A2 1 2018-style structure and function question · 5 marks
Explain how the structure of a leaf is adapted for efficient photosynthesis.
Marked instantly, no account needed
CCEA A2 Biology 2018 past paper and mark scheme
Summer 2018 is part of the current CCEA GCE Biology (2016) specification, so every question style in it is still examinable. First A2 series of the 2016 specification. The papers and mark schemes themselves are published by CCEA; BioCCEA is an independent study tool and is not endorsed by CCEA.
What you sit in A2 2018
A2 1: Physiology, Co-ordination and Control, and Ecosystems
A2 1 written paper (2 hours)
Homeostasis and the kidney, nervous transmission, muscle contraction, hormonal co-ordination, photosynthesis, respiration and ecosystems.
A2 1 topic list and exam focusA2 2: Biochemistry, Genetics and Evolutionary Trends
A2 2 written paper (2 hours)
Respiration and ATP, monohybrid and dihybrid inheritance, sex linkage, population genetics, natural selection, speciation and gene technology.
A2 2 topic list and exam focusA2 3: Practical Skills in A2 Biology
A2 3 practical booklet plus a written practical-skills paper (1 hour 15 minutes)
Designing valid investigations, precision and error, statistical treatment of biological data and evaluating limitations.
A2 3 topic list and exam focus
How to sit the 2018 paper properly
- 1Print the paper and sit it in one block, to time, with no notes. A paper done in pieces tells you nothing about your exam performance.
- 2Mark it with the official CCEA mark scheme in front of you and award a mark only where the scheme's point is actually made.
- 3Log every dropped mark by unit topic, not by question number, so patterns appear across papers.
- 4Rewrite the two worst answers in full, using mark-scheme wording, then re-mark them.
- 5Come back to the same paper three weeks later and redo only the questions you lost marks on.
Sit the A2 2018 style questions
Answer each one under exam conditions. Your wording is marked against every mark scheme point before the full-mark answer appears.
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These questions are written in the style of the CCEA GCE Biology (2016) papers and are not reproduced from any live paper. CCEA owns the copyright in its question papers and mark schemes — always download the official paper and mark scheme from CCEA and use these worked answers alongside them. BioCCEA is an independent study tool and is not endorsed by CCEA.
Worked A2 2018 questions with mark schemes
Question styles that carried the most marks in the Summer 2018 papers, each with the full list of creditworthy mark scheme points, a model answer at full marks, and the marking point candidates most often missed.
A2 1 2018-style structure and function question · 5 marks
Explain how the structure of a leaf is adapted for efficient photosynthesis.
Command word: Explain — how to answer itMark scheme — 6 creditworthy points
- Large surface area of the lamina to absorb more light
- Thin, so light penetrates to the palisade layer and diffusion distance for CO₂ is short
- Palisade mesophyll cells packed with chloroplasts near the upper surface to maximise light absorption
- Air spaces in the spongy mesophyll allow rapid diffusion of carbon dioxide to the mesophyll cells
- Stomata allow gas exchange; guard cells control opening to balance CO₂ uptake with water loss
- Xylem delivers water for photolysis and phloem removes sugars, maintaining the concentration gradient
Full-mark answer
The lamina has a large surface area to absorb as much light as possible, and it is thin so light penetrates to the palisade layer and the diffusion distance for carbon dioxide is short. Palisade mesophyll cells are packed with chloroplasts and arranged near the upper surface, maximising light absorption. The spongy mesophyll contains large air spaces that allow carbon dioxide to diffuse rapidly to all the photosynthesising cells. Stomata in the lower epidermis allow carbon dioxide to enter, with guard cells controlling their opening so that gas exchange is balanced against water loss. Xylem in the veins supplies water for photolysis and phloem removes the sugars produced, keeping the concentration gradient favourable.
Examiner insight: Each mark needs an adaptation plus its reason. A list of leaf features with no function attached scores at most 1.
A2 1 2018-style respiration question · 6 marks
Explain how oxidative phosphorylation produces ATP in a mitochondrion, and explain why the yield of ATP falls when a respiratory poison blocks the final electron carrier.
