2025 OCR A-Level Biology B (Advancing Biology) - H422 Practice Paper with Answers

Convert units: 36 mm = 36 000 \(\mu\)m. Magnification = image size / actual size = 36 000 / 12 = \(\times 3000\).

C is correct (1 mark). A, B and D arise from errors in unit conversion (mm to \(\mu\)m) or from inverting the equation. No half marks.

Rf = distance moved by solute / distance moved by solvent = 4.2 / 7.0 = 0.60.

B is correct (1 mark). C results from inverting the ratio; A and D are distractors from misreading the distances. Rf has no units.

By Chargaff's rules A = T, so A + T = 56%. G + C = 44%, and G = C, so cytosine = 22%.

D is correct (1 mark). A confuses cytosine with adenine; B gives G + C; C gives 100 - 28.

At about 95 \(^\circ\)C the strands separate. Cooling to about 55 \(^\circ\)C allows primers to anneal by hydrogen bonding to complementary base sequences. Extension by Taq polymerase occurs at about 72 \(^\circ\)C.

A is correct (1 mark). B describes the denaturing stage; C describes extension; D is not part of PCR.

In carbon fixation, RuBisCO catalyses the combination of CO2 with ribulose bisphosphate (RuBP), forming two molecules of glycerate 3-phosphate (GP). GP is then reduced to TP using ATP and reduced NADP, but this step is not catalysed by RuBisCO.

C is correct (1 mark). A is the light-dependent reaction; B misassigns the enzyme; D describes ATP production in the light-dependent reaction.

Companion cells use ATP to actively load sucrose into the sieve tube. This lowers the water potential, so water enters by osmosis and the hydrostatic pressure rises, driving mass flow towards sinks.

D is correct (1 mark). A, B and C each contain at least one incorrect statement about mechanism, ATP use or water potential.

In Type 2 diabetes, insulin is often still produced but target cells such as liver and muscle respond less to it, so blood glucose remains high. Autoimmune destruction of beta cells is characteristic of Type 1.

B is correct (1 mark). A describes Type 1; C is wrong because beta cells, not alpha cells, secrete insulin; D is incorrect.

After water uptake, the embryo produces gibberellin, which stimulates the aleurone layer to make amylase. Amylase hydrolyses stored starch in the endosperm to sugars that supply the growing embryo.

A is correct (1 mark). B, C and D each include a wrong source, tissue, enzyme or reaction type (condensation instead of hydrolysis).

Convert units: 36 mm = 36 000 µm. Magnification = image size / actual size = \(36000 \div 12 = 3000\), so ×3000.

B (×3000) = 1 mark. A: unit conversion error (factor of 10). C: multiplied rather than divided. D: inverted the formula. Only one correct response.

Volume delivered = 14.80 − 2.30 = 12.50 cm³. Two readings were taken, so total uncertainty = ±0.10 cm³. Percentage uncertainty = \(0.10 \div 12.50 \times 100 = 0.8\%\).

B (0.8%) = 1 mark. A: used a single ±0.05 uncertainty. C: used the wrong volume or total. D: decimal place error.

Adenine pairs with thymine, so thymine is also 22%; A + T = 44%. The remaining 56% is cytosine plus guanine, which are equal in amount, so cytosine = 28%.

B (28%) = 1 mark. A: assumed equal to adenine. C: gave A + T. D: gave C + G.

At about 95 °C the hydrogen bonds between complementary bases break, separating the strands. Primers anneal at a lower temperature (about 55 °C) and extension by a thermostable DNA polymerase occurs at about 72 °C.

A = 1 mark. B and C swap the temperatures for annealing and extension. D confuses denaturation with annealing.

Without light, ATP and reduced NADP are no longer produced. GP cannot be reduced to triose phosphate, so GP accumulates. RuBP is not regenerated but continues to be used in carbon fixation by RuBisCO, so RuBP falls.

A = 1 mark. B is the reverse (this pattern occurs when CO2 is removed). C and D do not match both consequences.

Companion cells actively load sucrose into sieve tubes. This lowers the water potential, water enters by osmosis, and the raised hydrostatic pressure at the source drives mass flow towards the sink, where pressure is lower.

A = 1 mark. B: water moves by osmosis, not active transport. C: translocation is mass flow, not driven by transpiration pull. D: water is not actively removed by companion cells.

On imbibition, the embryo produces gibberellin, which stimulates the aleurone layer to make amylase. Amylase hydrolyses stored starch in the endosperm to sugars used by the growing embryo.

A = 1 mark. B, C and D each contain an incorrect source, site or substrate.

The protein is synthesised on ribosomes at the rough ER, packaged into vesicles to the Golgi apparatus for modification, then packaged into secretory vesicles which fuse with the cell-surface membrane by exocytosis.

A = 1 mark. B reverses RER and Golgi. C includes a lysosome, which is not on the secretion pathway. D includes SER and nucleus, which are not part of the pathway.

Each step is a 1 in 10 dilution. Four steps give a dilution factor of \(10^{4}\). Original concentration = \(2.4 \times 10^{4} \times 10^{4} = 2.4 \times 10^{8}\) cells per cm\(^3\).

C (1 mark). Dilution factor \(10^4\) identified (method) and applied correctly (accuracy). Distractors: A/B from too few multiplications, D from five.

Actual size = image size / magnification = 24 mm / 12 000 = 0.0020 mm = 2.0 \(\mu\)m.

B (1 mark). Method: rearranging magnification = image size / actual size and converting mm to \(\mu\)m. D results from multiplying instead of dividing.

By Chargaff's rules A = T, so thymine is 30%. A + T = 60%, leaving 40% for G + C. G = C, so cytosine is 20%.

A (1 mark). Uses A = T and G = C. B ignores pairing, C gives G + C combined, D is 100% minus A.

DNA ligase seals the sugar-phosphate backbones of the plasmid and inserted gene by catalysing phosphodiester bond formation, producing recombinant plasmid. A describes a restriction enzyme, B reverse transcriptase.

C (1 mark).

Without light, ATP and reduced NADP are no longer made. GP cannot be reduced to triose phosphate, so GP accumulates. Triose phosphate is not available to regenerate RuBP, but RuBP continues to react with carbon dioxide, so RuBP falls.

B (1 mark). Both changes needed.

Active loading of sucrose lowers the water potential in the sieve tube. Water enters by osmosis from the xylem, raising hydrostatic pressure, which drives mass flow towards sinks.

D (1 mark). Other options misstate the loading mechanism or the direction of water movement.

Low blood glucose is detected by pancreatic islet cells. Alpha cells release glucagon, which binds to liver cell receptors and activates glycogenolysis, raising blood glucose (negative feedback).

A (1 mark). B has the wrong cell type, C and D name insulin.

RQ = carbon dioxide produced / oxygen consumed = 30 / 40 = 0.75.

D (1 mark). C results from inverting the ratio.

Percentage error = uncertainty / measured value x 100 = \(0.5 / 8.0 \times 100\) = 6.25%. A omits the x100; C uses the 1.0 cm\(^3\) range of the uncertainty; D is not derived from a valid calculation.

B (1 mark). Correct working: 0.5 / 8.0 x 100 = 6.25% (accept 6.3%). No mark for any other option.

Rf = distance moved by solute / distance moved by solvent = 5.6 / 8.0 = 0.70. B inverts the ratio; D uses the distance between spot and solvent front (2.4 cm) divided by 8.0.

C (1 mark). Rf = 5.6 / 8.0 = 0.70. Rf has no units.

By Chargaff's rules A = T = 132, so A + T = 264. G + C = 600 - 264 = 336. G = C so C = 168. C is the total G + C, not C alone; A ignores base pairing; B is not derived from a valid calculation.

D (1 mark). Method: A = T = 132; G + C = 336; C = 336 / 2 = 168.

Annealing occurs at about 55 \(^\circ\)C when primers form hydrogen bonds with complementary sequences. B is wrong because denaturation breaks hydrogen bonds, not phosphodiester bonds. C is wrong because a heat-stable DNA polymerase, not ligase, extends the primers. D gives the wrong temperature.

A (1 mark). No mark for any other option.

Volume = \(\pi r^2 l = \pi \times 0.50^2 \times 36 = 28.3\) mm\(^3\). Rate = 28.3 / 10 = 2.83 mm\(^3\) min\(^{-1}\). A omits \(\pi\); B uses r = 1.0 mm; C is the volume, not a rate.

D (1 mark). Method: volume = \(\pi r^2 l\) = 28.3 mm\(^3\); divide by 10 min = 2.83 mm\(^3\) min\(^{-1}\).

Actual size = image size / magnification = 24 mm / 12 000 = 0.002 mm = 2.0 \(\mu\)m. A inverts the calculation; B and D contain unit-conversion errors of a factor of ten.

C (1 mark). Method: 24 / 12 000 = 0.002 mm; x1000 = 2.0 \(\mu\)m.

(a)(i) Subtract the flanking sequence: 408 - 96 = 312 bp. Divide by repeat length: 312 / 24 = 13 repeat units.
(a)(ii) Only R has fragments of 408 bp and 264 bp, identical to S. P differs in the smaller fragment (312 bp) and Q differs in the larger fragment (432 bp). Other individuals in the population could have the same two alleles at this one locus, so several loci must be compared to reduce the probability of a chance match.
(b) PCR: DNA is heated to about 95 degrees C to break hydrogen bonds and separate the strands; cooled to about 55 to 65 degrees C so that primers anneal to the flanking sequences either side of the VNTR; heated to about 72 degrees C so that heat-stable DNA polymerase (e.g. Taq) extends the primers using free nucleotides; the cycle is repeated so the number of copies roughly doubles each time. Electrophoresis: products are loaded into wells in an agarose gel in buffer; a potential difference is applied; DNA is negatively charged (phosphate groups) so moves towards the positive electrode; smaller fragments move faster and further through the gel mesh; bands are made visible with a DNA stain or labelled primers; a DNA size ladder is run for comparison; samples with bands at identical positions match.

