IB Biology Quiz: Experimental Techniques
19 questions · exam conditions
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Experimental TechniquesQuestion 1 of 19

In a paper chromatography experiment, the solvent front travels 10.0 cm from the origin. The final positions of two pigments, P1 and P2, are 8.0 cm and 5.0 cm from the origin, respectively. What is the Rf value of pigment P1?

0.50
0.80
1.25
1.60
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IB Biology Quiz

IB Biology Quiz: Experimental Techniques

Practice Experimental Techniques in IB Biology with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Experimental Techniques, giving you a quick way to practice the rules, question types, and explanations that matter most for IB Biology.

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Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

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Question 1

In a paper chromatography experiment, the solvent front travels 10.0 cm from the origin. The final positions of two pigments, P1 and P2, are 8.0 cm and 5.0 cm from the origin, respectively. What is the Rf value of pigment P1?

  1. 0.50
  2. 0.80 (correct answer)
  3. 1.25
  4. 1.60
Explanation: The retention factor (Rf) is calculated as the ratio of the distance travelled by the solute (the pigment) to the distance travelled by the solvent front, both measured from the origin. For pigment P1, the distance is 8.0 cm. The solvent front distance is 10.0 cm. Therefore, Rf = 8.0 cm / 10.0 cm = 0.80. Distractor A is the Rf value for pigment P2. Distractor C is the inverse ratio (solvent/solute) for P1, which is an incorrect calculation as Rf values cannot exceed 1.0. Distractor D is the ratio of the distance of P1 to P2.

Question 2

An ecologist is comparing plant species distribution in two areas. Area 1 is a large, uniform grassland. Area 2 is a hillside with a noticeable moisture gradient from a stream at the bottom to a dry ridge at the top. Which sampling methods are most appropriate for this study?

  1. Systematic sampling using a transect in Area 1 and random sampling using quadrats in Area 2.
  2. Random sampling using quadrats in Area 1 and systematic sampling using a transect in Area 2. (correct answer)
  3. Random sampling using quadrats in both areas to eliminate any potential for researcher bias.
  4. Systematic sampling using transects in both areas to ensure that all parts of the habitats are sampled.
Explanation: For Area 1, a uniform habitat, random sampling with quadrats is appropriate to get a representative sample and avoid bias. For Area 2, which has a clear environmental gradient (abiotic factor change), systematic sampling along a transect from the stream to the ridge is the best method to investigate how species distribution correlates with this gradient. Distractor A reverses the correct methods. Distractor C is incorrect because random sampling in Area 2 could miss or poorly represent the environmental gradient. Distractor D is incorrect because systematic sampling is not necessary for the uniform habitat in Area 1 and may introduce bias.

Question 3

A micrograph of an organelle is produced at a total magnification of 8000x. On the micrograph, the organelle measures 40 mm in length. A scale bar on the same micrograph measures 5 mm and represents an actual length of 1 µm. What is the actual length of the organelle?

  1. 5.0 µm
  2. 8.0 µm (correct answer)
  3. 32.0 µm
  4. 50.0 µm
Explanation: The most reliable method is to use the scale bar, as the stated magnification can have printing errors. The ratio of actual length to image length is the same for the scale bar and the organelle. Let A be the actual length of the organelle. (Actual length of organelle) / (Image length of organelle) = (Actual length of scale bar) / (Image length of scale bar). So, A / 40 mm = 1 µm / 5 mm. Solving for A: A = (1 µm / 5 mm) * 40 mm = 8 µm. Alternatively, using magnification from the scale bar: Magnification = Image / Actual = 5 mm / 1 µm = 5000 µm / 1 µm = 5000x. Then, Actual organelle length = Image length / Magnification = 40 mm / 5000 = 40000 µm / 5000 = 8 µm. The 8000x value in the stem is a distractor.

Question 4

An experiment is designed to test the effect of light intensity on the rate of photosynthesis in an aquatic plant, measured by the volume of oxygen produced. Which of the following is an essential controlled variable for this experiment?

  1. The distance of the light source from the plant.
  2. The number of bubbles produced by the plant.
  3. The volume of oxygen produced per unit time.
  4. The concentration of carbon dioxide in the water. (correct answer)
Explanation: In this experiment, light intensity is the independent variable. The rate of photosynthesis (measured by oxygen production) is the dependent variable. To ensure a fair test, all other factors that could affect the rate of photosynthesis must be kept constant (controlled). Carbon dioxide is a substrate for photosynthesis, so its concentration directly affects the rate. Therefore, the concentration of CO2 in the water must be controlled. Distractor A is the independent variable itself, which is being changed. Distractors C and D are different ways of measuring the dependent variable.

