Summer 2023 is part of the current CCEA GCE Biology (2016) specification, so every question style in it is still examinable. The papers and mark schemes themselves are published by CCEA; BioCCEA is an independent study tool and is not endorsed by CCEA.
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 focusAnswer each one under exam conditions. Your wording is marked against every mark scheme point before the full-mark answer appears.
A2 2 2023-style genetics and statistics question · 6 marks
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.
Question styles that carried the most marks in the Summer 2023 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 2 2023-style genetics and statistics question · 6 marks
Both parents are BbSs, producing gametes BS, Bs, bS and bs. A 4 × 4 Punnett square gives the expected ratio 9 black short : 3 black long : 3 brown short : 1 brown long, which for 96 offspring is 54 : 18 : 18 : 6. To test the observed results I would carry out a chi-squared test, χ² = Σ (O − E)² / E, using these expected values. There are four phenotype classes, so degrees of freedom = 3 and the critical value at p = 0.05 is 7.82. If the calculated χ² is less than 7.82 the difference between observed and expected is not significant and is due to chance, so the null hypothesis is accepted.
Examiner insight: The expected numbers must be calculated from the actual total (96), not left as a ratio. That conversion is a discrete mark.
A2 2 2023-style speciation question · 5 marks
A barrier — for example a geographical barrier such as a river, or a behavioural difference in courtship — prevents the two populations from interbreeding, so there is no gene flow and no exchange of alleles between them. Different mutations arise in each population and different selection pressures act on them, so different alleles give greater reproductive success in each environment. The frequencies of alleles therefore change in different directions in the two populations over many generations. Eventually the populations are so genetically and phenotypically different that even if brought back together they could no longer interbreed to produce fertile offspring, so they are separate species.
Examiner insight: The definition mark requires 'fertile offspring'. Answers that stop at 'they look different' or 'cannot breed' lose the final mark.
A2 2 2023-style selection plus statistics question · 6 marks
Shell colour varies in the population because different alleles are present. If the environment changes so that the background is darker, the darker snails are better camouflaged and less likely to be seen by predators, so they are more likely to survive and reproduce. They pass the allele for dark shell to their offspring, so its frequency increases over successive generations - directional selection. To test whether the two phenotypes occur in the predicted ratio I would use the chi-squared test, calculating the sum of (observed - expected) squared divided by expected. With two phenotypes there is one degree of freedom, so I would compare the calculated value with the critical value at the 0.05 probability level. If the calculated value is less than the critical value, the difference between the observed and expected numbers is not significant and is due to chance.
Examiner insight: Selection answers must talk about alleles and reproduction, not 'the snails adapt'. For the statistics half, the degrees of freedom and the 0.05 level are each worth a mark.
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 schemeThese 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.
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.