(a)(i) Answer line first: 13 = 2 marks.
Method mark: 408 - 96 (= 312) / 312 / 24 [1]
Accuracy mark: 13 [1]
ALLOW 1 mark for 17 (408 / 24) or 12 (288 / 24 from subtracting the wrong value) only if working shows the method of dividing by 24; otherwise no ECF.

(a)(ii) R only [1] (DO NOT ALLOW P or Q) AND
other people / unrelated individuals may have the same two alleles / same fragment sizes at this locus; need to analyse more loci / probability of a match is too high [1]
IGNORE 'the test could be wrong' without reference to chance matches.

(b)* Level of response, 5 marks maximum (total for question 9).
Level 3 (5 marks): Gives a detailed, correct account of all three PCR stages with temperatures or the role of primers and heat-stable polymerase AND explains electrophoresis (charge, movement towards the positive electrode, size-dependent movement, visualisation, comparison of band positions). Ideas are linked logically.
Level 2 (3-4 marks): Describes PCR with at least two correct stages and some detail, AND gives some explanation of electrophoresis (e.g. smaller fragments move further) but with some omission or lack of linkage.
Level 1 (1-2 marks): Makes one or two correct points about PCR OR electrophoresis with little explanation.
Level 0 (0 marks): No creditworthy response.

Indicative content (non-exhaustive):
- heat to about 95 degrees C to denature / separate strands (break hydrogen bonds)
- cool to about 55-65 degrees C so primers anneal to flanking sequences either side of the VNTR
- primers complementary to the ends of the target sequence
- heat to about 72 degrees C; Taq / heat-stable DNA polymerase extends primers using free nucleotides
- repeated cycles double the amount of DNA each cycle (exponential)
- DNA placed in wells in agarose gel with buffer; potential difference applied
- DNA negatively charged (phosphate groups) so moves to the positive electrode
- smaller fragments move faster / further through the gel mesh; fragments separate by size
- bands stained / fluorescent label to make visible; size marker (ladder) run alongside
- bands at the same distance indicate fragments of the same length so samples can be matched
AVP for other valid points.

(a) Primers anneal by complementary base pairing to the 3' ends of the two strands flanking the target, giving DNA polymerase a starting point. Taq polymerase comes from a thermophilic bacterium and is heat stable, so it is not denatured during the strand-separation step at about 95 C, and does not have to be added again every cycle.
(b)(i) The control shows whether the reagents are contaminated with DNA; a band in the control would mean a false positive.
(b)(ii) Copies = 40 x 2^12 = 40 x 4096 = 163 840 = 1.6 x 10^5 (2 s.f.).
(c)(i) Let A = 450 bp and a = 300 bp. Mother has both bands, so Aa. Father has only 300 bp, so aa.
(c)(ii) Child 2 has only 300 bp, so is aa. This is possible because the mother can pass on a and the father passes on a. Child 1 has only 450 bp, so is AA. The mother can give A but the father has no A allele, so Child 1 cannot be the biological child of both. The statement is not supported for Child 1 (assuming no new mutation or sample mix-up).
(d) DNA has negatively charged phosphate groups, so it moves towards the positive electrode (anode) in the electric field. The gel is a mesh of agarose that slows the movement of molecules. Shorter fragments (300 bp) meet less resistance than longer fragments (450 bp), so in the same time they travel further towards the positive electrode.
(e) Indicative content: reverse transcriptase makes cDNA from mRNA; second strand made with DNA polymerase; restriction enzyme cuts the plasmid (and the gene is flanked by matching sites) leaving complementary sticky ends; DNA ligase joins the gene into the plasmid by forming phosphodiester bonds; bacteria made permeable (e.g. heat shock with calcium ions) so they take up plasmids; plasmid carries a marker gene such as antibiotic resistance; grow bacteria on agar containing the antibiotic so that only bacteria with the plasmid survive; if the gene is inserted in a second resistance gene, replica plate to find colonies that lose that resistance, which contain the recombinant plasmid; or a fluorescent/reporter marker.

(a) Primers: bind/anneal to (complementary) sequences at the ends of the target (DNA) / allow DNA polymerase to start synthesis (1). Taq polymerase is heat stable / does not denature at the high temperature used to separate strands (1). ALLOW 'thermostable'. IGNORE 'works at high temperature' without reference to denaturation.

(b)(i) To show whether the reagents/equipment are contaminated with DNA / to check that any product comes from the sample template (1). ALLOW 'negative control'. IGNORE 'to compare' alone.
(b)(ii) Check the answer line first. 1.6 x 10^5 = 2 marks. Method mark: 40 x 2^12 OR 163 840 seen (1). If the answer is not in standard form or is not rounded to 2 s.f., award 1 mark (163 840 or 1.64 x 10^5). DO NOT ALLOW 40 x 2 x 12 (= 960).

(c)(i) Mother Aa AND father aa (1). Both required. ALLOW any consistent symbol pair only if it is clear which band is which, but the question asks for A and a.
(c)(ii) Child 2 is aa so consistent / can inherit one a from each parent (1). Child 1 is AA but the father has no A allele (only aa), so Child 1 cannot be the child of both (1) AW. ALLOW ECF from (c)(i).

(d) DNA is negatively charged / has phosphate groups (1). Moves towards the positive electrode / anode (1). Smaller (shorter) fragments move through the agarose gel mesh faster / are hindered less, so move further in the same time (1) ORA. IGNORE 'lighter' without reference to size or resistance.

(e)* Level 3 (5-6 marks): a logical, coherent description covering both stages: making the recombinant plasmid (cDNA production using reverse transcriptase AND cutting/joining with named restriction enzyme and ligase, with sticky ends) AND uptake and identification of transformed bacteria (marker gene/selection with correct reasoning). Detail and correct sequence throughout.
Level 2 (3-4 marks): describes most of the stages of making the recombinant plasmid and attempts identification of bacteria, but with some gaps, such as omitting sticky ends or the reason selection works.
Level 1 (1-2 marks): a few relevant individual points with little linking; limited sequence or only one stage described.
Level 0 (0 marks): no response or no creditworthy content.
Indicative content (non-exhaustive, AVP): reverse transcriptase makes single-stranded cDNA from the mRNA; DNA polymerase makes the double strand; restriction enzyme cuts plasmid and gene ends at the same recognition site / leaves complementary sticky ends; DNA ligase joins sugar-phosphate backbones / forms phosphodiester bonds; bacteria treated (e.g. calcium ions and heat shock) so that they take up the plasmid; plasmid has an antibiotic resistance marker gene; culture on agar with the antibiotic so only transformed bacteria grow; use of a second marker/replica plating/fluorescent marker to find those with the inserted gene. DO NOT ALLOW 'cutting with ligase' or 'restriction enzymes join DNA'.

(a)(i) The phosphate groups in the DNA backbone are negatively charged, so the fragments are attracted towards the positive electrode.
(a)(ii) The ladder shows that each doubling of fragment size reduces the distance moved by 10 mm (48, 38, 28, 18). Plant 1 moved 33 mm, which lies between 38 mm (200 bp) and 28 mm (400 bp), halfway. Halfway on the log scale means one half of a doubling: size = 200 x 2^(5/10) = 200 x 1.414 = 283 bp, so about 280 bp.
(a)(iii) Plant 3 has two bands. One band at 28 mm is 400 bp, the same as plant 2, which is the full-length allele. The other band at 33 mm is the same as plant 1, about 280 bp, the allele with the deletion. So plant 3 has one of each allele, and is heterozygous.
(b) Denaturation at about 95 C breaks the hydrogen bonds so the DNA strands separate. Annealing at about 55 C lets primers bind to complementary sequences. Extension at about 72 C, where heat-stable DNA polymerase (Taq) adds nucleotides to the 3' end of each primer to make new complementary strands.
(c) Isolate the gene by using reverse transcriptase to make cDNA from mRNA (or by cutting it out using restriction enzymes). Cut the gene and the plasmid with the same restriction enzyme to give complementary sticky ends. Mix them and join with DNA ligase to make recombinant plasmid, including a promoter, and a marker gene. Bacteria are made to take up the plasmid (for example by heat shock with calcium ions). Identify transformed bacteria using the marker gene, for example antibiotic resistance. Culture them in a fermenter so that they transcribe and translate the gene, then purify the protein.

