates it cannot (e.g., starch without amylase) produce little CO₂.
INV 7
Cell Division: Mitosis and Meiosis
Units 4 & 5 — Cell Cycle / Heredity · Big Idea 3: Information
🎯 Big Question (Mitosis)
What proportion of time do cells in an onion root tip spend in each phase of the cell cycle? How does mitosis ensure genetic continuity?
🎯 Big Question (Meiosis)
How does meiosis generate genetic diversity? How does crossing over create new allele combinations? (Often simulated with pop-bead models)
🧪 Mitosis Method
Prepared slides of onion root tip (meristematic zone) or whitefish blastula. Count cells in each phase (prophase, metaphase, anaphase, telophase, interphase) in a fixed number of cells (~100). Calculate % time in each phase as a proxy for duration.
📊 Key Calculation
Mitotic Index = (cells in mitosis) / (total cells observed). Higher index → more rapid cell division. Cancer cells have unusually high mitotic indices.
📋 Phases and What to Look For
Interphase (~90% of time): No condensed chromosomes visible; nucleus clear; cell is growing and synthesizing DNA.
Prophase: Chromosomes condense and become visible; nuclear envelope disappearing; spindle forming.
Metaphase: Chromosomes aligned at the center (metaphase plate); most condensed; spindle fully formed.
Anaphase: Sister chromatids pulled apart to opposite poles; cell elongates; shortest phase.
Telophase + Cytokinesis: Two new nuclei forming; cell plate (plants) or cleavage furrow (animals) beginning to separate cytoplasm.
Prophase: Chromosomes condense and become visible; nuclear envelope disappearing; spindle forming.
Metaphase: Chromosomes aligned at the center (metaphase plate); most condensed; spindle fully formed.
Anaphase: Sister chromatids pulled apart to opposite poles; cell elongates; shortest phase.
Telophase + Cytokinesis: Two new nuclei forming; cell plate (plants) or cleavage furrow (animals) beginning to separate cytoplasm.
🔑 Key Concepts Tested
• Cell spends the most time in interphase — this is reflected by the large proportion of cells seen in interphase in the slide.
• Mitotic index as a cancer diagnostic: Abnormally high mitotic index in a tissue biopsy suggests cells are dividing without normal regulation → possible malignancy.
• Meiosis simulation: Pop-bead model allows tracking of specific allele combinations through Meiosis I (homolog separation) and Meiosis II (sister chromatid separation). Crossing over (exchange of bead segments between homologs in Prophase I) creates new allele combinations not present in either parent.
• Mitotic index as a cancer diagnostic: Abnormally high mitotic index in a tissue biopsy suggests cells are dividing without normal regulation → possible malignancy.
• Meiosis simulation: Pop-bead model allows tracking of specific allele combinations through Meiosis I (homolog separation) and Meiosis II (sister chromatid separation). Crossing over (exchange of bead segments between homologs in Prophase I) creates new allele combinations not present in either parent.
High-Frequency Exam Points
Why are most cells in interphase? Because cells spend the vast majority of their lifespan in interphase (G1, S, G2) — this is when the cell does its normal work. Mitosis itself is a brief event. The longer a phase lasts, the higher the probability that a randomly photographed cell will be in that phase.
Mitotic index calculation: If 8 of 100 cells are in mitosis, mitotic index = 8/100 = 0.08 = 8%. A tissue with a mitotic index of 25% is dividing much more rapidly than normal — important clinical indicator.
INV 8
Biotechnology: Bacterial Transformation
Unit 6 — Gene Expression & Regulation · Big Idea 3: Information
🎯 Big Question
How can foreign DNA be introduced into bacteria? What is the evidence that transformation was successful? What factors affect transformation efficiency?
🧪 Method
E. coli bacteria are made "competent" (permeable to DNA) by CaCl₂ treatment and heat shock (42°C for 90 sec). Bacteria take up a plasmid encoding antibiotic resistance (e.g., ampicillin resistance, amp^r) and/or GFP (green fluorescent protein). Bacteria are plated on LB (all bacteria grow) and LB+amp (only transformed bacteria survive).
📊 Four Plates Interpretation
LB (−plasmid, −amp): lawn of bacteria (positive control — shows bacteria are alive). LB (−plasmid, +amp): no colonies (shows amp kills untransformed bacteria). LB (+plasmid, −amp): lawn of bacteria. LB (+plasmid, +amp): only transformed colonies survive — these carry the amp-resistance gene.
