Biology by Bradford
D1.1 DNA replication. Practice questions with markscheme.
68 original IB-style questions on D1.1, written from the 2025 guide: 31 multiple-choice, 25 short-answer, 8 data-based, 2 extended-response parts, 1 labelling, 1 drawing. Below are two ready-made practice papers built from them, one at standard level (22 marks) and one at higher level (20 marks); switch between them above the paper. Hand one out as a class quiz or homework, or sit it yourself and mark against the scheme. Print it, project it, or build a fresh one on the same topic.
What the guide asks for
5 statements at SL and HL, 4 additional higher level.
- D1.1.1SL / HL DNA replication as production of exact copies of DNA with identical base sequences
- D1.1.2SL / HL Semi-conservative nature of DNA replication and role of complementary base pairing
- D1.1.3SL / HL Role of helicase and DNA polymerase in DNA replication
- D1.1.4SL / HL Polymerase chain reaction and gel electrophoresis as tools for amplifying and separating DNA
- D1.1.5SL / HL Applications of polymerase chain reaction and gel electrophoresis
- D1.1.6HL Directionality of DNA polymerases
- D1.1.7HL Differences between replication on the leading strand and the lagging strand
- D1.1.8HL Functions of DNA primase, DNA polymerase I, DNA polymerase III and DNA ligase in replication
- D1.1.9HL DNA proofreading
In the bank for D1.1
- 31 multiple-choice
- 25 short-answer
- 8 data-based
- 2 extended-response parts
- 1 labelling
- 1 drawing
- 27 higher level only
Every question is original and tagged to a guide statement. See the whole bank →
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The practice paper
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The gel shows DNA profiles of a mother (M), her child (C) and two men (F1, F2) who could be the father. Which conclusion is correct?
- F1 is the father, because he shares the 100 base-pair band with the child and the mother does not
- F2 is the father, because every band of the child that is not from the mother is present in F2
- Either man could be the father, because each shares exactly one band with the child's profile
- Neither man is the father, because the child has a band at 850 that is absent from both men
How does complementary base pairing ensure that DNA replication produces identical copies?
- The bases on the new strand are identical to those on the template
- Free nucleotides are only added where they pair with the template
- Hydrogen bonds transfer the base sequence directly to the new strand
- Any base can pair with any other base on the template strand
A cut in the skin heals as cells at the edge of the wound divide repeatedly to replace the lost tissue. Which statement about DNA replication in these cells is correct?
- Replication occurs once, before the first division, and the copies are shared out among all the cells produced later
- Replication occurs before every division so that each new cell receives a complete set of DNA with identical base sequences
- Replication occurs only in the cells that are about to die, to replace the DNA that they have lost
- Replication produces DNA with slightly altered base sequences so that the new cells can become specialized as skin cells rather than any other type
Outline the roles of helicase and DNA polymerase in DNA replication.
Two PCR tubes were set up with the same amount of template DNA, the same concentrations of the four DNA nucleotides and the same amount of Taq polymerase. The primers in both tubes bind at either end of the same region, 500 base pairs long. Tube 1 contained the standard concentration of primers and tube 2 contained one tenth of that concentration. Both tubes were run in the same PCR machine. After every second cycle from cycle 10 onwards, a small sample was taken from each tube and used to calculate the total mass of DNA product in the tube. The graph shows the results.
DNA from a mother, her child, and two men (X and Y) was amplified by PCR and separated by gel electrophoresis. The table shows the bands present in each profile at five variable sites; each person shows two bands per site, one inherited from each parent. Any band in the child's profile that is not in the mother's profile must have been inherited from the biological father.
| Site | Mother | Child | Man X | Man Y |
|---|---|---|---|---|
| 1 | a, b | a, d | c, e | d, e |
| 2 | f, h | f, g | g, i | g, h |
| 3 | j, l | j, k | l, m | k, n |
| 4 | p, r | p, q | q, r | o, q |
| 5 | s, v | s, u | t, v | u, w |
Original practice questions © Biology by Bradford · CC BY-NC-SA 4.0 · Not affiliated with or endorsed by the International Baccalaureate Organization.
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One mark per point; / separates alternative wording within a point, OR separates alternative answers, words in brackets are not required, underlined words are essential. OWTTE = or words to that effect.
- B: the child's bands at 850 and 500 match the mother, so the bands at 600 and 100 must come from the father; F2 has both, whereas F1 lacks the 600 band;
- B: pairing rules (A with T, G with C) mean the template dictates the new sequence exactly (new strand is complementary, so the copy of the partner strand is identical);
- B: growth and tissue replacement depend on repeated cell division, and every division must be preceded by replication so that both daughter cells receive an exact copy of the base sequence; A fails because a single copy could not supply every later cell with a full genome; C confuses replacement of cells with replacement of DNA; D describes a misconception, since specialization comes from which genes are expressed, not from changes to base sequences;
- helicase unwinds the (double) helix;
- helicase separates the two strands by breaking hydrogen bonds (between bases);
- DNA polymerase adds (free) nucleotides complementary to the template strand;
- (polymerase) links nucleotides (by covalent/phosphodiester bonds), forming the new strand (and proofreads), OWTTE;
- (a) [2 max]
- in both tubes the mass of product stays close to zero until about cycle 16, then rises steeply, then levels off;
- the two tubes give the same mass of product up to about cycle 16 to 18;
- tube 2 levels off earlier (at about cycle 24 to 26) than tube 1 (at about cycle 28 to 30);
- tube 2 levels off at a lower mass (about 600 ng) than tube 1 (about 2000 ng) / at about 30 % of the maximum in tube 1;
At least one similarity and one difference are needed for [2].
