Biology HL

1. 1.

   For written investigation preparation, hypothetical oxygen volumes at 0, 20, 40 and 60 s are 0.0, 4.0, 6.0 and 7.0 cm^3. Calculate the mean rate over the first 20 s and over all 60 s. Explain why these differ and why neither automatically equals the true instantaneous rate at time zero.

   [4 marks]

   Worked explanation

   1. Calculate change in volume divided by elapsed time for each stated interval.
   2. Compare successive 20-s increments of 4.0, 2.0 and 1.0 cm^3 before proposing any cause; the data alone do not identify why the rate declines.
   3. Do not label a 20-second average an exact time-zero derivative or claim a mechanism from the curve alone.

   Common mistake: A final volume is not a rate; it must be divided by a stated time interval with units retained.

   Marking points

   • First-20-s mean = 4.0/20 = 0.20 cm^3 s^-1.
   • Overall mean = 7.0/60 = 0.117 cm^3 s^-1 approximately.
   • Smaller successive volume increments show rate declines over the measured period, so the longer average is lower.
   • Both are finite-interval averages; more frequent early readings/a fitted early tangent with uncertainty would better estimate the initial instantaneous rate.

   Examiner tip: A final volume is not a rate; it must be divided by a stated time interval with units retained.

2. 2.

   A hypothetical enzyme investigation runs all low-pH trials on Monday with batch A and all high-pH trials on Friday with batch B. Five readings come from the same reaction vessel in each treatment. Identify the confounding and pseudoreplication risks and propose a design that addresses both.

   [4 marks] · no calculator

   Worked explanation

   1. A perfect association of day/batch with treatment prevents separating their effects.
   2. Repeated readings describe one reaction's trajectory rather than replicate treatment responses.
   3. Cross treatment with batch/day and randomize order while adding independent vessels; this targets the two separate design flaws.

   Common mistake: Calling each time point a replicate inflates the sample size and does not fix a batch-treatment confound.

   Marking points

   • pH is confounded with enzyme batch and day, so a difference cannot uniquely be attributed to pH.
   • Five readings from one vessel are repeated measures, not five independent reaction replicates.
   • Run pH treatments within each batch/day block in randomized order with matched temperature and reagent concentrations.
   • Use independently prepared reaction vessels for biological/experimental replication and retain repeated readings as nested time-series measurements.

   Examiner tip: Calling each time point a replicate inflates the sample size and does not fix a batch-treatment confound.

3. 3.

   In a hypothetical investigation, treatment A has mean rate 1.8 with sample standard deviation 0.4 units (n=4 independent vessels), and B has mean 2.1 with standard deviation 0.5 units (n=4). Evaluate the claim that overlapping standard-deviation bars prove no treatment effect. Suggest two next steps that improve the evidence without guaranteeing a preferred result.

   [4 marks] · no calculator

   Worked explanation

   1. Separate observed mean difference, within-group spread and uncertainty about the true treatment difference.
   2. The bar type matters: standard deviations are not interchangeable with confidence intervals or a hypothesis test.
   3. Plan better replication and a suitable analysis before deciding what the result must be; do not remove inconvenient observations without justification.

   Common mistake: Overlap of standard-deviation bars neither proves no effect nor automatically establishes statistical significance.

   Marking points

   • B's sample mean is 0.3 units higher, a descriptive difference rather than proof of a population effect.
   • Standard deviation describes sample spread; overlap alone is not a valid significance test and does not prove equality.
   • Increase well-controlled independent replication and report raw data/uncertainties to assess variation and anomalies.
   • Use an appropriate justified inferential analysis with its assumptions and/or estimate an effect with uncertainty; report the outcome even if inconclusive.

   Examiner tip: Overlap of standard-deviation bars neither proves no effect nor automatically establishes statistical significance.

4. 4.