Question 5

A student uses a colorimeter to investigate the effect of temperature on an enzyme that breaks down a coloured substrate. The student uses a 'blank' containing the buffer and substrate before measuring each sample. What is the purpose of using this blank?

  1. To set the maximum absorbance value to 100% so that changes can be measured easily.
  2. To cool the colorimeter to the same temperature as the sample being measured.
  3. To account for the absorbance of the coloured substrate at the start of the reaction.
  4. To calibrate the colorimeter to zero absorbance for the solution without enzyme activity. (correct answer)
Explanation: A 'blank' in colorimetry contains everything that the experimental cuvette contains except the substance whose absorbance is being measured (in this case, the product of the reaction, or the change in substrate). The purpose is to set the absorbance reading to zero. This ensures that any measured absorbance is due only to the substance of interest, by subtracting the background absorbance of the cuvette, solvent, buffer, and unreacted substrate. The student in this case should be measuring the disappearance of the substrate, so zeroing on a tube without enzyme is a way to set the initial absorbance reading. Thus, the purpose is calibration to zero. Option C is subtly different; the blank does account for substrate absorbance, but its function is to define this as the zero point for measuring change.

Question 6

A microbiologist performs a two-step dilution of a concentrated bacterial sample. First, 0.1 mL of the sample is added to 9.9 mL of sterile water (Tube A). Then, 1.0 mL of the mixture from Tube A is added to 9.0 mL of sterile water (Tube B). What is the total dilution factor in Tube B relative to the original sample?

  1. 1:100
  2. 1:200
  3. 1:1000 (correct answer)
  4. 1:10000
Explanation: The dilution factor at each step is calculated as (volume of sample) / (total volume). Step 1 (Tube A): The volume of the sample is 0.1 mL, and the total volume is 0.1 mL + 9.9 mL = 10.0 mL. The dilution factor is 0.1 / 10.0 = 1/100. Step 2 (Tube B): The volume of the sample from Tube A is 1.0 mL, and the total volume is 1.0 mL + 9.0 mL = 10.0 mL. The dilution factor is 1.0 / 10.0 = 1/10. The total dilution is the product of the individual dilutions: (1/100) * (1/10) = 1/1000. This is a 1:1000 dilution.

Question 7

The concentration of a product was measured over time in an enzyme-catalysed reaction. At t=0s, concentration was 0.0 M. At t=20s, concentration was 0.4 M. At t=40s, concentration was 0.6 M. At t=60s, concentration was 0.7 M. How is the initial rate of reaction best calculated from this data?

  1. By calculating the gradient between t=0s and t=20s, which is 0.020 M s⁻¹. (correct answer)
  2. By calculating the gradient between t=40s and t=60s, which is 0.005 M s⁻¹.
  3. By taking the final concentration (0.7 M) and dividing by the total time (60s), which is 0.012 M s⁻¹.
  4. By averaging the rates from all three intervals, which is approximately 0.013 M s⁻¹.
Explanation: The initial rate of reaction is the rate at the very beginning (close to t=0), when the substrate concentration is highest and the reaction is fastest. From discrete data points, the best estimate for the initial rate is the gradient over the first time interval. Rate = Δ\DeltaConcentration / Δ\DeltaTime = (0.4 M - 0.0 M) / (20s - 0s) = 0.4 / 20 = 0.020 M s⁻¹. Distractor A calculates the rate at a later stage when it has slowed. Distractor C calculates the average rate over the whole period, which underestimates the initial rate. Distractor D is also a form of averaging and is not the correct method for finding the initial rate.

Question 8

A student sets up a sealed mesocosm containing soil, small plants, and woodlice. It is kept in a location with a daily light cycle. What is a key limitation of using this mesocosm to model a larger forest ecosystem?

