HKDSE Biology Practice Paper: Cell Division, DNA And Protein Synthesis

HKDSE Biology Practice Paper: Cell Division, DNA and Protein Synthesis

2026-11-03

HKDSE Biology: Cell Division, DNA and Protein Synthesis

HKDSE Biology practice paper: 12 questions, 100 marks in 75 minutes on the cell cycle, mitosis, meiosis, DNA replication, transcription, translation and mutation, with worked answers.

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Paper details　Level: S4-S6 | Subject: Biology | Time: 75 minutes | Marks: 100 | Questions: 12

Section A (Questions 1 to 4, 32 marks)

Read each question, quote the formula, substitute the values and give the unit.

#	Question	Answer and marking notes
1	A micrograph shows an onion epidermal cell whose image diameter is 60 mm; the actual diameter of the cell is 30 micrometres. (a) Find the magnification of the micrograph. (b) In another field of view 0.5 mm in diameter, 25 cells fit across the diameter; find the mean cell diameter in micrometres. (c) If the image of the nucleus is 12 mm across, find the actual diameter of the nucleus. (d) Explain why an electron microscope, rather than a light microscope, is needed to observe mitochondria.	(a) Magnification = image diameter / actual diameter = 60 mm / 30 micrometres = 60,000 micrometres / 30 micrometres = 2,000 x [2]. (b) Field diameter = 0.5 mm = 500 micrometres, so mean cell diameter = 500 / 25 = 20 micrometres [2]. (c) Actual nuclear diameter = 12,000 micrometres / 2,000 = 6 micrometres [2]. (d) A mitochondrion is only about 0.5-1 micrometre across, close to the resolution limit of a light microscope (about 200 nanometres); an electron microscope uses a beam of electrons and resolves below 0.5 nanometre, so the cristae inside can be seen [2]. Marking: convert units first (1 mm = 1,000 micrometres); magnification has no unit.
2	The cell cycle of an animal cell lasts 24 hours: interphase occupies 22 hours (G1 10 h, S 8 h, G2 4 h) and mitosis occupies 2 hours. (a) Find mitosis as a percentage of the whole cycle. (b) Find S phase as a percentage of the whole cycle. (c) In a sample of 600 cells, how many would be expected in mitosis? (d) Explain why S phase is the best stage to label with radioactive thymidine.	(a) Mitosis as a percentage = 2 / 24 x 100 = 8.3 per cent [2]. (b) S phase as a percentage = 8 / 24 x 100 = 33.3 per cent [2]. (c) 600 x 2 / 24 = 50 cells [2]. (d) DNA replication occurs in S phase; thymine is found in DNA but is replaced by uracil in RNA, so labelled thymidine is built only into newly made DNA and only S phase cells become strongly labelled [2]. Marking: percentage is time in stage divided by total cycle time; S phase is the only stage of heavy DNA synthesis.
3	In a sample of DNA, adenine makes up 30 per cent of the bases. (a) Find the percentages of thymine, guanine and cytosine. (b) Find the ratio of purines to pyrimidines. (c) A second sample contains 22 per cent cytosine; find the percentage of adenine in it. (d) Explain why a DNA sample with a higher proportion of G and C needs a higher temperature to separate its two strands.	(a) Adenine is 30 per cent, so by A-T pairing thymine is 30 per cent; the remaining 40 per cent is shared equally, giving guanine 20 per cent and cytosine 20 per cent [2]. (b) Purines (A + G) = 50 per cent and pyrimidines (T + C) = 50 per cent, so the ratio is 1.00 [2]. (c) In the second sample cytosine is 22 per cent, so guanine is 22 per cent and A + T = 56 per cent, giving adenine 28 per cent [2]. (d) Each G-C pair has three hydrogen bonds while each A-T pair has two, so a DNA with more G and C needs more heat energy to separate its strands and has a higher melting temperature [2]. Marking: A = T and G = C; the number of hydrogen bonds sets strand stability.
4	A bacterium divides every 20 minutes. In another experiment a DNA molecule of 5,100 base pairs is replicated from two replication forks moving in opposite directions, each fork adding 50 nucleotides per second. (a) Starting from one bacterium, how many bacteria are present after 3 hours? (b) How many DNA molecules are present then? (c) Find the time needed to replicate this stretch of DNA in seconds. (d) Explain how the Meselson-Stahl experiment showed that replication is semi-conservative.	(a) 3 hours = 180 minutes, so the number of generations is 180 / 20 = 9 and the number of bacteria is 2 to the power 9 = 512 [2]. (b) Each bacterium has one DNA molecule, so there are 512 DNA molecules [2]. (c) Nucleotides to be added = 5,100 x 2 = 10,200; the two forks add 50 x 2 = 100 nucleotides per second, so the time is 10,200 / 100 = 102 seconds, about 1.7 minutes [2]. (d) Parent DNA was labelled with 15N and allowed to replicate once in 14N medium; all the DNA then had hybrid density, with one heavy and one light strand. After two generations both hybrid and fully light DNA appeared, which shows that each new molecule keeps one parental strand [2]. Marking: generations = time / generation time; two forks means doubling the rate.