- (b) [2 max]
- every new strand starts from a primer / Taq polymerase can only add nucleotides to a primer bound to the template;
- each primer becomes part of the new strand, so primers are used up as product is made;
- in tube 2 the primers run out after fewer cycles / when less product has been made, so no new strands can be started and the mass stops increasing;
- each double-stranded product molecule contains two primers, so there can be no more product molecules than there were pairs of primers, OWTTE;
- (c) [2 max]
- tube 1 stops increasing before its primers are used up / tube 1 is limited by some factor other than primers, so its maximum is lower than its primers would allow;
- the free DNA nucleotides may be used up (in tube 1);
- the Taq polymerase may lose some activity after many cycles of heating to about 95 °C;
- when there is a lot of product, the separated product strands pair with each other rather than with primers as the mixture cools, OWTTE;
Accept other reasonable factors that limit tube 1 but not tube 2.
- (d) [3 max]
- the bands from both tubes would be in the same position / the same distance from the wells, level with the 500 base-pair band of the ladder;
- (because) the same primers bind at the ends of the same region, so both products are the same length / 500 base pairs;
- fragments are separated by length only (shorter fragments moving further towards the positive electrode), so the amount of DNA does not change how far a band moves, OWTTE;
- the band from tube 1 would be brighter/thicker/more intense than the band from tube 2;
- (because) the same volume from tube 1 contains more DNA product (about 2000 ng against 600 ng in the whole tube / about three times as much);
- the lower primer concentration changes the amount of product made, not its length, OWTTE;
Award [2] max unless both the position and the brightness of the bands are compared. Do not accept that the tube 2 band moves further or less far because there is less DNA.
- (a) [1]
- band a;
- (b) [2]
- man Y;
- every paternal band in the child (d, g, k, q, u) appears in Y's profile, whereas X lacks d, k and u;
- (c) [2]
- the original sample (blood/saliva/hair) contains very little DNA;
- PCR amplifies the (target) sequences to amounts large enough to detect as visible bands, OWTTE;
- (d) [2]
- if everyone carried the same bands, profiles could not distinguish individuals;
- variable (short tandem repeat) sites give (combinations of) band patterns that are (effectively) unique to an individual, apart from identical twins, OWTTE;
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The diagram shows one cycle of the polymerase chain reaction. Why is Taq polymerase used rather than a human DNA polymerase?
- It adds nucleotides more accurately than any other DNA polymerase
- It can use RNA as well as DNA as the template for the new strands
- It is not denatured at the temperature used to separate the strands
- It can copy the target sequence without needing any primers
A single bacterium divides repeatedly by binary fission and forms a colony of about one million cells. The complete genomes of 100 cells from the colony are sequenced. In 99 of them the base sequence is identical to that of the original bacterium; in the other one, a single base differs. Which statement best accounts for these results?
- Replication almost always gives identical copies, but a rare copying error is then inherited by the cells descended from that copy
- Replication is always exact, so the cell that differs must have come from a second bacterium that got into the colony
- Replication alters a few bases at every division, but in most of the cells the changes were too small for sequencing to detect
- Replication copies only the genes that a cell is using, so the rest of its genome is free to change from one generation to the next
In a DNA profile, each short tandem repeat (STR) marker tested has a 1 in 10 probability of matching between two unrelated people by chance. A forensic laboratory increases the number of independent STR markers tested from two to four. What is the effect on the probability of a false match?
- It falls from 1 in 100 to 1 in 10 000, because the chance probabilities are multiplied
- It stays at 1 in 10, because each marker is tested independently of the others
- It halves, from 1 in 100 to 1 in 200, because twice as many markers are tested
- It falls from 1 in 20 to 1 in 40, because the chance probabilities of the separate markers are added together
Which enzyme activities are involved in replication on the lagging strand?
- I. Primase making RNA primers
- II. DNA polymerase I replacing primers with DNA
- III. DNA ligase joining the fragments
- I and II only
- I and III only
- II and III only
- I, II and III
Four tubes each contained a DNA template, free DNA nucleotides, primers and ATP at 37 °C. Different enzymes were added to each tube, and the tubes were then tested for newly made DNA. In tube 3 the template had first been heated to 95 °C and then cooled rapidly, before the enzyme was added. Which conclusion is supported by the results?
| Tube | Template | Enzyme(s) added | New DNA made |
|---|---|---|---|
| 1 | double-stranded | DNA polymerase | no |
| 2 | double-stranded | DNA polymerase and helicase | yes |
| 3 | heated to 95 °C, then cooled rapidly | DNA polymerase | yes |
| 4 | double-stranded | helicase | no |
- Heating damages the template DNA, so that only one of its two strands can then be copied
- DNA polymerase can build a new strand only on a template whose two strands are separated
- Helicase can link free nucleotides into a new strand of DNA when DNA polymerase is absent
- Helicase must first activate DNA polymerase before the polymerase can join any nucleotides
At a replication fork, one new strand is synthesized smoothly and continuously while the other is built in short sections that must later be stitched together.