  1. The glass container prevents the exchange of energy, so the mesocosm will quickly cool down.
  2. The amount of decomposition will be too high due to the small volume and lack of air flow.
  3. The system lacks large-scale nutrient cycling and the influence of migratory species. (correct answer)
  4. Photosynthesis cannot occur because the CO₂ produced by the woodlice is insufficient for the plants.
Explanation: Mesocosms are closed or semi-closed systems used to model larger ecosystems. A key limitation is their small scale and isolation. They lack the connectivity of a real ecosystem, including the immigration and emigration of organisms and the large-scale biogeochemical cycles (e.g., water cycle, large-scale nutrient deposition) that influence natural environments. Distractor A is incorrect; energy (light) can enter. Distractor B is speculative and not a fundamental limitation. Distractor D is incorrect; in a balanced mesocosm, gas exchange between organisms can sustain the system for some time.

Question 9

A researcher compares the mean biomass of algae in two different ponds and calculates a p-value of 0.04 using a t-test. What is the correct conclusion based on a significance level of α=0.05\alpha = 0.05?

  1. The difference between the means is not statistically significant, and the null hypothesis is accepted.
  2. The difference between the means is statistically significant, and the null hypothesis is rejected. (correct answer)
  3. There is a 4% probability that the null hypothesis is true.
  4. There is a 96% probability that the alternative hypothesis is true.
Explanation: The p-value represents the probability of observing the collected data (or more extreme data) if the null hypothesis were true. The significance level (α\alpha) is the threshold for rejecting the null hypothesis. If p ≤\le α\alpha, the result is considered statistically significant, and the null hypothesis is rejected. Here, p=0.04 is less than α=0.05\alpha=0.05, so the difference is significant. Distractors C and D represent a common misinterpretation of the p-value; it is not the probability of the hypothesis being true or false, but rather the probability of the data given the hypothesis.

Question 10

To estimate a population of woodlice, 30 woodlice were collected, marked with a small dot of paint, and released. Two days later, a second sample of 40 woodlice was collected, and 6 of these were found to be marked. What is the estimated population size based on the Lincoln index?

  1. 76
  2. 240
  3. 180
  4. 200 (correct answer)
Explanation: The Lincoln index formula for population estimation is: Population size (N) = (number captured and marked in first sample, n1) × (total number captured in second sample, n2) / (number of marked recaptures in second sample, m2). In this case, N = (30 × 40) / 6 = 1200 / 6 = 200. Therefore, the estimated population size is 200.

Question 11

A researcher performs a chi-squared test on the results of a genetic cross that was expected to yield four phenotypes in a 9:3:3:1 ratio. The calculated χ2\chi^2 value is 7.95. The critical value from a probability table for p=0.05 is 7.81 for 3 degrees of freedom and 9.49 for 4 degrees of freedom. What is the correct interpretation of this result?

  1. The null hypothesis is accepted because the calculated value (7.95) is less than the critical value for 4 degrees of freedom (9.49).
  2. The null hypothesis is rejected because the calculated value (7.95) is greater than the critical value for 3 degrees of freedom (7.81). (correct answer)
  3. The null hypothesis is accepted because the probability that the deviation is due to chance is greater than 5%.
  4. The null hypothesis is rejected because the calculated value (7.95) indicates a very poor fit with the expected ratio.
Explanation: First, the degrees of freedom (df) must be calculated. For a genetic cross with 'n' phenotypic classes, df = n - 1. Here, there are 4 phenotypes, so df = 4 - 1 = 3. Second, the calculated χ2\chi^2 value (7.95) must be compared to the critical value for 3 df at p=0.05, which is 7.81. Since 7.95 > 7.81, the calculated value falls in the rejection region. This means the deviation between observed and expected results is statistically significant, and the null hypothesis (that the results fit the 9:3:3:1 ratio) should be rejected. Distractor A uses the wrong degrees of freedom. Distractor C incorrectly interprets the p-value condition for acceptance. Distractor D is a vague statement; the rejection is based on the specific comparison to the critical value.

Question 12

A student measures the length of 25 cells and calculates a mean length of 45 µm. The student then finds a published value for the mean length of these cells is 48 µm. Which statement best describes the accuracy and precision of the student's measurements?

