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IB Biology D2.1: Cell and nuclear division worked solutions

Worked solutions to every practice question in D2.1: Cell and nuclear division of Photon Biology: Study Guide for the IB Diploma (SL & HL). Try each question, then compare your working with the full solution.

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Basic practice

Question A1

[1 mark]

Which description fits a pair of homologous chromosomes? (A) Two identical copies joined at one centromere (B) One maternal and one paternal chromosome with the same genes (C) Two chromosomes with identical alleles at every locus (D) The two chromosomes that decide sex

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B. Homologues are one maternal and one paternal chromosome with the same genes at the same loci, although the alleles may differ. A describes sister chromatids, C wrongly demands identical alleles, and D confuses homologues with the pair that decides sex.

Final answer

Question A2

[1 mark]

A diploid cell has 8 chromosomes. How many chromosomes and chromatids does it hold after DNA replication but before mitosis? (A) 8 chromosomes, 8 chromatids (B) 8 chromosomes, 16 chromatids (C) 16 chromosomes, 16 chromatids (D) 16 chromosomes, 32 chromatids

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B. Replication doubles the chromatids but does not change the number of chromosomes, because the sister chromatids stay joined at one centromere: 8 chromosomes made of 16 chromatids. A ignores replication, C counts every chromatid as a chromosome, and D doubles both.

Final answer

Question A3

[1 mark]

Which job needs meiosis rather than mitosis? (A) Healing a cut in the skin (B) Growth of a root tip (C) Making sperm in a testis (D) Budding of a yeast cell

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C. Haploid gametes are made by meiosis. Healing a wound (A), growth of a root tip (B) and budding of yeast (D) use mitosis to make identical diploid or genetically identical cells.

Final answer

Question A4

[1 mark]

In a dividing cell the chromosomes are in a single line across the equator, each attached to spindle microtubules from both poles. Which phase is this? (A) Prophase (B) Metaphase (C) Anaphase (D) Telophase

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B. A single line of chromosomes on the equator is metaphase. In prophase the chromosomes are scattered, in anaphase the chromatids are moving apart, and in telophase they are in two groups.

Final answer

Question A5

[1 mark]

What happens during cytokinesis in a plant cell? (A) A ring of actin and myosin pinches the cell in two (B) Vesicles build a new membrane and wall across the equator (C) Centrioles pull the cell membrane inwards (D) The nuclear membrane breaks down

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B. In a plant cell, vesicles bring sections of membrane and cell wall material to the equator, where they fuse to form the cell plate. A describes an animal cell, C names a structure that higher plants lack, and D is an event of prophase.

Final answer

Question A6

[1 mark]

Which division is an example of unequal cytokinesis? (A) Mitosis of a skin cell (B) The first division of a zygote (C) Budding of a yeast cell (D) Division of a bacterium

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C. A yeast bud receives far less cytoplasm than its parent. Mitosis of a skin cell (A) and the first division of a zygote (B) share the cytoplasm about equally, and bacteria (D) divide by binary fission, which is not mitosis at all.

Final answer

Question A7

[1 mark]

Which event occurs in both mitosis and meiosis? (A) Pairing of homologous chromosomes (B) Crossing over between non-sister chromatids (C) Condensation of chromosomes with histones (D) Reduction of the chromosome number

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C. Both divisions condense the chromosomes by supercoiling DNA with histones. Pairing of homologues (A), crossing over (B) and reduction of the chromosome number (D) occur only in meiosis.

Final answer

Question A8

[1 mark]

In which stage of meiosis do sister chromatids separate? (A) Prophase I (B) Anaphase I (C) Anaphase II (D) Telophase I

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C. Sister chromatids separate in anaphase II. In anaphase I (B) the homologues are pulled apart while the centromeres stay intact, and prophase I (A) and telophase I (D) do not separate chromatids.

Final answer

Question A9

[1 mark]

A human egg cell with 24 chromosomes is fertilised by a normal sperm. How many chromosomes does the zygote have, and what is the condition? (A) 47, trisomy (B) 47, polyploidy (C) 48, trisomy (D) 46, normal

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A. A normal sperm has 23 chromosomes, so the zygote has 24+23=47, with three copies of one chromosome: trisomy. It is not polyploidy (B), which means whole extra sets, and 48 (C) or 46 (D) are wrong counts.