(a)(i) Phosphate groups (in the sugar-phosphate backbone) are negatively charged / DNA is negatively charged (1) AND attracted to the positive electrode / anode (1). ALLOW opposite charges attract. DO NOT ALLOW DNA is polar.
(a)(ii) Answer 280 (bp) (accept 283 or 284 or 285 or 282 or 281 for 3 marks; accept range 270-290 for 2 marks if working shown). CHECK ANSWER LINE FIRST. Method: recognises each 10 mm = a doubling / halving of size, or plots log size against distance (1); 33 mm lies halfway between 200 and 400 bp, or 5 mm = half a doubling (1); 200 x 2^0.5 (1) = 283 so 280 bp. ALLOW an answer read from a correctly drawn log graph within 270-290 for 2 marks. Allow 300 bp for 2 marks (halfway on a linear scale, giving 300).
(a)(iii) Two bands present in plant 3 (1); AND these match the 400 bp allele (plant 2) and the shorter allele (plant 1) / two different alleles / one of each allele present (1). IGNORE just 'it has two bands'.
(b) Any three from: heated to about 95 C (accept 90-98) to separate / denature the strands by breaking hydrogen bonds (1); cooled to about 55 C (accept 50-65) so primers anneal / bind to complementary sequence (1); heated to about 72 C (accept 70-75) (1) for DNA polymerase / Taq polymerase to extend primers by adding complementary nucleotides (1); enzyme is heat-stable / does not denature at 95 C (1). Max 3.
(c) Levels of response, 6 marks.
Level 3 (5-6 marks): Detailed, logically ordered description covering all three areas (isolation of gene, insertion into plasmid with correct enzymes, and expression / identification in bacteria). Specific terms used correctly, including restriction enzyme, sticky ends and ligase, with correct link between steps.
Level 2 (3-4 marks): Describes most stages with some correct technical detail, but is missing one area or has gaps in the logic or detail.
Level 1 (1-2 marks): A few relevant points, general or poorly sequenced, with little technical detail.
0 marks: no creditworthy response.
Indicative content: mRNA from the human cells (e.g. pancreatic beta cells) and reverse transcriptase to make cDNA OR use restriction enzymes to cut out the gene; same restriction enzyme cuts the plasmid (vector) giving complementary sticky ends; DNA ligase joins the sugar-phosphate backbones / forms phosphodiester bonds; recombinant plasmid includes promoter and marker gene such as antibiotic resistance; bacteria take up the plasmid (calcium chloride, heat shock, electroporation); select transformed bacteria on medium containing antibiotic / using marker gene; culture in a fermenter; bacteria transcribe and translate the gene to make the protein (insulin); protein is extracted and purified. AVP.

(a)(i) Reverse transcriptase makes complementary DNA from mRNA. Using mRNA from beta cells means the sequence has introns already removed, so bacteria (which cannot splice) can express it. (ii) One enzyme leaves identical complementary single-stranded sticky ends on plasmid and gene, so they anneal by base pairing; ligase catalyses formation of phosphodiester bonds between adjacent nucleotides, sealing the sugar-phosphate backbone. (b) Copies = 3 x 2^8 = 3 x 256 = 768. (c)(i) Plasmid 2 gives a 1.8 kb fragment equal to the insert, plus the 4.2 kb vector fragment; Plasmid 1 has no insert. Smaller 1.8 kb fragments move through the gel pores faster and further from the wells than the 4.2 kb fragment. (ii) DNA is negatively charged because of its phosphate groups, so it is attracted to the anode. (d) Transformed bacteria are identified using a marker gene on the plasmid, e.g. antibiotic resistance, or a fluorescent marker; insertion into a second resistance gene can show which colonies carry the insert. Selected bacteria are cultured in a fermenter under optimum conditions, and insulin is extracted and purified. Evaluation: human insulin sequence so less immune rejection, large scale, consistent, faster, no animal slaughter, ethical or religious objections avoided; but costly equipment, purification needed, and some patients prefer or respond to animal insulin.

(a)(i) reverse transcriptase ✓; mRNA already has introns removed / spliced, so bacteria can express it (DO NOT ALLOW 'mRNA is easier to get' alone) ✓ [2]
(a)(ii) same enzyme produces complementary / matching sticky ends on both, which can anneal by base pairing ✓; ligase forms phosphodiester bonds / joins sugar-phosphate backbones (ALLOW seals the nicks) ✓ [2]
(b) Check answer line first. 768 ✓✓. If wrong, 3 x 2^8 OR 3 x 256 ✓ (method). ALLOW 1 method mark for 2^8 = 256 only. [2]
(c)(i) Plasmid 2 / lane 3 ✓; it has a 1.8 kb fragment (equal to the insert size) / two bands whereas plasmid 1 has only the vector band ✓. ALLOW reference to the 1.8 kb band matching the size of the insert [2]
(c)(ii) DNA is negatively charged / phosphate groups negative ✓ [1]
(d)* Level 3 (5-6 marks): clear description of identification AND production AND a balanced evaluation with at least two advantages and one limitation, logically structured. Level 2 (3-4 marks): description of identification or production with some evaluation, or a detailed account missing balance. Level 1 (1-2 marks): few relevant points, little linkage or no evaluation. 0 marks: no creditworthy response.
Indicative content: marker gene on plasmid (antibiotic resistance / fluorescence) ; grow on antibiotic agar / only transformed survive; insertional inactivation of a second marker; culture in fermenter; optimum temperature, pH, oxygen, nutrients; promoter needed for expression; extraction and purification of insulin; advantages: human sequence, lower immune response, large reliable supply, no animal slaughter, ethical or religious acceptability; limitations: capital cost, purification to remove bacterial contaminants, contamination risk, some patients prefer animal insulin. AVP. Total 15 marks.

(a) The polymerase is not denatured at the high temperature (about 95 C) used to separate DNA strands, and it works at a high temperature, so it does not need to be replaced in every cycle.

(b) Plot or calculate log10(length) against distance. Between 500 bp (24 mm) and 250 bp (36 mm): log10 500 = 2.699 and log10 250 = 2.398, so the change is 0.301 over 12 mm, a gradient of -0.0251 per mm. At 30 mm, which is 6 mm beyond 24 mm: log10(length) = 2.699 - (6 x 0.0251) = 2.548. Length = 10^2.548 = 353, so about 350 base pairs.

(c) The internal control band should always appear. Its absence shows the PCR did not work, so the lack of a transgene band is uninformative. Possible causes: no DNA, poor-quality or degraded DNA, an inhibitor carried over from extraction, or a missing reagent (primers, polymerase or nucleotides). The band at the bottom is probably primer or primer dimers.

(d) Indicative content is in the marking scheme.

(a) Any two, 1 mark each:
- not denatured / remains active at about 95 C / high temperatures (during strand separation) ;
- no need to add fresh enzyme in each cycle / works at the high temperature used for annealing and extension, AW ;
- optimum temperature is close to that of the extension step ;
IGNORE 'works at high temperature' alone. DO NOT ALLOW 'enzyme not killed' (enzymes are not alive).

(b) Check the answer line first. 350 (accept 330 to 370) = 3 marks.
- M1: uses log10 of length (or plots log scale, or reads from semi-log graph) ;
- M2: interpolation between 500 bp and 250 bp using the distances 24 mm and 36 mm (gradient -0.0251 per mm or equivalent) giving a log10 value of about 2.55 ;
- M3: converts back to the length, about 350 bp (answers 330 to 370, given to the nearest 10) ;
If the answer is not to the nearest 10 but is otherwise correct, allow 2 marks. ALLOW ECF from M2 for M3. A straight-line (non-log) interpolation giving 375 gets M2 at most, not M3.

(c) 1 mark for the reason and 1 mark for the cause:
- internal control band absent, so the PCR has failed / worked incorrectly (so the absence of the transgene band is not reliable) ;
- one cause: no DNA / too little or degraded DNA / inhibitor present / primers, polymerase or nucleotides missing / thermal cycler fault / incorrect annealing temperature / gel loading error, AVP ;
IGNORE 'contamination' alone.

(d)* Levels of response (6 marks)
Level 3 (5-6 marks): The method is described in a logical order and is detailed enough to be reproduced, covering both the PCR and the electrophoresis. It includes controls and a clear interpretation of results, including what the control band shows.
Level 2 (3-4 marks): A mostly correct method with some detail of PCR and electrophoresis. Controls or interpretation are either partly addressed or lack detail. May lack logical order.
Level 1 (1-2 marks): Limited, mostly isolated points about PCR or electrophoresis. Little or no reference to controls or interpretation.
Level 0 (0 marks): No creditworthy response.

Indicative content (non-exhaustive):
- extract / purify DNA from each batch (e.g. ground seed samples) ;
- mix DNA with primers complementary to the sequence flanking the transgene (two different primers, one for each strand), DNA polymerase, free nucleotides and buffer ;
- primers for a maize gene always present in the same tube, as an internal control ;
- thermal cycling: about 95 C to separate strands, about 55 to 65 C for primers to anneal (hydrogen bonds), about 72 C for the polymerase to extend ;
- repeat for about 30 cycles so the target is amplified exponentially ;
- positive control (DNA known to contain the transgene) and negative control (no DNA / water, or non-GM maize DNA) to check for contamination and that primers work ;
- load products into wells of agarose gel with buffer, with a DNA ladder, apply a potential difference ;
- DNA is negatively charged (phosphate groups) so moves to the positive electrode, smaller fragments move further ;
- stain / visualise the DNA (e.g. DNA-binding dye under UV light) ;
- interpretation: transgene band at the expected length (compare with the ladder and the positive control) = transgene present ; control band present but no transgene band = absent ; no control band = test invalid / repeat ; band in the negative control = contamination ;
- repeat tests / replicates to improve reliability, AVP.
DO NOT ALLOW 'DNA moves to the negative electrode'. IGNORE a requirement for specific times or voltages.

(a) Decrease = 8.1 - 5.6 = 2.5 mmol dm\(^{-3}\). Time = 90 - 30 = 60 min. Rate = 2.5 / 60 = 0.04167, which is 0.042 mmol dm\(^{-3}\) min\(^{-1}\) to 2 s.f.

(b) Q's fasting (0 min) concentration, 7.9 mmol dm\(^{-3}\), is much higher than P's 4.8. Q's concentration rises to a much higher peak (15.6) and is still 13.0 after 180 minutes, so it does not return to the fasting level, unlike P, which returns to about 4.7.

(c) Glucose absorbed from the drink raises blood glucose. Beta cells in the islets of Langerhans of the pancreas detect the rise and secrete insulin. Insulin binds to receptors on liver (and muscle) cells, increasing uptake of glucose and activating conversion of glucose to glycogen (glycogenesis) in the liver, so blood glucose falls. When it falls below the normal range, alpha cells secrete glucagon, which stimulates the liver to break down glycogen, so the concentration rises back; this is negative feedback and P returns to about 4.7 mmol dm\(^{-3}\). In Type 1 diabetes the beta cells are destroyed, so little or no insulin is secreted and Q's concentration would stay very high with little or no fall. In Type 2 diabetes insulin is secreted but target cells respond poorly (fewer or less responsive receptors), so there is a slow, partial fall from a high peak, as with the data shown.

