IB Biology command terms. What each one asks for, with marked examples.

• catalysis, e.g. amylase / any named enzyme;
• transport, e.g. haemoglobin (oxygen) / a named membrane pump or channel;
• structure/support, e.g. collagen / keratin;
• hormonal signalling, e.g. insulin;
• movement/contraction, e.g. actin / myosin;
• defence, e.g. antibodies / immunoglobulins;

Mark the first three function and example pairs only; the function and a correct example are both needed for each mark. Accept other valid pairs.

• energy / ATP;
• protein synthesis / ribosomes (and tRNA);
• nucleotides / amino acids / nutrition;
• enzymes for transcription/replication (RNA polymerase / DNA polymerase);

Mark the first two only.

• ligase joins (Okazaki) fragments / seals nicks in the sugar–phosphate backbone;
• primase synthesizes (short) RNA primers that give DNA polymerase a starting point;

• temperature (of the incubator);
• type or mass of flour in each jar / food supply;
• size of the jar / space available;
• age of the beetles when they are paired;
• time the pair is kept together / time allowed before the offspring are counted;
• humidity / light;

Mark the first two answers only.

• work in a dedicated clean room with positive air pressure, separate from where PCR products are handled;
• wear a full body suit, mask and gloves, changed frequently;
• sterilize surfaces and equipment with bleach and UV light;
• remove the outer surface of the bone before sampling;

Mark the first two only.

• during moderate exercise arterial O₂ hardly changes (13.3 to 13.1 kPa), yet ventilation rate rises more than fourfold (7.5 to 34.0 dm³ min⁻¹);
• breathing 15 % O₂ lowers arterial O₂ far more (to 8.4 kPa), yet ventilation rate hardly rises (7.5 to 8.1 dm³ min⁻¹ / by 8 %);
• breathing 5 % CO₂ raises arterial CO₂ (5.3 to 6.4 kPa) and lowers pH (7.40 to 7.33), and ventilation rate rises almost fourfold (to 28.0 dm³ min⁻¹) although arterial O₂ does not fall, OWTTE;

Each point needs the values or a clear comparison from the table. Mark the first two only.

• any two of: auxin / cytokinin / ethylene (ethene) / gibberellin / abscisic acid;

Mark the first two only; both required for the mark.

• found in the members of that clade but not in other groups / a derived trait shared only by that clade;
• inherited from the common ancestor of the clade, not evolved independently in unrelated groups (by convergent evolution);
• complex / made of many parts (e.g. the arrangement of the ear bones or the pattern of cusps on the teeth), so unlikely to have evolved more than once;
• not simply an adaptation to a diet or habitat that unrelated groups also share;
• clearly preserved in hard parts, so it can be seen and compared in the fossil, OWTTE;

Mark the first two features only.

• the hypothalamus stimulates the (anterior) pituitary gland (by secreting a releasing hormone / TRH);
• the pituitary gland secretes thyroid-stimulating hormone / TSH into the blood;
• TSH stimulates the thyroid gland to secrete (more) thyroxin;
• thyroxin raises the metabolic rate / rate of cell respiration in (most) body cells, so more heat is released, OWTTE;

Accept thyroxine.

• chemical: production of toxins / venom / distasteful chemicals, e.g. poison dart frogs with toxic skin, skunks spraying repellent;
• chemical adaptations are often combined with warning coloration (aposematism) so predators learn to avoid the prey, OWTTE;
• physical: hard shells / spines / armour, e.g. tortoise shell, porcupine quills, hedgehog spines;
• physical: camouflage / cryptic coloration or shape, e.g. stick insects, flatfish matching the sea floor;
• physical: mimicry of a harmful species by a harmless one, e.g. hoverflies resembling wasps;
• behavioural: fleeing / rapid escape, living in herds or shoals (dilution effect, more eyes for vigilance), e.g. zebra, sardines;
• behavioural: freezing / playing dead / being active only at night / alarm calls to warn others, OWTTE;

Mark the first three types only where a candidate gives many examples of one type; full marks require at least one chemical, one physical and one behavioural adaptation. Accept any correct named example.

