Biology SL

1. 1.

   Prepare a research question and method outline for how sucrose concentration affects potato-cylinder mass change over 60 minutes at 20 C. Give an operational research question, a measurable response, one important control and one safety or environmental precaution. This is planning only.

   [4 marks] · no calculator

   Worked explanation

   1. Make the independent variable a specified concentration range rather than an undefined stronger solution.
   2. Operationalize the dependent variable using paired masses and a reproducible blotting procedure.
   3. Control geometry and conditions, plan biological replication and follow local supervision and waste procedures; no experiment has been performed here.

   Common mistake: A broad aim such as investigate osmosis is not an operational question unless treatments, endpoint and conditions are defined.

   Marking points

   • A question relates a defined range of sucrose concentrations to mass change in potato cylinders after 60 minutes at 20 C.
   • Measure initial/final mass after a standardized blotting method and calculate percentage change.
   • Control cylinder dimensions/source, immersion time or temperature and use independent cylinders per treatment.
   • Manage cutting-tool risk with supervised safe cutting and minimize/dispose of biological and solution waste appropriately.

   Examiner tip: A broad aim such as investigate osmosis is not an operational question unless treatments, endpoint and conditions are defined.

2. 2.

   Hypothetical potato cylinders in one treatment change from 2.00 to 2.20 g, 2.50 to 2.70 g and 3.00 to 3.30 g. Calculate each percentage mass change and their mean. Explain why taking three balance readings of just one cylinder would not provide the same biological replication.

   [4 marks]

   Worked explanation

   1. Normalize by each initial mass: 0.20/2.00, 0.20/2.50 and 0.30/3.00.
   2. Average the three individual percentages, obtaining 9.33%, rather than using the mean absolute gain as a percentage.
   3. Repeated readings and biological replicates answer different uncertainty questions; neither removes systematic bias automatically.

   Common mistake: Equal absolute mass gains need not be equal percentage gains when initial masses differ.

   Marking points

   • Percentage change uses (final - initial)/initial x100, giving 10.0%, 8.0% and 10.0%.
   • Mean percentage change = 28/3 = 9.33% approximately.
   • Separate cylinders can reveal between-specimen biological variation.
   • Repeated balance readings of one cylinder primarily assess measurement repeatability, not independent biological responses.

   Examiner tip: Equal absolute mass gains need not be equal percentage gains when initial masses differ.

3. 3.

   A hypothetical student weighs wet potato cylinders immediately after removal, uses only one cylinder per concentration, and concludes that the concentration with zero measured mass change is the exact cell-sap concentration. Identify two methodological limitations with matched improvements and evaluate the exactness of the conclusion.

   [5 marks] · no calculator

   Worked explanation

   1. Pair each limitation with a targeted repair rather than a generic instruction to be more careful.
   2. Separate surface liquid bias from biological variability and from uncertainty in the zero-crossing estimate.
   3. A tissue endpoint includes heterogeneous cells and experimental conditions; the justified conclusion is an estimate, not an exact identity.

   Common mistake: Adding replicates does not remove a systematic surface-water bias unless the weighing method is also improved.

   Marking points

   • Surface solution adds mass unrelated to tissue water uptake.
   • Standardize gentle blotting before each weighing to reduce that bias.
   • One cylinder per concentration cannot estimate between-cylinder variation; use independent replicate cylinders from controlled tissue sources.
   • Zero measured mass change only estimates an isotonic condition under the tested time and tissue conditions, with measurement uncertainty.
   • More closely spaced concentrations near zero plus replication/uncertainty analysis improve estimation but do not establish an exact universal cell-sap concentration.

   Examiner tip: Adding replicates does not remove a systematic surface-water bias unless the weighing method is also improved.

4. 4.

   For written investigation preparation, a student proposes using time for leaf disks to float as an indicator of photosynthesis under different light intensities. Some disks already float before illumination because air remains inside them. Explain how this affects the endpoint, propose a standardisation step and a relevant control, and describe one reason flotation time is not a direct measurement of gross photosynthetic rate. Do not perform the procedure.

   [4 marks] · no calculator

   Syllabus placement evidence: Biology subject brief, first assessment 2025, p.4 Inquiry process, Exploring and designing / Concluding and evaluating; III Assessment objective 3, experimental procedures. Tests design and evaluation of an indirect biological measurement, with the flotation mechanism supplied. SL scientific-investigation preparation only; it neither performs practical work nor claims that this standard activity constitutes an authentic completed IA.

   Worked explanation

   1. The intended chain is photosynthetic oxygen accumulation followed by a buoyancy change. Initial trapped gas bypasses the first link, so the starting state must be standardised.
   2. A dark control checks whether floating can occur without photosynthetic illumination. Even with this control, reaching a threshold measures net retained gas and tissue buoyancy, not all oxygen produced; complementary direct measurements would be needed for a quantitative production claim.

   Common mistake: A disk that floats immediately has not necessarily photosynthesised fastest; its initial trapped gas may invalidate the intended comparison.

