Skill Align TASC Biology Level 3 - Free Online Pack 0
An independently authored five-section BIO315124 practice examination themed around Derwent estuary and kunanyi field evidence.
- Paper
- Five-booklet written practice examination
- Reading
- 15 minutes preparation
- Writing
- 3 hours
- Assessment
- 180 marks
The current TASC BIO315124 Biology Information Sheet is approved for use throughout the examination and is linked as a separate official resource.
Booklet A of 5 - Section A - Criterion 3: Biological inquiry and evaluation of data
Independent criterion booklet: 5 compulsory questions, 36 marks. Suggested working time: 36 minutes. This booklet assesses Criterion 3 and may be printed separately at its fresh-page boundary.
Question 1
6 marksDerwent programme investigated the effect of temperature on catalase D0 in extracts of native pepper. Equal extract and substrate volumes were reacted for 60 seconds at each temperature. The exact mean product rates were: 10 degrees C: 14 units per minute 20 degrees C: 23 units per minute 30 degrees C: 36 units per minute 40 degrees C: 19 units per minute Five independently prepared extracts were used at every temperature; pH, reaction time and substrate concentration were fixed. Before collecting the data, the researchers hypothesised that mean product formation rate would rise from 10 degrees C to 30 degrees C, then be lower at 40 degrees C than at 30 degrees C.
Question 2
6 marksResearchers propose temporarily capturing 18 healthy tasmanian pademelon individuals to compare resting body temperature before and after a short, mild heat exposure. The proposal uses implanted loggers under anaesthesia, releases each animal after recovery, but does not specify a veterinary stopping rule. One measurement is planned for each animal, and capture order is not randomised between morning and afternoon sessions.
Question 3
7 marksReplicate leaves of native pepper were held at four relative humidities for two hours under the same light and temperature. Mean percentages of stomata scored open were: 80% humidity: 72% open 60% humidity: 55% open 40% humidity: 31% open 20% humidity: 18% open Each mean came from six independently treated leaves. No humidity between the four tested values was examined.
Question 4
8 marksDerwent programme measured net oxygen production by illuminated pondweed. Dissolved oxygen in the same sealed chamber was 6.4 mg L-1 initially and 8.4 mg L-1 after 20 minutes according to a calibrated reference sensor. Four groups made repeated initial measurements with their own oxygen probes.
| Group | Reading 1 (mg L-1) | Reading 2 (mg L-1) | Reading 3 (mg L-1) | Reading 4 (mg L-1) |
|---|---|---|---|---|
| A | 5.0 | 5.0 | 5.0 | 5.0 |
| B | 6.2 | 6.5 | 6.4 | 6.5 |
| C | 4.4 | 8.5 | 5.0 | 8.1 |
| D | 7.4 | 7.4 | 7.5 | 7.4 |
Group A's final reading was 7.0 mg L-1. Its probe has a constant additive offset over this range. Temperature, light, plant mass and chamber volume were fixed; there was no gas exchange with the surroundings.
Question 5
9 marksDerwent programme wants to test whether soil-water availability changes growth of genetically similar native pepper seedlings over 28 days. The available growth chamber can hold 48 pots. Dry mass can be measured only by harvesting a plant; soil moisture, light, pot size, mineral supply and temperature can be controlled.
Booklet B of 5 - Section B - Criterion 5: Regulation of biological systems
Independent criterion booklet: 6 compulsory questions, 36 marks. Suggested working time: 36 minutes. This booklet assesses Criterion 5 and may be printed separately at its fresh-page boundary.
Question 6
7 marksIn the Derwent programme teaching laboratory, two E. coli cultures contain the same intact lac operon. Culture A has glucose available but no lactose. Culture B has little glucose and abundant lactose. In Culture A, lacZ and lacY transcripts are not detected. In Culture B, both transcripts and their protein products are abundant. Only repressor control, RNA polymerase, Lac Z and Lac Y are required for this question.
Question 7
5 marksA dividing cell copies a DWT1 DNA region. Helicase, primase, DNA polymerase and ligase are all active. After replication, each daughter DNA molecule contains one strand from the original molecule and one newly synthesised strand.
Question 8
4 marksInitial rates for catalase D0 were measured at four substrate concentrations with a fixed enzyme concentration. Exact rates were: 1 mmol L-1 substrate: control 10, inhibitor 4 units min-1 2 mmol L-1 substrate: control 19, inhibitor 11 units min-1 4 mmol L-1 substrate: control 29, inhibitor 23 units min-1 8 mmol L-1 substrate: control 35, inhibitor 33 units min-1 The inhibitor is reversible and resembles the substrate.
Question 9
9 marksTwo differentiated cell types from one organism contain the same DWT1 DNA sequence and the same total DNA quantity. Cell X contains 90 units of mature DWT1 mRNA and 85 units of its protein; cell Y contains 42 mRNA units and 18 protein units. Cell X performs the protein-associated specialised function at 82 units and cell Y at 28 units. The supplied model states that mature mRNA can be translated into protein; no mechanism controlling transcription is required.
Question 10
6 marksA short coding-region comparison for DWT1 gives mature mRNA codons AUG-GAA for the reference allele and AUG-GUA for the variant, shown in the order to be translated. The supplied codon excerpt is complete for this sequence. A downstream codon exists after the displayed pair, but translation stops immediately if a stop codon is encountered.
Question 11
5 marksIlluminated native pepper cells are supplied with carbon dioxide and water. During a dark interval, oxygen remains available and stored glucose is consumed. A biochemical audit records ATP production in chloroplasts only during illumination and ATP production in mitochondria during both intervals.
Booklet C of 5 - Section C - Criterion 6: Homeostatic regulation
Independent criterion booklet: 6 compulsory questions, 36 marks. Suggested working time: 36 minutes. This booklet assesses Criterion 6 and may be printed separately at its fresh-page boundary.
Question 12
5 marksIn a cell model derived from tasmanian pademelon and used by the Derwent programme, Compound S reduces opening of voltage-gated calcium channels in a presynaptic terminal. In a separate endocrine test, hormone H remains at its usual plasma concentration, but target cells have 70% fewer H receptors and show a weaker response. No specific neurotransmitter or hormone name is required.
Question 13
7 marksAfter a standard period without drinking, plasma osmolarity in tasmanian pademelon individuals rose from 288 to 301 mOsm L-1. Plasma ADH increased and urine flow fell from 2.6 to 0.70 mL min-1; urine became more concentrated. Kidney function was otherwise normal.
Question 14
6 marksDuring a controlled warm-air exposure, mean core temperature of resting tasmanian pademelon individuals rose from 36.2 degrees C to 37.1 degrees C. Peripheral blood flow and evaporative heat loss increased; individuals also moved towards a cooler shaded zone. Air temperature and exposure duration were stated before the trial. These synthetic temperatures use a species-appropriate range rather than a shared mammalian baseline.
Question 15
4 marksAn axon from a sensory neuron of tasmanian pademelon is at resting potential. A threshold stimulus opens voltage-gated sodium channels at one membrane region. Shortly afterwards those sodium channels inactivate, voltage-gated potassium channels open, and the sodium-potassium pump maintains the long-term ion gradients.
Question 16
5 marksAfter a standard carbohydrate-rich feed, mean blood glucose in healthy tasmanian pademelon individuals was 4.8 mmol L-1 before feeding, 7.8 mmol L-1 at 30 minutes and 5.2 mmol L-1 at 120 minutes. Insulin concentration rose after the feed; glucagon concentration fell during the same interval.
Question 17
9 marksReplicate shoots of native pepper were placed in calibrated potometers for 40 minutes at 30%, 50% or 70% relative humidity. Mean water uptake was 8.20, 5.90 and 3.40 mL per hour respectively. Light, temperature, air movement and leaf area were held constant, with eight independently prepared shoots per treatment. Potometer uptake is used as an estimate of transpiration but also includes water retained for growth and metabolism.