• peel/cut a single thin layer of epidermis, so that light can pass through and the cells lie in one layer;
• place the tissue flat and unfolded on a clean slide;
• add a drop of liquid (water) so that the specimen is mounted in liquid and does not dry out;
• add a drop of iodine solution to stain the tissue / to make nuclei and cell walls visible;
• lower the coverslip from one edge with a mounted needle, at an angle, to exclude air bubbles;
• blot excess liquid from the edge of the coverslip with filter paper;

Award [3 max]. Accept OWTTE throughout. Do not accept 'cut a section' with no reference to thinness.

• extract DNA from pure samples/cultures of each of the common harmless leaf fungi (and from healthy, uninfected ash leaves);
• carry out PCR on each with the same primers and under the same conditions (temperatures, number of cycles, concentrations of reagents and of DNA);
• include *H. fraxineus* DNA in the same run as a positive control, to show that the reaction works;
• include a tube with no DNA as a negative control, to detect contamination;
• separate the products on the same gel beside the ladder: the primers are specific if only the *H. fraxineus* DNA gives a band at 310 base pairs, OWTTE;

• synthesized by ribosomes on the rough endoplasmic reticulum (entering the rER as it is made);
• carried (from the rER) to the Golgi apparatus in vesicles;
• modified/processed and packaged in the Golgi apparatus;
• (secretory) vesicles move to and fuse with the plasma membrane, releasing the protein by exocytosis;

• flow to each organ is adjusted by vasodilation and vasoconstriction of the arterioles supplying it;
• during exercise, arterioles supplying skeletal muscles dilate, increasing delivery of oxygen/glucose for (aerobic) cell respiration;
• arterioles supplying the gut/kidneys constrict, diverting blood away from organs with a lower immediate demand;
• blood flow to the brain remains (relatively) constant in all states;

• fewer receptors are available for acetylcholine to bind to;
• fewer (ligand-gated) channels open in the membrane of the muscle fibre;
• fewer sodium / positive ions diffuse in, so the depolarization / excitatory postsynaptic potential is smaller;
• the depolarization is too small to reach the threshold / to trigger an action potential in some muscle fibres, so no action potential is generated in them;
• fewer muscle fibres are stimulated to contract, so the contraction is weaker;

• the primary structure is the sequence of amino acids (coded by the gene);
• the sequence determines where each type of R-group sits along the chain;
• R-group interactions (hydrogen bonds / ionic bonds / disulfide bridges / hydrophobic interactions) form between particular residues;
• (so) the chain folds into a specific/stable three-dimensional (tertiary) conformation, OWTTE;

• amphipathic molecules form bilayers/vesicles spontaneously in water, so bilayer membranes could have arisen independently in separate lineages;
• the assignment of codons to amino acids is (largely) arbitrary / many other codes would work equally well;
• (so) separate origins of life would be expected to produce different codes;
• the same code in all organisms is therefore most simply explained by inheritance from one ancestral population (LUCA), whereas shared membranes are consistent with it but do not require it, OWTTE;

Links A2.1.5 (self-assembly of bilayers) to A2.1.7. For full marks the answer must address both the membranes and the code.

• burning either fuel is combustion, which releases carbon dioxide into the atmosphere;
• the carbon in the wood was removed from the atmosphere recently by photosynthesis in the trees, so it is already part of the active carbon cycle;
• the replanted trees photosynthesize as they grow and absorb an equivalent quantity of carbon dioxide, so over the whole cycle there is no net addition to the atmosphere;
• the carbon in coal has been held in a long-term store for hundreds of millions of years, so burning it transfers carbon that was not in circulation into the atmosphere;

The justification must refer to both fuels. Accept an answer that also notes the delay while the trees regrow, provided the reasoning above is present.