  1. The measurements are accurate because the mean is close to the published value, but precision cannot be determined from the data provided. (correct answer)
  2. The measurements are precise because 25 cells were measured, but they are not accurate because the mean differs from the published value.
  3. The accuracy of the measurements cannot be determined without knowing the standard deviation, but they are precise because a mean was calculated.
  4. Neither accuracy nor precision can be determined, as accuracy requires comparison to a true value and precision requires knowing the spread of the data.
Explanation: Accuracy refers to how close a measurement is to the true or accepted value. The student's mean (45 µm) is reasonably close to the published value (48 µm), indicating good accuracy. Precision refers to how close multiple measurements are to each other (i.e., the spread or variability). The mean alone provides no information about the spread of the 25 individual measurements (e.g., the standard deviation or range). Therefore, the precision cannot be determined from the information given. Distractor B confuses a large sample size with precision. Distractor C incorrectly states that accuracy cannot be determined. Distractor D is too strong; a reasonable judgement about accuracy can be made by comparing the mean to the published value.

Question 13

A student uses a potometer to estimate the rate of transpiration in a plant cutting by measuring the movement of an air bubble in a capillary tube. Which statement best describes a fundamental limitation of this technique that affects the accuracy of the estimate?

  1. The rate of water uptake may not precisely equal the transpiration rate because a small amount of water is used for metabolic processes like photosynthesis. (correct answer)
  2. Air leaking into the apparatus at the joints will cause the measured rate of water uptake to be consistently lower than the actual rate.
  3. Changes in ambient temperature can cause the water in the capillary tube to expand or contract, but this random error does not affect the overall accuracy.
  4. Cutting the plant shoot underwater is difficult and often damages xylem vessels, preventing any reliable measurement of water uptake.
Explanation: The potometer directly measures water uptake, which is used as a proxy for transpiration. However, not all water taken up is transpired; a small fraction is used by the plant for photosynthesis and to maintain cell turgor. This creates a small, systematic error where the measured water uptake is slightly higher than the actual transpiration rate. Distractor B is incorrect because an air leak would allow water to escape the capillary tube, pushing the bubble along and causing an overestimation of the rate. Distractor C describes a valid source of error, but it is incorrect to state that it does not affect accuracy; it introduces random fluctuations. Distractor D describes a procedural failure, not an inherent limitation of a correctly performed technique.

Question 14

A student measures the rate of an enzyme-catalysed reaction by recording data every 15 seconds. The student consistently reads the volume on the measuring cylinder 0.5 mL higher than the true value at every measurement point due to parallax error. How would this error be classified and what is its effect on the calculated reaction rate?

  1. Random error; it decreases the precision of the measurements but does not affect the calculated rate.
  2. Systematic error; it increases the value of each volume measurement but has no effect on the calculated rate. (correct answer)
  3. Random error; it causes the calculated reaction rate to be unpredictably higher or lower than the true rate.
  4. Systematic error; it causes the calculated reaction rate to be consistently higher than the true rate.
Explanation: The error is systematic because it is consistent in magnitude and direction (always +0.5 mL). The reaction rate is calculated from the change in volume over a time interval (ΔV/Δt\Delta V / \Delta t). If each reading is 0.5 mL too high, the difference between any two readings (V2 - V1) will be ((V2_true + 0.5) - (V1_true + 0.5)) = V2_true - V1_true. The constant error cancels out when calculating the change in volume. Therefore, this specific systematic error has no effect on the calculated rate (the gradient of the volume vs. time graph). Distractor A and C incorrectly classify the error as random. Distractor D correctly classifies the error but incorrectly states its effect on the rate.

Question 15

A student investigates osmosis using dialysis tubing bags filled with different sucrose solutions and placed in beakers of distilled water. They measure the mass of each bag every 5 minutes for 30 minutes. What is the most significant flaw in this experimental design if the student concludes which solution has the highest water potential based only on the final mass?

  1. The temperature of the distilled water was not recorded during the experiment.
  2. The surface area to volume ratio of the dialysis bags may not have been identical.
  3. The initial masses of the bags were not measured, preventing calculation of percentage change. (correct answer)
  4. The experiment was not left long enough to allow the bags to reach osmotic equilibrium.
Explanation: To compare the effects of different solutions, one must compare the relative change in mass, not the absolute final mass. The initial masses of the bags could have been different due to slight variations in filling. Without knowing the initial mass, the student cannot calculate the percentage change in mass ((final-initial)/initial * 100), which is the standardized measure needed for a valid comparison. Distractors A, B, and D are all potential sources of error or limitations, but the failure to measure the initial mass is the most fundamental flaw that invalidates the proposed conclusion.