(a) Check the answer line first.
Correct answer 0.042 = 2 marks.
If the answer is incorrect: 8.1 - 5.6 = 2.5 (or 2.5 / 60) = 1 mark (method).
ALLOW 0.04 for 1 mark only if 2.5 / 60 is shown. DO NOT ALLOW 0.0417 for the second mark (not 2 s.f.) but allow the method mark.

(b) Any two from:
- fasting / initial concentration is high / 7.9 (compared with P's 4.8) (1)
- peak concentration is much higher / 15.6 (compared with 8.1) (1)
- concentration does not return to the fasting level / is still high / 13.0 at 180 minutes (1)
- concentration is still rising after 30 minutes / peaks later (at 60 min) than P (1)
Allow a comparative figure quoted from the table. IGNORE 'Q has more glucose' without data. Max 2.

(c)* Level of response (6 marks).
Level 3 (5-6 marks): Gives a clear, detailed and correctly sequenced explanation of the pancreas/liver negative feedback for P AND clearly distinguishes Type 1 from Type 2 in Q with correct reasoning. Uses correct terminology throughout.
Level 2 (3-4 marks): Describes the roles of insulin and the liver in lowering glucose, with some detail, and makes at least one correct statement about the difference between Type 1 and Type 2. There may be gaps in the sequence or reasoning.
Level 1 (1-2 marks): Makes simple, isolated statements about insulin or the liver or diabetes; little or no explanation or comparison.
0 marks: No relevant content.

Indicative content (non-exhaustive):
- glucose absorbed raises blood glucose; detected by beta cells in islets of Langerhans
- beta cells secrete insulin into the blood
- insulin binds to receptors on liver / muscle cells; increases glucose uptake (more channel proteins / transporters)
- liver converts glucose to glycogen (glycogenesis) so blood glucose falls
- as glucose falls below normal range, alpha cells secrete glucagon; liver converts glycogen to glucose (glycogenolysis); glucose rises back
- this is negative feedback, returning P to about 4.7 mmol dm^-3 / the normal range by 180 minutes
- Type 1: beta cells destroyed (autoimmune) so little / no insulin; blood glucose stays very high / falls very little or not at all; ORA
- Type 2: insulin produced but target cells have fewer / less responsive receptors (insulin resistance); glucose falls slowly / partially
- AVP, e.g. Type 1 would be corrected by insulin injection but Type 2 is not always.
IGNORE references to glucose being 'turned into' insulin. DO NOT ALLOW 'insulin converts glucose to glycogen' without reference to the liver/cells (enzymes activated).

(a) Cut the plasmid and the gene with the same restriction enzyme to give complementary sticky ends; mix so that the ends anneal by hydrogen bonding between bases; add DNA ligase to form phosphodiester bonds.

(b) Primers are short single-stranded DNA sequences complementary to the ends of the target sequence, so DNA polymerase can begin synthesis and only the target is copied. Heat-stable polymerase (e.g. Taq) is not denatured at 95 \(^\circ\)C so it can be reused each cycle. Heating to 95 \(^\circ\)C breaks hydrogen bonds between the strands so each strand can act as a template.

(c) P1, P3 and P4 show the 640 bp band. P3 lacks the 310 bp band because the wild-type sequence at this site has been replaced or disrupted by the insertion (so it is not amplified at this length).

(d) WT mean: \(\frac{208+215+210+219+208}{5}=\frac{1060}{5}=212\). D7 mean: \(\frac{151+148+160+145+156}{5}=\frac{760}{5}=152\). Percentage change \(=\frac{152-212}{212}\times100=-28.3\%\), a 28.3 % decrease.

(e) Fewer stomata means that CO2 diffuses into the leaf more slowly (a lower diffusion rate / greater diffusion resistance). The concentration of CO2 in the mesophyll is lower, so less CO2 is combined with RuBP by RuBisCO and less GP/TP is made in the light-independent reaction. The rate of photosynthesis is therefore lower, if CO2 is limiting (e.g. in high light).

(f) Fewer stomata gives lower water loss by transpiration. Less evaporation from mesophyll cell walls lowers the water potential gradient/tension pulling water up the xylem. Cohesion-tension means the transpiration stream/mass flow in the xylem is slower.

(g) See marking scheme levels.

(a) [3] Max 3:
- restriction enzyme cuts both plasmid and gene/DNA (with the same enzyme) ✓
- leaving (complementary) sticky ends ✓
- sticky ends anneal / base pair by hydrogen bonds ✓
- DNA ligase joins / seals the sugar-phosphate backbone / forms phosphodiester bonds ✓
DO NOT ALLOW ligase 'sticks' hydrogen bonds. IGNORE 'cuts at the recognition site' alone.

(b) [3] One mark each:
- primers are complementary to the ends/start of the target sequence AND allow DNA polymerase to start/bind OR make only target copied ✓
- polymerase heat-stable / not denatured at high temperature AND so does not need replacing each cycle ✓
- heating separates / denatures the two strands / breaks hydrogen bonds AND so each strand acts as a template ✓
ALLOW Taq. DO NOT ALLOW 'kills' / 'breaks the DNA'.

(c) [2]
- P1, P3 and P4 (all three needed) ✓
- the wild-type 310 bp sequence is absent / disrupted / replaced by the insert OR the primer-binding site is lost OR AVP ✓
IGNORE P2 inclusion for first mark (cancels it).

(d) [3] Check the answer line first. If the answer is -28.3 (or 28.3 decrease) award 3 marks.
- WT mean = 212 ✓ (method/accuracy)
- D7 mean = 152 ✓
- \(\frac{152-212}{212}\times100\) evaluated as -28.3 / 28.3 % decrease ✓
ALLOW -28.3 or 28.3 with 'decrease' stated. ALLOW 28.30. If sign missing and no 'decrease', max 2. ECF for the third mark from incorrect means.

(e) [4] Max 4:
- lower rate / rate of photosynthesis may be reduced ✓
- fewer stomata so slower diffusion of CO2 into the leaf ✓
- lower concentration of CO2 in the leaf/at RuBisCO ✓
- less CO2 fixed by RuBisCO / less carboxylation of RuBP in the Calvin cycle ✓
- less GP/TP made OR fewer products of light-independent reaction ✓
- only if CO2 is the limiting factor / no effect if another factor (light/temperature) is limiting ✓
IGNORE 'less water'. DO NOT ALLOW 'less oxygen' as the cause.

(f) [3] Max 3:
- less transpiration / less water lost from the leaf ✓
- less tension / smaller water potential gradient (pulling water up xylem) ✓
- cohesion of water molecules means a slower transpiration stream / mass flow ✓
- less water taken up by roots ✓
ALLOW ORA for the wild-type.

(g)* [6] Levels of response.

- Level 3 (5–6 marks): a coherent, reproducible plan that includes the independent variable (genotype), at least two correctly measured dependent variables (e.g. water loss AND biomass/growth), a drought treatment applied identically, at least three control variables, replication, and a statement of how results will be analysed or compared. Logical sequence.
- Level 2 (3–4 marks): a plan with a measure of water loss or growth and some control of variables, with replication OR analysis included, but with some lack of detail or sequence.
- Level 1 (1–2 marks): a limited description with one or two relevant points (e.g. grow both lines and measure) and little control or detail.
- 0 marks: no creditworthy response.

Indicative content (non-exhaustive):
- grow equal numbers of seeds of D7 and wild-type in identical pots/soil mass/compost, same light intensity, temperature and humidity, same starting seedling height/age
- water all plants equally to start, then withhold/limit water to a fixed volume per day for the dry treatment; include a well-watered control group for each line
- measure water loss using a potometer on leafy shoots OR mass loss of potted plants (sealed soil surface) at fixed time intervals, OR volume of water supplied
- measure growth: dry mass at the end (dried to constant mass), leaf area or height, at regular intervals
- calculate water use efficiency, e.g. dry mass gained per unit volume of water used
- use at least 10 replicate plants per group; calculate means and standard deviation; compare using a t-test
- randomise positions of pots to reduce bias; risk assessment (e.g. handling the GM plants: follow containment rules, wash hands, dispose of waste correctly)
- AVP, e.g. count stomata/measure stomatal conductance with a porometer

(a)(i) Fasting ensures the starting concentration reflects regulation by the body and is not raised by glucose from a recent meal, so results are comparable.
(a)(ii) Increase = 7.9 - 4.8 = 3.1 mmol dm\(^{-3}\). Percentage increase = \( \frac{3.1}{4.8} \times 100 = 64.583...\) = 64.6 % (3 s.f.).
(b) The rise in glucose is detected by beta cells in the islets of Langerhans of the pancreas. They secrete more insulin, which binds to receptors on liver and muscle cells. Cells take up more glucose (more glucose transporters in membranes) and the liver converts glucose to glycogen. Blood glucose therefore falls; as it approaches the normal range insulin secretion falls (negative feedback).
(c) Y has a raised fasting concentration (8.1, above the normal fasting value of about 4-6), peaks much higher (15.6) and later (60 min vs 30 min), and is still very high at 150 min (12.5), showing poor uptake/clearance of glucose, i.e. cells are less responsive to insulin. In Type 1 the immune system destroys beta cells so little or no insulin is made and insulin injections are essential; in Type 2 insulin is still produced but target cells have reduced sensitivity, usually linked to obesity, age, inactivity and genetics, and onset is usually later. Management of Type 2: weight loss, diet with controlled carbohydrate intake, regular exercise, monitoring glucose with biosensor, drugs that increase insulin sensitivity or secretion, and insulin only if needed; supported by the diabetes care team.