• decomposition by soil microorganisms is at present very slow in the cold, often waterlogged soils, so dead organic matter and peat accumulate;
• warming raises the rate of decomposition and of soil respiration more than it raises the rate of photosynthesis, so more carbon dioxide is released than is fixed;
• thawing of permafrost exposes organic carbon that has been frozen for thousands of years to decomposers, releasing carbon dioxide and methane;
• warming increases the frequency and severity of forest fires and of insect outbreaks such as bark beetle, which kill trees and release stored carbon by combustion and decay;
• the released carbon dioxide and methane cause further warming, so the change is reinforced by positive feedback;

Answers must link warming to a named process that releases carbon. A bare assertion that 'more carbon will be released' gains no mark.

• the 5′ end carries a (terminal) phosphate group; the 3′ end carries a free hydroxyl group;
• the two strands of the double helix run antiparallel (5′→3′ opposite 3′→5′);
• DNA polymerase can only add new nucleotides to a free 3′-OH, so a new strand is always built 5′→3′;

Beyond this subtopic. Award if given; the marks can be earned without it.

• D1.1 (as a result) synthesis is continuous on one template strand but discontinuous (as Okazaki fragments) on the other, OWTTE;

• a lung cell is a body (somatic) cell, and mitosis passes the mutation only to its daughter cells within that tissue;
• an accumulation of such mutations in a cell can lead to uncontrolled division, i.e. a tumour;
• somatic cells do not contribute DNA to gametes, so the mutation cannot be passed to the next generation, OWTTE;

• people killed eagles directly, e.g. because they were a danger to people / for their feathers;
• burning and clearing of forest and scrub destroyed the eagle's habitat (nesting and hunting areas);
• people also hunted other large birds that the eagle could have eaten instead of moa;
• animals introduced by people (dogs, rats) took eggs or young / competed for prey, OWTTE;

Mark the first two only.

• lymph from the infected hand carries the bacteria/their antigens to the nearest lymph nodes;
• lymphocytes specific to these antigens are activated in these nodes;
• activated lymphocytes divide / form clones, so the number of cells in the node increases and it swells, OWTTE;
• lymph nodes elsewhere receive no antigen from this infection, so no lymphocytes are activated there;

• the four polypeptides would separate / the quaternary structure would be lost;
• each polypeptide would keep its own (secondary and) tertiary structure and its haem group, because the bonds within each chain are unaffected;
• each separated chain would resemble a myoglobin molecule / could still bind one oxygen, OWTTE;

• older / the divergence is overestimated / the split appears to have happened longer ago than it did;
• in the dark the opsin protein has no function, so changes to the opsin gene are no longer harmful / selective pressure on the gene is removed in the cave lineage;
• (so) mutations in the opsin gene of the cave lineage are no longer removed by natural selection and accumulate faster than in the surface species used for calibration;
• there are more differences than expected for the time that has passed, and the clock converts the extra differences into extra time, OWTTE;

Accept "the mutation rate / rate of change increases" for the third point only when it is linked to the loss of selective pressure on the opsin gene in the cave lineage. Award [1 max] for the prediction with no reason.

• white-flowered plants appear in the F1 (some or all of the plants, with any others purple-flowered);
• the white-flowered plants grew from seeds formed by self-fertilization, when the flower's own pollen reached its stigma (pp × pp gives pp);
• the purple-flowered plants grew from seeds of the intended cross, which are all Pp;
• no plant from the intended cross can be white, because every one of its seeds receives P from the purple (PP) parent, OWTTE;

Accept a mixture of purple and white plants, or all white plants, when justified. Award [1 max] for a correct prediction with no reason.

