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

These revision notes condense IB Diploma Programme Biology A1.1 (Water) and A1.2 (Nucleic acids) from the IB Biology guide, first assessment 2025. For full explanations and worked examples, read the study guide. Test yourself with the practice set. Statements marked “HL only” are additional higher level; everything else is SL and HL.

Course pages: IB DP Biology hub, printable checklist, free 10-minute diagnostics and Theme A revision notes.

Water at a glance (A1.1.1 to A1.1.6)

Property	Cause	Biological consequence
Polarity	Unequal sharing of electrons in O-H bonds	Hydrogen bonds between molecules; dissolves polar substances
Cohesion	Hydrogen bonds between water molecules	Continuous column in xylem; surface tension supports pond skaters
Adhesion	Attraction to polar or charged surfaces	Capillary action in soil and cell walls
Solvent	Water clusters round ions and polar molecules	Metabolism and transport in solution
High specific heat capacity	Many hydrogen bonds must be disturbed	Stable aquatic temperatures
Density, viscosity	Close-packed molecules	Buoyancy and drag for aquatic animals

Key points

• The first cells arose in water, and most life processes still happen in water.
• A hydrogen bond is an attraction between a δ+ hydrogen on one molecule and a δ- oxygen on another. It is not a covalent bond.
• Cohesion plus transpiration pull gives water under tension in xylem. Adhesion to cellulose walls supports the column.
• Surface tension comes from cohesion at the surface.
• Hydrophilic substances (ions, glucose, amino acids) dissolve. Hydrophobic substances (fats, phospholipid tails) do not, and their functions depend on that.

Water against air

Property	Water	Air	Effect on animals
Density	much higher	low	Strong buoyancy in water; less support needed
Viscosity	higher	low	More drag in water; streamlining
Thermal conductivity	much higher	low	Faster heat loss in water; insulation such as blubber
Specific heat capacity	higher	lower	Slow temperature change in water habitats

Use the loon (diving, swimming and flying) and the ringed seal as paired examples, or any other pair you can describe accurately.

Method: energy to heat a mass

1. Write energy = mass x specific heat capacity x temperature change.
2. Check units: grams with J per g per K.
3. Substitute and calculate.
4. Compare substances with a ratio, such as c(water) / c(air).

Reminder: with c = 4.18 J per g per K, heating 500 g of water by 3 K needs 500 x 4.18 x 3 = 6270 J.

Water beyond Earth (HL only: A1.1.7, A1.1.8)

• Asteroids are the only hypothesis you need for the origin of Earth’s water.
• Retention: gravity holds water, and temperatures are low enough for vapour to condense.
• Goldilocks zone: orbital region where a planet’s surface could hold liquid water. It is a guide for where to search, not proof of life.

Nucleic acid structure (A1.2.1 to A1.2.7)

Feature	DNA	RNA
Strands	two, antiparallel, in a double helix	one, usually
Pentose sugar	deoxyribose	ribose
Bases	A, T, G, C	A, U, G, C
Examples	chromosomal DNA	mRNA, tRNA, rRNA

• DNA carries the genetic information of every living organism. A few viruses store theirs as RNA, but viruses are not classed as living.
• Nucleotide = phosphate (circle) + pentose sugar (pentagon) + base (rectangle).
• Condensation reactions join the sugar of one nucleotide to the phosphate of the next, releasing water. This forms the continuous covalent sugar-phosphate backbone, which is strong.
• Pairing: A with T (A with U in RNA), G with C. Pairs are joined by hydrogen bonds, so strands can be separated for copying.
• Sketch of the sugars: ribose has OH on carbon 2; deoxyribose has H.

Method: writing a complementary strand

1. Label the given strand 5’ and 3’ ends.
2. Pair each base (A-T, G-C; use U in RNA).
3. The new strand runs antiparallel, so its 3’ end is under the 5’ end of the first.
4. If asked for the sequence 5’ to 3’, reverse what you wrote.

Reminder: 5’-TTGACA-3’ pairs to give 3’-AACTGT-5’, which is 5’-TGTCAA-3’.

Information and ancestry (A1.2.8 to A1.2.10)

• Complementary base pairing, based on hydrogen bonding, makes replication (copying) and expression (transcription) possible.
• Sequence of n bases: 4ⁿ possible sequences. Any length and any order are possible, so capacity is limitless.
• DNA stores a lot of information in a very small volume.
• The genetic code is shared across nearly all life, which supports common ancestry.