📋 Transformation Efficiency Calculation
Transformation efficiency = (number of colonies on +amp plate) / (μg of plasmid DNA used)
Units: colonies per μg DNA
Higher transformation efficiency = more bacteria successfully took up the plasmid. Factors affecting efficiency include: CaCl₂ concentration, heat shock duration/temperature, DNA concentration, bacterial growth phase (log phase is optimal), and recovery time before plating.
Units: colonies per μg DNA
Higher transformation efficiency = more bacteria successfully took up the plasmid. Factors affecting efficiency include: CaCl₂ concentration, heat shock duration/temperature, DNA concentration, bacterial growth phase (log phase is optimal), and recovery time before plating.
🔑 Key Concepts Tested
• Transformation = uptake of foreign DNA — one of four horizontal gene transfer mechanisms in prokaryotes (transformation, transduction, conjugation, transposition).
• Selectable marker strategy: The antibiotic resistance gene allows selection of only those cells that successfully took up the plasmid — without selection, you cannot distinguish transformed from untransformed cells.
• GFP expression = gene expression evidence: If the plasmid also contains a GFP gene under a strong promoter, transformed colonies glow under UV light — direct visual confirmation that the foreign gene is being transcribed and translated.
• Recombinant DNA technology: A gene of interest can be inserted into the plasmid alongside the amp resistance gene → bacteria become "protein factories" producing the gene of interest (e.g., insulin, growth hormone).
• Selectable marker strategy: The antibiotic resistance gene allows selection of only those cells that successfully took up the plasmid — without selection, you cannot distinguish transformed from untransformed cells.
• GFP expression = gene expression evidence: If the plasmid also contains a GFP gene under a strong promoter, transformed colonies glow under UV light — direct visual confirmation that the foreign gene is being transcribed and translated.
• Recombinant DNA technology: A gene of interest can be inserted into the plasmid alongside the amp resistance gene → bacteria become "protein factories" producing the gene of interest (e.g., insulin, growth hormone).
High-Frequency Exam Points
What would happen if you forgot the heat shock step? Bacterial cell membranes would remain relatively impermeable to the plasmid DNA → very few or no transformants on the +amp plate. Heat shock transiently increases membrane permeability, allowing DNA to enter.
Why do untransformed bacteria die on +amp plates? Ampicillin inhibits cell wall synthesis — bacteria without the ampicillin-resistance gene (which encodes β-lactamase, an enzyme that degrades ampicillin) cannot neutralize the antibiotic → cell wall synthesis fails → bacteria lyse (burst). Only bacteria expressing β-lactamase (from the transformed plasmid) survive. This is a direct demonstration of gene expression → phenotype.
INV 9
Biotechnology: Restriction Enzyme Analysis of DNA (Gel Electrophoresis)
Unit 6 — Gene Expression & Regulation · Big Idea 3: Information
🎯 Big Question
How can restriction enzymes and gel electrophoresis be used to analyze DNA? How can restriction fragment patterns be used for identification and comparison of DNA samples?
🧪 Method
Restriction enzymes cut DNA at specific palindromic recognition sequences (4–8 base pairs). Different DNA samples cut with the same enzyme(s) produce characteristic fragment patterns — a "DNA fingerprint." Fragments separated by size on agarose gel using electric current.
📊 Independent Variable
Type of restriction enzyme(s) used; which DNA sample is cut; combination of enzymes used (double digestion)
📈 Dependent Variable
Number and size (in base pairs) of restriction fragments produced — read from gel by comparing to DNA ladder bands
📋 Gel Electrophoresis Setup & Reading
Setup: DNA samples mixed with loading dye (adds density + color) are loaded into wells. Electric current applied: DNA migrates toward positive electrode (DNA is negatively charged from phosphate groups). Smaller fragments travel farther from the well in a given time.
Reading the gel: Each band = a population of fragments of the same size. Compare band positions to DNA ladder (known size standards) to estimate fragment sizes. Matching band patterns between samples = same restriction sites present = likely same or closely related DNA sequences.
Creating a restriction map: From single and double digestion patterns, determine the relative positions of restriction sites along the DNA molecule. Fragment sizes must add up to the total plasmid/DNA size.