(a)(i) Fasting gives a baseline / starting glucose concentration not affected by recent food intake (AW) [1]. ALLOW so that the only glucose absorbed is from the drink. IGNORE 'to make it a fair test' alone.

(a)(ii) Check answer line first. Correct answer 64.6 (%) = 2 marks. Method mark: (7.9 - 4.8) / 4.8 x 100 OR 3.1 / 4.8 x 100 [1]. 64.58 or 65 or 64.5833 with no rounding to 3 s.f. = 1 mark (method) only. [2]

(b) Any three from: (rise in glucose) detected by beta cells / cells in islets of Langerhans in pancreas [1]; more insulin secreted [1]; insulin binds to receptors on target cells (liver, muscle) [1]; increased uptake of glucose by cells / more glucose transporters in membrane [1]; liver converts glucose to glycogen (glycogenesis) [1]; blood glucose falls, so insulin secretion reduced - negative feedback [1]. DO NOT ALLOW glucagon as the cause of the fall. MAX 3. [3]

(c)* Level 3 (5-6 marks): Uses data from both X and Y with comparison (e.g. quoted values or timing) to support Type 2 diagnosis, gives a clear and accurate comparison of Type 1 and Type 2 including cause and insulin status, and describes at least two management methods with reasons. Coherent, logically structured response.
Level 2 (3-4 marks): Addresses at least two of the three strands (data, type difference, management) with some detail or correct data use; some reasoning, may lack linkage or have minor omissions.
Level 1 (1-2 marks): Makes simple statements about one or two strands with little detail or data use, or lists points without explanation.
Level 0 (0 marks): No response or no creditworthy content.

Indicative content (non-exhaustive): Data: Y has higher fasting concentration (8.1 vs 4.8); higher peak (15.6 vs 7.9); peak later (60 vs 30 min); does not return to the fasting level / remains high at 150 min (12.5) whereas X is back to 4.8 by 150 min; suggests glucose not taken up/cleared effectively - target cells less responsive to insulin / insufficient insulin. Comparison: Type 1 - autoimmune destruction of beta cells, no/little insulin, early onset, needs insulin injections; Type 2 - insulin produced but receptors/cells less responsive, risk factors obesity, inactivity, age, genetics, later onset, often controlled by diet/exercise/drugs. Management: diet control (low sugar/carbohydrate, controlled intake), weight loss, exercise, monitoring blood glucose with biosensor/test strips, medication (e.g. to increase insulin sensitivity/secretion), insulin if required, diabetes care team/regular screening for complications. AVP. ALLOW ECF for quoted values. [6]

(a) Reverse transcriptase makes complementary DNA from an mRNA template.

(b) Copies after n cycles = \( 4 \times 2^{n} \). For n = 8: \( 4 \times 256 = 1024 \) copies.

(c) Each PCR cycle includes heating to about 95 \(^{\circ}\)C to break hydrogen bonds between the DNA strands. A DNA polymerase from a thermophilic bacterium is thermostable, so its tertiary structure and active site are not denatured. It stays active through repeated cycles and does not need to be added again.

(d) Cut the plasmid and the gene with the same restriction enzyme. This leaves complementary sticky ends. Mix them so the sticky ends form hydrogen bonds (anneal). DNA ligase joins the sugar-phosphate backbones with phosphodiester bonds. The recombinant plasmids are mixed with bacteria and taken up (e.g. after heat shock). The plasmid carries marker genes. If the insulin gene is inserted into an antibiotic resistance gene, that gene is disrupted. Bacteria are grown on agar with the first antibiotic, to show the plasmid was taken up. Colonies are then replica plated onto agar containing the second antibiotic. Colonies that grow on the first but not the second contain the insulin gene. Alternative: a fluorescent marker gene is disrupted and non-fluorescent colonies are chosen.

(a) reverse transcriptase ✓ [1]. DO NOT ALLOW DNA polymerase / RNA polymerase / transcriptase alone.

(b) Check the answer line first. 1024 ✓✓ [2]. If the answer is incorrect, award 1 mark for \( 4 \times 2^{8} \) OR \( 4 \times 256 \) OR evidence of doubling 8 times ✓ [1]. A final answer of 512 or 1000 scores 0 unless the method is shown.

(c) Max 2:
- it is thermostable / heat tolerant / from a thermophile ✓
- it does not denature / the active site keeps its shape at the high temperature (about 95 degrees C) used to separate the DNA strands ✓
- AW: it does not have to be replaced after each cycle / it can be reused through the cycles ✓
IGNORE 'it works at high temperature' without reference to denaturation.

(d)* Level of response (6 marks)
Level 3 (5-6 marks): Describes both the insertion of the gene into the plasmid AND a logical method of identifying bacteria with the recombinant plasmid. Uses correct terminology (e.g. restriction enzyme, sticky ends, ligase, marker gene) and a logically ordered account with clear detail on both parts.
Level 2 (3-4 marks): Describes the insertion of the gene with some correct detail AND makes an attempt at identification, OR gives a detailed account of one part with a limited account of the other. Some terminology correct.
Level 1 (1-2 marks): Limited statements about one part only, or a vague account of both. Little use of correct terminology.
Level 0 (0 marks): No response or no response worthy of credit.

Indicative content (non-exhaustive):
Insertion:
- cut plasmid and gene with the same restriction enzyme / endonuclease
- leaves complementary sticky ends / unpaired bases
- mix plasmid and gene so sticky ends anneal / base pair by hydrogen bonding
- DNA ligase joins the sugar-phosphate backbones / forms phosphodiester bonds
- recombinant plasmid taken up by bacteria (e.g. heat shock / calcium ions / electroporation)
Identification:
- plasmid contains marker genes, e.g. antibiotic resistance genes / fluorescent marker gene
- the insulin gene inserted into one marker gene disrupts it
- grow bacteria on agar containing the antibiotic whose gene is intact / replica plate onto agar with the second antibiotic
- colonies that grow on the first but not on the second contain the recombinant plasmid
- OR colonies that do not fluoresce contain the recombinant plasmid
- AVP e.g. gene probes / PCR to detect the insulin gene
ALLOW 'cut' for restriction enzyme action. IGNORE 'enzyme' alone without naming restriction / ligase.

(a) Taq polymerase catalyses synthesis of a new complementary strand by adding nucleotides to the primers at the extension stage. It comes from a thermophilic bacterium, so it is heat-stable and is not denatured during the strand-separation step at about 95 degrees C. Human DNA polymerase would denature and would need replacing every cycle.
(b) Copies = 4 x 2^25 = 4 x 33 554 432 = 134 217 728 = 1.34 x 10^8 (3 s.f.).
(c)(i) The child has 4.2 kb and 1.9 kb. 4.2 kb matches the mother, so the 1.9 kb fragment must be paternal. Man A has 1.9 kb; Man B (2.6 and 1.4 kb) does not, so Man A could be the father.
(ii) Smaller fragments are impeded less by the agarose mesh, so they migrate faster and further in the same time.
(d) The phosphate groups of the DNA backbone are negatively charged, so in an electric field the fragments are attracted to the positive electrode (anode).
(e) Two different people may share the same fragment sizes at one locus by chance; several loci make a coincidental match much less probable (probabilities multiply), and so increase discrimination. It also allows for mutation or a faulty/failed result at one locus to be checked against the others.

(a) [2 marks]
- adds (complementary) nucleotides to primer / synthesises new DNA strand / extension stage; ✓
- heat-stable / does not denature at (about) 95 degrees C / from thermophilic bacterium AND so does not need to be added again each cycle ✓
DO NOT ALLOW 'enzyme not killed' without reference to temperature or denaturation. IGNORE 'joins Okazaki fragments'.

(b) [2 marks] Check answer line first.
- Correct answer 1.34 x 10^8 = 2 marks.
- Method mark: 4 x 2^25 (or 4 x 33 554 432 / 134 217 728 seen) ✓
- Accuracy mark: 1.34 x 10^8 in standard form to 3 s.f. ✓
ALLOW 1 mark for 1.342... x 10^8 or 134 217 728 not in standard form / not rounded. Allow 1 mark for correct method with incorrect power of 2 (ECF for the final form only).
DO NOT ALLOW 2^27 miscalculated as a different value without working.

(c)(i) [2 marks]
- Man A ✓
- the child's 4.2 kb band matches the mother, so 1.9 kb must be from the father AND Man A has 1.9 kb (Man B does not) ✓
The second mark requires both data references. ALLOW 'Man B has neither band the child needs' AW.

(c)(ii) [1 mark]
- smaller fragments are less impeded / meet less resistance in the agarose gel (mesh/matrix) AW ✓
DO NOT ALLOW 'smaller fragments are lighter' without reference to resistance/mesh. ORA for larger fragments.

(d) [2 marks]
- DNA has phosphate groups / sugar-phosphate backbone ✓
- (which are) negatively charged so attracted to positive electrode / anode ✓
DO NOT ALLOW 'DNA is negative' without a reason for the first mark to be credited; ALLOW first mark if reason is given in context.

(e) [2 marks, max 2]
- two (unrelated) people could have the same band(s) at one locus by chance ✓
- using several loci reduces probability of a coincidental match / increases discrimination between individuals ✓
- allows a mutation / anomalous result at one locus to be checked against others ✓
- AVP e.g. confirms results are reproducible / greater statistical power ✓
IGNORE 'more accurate' alone.