   Marking points

   • Pre-existing trapped gas can make disks float independently of newly produced oxygen, giving spuriously short or zero flotation times.
   • Use a consistent, locally approved infiltration procedure and predefined criterion that disks begin submerged; standardise disk size/source and document any exclusions before comparing treatments.
   • Include an otherwise matched dark control to test whether flotation occurs without light-driven oxygen production, keeping temperature and solution conditions matched.
   • Flotation is a buoyancy threshold affected by gas retention, respiration and tissue properties; it does not directly quantify total oxygen production per unit time.

   Examiner tip: A disk that floats immediately has not necessarily photosynthesised fastest; its initial trapped gas may invalidate the intended comparison.

5. 5.

   In hypothetical germination data for written investigation preparation, 20 seeds are allocated to each of two treatments. Treatment A has 18 germinated seeds with mean root length 12 mm among those germinated; B has 12 germinated seeds with mean 15 mm among those germinated. Calculate germination percentages and mean root length per originally allocated seed if ungerminated seeds are assigned zero by a stated composite-endpoint rule. Evaluate the claim that B is simply 'better for growth', and state what should be predefined in a follow-up plan.

   [5 marks]

   Syllabus placement evidence: Biology subject brief, first assessment 2025, p.4 Tools/Mathematics; Inquiry process, Collecting and processing data / Concluding and evaluating. Applies percentages and means to evaluate a specified germination endpoint and plan transparent analysis. This is SL written preparation with hypothetical counts, not real germination evidence, a required zero-imputation convention or an assessed investigation.

   Worked explanation

   1. The original denominator is twenty in each group. Conditional means describe different selected subsets, whereas the explicitly defined zero-assignment endpoint averages over all allocated seeds.
   2. Reconstruct length totals before dividing by twenty: 216/20 versus 180/20. These endpoints rank the treatments differently because they measure different outcomes; assigning zero is a declared composite rule, not proof that ungerminated seeds are dead or incapable of later growth.

   Common mistake: A mean calculated only among germinated seeds cannot automatically represent every seed assigned to the treatment.

   Marking points

   • Germination is 90% in A and 60% in B.
   • A total root length=18(12)=216 mm, composite mean=216/20=10.8 mm per allocated seed.
   • B total root length=12(15)=180 mm, composite mean=180/20=9.0 mm per allocated seed.
   • B has longer roots among germinated seeds but lower germination and lower composite mean; no unqualified best treatment follows without an agreed objective and variability information.
   • Predefine the germination criterion, measurement time, primary endpoint and handling of ungerminated seeds before inspecting outcomes; retain both germination and conditional-length summaries.

   Examiner tip: A mean calculated only among germinated seeds cannot automatically represent every seed assigned to the treatment.

6. 6.

   For written investigation preparation, a potato cylinder has initial mass 2.00 g and final mass 2.04 g. Each balance reading has a stated absolute uncertainty of +/-0.01 g. For this task use the conservative rule that absolute uncertainties add when two readings are subtracted. Calculate the mass gain and its absolute uncertainty, then express that uncertainty as a percentage of the mass gain. Explain why a small percentage increase in the cylinder's mass is not the same quantity as a small percentage uncertainty in its gain. These are hypothetical readings, not a performed experiment.

   [4 marks]

   Syllabus placement evidence: Biology guide, first assessment 2025, updated March 2024: Skills in the study of biology, Tool 3 Mathematics, Applying general mathematics and Processing uncertainties, printed pp.30-31 (PDF pp.36-37); Inquiry 3, p.33. SL written preparation applies the supplied conservative addition-of-bounds rule within prescribed percentage calculations and raw/processed uncertainty interpretation. The guide supports the skill, not a requirement to memorize this one propagation rule. No actual experiment, assessed IA or complete statement coverage is established.

   Worked explanation

   1. Subtract the two mass readings to obtain the response, then add their stated absolute uncertainty bounds as instructed. Subtraction makes the response small but does not cancel the supplied bounds.
   2. Divide 0.02 g by the measured gain of 0.04 g to get 50%. By contrast, dividing the gain by the original 2.00 g gives a 2% mass increase. The supplied bound is not a confidence interval or an estimate of between-cylinder variation.

   Common mistake: Dividing 0.01 by 2.00 calculates neither the combined absolute uncertainty nor the percentage uncertainty of the mass difference.

   Marking points

   • Mass gain = 2.04 - 2.00 = 0.04 g.
   • Conservative absolute uncertainty = 0.01 + 0.01 = 0.02 g, so gain = 0.04 +/-0.02 g under the supplied rule.
   • Percentage uncertainty in the gain = (0.02/0.04) x 100 = 50%.
   • Mass increase relative to initial mass is (0.04/2.00) x 100 = 2%; the 2% change and 50% uncertainty use different numerators and denominators and must not be interchanged.

   Examiner tip: Dividing 0.01 by 2.00 calculates neither the combined absolute uncertainty nor the percentage uncertainty of the mass difference.

Original, unofficial practice questions from SubjectScout — not official past-paper questions or complete syllabus coverage. Added English teaching explanations are AI-authored; independent teacher review is pending.