Booklet D of 5 - Section D - Criterion 7: Pathogens and immune responses
Independent criterion booklet: 6 compulsory questions, 36 marks. Suggested working time: 36 minutes. This booklet assesses Criterion 7 and may be printed separately at its fresh-page boundary.
Question 18
9 marksA controlled field study evaluated an intervention designed specifically to interrupt airborne droplets and close direct contact, the stated route of an enveloped respiratory virus. During the same eight-week period, the matched control cohort had 60 cases among 1000 individuals, equal to 60.0 cases per 1000. The intervention cohort had 30 cases among 1000 individuals, equal to 30.0 cases per 1000. Cohorts were matched for age distribution and baseline immunity, but allocation occurred by community rather than individual.
Question 19
6 marksIn the Derwent programme, host cells infected by an intracellular strain labelled DWT1 display pathogen-derived antigen at their surface. Antigen-specific helper T lymphocytes have been activated by antigen-presenting cells. A treatment selectively prevents cytotoxic T-lymphocyte proliferation; infected-cell survival then increases, while antibody concentration remains unchanged.
Question 20
5 marksDuring the Derwent programme, infected droplets reach and breach a locally damaged respiratory epithelium. Local cells release inflammatory signals; nearby vessels dilate and become more permeable. Phagocytes accumulate, engulf particles and some display pathogen-derived antigen fragments to lymphocytes.
Question 21
7 marksThe Derwent programme compares four cohorts relevant to an enveloped respiratory virus: (1) recovery after natural infection, (2) an antigen-containing vaccine, (3) antibody transferred across a placenta, or (4) an injection of purified antibody after exposure. Only cohorts 1 and 2 later show a rapid response generated by their own memory cells.
Question 22
4 marksA field investigation detects an enveloped respiratory virus in symptomatic hosts. Its stated route is airborne droplets and close direct contact. The agent cannot reproduce independently outside a host when it is a virus; bacterial and fungal agents in the comparison possess cells, while a prion is infectious misfolded protein without nucleic acid.
Question 23
5 marksAfter first exposure to an antigen from an enveloped respiratory virus, antibody concentration reached 12 units on day 10. Following a second exposure 60 days later, it reached 44 units by day 5. The same assay and antigen dose were used, and no maternal or injected antibodies were supplied.
Booklet E of 5 - Section E - Criterion 8: Biological persistence and diversity
Independent criterion booklet: 6 compulsory questions, 36 marks. Suggested working time: 36 minutes. This booklet assesses Criterion 8 and may be printed separately at its fresh-page boundary.
Question 24
6 marksThe human pedigree shows a single Mendelian trait with complete penetrance. Assume no new mutations or phenocopies, independent births and equal probabilities of sons and daughters. Symbols and family connections are the evidence; identifiers give generation and individual number. Consider autosomal dominant, autosomal recessive, X-linked dominant and X-linked recessive models. Use A for the normal allele and a for an autosomal recessive trait allele.
Question 25
9 marksIn the estuary skink population, a permanent tidal channel and broad exposed mudflat split two dune habitats, and tagged skinks do not cross the unsuitable open zone. Gene flow between the separated breeding groups is therefore undetected. After 90 generations, neutral-marker divergence is 0.18 and reciprocal mating trials produce only 12% viable offspring, compared with 86% within groups. The marker values and mating results are exact printed values, not estimates to be read from a coarse graph.
Question 26
5 marksIn an estuary plant, allele D is completely dominant to d. Two heterozygous parents Dd are crossed.
Question 27
5 marksA diploid germ cell in the estuary skink population has 2n = 30. Homologous chromosomes pair in meiosis I, crossing over occurs, and homologous pairs orient independently at metaphase I. Meiosis is completed normally.
Question 28
7 marksIn the estuary skink population, allele D had frequency 0.18 before four unusually dry seasons and 0.42 afterwards. During those seasons, D-carriers produced a mean of 1.60 surviving juveniles per adult and non-carriers produced 0.80. Marked immigrants also entered after two storms; their allele frequency was not measured.
Question 29
4 marksA diploid precursor cell from the estuary skink population has 30 chromosomes in G1. DNA replication occurs during S phase. The cell then completes mitosis and cytokinesis without chromosome loss or mutation.
Worked Solutions And Marking Guide
General marking principles
- Apply each question-specific criterion to the biological evidence shown and accept scientifically equivalent wording.
- Award each mark only once and against the mapped external criterion and the evidence supplied for that item.
- Accept a later conclusion that correctly follows from a clearly shown earlier calculation error unless the later item independently assesses that calculation.
- Do not infer hidden scenario details or require exact graph readings that the candidate stimulus does not display.
Question 1
(a) The independent variable is reaction temperature. The operational dependent variable is mean product formation rate in units per minute.
(b) The results support the supplied hypothesis: mean product formation rate rises from 14 units per minute at 10 degrees C to 36 units per minute at 30 degrees C, then falls to 19 units per minute at 40 degrees C.
(c) pH can alter active-site shape and enzyme activity, so fixing it prevents pH from confounding the temperature comparison.
Detailed marking criteria
Part a (1 mark)
Identifies reaction temperature as the independent variable.
Acceptable alternatives: Accept scientifically equivalent wording that preserves the assessed biological proposition.
Part a (1 mark)
Identifies mean product formation rate in units per minute as the dependent variable.
Acceptable alternatives: Accept scientifically equivalent wording that preserves the assessed biological proposition.
Part b (1 mark)
Concludes that the results support the supplied hypothesis.
Acceptable alternatives: Accept scientifically equivalent wording that preserves the assessed biological proposition.
Part b (1 mark)
Uses the displayed increase from 14 units per minute at 10 degrees C to 36 units per minute at 30 degrees C.
Acceptable alternatives: Accept scientifically equivalent wording that preserves the assessed biological proposition.
Part b (1 mark)
Uses the displayed decrease from 36 units per minute at 30 degrees C to 19 units per minute at 40 degrees C.
Acceptable alternatives: Accept scientifically equivalent wording that preserves the assessed biological proposition.
Part c (1 mark)
Explains that fixed pH prevents a second influence on enzyme activity from confounding the temperature effect.
Acceptable alternatives: Accept scientifically equivalent wording that preserves the assessed biological proposition.
Question 2
(a) Capture, implantation and heat exposure may cause stress, pain or thermal harm. Use independent animal-ethics and veterinary oversight, analgesia and refined handling, and a predefined temperature or distress endpoint that stops exposure and triggers treatment or withdrawal.
(b) Randomly allocate exposure order across morning and afternoon, take repeated baseline and post-exposure measurements, and use enough independently sampled animals while standardising chamber temperature, activity and measurement timing. These changes reduce time-of-day confounding and quantify repeatability and biological variation.
Detailed marking criteria
Part a (1 mark)
Identifies stress, pain, implantation injury or thermal harm as a protocol-specific ethical risk.
Acceptable alternatives: Accept scientifically equivalent wording that preserves the assessed biological proposition.
Part a (1 mark)
Outlines independent ethics or veterinary oversight with anaesthesia, analgesia or refined handling.
Acceptable alternatives: Accept scientifically equivalent wording that preserves the assessed biological proposition.
Part a (1 mark)
Outlines a predefined humane endpoint that stops exposure and provides treatment or withdrawal.
Acceptable alternatives: Accept scientifically equivalent wording that preserves the assessed biological proposition.
Part b (1 mark)
Randomises or counterbalances session order to prevent time of day being confounded with exposure.
Acceptable alternatives: Accept scientifically equivalent wording that preserves the assessed biological proposition.
Part b (1 mark)
Uses repeated measurements or a verified logger baseline to assess consistency.
