D1.1 DNA replication. Practice questions with markscheme.
45 original IB-style questions on D1.1, written from the 2025 guide: 19 multiple-choice, 17 short-answer, 6 data-based, 1 extended-response part, 1 labelling, 1 drawing. Below is a 20-mark standard-level practice paper built from them, ready to hand out as a class quiz or homework, or to sit 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
- 19 multiple-choice
- 17 short-answer
- 6 data-based
- 1 extended-response part
- 1 labelling
- 1 drawing
- 17 higher level only
Every question is original and tagged to a guide statement. See the whole bank →
Make your own
The practice paper
Paper code: this paper was generated, so the code is its recipe. Enter it at biologybybradford.com/exam-maker to rebuild this exact paper and its markscheme.
A PCR reaction starts with 10 copies of a target DNA sequence. Approximately how many copies are present after 10 complete cycles?
- About 100
- About 10 000
- About 10 million
- About 1000
Why must DNA replication occur before a cell divides?
- To remove mutations from the DNA
- So each daughter cell gets a full copy
- To halve the cell's chromosome number
- To provide energy for cell division
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 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 100 to 1 in 10 000, because the chance probabilities are multiplied
- It falls from 1 in 20 to 1 in 40, because the chance probabilities of the separate markers are added together
A drug binds to helicase and prevents it from functioning, but has no direct effect on DNA polymerase. Predict, with reasons, the effect of this drug on DNA replication in a cell.
| 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 |
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 fragment bands present in each profile at five variable sites; each person shows two bands per site (one inherited from each parent). For simplicity, one representative band from each person's pair is compared: the band inherited from the biological father must appear in the child's profile.
| Site | Mother | Child | Man X | Man Y |
|---|---|---|---|---|
| 1 | a | a, d | c | d |
| 2 | f | f, g | g | g |
| 3 | j | j, k | m | k |
| 4 | p | p, q | q | q |
| 5 | s | s, u | t | u |
Original practice questions © Biology by Bradford · CC BY-NC-SA 4.0 · Not affiliated with or endorsed by the International Baccalaureate Organization.
Rebuild or edit this exact paper (and its markscheme): biologybybradford.com/exam-maker?code=BbB-EAAAAAAQABQAO8O-
Show the markscheme
Markscheme BbB-EAAAAAAQABQAO8O-
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: each cycle doubles the target: 10 × 2¹⁰ ≈ 10 240;
- B: replication doubles the DNA so division can give both daughter cells a complete genome, maintaining continuity of genetic information;
- C: 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;
- helicase cannot unwind the double helix / cannot break the hydrogen bonds between the (complementary) base pairs;
- (so) the two strands are not separated / no single-stranded template is exposed;
- DNA polymerase cannot add (free) nucleotides by complementary base pairing without a single-stranded template to copy;
- (so) replication stops / no new DNA strands are made, even though DNA polymerase itself is unaffected, OWTTE;
- (so) the cell cannot produce daughter cells with a complete set of DNA / cannot divide normally;
Do not accept "helicase breaks the DNA strands" or "breaks covalent/phosphodiester bonds". The prediction that replication stops must be linked to the lack of a template to gain full marks.
- (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.
- (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;
More in Theme D · Continuity and change
- D1.2 Protein synthesis 48
- D1.3 Mutation and gene editing 45
- D2.1 Cell and nuclear division 46
- D2.2 Gene expression 45
- D2.3 Water potential 47
- D3.1 Reproduction 53
- D3.2 Inheritance 76
- D3.3 Homeostasis 45
- D4.1 Natural selection 46
- D4.2 Stability and change 45
- D4.3 Climate change 46
All 40 IB Biology subtopics → · AP Biology units → · Open the exam maker →