(a) Taq polymerase catalyses synthesis of the new DNA strand by adding complementary nucleotides to the primers. It comes from a thermophilic bacterium, so it is not denatured at the c. 95 °C used to separate the strands, whereas a human enzyme would be denatured.
(b) Copies = \( 2^{25} = 33\,554\,432 \approx 3.36 \times 10^{7} \).
(c) X has only the 480 bp band so is homozygous RR; Y has both bands so is heterozygous Rr; Z has only the 360 bp band so is rr.
(d) The phosphate groups make DNA negatively charged, so it moves to the positive electrode (anode). The gel is a mesh of agarose; shorter fragments are impeded less and so move further in the same time.
(e)(i) Number of r alleles = 20 + 2 x 10 = 40 out of 100 alleles, so \( q = 40/100 = 0.40 \).
(e)(ii) \( p = 0.60 \); \( 2pq = 2 \times 0.60 \times 0.40 = 0.48 \); \( 0.48 \times 50 = 24 \) heterozygotes.
(f) Selection by PCR genotyping: fast, can test seedlings, no foreign DNA, usually more publicly acceptable, retains genetic variation from crossing; but several generations of breeding needed, other (linked) alleles may be transferred, and only existing alleles can be used. GM: rapid transfer of R directly into a high-yielding variety, may also add other traits; but costly, regulation and public acceptance issues, risk of gene flow to wild relatives, insertion position may disrupt other genes, and the fungus may evolve to overcome a single resistance gene.

(a) Max 2:
- adds (complementary) nucleotides to primer / synthesises new DNA strand / extends the DNA ✓
- heat stable / not denatured at high temperature (about 95 °C) used to separate strands ✓ ALLOW human enzyme would be denatured. IGNORE 'works at high temperature' alone.

(b) Check answer line first. Correct answer 3.36 x 10^7 = 2 marks. ALLOW 3.4 x 10^7 for 1 mark only (wrong significant figures). Method mark: \( 2^{25} \) / 33 554 432 seen ✓. Answer in non-standard form with correct value (33 600 000) = 1 mark.

(c) Max 2: all three correct = 2 marks; two correct = 1 mark. X = RR; Y = Rr; Z = rr. ALLOW homozygous resistant / heterozygous / homozygous non-functional for genotype descriptions only if the symbols are also correct.

(d) Max 2:
- DNA negatively charged / phosphate groups negatively charged AND moves to anode / positive electrode ✓
- smaller/shorter fragments move through the gel (mesh) more easily / are impeded less / move faster ✓ ORA. DO NOT ALLOW 'lighter' without reference to size/length impeded in gel.

(e)(i) Check answer line first. 0.4 / 0.40 = 2 marks. Method: (20 + 2 x 10) / 100 OR 40/100 ✓ (1 mark). Answer 0.2 (counting 10/50) = 0 marks. ALLOW 40%.
(e)(ii) 24 ✓ (p = 0.6, 2pq = 0.48, x 50). ALLOW ECF from (e)(i) for 2 x (1 - q) x q x 50 only.

(f) Levels of response (6 marks). Indicative content, non-exhaustive:
Genotyping/conventional breeding advantages: only existing species DNA used / no foreign DNA / fewer ethical or regulatory concerns; PCR is quick and sensitive and can test seedlings so only plants with R are grown on; maintains variation.
Genotyping/conventional breeding limitations: many generations needed; linked undesirable alleles carried over; limited to alleles already in the species; cost of PCR and electrophoresis equipment for many plants.
GM advantages: R transferred directly into a chosen high-yielding variety quickly; can use R from a different species; can be combined with other traits.
GM limitations: cost of development and regulation; public concern; possible gene flow to wild relatives; insertion site may disrupt other genes; fungus may evolve virulence against single gene; seeds may be patented.
AVP (e.g. both could be combined; durability of resistance; need to monitor).

Level 3 (5-6 marks): Gives advantages AND limitations of BOTH approaches, with clear detail and reasoned comparison/links to the context. A coherent, logical argument.
Level 2 (3-4 marks): Gives advantages/limitations of both approaches but with limited detail or comparison, OR full detail for one approach and some for the other. Mostly logical.
Level 1 (1-2 marks): Some relevant points about one or both approaches, descriptive only, little or no discussion. Limited structure.
Level 0 (0 marks): No creditworthy response.

Total 17 marks.

(a) Reverse transcriptase makes complementary DNA (cDNA) from mRNA.

(b) Recombinant plasmid length = 3.60 + 0.45 = 4.05 kb. Percentage = \( \frac{0.45}{4.05} \times 100 = 11.111... \) = 11.1 % (3 s.f.).

(c) The same restriction enzyme cuts plasmid and the DNA fragment, leaving complementary sticky ends; these anneal by hydrogen bonding and DNA ligase forms phosphodiester bonds. Bacteria are made competent (e.g. calcium chloride and heat shock) to take up plasmids. Plasmids carry marker genes (e.g. antibiotic resistance), so bacteria are grown on agar containing the antibiotic; only those that took up a plasmid survive. If the insulin gene is inserted into a second marker gene (e.g. a second antibiotic resistance gene), replica plating shows that colonies growing on the first antibiotic but not the second contain recombinant plasmids. These bacteria are cultured in a fermenter, and the insulin is extracted, purified and tested.

(a) reverse transcriptase ✓ [1]. DO NOT ALLOW DNA polymerase / reverse transcription alone.

(b) Check answer line first. 4.05 (kb) seen ✓ (method); 11.1 ✓ (accuracy) [2]. Correct answer to 3 s.f. = 2 marks. ALLOW 1 mark for 11.1111 / 11 (wrong s.f.) provided working is shown. 12.5 (0.45/3.60 x 100) scores 0 as the denominator is not the recombinant plasmid length. ECF for an incorrect sum of lengths: 1 mark for correct division and multiplication.

(c)* Level of response [6]
Level 3 (5-6 marks): Describes insertion of the gene with correct detail of enzymes and sticky ends AND explains how a marker is used to identify transformed/recombinant bacteria AND outlines how insulin is then obtained. Clear, coherent, logically structured.
Level 2 (3-4 marks): Describes some insertion detail and some identification detail, but one of the three areas (insertion, identification, production) is missing or lacks detail. Mostly logical.
Level 1 (1-2 marks): Makes a few simple, relevant statements from one or two areas, with limited detail or explanation.
Level 0 (0 marks): No response worthy of credit.

Indicative content (non-exhaustive):
Insertion: same restriction enzyme / endonuclease cuts plasmid and gene ✓; complementary sticky ends ✓; anneal by complementary base pairing / hydrogen bonds ✓; DNA ligase joins sugar-phosphate backbones / forms phosphodiester bonds ✓.
Uptake: bacteria made competent / treated with calcium ions and heat shock ✓; only a few take up plasmid ✓.
Identification: plasmid contains marker gene, e.g. antibiotic resistance ✓; grow on agar with antibiotic, only bacteria with plasmid survive ✓; insertion inactivates a second marker gene (insertional inactivation) ✓ AND replica plating / colonies that grow on one antibiotic but not the other contain recombinant plasmid ✓; OR fluorescent / GFP marker, colonies viewed under UV ✓.
Production: culture identified bacteria in fermenter / large-scale culture ✓; promoter lets gene be expressed ✓; bacteria transcribe and translate gene to make insulin ✓; extract and purify insulin ✓.
AVP for other valid points, e.g. plasmid has origin of replication so it is copied.
IGNORE references to 'cutting out the gene with scissors'.

(a)(i) A fasting state means glucose from recent food is absent, so the 0 min value is a standard baseline and any rise is due to the test drink.

(a)(ii) Increase = 13.0 - 7.8 = 5.2 mmol dm-3. Percentage increase = 5.2 / 7.8 x 100 = 66.67, so 66.7 % to 3 s.f.

(b) Between 60 and 120 min A falls from 6.2 to 4.7, back to about the fasting level, while B only falls from 13.0 to 11.6 and stays far above the fasting level. In A, the raised glucose stimulates beta cells to secrete insulin, which binds to receptors on liver/muscle cells so glucose is taken up and converted to glycogen. In B, target cells respond poorly to insulin (fewer or less sensitive receptors), so uptake is slow and glucose stays high.

(c) Glucose oxidase is immobilised on an electrode. Glucose in the blood sample is oxidised by the enzyme (specific to glucose). This transfers electrons, giving a small current that is proportional to glucose concentration. The meter converts this signal to a numerical reading.

(d) A good response compares causes (autoimmune destruction of beta cells, no insulin, versus insulin resistance / reduced response with lifestyle and genetic risk factors such as obesity) and management (insulin injections/pump with monitoring and carbohydrate counting versus diet, exercise, weight loss, then drugs such as metformin, and sometimes insulin later), with linked reasoning.

(a)(i) 1 mark: so that glucose concentration at the start is a baseline / glucose from food does not affect the result / the rise can be attributed to the test drink AW. IGNORE 'so test is fair' alone.

(a)(ii) Check answer line first. Correct answer 66.7 (%) = 2 marks. ALLOW 66.67 or 67 for 1 mark only if working shown (the third s.f. is required for the accuracy mark). Method mark (1): (13.0 - 7.8) / 7.8 x 100 OR 5.2 / 7.8 OR 0.667 shown. 0.667 / 1.667 without x 100 gets 1 mark max. ECF not applicable.

(b) Max 3:
1. A returns to (near) fasting level / 4.7 by 120 min, B remains high / 11.6 (use of data) ✓
2. in A, insulin secreted by beta cells (of islets of Langerhans) / in response to raised glucose ✓
3. insulin binds to receptors on target cells (liver, muscle) so glucose taken up / converted to glycogen (glycogenesis) ✓
4. in B, target cells do not respond to insulin / fewer or less sensitive insulin receptors / insulin resistance ✓
5. so less glucose uptake / glucose removed more slowly in B ORA ✓
DO NOT ALLOW 'no insulin' for B. ALLOW glucose channels / GLUT4 insertion into membrane.