Acceptable alternatives: Accept scientifically equivalent wording that preserves the assessed biological proposition.
Part b (1 mark)
Uses an adequate independent sample while standardising chamber, activity and measurement conditions.
Acceptable alternatives: Accept scientifically equivalent wording that preserves the assessed biological proposition.
Question 3
(a) The percentage of stomata open increases as relative humidity increases: it rises from 18% at 20% humidity to 72% at 80% humidity.
(b) The change is 54 percentage points. This subtracts two percentages; it is not the relative percentage increase calculated against the 20% humidity value.
(c) They do not establish an exact threshold because only four humidity values were tested and the means do not show individual-leaf variation. Test more closely spaced humidities with independent leaves and report the variation around each mean.
Detailed marking criteria
Part a (1 mark)
Describes the positive association between humidity and percentage of stomata open.
Acceptable alternatives: Accept scientifically equivalent wording that preserves the assessed biological proposition.
Part a (1 mark)
Supports the description with the displayed endpoints of 18% and 72%.
Acceptable alternatives: Accept scientifically equivalent wording that preserves the assessed biological proposition.
Part b (1 mark)
Calculates 54 percentage points.
Acceptable alternatives: Accept scientifically equivalent wording that preserves the assessed biological proposition.
Part b (1 mark)
Distinguishes a percentage-point difference from a relative percentage change.
Acceptable alternatives: Accept scientifically equivalent wording that preserves the assessed biological proposition.
Part c (1 mark)
Concludes that an exact threshold is not established.
Acceptable alternatives: Accept scientifically equivalent wording that preserves the assessed biological proposition.
Part c (1 mark)
Explains that only four separated humidity treatments were tested or that variation is not displayed.
Acceptable alternatives: Accept scientifically equivalent wording that preserves the assessed biological proposition.
Part c (1 mark)
Proposes closer humidity intervals with independent replication and reporting of variation.
Acceptable alternatives: Accept scientifically equivalent wording that preserves the assessed biological proposition.
Question 4
(a) Group A is most precise because all four initial readings are identical, giving a range of 0.0 mg L-1. This repeatability does not establish accuracy or biological replication.
(b) Group B is most accurate: its mean is (6.2 + 6.5 + 6.4 + 6.5) / 4 = 6.4 mg L-1, matching the calibrated initial reference.
(c) Group A under-reads dissolved oxygen by 1.4 mg L-1, a systematic additive error. Its net rate is (7.0 - 5.0) / 20 = 0.10 mg L-1 min-1, equal to the reference rate because the constant offset cancels in the difference. Check the probe against at least two oxygen standards spanning the measured range, recalibrate or replace it, and repeat the pondweed trial.
Detailed marking criteria
Part a (1 mark)
Identifies Group A as most precise.
Acceptable alternatives: Accept scientifically equivalent wording that preserves the assessed biological proposition.
Part a (1 mark)
Justifies the choice using its zero range or identical repeated readings.
Acceptable alternatives: Accept scientifically equivalent wording that preserves the assessed biological proposition.
Part b (1 mark)
Identifies Group B as most accurate.
Acceptable alternatives: Accept scientifically equivalent wording that preserves the assessed biological proposition.
Part b (1 mark)
Justifies the choice using the 6.4 mg L-1 group mean and calibrated initial reference.
Acceptable alternatives: Accept scientifically equivalent wording that preserves the assessed biological proposition.
Part c (1 mark)
Identifies a systematic measurement or calibration error rather than random error.
Acceptable alternatives: Accept scientifically equivalent wording that preserves the assessed biological proposition.
Part c (1 mark)
Quantifies the under-reading as 1.4 mg L-1.
Acceptable alternatives: Accept scientifically equivalent wording that preserves the assessed biological proposition.
Part c (1 mark)
Calculates 0.10 mg L-1 min-1 and explains cancellation of the constant additive offset.
Acceptable alternatives: Accept scientifically equivalent wording that preserves the assessed biological proposition.
Part c (1 mark)
Checks at least two oxygen standards and recalibrates or replaces the probe before repeating the biological trial.
Acceptable alternatives: Accept scientifically equivalent wording that preserves the assessed biological proposition.
Question 5
(a) Seedlings maintained at the higher target soil-water content will have a greater mean final dry mass after 28 days than seedlings maintained at the lower target soil-water content.
(b) Randomly allocate 16 genetically similar seedlings to each of three prespecified soil-water targets, including an intermediate treatment. Maintain target moisture by weighing or sensing pots and adding measured water. Keep pot, soil, light, minerals and temperature fixed; rotate pot positions. Harvest all plants at day 28, dry them to constant mass, calculate each treatment mean and variation, graph the results and compare the independently treated pots. Repeat the experiment at another time if feasible.
Detailed marking criteria
Part a (1 mark)
Identifies target soil-water content as the independent variable.
Acceptable alternatives: Accept scientifically equivalent wording that preserves the assessed biological proposition.
Part a (1 mark)
Identifies mean final dry mass after 28 days as the operational dependent variable.
Acceptable alternatives: Accept scientifically equivalent wording that preserves the assessed biological proposition.
Part a (1 mark)
Predicts that the higher-water group will have greater mean dry mass than the lower-water group.
Acceptable alternatives: Accept scientifically equivalent wording that preserves the assessed biological proposition.
Part b (1 mark)
Uses three prespecified water treatments with 16 independently treated pots per treatment.
Acceptable alternatives: Accept scientifically equivalent wording that preserves the assessed biological proposition.
Part b (1 mark)
Randomly allocates comparable seedlings to treatments.
Acceptable alternatives: Accept scientifically equivalent wording that preserves the assessed biological proposition.
Part b (1 mark)
Measures and maintains soil moisture by a defined method while keeping relevant variables fixed.
Acceptable alternatives: Accept scientifically equivalent wording that preserves the assessed biological proposition.
Part b (1 mark)
Measures final dry mass by a consistent harvest-and-dry procedure after 28 days.
Acceptable alternatives: Accept scientifically equivalent wording that preserves the assessed biological proposition.
Part b (1 mark)
Calculates and graphs treatment means with variation using pots as the independent replicates.
Acceptable alternatives: Accept scientifically equivalent wording that preserves the assessed biological proposition.
Part b (1 mark)
Includes position rotation and a repeat or other specific safeguard against chamber or time effects.
Acceptable alternatives: Accept scientifically equivalent wording that preserves the assessed biological proposition.
Question 6
(a) Without lactose, the active repressor binds the operator and prevents productive transcription by RNA polymerase, so lacZ and lacY mRNA is not produced at a detectable level.
(b) Lactose-derived inducer binds the repressor and changes its shape so it no longer remains bound to the operator. RNA polymerase can transcribe lacZ and lacY. Lac Y permease increases lactose transport into the cell, while Lac Z beta-galactosidase breaks lactose into usable monosaccharides.
Detailed marking criteria
Part a (1 mark)
States that the active repressor binds the operator when lactose is absent.
Acceptable alternatives: Accept scientifically equivalent wording that preserves the assessed biological proposition.
Part a (1 mark)
Links operator occupancy to blocked productive transcription by RNA polymerase.
Acceptable alternatives: Accept scientifically equivalent wording that preserves the assessed biological proposition.
Part a (1 mark)
Links repression to absent or negligible lacZ and lacY transcripts.
Acceptable alternatives: Accept scientifically equivalent wording that preserves the assessed biological proposition.
Part b (1 mark)
Explains that lactose-derived inducer changes the repressor so it releases or does not bind the operator.
Acceptable alternatives: Accept scientifically equivalent wording that preserves the assessed biological proposition.
Part b (1 mark)
Explains that operator release permits RNA polymerase to transcribe lacZ and lacY.