Final answer

Question A10

[1 mark]

Which process produces new mixtures of maternal and paternal chromosomes in gametes? (A) Crossing over in prophase I (B) Random orientation in metaphase I (C) DNA replication in S phase (D) Cytokinesis after meiosis II

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B. Random orientation of bivalents in metaphase I decides which chromosomes go to which pole. Crossing over (A) mixes alleles on chromatids, replication (C) copies DNA and cytokinesis (D) shares the cytoplasm.

Final answer

Question A11

HL[1 mark]

A cell in G1 has 4.0 pg of DNA. How much DNA does it hold in G2, and how much does each daughter cell hold just after mitosis? (A) 4.0 pg; 4.0 pg (B) 8.0 pg; 4.0 pg (C) 8.0 pg; 8.0 pg (D) 4.0 pg; 2.0 pg

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B. S phase doubles the DNA to 8.0 pg, and mitosis shares it equally between two cells, so each daughter has 4.0 pg again. A forgets replication, C forgets that mitosis divides the DNA, and D halves the DNA.

Final answer

Question A12

HL[1 mark]

A mutation converts a proto-oncogene into an oncogene. What is the likely effect on the cell? (A) It stops dividing (B) It divides when division is not needed (C) It cannot replicate DNA (D) It becomes haploid

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B. An oncogene stimulates division when it is not needed, which can lead to a tumour. A cell with an oncogene does not stop dividing (A), does not lose DNA replication (C) and does not become haploid (D).

Final answer

Tutorial practice

Question B1

[4 marks]

Outline how mitosis produces two daughter nuclei that are genetically identical to the parent nucleus.

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Any four (1 each): DNA is replicated before mitosis, in S phase, so each chromosome has two identical sister chromatids; chromosomes condense in prophase and attach to spindle microtubules; chromosomes line up on the equator in metaphase; centromeres divide and microtubules pull sister chromatids to opposite poles in anaphase; each pole receives a copy of every chromosome, so each new nucleus has the same genome; nuclear membranes re-form in telophase. Marks are for events that explain identity, not for naming phases only.

Final answer

DNA is replicated in S phase so each chromosome is two identical sister chromatids; chromosomes condense and attach to the spindle, line up in metaphase; centromeres divide and sister chromatids are pulled to opposite poles in anaphase; each pole receives one copy of every chromosome and a nuclear membrane forms round each group.

Question B2

[4 marks]

Five cells from an onion root tip are described in the table. They are lettered P to T. Cell P: nuclear membrane absent, chromosomes short and thick and scattered. Cell Q: nuclear membrane round each group, chromosomes in two groups, uncoiling. Cell R: membrane absent, chromosomes in one line across the centre. Cell S: membrane absent, V-shaped chromosomes moving to opposite ends. Cell T: membrane present, chromosomes thin threads and not separate. (a) Identify the phase of cell R. [1] (b) State the order of the five cells in the cell cycle, starting with T. [1] (c) Outline what happens next in cell Q. [2]

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(a) Cell R is in metaphase (1): the chromosomes are in a single line across the equator. (b) T, P, R, S, Q (1): interphase, prophase, metaphase, anaphase, telophase. (c) Cell Q is in telophase, so cytokinesis follows and splits the cytoplasm (1) so that each of the two nuclei is in its own daughter cell, and the chromosomes continue to uncoil (1).

Final answer

(a) Metaphase. (b) T, P, R, S, Q. (c) Cytokinesis divides the cytoplasm so that each new nucleus ends up in its own cell; the chromosomes uncoil.

Question B3

[5 marks]

A dividing animal cell and a dividing plant cell are both at the end of mitosis. (a) Compare how the cytoplasm divides in the two cells. [3] (b) Explain why each daughter cell must receive at least one mitochondrion. [2]

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(a) Animal cell: a ring of actin and myosin contracts (1) and pinches the plasma membrane inwards, forming a cleavage furrow (1). Plant cell: vesicles assemble sections of membrane and wall at the equator, forming a cell plate (1). Award at most 3. (b) Mitochondria can only be made by division of existing mitochondria (1), so a daughter cell with none could not make ATP by aerobic respiration or make new mitochondria (1).

Final answer

(a) Animal: a contractile ring of actin and myosin pinches the plasma membrane in, a cleavage furrow. Plant: vesicles bring membrane and wall material to the equator and form a cell plate. (b) A mitochondrion arises only by division of an existing one, so a daughter without one cannot make ATP aerobically or make more mitochondria.

Question B4

[4 marks]

Distinguish between mitosis and meiosis.

