2022 May/June · Paper 13 · Q211 markMitosis & the cell cycle
2022 May/June · Paper 22 · Q1 (d)4 marksStem cells & cancer
2022 May/June · Paper 23 · Q3 (b(i))3 marksStem cells & cancer
2022 May/June · Paper 23 · Q5 (c(ii))2 marksStem cells & cancer
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References only, not question text. Open Official paper ↗ where Cambridge publishes the paper,
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school. “Paper 42” means Paper 4, variant 2.
Practice questions
Original questions written by HomiLabs in the style of the exam, with answers and mark schemes. They are not
Cambridge past paper questions and have not been reviewed by Cambridge. Try each one before opening the answer.
Q1Mitosis & the cell cycle · Core · Paper 1 style · 1 mark
(a)During which stage of mitosis do the centromeres divide and the sister chromatids move to opposite poles of the cell?[1]
Show answer
(a) C anaphase
In anaphase the centromeres split and spindle microtubules shorten, pulling the chromatids (now chromosomes) to opposite poles. B is wrong: at metaphase the chromosomes are lined up on the equator but the chromatids are still joined.
Q2Mitosis & the cell cycle · Core · Paper 1 style · 1 mark
(a)Which statement about mitosis is correct?[1]
Show answer
(a) B The daughter cells are genetically identical to the parent cell.
Mitosis produces two genetically identical nuclei with the same chromosome number as the parent. A, C and D all describe meiosis.
Q3Mitosis & the cell cycle · Core · Paper 2 style · 5 marks
A farmer in Punjab cuts his hand on a sickle. Over the next ten days the skin heals.
(a)State three roles of mitosis in multicellular organisms.[3]
(b)Explain why it is important that the daughter cells produced by mitosis are genetically identical to the parent cell when skin is repaired after a cut.[2]
Show answer
(a) Any three of: growth; repair of damaged tissue; replacement of worn-out cells; asexual reproduction; clonal expansion of lymphocytes in an immune response.
1 mark each for any three of: growth (increase in cell number); repair / wound healing; replacement of cells such as skin or gut lining or red blood cell precursors; asexual reproduction (e.g. potato tubers, Hydra budding); clonal expansion of B and T lymphocytes.
(b) The new cells must have the same genes so they can produce the same proteins and function as skin cells; genetic stability keeps the tissue working and avoids harmful mutations.
B1 same genetic information / same alleles as the parent cells; B1 so new cells can carry out the same function / make the same proteins as the cells they replace.
Q4Mitosis & the cell cycle · Core · Paper 3 style · 4 marks
A student looks at a stained onion root tip squash under high power and describes three cells.
(a)Identify the stage shown by each cell. Give one reason for each.Cell P: chromosomes are short and thick and lie in a line across the middle of the cell; no nuclear envelope is visible.Cell Q: two groups of V-shaped chromosomes are near opposite ends of the cell, with arms trailing behind.Cell R: a nucleus with a clear nuclear envelope and a dark nucleolus; no individual chromosomes visible.[3]
(b)Suggest why most cells in the field of view look like cell R.[1]
Show answer
(a) P metaphase (chromosomes on the equator); Q anaphase (chromatids being pulled to the poles, centromeres leading); R interphase (intact nucleus, chromatin not condensed).
1 mark per correct stage with reason: P metaphase – on equator/metaphase plate; Q anaphase – chromatids moving to opposite poles, V shape as centromere leads; R interphase – nuclear envelope and nucleolus visible / chromosomes not condensed.
(b) Interphase is the longest stage of the cell cycle.
B1 cells spend most of the cell cycle in interphase.
Q5Mitosis & the cell cycle · Standard · Paper 1 style · 1 mark
(a)Tomato cells have a diploid number of 24 chromosomes. How many chromatids are present in one tomato cell at metaphase of mitosis?[1]
Show answer
(a) C 48
DNA was replicated in S phase, so each of the 24 chromosomes consists of two sister chromatids: 24 × 2 = 48. B is the number of chromosomes, not chromatids.
