Questions on the cell cycle and mitosis

Question 1

Describe what happens to the chromosomes during each of the four stages of mitosis, in the order the stages occur.

Accepted Answer

B1 prophase: chromosomes condense and become visible, each already two sister chromatids B1 metaphase: chromosomes line up along the equator, attached to spindle fibres at their centromeres B1 anaphase: centromeres divide and the sister chromatids are pulled to opposite poles B1 telophase: chromosomes reach the poles, decondense, and a nuclear envelope re-forms around each group

Question 2

Explain how a mutation in a tumour suppressor gene can lead to the formation of a tumour.

Accepted Answer

B1 tumour suppressor genes normally halt the cell cycle at a checkpoint B1 a mutation can switch the gene off, so the protein it codes for is not produced or does not function B1 the checkpoint no longer stops cells with damaged DNA or incomplete replication from dividing A1 division continues unchecked and a mass of cells accumulates, which is a tumour

Question 3

Each chromosome carries a telomere at each of its ends. Explain why a short length of DNA is lost from the end of a chromosome every time a cell divides, and explain the two things a telomere protects.

Accepted Answer

B1 DNA polymerase builds a new strand only in the 5 prime to 3 prime direction and can only add nucleotides to an existing 3 prime end, so each fragment of the lagging strand has to be started with an RNA primer B1 when the last primer at the very end of the lagging strand is removed there is no fragment beyond it to extend from, so that gap cannot be filled and the end is left uncopied B1 a telomere is a short sequence repeated many times and codes for nothing, so the shortening removes repeats rather than genes and the cell can divide many times before a transcribed sequence is reached A1 the capped telomere also stops the end being treated as a broken chromosome, which would otherwise be joined to another end by repair enzymes or would halt the cycle at a damage checkpoint

Question 4

Compare cytokinesis in an animal cell with cytokinesis in a plant cell, referring to what forms in each case.

Accepted Answer

B1 in an animal cell a ring of protein filaments contracts beneath the cell surface membrane and pinches the membrane inwards, whereas in a plant cell the rigid cell wall prevents this pinching B1 in the plant cell, instead, vesicles line up along the middle of the cell and fuse to form a cell plate, which does not happen in an animal cell B1 the cell plate is where a new cell wall is built outwards from, separating the two daughter cells, whereas the animal cell simply pinches apart with no wall formed between them A1 in both cases cytokinesis divides the cytoplasm and follows the division of the nucleus in mitosis, rather than being part of mitosis itself

Question 5

Compare binary fission in a prokaryote with mitosis in a eukaryote, referring to the spindle, the nuclear envelope and the alignment of the chromosomes.

Accepted Answer

B1 mitosis uses a spindle of protein microtubules to move the chromosomes, whereas binary fission uses no spindle at all B1 mitosis breaks down and re-forms a nuclear envelope around each set of chromosomes, whereas a prokaryote has no nuclear envelope to break down or re-form B1 in mitosis the chromosomes line up at an equator before separating, whereas in binary fission the circular chromosome is not aligned at an equator, and dedicated proteins move the two copies apart A1 both processes produce two daughter cells, each carrying a complete copy of the parental DNA, but binary fission is far faster, dividing in about twenty minutes rather than about a day

Question 6

Outline how a root tip is prepared and stained so that its mitotic index can be counted reliably.

Accepted Answer

B1 a root tip within about 5 mm of the tip is used, because that is the meristem where cells are still dividing B1 the tip is warmed in dilute hydrochloric acid, which breaks down the middle lamella holding the cells together B1 it is stained with a dye such as toluidine blue or acetic orcein, which binds to DNA and makes the chromosomes visible B1 it is then squashed under a coverslip to a single layer of cells, because overlapping layers cannot be counted reliably

Question 7

A class prepared root tip squashes from one batch of garlic roots. Every student but one obtained a mitotic index between 0.15 and 0.20; that student obtained 0.05, having cut 15 mm from the end of each root, pressed the coverslip down lightly, and scored cells in early prophase as interphase. Explain how each of these three choices would lower the mitotic index, and suggest how the student could show that the low value was not caused by the way the cells were scored.

Accepted Answer

B1 a 15 mm piece includes tissue behind the meristem, where cells have stopped dividing and are elongating, so non-dividing cells are added to the total B1 a light squash leaves cells in overlapping layers, so chromosomes cannot be distinguished and dividing cells are missed or scored as interphase B1 cells in early prophase are in mitosis, so scoring them as interphase removes them from the number in mitosis; prophase is the longest stage, so many cells are affected B1 agree the criteria for each stage with the class and have a second person recount the same fields; accept scoring against reference photographs of each stage

Question 8

A cell from a mouse has 40 chromosomes. Explain why the cell still has 40 chromosomes at the end of G2, even though the quantity of DNA in it has doubled.