Reading the gel: Each band = a population of fragments of the same size. Compare band positions to DNA ladder (known size standards) to estimate fragment sizes. Matching band patterns between samples = same restriction sites present = likely same or closely related DNA sequences.
Creating a restriction map: From single and double digestion patterns, determine the relative positions of restriction sites along the DNA molecule. Fragment sizes must add up to the total plasmid/DNA size.
🔑 Key Concepts Tested
• Restriction enzymes are molecular scissors: Each enzyme recognizes a specific palindromic sequence (reads the same 5'→3' on both strands). Cut frequency depends on recognition sequence length — 4-base cutters cut more frequently than 6-base cutters.
• Forensic DNA profiling: VNTR (variable number tandem repeat) regions differ among individuals → different fragment sizes after restriction digestion → unique band pattern for each person (except identical twins).
• Phylogenetic analysis: DNA from different species cut with same restriction enzyme → species with more similar band patterns are more closely related.
• Restriction mapping as gene expression tool: Restriction fragment length polymorphisms (RFLPs) can be used to track alleles through families (genetic linkage analysis) and diagnose genetic diseases.
• Forensic DNA profiling: VNTR (variable number tandem repeat) regions differ among individuals → different fragment sizes after restriction digestion → unique band pattern for each person (except identical twins).
• Phylogenetic analysis: DNA from different species cut with same restriction enzyme → species with more similar band patterns are more closely related.
• Restriction mapping as gene expression tool: Restriction fragment length polymorphisms (RFLPs) can be used to track alleles through families (genetic linkage analysis) and diagnose genetic diseases.
A linear DNA molecule of 3,600 bp is digested with restriction enzymes. HindIII alone produces fragments of 2,400 bp and 1,200 bp. EcoRI alone produces fragments of 2,800 bp and 800 bp. A double digest with both enzymes simultaneously produces fragments of 1,600 bp, 1,200 bp, and 800 bp. (a) How many cut sites does each enzyme have? (b) Construct a restriction map showing the positions of both sites on the linear molecule. (c) Verify your map is consistent with all three digests.
(a) Number of cut sites:
HindIII: 2 fragments from a linear molecule → 1 cut site
EcoRI: 2 fragments from a linear molecule → 1 cut site
Double digest: 3 fragments → 2 cuts total (one from each enzyme) ✓
(b) Restriction map construction:
Step 1 — Place HindIII. It creates fragments of 2,400 and 1,200. So HindIII is located 2,400 bp from one end (or 1,200 bp from the other).
Map so far: |←—— 2,400 ——→[H]←— 1,200 —→|
Step 2 — Place EcoRI using the double digest. The double digest produces 800, 1,600, 1,200. The 1,200-bp fragment matches the HindIII right fragment exactly — meaning EcoRI does NOT cut within the 1,200-bp region. EcoRI must cut within the 2,400-bp region.
Step 3 — If EcoRI cuts inside the 2,400-bp region, it divides it into two pieces. From the double digest: 800 and 1,600 (since 800 + 1,600 = 2,400 ✓). So EcoRI is 800 bp from the left end and 1,600 bp from HindIII.
Final restriction map:
(c) Verification:
EcoRI alone: cuts at position 800 → fragments: 800 and (1,600 + 1,200) = 800 + 2,800 = 3,600 ✓
HindIII alone: cuts at position 2,400 → fragments: (800 + 1,600) + 1,200 = 2,400 + 1,200 = 3,600 ✓
Double digest: 800 + 1,600 + 1,200 = 3,600 ✓ — all three sizes match the given data.
Key principle: The double digest "breaks apart" each single-enzyme fragment. The 1,200-bp HindIII fragment is unchanged in the double digest → EcoRI has no site within it. The 2,400-bp HindIII fragment splits into 800 + 1,600 in the double digest → EcoRI site is 800 bp from the left end.
HindIII: 2 fragments from a linear molecule → 1 cut site
EcoRI: 2 fragments from a linear molecule → 1 cut site
Double digest: 3 fragments → 2 cuts total (one from each enzyme) ✓
(b) Restriction map construction:
Step 1 — Place HindIII. It creates fragments of 2,400 and 1,200. So HindIII is located 2,400 bp from one end (or 1,200 bp from the other).
