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

This study guide teaches IB Diploma Programme Biology topics A1.1 (Water) and A1.2 (Nucleic acids) in depth. It follows the IB Biology guide, first assessment 2025. Statements A1.1.7, A1.1.8 and A1.2.11 to A1.2.15 are additional higher level and are marked “HL only”; the rest is SL and HL.

Use it with the revision notes and the practice set. Wider theme: Theme A: Unity and Diversity. Course pages: IB DP Biology hub, printable checklist and free 10-minute diagnostics.

What this unit covers

Section	What you must be able to do	Tier
A1.1.1 to A1.1.6	Water as a medium; polarity and hydrogen bonds; cohesion; adhesion; solvent properties; water compared with air	SL and HL
A1.1.7, A1.1.8	Origin and retention of water on Earth; the “Goldilocks zone”	HL only
A1.2.1 to A1.2.10	Nucleotides, backbone, bases, RNA, double helix, DNA versus RNA, base pairing, information capacity, universal code	SL and HL
A1.2.11 to A1.2.13	Directionality, purine-pyrimidine pairing, nucleosome	HL only
A1.2.14, A1.2.15	Hershey-Chase and Chargaff evidence	HL only

A1.1.1 Water and the origin of life

The first cells appeared in water. Most cell reactions still happen in aqueous solution, in cytoplasm, blood plasma and tissue fluid.

A1.1.2 Polarity and hydrogen bonds

In an O-H bond, oxygen attracts the shared electrons more strongly than hydrogen does. The electrons are shared unequally, so the bond is polar. The oxygen carries a small negative charge (δ-) and each hydrogen a small positive charge (δ+). The molecule is bent, so the charges do not cancel.

A δ+ hydrogen on one molecule is drawn towards the δ- oxygen of the molecule next to it, and this pull is called a hydrogen bond. One hydrogen bond is weak, but very many together are strong.

In a diagram, label δ- on each oxygen and δ+ on each hydrogen, and draw hydrogen bonds as dashed lines between molecules.

A1.1.3 Cohesion

Cohesion is the attraction between water molecules, caused by hydrogen bonding.

• Xylem transport. Water evaporates from leaf cells (transpiration) and pulls the water column upwards. Cohesion stops the column breaking. The water is under tension: it is pulled up, not pushed.
• Surface tension. Surface molecules have water only below and beside them, so they are drawn inwards and the surface acts like a thin skin. Pond skaters use it as a habitat.

A1.1.4 Adhesion

Adhesion is the attraction between water and other substances that are polar or charged. Cellulose in plant cell walls has polar hydroxyl groups, and soil particles such as clay carry charges.

• Capillary action is the movement of water through narrow spaces. Adhesion pulls water along the surface and cohesion drags more water behind it. This happens in the narrow gaps between soil particles, which holds water against drainage, and in plant cell walls, which water moves through.

Worked example 1: cohesion and adhesion together

On a hot day a tall tree loses water through its leaves. Explain why the xylem water column does not break.

1. Evaporation from leaf cell walls pulls water out of the xylem, creating tension.
2. Hydrogen bonds between water molecules (cohesion) hold the column together, so it is pulled up as one.
3. Adhesion to the polar cellulose walls of the vessels supports the column.

A1.1.5 Solvent properties

Polar molecules and ions dissolve because water molecules cluster round them; they are hydrophilic. Glucose, amino acids, sodium ions and chloride ions are examples. Most enzymes work in solution, so substrates must dissolve to reach active sites.

Plasma carries dissolved glucose and urea, phloem sap carries sucrose and xylem sap carries mineral ions.

Some molecules are hydrophobic: nonpolar, insoluble in water. Their roles depend on this. Phospholipid tails avoid water and form the core of the bilayer, so membranes separate watery regions. Fats store energy without dissolving in cell fluid. Poorly soluble substances such as oxygen need carriers, for example haemoglobin.

A1.1.6 Water compared with air

Property	Water	Air
Density (kg per m³)	about 1000	about 1.2
Viscosity (mPa s)	about 1.0	about 0.018
Thermal conductivity (W per m per K)	about 0.6	about 0.026
Specific heat capacity (J per g per K)	about 4.18	about 1.0

• Buoyancy. Water is far denser than air and gives strong upthrust, so aquatic animals need less support.
• Viscosity. Water resists movement, so aquatic animals are streamlined. The black-throated loon has its feet set far back, which suits swimming.
• Thermal conductivity. Water takes heat away from a body much faster than air does. Ringed seals have insulating blubber.
• Specific heat capacity. Water must absorb a lot of energy to change temperature, so aquatic habitats change temperature slowly. Land and air habitats change faster.

Worked example 2: comparing water and air

Compare warming 2000 g of water by 5 °C with warming 2000 g of air, and compare heat conduction.

Energy = mass x specific heat capacity x temperature change.

• Water: 2000 x 4.18 x 5 = 41 800 J.
• Air: 2000 x 1.0 x 5 = 10 000 J.
• Water needs 4.18 times as much energy for the same mass.

Per cubic metre the contrast is greater: 1 m³ of water has a mass of about 1 000 000 g and 1 m³ of air about 1200 g, so water holds about 3500 times as much heat per degree.

Conduction: 0.6 / 0.026 is about 23, so water conducts heat roughly 23 times as fast as air. A swimming mammal loses body heat far faster, so insulation matters more.

A1.1.7 and A1.1.8 (HL only): water beyond Earth

Origin and retention (A1.1.7). One hypothesis is that water arrived on Earth from space, carried by asteroids. Earth kept it because gravity holds it and because temperatures are low enough for water vapour to condense. Abundant water over billions of years allowed life to evolve.