Accepted Answer

B1 DNA is replicated during S phase, so each chromosome is copied B1 the two copies stay joined at a single centromere as sister chromatids A1 chromosome number is counted by centromeres, so it is unchanged at 40 while the DNA quantity is doubled

Question 9

A student counts 240 cells in a stained root tip squash and finds 36 of them are visibly in mitosis. Calculate the mitotic index, and give your answer as a proportion.

Accepted Answer

M1 mitotic index = number of cells in mitosis divided by the total number of cells counted M1 36 divided by 240 A1 0.15, or equivalently 15 per cent

Question 10

A drug being tested as a cancer treatment binds to tubulin and prevents spindle fibres from forming. Suggest why this stops a tumour growing, and why the patient loses their hair.

Accepted Answer

B1 without spindle fibres the chromosomes cannot be attached and aligned at the equator, so the cell cannot pass from metaphase into anaphase B1 division stops, so the tumour cannot grow by producing more cells B1 the drug cannot distinguish tumour cells from any other dividing cells, and hair follicle cells divide rapidly, so hair is not replaced as it is shed

Question 11

Telomerase is active in cells of the germ line and in stem cells, but is switched off in most other body cells. Most cancer cells switch it back on. Suggest why telomerase activity is needed in the germ line, and suggest how switching it back on helps a tumour to grow.

Accepted Answer

B1 telomerase adds telomere repeats back onto the end of a chromosome, using an RNA template it carries, so it restores what replication fails to copy B1 germ line cells give rise to gametes and so to the next generation, which must receive full-length chromosomes; without telomerase the telomeres would be shorter in every generation A1 a body cell whose telomeres become too short stops dividing or dies, so a tumour cell with telomerase restored can go on dividing past the point where its divisions would otherwise have run out

Question 12

A mature muscle fibre stops dividing and enters G0. Explain what happens to a cell that enters G0, and explain why the fibre does not replicate its DNA again.

Accepted Answer

B1 a cell in G0 has left the cell cycle at the end of G1 and is not preparing to divide B1 it does not pass through S phase while it remains in G0, so its DNA is not replicated A1 many cells, such as neurones and mature muscle fibres, remain in G0 for the rest of their lives

Question 13

Explain how a mutation in a proto-oncogene can contribute to the formation of a tumour.

Accepted Answer

B1 a proto-oncogene normally codes for a protein that stimulates the cell cycle to proceed B1 a mutation can leave the gene, or the protein it codes for, permanently switched on rather than active only when division is appropriate A1 the cell cycle is no longer restrained by the normal signals, so the cell divides repeatedly and a mass of cells accumulates as a tumour

Question 14

A tissue has a cell cycle lasting 20 hours and a mitotic index of 0.15. Calculate the length of time, in hours, that a cell in this tissue spends in mitosis.

Accepted Answer

M1 the mitotic index gives the fraction of the cycle spent in mitosis, since it is the fraction of cells caught in that stage in a snapshot M1 multiply the cycle length by the index: 20 × 0.15 A1 3.0 hours (accept 3 hours) spent in mitosis

Question 15

Suggest why counting every cell in the microscope field, rather than a sample of cells chosen by eye, gives a more reliable mitotic index.

Accepted Answer

B1 a sample chosen by eye is likely to be biased towards the cells that are easiest to notice, which are the ones visibly in mitosis with condensed chromosomes B1 counting every cell in the field, dividing and non-dividing alike, avoids that bias and gives a count representative of the whole tissue B1 several fields should be counted and the totals added, because a single field may not be a representative sample of the whole root tip

Question 16

A student scored 250 cells in a stained root tip squash: 205 were in interphase, 20 in prophase, 8 in metaphase, 10 in anaphase and 7 in telophase. The counts are idealised for this question. The cell cycle in this root tip lasts 18 hours. Calculate how long anaphase lasts, in minutes.

Accepted Answer

M1 proportion of cells in anaphase = 10 ÷ 250 = 0.04, which is the fraction of the cycle spent in anaphase M1 0.04 × 18 hours = 0.72 hours A1 0.72 × 60 = 43 minutes; accept 43.2

Question 17

State what the G1 checkpoint checks and what the G2 checkpoint checks, before each allows the cell cycle to proceed.

Accepted Answer

B1 the G1 checkpoint checks that the cell is large enough, that nutrients are available and that the DNA is undamaged, before allowing entry to S phase B1 the G2 checkpoint checks that DNA replication is complete and that no errors or damage remain, before allowing entry to mitosis