HL only: details of nucleic acids (A1.2.11 to A1.2.15)

Statement	Remember
Directionality	5’ to 3’ linkages; nucleotides are added at the 3’ end; replication and transcription run 5’ to 3’; translation reads mRNA 5’ to 3’
Purine-pyrimidine pairing	One purine (two rings) with one pyrimidine (one ring), so equal pair length and a regular helix
Nucleosome	DNA round eight histones, with another histone on linker DNA
Hershey-Chase	32P (DNA label) mostly in bacteria; 35S (protein label) mostly outside; DNA is the genetic material
Chargaff	A about T, G about C; (A+T)/(G+C) varies between species; falsified the tetranucleotide hypothesis

NOS: radioisotopes becoming available as research tools made Hershey-Chase possible. Many confirmations cannot prove a rule (problem of induction), but one reliable counterexample can disprove it (falsification).

Method: reading experiment data (HL only)

1. Say what each label marks: 32P marks DNA, 35S marks protein.
2. Convert counts to percentages of the total for each label.
3. Compare where each label ends up (pellet or supernatant).
4. Conclude cautiously: the data support DNA as the genetic material.

For Chargaff-style tables, check three things in turn: whether A is close to T, whether G is close to C, and whether (A+T)/(G+C) is the same or different between species. Then state what each pattern shows. Roughly equal A and T, with a ratio that differs between species, shows that the four bases are not present in equal amounts, which falsifies a fixed-repeat hypothesis.

Command terms to match

• Explain: give the cause as well as the fact, such as polar bonds giving hydrogen bonds.
• Compare: state both sides and the direction of each difference, such as water having higher conductivity than air.
• Outline: give the main points briefly.
• Draw: label every part, such as the 5’ and 3’ ends and the base-pair link.
• Discuss: set out evidence on more than one side before a conclusion.

Must-know distinctions

• Cohesion (water to water) against adhesion (water to other surfaces).
• Hydrogen bond (between molecules, weak) against covalent bond (within molecules, strong).
• Purine (A, G) against pyrimidine (C, T, U).
• DNA against RNA in strands, sugar and bases.
• Deoxyribose against ribose.

Quick self-test

1. Name the type of bond between the oxygen of one water molecule and a hydrogen of another.
2. State what makes the O-H bond polar.
3. Why does the water column in a xylem vessel not break under tension?
4. Give one example of a hydrophobic molecule and say how its function depends on this.
5. A mass of 800 g of water is warmed by 4 K. Using 4.18 J per g per K, calculate the energy needed.
6. Use thermal conductivity to explain why a mammal loses body heat faster in water than in air at the same temperature.
7. Write the complementary DNA strand for 5’-CCATGA-3’, in the 5’ to 3’ direction.
8. A DNA sample has 27% guanine. State the percentage of adenine.
9. How many different sequences are possible for a 4-base stretch of DNA?
10. Which bases are purines?
11. (HL only) In a Hershey-Chase type result, 9200 of 10 000 counts of the DNA label are in the pellet. State the percentage and the conclusion.
12. (HL only) State what Chargaff’s data falsified.

Answers

1. A hydrogen bond.
2. Oxygen attracts the shared electrons more strongly than hydrogen, so the electrons are shared unequally.
3. Cohesion between water molecules (hydrogen bonds) holds it together, and adhesion to the walls supports it.
4. For example, phospholipid tails: they avoid water and so form the inner layer of a membrane.
5. 800 x 4.18 x 4 = 13 376 J, about 13 400 J.
6. Water conducts heat much faster than air, so more heat leaves the body surface each second in water.
7. Complement is 3’-GGTACT-5’, so written 5’ to 3’ it is 5’-TCATGG-3’.
8. Cytosine is also 27%, so A + T = 46%, and adenine is 23%.
9. 4⁴ = 256.
10. Adenine and guanine.
11. 92%. The DNA label entered the bacteria, which supports DNA as the genetic material.
12. The tetranucleotide hypothesis, that the four bases repeat in a fixed sequence.

Where marks are usually lost

• Writing that hydrogen bonds are inside the water molecule.
• Describing cohesion in xylem without naming transpiration or tension.
• Mixing up cohesion and adhesion, or naming only one.
• Saying water “dissolves everything” instead of polar and charged substances.
• Comparing water and air without stating the direction of the difference.
• Giving the complementary strand in the wrong direction.
• Writing thymine in RNA or uracil in DNA.
• Saying A pairs with G, or confusing purines and pyrimidines.
• Describing the nucleosome as histone “inside” DNA rather than DNA wound round histones.
• Claiming Hershey-Chase results prove DNA is genetic material without referring to the data.

Official syllabus

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