Q6Mitosis & the cell cycle · Standard · Paper 1 style · 1 mark
(a)[1]
Show answer
(a) B end of G2 32; each daughter cell 16
DNA is replicated in S phase so G2 has 2 × 16 = 32 units; mitosis shares this equally, so each daughter has 16 units, the same as the parent in G1. A halves the DNA as if this were meiosis.
Q7Mitosis & the cell cycle · Standard · Paper 2 style · 5 marks
A student made a squash of a stained garlic root tip and counted the cells in each stage of the cell cycle.Interphase 412; prophase 46; metaphase 18; anaphase 9; telophase 15.
(a)Calculate the mitotic index of this root tip. Give your answer to 3 decimal places.[2]
(b)The cell cycle in this root tip lasts 24 hours. Assuming the number of cells seen in a stage is proportional to the time spent in that stage, calculate how long the cells spend in metaphase. Give your answer in minutes to the nearest minute.[2]
(c)Suggest why a root tip, rather than a region 2 cm behind the tip, was used to find the mitotic index.[1]
Show answer
(a) 0.176
M1 cells in mitosis = 46 + 18 + 9 + 15 = 88; total = 500. A1 mitotic index = 88/500 = 0.176.
(b) 52 minutes
M1 18/500 × 24 × 60; A1 = 51.8 ≈ 52 min.
(c) The root tip contains a meristem where cells are actively dividing, so many cells are in mitosis.
B1 meristem / zone of cell division at the tip; cells further back are elongating or differentiating and no longer divide.
Q8Mitosis & the cell cycle · Standard · Paper 2 style · 5 marks
(a)Describe the behaviour of the chromosomes during prophase and metaphase of mitosis.[3]
(b)Describe what happens during telophase.[2]
Show answer
(a) Prophase: chromosomes condense / shorten and thicken and become visible as two sister chromatids joined at a centromere; nuclear envelope breaks down. Metaphase: chromosomes attach to spindle microtubules at their centromeres/kinetochores and line up on the equator.
1 mark each for any three of: chromosomes condense / coil / become visible (prophase); each seen as two chromatids joined at the centromere; nuclear envelope / nucleolus disappears so chromosomes lie in the cytoplasm; spindle fibres attach at centromere / kinetochore; chromosomes lined up on the equator / metaphase plate.
(b) Chromosomes arrive at the poles and uncoil / decondense; a nuclear envelope reforms around each group, giving two nuclei; spindle breaks down.
1 mark each for any two of: chromatids reach poles; chromosomes uncoil / become less visible; nuclear envelope and nucleolus reform; spindle disappears.
Q9Mitosis & the cell cycle · Standard · Paper 2 style · 4 marks
(a)Describe how cytokinesis differs in an animal cell and a plant cell.[2]
(b)Suggest why a plant cell cannot divide its cytoplasm in the same way as an animal cell.[1]
(c)A student observed a cell in a stained root tip squash with two nuclei and a thin line between them across the middle of the cell. State the stage the cell was in.[1]
Show answer
(a) Animal: cell membrane pinches in to form a cleavage furrow that divides the cytoplasm. Plant: vesicles from Golgi form a cell plate across the equator, which becomes the new cell wall.
B1 animal: cleavage furrow / membrane constricts inwards; B1 plant: cell plate / new cell wall built from the centre outwards (from vesicles).
(b) The rigid cellulose cell wall prevents the cell surface membrane from pinching inwards.
B1 rigid / cellulose cell wall cannot be constricted.
(c) Late telophase / cytokinesis.
B1 telophase / cytokinesis (cell plate forming).
Q10Mitosis & the cell cycle · Standard · Paper 2 style · 4 marks
During a typhoid infection in Karachi, a patient's immune response involves rapid division of lymphocytes.
(a)A single activated B-lymphocyte divides by mitosis every 8 hours. Calculate the number of identical cells produced from this one cell after 48 hours, assuming every cell divides each cycle.[2]
(b)Explain why it is important that these lymphocytes are produced by mitosis rather than meiosis.[2]
Show answer
(a) 64 cells
M1 number of cycles = 48/8 = 6; A1 2^6 = 64.