Acceptable alternatives: Accept scientifically equivalent wording that preserves the assessed biological proposition.
Part b (1 mark)
States that Lac Y is a permease that transports lactose into the cell.
Acceptable alternatives: Accept scientifically equivalent wording that preserves the assessed biological proposition.
Part b (1 mark)
States that Lac Z beta-galactosidase catalyses lactose breakdown.
Acceptable alternatives: Accept scientifically equivalent wording that preserves the assessed biological proposition.
Question 7
(a) Each separated parental strand acts as a template because A pairs with T and C pairs with G. Each daughter molecule therefore retains one parental strand and contains one complementary new strand.
(b) Helicase separates the parental strands by breaking hydrogen bonds. Primase synthesises short RNA primers. DNA polymerase adds complementary DNA nucleotides to extend the new strands; ligase later joins discontinuities.
Detailed marking criteria
Part a (1 mark)
Explains that each parental strand acts as a template through specific A-T and C-G pairing.
Acceptable alternatives: Accept scientifically equivalent wording that preserves the assessed biological proposition.
Part a (1 mark)
Explains that each daughter DNA contains one parental and one newly synthesised strand.
Acceptable alternatives: Accept scientifically equivalent wording that preserves the assessed biological proposition.
Part b (1 mark)
States that helicase separates the strands by breaking hydrogen bonds.
Acceptable alternatives: Accept scientifically equivalent wording that preserves the assessed biological proposition.
Part b (1 mark)
States that primase produces RNA primers.
Acceptable alternatives: Accept scientifically equivalent wording that preserves the assessed biological proposition.
Part b (1 mark)
States that DNA polymerase adds complementary DNA nucleotides to extend the new strand.
Acceptable alternatives: Accept scientifically equivalent wording that preserves the assessed biological proposition.
Question 8
(a) At 1 mmol L-1 the inhibitor lowers rate by 6 units min-1, whereas at 8 mmol L-1 it lowers rate by only 2 units min-1; increasing substrate reduces the inhibitor's relative effect.
(b) A substrate-like inhibitor can reversibly occupy the active site. At higher substrate concentration, substrate molecules more often occupy that site, so the reaction rate approaches the uninhibited rate.
Detailed marking criteria
Part a (1 mark)
Uses the displayed low-substrate difference of 6 units min-1.
Acceptable alternatives: Accept scientifically equivalent wording that preserves the assessed biological proposition.
Part a (1 mark)
Uses the displayed high-substrate difference of 2 units min-1 and states the effect becomes smaller.
Acceptable alternatives: Accept scientifically equivalent wording that preserves the assessed biological proposition.
Part b (1 mark)
Explains reversible competition between substrate and inhibitor for the active site.
Acceptable alternatives: Accept scientifically equivalent wording that preserves the assessed biological proposition.
Part b (1 mark)
Links increased substrate concentration to the near-restoration of reaction rate.
Acceptable alternatives: Accept scientifically equivalent wording that preserves the assessed biological proposition.
Question 9
(a) Cell X has 48 more mRNA units and 67 more protein units than cell Y. The higher mRNA abundance is associated with higher protein abundance.
(b) Both cell types retain the same DWT1 sequence, so their different functions do not require different genomes. More mature mRNA in X provides more templates for translation, consistent with more protein and the stronger protein-associated specialised function. Differential protein production can therefore contribute to different cell phenotypes.
(c) The comparison shows association, not causation, and may reflect other differences between the cell types. Measure mRNA, protein and the specialised function in independently sampled cells across several differentiation stages, using the same assays and reporting replicate variation. A repeatable sequence in which expression and protein changes precede the functional change would strengthen, but not by itself prove, the proposed contribution.
Detailed marking criteria
Part a (1 mark)
Calculates the mature-mRNA difference as 48 units.
Acceptable alternatives: Accept scientifically equivalent wording that preserves the assessed biological proposition.
Part a (1 mark)
Calculates the protein-abundance difference as 67 units.
Acceptable alternatives: Accept scientifically equivalent wording that preserves the assessed biological proposition.
Part a (1 mark)
Describes the positive association between mature mRNA and protein abundance.
Acceptable alternatives: Accept scientifically equivalent wording that preserves the assessed biological proposition.
Part b (1 mark)
Uses the identical DNA evidence to explain that differentiated cells need not have different genomes.
Acceptable alternatives: Accept scientifically equivalent wording that preserves the assessed biological proposition.
Part b (1 mark)
Links more mature mRNA to more translation templates and greater protein abundance.
Acceptable alternatives: Accept scientifically equivalent wording that preserves the assessed biological proposition.
Part b (1 mark)
Links unequal protein abundance to the different specialised cell functions or phenotypes.
Acceptable alternatives: Accept scientifically equivalent wording that preserves the assessed biological proposition.
Part c (1 mark)
Recognises that a two-cell-type association does not establish causation.
Acceptable alternatives: Accept scientifically equivalent wording that preserves the assessed biological proposition.
Part c (1 mark)
Proposes independently replicated measurements across multiple differentiation stages with consistent methods.
Acceptable alternatives: Accept scientifically equivalent wording that preserves the assessed biological proposition.
Part c (1 mark)
Interprets temporal expression, protein and function evidence cautiously rather than claiming proof.
Acceptable alternatives: Accept scientifically equivalent wording that preserves the assessed biological proposition.
Question 10
(a) Reference: Met-glutamate. Variant: Met-valine.
(b) This is a missense substitution because a base substitution changes glutamate to valine. The different side chain may alter local interactions, folding or an active site, but the effect could also be small; protein-activity and phenotype data are needed.
Detailed marking criteria
Part a (1 mark)
Translates the reference sequence as Met-glutamate.
Acceptable alternatives: Accept scientifically equivalent wording that preserves the assessed biological proposition.
Part a (1 mark)
Translates the variant sequence as Met-valine and does not translate beyond STOP.
Acceptable alternatives: Accept scientifically equivalent wording that preserves the assessed biological proposition.
Part b (1 mark)
Classifies the change as a missense substitution caused by a nucleotide substitution.
Acceptable alternatives: Accept scientifically equivalent wording that preserves the assessed biological proposition.
Part b (1 mark)
Explains that the amino-acid substitution changes glutamate to valine.
Acceptable alternatives: Accept scientifically equivalent wording that preserves the assessed biological proposition.
Part b (1 mark)
Links the sequence change to a possible alteration of folding, stability or active-site structure.
Acceptable alternatives: Accept scientifically equivalent wording that preserves the assessed biological proposition.
Part b (1 mark)
Qualifies the prediction because protein-function and phenotype evidence are not supplied.
Acceptable alternatives: Accept scientifically equivalent wording that preserves the assessed biological proposition.
Question 11
(a) Light-dependent reactions occur on thylakoid membranes; light-independent reactions occur in the stroma; photolysis produces oxygen as well as hydrogen ions and electrons.
(b) Stored glucose can undergo glycolysis in the cytosol, followed by the Krebs cycle and electron transport in mitochondria while oxygen is available. Oxygen acts as the final electron acceptor, allowing the electron-transport chain to generate most ATP even when photosynthesis stops in darkness.
Detailed marking criteria
Part a (1 mark)
Locates the light-dependent reactions on thylakoid membranes and light-independent reactions in the stroma.
Acceptable alternatives: Accept scientifically equivalent wording that preserves the assessed biological proposition.
Part a (1 mark)
Identifies oxygen as a product of photolysis.
Acceptable alternatives: Accept scientifically equivalent wording that preserves the assessed biological proposition.
Part b (1 mark)
Links stored glucose to glycolysis and subsequent aerobic respiration.
Acceptable alternatives: Accept scientifically equivalent wording that preserves the assessed biological proposition.
Part b (1 mark)
Locates the later aerobic stages in mitochondria and identifies electron transport as producing most ATP.