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Any four pairs (1 each; both sides needed): one division in mitosis, two in meiosis; two cells formed, four cells formed; daughters diploid, daughters haploid; daughters genetically identical, daughters genetically different; no pairing or crossing over of homologues, pairing and crossing over in prophase I; produces body cells for growth and repair, produces gametes.

Final answer

Mitosis has one division and meiosis two; mitosis gives two cells, meiosis four; mitosis keeps the chromosome number (diploid), meiosis halves it (haploid); mitosis gives identical cells, meiosis genetically different cells; homologues pair and cross over only in meiosis.

Question B5

[5 marks]

A plant species has 2n=12. (a) Calculate the number of different chromosome combinations in gametes from random orientation, ignoring crossing over. [1] (b) Explain how meiosis and fertilisation produce genetic variation. [4]

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(a) n=6 pairs, so 2^6=64 combinations (1). (b) Any four (1 each): crossing over occurs in prophase I; between non-sister chromatids of homologues; this gives new combinations of alleles on a chromatid; random orientation of bivalents in metaphase I; so gametes receive different mixtures of maternal and paternal chromosomes; any male gamete may fuse with any female gamete at fertilisation; so each zygote has a new combination of alleles.

Final answer

(a) 2^6=64. (b) Crossing over in prophase I between non-sister chromatids gives new allele combinations on chromatids; random orientation of bivalents in metaphase I gives different mixes of maternal and paternal chromosomes; random fusion of gametes from two parents gives new combinations in the zygote.

Question B6

[5 marks]

A karyogram of a girl shows 47 chromosomes, with three copies of chromosome 21. (a) State the number of chromosomes in the egg, assuming the sperm was normal and the extra chromosome came from the egg. [1] (b) Explain how non-disjunction can lead to this karyogram. [3] (c) State one factor that increases the frequency of this condition. [1]

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(a) 24 (1), since 47-23=24. (b) Non-disjunction (1): the two chromosomes 21 failed to separate, either homologues in meiosis I or sister chromatids in meiosis II (1); the egg had two copies of chromosome 21 and fused with a normal sperm carrying one, so the zygote had three (1). (c) The age of the mother, with the risk rising as age increases (1).

Final answer

(a) 24 chromosomes if the extra chromosome came from the egg and the sperm was normal with 23. (b) Homologues failed to separate in meiosis I (or sister chromatids in meiosis II), so one gamete had two copies of chromosome 21; fertilisation with a normal gamete gave three copies. (c) Maternal age.

Question B7

[5 marks]

A cell is heterozygous for two linked genes. One chromosome of a homologous pair carries alleles A and B, and the other carries a and b. (a) State the stage of meiosis in which crossing over occurs. [1] (b) Name the point at which chromatids exchange segments, and state which chromatids take part. [2] (c) State the allele combinations on the four chromatids after one crossover between the two genes, and identify the recombinants. [2]

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(a) Prophase I (1). (b) The chiasma (1) forms between non-sister chromatids, one from each homologue (1). (c) After one crossover between the genes the four chromatids carry AB, Ab, aB and ab (1); Ab and aB are the recombinants because they carry new combinations of alleles (1).

Final answer

(a) Prophase I. (b) A chiasma, between non-sister chromatids, one from each homologue. (c) AB, Ab, aB, ab; the recombinants are Ab and aB.

Question B8

[4 marks]

A diploid animal species has 2n=20. (a) State the number of chromosomes in a gamete. [1] (b) State the number of chromosomes in a zygote. [1] (c) Explain what would happen to the number in later generations if gametes were made by mitosis. [2]

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(a) 10 (1). (b) 20 (1), since 10+10. (c) Without meiosis the gametes would have 20 chromosomes, so each fertilisation would double the number: 40, then 80 (1); meiosis halves the number, so fertilisation restores 20 and the number stays constant between generations (1).

Final answer

(a) 10. (b) 20. (c) Fertilisation would double the number in each generation (40, 80, ...); meiosis halves the number in gamete formation so it stays at 20.