Search for life (A1.1.8). Liquid water is the key requirement when looking for life elsewhere. The “Goldilocks zone” is the region around a star where a planet’s surface could hold liquid water: not too hot and not too cold. Being in the zone does not prove life exists.

A1.2.1 to A1.2.5 Nucleotides, bases and polymers

DNA carries the genetic information of every living organism. A few viruses store theirs as RNA, but viruses are not classed as living.

A nucleotide has a phosphate group, a pentose (five-carbon) sugar and a nitrogenous base. In diagrams use a circle for phosphate, a pentagon for sugar and a rectangle for the base.

The bases are adenine (A), cytosine (C), guanine (G), thymine (T, DNA only) and uracil (U, RNA only). Their order forms a code.

Nucleotides join by condensation reactions. A covalent link joins the phosphate of each incoming nucleotide to the sugar of the last one, and water is released. Repeating this gives a sugar-phosphate backbone of continuous covalent bonds, which is strong. RNA is made as a polymer in this way, usually as a single strand.

A1.2.6 to A1.2.8 The double helix and base pairing

DNA is two strands wound into a double helix. The strands run in opposite directions (antiparallel). Bases on one strand pair with bases on the other through hydrogen bonds: A with T and G with C. These are complementary base pairs. When drawing, show two antiparallel strands; you need not draw the spiral.

Complementary base pairing allows information to be copied and used. In replication, each strand is a template for a new strand. In transcription, one DNA strand is the template for mRNA, in which U pairs with A. Pairing depends on hydrogen bonding, so strands can separate and re-form.

Worked example 3: writing complementary strands

A DNA strand reads 5’-GATTCAGC-3’. Write its complementary strand, then the mRNA made if the second strand is the template.

1. Pair each base directly under the first strand: G-C, A-T, T-A, T-A, C-G, A-T, G-C, C-G. The new strand reads 3’-CTAAGTCG-5’.
2. Written in the usual 5’ to 3’ direction it is 5’-GCTGAATC-3’. Reversing is the step most often missed.
3. With 3’-CTAAGTCG-5’ as the template, the mRNA pairs base by base (C gives G, T gives A, A gives U, G gives C): 5’-GAUUCAGC-3’.

If the original strand is the template instead, the mRNA is 5’-GCUGAAUC-3’. Always check which strand is the template.

A1.2.7 DNA compared with RNA

Feature	DNA	RNA
Strands	two (double)	one (single)
Sugar	deoxyribose	ribose
Bases	A, T, G, C	A, U, G, C

Ribose has an OH group on carbon 2; deoxyribose has only H there. Examples: chromosomal DNA, mRNA, tRNA and rRNA.

A1.2.9 and A1.2.10 Information and common ancestry

Any base can follow any other and a molecule can be any length, so the possible sequences are limitless. A sequence of n bases has 4ⁿ possibilities: 4⁵ = 1024, 4¹⁰ = 1 048 576 and 4¹² = 16 777 216. Each base carries 2 bits, so 1000 base pairs carry 2000 bits. One million base pairs span only 0.34 mm at 0.34 nm per base pair (supplied value), so DNA stores data very economically.

The genetic code is the same in almost all life: the same triplets specify the same amino acids. This is evidence that all organisms share a common ancestor. Specific examples need not be memorised.

A1.2.11 to A1.2.13 (HL only) Structure details

Directionality (A1.2.11). A phosphate joins the 5’ carbon of one sugar to the 3’ carbon of the next, so each strand has a 5’ end and a 3’ end. New nucleotides are added to the 3’ end, so replication and transcription run 5’ to 3’, and ribosomes read mRNA 5’ to 3’.

Purine and pyrimidine pairing (A1.2.12). Purines (A, G) are double-ring bases, and pyrimidines (C, T, U) are single-ring bases. Every pair is one purine with one pyrimidine, so all pairs have equal length and the helix has the same shape whatever the sequence.

Nucleosome (A1.2.13). DNA wraps round a core of eight histone proteins. A further histone attaches to the linker DNA. In molecular visualisation software, find the DNA wound round the histone core and note where the proteins contact the phosphate groups.

A1.2.14 and A1.2.15 (HL only) Evidence

Hershey-Chase. A bacteriophage is a virus made of DNA and protein that infects bacteria. Phosphorus is in DNA but not protein; sulfur is in protein but not DNA. One batch was labelled with radioactive 32P and another with 35S. Each infected bacteria; the coats were shaken off and the mixture spun to give a pellet (bacteria) and supernatant (coats). Radioisotopes becoming available as research tools made the experiment possible.

Worked example 4: interpret the results

Here are invented counts (counts per minute).

Label	Pellet (bacteria)	Supernatant (coats)
32P	8500	1500
35S	1800	6200

Percentage of 32P in pellet = 8500 / 10 000 x 100 = 85%. Percentage of 35S in supernatant = 6200 / 8000 x 100 = 77.5%. The label from DNA mostly entered the bacteria while the protein label mostly stayed outside. This supports DNA being the genetic material. The results are not perfectly clean, so they support the conclusion without proving it.

Chargaff. In samples from many species, A is about equal to T and G to about equal C, so purines and pyrimidines are about equal in amount, but the ratio (A+T)/(G+C) differs between species. This falsified the tetranucleotide hypothesis of a repeating sequence of the four bases. The NOS point: no number of observations can prove a general rule (the problem of induction), but one reliable counterexample can disprove it (falsification).

Common errors

• Placing hydrogen bonds within a water molecule. They are between molecules.
• Forgetting the 5’ to 3’ reversal.
• Putting thymine in RNA.

Official syllabus

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