Map so far: |←—— 2,400 ——→[H]←— 1,200 —→|
Step 2 — Place EcoRI using the double digest. The double digest produces 800, 1,600, 1,200. The 1,200-bp fragment matches the HindIII right fragment exactly — meaning EcoRI does NOT cut within the 1,200-bp region. EcoRI must cut within the 2,400-bp region.
Step 3 — If EcoRI cuts inside the 2,400-bp region, it divides it into two pieces. From the double digest: 800 and 1,600 (since 800 + 1,600 = 2,400 ✓). So EcoRI is 800 bp from the left end and 1,600 bp from HindIII.
Final restriction map:
|←— 800 —→[EcoRI]←———— 1,600 ————→[HindIII]←— 1,200 —→|(c) Verification:
EcoRI alone: cuts at position 800 → fragments: 800 and (1,600 + 1,200) = 800 + 2,800 = 3,600 ✓
HindIII alone: cuts at position 2,400 → fragments: (800 + 1,600) + 1,200 = 2,400 + 1,200 = 3,600 ✓
Double digest: 800 + 1,600 + 1,200 = 3,600 ✓ — all three sizes match the given data.
Key principle: The double digest "breaks apart" each single-enzyme fragment. The 1,200-bp HindIII fragment is unchanged in the double digest → EcoRI has no site within it. The 2,400-bp HindIII fragment splits into 800 + 1,600 in the double digest → EcoRI site is 800 bp from the left end.
INV 10
Energy Dynamics
Unit 8 — Ecology · Big Idea 2: Energetics
🎯 Big Question
How efficiently is energy transferred between trophic levels? What is the ecological efficiency of energy transfer from producers to consumers?
🧪 Common Model System
Wisconsin Fast Plants (Brassica) as producers → cabbage white caterpillars (Pieris rapae) as primary consumers. Measure energy content and biomass at each level using a calorimeter or bomb calorimeter.
📊 Key Measurements
Mass of food consumed by caterpillars; mass gained by caterpillars; mass of frass (waste) produced; biomass or caloric content of plants before/after feeding
📋 Key Calculations
• Assimilation efficiency = (energy assimilated) / (energy ingested) × 100
• Net production efficiency = (energy stored in biomass) / (energy assimilated) × 100
• Ecological efficiency (trophic efficiency) = energy at trophic level n+1 / energy at trophic level n × 100 ≈ 10%
• Energy balance: Energy ingested = Energy assimilated + Energy in frass. Energy assimilated = Energy in biomass gain + Energy lost to respiration.
• Net production efficiency = (energy stored in biomass) / (energy assimilated) × 100
• Ecological efficiency (trophic efficiency) = energy at trophic level n+1 / energy at trophic level n × 100 ≈ 10%
• Energy balance: Energy ingested = Energy assimilated + Energy in frass. Energy assimilated = Energy in biomass gain + Energy lost to respiration.
🔑 Key Concepts Tested
• Why ~90% of energy is lost at each trophic transfer: Energy is lost to (1) cellular respiration (heat), (2) undigested material (frass/feces), (3) excretion, and (4) not all biomass is consumed. Only the energy actually incorporated into consumer biomass is available to the next trophic level.
• Implications for food chains: The 10% rule explains why food chains rarely have more than 4–5 trophic levels — too little energy remains at higher levels to support predator populations. It also explains why eating plants is more energy-efficient than eating meat (fewer trophic transfers).
• Connect to Unit 3: The energy lost as heat is the result of cellular respiration — second law of thermodynamics (entropy). Every metabolic process releases some energy as heat, which cannot be recaptured for biological work.
• Implications for food chains: The 10% rule explains why food chains rarely have more than 4–5 trophic levels — too little energy remains at higher levels to support predator populations. It also explains why eating plants is more energy-efficient than eating meat (fewer trophic transfers).
• Connect to Unit 3: The energy lost as heat is the result of cellular respiration — second law of thermodynamics (entropy). Every metabolic process releases some energy as heat, which cannot be recaptured for biological work.
INV 11
Transpiration
Units 1 & 2 — Chemistry of Life / Cells · Big Idea 2: Energetics
🎯 Big Question
What environmental conditions (humidity, wind, light, temperature) affect the rate of transpiration in plants? How do cohesion-tension and stomatal opening/closing regulate water movement?