Outline how PCR and gel electrophoresis are used together to examine a tiny DNA sample from a crime scene.
Bacteria were grown for many generations in a medium whose only nitrogen source was heavy nitrogen (¹⁵N), so that all their DNA contained ¹⁵N. They were then transferred to a medium containing only light nitrogen (¹⁴N). After each round of replication, DNA was extracted and centrifuged in a density gradient. In the table, H is the density of DNA with ¹⁵N in both strands, L is the density of DNA with ¹⁴N in both strands and I is a density halfway between them. A third model of replication, the dispersive model, proposes that the parental double helix is broken into short stretches, so that both strands of every new molecule are a mixture of old and new DNA.
| Replications in ¹⁴N | Density band(s) observed |
|---|---|
| 0 | one band at H |
| 1 | one band at I |
| 2 | two bands: ½ at I and ½ at L |
| 3 | two bands: ¼ at I and ¾ at L |
Original practice questions © Biology by Bradford · CC BY-NC-SA 4.0 · Not affiliated with or endorsed by the International Baccalaureate Organization.
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- C: Taq polymerase comes from a bacterium of hot springs and survives the 95 °C denaturation step, so fresh enzyme need not be added every cycle;
- A: replication produces copies with identical base sequences, which is why 99 genomes match the original and each other; the one difference is best explained by a rare error in one replication, which later replications then copy into every descendant of that cell; a contaminating bacterium (A) would be expected to differ from the original at many bases, not at a single one, C is contradicted by the 99 identical genomes, which sequencing resolves to the single base, and the whole genome is copied before every division (D);
- A: the markers are independent, so the probability that all of them match by chance is the product of the separate probabilities, (1/10)² = 1/100 for two markers and (1/10)⁴ = 1/10 000 for four, which is why more markers make a false match far less likely (NOS: more measurements increase reliability); A ignores the combination of markers; B and D treat the probabilities as adding or halving rather than multiplying;
- D: each Okazaki fragment starts from an RNA primer (primase), primers are excised/replaced (DNA pol I), and fragments are sealed (ligase);
- B: DNA polymerase alone made no DNA on a double-stranded template (tube 1) but did once the strands had been separated, whether by helicase (tube 2) or by heating to 95 °C (tube 3); helicase alone made no new DNA (tube 4), so it does not link nucleotides (A); tube 3 made DNA with no helicase at all (B); and the results say nothing about which strands were copied (D);
- (a) [1]
- helicase;
- (b) [2 max]
- the template strands are antiparallel;
- DNA polymerase (III) can only synthesize 5′ → 3′ / add nucleotides to a 3′ end;
- (so) the strand running the "wrong" way must be made backwards, in sections, as the fork opens, OWTTE;
- (c) [2]
- DNA polymerase I removes the RNA primers and replaces them with DNA;
- DNA ligase joins/seals the fragments (covalently linking the sugar–phosphate backbone);
- PCR amplifies the chosen (variable) sequences, making millions of copies from a trace sample;
- (heat-stable Taq) polymerase and primers copy the target through repeated temperature cycles;
- gel electrophoresis then separates the amplified fragments by size (DNA moves to the positive electrode);
- the resulting band pattern (DNA profile) is compared with reference samples (suspects), OWTTE;
- (a) [2]
- dispersive predicts a single band of intermediate density (half old, half new DNA mixed in every molecule);
- this is the same as the semi-conservative prediction, so the generation-1 result cannot distinguish the two models;
- (b) [3]
- the dispersive model predicts one band that becomes progressively lighter each generation (never two separate bands), as old DNA is diluted evenly through all molecules;
- the data instead show two discrete bands (a constant I band plus a growing L band), which the dispersive model cannot explain;
- (so) the dispersive model is falsified/rejected, and the two-band pattern supports the semi-conservative model (intact parental strands conserved in half the molecules), OWTTE;
- (c) [2]
- a testable/falsifiable model makes a specific prediction that an experiment could contradict;
- because the models made different predictions at generation 2, the experiment could (and did) rule some out; a model that predicted every possible outcome could never be tested, OWTTE;
Nature of Science: falsifiability and competing models.
More in Theme D · Continuity and change
- D1.2 Protein synthesis 64
- D1.3 Mutation and gene editing 82
- D2.1 Cell and nuclear division 80
- D2.2 Gene expression 77
- D2.3 Water potential 68
- D3.1 Reproduction 70
- D3.2 Inheritance 109
- D3.3 Homeostasis 63
- D4.1 Natural selection 64
- D4.2 Stability and change 66
- D4.3 Climate change 66
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