(c) Max 3:
1. enzyme glucose oxidase (immobilised) on electrode / test strip ✓
2. specific to glucose / glucose oxidised (to gluconic acid) ✓
3. electrons transferred / electrical current / voltage generated (the transducer) ✓
4. size of current proportional to glucose concentration, converted to reading on display ✓
IGNORE 'colour change' (this describes test strips, not biosensors).

(d)* Level of response marking (6 marks). Indicative content (non-exhaustive):
Causes: Type 1 – autoimmune destruction of beta cells, little or no insulin, usually early onset, not linked to lifestyle; Type 2 – target cells less responsive to insulin / fewer receptors, linked to obesity, inactivity, age, genetic risk, insulin still produced (may fail later).
Management: Type 1 – insulin injections / pump, monitoring glucose with biosensors, matching dose to carbohydrate intake and exercise; Type 2 – diet control, weight loss, exercise, drugs (e.g. metformin / increase sensitivity or insulin secretion), insulin only if needed later; care team / regular review in both; reasons e.g. insulin cannot be taken orally as it is a protein digested.

Level 3 (5–6 marks): Detailed comparison covering BOTH causes and management, with similarities/differences clearly linked and correct biological reasoning (e.g. why insulin injection works in Type 1 but not sufficient alone in Type 2). Well structured and coherent.
Level 2 (3–4 marks): Some detail on causes and management of both types, but comparison or reasoning limited or one area less developed.
Level 1 (1–2 marks): Few relevant points, mostly about one type or only description with no comparison; limited reasoning.
Level 0 (0 marks): No response or no creditworthy content.

Total 16 marks.

(a)(i) Mean = (0.52 + 0.48 + 0.55) / 3 = 1.55 / 3 = 0.5167 = 0.52.
(a)(ii) Increase = 0.82 - 0.20 = 0.62. Percentage increase = 0.62 / 0.20 x 100 = 310%.
(b) Red pigment absorbs blue-green light most strongly, so the absorbance change per unit pigment concentration is greatest. A distilled water blank removes absorbance due to the cuvette and the solvent, so readings reflect only the pigment released.
(c) Absorbance rises with ethanol concentration (slowly up to 10%, then steeply from 20% to 40%). Ethanol dissolves/disrupts the phospholipid bilayer of the cell surface membrane and tonoplast and denatures channel and carrier proteins. The membranes become more permeable, so more pigment diffuses out of the vacuole and cells into the solution. More pigment means more light is absorbed.
(d) Cutting damages cells at the disc surface, releasing pigment. Rinsing removes this so the pigment measured is only that released during the treatment.
(e) A good plan: water baths at at least five temperatures across about 10-80 degrees C; identical discs (same cork borer, thickness, number, same beetroot); equal volume of distilled water; pre-equilibrate; same time (e.g. 30 min); thermometer in each bath to monitor; remove discs, measure absorbance with blue-green filter after zeroing with distilled water; repeat at least three times and calculate means; plot absorbance against temperature; consider range/standard deviation and anomalies; expect more leakage at higher temperature as the bilayer becomes more fluid and proteins denature.

(a)(i) 0.52 (1). ALLOW 0.517 if rounding not applied only if 0.52 is also stated. Check the answer line first.
(a)(ii) 310 (%) (2). If the answer is incorrect, award 1 mark for (0.82 - 0.20) / 0.20 OR 0.62 / 0.20 OR 3.1. ALLOW 300 to 310 if working shown. DO NOT ALLOW 410 (%) or 76 (%) (ratio of final to initial or wrong base) for 2 marks.
(b) Max 2: filter - colour of the pigment is red so it absorbs blue-green light (most) / complementary colour is absorbed (1); blank - to remove absorbance due to the solvent/cuvette / so that the reading is due to the pigment only / calibrates the colorimeter to zero (1). IGNORE 'to make it a fair test'.
(c) Max 3: ethanol dissolves/disrupts/damages the phospholipid bilayer / membrane (1); denatures channel/carrier/membrane proteins (1); (cell membrane and tonoplast) more permeable so more pigment leaks out / diffuses out (1); more pigment means more light absorbed / higher absorbance (1). ALLOW credit for a comparison, e.g. little change at 0-10% and large increase at 20-40%, as a supporting statement only. DO NOT ALLOW 'membrane is destroyed/dissolved' without reference to the lipid or proteins.
(d) Removes pigment released from cells damaged during cutting / on the surface (1). AW. IGNORE 'to clean the discs'.
(e) QWC - levels of response.
Level 3 (5-6 marks): Detailed, logical and reproducible plan. Independent variable (at least five temperatures over a suitable range, with a method to maintain them, e.g. water baths) AND dependent variable measured with the colorimeter correctly (filter, blank) AND at least three controls with a reason or method given AND replicates with a means-based or graphical treatment of the data. The conclusion is linked to the data processing.
Level 2 (3-4 marks): A sound plan covering most stages, but with some omissions or lack of detail, e.g. controls named without how they are kept constant, or data processing limited or unclear. The plan could mostly be followed.
Level 1 (1-2 marks): A basic outline with few correct features, e.g. states different temperatures and measures colour/absorbance. Little control or no repeats or processing. The plan lacks coherence.
0 marks: No response or no relevant content.
Indicative content (non-exhaustive): at least five temperatures, e.g. 10, 20, 40, 60, 80 degrees C; thermostatically controlled water baths, with a thermometer to check the temperature; equilibrate the tube of water at temperature before adding the discs; same size discs (same cork borer, same thickness), same number of discs, same beetroot, rinsed; same volume of distilled water, same time of immersion (e.g. 30 min); stir/shake equally; remove discs/filter before measurement; blue-green filter; zero the colorimeter with distilled water; repeat at least three times at each temperature and calculate the mean; plot a graph of mean absorbance against temperature; use range bars or standard deviation to judge reliability, identify anomalies; optional: use a calibration curve with serial dilutions of a pigment extract to convert absorbance to concentration; conclusion in terms of increased fluidity of the bilayer at higher temperature and denaturation of membrane proteins; safety: hot water, scalpel, AVP.
Total 15 marks.

(a)(i) Exclude 0.35. Mean = (0.21 + 0.23) / 2 = 0.22.
(a)(ii) Percentage increase = (0.95 − 0.22) / 0.22 × 100 = 331.8...% = 332 % (nearest whole number).
(b) Between 40 and 50 °C the absorbance rises from 0.22 to 0.59. Heat gives phospholipids more kinetic energy so the bilayer becomes more fluid and gaps appear. Channel and carrier proteins and other membrane proteins are denatured as hydrogen bonds break. The membrane (and tonoplast) is more permeable, so more red pigment leaks out of the vacuoles and cells into the water and more light is absorbed.
(c) A good plan uses narrower temperature intervals (e.g. every 2 °C between 40 and 50 °C), the same size and number of cylinders and equal volumes of water, a fixed time, a colorimeter zeroed with distilled water, a repeated set of replicates at each temperature, calculation of mean and a measure of spread, identification of anomalies and a graph to find the temperature where absorbance first rises above that of the lower temperatures.

(a)(i) 0.22 ✓ [1] (ALLOW 0.220)

(a)(ii) [2] Check the answer line first. Answer 332 (%) = 2 marks. ALLOW 331.8 for 1 mark (correct working, not rounded as required).
Method mark: (0.95 − 0.22) / 0.22 × 100 ✓ (0.73 / 0.22 seen)
Accuracy mark: 332 ✓
ECF: ALLOW a correct percentage calculation from an incorrect value in (a)(i) for the method mark and the answer mark.

(b) [3] max 3
1. high temperature / above 40 °C denatures membrane proteins / changes their tertiary structure (by breaking hydrogen bonds) ✓
2. phospholipids gain kinetic energy / membrane becomes more fluid / bilayer disrupted / gaps form ✓
3. membrane / tonoplast more permeable / loses its selective barrier function ✓
4. more pigment / betalain diffuses out (down concentration gradient) into the water ✓
5. more light absorbed / greater absorbance ✓
IGNORE 'enzymes denatured'. DO NOT ALLOW 'membrane melts' or 'cells burst' unless linked to protein/lipid damage.

(c)* [6] Level of response.
Level 3 (5–6 marks): A coherent, logical plan which covers range and intervals, control of several variables, replication/calculation and data assessment. The method is detailed enough to be reproduced, and the response explains why the steps give valid results.
Level 2 (3–4 marks): A plan with several relevant steps covering most of: narrower temperature intervals, control variables, replicates and a means of assessing data. Some detail or explanation is missing or sequencing is unclear.
Level 1 (1–2 marks): Limited plan with one or two relevant points (e.g. repeat readings, keep cylinders the same size) with little detail.
Level 0 (0 marks): No response or no creditworthy content.

Indicative content (not exhaustive):
• narrower temperature range / smaller intervals (e.g. 2 °C or 5 °C) between 30 and 50 °C, with a thermometer or thermostatic water bath to keep each temperature constant
• equilibrate the water and tubes at each temperature before adding the cylinders
• control variables: same beetroot (source / age), cylinder length and diameter (cut with a cork borer and ruler / scalpel), same volume of water, same time in water, same rinsing time to remove pigment from cut cells
• colorimeter: same filter, zero / calibrate with distilled water as a blank, same cuvette type, mix before measuring
• at least three replicates per temperature; calculate the mean
• identify anomalies by comparison with the other replicates and exclude / repeat them
• calculate standard deviation / range, or use error bars, to assess the spread of the data and the reliability
• plot absorbance against temperature and identify the temperature at which absorbance first rises above the baseline (e.g. above the range of the replicates at lower temperatures)
• use a calibration curve (known pigment concentrations) to convert absorbance to concentration
• safety: hot water, take care when handling a scalpel / wear eye protection
• AVP
Allow credit for a clear sketch of the planned results table.