• collect a sample of several individuals from each lake (and one individual of a related species as an outgroup);
• extract DNA and sequence the same gene/region (e.g. a mitochondrial gene) from every individual;
• align the sequences and count the number of base differences between every pair of individuals;
• compare the mean number of differences within each population with the mean number between the two populations;
• use the differences to construct a cladogram, and repeat with a second, independent gene to check the result, OWTTE;

• independent variable: leaf toughness, measured quantitatively as the force needed to push a probe through the lamina (penetrometer), for a range of plant species;
• dependent variable: percentage of the leaf area removed by herbivores, estimated using a transparent grid or by image analysis;
• controlled variables: height and position of the leaf on the plant, age of the leaf, degree of shading/aspect, the same woodland site, and the same period over which damage has accumulated;
• replication: at least ten leaves sampled at random per species, with a mean toughness and a mean percentage damage calculated for each species;
• the relationship between the two means is then tested, e.g. by calculating a correlation coefficient;

Accept other quantitative measures of toughness or of damage. Award marks only where a variable is clearly identified as independent, dependent or controlled.

• use CRISPR-Cas9 with a guide RNA complementary to a sequence within Zx1, so that Cas9 cuts and disrupts the gene;
• carry out the editing in fertilized eggs/early embryos and breed the mice to obtain individuals homozygous for the knockout allele;
• compare the knockout mice with unedited mice of the same strain (the control group), examining kidney structure and function;
• keep age, sex, diet and housing conditions the same for both groups, and use enough individuals for the difference to be tested statistically;
• confirm by sequencing that the gene is disrupted, and attribute any difference between the groups to the loss of Zx1, OWTTE;

• both are forms of nuclear division / both are preceded by DNA replication (S phase);
• both involve spindle microtubules moving chromosomes (through similar phases);
• mitosis has one division, whereas meiosis has two;
• mitosis produces two cells, whereas meiosis produces four;
• mitotic daughter cells are diploid/same ploidy as the parent, whereas meiotic cells are haploid (half);
• mitotic cells are genetically identical, whereas meiotic cells differ (crossing over, random orientation);
• mitosis serves growth/repair/asexual reproduction, whereas meiosis makes gametes/spores for sexual reproduction;

Credit only explicitly comparative points for differences; award converse.

• in both cases humans hunted the birds directly and destroyed their habitat;
• both species were easy to kill: the dodo was flightless and unafraid of people, whereas passenger pigeons nested and roosted in huge, predictable colonies where many could be killed at once, OWTTE;
• the dodo was also affected by introduced species (pigs, rats, monkeys) eating eggs and young, whereas the passenger pigeon was not;
• the passenger pigeon was killed on an industrial scale for city markets, with flocks located by telegraph and the birds shipped by rail, whereas the dodo was taken by small numbers of sailors and settlers;

Award marks only for explicitly comparative statements.

• both are polypeptides built from amino acids joined by peptide bonds;
• in both, the three-dimensional form is determined by the sequence of amino acids (the primary structure);
• both contain secondary structure, α-helix and/or β-pleated sheet, stabilized by hydrogen bonds;
• fibrous proteins are long, narrow and highly repetitive in structure, whereas globular proteins are compact and rounded with an irregular tertiary structure;
• fibrous proteins are usually insoluble in water, whereas globular proteins are usually soluble;
• fibrous proteins usually have structural roles, e.g. collagen and keratin, whereas globular proteins usually have metabolic roles, e.g. haemoglobin and enzymes;
• globular proteins are more readily denatured by heat or a change in pH than fibrous proteins are;

Each point must be an explicitly paired statement. Award converse.

• both supply energy that drives the formation of organic molecules: heat and chemical gradients at the vent, electrical discharge in the apparatus;
• in both, only small, simple molecules such as CH₄, NH₃, H₂, CO₂ and H₂O are present at the start, with no larger organic molecules;
• both are free of molecular oxygen, so any organic molecules formed are not oxidized;
• both have been shown to yield organic molecules, including amino acids;
• in both, water is the medium in which the reactions take place;

Compare requires similarities only; each point must refer to both settings. Do not credit differences such as the catalytic mineral surfaces present at vents.