Acceptable alternatives: Accept scientifically equivalent wording that preserves the assessed biological proposition.
Part b (1 mark)
Explains that available oxygen permits electron transport during darkness even though photosynthesis has stopped.
Acceptable alternatives: Accept scientifically equivalent wording that preserves the assessed biological proposition.
Question 12
(a) Reduced calcium entry causes fewer synaptic vesicles to fuse with the presynaptic membrane, so less neurotransmitter enters the cleft and the postsynaptic response is reduced.
(b) Only cells with complementary receptors respond to hormone H. Fewer receptors reduce hormone-receptor binding and activation of intracellular signalling, so target-cell activity changes less even though delivery in plasma is unchanged.
Detailed marking criteria
Part a (1 mark)
Links reduced calcium entry to reduced synaptic-vesicle fusion.
Acceptable alternatives: Accept scientifically equivalent wording that preserves the assessed biological proposition.
Part a (1 mark)
Links reduced neurotransmitter release to a smaller postsynaptic response.
Acceptable alternatives: Accept scientifically equivalent wording that preserves the assessed biological proposition.
Part b (1 mark)
States that receptor specificity determines which target cells respond.
Acceptable alternatives: Accept scientifically equivalent wording that preserves the assessed biological proposition.
Part b (1 mark)
Links fewer receptors to less hormone-receptor binding or signalling.
Acceptable alternatives: Accept scientifically equivalent wording that preserves the assessed biological proposition.
Part b (1 mark)
Explains that reduced target-cell response does not require reduced plasma hormone concentration.
Acceptable alternatives: Accept scientifically equivalent wording that preserves the assessed biological proposition.
Question 13
(a) Plasma osmolarity rose by 13 mOsm L-1 while urine flow fell by 1.90 mL min-1. The simultaneous ADH rise, lower urine volume and higher urine concentration are consistent with water conservation.
(b) Hypothalamic osmoreceptors detect raised osmolarity. The posterior pituitary releases more ADH, which acts on the kidney collecting ducts and increases collecting-duct water permeability. More water is reabsorbed into the blood, reducing urine volume and tending to lower plasma osmolarity, which reduces ADH stimulation.
Detailed marking criteria
Part a (1 mark)
Calculates the osmolarity rise as 13 mOsm L-1.
Acceptable alternatives: Accept scientifically equivalent wording that preserves the assessed biological proposition.
Part a (1 mark)
Calculates the urine-flow decrease as 1.90 mL min-1.
Acceptable alternatives: Accept scientifically equivalent wording that preserves the assessed biological proposition.
Part a (1 mark)
Links increased ADH, reduced urine flow and concentrated urine to water conservation.
Acceptable alternatives: Accept scientifically equivalent wording that preserves the assessed biological proposition.
Part b (1 mark)
Identifies hypothalamic osmoreceptors as detecting increased plasma osmolarity.
Acceptable alternatives: Accept scientifically equivalent wording that preserves the assessed biological proposition.
Part b (1 mark)
Links increased posterior-pituitary ADH release to receptor binding in collecting-duct cells.
Acceptable alternatives: Accept scientifically equivalent wording that preserves the assessed biological proposition.
Part b (1 mark)
Explains that ADH increases collecting-duct water permeability and therefore increases water reabsorption.
Acceptable alternatives: Accept scientifically equivalent wording that preserves the assessed biological proposition.
Part b (1 mark)
Links water retention to reduced urine volume and negative-feedback reduction of the osmotic stimulus.
Acceptable alternatives: Accept scientifically equivalent wording that preserves the assessed biological proposition.
Question 14
(a) Moving to shade is behavioural and reduces environmental heat gain. Increased peripheral blood flow is physiological and increases heat transfer from the body core towards heat-exchange surfaces. Increased evaporative loss is physiological and removes latent heat as water evaporates.
(b) Thermoreceptors detect the rise and the thermoregulatory control centre coordinates effectors that increase heat loss. The response opposes the rise, so its stimulus declines as core temperature returns towards the regulated range. Because only means and one exposure condition are supplied, individual variation and responses at other temperatures remain unknown.
Detailed marking criteria
Part a (1 mark)
Classifies shade seeking as behavioural and links it to reduced heat gain.
Acceptable alternatives: Accept scientifically equivalent wording that preserves the assessed biological proposition.
Part a (1 mark)
Classifies increased peripheral blood flow as physiological and links it to increased heat transfer towards exchange surfaces.
Acceptable alternatives: Accept scientifically equivalent wording that preserves the assessed biological proposition.
Part a (1 mark)
Classifies evaporative cooling as physiological and links evaporation to heat loss.
Acceptable alternatives: Accept scientifically equivalent wording that preserves the assessed biological proposition.
Part b (1 mark)
Links thermoreceptor detection and a control centre to coordinated heat-loss effectors.
Acceptable alternatives: Accept scientifically equivalent wording that preserves the assessed biological proposition.
Part b (1 mark)
Explains that the response opposes the temperature rise and reduces the original stimulus.
Acceptable alternatives: Accept scientifically equivalent wording that preserves the assessed biological proposition.
Part b (1 mark)
Identifies a supported limit such as mean-only data, one condition or unknown individual variation.
Acceptable alternatives: Accept scientifically equivalent wording that preserves the assessed biological proposition.
Question 15
(a) Opening voltage-gated sodium channels allows sodium ions to move into the axon and depolarise the membrane. Sodium-channel inactivation and opening potassium channels allow potassium ions to move out, repolarising the membrane.
(b) Local current brings the next membrane region to threshold, opening its voltage-gated sodium channels. The region just activated is refractory because its sodium channels are inactivated, so immediate backward re-excitation is prevented.
Detailed marking criteria
Part a (1 mark)
Links sodium-channel opening and sodium entry to depolarisation.
Acceptable alternatives: Accept scientifically equivalent wording that preserves the assessed biological proposition.
Part a (1 mark)
Links sodium-channel inactivation plus potassium-channel opening and potassium exit to repolarisation.
Acceptable alternatives: Accept scientifically equivalent wording that preserves the assessed biological proposition.
Part b (1 mark)
Explains that local current depolarises the next membrane region to threshold.
Acceptable alternatives: Accept scientifically equivalent wording that preserves the assessed biological proposition.
Part b (1 mark)
Links the refractory region and sodium-channel inactivation to one-way propagation.
Acceptable alternatives: Accept scientifically equivalent wording that preserves the assessed biological proposition.
Question 16
(a) The peak rise was 3.0 mmol L-1. At 120 minutes only 0.4 mmol L-1 remained above baseline, so 2.6 mmol L-1 of the rise had been removed.
(b) The pancreas detects elevated blood glucose and increases insulin release. Insulin binds receptors on target cells, increasing glucose uptake, while the liver converts glucose to glycogen. As blood glucose falls towards baseline, the original stimulus to insulin release is reduced.
Detailed marking criteria
Part a (1 mark)
Calculates the peak rise above baseline as 3.0 mmol L-1.
Acceptable alternatives: Accept scientifically equivalent wording that preserves the assessed biological proposition.
Part a (1 mark)
Calculates the amount removed by 120 minutes as 2.6 mmol L-1.
Acceptable alternatives: Accept scientifically equivalent wording that preserves the assessed biological proposition.
Part b (1 mark)
Identifies the pancreas as detecting elevated blood glucose and increasing insulin secretion.
Acceptable alternatives: Accept scientifically equivalent wording that preserves the assessed biological proposition.
Part b (1 mark)
Links insulin action to increased target-cell glucose uptake and liver glycogen formation.
Acceptable alternatives: Accept scientifically equivalent wording that preserves the assessed biological proposition.
Part b (1 mark)
Explains that the response lowers glucose and thereby reduces the original insulin-releasing stimulus.