🧪 Method — Potometer
A cut plant shoot is attached to a water-filled tube with a capillary tube. As water is lost by transpiration through leaves, it is replaced by water from the capillary — the movement of an air bubble in the capillary tube measures the rate of water uptake (proxy for transpiration rate). Some labs use a balance to directly measure mass loss from potted plants.
📊 Independent Variable
Environmental conditions: fan (wind) vs. still air; high humidity vs. low humidity; light vs. dark; high temperature vs. low temperature; vaseline-coated leaves (stomata blocked)
📈 Dependent Variable
Transpiration rate — mL of water lost per cm² leaf area per minute (or bubble distance per minute in a potometer)
🔑 Key Concepts Tested
• Transpiration pull (cohesion-tension mechanism): Water evaporates from mesophyll cell surfaces through stomata → creates tension (negative pressure / lower water potential) in the leaf → pulls water up through xylem by cohesion (water molecules cling together via H-bonds). This is entirely passive — no ATP required.
• Stomata as regulators: Stomata open when guard cells are turgid (high water potential, active K⁺ uptake → water follows osmotically → guard cells bow outward). Stomata close in drought (water stress → ABA hormone → K⁺ leaves guard cells → guard cells lose water → become flaccid → pores close).
• Environmental effects: Wind increases transpiration (removes humid air from leaf surface, steepening the water potential gradient). High temperature increases transpiration. Low humidity increases transpiration. Darkness decreases transpiration (stomata close). High humidity decreases transpiration rate.
• Vaseline experiment: Blocking abaxial (lower) stomata has a larger effect than blocking adaxial (upper) — most stomata are on the lower leaf surface.
• Stomata as regulators: Stomata open when guard cells are turgid (high water potential, active K⁺ uptake → water follows osmotically → guard cells bow outward). Stomata close in drought (water stress → ABA hormone → K⁺ leaves guard cells → guard cells lose water → become flaccid → pores close).
• Environmental effects: Wind increases transpiration (removes humid air from leaf surface, steepening the water potential gradient). High temperature increases transpiration. Low humidity increases transpiration. Darkness decreases transpiration (stomata close). High humidity decreases transpiration rate.
• Vaseline experiment: Blocking abaxial (lower) stomata has a larger effect than blocking adaxial (upper) — most stomata are on the lower leaf surface.
High-Frequency Exam Points
Why does wind increase transpiration? Moving air replaces the humid air layer surrounding the leaf with drier air, increasing the water vapor pressure gradient between the leaf interior and external air. Steeper gradient → faster diffusion of water vapor out of stomata → faster transpiration. This connects to Fick's Law of diffusion: rate ∝ concentration gradient × surface area / distance.
Connect water properties to transpiration: Cohesion (H-bonding between water molecules) allows the continuous water column in xylem to be pulled upward without breaking. Adhesion (water to xylem walls) helps maintain the column. These properties from Unit 1 are directly tested in this lab context.
INV 12
Fruit Fly Behavior
Unit 8 — Ecology · Big Idea 2/3: Energetics / Information
🎯 Big Question
Do Drosophila demonstrate innate behavioral preferences for certain stimuli? Can we detect statistically significant behavioral differences between fly strains or conditions?
🧪 Method — Choice Chamber
Flies placed in a T-maze or choice chamber where they can choose between two environments (e.g., light vs. dark; moist vs. dry; different odorants; ethanol gradient). Count flies in each compartment at regular intervals. Use chi-square test to determine if distribution differs significantly from 50:50 (random).
📊 Independent Variable
The environmental stimulus being tested (light/dark, humidity gradient, odorant type, ethanol concentration, fly genotype)
📈 Dependent Variable
Number (or proportion) of flies in each chamber at defined time intervals
📋 Statistical Analysis — Chi-Square
Null hypothesis (H₀): Flies show no preference — equal numbers are expected in each compartment (50:50 ratio).
Expected number per side = total flies / 2 (if two compartments).
χ² = Σ [(Observed − Expected)² / Expected]
Compare calculated χ² to critical value (df = 1, p = 0.05: critical value = 3.84). If χ² > 3.84 → reject H₀ → the distribution is not random → flies show a statistically significant preference.