Section B (Questions 5 to 8, 36 marks)

Answer the following, showing full working and the formula used.

#	Question	Answer and marking notes
5	An mRNA is 1,200 nucleotides long, with a 60-nucleotide 5 prime untranslated region, a 90-nucleotide 3 prime untranslated region and one stop codon of 3 nucleotides. (a) Find the number of codons that are translated. (b) Find the number of amino acids in the polypeptide. (c) What is the minimum number of tRNA molecules used in translation, counting each binding event once? (d) Explain the degeneracy of the genetic code and why it matters.	(a) Translated nucleotides = 1,200 – 60 – 90 – 3 = 1,047, so the number of codons is 1,047 / 3 = 349 [2]. (b) Each codon specifies one amino acid, so there are 349 amino acids [2]. (c) One tRNA carries each amino acid, so at least 349 tRNA binding events are needed; the same tRNA can be reused [2]. (d) Of the 64 codons, three are stop codons and the other 61 code for only 20 amino acids, so several codons share one amino acid. This degeneracy means a point mutation in the third base often leaves the amino acid unchanged, so the protein is less likely to be damaged [3]. Marking: subtract the untranslated regions and the stop codon before dividing by three.
6	A human body cell has 46 chromosomes (2n = 46). (a) Find the number of chromosomes and chromatids in one cell at metaphase II. (b) Find the number of chromatids in a primary spermatocyte at metaphase I. (c) Ignoring crossing over, how many genetically different gametes can independent assortment produce in humans? (d) Explain how crossing over increases genetic variation further.	(a) At metaphase II there are 23 chromosomes, each with two chromatids, so 46 chromatids [2]. (b) At metaphase I the primary spermatocyte has 46 chromosomes, each with two chromatids, so 92 chromatids [2]. (c) Independent assortment gives 2 to the power 23 = 8,388,608 kinds of gamete [3]. (d) In prophase I non-sister chromatids of homologous chromosomes exchange segments, so maternal and paternal alleles are recombined on the same chromosome; the number of genetically different gametes then far exceeds that from independent assortment alone [2]. Marking: chromosome number is already halved by metaphase II; crossing over happens in prophase I.
7	A squash of onion root tip shows 45 cells in mitosis out of 600 counted; the cell cycle of these cells lasts about 20 hours. (a) Find the mitotic index. (b) Express the index as a percentage. (c) Estimate how long an average cell spends in mitosis, in hours. (d) Explain why the root tip is good material for studying mitosis and what the hydrochloric acid does before squashing.	(a) Mitotic index = 45 / 600 = 0.075 [2]. (b) As a percentage, 0.075 x 100 = 7.5 per cent [2]. (c) Time in mitosis = 0.075 x 20 = 1.5 hours [2]. (d) Cells in the root tip meristem divide continuously, so a high proportion are in mitosis and every stage is easy to find; hydrochloric acid breaks down the middle lamella of the cell walls so the cells separate and a single layer can be squashed for counting [3]. Marking: index = dividing cells / total cells; the index measures how actively cells divide.
8	A gene is 1,800 base pairs long and the mutation rate is 5 x 10 to the power -6 per base pair per replication. (a) Find the probability that this gene carries at least one mutation after one replication. (b) Express it as a percentage. (c) Find the probability of at least one mutation after 100 generations. (d) Explain why most point mutations do not change the amino acid sequence of the protein.	(a) Probability per replication = 1,800 x 5 x 10 to the power -6 = 0.009 [2]. (b) As a percentage, 0.9 per cent [2]. (c) Probability of at least one mutation after 100 generations = 1 – (1 – 0.009) to the power 100 = 59.5 per cent [3]. (d) The code is degenerate, so a change in the third base often still codes for the same amino acid; mutations in non-coding regions or introns, and synonymous mutations, leave the amino acid sequence and protein function unchanged [2]. Marking: use 1 – (1 – p) to the power n for many generations, never multiply 0.009 by 100.