• both receive low annual precipitation, so liquid water is scarce in both;
• both have sparse, low-growing vegetation and low rates of primary production;
• both experience extreme temperatures for part of the year;
• both support low species diversity compared with forest biomes;
• in both, decomposition is slow: limited by lack of water in the desert and by low temperature in the tundra;
• organisms in both show adaptations for conserving water, since liquid water is either scarce or frozen;

Compare requires similarities only; each point must refer to both biomes. Do not credit differences.

• both are carried in the blood and so reach cells throughout the body;
• both are effective at very low concentrations;
• both bind to a specific receptor whose binding site is complementary in shape to the hormone;
• both act only on cells that possess the appropriate receptor, so cells without it give no response;
• both change the activity of the target cell, e.g. by altering the quantity of a protein that the cell makes;
• both are broken down or removed, so the response does not continue indefinitely;

Compare requires similarities only; each point must refer to both types of hormone. A statement that the receptors are in different locations is a contrast and gains no credit here.

• monosaccharides are single sugar units, e.g. glucose/fructose/ribose;
• disaccharides are two units joined (by a glycosidic bond), e.g. maltose/sucrose/lactose;
• polysaccharides are many units / polymers, e.g. starch/glycogen/cellulose;
• (joined/split by condensation/hydrolysis), OWTTE;

• DNA contains deoxyribose, whereas RNA contains ribose;
• DNA contains thymine, whereas RNA contains uracil (in its place);
• DNA is (usually) double-stranded / a double helix, whereas RNA is single-stranded;
• DNA molecules are (generally) longer than RNA molecules, OWTTE;

Comparative phrasing required; award converse.

• excretion is the removal from the body of (toxic) waste products of metabolism, such as urea;
• osmoregulation is the control of the osmotic concentration (of water and solutes) of the blood / body fluids;
• excretion removes substances made by the body's own cells, whereas osmoregulation adjusts the amounts of water and salts, which mostly come from food and drink, OWTTE;

Do not accept 'removal of water' alone as osmoregulation.

• the biological species concept groups organisms that interbreed / breed together to produce fertile offspring;
• these dandelions do not interbreed: seeds form without fertilization, so genes are not exchanged between lines;
• (so) it could not: every line is reproductively isolated and would count as a separate species / the concept gives no way of grouping lines, OWTTE;

A conclusion that the concept cannot be used earns no mark without a reason.

• the thin filaments at both ends of the sarcomere slide towards its centre, so each half shortens by 0.25 µm and one thin filament slides 0.25 µm / 250 nm;
• 0.25 ÷ 0.010 = 25 power strokes;
• 25 ATP, because one ATP is hydrolysed in each cross-bridge cycle;

A candidate who divides the whole 0.5 µm by 0.010 µm (50 power strokes) can score the ATP mark by ECF (50 ATP).

• when the plasma membrane is just beginning to pull away from the wall, the protoplast no longer presses on the wall, so ψp = 0;
• there is no net movement of water, so the water potential of the cells equals that of the solution, −1100 kPa;
• ψs = ψw − ψp = −1100 − 0 = −1100 kPa;
• cells vary: those that plasmolysed had a less negative ψs than −1100 kPa and those that did not had a more negative ψs, so −1100 kPa is a mean value;
• the value is slightly more negative than the ψs of the cells before immersion, because the water lost in reaching this point concentrated the cell sap, OWTTE;

• a eukaryotic pre-mRNA contains coding exons separated by non-coding introns;
• during processing, introns are removed and exons joined together (splicing) to make the mature mRNA;
• in alternative splicing, different combinations or subsets of exons are joined from the same pre-mRNA;
• (so) one gene can yield several different mature mRNAs and hence several different polypeptides;
• the number of possible proteins can therefore greatly exceed the number of genes;
• different cell types or conditions splice the same transcript differently (a regulated process), further expanding the proteome, OWTTE;