Question B9

HL[6 marks]

The table shows the DNA content and the concentration of one cyclin in a dividing cell, measured at hourly intervals. The cell completes mitosis between 12 and 13 hours. Time in hours 0, 2, 4, 6, 8, 10, 11, 12, 13. DNA content: 1.0, 1.0, 1.0, 1.5, 2.0, 2.0, 2.0, 2.0, 1.0. Cyclin concentration (arbitrary units): 0.2, 0.3, 0.4, 0.6, 0.8, 1.0, 1.2, 1.3, 0.2. (a) Identify the phase between 4 and 8 hours and give the evidence. [2] (b) Explain the change between 12 and 13 hours. [1] (c) Explain how cyclin concentration controls the cell cycle. [3]

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(a) S phase (1); the DNA content doubles from 1.0 to 2.0 between hours 4 and 8 because DNA is replicated (1). (b) The DNA content falls from 2.0 to 1.0 because the DNA is shared between the two daughter cells as mitosis and cytokinesis end (1). (c) The cyclin concentration rises through interphase to a peak of 1.3 just before mitosis ends (1); a threshold concentration of a specific cyclin has to be reached for the cell to pass a checkpoint (1); the concentration then falls sharply, to 0.2 at 13 hours, so the cell leaves mitosis and the cycle can restart (1).

Final answer

(a) S phase: the DNA content rises from 1.0 to 2.0 as the DNA is replicated. (b) 2.0 to 1.0: the DNA is shared between the two daughter cells at the end of mitosis. (c) The cyclin concentration rises to a peak just before mitosis ends; a threshold concentration of a specific cyclin must be reached to pass a checkpoint; the fall then lets the cell leave mitosis.

Question B10

HL[5 marks]

A smoker develops a lung tumour. Some of its cells are found in a lymph node near the lung. (a) Distinguish between the primary tumour and a secondary tumour. [2] (b) Explain how a mutation in a tumour suppressor gene can lead to a tumour. [2] (c) State one mutagen. [1]

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(a) The primary tumour is the first tumour, where the cancer started (1); a secondary tumour forms when cells spread through blood or lymph (metastasis) and divide at a new site (1). (b) A tumour suppressor gene normally slows or stops the cell cycle (1); a mutation that disables it removes the brake, so mitosis continues without control and a tumour forms (1). (c) A mutagen such as a chemical in tobacco smoke, ultraviolet light or X-rays (1).

Final answer

(a) The primary tumour is where the cancer started (the lung); a secondary tumour forms when cells metastasise and divide in another place (the lymph node). (b) A mutation in a tumour suppressor gene removes its brake on division, so cells divide uncontrollably. (c) A chemical in tobacco smoke, or ultraviolet light or X-rays.

Question B11

HL[5 marks]

Students treated onion roots with compound X and compared them with untreated roots. They counted 200 cells on each of three slides per group. Cells in mitosis out of 200 on each slide. Control: 28, 30, 26. Compound X: 12, 10, 8. (a) Calculate the mitotic index of the control group as a percentage. [2] (b) State the mitotic index of the treated group and suggest what it shows. [2] (c) Explain why the students counted several fields. [1]

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(a) Total cells =600 (1); control index =(28+30+26)/600=84/600=0.14, or 14% (1). (b) Treated index =(12+10+8)/600=0.05, or 5% (1); a smaller proportion of cells is in mitosis, so compound X may slow cell division (1). (c) Different fields contain different cells, so counting several fields gives a better estimate of the proportion in mitosis (1).

Final answer

(a) 84/600=0.14, or 14%. (b) The treated roots have a lower index, 5%: fewer cells are dividing, so X may slow or stop mitosis; three slides and one root species limit the conclusion. (c) Different fields contain different cells, so more fields give a better estimate of the true proportion.

Advanced practice

Question C1

HL[6 marks]

A deep graze removes the surface cells from a patch of skin. Cells in the deeper layer at the wound edge begin dividing to cover it. Plant roots also grow by cell proliferation. (a) State where cell division occurs in a growing root and explain why it continues there. [2] (b) Assume that 500 wound-edge cells all divide and every daughter divides again, with no loss. Calculate the number of rounds of division needed to produce at least 12 000 cells. [2] (c) Each cell cycle takes 20 hours. Calculate the time to close the wound. [1] (d) Suggest why a real wound takes longer than this value. [1]

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(a) The meristem at the root tip (1), whose cells divide by mitosis to add new cells, some staying undifferentiated (1). (b) 500\times2^n\geq12\,000, so 2^n\geq24 (1); 2^4=16 is too small and 2^5=32 is enough, so 5 divisions, rounding up (1). (c) 5\times20=100 hours (1). (d) Not every cell divides on schedule, and some cells die or stay still, so the wound takes longer than the model predicts (1).

Final answer

(a) Meristem at the root tip: cells divide there by mitosis, adding cells for growth. (b) 500\times2^{n}\geq12\,000 gives 2^n\geq24, so n=5. (c) 5\times20=100 hours. (d) Not all cells divide at the same rate, so a real wound takes longer.