Section C (Questions 9 to 12, 32 marks)

Show fuller working and explanation, with units in every answer.

#	Question	Answer and marking notes
9	Onion roots were treated with different concentrations of colchicine and 1,000 cells were counted in each group. The mitotic index was 0.06 in the control, 0.13 at 0.01 per cent, 0.28 at 0.1 per cent and 0.31 at 1.0 per cent. (a) Comparing 0.31 with 0.06, find the fold increase in mitotic index. (b) Find the percentage increase of the 0.1 per cent group over the control. (c) Explain why colchicine raises the mitotic index. (d) Explain why the index at 1.0 per cent is only slightly above that at 0.1 per cent.	(a) Fold increase = 0.31 / 0.06 = 5.17 [2]. (b) Percentage increase = (0.28 – 0.06) / 0.06 x 100 = 366.7 per cent [2]. (c) Colchicine blocks the formation of spindle fibres, so chromosomes that have already replicated cannot be pulled to the poles and cells arrest at metaphase; dividing cells accumulate, so the mitotic index rises [2]. (d) Once most cells that enter mitosis are arrested the index is near its ceiling, so a higher concentration cannot keep more cells in mitosis; very high doses may also stop division starting or damage the cells, so the rise flattens [2]. Marking: a rising index shows metaphase arrest, not faster division.
10	A polypeptide chain is made of 349 amino acids; the mean relative molecular mass of an amino acid is 110 and that of water is 18. (a) Find the number of peptide bonds. (b) Find the number of water molecules released. (c) Find the relative molecular mass of the chain in Da and in kDa (1 kDa = 1,000 Da). (d) State where the condensation reaction occurs and what the enzyme does.	(a) Number of peptide bonds = amino acids – 1 = 349 – 1 = 348 [2]. (b) One water molecule is released per peptide bond, so 348 water molecules [2]. (c) Relative molecular mass = 349 x 110 – 348 x 18 = 38,390 – 6,264 = 32,126 Da, about 32.1 kDa [2]. (d) The condensation reaction takes place on the ribosome; peptidyl transferase links the carboxyl group of one amino acid to the amino group of the next, forming a peptide bond and releasing water, so the chain grows one residue at a time [2]. Marking: n amino acids give n – 1 peptide bonds; subtract the water released when finding the mass.
11	Compare mitosis and meiosis in at least four respects and explain the role of each in living organisms.	(a) Mitosis occurs in body cells, gives two daughter cells, keeps the chromosome number the same as the parent (2n), and the daughter cells are genetically identical to the parent [2]. (b) Meiosis occurs in germ cells, involves two successive divisions giving four cells with half the chromosome number (n), and the cells differ genetically because of independent assortment and crossing over [2]. (c) Mitosis gives growth, tissue repair and asexual reproduction with a constant chromosome number; meiosis gives gametes so that fertilisation restores 2n and it generates the variation on which natural selection acts [2]. (d) Mitosis copies the chromosomes once and divides once, while meiosis copies once and divides twice, so the chromosome number is halved [2]. Marking: the comparison must cover site, number of daughter cells, chromosome number and genetic make-up.
12	(a) Explain how cell cycle checkpoints prevent uncontrolled division. (b) Explain how mutations in proto-oncogenes and tumour suppressor genes lead to cancer. (c) Give one medical application that inhibits cell division and explain its principle.	(a) The G1 checkpoint checks cell size, nutrients and DNA damage and blocks entry to S phase if conditions are not met; the G2 checkpoint confirms that DNA replication is complete and undamaged; the metaphase checkpoint confirms that every chromosome is attached to the spindle before anaphase, preventing mis-segregation [3]. (b) A mutated proto-oncogene stays active and keeps driving division, while a mutated tumour suppressor gene such as p53 loses its braking role; accumulating mutations let cells divide uncontrollably, escape apoptosis and invade other tissues, forming a cancer [3]. (c) For example colchicine or paclitaxel blocks spindle fibres so that rapidly dividing cancer cells arrest at metaphase and die; cancer cells divide more often than normal cells, so they are hit harder [2]. Marking: checkpoints are pass or fail gates; cancer comes from the accumulation of several mutations.

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