Command word: Explain — how to answer itMark scheme — 7 creditworthy points
- Reduced NAD and reduced FAD are oxidised, releasing hydrogen atoms that split into protons and electrons
- Electrons pass along the electron transport chain of carriers in the inner mitochondrial membrane
- Energy released is used to pump protons from the matrix into the intermembrane space
- A proton gradient (chemiosmotic gradient) is established
- Protons pass back into the matrix through ATP synthase, producing ATP (chemiosmosis)
- Oxygen is the final electron acceptor, combining with electrons and protons to form water
- If the final carrier is blocked, electrons cannot pass to oxygen, the chain backs up, NAD stays reduced so the Krebs cycle stops and only glycolysis yields ATP
Full-mark answer
Reduced NAD and reduced FAD from glycolysis and the Krebs cycle are oxidised, and the hydrogen atoms released split into protons and electrons. The electrons pass along a chain of electron carriers in the inner mitochondrial membrane, and the energy released at each transfer is used to pump protons from the matrix into the intermembrane space, establishing a proton gradient. Protons then diffuse back into the matrix through ATP synthase, and this flow drives the phosphorylation of ADP to ATP - chemiosmosis. Oxygen acts as the final electron acceptor, combining with electrons and protons to form water. If a poison blocks the final carrier, electrons cannot be passed to oxygen, so the whole chain becomes reduced and no more protons are pumped. Reduced NAD cannot be reoxidised, so the Krebs cycle and the link reaction stop, and the cell is left with only the small ATP yield from glycolysis.
Examiner insight: The poison half of the question is answered by working backwards along the chain: no oxygen acceptor, no reoxidised NAD, no Krebs cycle. Candidates who stop at 'no ATP is made' get one mark.
A2 2 2018-style dihybrid and linkage question · 6 marks
A student crossed a pure-breeding tall red-flowered plant with a pure-breeding dwarf white-flowered plant and self-pollinated the F1. Explain the expected F2 ratio and explain how linkage would change it.
Command word: Explain — how to answer itMark scheme — 6 creditworthy points
- F1 are all heterozygous for both genes and show both dominant phenotypes
- The F1 produce four types of gamete in equal numbers because the genes assort independently
- Random fertilisation gives a 9:3:3:1 phenotypic ratio in the F2
- 9 tall red : 3 tall white : 3 dwarf red : 1 dwarf white
- If the genes are linked they are on the same chromosome, so parental combinations of alleles stay together
- Far more parental phenotypes and fewer recombinants than 9:3:3:1, with recombinants only produced by crossing over
Full-mark answer
The F1 plants are heterozygous for both genes and show the dominant phenotypes, so all are tall with red flowers. Because the two genes are on different chromosomes they assort independently during meiosis, so each F1 plant produces four types of gamete in equal proportions. Random fertilisation between these gametes gives sixteen equally likely combinations and a phenotypic ratio of 9 tall red : 3 tall white : 3 dwarf red : 1 dwarf white in the F2. If the two genes were linked, that is carried on the same chromosome, the parental combinations of alleles would tend to be inherited together, so the F2 would contain a large excess of the two parental phenotypes and only a small number of recombinant phenotypes, produced when crossing over occurs between the two loci during prophase I.
Examiner insight: Say why independent assortment gives four equal gamete types - the ratio alone is one mark. For linkage, the recombinants must be attributed to crossing over.
These questions are written in the style of the CCEA GCE Biology (2016) papers and are not reproduced from any live paper. CCEA owns the copyright in its question papers and mark schemes — always download the official paper and mark scheme from CCEA and use these worked answers alongside them. BioCCEA is an independent study tool and is not endorsed by CCEA.
Download the official CCEA paper and mark schemePractise the A2 2018 paper in the app
These open BioCCEA with A2 already selected, so you can sit questions in this paper's style and have them marked against the CCEA mark scheme.
Get your 2018 answers marked
Photograph a written answer from this paper and see it marked line by line in CCEA mark-scheme language, with the exact phrase you needed to earn each missing mark.