Question C2

HL[6 marks]

A student suspects that caffeine solution slows cell division in onion root tips. (a) Outline a method to test this, naming the independent variable and two variables to keep constant. [3] (b) Explain how the student should count cells and calculate the result. [2] (c) A single slide gave a mitotic index of 15% for water and 9% for caffeine. Evaluate whether this proves that caffeine slows division. [1]

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(a) Independent variable: concentration of caffeine, with a water control (1); two controlled variables, such as exposure time, temperature, onion variety or root length (1); root tips stained and squashed, and viewed at the same magnification (1). (b) Count the cells in mitosis and all cells in several randomly chosen fields on each slide (1); mitotic index = cells in mitosis divided by all cells counted, found for each slide, then the means and ranges compared (1). (c) One slide per treatment gives no range, so a difference of 15% against 9% is suggestive but not firm; more slides are needed (1).

Final answer

(a) Independent: caffeine concentration (water as control); controlled: exposure time, temperature, onion variety, root length, stain and magnification. Stain and squash the root tip, view at the same magnification. (b) Count the cells in prophase, metaphase, anaphase and telophase and all cells in several random fields, then divide: mitotic index = cells in mitosis / total cells. (c) One slide each gives no measure of spread, so the difference of 15% against 9% is not firm evidence.

Question C3

[5 marks]

A single-celled alga has 4 mitochondria. During cytokinesis each mitochondrion passes at random to one daughter cell or the other, each with probability \frac12, independently. (a) Calculate the probability that daughter cell X receives no mitochondrion. [1] (b) Calculate the probability that one of the two daughters receives none. [2] (c) Explain why a daughter cell without a mitochondrion is unlikely to survive. [2]

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(a) Daughter X receives none only if all four go to Y: (\tfrac12)^4=\tfrac1{16} (1). (b) Daughter Y is empty with probability \tfrac1{16}, and the two events cannot both happen because the four mitochondria must go somewhere (1), so the probability that one daughter is empty is \tfrac1{16}+\tfrac1{16}=\tfrac18 (1). (c) A new mitochondrion can arise only by division of an existing one (1), so a daughter with none cannot make ATP by aerobic respiration or rebuild its mitochondria and is unlikely to survive (1).

Final answer

(a) (\tfrac12)^4=\tfrac1{16}. (b) \tfrac1{16}+\tfrac1{16}=\tfrac18, since both daughters cannot be empty. (c) Mitochondria arise only from existing mitochondria and cannot be made new from DNA in the daughter, so a daughter with none cannot make ATP by aerobic respiration and cannot recover.

Question C4

[6 marks]

A species has 2n=10. (a) Calculate the number of different chromosome combinations in a gamete from random orientation, ignoring crossing over. [1] (b) Calculate the number of different chromosome combinations in a zygote from two unrelated parents. [1] (c) Explain why the real number of different gametes is greater than your answer to (a). [3] (d) State one reason why this variation helps a species. [1]

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(a) n=5 pairs, so 2^5=32 (1). (b) Two parents: 32\times32=1024 (1). (c) Crossing over occurs in prophase I (1) between non-sister chromatids, so each chromatid carries a new combination of alleles (1), and the number of different gametes is far greater than 32 (1). (d) Variation allows natural selection to act, so some offspring survive when conditions change (1).

Final answer

(a) 2^5=32. (b) 32\times32=1024. (c) Crossing over in prophase I swaps segments between non-sister chromatids, so each chromatid carries new allele combinations and the number of different gametes is far greater than 32. (d) Evolution: variation allows natural selection to act, so the population can adapt to change.

Question C5

[6 marks]

The table gives illustrative values for the risk of a baby with Down syndrome at different ages of the mother. Illustrative risk of Down syndrome by age of mother: age 25, 1 in 1250; 30, 1 in 950; 35, 1 in 350; 40, 1 in 100; 45, 1 in 30. (a) Describe the trend shown by the data. [2] (b) Calculate the risk at age 40 as a percentage. [1] (c) Calculate how many times greater the risk is at age 45 than at age 25. [1] (d) Outline how an error in meiosis in the mother can lead to Down syndrome. [2]

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(a) The risk rises as the mother's age increases (1), from 1 in 1250 at 25 to 1 in 30 at 45, and more steeply after 35 (1). (b) 1/100=0.01, or 1.0% (1). (c) 1250/30=41.7, so about 42 times higher at 45 than at 25 (1). (d) The two chromosomes 21 fail to separate in meiosis, either homologues in meiosis I or sister chromatids in meiosis II, so the egg has two copies (1); fertilisation by a normal sperm gives a zygote with three copies of chromosome 21, trisomy, and 47 chromosomes (1).

