2025 Pure Biology TYS Answers

MCQ 1 2 3 4 5 6 7 8 9 10 A C C D B A B B A A 11 12 13 14 15 16 17 18 19 20 B B C D D D D B C B 21 22 23 24 25 26 27 28 29 30 A C B C A C A C B D 31 32 33 34 35 36 37 38 39 40 B B A D C A C B C B Qn Answers Remarks 1a 3 Candidates should identify that 3 is protein based on the peptide bonds present. 1b Production of protein-digesting enzymes: C, D and H. Protein digestion: C and H. Students need to identify all the correct parts of the digestive system. The two sites of protein digestion were more well known than the three sites of production of protein-digesting enzymes. 1c Protease. 1d Assimilation: amino acids; are absorbed by body cells and used to synthesise proteins for growth/repair and form protein such as enzymes/hormones; Removal: excess amino acids are deaminated in the liver; the amino group is converted to urea; urea is carried to kidneys and excreted in urine. Candidates need to recognise that the products of protein digestion were amino acids. Candidates should not spend unnecessary time and space describing digestion and absorption, as the question indicated that only events post-absorption were required. Candidates need to show understanding that the

deamination of amino acids occurs in the liver with the production of urea. They do not need to provide detail about the route of urea around the circulatory system between liver and kidney. Students should also show understanding of the excretion of urea by the kidney in urine. Students should also show understanding of how amino acids are being used in the production of new proteins or of named proteins such as enzymes. 2a M: G = 21, T = 14, A = 11. N: G = 36, T = 11. The numbers of bases for each strand should both add up to 82 (million). Candidates need to recognise the number of cytosine bases on the N strand should be the same as the number of guanine bases on the M strand using understanding of complementary base pairing. Hence, the number of thymine bases on the N strand must be 11, so the total on that strand could be 82. 2b Different DNA base sequences give different base triplets/codons (and mRNA codons); As sequence of base codes for a sequence of amino acids; so different amino acids may be joined in a different sequence, producing a different polypeptide. Candidates should go beyond identifying that the two base sequences were different. Candidates should also describe the idea that the sequence of bases codes for a sequence of amino acids; hence the different amino acid sequences equate to different polypeptides. Description of DNA replication would not answer the question. 3ai P: cell surface membrane Q: cytoplasm Candidates should only name the structures in the cell body.

There is no specialised names required. Candidates should not repeat labels already on Fig. 3.1, such as cell body or dendrites. 3aii Myelin sheath insulates the nerve fibre and increases the speed of nerve impulse transmission; It has a long nerve fibre to carry the nerve impulses away from cell body across long distances in the body; There are highly branched dendrites attached to cell body that provide large surface area to receive incoming signals or neurotransmitters from neighbouring neurones; Candidates should use the information in the introduction or on Fig. 3.1 and instead suggested structures which were not in the figure. Candidates need to go beyond naming the structural adaptation, and provide explanation to link the structure correctly to an explanation. 3bi Any two: retina/ eye ; nose /olfactory epithelium; taste buds on tongue . Candidates should the names of two appropriate sense organs; Some gave names of components of those sense organs, such as ‘retina’. Answers such as paws, whiskers and skin are acceptable as none could be used to detect food. Candidates should also not describe type of receptor, rather than its location. 3bii Leg muscle / skeletal muscle . The term effector should be understood by candidates. Candidates should not just state ‘legs’ or give other parts of the overall response, such as the stimulus or the cat’s response 3biii sensory neurone -> relay neurone -> motor neurone Students should include the sensory neurone in the sequence of neurones. Other additional and unnecessary detail do not need to be given including receptors, the location of the relay neurone and the effect; when only the three neurones, in sequence were required.

3ci Factor 1: myelin sheath - Reason fibres 1-3 are much faster than unmyelinated fibre 4 which has the lowest speed of nerve impulse at 2 m per seconds; Factor 2: nerve-fibre diameter Reason: among myelinated fibres, larger diameter is associated with higher speed (1 > 2 > 3), where nerve fibre 1 with thickest diameter followed by fibre 2 and 3 has speed from 100m per seconds for nerve fibre one which is faster than nerve fibre 2 by 20m per seconds and fibre 3 by 60m per seconds; Candidates need to describe the evidence for their two selected factors, using the neurone numbers or describing the variation in the feature and how this affected impulse speed. 3cii Myelination has the greater effect. Fibre 3 is myelinated and conducts at 40 m s ⁻ ¹, whereas similar/small unmyelinated fibre 4 conducts at 2 m s ⁻ ¹ at about 20× difference; Diameter differences among 1-3 give a smaller change. Candidates should consider the two factors they had identified in the previous question and select the one which had the most impact. The explanation of the choice should compare the evidence for the two factors previously given. 4a Nucleus. Candidates should show understanding of the term organelle. 4b Condition is caused by dominant allele; Persons 7 and 8 are unaffected, so both only have recessive alleles; Hence, every child only receive recessive alleles from each parent and are unaffected. Candidates need not explain how person 7 came to be homozygous recessive by considering her parental genotypes and inheritance of that genotype. Instead, candidates should include the idea that the offspring of parents 7 and 8 would have to inherit, or be passed down, recessive alleles from both parents. Candidates should not simply state that as 7 and 8 were both homozygous recessive, then their children would also have this genotype, without explaining how they came to get these alleles.

4ci Person 16: dd. Person 20: Dd. Gametes: d × (D or d). Children: Dd affected and dd normal, 1:1. Probability affected = 1/2 = 50%. Geenetic diagram completed well should show correct haploid gametes and incorrect parental genotypes. 4cii Each fertilisation is an independent random event; A 50% probability does not mean exactly half of a small family must be affected and one of six can occur by chance as the chance of inheriting particular alleles