(b) Mitosis gives genetically identical cells, so all of the clone have the same genes for the same specific antibody / receptor that binds the antigen.
B1 genetically identical / clones; B1 all make the same antibody / have the same receptor complementary to the pathogen's antigen (meiosis would give variation and half the chromosome number).
Q11Mitosis & the cell cycle · Challenge · Paper 2 style · 6 marks
(a)Colchicine prevents tubulin molecules joining together to form microtubules. Explain why cells treated with colchicine do not complete mitosis.[3]
(b)Onion root tips were placed in colchicine for 4 hours and then squashed. Suggest and explain how the mitotic index and the percentage of dividing cells in metaphase would differ from untreated root tips.[2]
(c)Some cancer drugs act in a similar way to colchicine. Suggest why patients receiving these drugs often lose hair.[1]
Show answer
(a) No spindle forms, so chromosomes cannot attach to spindle fibres, cannot line up at the equator, and chromatids cannot be pulled apart to the poles in anaphase.
B1 spindle (fibres) made of microtubules cannot form; B1 chromosomes not attached at centromeres / cannot line up at equator; B1 sister chromatids not separated / pulled to opposite poles (so mitosis halts at metaphase).
(b) Mitotic index higher and a higher proportion of dividing cells in metaphase, because cells keep entering mitosis but become stuck at metaphase and cannot leave it.
B1 higher proportion of dividing cells in metaphase (and mitotic index increases); B1 cells continue to enter mitosis but cannot progress beyond metaphase so accumulate.
(c) Hair follicle cells divide rapidly, so their mitosis is also blocked.
B1 hair follicle cells divide frequently by mitosis and are affected as well as cancer cells.
Q12Mitosis & the cell cycle · Challenge · Paper 2 style · 6 marks
Mammalian cells growing in culture were synchronised so that they all began the cell cycle at the same time. The mean mass of DNA per cell (arbitrary units) was measured.time 0 h: 10time 4 h: 10time 8 h: 10time 10 h: 13time 12 h: 17time 14 h: 20time 16 h: 20time 18 h: 20time 18.5 h: 10time 22 h: 10
(a)Identify the time interval during which S phase occurs. Explain your answer.[2]
(b)Explain the change in DNA mass between 18 and 18.5 hours.[2]
(c)Using the data, calculate the length of G1 for these cells, assuming the next cycle begins at 18.5 hours and is the same as the first. State one other assumption you have made.[2]
Show answer
(a) 8 to 14 hours; DNA mass doubles from 10 to 20 a.u. as DNA is replicated.
B1 8–14 h; B1 DNA content rises / doubles, showing DNA replication.
(b) The DNA mass halves because the cell divides (telophase / cytokinesis): each daughter cell receives one copy of each chromosome, i.e. half the replicated DNA.
B1 DNA halves from 20 to 10; B1 sister chromatids separated into two nuclei / cytokinesis produces two cells, each with the original DNA content.
(c) 8 hours; assumes all cells are synchronised / DNA starts to rise immediately at the end of G1.
B1 G1 = time from start of cycle to start of DNA replication = 8 − 0 = 8 h; B1 assumption: cells synchronised / readings frequent enough to show exact start of S phase.
Q13Mitosis & the cell cycle · Challenge · Paper 2 style · 6 marks
(a)Paclitaxel binds to microtubules and stops them from shortening. Explain why this prevents a cell from completing anaphase.[2]
(b)In a trial, the mitotic index in a breast tumour fell from 0.21 to 0.06 after treatment, while in the lining of the small intestine it fell from 0.15 to 0.11. Calculate the percentage decrease in mitotic index in the tumour. Give your answer to the nearest whole number.[2]
(c)Suggest why the drug also reduced the mitotic index in the small intestine and explain one consequence for the patient.[2]
Show answer
(a) Chromatids are separated by shortening of spindle microtubules attached at their centromeres; if microtubules cannot shorten, chromatids are not pulled to opposite poles.
B1 spindle fibres attached to centromeres / kinetochores must shorten (depolymerise); B1 so chromatids cannot be pulled apart to the poles.