Acceptable alternatives: Accept scientifically equivalent wording that preserves the assessed biological proposition.
Question 17
(a) Mean uptake falls by 4.80 mL per hour as humidity rises from 30% to 70%, supporting lower transpiration at higher humidity. Uptake is only an estimate because some absorbed water can be retained for growth or metabolism rather than transpired.
(b) Water evaporates from moist mesophyll surfaces and water vapour diffuses through open stomata to the air. Closing stomata reduces the opening available for diffusion, so transpiration and potometer water uptake decrease; carbon-dioxide entry can also fall.
(c) A thick waxy cuticle reduces evaporation through the epidermis. Sunken stomata or leaf hairs trap a humid, still boundary layer, reducing diffusion of water vapour. Reduced leaf area would also reduce the surface available for evaporation.
Detailed marking criteria
Part a (1 mark)
Describes lower mean water uptake as relative humidity increases.
Acceptable alternatives: Accept scientifically equivalent wording that preserves the assessed biological proposition.
Part a (1 mark)
Calculates the 30%-to-70% decrease as 4.80 mL per hour.
Acceptable alternatives: Accept scientifically equivalent wording that preserves the assessed biological proposition.
Part a (1 mark)
Explains that potometer uptake is not identical to water-vapour loss because some water is retained or used.
Acceptable alternatives: Accept scientifically equivalent wording that preserves the assessed biological proposition.
Part b (1 mark)
Explains evaporation from moist internal leaf or mesophyll surfaces.
Acceptable alternatives: Accept scientifically equivalent wording that preserves the assessed biological proposition.
Part b (1 mark)
Explains diffusion of water vapour through stomata to the surrounding air.
Acceptable alternatives: Accept scientifically equivalent wording that preserves the assessed biological proposition.
Part b (1 mark)
Links stomatal closure to lower transpiration and identifies reduced carbon-dioxide entry as a trade-off.
Acceptable alternatives: Accept scientifically equivalent wording that preserves the assessed biological proposition.
Part c (1 mark)
Explains how a thick waxy cuticle reduces epidermal evaporation.
Acceptable alternatives: Accept scientifically equivalent wording that preserves the assessed biological proposition.
Part c (1 mark)
Explains how sunken stomata or leaf hairs retain a humid, still boundary layer.
Acceptable alternatives: Accept scientifically equivalent wording that preserves the assessed biological proposition.
Part c (1 mark)
Explains a second valid adaptation such as reduced leaf area, rolled leaves or fewer stomata in terms of reduced water loss.
Acceptable alternatives: Accept scientifically equivalent wording that preserves the assessed biological proposition.
Question 18
(a) Rate ratio = 30.0 / 60.0 = 0.50. Relative reduction = 1 - 0.50 = 0.50, or 50%.
(b) The intervention is designed to reduce exposure through airborne droplets and close direct contact. Reducing that contact lowers the probability that the pathogen reaches a susceptible host, so fewer new infections and a lower incidence rate are expected.
(c) The equal denominators, matched baseline variables and lower intervention rate support effectiveness in these cohorts. However, community-level rather than individual allocation allows community differences or clustering to confound precision and causality. Replicate randomised communities, account for clustering, verify intervention adherence and measure exposure and immunity before generalising.
Detailed marking criteria
Part a (1 mark)
Calculates the intervention-to-control rate ratio as 0.50.
Acceptable alternatives: Accept scientifically equivalent wording that preserves the assessed biological proposition.
Part a (1 mark)
Calculates the relative rate reduction as 0.50.
Acceptable alternatives: Accept scientifically equivalent wording that preserves the assessed biological proposition.
Part a (1 mark)
Expresses the relative reduction as 50%.
Acceptable alternatives: Accept scientifically equivalent wording that preserves the assessed biological proposition.
Part b (1 mark)
States the supplied transmission route as airborne droplets and close direct contact.
Acceptable alternatives: Accept scientifically equivalent wording that preserves the assessed biological proposition.
Part b (1 mark)
Explains that the intervention interrupts host exposure through that route.
Acceptable alternatives: Accept scientifically equivalent wording that preserves the assessed biological proposition.
Part b (1 mark)
Links reduced exposure to fewer new infections or a lower incidence rate.
Acceptable alternatives: Accept scientifically equivalent wording that preserves the assessed biological proposition.
Part c (1 mark)
Uses the population-standardised rates and matching as evidence supporting effectiveness in these cohorts.
Acceptable alternatives: Accept scientifically equivalent wording that preserves the assessed biological proposition.
Part c (1 mark)
Identifies community allocation, clustering or an unmeasured community difference as a causal limitation.
Acceptable alternatives: Accept scientifically equivalent wording that preserves the assessed biological proposition.
Part c (1 mark)
Proposes replicated randomised communities with adherence or exposure measurement and analysis that accounts for clustering.
Acceptable alternatives: Accept scientifically equivalent wording that preserves the assessed biological proposition.
Question 19
(a) An antigen-presenting cell activates a complementary helper T lymphocyte. Helper-cell signalling supports clonal expansion of antigen-specific cytotoxic T lymphocytes. Cytotoxic cells recognise infected cells displaying the antigen and induce their death.
(b) Increased infected-cell survival after cytotoxic proliferation is blocked supports a cytotoxic contribution to clearance. Unchanged antibody concentration shows that the manipulation did not directly reduce the measured humoral response, which is mediated by B-cell-derived plasma cells and antibodies rather than target-cell killing.
Detailed marking criteria
Part a (1 mark)
Links antigen presentation to activation of a complementary helper T lymphocyte.
Acceptable alternatives: Accept scientifically equivalent wording that preserves the assessed biological proposition.
Part a (1 mark)
Links helper-cell signalling to clonal expansion of antigen-specific cytotoxic T lymphocytes.
Acceptable alternatives: Accept scientifically equivalent wording that preserves the assessed biological proposition.
Part a (1 mark)
Links cytotoxic recognition to killing of infected target cells.
Acceptable alternatives: Accept scientifically equivalent wording that preserves the assessed biological proposition.
Part b (1 mark)
Uses increased infected-cell survival to support a cytotoxic T-cell role.
Acceptable alternatives: Accept scientifically equivalent wording that preserves the assessed biological proposition.
Part b (1 mark)
Uses unchanged antibody concentration to distinguish the measured humoral response from the manipulated pathway.
Acceptable alternatives: Accept scientifically equivalent wording that preserves the assessed biological proposition.
Part b (1 mark)
Identifies B lymphocytes or plasma cells and antibodies as central to humoral immunity.
Acceptable alternatives: Accept scientifically equivalent wording that preserves the assessed biological proposition.
Question 20
(a) Vasodilation increases blood flow, while increased permeability allows plasma proteins and immune cells to enter the damaged tissue, helping contain and remove pathogens.
(b) A phagocyte surrounds and internalises a pathogen, encloses it in a vesicle and digests it. Displayed pathogen antigen can be recognised by a complementary helper T lymphocyte, contributing to activation of antigen-specific adaptive clones.
Detailed marking criteria
Part a (1 mark)
Links vasodilation to increased local blood flow.
Acceptable alternatives: Accept scientifically equivalent wording that preserves the assessed biological proposition.
Part a (1 mark)
Links increased permeability to movement of immune cells or plasma components into tissue.
Acceptable alternatives: Accept scientifically equivalent wording that preserves the assessed biological proposition.
Part b (1 mark)
Describes engulfment and internalisation of the pathogen by a phagocyte.
Acceptable alternatives: Accept scientifically equivalent wording that preserves the assessed biological proposition.
Part b (1 mark)
Describes intracellular digestion of the engulfed material.
Acceptable alternatives: Accept scientifically equivalent wording that preserves the assessed biological proposition.