Example: 30 flies placed in light/dark chamber. After 5 minutes: 22 in dark, 8 in light. Expected: 15 each. χ² = (22−15)²/15 + (8−15)²/15 = 49/15 + 49/15 = 6.53. Since 6.53 > 3.84, reject H₀ — flies significantly prefer darkness.
Expected number per side = total flies / 2 (if two compartments).
χ² = Σ [(Observed − Expected)² / Expected]
Compare calculated χ² to critical value (df = 1, p = 0.05: critical value = 3.84). If χ² > 3.84 → reject H₀ → the distribution is not random → flies show a statistically significant preference.
Example: 30 flies placed in light/dark chamber. After 5 minutes: 22 in dark, 8 in light. Expected: 15 each. χ² = (22−15)²/15 + (8−15)²/15 = 49/15 + 49/15 = 6.53. Since 6.53 > 3.84, reject H₀ — flies significantly prefer darkness.
🔑 Key Concepts Tested
• Innate (instinctive) vs. learned behaviors: The preference demonstrated is innate — hardwired by genetics, not learned through experience.
• Taxis type: If flies actively move toward/away from light → phototaxis (positive = toward, negative = away from light). If movement speed changes without orientation → kinesis.
• Fitness connection: Behavioral preferences have evolved because they enhance fitness. Fruit flies preferring dim environments may avoid predators or desiccation.
• Taxis type: If flies actively move toward/away from light → phototaxis (positive = toward, negative = away from light). If movement speed changes without orientation → kinesis.
• Fitness connection: Behavioral preferences have evolved because they enhance fitness. Fruit flies preferring dim environments may avoid predators or desiccation.
INV 13
Enzyme Activity (Catecholase / Peroxidase)
Unit 3 — Cellular Energetics · Big Idea 2: Energetics
🎯 Big Question
How do environmental factors (temperature, pH, substrate concentration, inhibitors) affect enzyme activity? How can spectrophotometry quantify enzyme reaction rates?
🧪 Enzyme System
Catecholase (from banana/potato extract) converts catechol → benzoquinone (brown). Measured by absorbance at 420 nm with spectrophotometer. Or: Peroxidase from turnip/horseradish — converts H₂O₂ + guaiacol → colored product. Or: Catalase — converts H₂O₂ → H₂O + O₂ (measure O₂ production or disc displacement).
📊 Independent Variable
Temperature; pH (different buffers); substrate concentration; enzyme concentration; presence/type of inhibitor (competitive vs. noncompetitive)
📈 Dependent Variable
Enzyme activity = rate of reaction = rate of color change measured as ΔAbsorbance/minute (spectrophotometer) OR rate of O₂ production
⚠️ Controls
Blank = reaction mixture without enzyme (shows background absorbance change); denatured enzyme control (boiled enzyme) = shows rate without any enzymatic activity
📋 Key Measurements & Calculations
1. Record absorbance (A₄₂₀) at regular time intervals (e.g., every 30 seconds for 3 minutes).
2. Plot absorbance vs. time → slope of the initial linear portion = initial reaction rate.
3. Compare initial rates across different conditions (temperature, pH, concentration).
4. To determine Vmax and Km: vary substrate concentration while keeping enzyme constant → plot rate vs. [substrate] → find plateau (Vmax) and substrate concentration at ½Vmax (= Km).
2. Plot absorbance vs. time → slope of the initial linear portion = initial reaction rate.
3. Compare initial rates across different conditions (temperature, pH, concentration).
4. To determine Vmax and Km: vary substrate concentration while keeping enzyme constant → plot rate vs. [substrate] → find plateau (Vmax) and substrate concentration at ½Vmax (= Km).
🔑 Key Concepts Tested
• Optimal temperature and pH: Each enzyme has a specific temperature and pH at which it is most active. Above optimal temperature → denaturation (loss of 3D structure) → irreversible loss of activity. Wrong pH → changes ionization of active site R-groups → distorted active site → reduced activity.
• Saturation kinetics (Km and Vmax): As [substrate] increases, rate increases until enzyme becomes saturated → plateau at Vmax. Km = substrate concentration at ½Vmax — measure of enzyme affinity (lower Km = tighter binding).