• variant G reaches half of Vmax only at a much higher substrate concentration (6.8 vs 2.1 mmol dm⁻³), so it binds substrate less tightly / has a lower affinity;
• the two Vmax values are almost equal (~47–48), so the maximum catalytic rate is essentially unchanged, OWTTE;

• the coastal town has a much smaller daily temperature range (than the inland town);
• (consistent with the idea that) the sea warms and cools slowly / buffers/moderates the surrounding air temperature, OWTTE;

• the three examples show eukaryotic cells with no nucleus and with many nuclei, so the statement is not correct as a general rule;
• the statement does describe the typical eukaryotic cell, and the great majority of eukaryotic cells do contain a single nucleus;
• the exceptions are specialized cells in which the loss or multiplication of nuclei is related to function, e.g. loss of the nucleus leaves more room for haemoglobin in a red blood cell, and a long muscle fibre formed by cell fusion needs many nuclei to control its large volume of cytoplasm;
• a better statement would be that eukaryotic cells usually contain a nucleus, with named exceptions, OWTTE;

Both an interpretation of the examples and a judgement on the statement are needed for full marks.

• inland and dune soils may differ (e.g. in drainage, salinity or nutrients), which could affect marram growth independently of competition;
• this could be dealt with by also transplanting marram into competitor-free inland soil / analysing and matching the soils / using a reciprocal transplant design, OWTTE;

• before the severe drought the two groups differ very little (1.0 compared with 1.3 arbitrary units), so priming does not simply leave the gene switched on;
• after 24 hours of severe drought the primed plants contain almost three times as much mRNA as the untreated plants (11.6 compared with 4.2 arbitrary units);
• both groups increase expression in response to drought, but the primed plants respond far more strongly;
• this means the earlier mild drought changed the way the gene responds weeks later, a memory of the first drought is retained, and the stronger response is associated with much higher survival (78 % against 31 %);

A value quoted with its unit is required for the first mark. Both a description and an interpretation are needed.

• a knockout removes the product of one gene only, so a change in phenotype can be attributed to that gene, giving direct evidence of its function;
• gene editing with CRISPR-Cas9 allows a chosen sequence to be targeted precisely, and knockouts can be produced quickly in organisms with short life cycles;
• unedited organisms of the same strain provide a control, so the comparison is valid;
• however, other genes may have overlapping functions, so a knockout may produce no visible phenotype even when the gene is important (functional redundancy);
• knocking out an essential gene may kill the embryo, so any later function of the gene cannot be observed;
• loss of one gene can alter the expression of others, so the phenotype seen may be an indirect consequence rather than the normal role of the gene;
• conclusion: the approach yields useful, testable evidence but its results need confirming by other methods, such as measuring where and when the gene is normally expressed, OWTTE;

A judgement is required for the final mark. Both strengths and weaknesses must appear.

• it is true that every animal is heterotrophic: animals cannot make their own carbon compounds from inorganic sources, so they obtain them from other organisms;
• the tapeworm is heterotrophic too, as it takes its carbon compounds (glucose and amino acids) from the rat's food;
• most animals are holozoic: they ingest food, digest it internally, then absorb and assimilate the products;
• but the tapeworm has no mouth or gut, so it does not ingest food or digest it internally / its food has already been digested by the rat;
• it only absorbs small molecules across its body surface (and assimilates them), so its nutrition is not holozoic;
• so the first part of the claim is correct but the conclusion does not follow: being heterotrophic does not make an animal holozoic, and the claim is false, OWTTE;

Full marks require a judgement on the claim. Award a maximum of two marks if the tapeworm is not used as evidence.

• (for) while a sufficient percentage of the population is immune, transmission is greatly impeded, so the unvaccinated child is unlikely to meet the pathogen;
• (against) if many parents reason this way, the proportion immune falls below the level needed for herd immunity;
• transmission chains are then re-established and outbreaks can occur, putting the child and others at risk;
• members of a population are interdependent: people who cannot be vaccinated (e.g. infants, the immunocompromised) rely on others being immune;
• the child remains susceptible and has no memory cells if exposed, e.g. when travelling to an area with the disease, OWTTE;

For full marks the answer must include at least one point for and one point against the argument.

