IB DP Biology: Water and nucleic acids: properties of water, DNA and RNA structure (DP Biology) – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – the IB holds copyright in its own papers. Use these alongside the official past papers available through your school or the IB store.

These questions practise IB Diploma Programme Biology A1.1 (Water) and A1.2 (Nucleic acids) from the IB Biology guide, first assessment 2025. Questions 10 to 12 are marked “HL only” because they test additional higher level statements (A1.1.7, A1.1.8 and A1.2.11 to A1.2.15).

Revise with the study guide and the revision notes. Course pages: IB DP Biology hub, printable checklist and free 10-minute diagnostics.

Questions

1. Explain how the polar covalent bonds in a water molecule lead to hydrogen bonding between water molecules. [3]

2. A tall tree loses water from its leaves on a dry, windy day. Explain how water moves up the xylem without the water column breaking. [4]

3. Soil that contains a lot of clay stays damp for longer after rain than coarse sand does. Explain this using the properties of water. [3]

4. Glucose dissolves in blood plasma but cholesterol-rich fat droplets do not.

(a) Explain why glucose dissolves in water. [2] (b) Explain one way in which the hydrophobic nature of phospholipid tails is important for cells. [2]

5. A diver’s body is cooled by surrounding water. Use these values supplied for the question: specific heat capacity of water 4.18 J per g per K and of air 1.0 J per g per K; thermal conductivity of water 0.6 W per m per K and of air 0.026 W per m per K.

(a) Calculate the energy needed to warm 750 g of water by 6 K. [2] (b) Calculate how many times more energy is needed than for 750 g of air warmed by the same amount. [1] (c) Calculate how many times faster water conducts heat than air, to 2 significant figures. [1] (d) The ringed seal has a thick layer of blubber. Explain how this suits its habitat, with reference to water and air. [2]

6. Outline how a molecule of RNA is built from nucleotides and why its backbone is strong. Include a description of one nucleotide. [5]

7. State three differences between DNA and RNA. [3]

8. A section of one DNA strand reads 5’-ATGGCTAC-3’.

(a) Write the sequence of the complementary strand, in the 5’ to 3’ direction. [2] (b) The strand given above is used as the template for transcription. Write the mRNA sequence in the 5’ to 3’ direction. [2] (c) Explain how hydrogen bonding between bases allows the information in DNA to be both copied and used. [3]

9.

(a) Calculate how many different base sequences are possible for a stretch of DNA 9 bases long. [2] (b) Explain why the information storage capacity of DNA is described as limitless. [2] (c) Explain how the conserved genetic code is evidence for universal common ancestry. [2]

10. (HL only) Outline the evidence and reasoning that link liquid water to the search for life beyond Earth, and state why Earth has kept its water. [4]

11. (HL only)

(a) Explain what is meant by the 5’ end and 3’ end of a nucleic acid strand, and why new nucleotides are added at one end only. [2] (b) Explain why every base pair in DNA has the same length. [2] (c) Describe the structure of a nucleosome. [3]

12. (HL only) A virus that infects bacteria was grown with either radioactive 32P or radioactive 35S. Each batch was mixed with bacteria, shaken to remove viral coats, and spun to separate a pellet (bacteria) from the supernatant (coats). Counts per minute are shown.

Label	Pellet	Supernatant
32P	7400	800
35S	1100	7900

(a) Calculate the percentage of the 32P label in the pellet and the percentage of the 35S label in the supernatant, to 3 significant figures. [2] (b) Explain what these results suggest about the genetic material, and why 32P and 35S were chosen. [2]

Two species gave these base percentages: species W had A 31.2, T 31.0, G 18.9, C 18.9; species X had A 22.4, T 22.6, G 27.5, C 27.5.

(c) Calculate (A+T)/(G+C) for each species, to 2 decimal places. [2] (d) Explain how data like these falsified the tetranucleotide hypothesis. [2]

Answers

1. Oxygen attracts the shared electrons more strongly than hydrogen does [1], so the electrons are shared unequally and oxygen is slightly negative (δ-) while each hydrogen is slightly positive (δ+) [1]. The δ+ hydrogen of one molecule is attracted to the δ- oxygen of another, forming a hydrogen bond between molecules [1]. [3] Examiner insight: Answers lose credit by placing the hydrogen bond inside one molecule. Say the bond is between molecules and name both charges.

2. Evaporation (transpiration) from leaf cells pulls water out of the xylem [1], which puts the water column under tension [1]. Hydrogen bonds between water molecules (cohesion) hold the column together so it is pulled up as a continuous stream [1]. Adhesion to the polar cellulose walls of the xylem supports the column [1]. [4] Examiner insight: Four marks need four separate ideas, so link cohesion to tension and include adhesion.

3. Water is attracted to the charged clay particles (adhesion) [1]. The gaps between clay particles are very narrow, so capillary action holds water in them [1], and cohesion pulls further water along behind, which resists drainage; sand has larger gaps [1]. [3] Examiner insight: A comparison needs both sides, so say why coarse sand differs.

4. (a) Glucose is polar [1], so water molecules cluster round it through attraction between opposite partial charges, and it dissolves [1]. (b) Phospholipid tails are nonpolar and avoid water [1], so in a membrane they form the inner layer of the bilayer, separating the watery solutions on either side [1]. [4] Examiner insight: “Hydrophobic” must be linked to a function. State the function, not just that the tails repel water.