(b) 71 %
M1 (0.21 − 0.06)/0.21 × 100; A1 = 71.4 ≈ 71 %.
(c) Intestinal lining cells divide rapidly (stem cells constantly replace epithelium), so they are also affected; the gut lining is not replaced, causing poor absorption / diarrhoea / sores.
(a) B They are repeated, non-coding base sequences at the ends of chromosomes.
Telomeres are many repeats of a short non-coding sequence at chromosome ends. A describes the centromere/kinetochore; D describes histones.
Q15Chromosome structure & telomeres · Core · Paper 2 style · 4 marks
(a)Describe the structure of a chromosome as it appears at the start of mitosis.[3]
(b)State the role of histone proteins in the nucleus.[1]
Show answer
(a) Two identical sister chromatids joined at a centromere; each chromatid is one long DNA molecule wound around histone proteins (chromatin), tightly coiled / condensed; telomeres at the ends.
1 mark each for any three of: two (sister) chromatids; joined at centromere; each chromatid contains one DNA molecule (double helix); DNA associated with / wound round histones; condensed / supercoiled; telomeres at ends.
(b) They package / coil the DNA so the long molecules fit into the nucleus (and help control access to genes).
B1 DNA wound around histones for packaging / condensing (accept: regulating gene expression).
Q16Chromosome structure & telomeres · Standard · Paper 2 style · 4 marks
(a)Explain why the ends of a DNA molecule become shorter each time the DNA is replicated, and why this does not normally cause the loss of genes.[3]
(b)Stem cells in bone marrow continue to divide throughout life. Suggest how their telomeres are kept long.[1]
Show answer
(a) DNA polymerase cannot replicate the very end of the lagging strand, so a short length of DNA is lost each replication; the lost DNA comes from the telomere, which is non-coding, so genes are protected.
B1 the extreme end of the (lagging) strand is not copied; B1 so the chromosome loses a short length each division; B1 the lost section is telomere / non-coding repeats, so no genes are lost (telomeres act as a buffer).
(b) They have active telomerase, an enzyme that adds repeats back onto the telomeres.
Q17Chromosome structure & telomeres · Challenge · Paper 2 style · 5 marks
The mean telomere length in white blood cells was measured in healthy volunteers in Lahore.age 20 years: 9.8 kbage 35 years: 9.2 kbage 50 years: 8.5 kbage 65 years: 7.9 kbage 80 years: 7.4 kb(1 kb = 1000 base pairs)
(a)Calculate the mean rate of telomere shortening between ages 20 and 80, in base pairs per year. Give your answer to the nearest whole number.[2]
(b)Suggest why telomere shortening may limit the number of times a body cell can divide.[2]
(c)Cancer cells can divide an unlimited number of times. Suggest one feature of telomeres in cancer cells that explains this.[1]
Show answer
(a) 40 bp per year
M1 (9.8 − 7.4) kb / (80 − 20) years = 0.04 kb per year; A1 × 1000 = 40 bp per year.
(b) When telomeres become very short, further replication would remove coding DNA / genes, so the cell stops dividing (or dies) to avoid losing genes.
B1 telomeres eventually too short to protect chromosome ends; B1 genes would be lost / chromosomes unstable, so cell stops dividing (senescence) or undergoes apoptosis.
(c) Telomerase is active, so telomeres are maintained at a constant length.
B1 telomerase active / telomere length maintained.
Q18Chromosome structure & telomeres · Challenge · Paper 2 style · 4 marks
(a)The telomere on one chromosome end is 10000 base pairs long and consists of repeats of the sequence TTAGGG. Calculate how many repeats are lost before the telomere reaches a critical length of 4000 base pairs.[1]
(b)The telomere loses 60 base pairs at each division. Calculate the number of divisions possible before the critical length is reached.[1]
(c)Suggest why a short repeated sequence is better at protecting chromosome ends than a gene would be.[2]
Show answer
(a) 1000 repeats
B1 (10000 − 4000) / 6 = 1000.
(b) 100 divisions
B1 6000/60 = 100.