• benefit: the enzymes and substrates of the Krebs cycle are concentrated together in the small volume of the matrix, so the reactions are faster;
• benefit: the inner membrane is folded into cristae, giving a large area for electron transport chains and ATP synthase;
• benefit: the intermembrane space is small, so a proton gradient builds up quickly across the inner membrane;
• benefit: these reactions are kept separate from incompatible processes in the cytoplasm, OWTTE;
• cost: substrates and ADP must be moved into the mitochondrion and ATP moved out across its membranes, which needs transport proteins / energy;
• cost: the cell must make and maintain two extra membranes for every mitochondrion, OWTTE;
• judgement: the much faster / greater production of ATP outweighs the costs of transport and of making the membranes, OWTTE;

A balanced answer needs at least one cost and a judgement for [5]; benefits alone score a maximum of [4].

• benefit: allows (otherwise) infertile couples to have children, e.g. blocked oviducts / low sperm counts;
• (benefit developed) embryos can be screened for serious genetic disease before transfer;
• concern: surplus embryos are created and may be discarded (ethical objections) / multiple pregnancies from multiple transfers;
• concern: cost/access inequality / emotional and physical burden of hormone treatment, OWTTE;

At least one benefit AND one concern for full marks.

• for: the basis of classification changed from overall (morphological) similarity to common ancestry;
• for: a group is now accepted only if it contains an ancestor and all of its descendants, so some long-accepted taxa were redefined or abandoned;
• for: a fixed series of ranks gave way to unranked, nested clades, since ranks are arbitrary and do not reflect the gradation of variation;
• for: new methods (comparison of base sequences, parsimony analysis) and new questions (is this group a clade?) replaced the old ones;
• against: the traditional ranks and names are still widely used alongside clades, so the old framework was not fully replaced;
• against: the change happened gradually over decades rather than as a sudden revolution / many traditional taxa (e.g. mammals) turned out to be clades and were kept;
• against: the underlying theory of common descent did not change, cladistics applied it more consistently, OWTTE;

Award [3 max] if only one side is considered. Accept a reasoned conclusion either way.

• (0.33 − 0.15) ÷ 0.15 (× 100);
• 120 %;

Award [1] for 220 % or for 2.2 times the length.

• N = (84 × 63) ÷ 12;
• = 441 (woodlice);

Accept 440–441. Award ECF for correct method.

• q² = 1/10 000 = 0.0001, so q = √0.0001 = 0.01 (and p = 0.99);
• carrier frequency 2pq = 2 × 0.99 × 0.01 = 0.0198 ≈ 2 % (about 1 in 50);
• carriers (≈1 in 50) far outnumber affected individuals (1 in 10 000), so most recessive alleles are hidden in unaffected heterozygotes;
• (so) preventing affected individuals from reproducing would remove the allele extremely slowly / barely change its frequency, because the carriers are unaffected and continue to pass it on, OWTTE;

Hardy–Weinberg: p + q = 1, q = √q²; numerically exact.

• (132 - 100) / 4;
• 8 (repeats);

Award [2] for the correct answer with no working. ECF.

• untreated slice: 70 µm (accept 65 to 75 µm);
• slice treated with the drug: 175 µm (accept 165 to 185 µm);

• reads from the diagram that 40 eyepiece divisions align with 0.5 mm of the stage micrometer;
• converts to 500 µm and divides by 40;
• 12.5 (µm);

Award full marks for a correct answer with no working. Accept 12.5 µm per division. ECF from a misread alignment.

