A1.2 Nucleic acids. Practice questions with markscheme.
56 original IB-style questions on A1.2, written from the 2025 guide: 22 multiple-choice, 17 short-answer, 8 data-based, 4 drawing, 3 extended-response part, 2 labelling. 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
10 statements at SL and HL, 5 additional higher level.
- A1.2.1SL / HL DNA as the genetic material of all living organisms
- A1.2.2SL / HL Components of a nucleotide
- A1.2.3SL / HL Sugar–phosphate bonding and the sugar–phosphate “backbone” of DNA and RNA
- A1.2.4SL / HL Bases in each nucleic acid that form the basis of a code
- A1.2.5SL / HL RNA as a polymer formed by condensation of nucleotide monomers
- A1.2.6SL / HL DNA as a double helix made of two antiparallel strands of nucleotides with two strands linked by hydrogen bonding between complementary base pairs
- A1.2.7SL / HL Differences between DNA and RNA
- A1.2.8SL / HL Role of complementary base pairing in allowing genetic information to be replicated and expressed
- A1.2.9SL / HL Diversity of possible DNA base sequences and the limitless capacity of DNA for storing information
- A1.2.10SL / HL Conservation of the genetic code across all life forms as evidence of universal common ancestry
- A1.2.11HL Directionality of RNA and DNA
- A1.2.12HL Purine-to-pyrimidine bonding as a component of DNA helix stability
- A1.2.13HL Structure of a nucleosome
- A1.2.14HL Evidence from the Hershey–Chase experiment for DNA as the genetic material
- A1.2.15HL Chargaff’s data on the relative amounts of pyrimidine and purine bases across diverse life forms
In the bank for A1.2
- 22 multiple-choice
- 17 short-answer
- 8 data-based
- 4 drawing
- 3 extended-response part
- 2 labelling
- 17 higher level only
Every question is original and tagged to a guide statement. See the whole bank →
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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.
In a sample of double-stranded DNA, 30 % of the bases are adenine. What percentage are cytosine?
- 20 %
- 30 %
- 40 %
- 70 %
RNA is described as a polymer formed by condensation of nucleotide monomers. What is released each time two nucleotides are joined?
- A molecule of carbon dioxide
- A molecule of ammonia
- A molecule of water
- A hydrogen ion
Which components make up a single nucleotide?
- Two pentose sugars joined by a nitrogenous base
- A pentose sugar, a phosphate group and a nitrogenous base
- A pentose sugar, a phosphate group and an amino acid
- A hexose sugar, a phosphate group and a nitrogenous base
A human gene can be transferred into the bacterium Escherichia coli, which then synthesizes the human protein correctly. Explain what this reveals about the genetic code and about the ancestry of living organisms.
Outline how nucleotides are linked together to form a single strand of DNA.
Students extracted the nucleic acid–protein mixture from onion cells and divided it into three equal samples. Sample 1 was untreated. Sample 2 was treated with DNase, an enzyme that hydrolyses DNA into short fragments. Sample 3 was treated with protease, an enzyme that hydrolyses proteins into amino acids. Each sample was then mixed with bacterial cells that had lost the ability to make a particular enzyme, to see whether the extract could restore this ability (a sign that heritable information had been taken up). The table shows whether the bacteria regained the missing function.
| Sample | Treatment | Bacteria regain function? |
|---|---|---|
| 1 | none (untreated extract) | yes |
| 2 | DNase | no |
| 3 | protease | yes |
Explain how the structure of DNA is related to its function as the genetic material.
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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Markscheme BbB-EAACAAAAABQAOh6c
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.
- A: A = T = 30 %, so G + C = 40 % and C = 20 % (complementary base pairing);
- C — condensation reactions link monomers and release a water molecule for each bond formed;
- B: pentose + phosphate + base;
- the genetic code is (virtually) universal / the same codons specify the same amino acids in (almost) all organisms;
- (so) the bacterium transcribes and translates the human gene into the same amino acid sequence / the same protein;
- conservation of the code across all life forms is evidence of (universal) common ancestry / the code was inherited from a shared ancestral form;
- (the code has been conserved because) any change to it would alter (almost) every protein made, so such mutations are (almost always) lethal, OWTTE;
- (condensation reactions form) covalent bonds between nucleotides;
- the phosphate (group) of one nucleotide bonds to the sugar/pentose of the next;
- forming a (continuous) sugar–phosphate backbone (with bases projecting from it);
Accept phosphodiester bond. Do not accept hydrogen bonds.
- (a) [1]
- sample 2 (the DNase-treated extract);
- (b) [2]
- DNA carries the heritable information;
- destroying DNA (sample 2) abolished the transforming effect, whereas destroying protein (sample 3) did not, OWTTE;
- (c) [2]
- sample 3 acts as a control for the effect of enzyme treatment itself (rather than loss of a specific molecule);
- it shows that removing protein does not abolish the effect, supporting the conclusion that protein is not the genetic material, OWTTE;
- (d) [2 max]
- repeat the experiment (several times / with several bacterial strains) to check the result is reproducible;
- check that the enzymes themselves do not directly restore the missing function (enzyme-only control);
- check that DNase/protease preparations are not contaminated with the other enzyme, OWTTE;
- (e) [1]
- (bacterial) transformation;
- the base sequence stores/encodes (genetic) information;
- (any) sequence and length are possible, giving unlimited capacity/diversity;
- complementary base pairing (A–T, G–C) allows each strand to act as a template;
- (so) DNA can be replicated accurately (and information passed on);
- hydrogen bonds between strands are (individually) weak, so the strands can be separated (for replication/transcription);
- the covalent sugar–phosphate backbone makes the molecule stable, protecting the information;
More in Theme A · Unity and diversity
- A1.1 Water 45
- A2.1 Origins of cells 45
- A2.2 Cell structure 58
- A2.3 Viruses 45
- A3.1 Diversity of organisms 46
- A3.2 Classification and cladistics 45
- A4.1 Evolution and speciation 46
- A4.2 Conservation of biodiversity 46
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