(c) The repeats do not code for anything, so losing some of them has no effect on protein synthesis; the many repeats provide a long buffer that can be shortened many times.
B1 non-coding, so loss does not change any polypeptide; B1 many repeats so a long length can be lost over many divisions before genes are reached.
Q19Stem cells & cancer · Core · Paper 1 style · 1 mark
(a)Which statement describes a stem cell?[1]
Show answer
(a) B a cell that can divide by mitosis to give cells that can differentiate into specialised cells
Stem cells self-renew by mitosis and their daughter cells can differentiate. A describes a fully differentiated cell; D a red blood cell.
Q20Stem cells & cancer · Core · Paper 2 style · 4 marks
(a)State where stem cells that produce blood cells are found in an adult.[1]
(b)Children with thalassaemia major in Pakistan may be treated with a bone marrow transplant from a matched sibling. Explain how the donor stem cells can cure the condition.[3]
Show answer
(a) Bone marrow (red marrow).
B1 bone marrow.
(b) Donor stem cells settle in the bone marrow, divide by mitosis to self-renew, and differentiate into red blood cells containing normal haemoglobin, so regular blood transfusions are no longer needed.
B1 donor stem cells divide by mitosis / self-renew; B1 differentiate into (red) blood cells; B1 these have normal genes / make normal haemoglobin (for life, so transfusions no longer needed).
Q21Stem cells & cancer · Standard · Paper 2 style · 4 marks
Lung cancer is one of the most common cancers in men in Pakistan, and most cases are linked to smoking cigarettes, bidis or shisha.
(a)Explain how a mutation in a gene that controls cell division can lead to the formation of a tumour.[3]
(b)Tobacco smoke contains carcinogens. State what is meant by a carcinogen.[1]
Show answer
(a) The mutated gene produces a faulty protein (or no protein), so mitosis is no longer controlled; the cell divides repeatedly and uncontrollably; its daughter cells inherit the mutation, forming a mass of cells (tumour).
1 mark each for any three of: mutation changes the protein that regulates the cell cycle; cell divides by mitosis without control / continuously; daughter cells have the same mutation (clone); mass of abnormal cells = tumour; cells do not respond to signals to stop dividing / undergo apoptosis.
(b) A substance or agent that causes cancer (by increasing the rate of mutation of DNA).
B1 agent / substance that causes cancer / mutations leading to cancer.
Q22Stem cells & cancer · Standard · Paper 2 style · 4 marks
(a)Describe what is meant by cell differentiation.[2]
(b)Explain why stem cells taken from an early embryo can produce more types of cell than stem cells taken from adult bone marrow.[2]
Show answer
(a) The process by which an unspecialised cell becomes specialised for a particular function, by switching on (expressing) some genes and not others.
B1 cell becomes specialised (changes structure for a function); B1 only certain genes expressed.
(b) Embryonic stem cells have not yet started to differentiate, so they can express any genes and form any cell type; adult marrow stem cells are already partly committed and can form only blood cell types.
B1 embryonic cells can form (almost) all cell types / are totipotent or pluripotent; B1 adult stem cells only form a limited range (multipotent), e.g. blood cells.
Q23Stem cells & cancer · Challenge · Paper 2 style · 5 marks
(a)Suggest two differences between the behaviour of the cells in a malignant tumour and the cells of normal tissue.[2]
(b)Explain why exposure to radiation, such as UV light in a sunny region, increases the risk of skin cancer.[2]
(c)Explain why cancer is not an infectious disease.[1]
Show answer
(a) Malignant cells divide uncontrollably/more often and do not stop dividing on contact or by apoptosis; they can spread (metastasise) into other tissues; they do not differentiate normally.
1 mark each for any two of: divide more often / without control; do not undergo apoptosis; not specialised / undifferentiated; invade other tissues / spread through blood or lymph; telomerase active / unlimited divisions.
(b) UV damages DNA in skin cells, causing mutations; if the mutation affects genes controlling the cell cycle, cells divide uncontrollably.
B1 UV is a mutagen / damages DNA (base sequence changed); B1 in genes controlling cell division, causing uncontrolled mitosis.