• Inhibitor type identification: Competitive inhibitor → Vmax unchanged, apparent Km increases (more substrate needed to overcome inhibitor). Noncompetitive inhibitor → Vmax decreases, Km unchanged (adding substrate cannot overcome inhibitor effect).
• Initial rate measurement: Only use the initial linear portion of the curve (before substrate is depleted and products accumulate). Later time points underestimate true enzyme activity.
• Saturation kinetics (Km and Vmax): As [substrate] increases, rate increases until enzyme becomes saturated → plateau at Vmax. Km = substrate concentration at ½Vmax — measure of enzyme affinity (lower Km = tighter binding).
• Inhibitor type identification: Competitive inhibitor → Vmax unchanged, apparent Km increases (more substrate needed to overcome inhibitor). Noncompetitive inhibitor → Vmax decreases, Km unchanged (adding substrate cannot overcome inhibitor effect).
• Initial rate measurement: Only use the initial linear portion of the curve (before substrate is depleted and products accumulate). Later time points underestimate true enzyme activity.
A student tests catecholase activity at five temperatures (5°C, 15°C, 25°C, 35°C, 45°C) and measures initial reaction rates. The rates are: 0.008, 0.022, 0.045, 0.062, and 0.004 absorbance units/minute, respectively. (a) At what temperature is catecholase most active? (b) What most likely explains the dramatic drop in activity at 45°C? (c) Design a follow-up experiment to determine whether the reduced activity at 45°C is reversible.
(a) Optimal temperature: Catecholase is most active at 35°C, which shows the highest initial rate (0.062 absorbance units/min). As temperature increased from 5°C to 35°C, activity increased because higher temperatures increased molecular collision frequency between enzyme and substrate, increasing reaction rate.
(b) Activity drop at 45°C: The dramatic drop from 0.062 (at 35°C) to 0.004 (at 45°C) is most likely due to thermal denaturation of the enzyme. Temperatures above the optimum disrupt the hydrogen bonds, ionic bonds, and hydrophobic interactions that maintain the enzyme's tertiary structure. The active site loses its precise shape, and substrate can no longer bind or be catalyzed effectively. At 45°C, nearly all catecholase molecules are denatured and non-functional.
(c) Experiment to test reversibility:
1. Prepare two sets of catecholase samples: Set A (heated to 45°C for 5 minutes, then returned to 35°C) and Set B (kept at 35°C throughout as a positive control).
2. After cooling Set A back to 35°C, measure the enzyme activity of both sets under identical conditions (same substrate concentration, same pH, same time intervals) using a spectrophotometer.
3. If Set A shows recovered activity comparable to Set B → denaturation was reversible (the enzyme refolded). If Set A remains at very low activity (similar to the 45°C result) → denaturation was irreversible (the enzyme cannot refold properly after heating).
4. A negative control (denatured enzyme not cooled) confirms that any activity recovery is due to cooling, not spontaneous activity of the denatured form.
(b) Activity drop at 45°C: The dramatic drop from 0.062 (at 35°C) to 0.004 (at 45°C) is most likely due to thermal denaturation of the enzyme. Temperatures above the optimum disrupt the hydrogen bonds, ionic bonds, and hydrophobic interactions that maintain the enzyme's tertiary structure. The active site loses its precise shape, and substrate can no longer bind or be catalyzed effectively. At 45°C, nearly all catecholase molecules are denatured and non-functional.
(c) Experiment to test reversibility:
1. Prepare two sets of catecholase samples: Set A (heated to 45°C for 5 minutes, then returned to 35°C) and Set B (kept at 35°C throughout as a positive control).
2. After cooling Set A back to 35°C, measure the enzyme activity of both sets under identical conditions (same substrate concentration, same pH, same time intervals) using a spectrophotometer.
3. If Set A shows recovered activity comparable to Set B → denaturation was reversible (the enzyme refolded). If Set A remains at very low activity (similar to the 45°C result) → denaturation was irreversible (the enzyme cannot refold properly after heating).
4. A negative control (denatured enzyme not cooled) confirms that any activity recovery is due to cooling, not spontaneous activity of the denatured form.
Science Practices
The 6 Science Practices — Tested in Every Lab
Practice 1
Concept Explanation: Describe and explain biological concepts in written format. In labs: explain why your results make biological sense (connect data to Unit concepts).