   For written HL investigation preparation, two hypothetical enzyme preparations are measured at matched substrate concentration, pH and temperature. A produces 0.60 micromol product min^-1 using 0.20 mg total preparation protein; B produces 0.90 micromol min^-1 using 0.50 mg. Assume rates are measured in a range proportional to enzyme amount. Calculate activity per mg total protein and the percentage difference of B relative to A on this basis. Explain why the raw rate and normalised rate rank differently, and why neither alone establishes a difference in intrinsic catalytic activity of the enzyme molecules.

   [5 marks]

   Syllabus placement evidence: Biology subject brief, first assessment 2025, p.3 Enzymes and metabolism; p.4 Tools/Mathematics and Inquiry process, Concluding and evaluating. Applies explicitly supplied proportionality to enzyme-rate normalisation and an evidence limit. HL investigation preparation requires no unsupplied turnover-number formula and does not establish real enzyme purity, molecular causation or a completed IA.

   Worked explanation

   1. Divide each observed rate by its own protein mass, retaining the denominator's meaning: total preparation protein is not necessarily purified active enzyme.
   2. Compare normalised B with A using A as the percentage baseline. The result answers an operational preparation-efficiency question; a molecular turnover comparison would require a justified active-enzyme amount and comparable assay conditions.

   Common mistake: Total protein mass is not automatically the mass of active enzyme, so a lower activity per mg preparation does not prove slower individual enzyme molecules.

   Marking points

   • A activity=0.60/0.20=3.0 micromol min^-1 mg^-1 total protein.
   • B activity=0.90/0.50=1.8 micromol min^-1 mg^-1 total protein.
   • B is (1.8-3.0)/3.0 x100=-40%, or 40% lower on this basis.
   • B's higher raw rate uses more total protein; normalisation changes the comparison to equal preparation-protein mass.
   • Different purity or active-enzyme fraction can explain the difference; enzyme amount/purity and rate uncertainty need assessment before inferring intrinsic molecular activity.

   Examiner tip: Total protein mass is not automatically the mass of active enzyme, so a lower activity per mg preparation does not prove slower individual enzyme molecules.

5. 5.

   A hypothetical written enzyme investigation reports A=2.00 rate units at 20 degrees C and B=3.00 at 25 degrees C. For a sensitivity exercise only, assume both preparations obey the supplied exponential temperature model r(T)=r(20)*Q10^((T-20)/10), with Q10=2 and no other temperature effects across this range. Convert B to a model-equivalent rate at 20 degrees C and compare it with A as a percentage. Repeat the comparison using Q10=1.5. Explain what these calculations imply for a claim that B has a temperature-independent advantage, and propose a stronger comparison.

   [5 marks]

   Syllabus placement evidence: Biology subject brief, first assessment 2025, p.4 Tools/Mathematics; Inquiry process, Exploring and designing / Concluding and evaluating. A supplied Q10 rule is used as a bounded model-sensitivity application, not required recall or a universal biological law. HL written investigation preparation only; actual controlled trials, significance and assessed scientific-investigation completion remain unclaimed.

   Worked explanation

   1. Five degrees is half a ten-degree interval, so the assumed factor is sqrt(Q10), not Q10/2. Divide the warmer rate by that factor to express a conditional cooler equivalent.
   2. The same raw 50% difference becomes about 6.07% or 22.47% under two assumed multipliers. This is model sensitivity, not an experimentally established correction; matched-temperature measurements target the confounding directly.

   Common mistake: An assumed Q10 is not a measured universal enzyme constant, and a positive adjusted difference is not proof of statistical significance.

   Marking points

   • At Q10=2, B_20=3.00/(2^0.5)=2.12132... rate units.
   • Relative to A, this is approximately 6.07% higher.
   • At Q10=1.5, B_20=3.00/sqrt(1.5)=2.44949..., approximately 22.47% higher than A.
   • The adjusted difference depends on an unverified temperature-response assumption; neither scenario establishes an intrinsic advantage or statistical significance without variability data.
   • Plan replicated A and B measurements at the same controlled temperature, ideally across a shared temperature series to test rather than assume their response.

   Examiner tip: An assumed Q10 is not a measured universal enzyme constant, and a positive adjusted difference is not proof of statistical significance.

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.