(c) It is not caused by a pathogen and cannot be passed from person to person.
B1 not caused by a pathogen / not transmissible.
Q24Skill: measurement, graticule & magnification · Standard · Paper 3 style · 4 marks
A student calibrated an eyepiece graticule using a stage micrometer in which each division is 0.01 mm. Using the ×40 objective, 40 eyepiece units lined up exactly with 25 stage micrometer divisions.
(a)Calculate the length of one eyepiece graticule unit, in µm.[2]
(b)A cell in metaphase measured 4 eyepiece units from one end to the other. Calculate its actual length in µm.[1]
(c)Explain why the eyepiece graticule must be calibrated again if the student changes to a higher-power objective lens.[1]
Show answer
(a) 6.25 µm
M1 25 × 0.01 mm = 0.25 mm = 250 µm; A1 250/40 = 6.25 µm.
(b) 25 µm
B1 4 × 6.25 = 25 µm (accept 25.0 µm).
(c) The graticule scale stays the same size but the specimen is magnified more, so each eyepiece unit represents a smaller actual length.
B1 each eyepiece unit corresponds to a different (smaller) distance on the specimen at a different magnification.
Q25Skill: conclusions & interpreting data · Core · Paper 5 style · 4 marks
Sections of a mung bean root were cut at different distances behind the tip, and the mitotic index of each section was found.1 mm: 0.182 mm: 0.123 mm: 0.054 mm: 0.015 mm: 0.00
(a)Describe the relationship shown by the data.[2]
(b)A student concluded that no cell division takes place more than 4 mm from the root tip. Evaluate this conclusion.[2]
Show answer
(a) As distance from the tip increases, mitotic index decreases; it falls from 0.18 at 1 mm to 0 at 5 mm, falling most steeply between 2 and 3 mm.
B1 negative trend / mitotic index decreases with distance; B1 data quote with units, e.g. 0.18 at 1 mm to 0.00 at 5 mm.
(b) Partly supported: mitotic index is 0 at 5 mm, but 0.01 at 4 mm and only one root/one sample was used and only a limited number of cells counted, so some dividing cells may be missed beyond 4 mm.
B1 supported because MI is 0.00 at 5 mm / 0.01 at 4 mm; B1 limitation: only one root / small sample / no readings between 4 and 5 mm / no repeats, so cannot be sure.
Q26Skill: errors, reliability & improvements · Standard · Paper 3 style · 4 marks
A student wanted to find the mitotic index of onion root tips. She cut 1 cm from the end of one root, heated it in hydrochloric acid, stained it, placed it on a slide under a coverslip and pressed gently. She counted the dividing and non-dividing cells in one field of view at ×100 magnification.
(a)Identify two sources of error in this procedure and suggest an improvement for each.[4]
Show answer
(a) e.g. (1) cells overlapping because the squash was too thick → press harder / thinner squash so cells form a single layer; (2) only one field of view counted → count several fields of view / at least 200 cells from several root tips; (3) difficult to decide the stage of some cells (e.g. early prophase vs interphase) → use agreed criteria / higher magnification; (4) different root tips from different bulbs / different ages → use roots of the same length taken at the same time.
For each of two errors: B1 error stated in terms of this method; B1 matching improvement. Credit: overlapping cells / thick squash; small sample of cells / one field of view; subjective identification of stages; cells counted twice when moving the slide; roots of different lengths.
Q27Skill: planning a method · Challenge · Paper 5 style · 6 marks
(a)Describe a method the student could use to investigate the effect of temperature on the mitotic index of garlic root tips. Your method should be detailed enough for another person to follow.[6]
Show answer
(a) Grow garlic cloves with roots in water at 5 different temperatures (e.g. 10, 15, 20, 25, 30 °C) in incubators/water baths for the same time; cut 5 mm tips at the same time of day; hydrolyse in 1 mol dm⁻³ HCl at 60 °C for a fixed time; stain (e.g. acetic orcein / toluidine blue); squash; count cells in mitosis and total cells (at least 200 per tip, several fields); MI = dividing ÷ total; at least 3 roots per temperature; calculate mean MI.