5. (a) Energy = mass x specific heat capacity x temperature change = 750 x 4.18 x 6 [1] = 18 810 J (18 800 J to 3 significant figures) [1]. (b) Air: 750 x 1.0 x 6 = 4500 J, so 18 810 / 4500 = 4.18 times [1]. (c) 0.6 / 0.026 = 23.08, so 23 times faster [1]. (d) Water takes heat from the body much faster than air does [1], so blubber insulates the seal and reduces heat loss in cold water [1]. [6] Examiner insight: Show the substitution as well as the result, and give units. In part (d) state which medium loses heat faster.

6. A nucleotide has a phosphate group, a pentose sugar (ribose) and a nitrogenous base [1]. Nucleotides join by condensation reactions [1], which release water [1]. A covalent bond forms between the phosphate of one nucleotide and the sugar of the next [1], forming a sugar-phosphate backbone of continuous covalent bonds, which is strong [1]. [5] Examiner insight: In a diagram, show relative positions with a circle, a pentagon and a rectangle, and label each.

7. Any three of: DNA is double-stranded, RNA is single-stranded [1]; DNA contains deoxyribose, RNA contains ribose [1]; DNA contains thymine, RNA contains uracil instead [1]; [3] Examiner insight: A difference needs both sides stated. “RNA has uracil” alone is weaker than “RNA has uracil where DNA has thymine”.

8. (a) Pair the bases: A-T, T-A, G-C, G-C, C-G, T-A, A-T, C-G, so the complement is 3’-TACCGATG-5’ [1]. Written 5’ to 3’ it is 5’-GTAGCCAT-3’ [1]. (b) mRNA pairs with the template, using U for T: 3’-UACCGAUG-5’ [1], which is 5’-GUAGCCAU-3’ [1]. (c) Bases pair by hydrogen bonds: A-T (A-U in RNA) and G-C [1]. The bonds are weak, so the strands can separate [1], and each strand can act as a template for a new strand or for mRNA, with the same pairing rules giving an accurate copy [1]. [7] Examiner insight: Reverse the order after pairing if 5’ to 3’ is asked, and use U, not T, in mRNA.

9. (a) Each position has 4 possible bases, so 4⁹ [1] = 262 144 [1]. (b) A DNA molecule can be of any length [1] and any base can follow any other, so the number of possible sequences has no upper limit [1]. (c) Almost all organisms use the same triplet code for the same amino acids [1], which is most simply explained by inheritance of the code from a common ancestor [1]. [6] Examiner insight: For common ancestry, link the observation (shared code) to the conclusion. Specific examples need not be memorised.

10. (HL only) Liquid water is needed by all known life, so planets with liquid water are the best candidates in the search for life [1]. The “Goldilocks zone” is the region around a star where temperatures could allow liquid water at a planet’s surface [1]. Earth kept its water because gravity holds it [1] and temperatures are low enough for water vapour to condense [1]. [4] Examiner insight: Say a planet in the zone could support liquid water, not that it has life.

11. (HL only) (a) Each strand has a 5’ end and a 3’ end, named for the carbon of the sugar at that end [1]. Nucleotides are added to the 3’ end, so strands are built 5’ to 3’ [1]. (b) Every pair joins one purine (two rings) to one pyrimidine (one ring) [1], so A-T and C-G pairs have equal length and the helix has the same shape whatever the sequence [1]. (c) DNA is wrapped round a core of eight histone proteins [1]; an additional histone is attached to linker DNA [1], which joins one nucleosome to the next [1]. [7] Examiner insight: Say the DNA wraps round the histones, not that histones lie inside the DNA.

12. (HL only) (a) 32P in pellet: 7400 / (7400 + 800) x 100 [1] = 90.2%. 35S in supernatant: 7900 / (7900 + 1100) x 100 = 87.8% [1]. (b) 32P marks DNA, which is absent from protein, and 35S marks protein, which is absent from DNA [1]. Most DNA label entered the bacteria while most protein label stayed outside, which supports DNA as the genetic material [1]. (c) Species W: (31.2 + 31.0) / (18.9 + 18.9) = 62.2 / 37.8 = 1.65 [1]. Species X: (22.4 + 22.6) / (27.5 + 27.5) = 45.0 / 55.0 = 0.82 [1]. (d) The tetranucleotide hypothesis predicted that the four bases repeat in a fixed order, giving equal amounts [1]. The ratios differ between species and are not 1, so the bases are not present in equal amounts, which contradicts it [1]. [8] Examiner insight: Quote the data, and say the results support the conclusion rather than prove it.

Where marks are usually lost

• Writing that hydrogen bonds are inside the water molecule rather than between molecules.
• Describing xylem transport without naming tension, cohesion and adhesion.
• Stating a comparison of water and air without the direction of the difference.
• Giving a complementary strand in the wrong direction, or leaving it unreversed.
• Using thymine in RNA or uracil in DNA.
• Using the wrong total when converting counts to a percentage.
• Claiming that experimental data prove a conclusion rather than support it.

Next steps

• Reread the study guide for any question you missed.
• Use the revision notes for a final recap.
• Go to the IB DP Biology hub and the printable checklist.
• Try the free 10-minute diagnostics.
• Book a free trial class.

Official syllabus

IB Diploma Programme Biology guide, first assessment 2025, published by the International Baccalaureate Organization (topics A1.1 and A1.2).