Part b (1 mark)
Links displayed antigen to activation of an antigen-specific helper T lymphocyte and the adaptive response.
Acceptable alternatives: Accept scientifically equivalent wording that preserves the assessed biological proposition.
Question 21
(a) Recovery after infection is naturally acquired active immunity. Vaccination is artificially acquired active immunity. Placental antibody is naturally acquired passive immunity. Purified-antibody injection is artificially acquired passive immunity.
(b) Passive immunity supplies ready-made antibody, so protection can begin quickly but declines as antibody is broken down and normally creates no recipient memory cells. Active exposure requires clonal selection and expansion, so initial protection develops more slowly but memory cells can support a faster, longer-lasting response on re-exposure.
Detailed marking criteria
Part a (1 mark)
Classifies recovery after infection as natural active and vaccination as artificial active immunity.
Acceptable alternatives: Accept scientifically equivalent wording that preserves the assessed biological proposition.
Part a (1 mark)
Classifies placental transfer as natural passive immunity.
Acceptable alternatives: Accept scientifically equivalent wording that preserves the assessed biological proposition.
Part a (1 mark)
Classifies antibody injection as artificial passive immunity.
Acceptable alternatives: Accept scientifically equivalent wording that preserves the assessed biological proposition.
Part b (1 mark)
Explains that passive immunity supplies ready-made antibody and can act rapidly.
Acceptable alternatives: Accept scientifically equivalent wording that preserves the assessed biological proposition.
Part b (1 mark)
Explains that passive antibody declines and normally does not generate recipient memory cells.
Acceptable alternatives: Accept scientifically equivalent wording that preserves the assessed biological proposition.
Part b (1 mark)
Explains that active immunity initially requires clonal selection and expansion.
Acceptable alternatives: Accept scientifically equivalent wording that preserves the assessed biological proposition.
Part b (1 mark)
Links active-immunity memory cells to faster secondary response and longer-lasting protection.
Acceptable alternatives: Accept scientifically equivalent wording that preserves the assessed biological proposition.
Question 22
(a) The named category is an enveloped respiratory virus. Its distinguishing features must match that named category; for example, a virus is acellular and replicates only using host-cell machinery, whereas a bacterium is a prokaryotic cell.
(b) The stated route is airborne droplets and close direct contact. A route-specific intervention must interrupt that exposure, such as ventilation and reduced close contact for droplets, safe water treatment for water transmission, food hygiene for food transmission, vector control for vector transmission, or barrier precautions for body fluids.
Detailed marking criteria
Part a (1 mark)
Identifies the category stated as an enveloped respiratory virus.
Acceptable alternatives: Accept scientifically equivalent wording that preserves the assessed biological proposition.
Part a (1 mark)
States one accurate structural or replicative feature of that named pathogen category.
Acceptable alternatives: Accept scientifically equivalent wording that preserves the assessed biological proposition.
Part b (1 mark)
States the supplied route as airborne droplets and close direct contact.
Acceptable alternatives: Accept scientifically equivalent wording that preserves the assessed biological proposition.
Part b (1 mark)
Names an intervention that specifically interrupts the supplied route rather than a generic treatment.
Acceptable alternatives: Accept scientifically equivalent wording that preserves the assessed biological proposition.
Question 23
(a) The secondary peak is 32 units higher and is reached 5 days sooner than the primary day-10 peak.
(b) The first exposure selected antigen-specific B and helper T lymphocytes for clonal expansion, producing plasma cells and long-lived memory cells. On re-exposure, memory cells activate and proliferate more rapidly, producing more plasma cells and antigen-specific antibody.
Detailed marking criteria
Part a (1 mark)
Calculates that the secondary measured peak is 32 units higher.
Acceptable alternatives: Accept scientifically equivalent wording that preserves the assessed biological proposition.
Part a (1 mark)
States that the secondary measured response reaches its peak 5 days sooner.
Acceptable alternatives: Accept scientifically equivalent wording that preserves the assessed biological proposition.
Part b (1 mark)
Links first exposure to clonal selection and expansion of antigen-specific lymphocytes.
Acceptable alternatives: Accept scientifically equivalent wording that preserves the assessed biological proposition.
Part b (1 mark)
Identifies formation and persistence of memory cells.
Acceptable alternatives: Accept scientifically equivalent wording that preserves the assessed biological proposition.
Part b (1 mark)
Links memory-cell reactivation to faster expansion and greater plasma-cell or antibody production.
Acceptable alternatives: Accept scientifically equivalent wording that preserves the assessed biological proposition.
Question 24
(a) Autosomal recessive inheritance fits. Unaffected I-1 and I-2 have affected children II-2 and II-3, excluding a fully penetrant dominant model. Affected daughter II-3 has an unaffected father I-1, excluding X-linked recessive inheritance.
(b) Both parents are Aa because they are unaffected but each supplied a to an affected child. Aa by Aa gives aa with probability 1/4 or 25% in every independent pregnancy; the earlier children's phenotypes do not change it.
Detailed marking criteria
Part a (1 mark)
Identifies autosomal recessive inheritance.
Acceptable alternatives: Accept scientifically equivalent wording that preserves the assessed biological proposition.
Part a (1 mark)
Uses unaffected I-1 and I-2 with affected offspring to exclude dominance.
Acceptable alternatives: Accept scientifically equivalent wording that preserves the assessed biological proposition.
Part a (1 mark)
Uses affected II-3 and unaffected father I-1 to exclude X-linked recessive inheritance.
Acceptable alternatives: Accept scientifically equivalent wording that preserves the assessed biological proposition.
Part b (1 mark)
Infers Aa for both I-1 and I-2 from their affected offspring.
Acceptable alternatives: Accept scientifically equivalent wording that preserves the assessed biological proposition.
Part b (1 mark)
Derives a 1/4 or 25% affected-child probability.
Acceptable alternatives: Accept scientifically equivalent wording that preserves the assessed biological proposition.
Part b (1 mark)
Explains independence of births rather than a guaranteed family ratio.
Acceptable alternatives: Accept scientifically equivalent wording that preserves the assessed biological proposition.
Question 25
(a) The evidence shows geographic isolation. The separated habitats and observed absence of movement provide geographic-isolation evidence appropriate to the estuary skink population. Mutation, selection and drift can then change allele frequencies independently, allowing divergence to accumulate.
(b) Neutral-marker divergence shows genetic differentiation, while between-group viable offspring are much lower than within-group offspring (12% versus 86%). Together they support developing reproductive isolation, but non-zero viability shows isolation is incomplete and the data do not by themselves establish separate species under every species concept.
(c) The measured marker-frequency divergence within the population lineage is microevolution. Persistent lineage splitting and the origin of reproductively isolated species would contribute to macroevolution. Repeat reciprocal crosses across several generations, test mating behaviour and hybrid fertility, and sample multiple independent genetic markers while directly measuring any remaining migration.
Detailed marking criteria
Part a (1 mark)
Identifies geographic isolation from the supplied evidence.
Acceptable alternatives: Accept scientifically equivalent wording that preserves the assessed biological proposition.
Part a (1 mark)
The separated habitats and observed absence of movement provide geographic-isolation evidence appropriate to the estuary skink population.
Acceptable alternatives: Accept scientifically equivalent wording that preserves the assessed biological proposition.
Part a (1 mark)
Explains that reduced gene flow permits mutation, selection or drift to produce independent divergence.
Acceptable alternatives: Accept scientifically equivalent wording that preserves the assessed biological proposition.
Part b (1 mark)
Uses neutral-marker divergence as evidence of genetic differentiation.
Acceptable alternatives: Accept scientifically equivalent wording that preserves the assessed biological proposition.
Part b (1 mark)
Uses the exact 12% between-group versus 86% within-group viability comparison.