Final answer

(a) The risk rises with maternal age, slowly at first and then steeply after 35. (b) 1/100=1.0\%. (c) 1250/30\approx42 times higher. (d) Non-disjunction in meiosis in the mother gives an egg with two copies of chromosome 21; fertilisation by a normal sperm gives a zygote with three copies.

Connected questions

Question X1

Connected questions[6 marks]

D2.1 + D1.1 + A1.2. A chromosome in a cell grown in heavy nitrogen (^{15}N) has both DNA strands labelled. The cell is moved to light nitrogen (^{14}N) and goes through one S phase and mitosis, then a second S phase and mitosis. (a) State the make-up of each chromatid after the first S phase. [1] (b) Explain why the two daughter cells receive identical DNA. [2] (c) After the second mitosis, state the fraction of the DNA molecules for this chromosome that contain only light nitrogen. [2] (d) State the role of histones in preparing the chromosome for division. [1]

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(a) Each of the two sister chromatids has one heavy (parental) strand and one light (new) strand (1), because replication is semi-conservative. (b) Each new strand is made by complementary base pairing, so both chromatids are identical (1); they separate in anaphase and one goes to each daughter (1). (c) In the second S phase each hybrid molecule gives one hybrid and one all-light molecule (1), so \tfrac12 of the molecules are all light (1). (d) Histones: DNA is wound round them and supercoiled to condense the chromosome (1).

Final answer

(a) Each chromatid has one old heavy strand and one new light strand (semi-conservative replication). (b) The sister chromatids are identical copies made by complementary base pairing and are pulled to opposite poles. (c) \tfrac12 of the DNA molecules are all light and \tfrac12 are hybrid. (d) Histones: DNA is wound round them to condense the chromosome.

Question X2

Connected questions · HL[6 marks]

D2.1 + D1.3 + D3.1. Ultraviolet light changes one base in a tumour suppressor gene in a skin cell. The cell divides without control and forms a small mole. Later some cells of the mole invade the skin around it. (a) State what is meant by a point mutation. [1] (b) Explain how the mutation can lead to a tumour. [2] (c) Distinguish a mole that stays in place from one that has become malignant. [2] (d) Explain why the mutation is not passed to the person's children. [1]

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(a) A point mutation changes a single base in the DNA sequence (1). (b) The mutated tumour suppressor gene no longer slows or stops the cell cycle (1), so mitosis continues and a tumour forms (1). (c) A mole that stays in one place is benign (1); invading the surrounding tissue shows it has become malignant (1). (d) A mutation in a somatic skin cell is not in a gamete, so it is not inherited by offspring (1).

Final answer

(a) A base substitution is a change of one base for another. (b) The gene no longer slows the cell cycle, so mitosis continues. (c) A mole that stays in place is benign; invasion is a sign of malignancy. (d) The mutation is in a skin cell, not a gamete, so it is not passed to offspring.

Question X3

Connected questions[6 marks]

D2.1 + D3.1 + D4.1. A fungus reproduces both by budding and by sexual spores formed after meiosis. (a) Explain why a bud has exactly the same alleles as its parent cell. [2] (b) Explain why spores formed by meiosis and fusion give genetically varied offspring. [2] (c) Suggest an advantage of sexual spores if the environment changes. [2]

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(a) Budding involves mitosis and gives offspring genetically identical to the parent (1), since sister chromatids are copies of one another (1). (b) Meiosis shuffles alleles by crossing over and random orientation (1) and fusion of spores or gametes is random, so offspring differ from each other and from the parents (1). (c) After an environmental change, identical offspring are all equally vulnerable (1), whereas variation gives some individuals an advantage that natural selection can favour, so their alleles become more frequent (1).

Final answer

(a) Budding is mitosis: offspring are genetically identical to the parent. (b) Meiosis and random fertilisation give genetically different offspring. (c) If the environment changes, identical offspring all share the same weakness, while varied offspring include individuals that survive and pass on alleles.

From the book

Photon Biology: Study Guide for the IB Diploma (SL & HL)

Cell and nuclear division from the core ideas to full exam questions, with question types, worked examples and graded practice.

About the book