1 mark each for any six of: IV with at least 5 temperatures and stated range; method of keeping temperature constant (incubator / thermostatic water bath); time roots grown at each temperature standardised; same variety / size of clove; tips cut to same length at same time of day; hydrolysis with HCl, fixed concentration, temperature and time; stain named; squash to a single layer; count number in mitosis and total (≥ 200 cells, random fields); calculate mitotic index; replicates ≥ 3 roots per temperature and mean.
Q28Skill: variables & controls · Core · Paper 5 style · 6 marks
A student investigated the effect of the concentration of caffeine solution on the mitotic index of onion root tips.
(a)Identify the independent variable and the dependent variable in this investigation.[2]
(b)State three variables that should be standardised, and explain why one of them must be kept constant.[3]
(c)Describe a suitable control for this investigation.[1]
Show answer
(a) Independent: concentration of caffeine. Dependent: mitotic index (number of cells in mitosis ÷ total cells counted).
B1 IV caffeine concentration; B1 DV mitotic index.
(b) e.g. temperature; time in solution; length of root tip cut; type/age of onion; hydrolysis time and acid concentration; stain and time in stain. Explanation e.g. temperature affects enzyme activity and the rate of mitosis, so would change mitotic index.
B1 × 2 for any three valid standardised variables (two correct = 1 mark, three = 2 marks); B1 valid explanation for one.
(c) Roots grown in distilled water with no caffeine, treated in the same way.
B1 0 % caffeine / distilled water, all other conditions the same.
Q29Skill: hypotheses & predictions · Standard · Paper 5 style · 3 marks
A researcher compares the mitotic index of bean root tips grown in a plant growth regulator solution with those grown in water.
(a)State a null hypothesis for this investigation.[1]
(b)The plant growth regulator is thought to stimulate cell division. Make a prediction about the results and explain it.[2]
Show answer
(a) There is no significant difference between the mean mitotic index of root tips grown in the plant growth regulator and those grown in water.
B1 no significant difference in mean mitotic index between the two treatments (must name both groups and the measured variable).
(b) Mean mitotic index will be higher in the treated roots, because more cells will enter mitosis at any one time if division is stimulated.
B1 prediction: higher mitotic index with regulator; B1 explanation: more cells dividing / shorter cell cycle so greater proportion in mitosis.
Q30Skill: statistics (SD, SE, CI, t-test, Spearman, chi-squared) · Challenge · Paper 5 style · 6 marks
Mitotic index (%) of 10 garlic root tips grown at each of two temperatures.25 °C: 14.2, 15.8, 13.9, 16.1, 15.0, 14.7, 15.5, 16.4, 14.9, 15.315 °C: 11.6, 12.9, 12.1, 13.4, 11.2, 12.5, 13.0, 11.9, 12.7, 12.2
(a)The standard deviation for root tips grown at 25 °C is 0.80 and for those at 15 °C is 0.68. Calculate the mean mitotic index (as a percentage) for each group.[2]
(b)Use the formula t = |x̄₁ − x̄₂| ÷ √(s₁²/n₁ + s₂²/n₂) to calculate the value of t. Give your answer to 2 decimal places.[2]
(c)The critical value of t at p = 0.05 for 18 degrees of freedom is 2.10. State what the student can conclude.[2]
Show answer
(a) 25 °C: 15.18 %; 15 °C: 12.35 %
B1 25 °C mean = 151.8/10 = 15.18; B1 15 °C mean = 123.5/10 = 12.35.
(c) t (about 8.5) is greater than 2.10, so the null hypothesis is rejected: the difference in mean mitotic index between 25 °C and 15 °C is significant (probability of it being due to chance is less than 5 %).
B1 calculated t > critical value 2.10 (df = 10 + 10 − 2 = 18); B1 reject null hypothesis / difference is significant, not due to chance (p < 0.05).
Similar questions
Questions that test the same subtopic in different years, grouped so you can work on one skill at a time.
References only: find each question in its paper (official link where available, otherwise through your school).
Mitosis & the cell cycleBanker87 questions · 9/9 sessions