Question 16

When using a light microscope to view a specimen, a student first focuses the image using the 4x objective lens and then switches to the 10x objective lens. Which is the correct procedure to follow immediately after switching to the higher power objective?

  1. Adjust the diaphragm to increase the amount of light passing through the specimen. (correct answer)
  2. Move the stage down using the coarse focus knob before adjusting the fine focus.
  3. Use only the coarse focus knob to bring the image into sharp focus quickly.
  4. Apply immersion oil to the slide to improve the resolution of the image.
Explanation: When switching to a higher power objective, the field of view becomes smaller and darker. It is therefore necessary to increase the illumination by opening the diaphragm or adjusting the light source intensity to see the specimen clearly. Since most modern microscopes are parfocal, only minor adjustments with the fine focus knob should be needed. Using the coarse focus knob (A, C) risks crashing the objective into the slide. Immersion oil (D) is only used with a specific high-power oil immersion lens (typically 100x), not the 10x or 40x lenses.

Question 17

An experiment investigates if carbon dioxide is a limiting factor for photosynthesis in an aquatic plant, which is placed in a beaker of water and illuminated. The rate is measured by counting bubbles of oxygen produced. Which of the following setups would serve as the best control?

  1. An identical setup placed in a dark cupboard for the same amount of time.
  2. An identical setup but with a plastic plant of the same size and shape.
  3. An identical setup with additional sodium hydrogencarbonate dissolved in the water. (correct answer)
  4. An identical setup with the water first boiled to remove dissolved gases and then cooled.
Explanation: The question asks if CO₂ is a limiting factor. This means the experiment needs to show that adding more CO₂ increases the rate. The initial setup is the baseline. The experimental group would be the one where the potential limiting factor is increased. Therefore, adding sodium hydrogencarbonate (which provides a source of CO₂) creates the condition to test the hypothesis. This setup is then compared to the baseline (the control) to see if the rate increases. Distractor A is a control for light dependence. Distractor B is a control for physical effects. Distractor D would be used to test if CO₂ is necessary, not if it is limiting.

Question 18

In an experiment measuring the rate of oxygen consumption by respiring peas, a control respirometer is set up containing glass beads with the same mass and volume as the peas. Both setups contain potassium hydroxide (KOH). What is the primary function of this control?

  1. To measure the rate of oxygen consumption by the potassium hydroxide.
  2. To demonstrate that non-living glass beads do not undergo cellular respiration.
  3. To provide a baseline for calibrating the equipment before measuring the respiration of the peas.
  4. To compensate for any changes in gas volume caused by fluctuations in ambient temperature or atmospheric pressure. (correct answer)
Explanation: The ideal gas law (PV=nRT) shows that the volume of a gas is affected by both temperature (T) and pressure (P). Any change in the ambient temperature or atmospheric pressure during the experiment will cause the gas volume inside both respirometers to change. The control setup with inert glass beads allows the experimenter to measure these non-biological volume changes. This value can then be subtracted from the change observed in the experimental tube to isolate the volume change due solely to oxygen consumption by the peas. While it's true that glass beads don't respire (B), this is not the primary reason for the control's design. A and C are incorrect functions.

Question 19

A DNA sample containing fragments of different sizes is analysed using gel electrophoresis. Which statement correctly explains the separation mechanism?

  1. Smaller DNA fragments have a greater negative charge than larger fragments, so they are attracted more strongly to the positive electrode and move further.
  2. Larger DNA fragments experience more resistance when moving through the agarose gel matrix, so they travel a shorter distance than smaller fragments. (correct answer)
  3. All DNA fragments have the same charge-to-mass ratio, so they move at the same speed, but are separated later by a chemical stain.
  4. The electric field causes larger fragments to denature into single strands which move more easily through the gel, while smaller fragments remain double-stranded.
Explanation: During gel electrophoresis, DNA fragments (which are negatively charged due to their phosphate backbone) migrate towards the positive electrode. The agarose gel acts as a molecular sieve. Smaller fragments can navigate the pores of the gel more easily and thus travel further in a given amount of time. Larger fragments are impeded more by the matrix and travel shorter distances. Distractor A is incorrect; while larger fragments do have more total charge, the charge-to-mass ratio is essentially constant, and size-based sieving is the primary separation principle. Distractor C correctly states the charge-to-mass ratio is constant but incorrectly concludes this leads to no separation. Distractor D is incorrect; denaturation is not the separation mechanism.