• no inhibitor: 2 mmol dm⁻³ and inhibitor P: 6 mmol dm⁻³ (both needed, accept ±0.5);
• P competes with the substrate for the active site, so more substrate is needed to occupy the same proportion of active sites / to outcompete the inhibitor;
• (so) P is a competitive inhibitor, OWTTE;

• epinephrine shown binding to a (G protein-coupled) transmembrane receptor on the outer surface of the plasma membrane;
• G protein shown alongside the receptor on the inner side of the membrane, activated by exchanging GDP for GTP;
• activated G protein shown activating adenylyl cyclase (in the membrane);
• adenylyl cyclase shown converting ATP to cyclic AMP (cAMP), the second messenger, in the cytoplasm;
• cAMP shown activating protein kinase A, which (phosphorylates and) activates enzymes for glycogen breakdown / release of glucose;

Accept a schematic with boxes and arrows provided the membrane is shown and the components are in the correct compartments (epinephrine outside, cAMP and kinase inside).

• testis, annotated as producing sperm (by meiosis) and secreting testosterone;
• epididymis, annotated as the site where sperm mature and are stored;
• sperm duct (vas deferens), annotated as carrying sperm from the epididymis to the urethra during ejaculation;
• seminal vesicle and/or prostate gland, annotated as secreting fluid (containing nutrients / alkaline fluid) that makes up most of the semen;
• urethra, annotated as carrying semen (and urine) through the penis to the outside;
• penis (erectile tissue) and/or scrotum, annotated as delivering semen into the vagina / holding the testes below core body temperature;

Each mark needs a correctly placed label and a correct function. Positions must be anatomically reasonable: testis in the scrotum, epididymis on the testis, sperm duct looping up to join the urethra.

• each backbone drawn as alternating pentagons (sugar) and circles (phosphate), with each phosphate joining the sugar of one nucleotide to the sugar of the next;
• bases drawn as rectangles, each attached to a sugar (pentagon) and pointing inwards towards the other strand;
• the two strands drawn antiparallel: one strand inverted relative to the other, so that the sugar–phosphate sequence runs in opposite directions (5′ and 3′ ends may be labelled);
• three base pairs shown, correctly paired as adenine with thymine and guanine with cytosine (at least one A–T and one G–C);
• hydrogen bonds shown as dashed or dotted lines between the paired bases only, not as part of the backbone;
• (bases named, or initials A, T, G, C used consistently);

A1.2.2 requires circles, pentagons and rectangles for phosphates, sugars and bases, so do not award the first mark for a plain line drawn as the backbone. The helical shape is not required (A1.2.6). Students are not required to show the numbers of hydrogen bonds, so accept any number of dashed lines per pair. Do not accept bases joined directly to each other without a backbone.

• axes labelled: pH of the seawater on the horizontal axis, with the direction of the scale shown, and rate of calcification on the vertical axis;
• the rate of calcification is highest at the highest (present-day) pH;
• the rate falls as pH falls / as the seawater becomes more acidic;
• the line reaches zero, or becomes negative (net dissolution of calcium carbonate), at the lowest pH values;

Accept a straight line or a curve. No numerical values are required; credit shape and trend only.

• axes labelled: relative humidity of the air / % on the horizontal axis and rate of transpiration (or water uptake) per unit time on the vertical axis;
• the rate is highest at the lowest relative humidity;
• the rate falls as relative humidity rises;
• the line reaches, or approaches, zero as relative humidity approaches 100 %;

Accept a smooth curve or a straight line of negative gradient. No numerical values are required; credit shape and trend only.

• δ− on the oxygen atom and δ+ on the hydrogen atom of the hydroxyl group;
• δ− on the oxygen atom and δ+ on both hydrogen atoms of W;

Both hydrogen atoms of W must be marked δ+ for the second mark.

• gametes of Rr parent shown as R and r; gametes of rr parent shown as r and r;
• grid correctly filled to give offspring genotypes Rr, Rr, rr, rr: a 1:1 ratio of round (Rr) to wrinkled (rr);