Acceptable alternatives: Accept scientifically equivalent wording that preserves the assessed biological proposition.
Part b (1 mark)
Qualifies the inference because reproductive isolation is incomplete or species concepts differ.
Acceptable alternatives: Accept scientifically equivalent wording that preserves the assessed biological proposition.
Part c (1 mark)
Identifies within-population allele-frequency or marker change as microevolution.
Acceptable alternatives: Accept scientifically equivalent wording that preserves the assessed biological proposition.
Part c (1 mark)
Links persistent lineage splitting or species formation to macroevolution.
Acceptable alternatives: Accept scientifically equivalent wording that preserves the assessed biological proposition.
Part c (1 mark)
Proposes reciprocal multi-generation mating or hybrid-fertility tests with multiple markers and direct gene-flow measurement.
Acceptable alternatives: Accept scientifically equivalent wording that preserves the assessed biological proposition.
Question 26
(a) One of the four equally likely genotypes is dd, so the probability of a recessive-phenotype offspring is 1/4 or 25%.
(b) DD and Dd express the dominant phenotype because one D allele is sufficient, while only dd expresses the recessive phenotype. The expected phenotype probability is therefore three dominant to one recessive.
Detailed marking criteria
Part a (1 mark)
Derives dd in one of four equally likely fertilisation outcomes, giving 1/4 or 25%.
Acceptable alternatives: Accept scientifically equivalent wording that preserves the assessed biological proposition.
Part a (1 mark)
States that each listed fertilisation outcome is an independent probability rather than a guaranteed family ratio.
Acceptable alternatives: Accept scientifically equivalent wording that preserves the assessed biological proposition.
Part b (1 mark)
States that DD and Dd express the dominant phenotype.
Acceptable alternatives: Accept scientifically equivalent wording that preserves the assessed biological proposition.
Part b (1 mark)
States that only dd expresses the recessive phenotype.
Acceptable alternatives: Accept scientifically equivalent wording that preserves the assessed biological proposition.
Part b (1 mark)
Derives a 3:1 expected dominant-to-recessive phenotype ratio.
Acceptable alternatives: Accept scientifically equivalent wording that preserves the assessed biological proposition.
Question 27
(a) Each gamete has n = 15 chromosomes. Homologous chromosomes separate in meiosis I, reducing the chromosome number from diploid to haploid.
(b) Crossing over exchanges DNA between non-sister chromatids and creates new allele combinations within chromosomes. Independent orientation distributes maternal and paternal homologues into gametes in different combinations. Random fusion of genetically varied gametes combines two independently sampled haploid genomes.
Detailed marking criteria
Part a (1 mark)
States that each gamete has 15 chromosomes.
Acceptable alternatives: Accept scientifically equivalent wording that preserves the assessed biological proposition.
Part a (1 mark)
Links the reduction to separation of homologous chromosomes in meiosis I.
Acceptable alternatives: Accept scientifically equivalent wording that preserves the assessed biological proposition.
Part b (1 mark)
Explains that crossing over between non-sister chromatids creates new within-chromosome allele combinations.
Acceptable alternatives: Accept scientifically equivalent wording that preserves the assessed biological proposition.
Part b (1 mark)
Explains that independent orientation distributes homologues into different gamete combinations.
Acceptable alternatives: Accept scientifically equivalent wording that preserves the assessed biological proposition.
Part b (1 mark)
Explains that random fertilisation combines independently varied gametes.
Acceptable alternatives: Accept scientifically equivalent wording that preserves the assessed biological proposition.
Question 28
(a) Allele D rose by 0.24. Its carriers produced 2.00 times as many surviving juveniles as non-carriers. If that reproductive difference is heritable and related to dry-season performance, carriers contribute a larger share of alleles to the next generation, increasing D frequency.
(b) Gene flow could raise or lower the allele frequency depending on the immigrants' allele frequency, which is missing. Genetic drift could also change frequency by chance, especially if effective population size was small. Genotype immigrants and residents before and after storms, estimate effective population size, track genotype-specific survival and reproduction, and compare replicated populations across wet and dry seasons.
Detailed marking criteria
Part a (1 mark)
Calculates the allele-frequency increase as 0.24.
Acceptable alternatives: Accept scientifically equivalent wording that preserves the assessed biological proposition.
Part a (1 mark)
Calculates or states the carrier-to-non-carrier juvenile-output ratio as 2.00.
Acceptable alternatives: Accept scientifically equivalent wording that preserves the assessed biological proposition.
Part a (1 mark)
Links heritable differential reproductive contribution to an increased allele frequency across generations.
Acceptable alternatives: Accept scientifically equivalent wording that preserves the assessed biological proposition.
Part b (1 mark)
Explains that gene-flow direction depends on the unmeasured immigrant allele frequency.
Acceptable alternatives: Accept scientifically equivalent wording that preserves the assessed biological proposition.
Part b (1 mark)
Explains that drift is a chance process whose effect is stronger in a small effective population.
Acceptable alternatives: Accept scientifically equivalent wording that preserves the assessed biological proposition.
Part b (1 mark)
Proposes genotyping immigrants and residents or measuring migrant reproductive contribution.
Acceptable alternatives: Accept scientifically equivalent wording that preserves the assessed biological proposition.
Part b (1 mark)
Proposes effective-size and replicated survival or reproduction monitoring across seasons to distinguish processes.
Acceptable alternatives: Accept scientifically equivalent wording that preserves the assessed biological proposition.
Question 29
(a) DNA is replicated in S phase. Immediately afterwards the cell still has 30 chromosomes, but each chromosome consists of two sister chromatids, so DNA amount has doubled.
(b) Each daughter cell has 30 chromosomes and, excluding new mutation, is genetically equivalent to the parent cell and the other daughter cell because sister chromatids were separated.
Detailed marking criteria
Part a (1 mark)
States that DNA is replicated during S phase while chromosome number remains unchanged.
Acceptable alternatives: Accept scientifically equivalent wording that preserves the assessed biological proposition.
Part a (1 mark)
States that each of the 30 chromosomes has two sister chromatids and DNA amount has doubled.
Acceptable alternatives: Accept scientifically equivalent wording that preserves the assessed biological proposition.
Part b (1 mark)
States that each daughter cell has 30 chromosomes.
Acceptable alternatives: Accept scientifically equivalent wording that preserves the assessed biological proposition.
Part b (1 mark)
Explains that sister-chromatid separation produces two genetically equivalent daughter cells when mutation is excluded.
Acceptable alternatives: Accept scientifically equivalent wording that preserves the assessed biological proposition.
Diagnostic Checklist
| Topic | Questions | Marks | Marks Lost | Action |
|---|---|---|---|---|
| Criterion 3 - Biological inquiry and evaluation of data | Q1, Q2, Q3, Q4, Q5 | 36 | ___ | Review hypotheses, variables, controls, ethics, data interpretation, accuracy, reliability, validity and method evaluation. |
| Criterion 5 - Regulation of biological systems | Q6, Q7, Q8, Q9, Q10, Q11 | 36 | ___ | Review enzymes, photosynthesis, respiration, DNA replication, gene expression, the lac operon and mutation. |
| Criterion 6 - Homeostatic regulation | Q12, Q13, Q14, Q15, Q16, Q17 | 36 | ___ | Review neurons, synapses, hormones, negative feedback, glucose control, thermoregulation and osmoregulation. |
| Criterion 7 - Pathogens and immune responses | Q18, Q19, Q20, Q21, Q22, Q23 | 36 | ___ | Review pathogen types and transmission, innate and adaptive responses, immunity, vaccination and outbreak evidence. |
| Criterion 8 - Biological persistence and diversity | Q24, Q25, Q26, Q27, Q28, Q29 | 36 | ___ | Review reproduction, cell division, inheritance, pedigrees, evolution, isolation and speciation. |