Practice 2
Visual Representations: Analyze, construct, and interpret graphs, diagrams, and models. In labs: draw correctly labeled graphs with appropriate type (bar vs. line vs. scatter), axes, units, scaling, error bars.
Practice 3
Scientific Questions & Methods: Identify variables, controls, and justify experimental design. In labs: identify independent/dependent variables, explain why the control is appropriate, propose follow-up experiments.
Practice 4
Representing & Describing Data: Accurately plot data; describe trends and patterns; identify relationships between variables. In labs: correctly graph data, label axes with units, add error bars, describe what the graph shows.
Practice 5
Statistical Tests & Calculations: Perform chi-square, calculate means/rates/percent change, use error bars to compare means. In labs: chi-square for genetics/behavior; % mass change for osmosis; error bars for comparing experimental groups.
Practice 6
Scientific Argumentation: Make claims, support with evidence, provide reasoning connecting evidence to biological theory. In labs: state a conclusion as a claim → cite your data as evidence → explain WHY the data supports the claim using biological theory.
FRQ Lab Question Format — How to Earn Full Points
Claim: State your conclusion directly. ("The data support the hypothesis that...").
Evidence: Cite specific data. ("At 35°C, the reaction rate was 0.062 AU/min, compared to 0.004 AU/min at 45°C — a 15× decrease")
Reasoning: Explain the mechanism. ("This decrease is due to thermal denaturation — high temperature disrupts the hydrogen bonds maintaining the enzyme's 3D structure, distorting the active site so substrate can no longer bind...")
For experimental design questions: Always identify (1) independent variable, (2) dependent variable, (3) controlled variables (held constant), (4) control group, (5) how you would measure the dependent variable, and (6) expected results if the hypothesis is supported.
Common Errors
High-Frequency Lab Errors to Avoid
- 📊Using absolute mass change instead of percent mass change (Osmosis lab)Samples start with different initial masses — absolute change is meaningless for comparison. Always calculate % mass change = [(final − initial)/initial] × 100. This standardizes across samples regardless of initial size.
- 🎯Forgetting to subtract the bead (non-living) control in respirometryThe non-living control accounts for physical changes in gas volume due to temperature or pressure fluctuations in the water bath. The corrected rate = experimental rate minus control rate gives the TRUE biological rate of O₂ consumption.
- 🔬Measuring enzyme activity at only one time point instead of initial rateEnzyme activity is greatest at the beginning of the reaction, before substrate is depleted and product inhibition builds up. The initial rate (slope of the first few time points) is the most accurate measure. The curve flattens over time — measuring total change at a fixed endpoint underestimates Vmax and is not a rate.
- 🧫Saying "no colonies on +amp plate" proves no bacteria are presentNo colonies on the +amp plate simply means NO TRANSFORMED bacteria were present — but untransformed bacteria could still be there (they just die on the amp medium). The LB plate (no amp) should show a bacterial lawn confirming bacteria were viable. Always include both plates in your conclusion.
- ➡Drawing food web arrows in the wrong directionArrows point in the direction of ENERGY FLOW — from prey to predator, from eaten organism to organism that eats it. "Grass → Grasshopper" means energy flows FROM grass TO grasshopper. The arrow does NOT show who eats whom in the conventional sense of pointing at the prey.
- 🧮Chi-square: writing "accept H₀" instead of "fail to reject H₀"In statistics, you NEVER "accept" or "prove" a null hypothesis — you either reject it (evidence against it is strong enough) or fail to reject it (insufficient evidence to reject it). The correct phrasing when χ² < critical value is "fail to reject H₀ — the data are consistent with the expected ratio."
- 🌿Photosynthesis: forgetting to include a dark control in the floating leaf disk labThe dark control (disks submerged in sodium bicarbonate but in complete darkness) demonstrates the base rate — in darkness, O₂ is consumed by cellular respiration but not produced by photosynthesis, so disks should remain sunken or re-sink if they were floating. Without this control, you cannot attribute disk floating specifically to photosynthetic O₂ production.
- 🦠HW equilibrium: starting calculation from dominant phenotype frequency instead of recessiveThe critical starting point for HWE is q² = frequency of homozygous recessive individuals (the only genotype identifiable directly from phenotype). Never start from dominant phenotype frequency — that combines both AA and Aa genotypes and you cannot separate them without further calculation.