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IB Biology 2025 · Theme A · Unity and diversity

A1.2 Nucleic acids. Practice questions with markscheme.

72 original IB-style questions on A1.2, written from the 2025 guide: 30 multiple-choice, 26 short-answer, 8 data-based, 4 drawing, 3 extended-response parts, 1 labelling. Below are two ready-made practice papers built from them, one at standard level (20 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

10 statements at SL and HL, 5 additional higher level.

  1. A1.2.1SL / HL DNA as the genetic material of all living organisms
  2. A1.2.2SL / HL Components of a nucleotide
  3. A1.2.3SL / HL Sugar–phosphate bonding and the sugar–phosphate “backbone” of DNA and RNA
  4. A1.2.4SL / HL Bases in each nucleic acid that form the basis of a code
  5. A1.2.5SL / HL RNA as a polymer formed by condensation of nucleotide monomers
  6. 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
  7. A1.2.7SL / HL Differences between DNA and RNA
  8. A1.2.8SL / HL Role of complementary base pairing in allowing genetic information to be replicated and expressed
  9. A1.2.9SL / HL Diversity of possible DNA base sequences and the limitless capacity of DNA for storing information
  10. A1.2.10SL / HL Conservation of the genetic code across all life forms as evidence of universal common ancestry
  11. A1.2.11HL Directionality of RNA and DNA
  12. A1.2.12HL Purine-to-pyrimidine bonding as a component of DNA helix stability
  13. A1.2.13HL Structure of a nucleosome
  14. A1.2.14HL Evidence from the Hershey–Chase experiment for DNA as the genetic material
  15. 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

  • 30 multiple-choice
  • 26 short-answer
  • 8 data-based
  • 4 drawing
  • 3 extended-response parts
  • 1 labelling
  • 27 higher level only

Every question is original and tagged to a guide statement. See the whole bank →

Make your own

The practice paper

20 marks at SL · 20 at HL · each has its own code
Take it online → Build a fresh paper Paper code BbB-EAACAAAAABQAOh6c
Biology · topic quiz
Standard level · topic practice, not an exam format
30 minutes20 marks

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.

Covers A1.2 Nucleic acids · SL
Name:
1.

The diagram shows a section of a DNA molecule containing two base pairs. Explain how the diagram shows that the two strands are antiparallel, and how the two strands are held together.

[4]
CytosineGuanineThymineAdenine 5′3′3′5′
2.

State the roles of DNA and of RNA in cells.

[2]
3.

Compare the nitrogenous bases found in a molecule of DNA with those found in a molecule of RNA.

[2]
4.

In any double-stranded DNA molecule, the amount of adenine equals the amount of thymine. Explain why.

[2]
5.

Draw and label a diagram of a single DNA nucleotide.

[3]
6.

The base composition of the genomes of four viruses was determined by chemical analysis. The results are shown as percentages of the four bases.

VirusAdenine / %Thymine / %Guanine / %Cytosine / %
P30302020
Q35241922
R21212929
S28163422
(a)Identify the viruses whose genomes obey the rule adenine = thymine and guanine = cytosine.[1]
(b)The genome of virus P is 48 000 base pairs long. Calculate the number of guanine bases in the genome of virus P.[2]
(c)Deduce, with a reason, whether the genomes of these viruses are made of DNA or of RNA.[2]
(d)Viruses P and R differ in base composition. Outline why two viruses whose genomes had exactly the same base composition could still carry different genetic information.[2]

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-EAACAAAAABQAOh6c

Show the markscheme

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.

1. A1.2.6, A1.2.2 [4 max]
  • each backbone is built of alternating phosphates (circles) and sugars (pentagons), so each strand has a direction set by the order of these groups;
  • the right-hand strand is drawn inverted relative to the left, so its sugars and phosphates run in the opposite order down the molecule;
  • the strand ends are therefore labelled 5′ and 3′ at opposite ends of the molecule / one strand runs 5′ to 3′ downwards while the other runs 3′ to 5′ downwards;
  • the strands are joined by hydrogen bonds (dashed lines) between bases on opposite strands;
  • the pairing is complementary: cytosine with guanine and thymine with adenine, as shown;
  • hydrogen bonds are individually weak, so the strands can be separated, but together they hold the double helix stably, OWTTE;

Accept 'the two strands run in opposite directions' for the second point if the candidate refers to the diagram. Students are not required to state the numbers of hydrogen bonds (A1.2.6), so accept any number of dashed lines per pair.

2. A1.2.1 [2]
  • DNA is the genetic material / stores hereditary information (in all cells);
  • RNA carries/transfers information for protein synthesis (mRNA) / has roles in expressing genes (tRNA/rRNA), OWTTE;
3. A1.2.4 [2 max]
  • both contain adenine, guanine and cytosine;
  • DNA contains thymine, whereas RNA contains uracil in its place;
  • in both, adenine pairs with its complementary pyrimidine (thymine in DNA, uracil in RNA) and guanine pairs with cytosine;
4. A1.2.6 [2 max]
  • the two strands are complementary;
  • every A on one strand pairs with a T on the other (by hydrogen bonding);
  • (so) A and T occur in equal numbers (and likewise G and C), OWTTE;
5. A1.2.2 [3]
  • pentose sugar (shown as a pentagon) labelled deoxyribose/sugar;
  • phosphate (group) attached to the sugar, labelled;
  • (nitrogenous) base attached to the sugar (on a different corner from the phosphate), labelled;

Phosphate and base must be attached to the sugar, not to each other. Accept circle for phosphate and rectangle for base.

6. A1.2.6, A1.2.7, A1.2.9
  • (a) [1]
    • P and R;
  • (b) [2]
    • 48 000 base pairs = 96 000 bases;
    • 20 % of 96 000 = 19 200 (guanine bases);

    Award [2] for 19 200 with or without working. Award [1] for 9 600 (20 % of 48 000).

  • (c) [2]
    • DNA;
    • (all four) genomes contain thymine, which occurs only in DNA / RNA would contain uracil in place of thymine, OWTTE;
  • (d) [2 max]
    • genetic information is carried by the sequence/order of the bases, not by their proportions;
    • the same percentages of A, T, G and C can be arranged in an enormous number of different sequences;
    • (so) the genomes can contain different genes / code for different proteins, OWTTE;

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