(a) Radius = 0.5 mm. Volume moved = \(\pi \times 0.5^2 \times 62 = 48.69\) mm\(^3\). Rate = \(48.69 \div 4 = 12.17\), which is 12.2 mm\(^3\) min\(^{-1}\) (3 s.f.).
(b)(i) The still-air result for trial 4 (12.9) is far from the others (7.8 to 8.4). A possible cause is a draught across the leaves during that trial.
(b)(ii) \((8.1 + 7.8 + 8.4 + 8.0) \div 4 = 32.3 \div 4 = 8.075\), which is 8.1 mm\(^3\) min\(^{-1}\) (2 s.f.).
(b)(iii) \(((14.0 - 8.1) \div 8.1) \times 100 = 72.8\), which is 73 %.
(c) The plan should vary light intensity by changing the distance of a lamp from the shoot, keep temperature, humidity, air movement and the shoot constant, and use repeats with a mean.

(a) Check the answer line first. 12.2 = 3 marks.
- Volume = \(\pi \times 0.5^2 \times 62\) (= 48.7 mm\(^3\)) or radius of 0.5 mm used [1]
- Divided by 4 minutes [1]
- 12.2 (3 s.f.) [1]
ALLOW 12.17 or 12.1 for 2 marks only if the answer is not to 3 s.f. DO NOT ALLOW use of the diameter as the radius (maximum 2 marks, ECF).

(b)(i) [2]
- Trial 4 in still air / 12.9 [1]
- Cause, e.g. draught / fan on / door opened / higher temperature / air bubble not in the capillary tube / leak in the apparatus / bubble start position misread [1] (AVP)
IGNORE human error unqualified.

(b)(ii) 8.1 [1]. ALLOW 8.075 for 0 marks if not given to 2 s.f. ALLOW 8.1 only if anomaly excluded.

(b)(iii) Check the answer line first. 73 = 2 marks.
- \((14.0 - 8.1) \div 8.1 \times 100\) [1]
- 73 [1]
ALLOW 72 to 74 if working is shown (rounding of unrounded means). ALLOW ECF from (b)(ii) for the second mark.

(c) Indicative content (level of response, 6 marks):
- Cut shoot under water and assemble the potometer under water, so no air enters the xylem; seal joints with petroleum jelly / check airtight.
- Dry the leaves and let the shoot equilibrate before the readings.
- Introduce an air bubble and measure the distance moved in a fixed time, e.g. 4 minutes, using a scale on the capillary tube.
- Change light intensity by moving a lamp to measured distances from the shoot, e.g. 10, 20, 30, 40, 50 cm (light meter may be used; light intensity is proportional to 1/distance squared).
- Use a heat shield / water bath between the lamp and the shoot, or LED lamp, so that temperature does not change.
- Control the same shoot / leaf area, humidity, air movement, temperature, time and starting position of the bubble.
- Reset the bubble with the reservoir / syringe and leave time for the shoot to adjust to each new intensity.
- Repeat at least three times at each intensity and calculate a mean; exclude anomalies.
- Convert the distance to volume using \(\pi r^2 l\), and optionally express the rate per unit leaf area.
- Include a control at dark / zero light.
- Plot a graph of rate against light intensity.

Level 3 (5 to 6 marks): A detailed, logical plan that includes how light intensity is varied with a range of values, how at least two other variables are controlled with the method for doing so, a measurement and calculation of the rate, and repeats with a mean. The plan could be followed by another person and is well structured.
Level 2 (3 to 4 marks): A plan with some detail on the independent variable and the measurement of uptake, and mention of some control variables or repeats. There may be some omissions in the detail or the sequence.
Level 1 (1 to 2 marks): A basic outline with few correct points, e.g. moving a lamp and measuring bubble movement, but little detail on controls or reliability. The plan may be hard to follow.
Level 0 (0 marks): No response or no response worthy of credit.

(a) Cutting damages cells at the disc surface, releasing pigment. Rinsing removes this so that absorbance only reflects the effect of the ethanol.

(b) The pigment is red, so it absorbs green/blue-green light most strongly. A green (complementary) filter gives the largest change in absorbance with pigment concentration, giving the greatest sensitivity.

(c)(i) 0.60 at 30% ethanol; the other two repeats are 0.42 and 0.44, so 0.60 is far from the rest.
(ii) \( (0.42 + 0.44) / 2 = 0.43 \).

(d) \( \frac{0.80 - 0.16}{0.16} \times 100 = \frac{0.64}{0.16} \times 100 = 400\% \).

(e) Ethanol is a lipid solvent and disrupts/dissolves the phospholipid bilayer (and denatures membrane proteins) of the cell surface membrane and tonoplast. The membranes become more permeable, so more red pigment diffuses out into the solution, raising absorbance. The rise is small up to 20% and then steep, as more membrane is damaged at higher concentrations.

(f) Use a narrower range of ethanol concentrations, e.g. 10% to 30% in 2% steps, made by diluting stock ethanol with distilled water using measuring cylinders or pipettes. Use discs of identical size cut from the same beetroot, rinsed until the water is clear, with the same number per tube, the same volume of solution, the same temperature (water bath) and the same time. Measure absorbance with a colorimeter using the green filter, zeroed with distilled water. Repeat at least three times at each concentration, exclude anomalies and calculate means. Plot mean absorbance against ethanol concentration and identify the concentration at which the curve begins to rise steeply (or use a control tube with no ethanol).

(a) Removes pigment released from cells damaged by cutting / on the surface (of the discs) / AW ✓ [1]. IGNORE 'to clean' alone.

(b) Green / blue-green / blue filter ✓ AND the pigment (red) absorbs this colour / this colour is absorbed most (by the pigment) / complementary colour, so absorbance changes most with pigment concentration ✓ [2]. DO NOT ALLOW red filter. Second mark depends on a correct colour.

(c)(i) 0.60 (at 30%) ✓ AND it is much higher than/not close to the other repeats (0.42, 0.44) / AW [1]. Both needed.
(ii) 0.43 ✓ [1]. ALLOW 0.430. Anomaly must be ignored.

(d) Check answer line first. 400 (%) = 2 marks. Method: \( (0.80 - 0.16) / 0.16 \times 100 \) ✓ (1 mark). Answer 400 ✓. If 0.80/0.16 × 100 = 500 is given, award 0 for the answer and 1 for the correct difference 0.64 only if shown. ALLOW 4 × 10\(^2\).

(e) Max 2:
1. ethanol dissolves / disrupts phospholipids (in the bilayer) / denatures membrane proteins ✓
2. membranes (cell surface membrane and tonoplast) become more permeable / leaky ✓
3. more pigment diffuses out, so greater absorbance (AW) ✓
IGNORE 'cells burst'. AVP: more cells damaged at higher concentration so rise is steeper after 20%.

(f)* Level descriptors (indicative content below is not exhaustive):
Level 3 (5-6 marks): a clear, detailed and logically sequenced plan that could be followed by another student. It includes a narrower range of concentrations, repeats and a range of correctly justified control variables, and states how the results are processed. Few or no omissions.
Level 2 (3-4 marks): a plan with some detail covering most of the aspects (method, controls, processing), but lacking clarity or justification in places, or with some steps missing.
Level 1 (1-2 marks): a limited plan with few relevant points, mostly lacking detail, e.g. only stating that ethanol concentration is changed and absorbance measured.
Level 0 (0 marks): no response or no creditworthy content.

Indicative content:
- narrower range of ethanol concentrations around the steep rise, e.g. 10-30% in steps of 2-5%, prepared by dilution of stock with distilled water
- independent variable: ethanol concentration; dependent variable: absorbance of the surrounding solution
- controls: disc size/surface area (same cork borer), same beetroot, number of discs, volume of solution, time, temperature (water bath), rinsing
- colorimeter with green filter, zeroed with distilled water/blank or 0% ethanol control
- at least three repeats per concentration, calculate means, identify and exclude anomalies
- plot graph of mean absorbance against concentration / calculate rate of change to find the concentration at which absorbance begins to rise
- calibration with known pigment dilutions to convert absorbance to concentration
- safety: ethanol is flammable, keep away from naked flames; sharp scalpel
AVP.

(a) Increase = 0.92 - 0.21 = 0.71. Percentage increase = \( \frac{0.71}{0.21} \times 100 = 338.1 \) %, which is 338 % to the nearest whole number.
(b) Zeroing against distilled water (a blank) means that any absorbance of the water or tube is subtracted, so the readings show only the pigment that has leaked from the beetroot.
(c) Only one disc was used at each temperature, so repeating the investigation and calculating a mean reduces the effect of variation between discs and allows anomalies to be identified.

(a) Check the answer line first. Correct answer 338 (%) = 2 marks. ALLOW 338.1 or 338.095... for 1 mark only if not rounded as instructed, otherwise full marks if the value is rounded correctly to 338.
If the answer is incorrect, award 1 mark for method: (0.92 - 0.21) / 0.21 (x 100) OR 0.71 / 0.21 OR 3.38 / 3.4 (ratio). DO NOT ALLOW 77 % (0.71 / 0.92) for the method mark.

(b) Max 1:
- to calibrate / set a baseline / remove the absorbance of water or the tube / so that only the pigment from the beetroot is measured ✓ (AW)
IGNORE 'to make it a fair test' or 'to make it accurate' without reference to the blank.

(c) Max 1:
- repeat each temperature (at least 3 times) AND calculate a mean ✓
- use several discs at each temperature AND calculate a mean ✓
- use a larger number of temperatures / narrower intervals between 40 and 60 \(^\circ\)C ✓ (AVP)
IGNORE 'control variables' unless a named variable is given with a method of control (e.g. same disc surface area, same volume of water, same time).