IB DP Biology: DNA replication, mutation and gene editing (DP Biology)

This study guide teaches IB Diploma Programme Biology sections D1.1 (DNA replication) and D1.3 (Mutation and gene editing) from scratch. It follows the IB Biology guide, first assessment 2025. Statements D1.1.6 to D1.1.9 and D1.3.8 to D1.3.10 are additional higher level, so each is labelled “HL only”; everything else is SL and HL.

Use it with the revision notes and the practice set. Course pages: IB DP Biology hub, printable checklist and free 10-minute diagnostics. You will need the genetic code from D1.2 for the mutation work.

What this unit covers

Section	What you must be able to do	Tier
D1.1.1 to D1.1.3	Replication as exact copying; semi-conservative copying and base pairing; helicase and DNA polymerase	SL and HL
D1.1.4, D1.1.5	PCR and gel electrophoresis; their applications, including DNA profiling	SL and HL
D1.1.6 to D1.1.9	Directionality; leading and lagging strands; primase, polymerase I and III, ligase; proofreading	HL only
D1.3.1 to D1.3.3	Substitutions, insertions, deletions and their effects	SL and HL
D1.3.4 to D1.3.7	Causes, randomness, germ and somatic cells, variation	SL and HL
D1.3.8 to D1.3.10	Gene knockout; CRISPR and Cas9; conserved sequences	HL only

D1.1.1 and D1.1.2 Exact copies, semi-conservatively

Replication makes two DNA molecules with the same base sequence as the original. Organisms need it to reproduce, and multicellular ones also need it to grow and replace worn-out tissue, because every new cell needs a full set of genes.

Replication is semi-conservative: each new molecule has one old strand and one new strand. The old strands act as templates. Free nucleotides pair by hydrogen bonds with the exposed bases: A with T, G with C. Because only one base fits each position, the sequence of the template fixes the sequence of the new strand. That specificity is what gives high accuracy.

Template read as 3’ TACGGA 5’ gives a new strand 5’ ATGCCT 3’. Flip the direction for the partner.

Worked example. A DNA molecule has both strands labelled with a heavy isotope. It replicates in light medium for 4 rounds. Find the fraction of molecules that are hybrid (one heavy strand, one light).

1. Molecules after 4 rounds = 2⁴ = 16.
2. The two original heavy strands are never destroyed, so they sit in two molecules only.
3. Each of those two molecules is a hybrid: heavy template plus light new strand.
4. The other 14 contain only light strands.
5. Hybrid fraction = 2/16 = 1/8 (12.5%).

D1.1.3 Helicase and DNA polymerase

At a replication fork, helicase unwinds the double helix by breaking hydrogen bonds between the strands. This exposes the bases as templates. DNA polymerase builds the new strand by joining nucleotides that pair with the template. At SL, keep to these general roles.

D1.1.4 PCR and gel electrophoresis

The polymerase chain reaction (PCR) copies a chosen DNA region in a tube, so a tiny sample gives enough DNA to study. Ingredients: template DNA, two primers (short single strands complementary to the ends of the target), free nucleotides, buffer and Taq polymerase. Each cycle has three temperature steps (typical values):

Step	Temperature	What happens
Denaturation	about 95 °C	Hydrogen bonds break; strands separate
Annealing	about 55 to 60 °C	Primers bind to their complementary sequences
Extension	about 72 °C	Taq polymerase extends each primer

Taq polymerase comes from a heat-tolerant bacterium, so it is not denatured by the 95 °C step and need not be added again each cycle. Primers are needed because DNA polymerase can only add to an existing end. The number of copies of the target roughly doubles each cycle: copies = 2ⁿ after n cycles, an ideal upper limit.

Worked example. One target molecule goes through 25 cycles. Ideal copies = 2²⁵ = 33,554,432. Real efficiency is lower.

Gel electrophoresis separates DNA fragments by size. DNA is loaded into wells in a gel and an electric current is applied. DNA is negatively charged because of its phosphate groups, so it moves towards the positive electrode. The gel acts as a mesh: small fragments pass through more easily and travel further in a given time. A ladder of known sizes runs alongside.

Worked example. Ladder bands: 1000 bp at 12 mm, 500 bp at 22 mm, 250 bp at 32 mm from the well. A sample band is at 27 mm. Estimate its size.

1. Each 10 mm of travel halves the size, so size falls by a factor of 2 per 10 mm in this range.
2. The sample is 5 mm beyond the 500 bp band.
3. Size = 500 × 2^(−5/10) = 500 ÷ 1.414 = 353.6.
4. Answer: about 350 bp.

D1.1.5 Applications

PCR and electrophoresis together support DNA profiling: DNA from a sample is amplified at several variable marker regions, cut or sized, and run on a gel to give a pattern. Uses include forensic investigations (matching crime-scene DNA to a suspect) and paternity testing (each band in a child must come from the mother or father).

Nature of science. Using more markers makes a profile more reliable because the chance of two unrelated people matching at all of them falls. If four independent markers have match chances of 1 in 10, 1 in 8, 1 in 20 and 1 in 12, the combined chance is 1/(10 × 8 × 20 × 12) = 1 in 19,200. A fifth marker at 1 in 15 gives 1 in 288,000. This needs independent markers.

D1.1.6 to D1.1.9 Detail of replication (HL only)

Directionality (HL only). A nucleotide strand has a 5’ end (free phosphate on carbon 5’) and a 3’ end (free hydroxyl on carbon 3’). DNA polymerases add a nucleotide’s 5’ phosphate to the 3’ end of the growing strand, so new strands are made 5’ to 3’.

Leading and lagging strands (HL only). At a fork, the strand whose template is read towards the fork is made continuously (leading strand), needing one RNA primer. The other template runs the wrong way, so it is copied discontinuously in short pieces called Okazaki fragments, each needing its own primer, made repeatedly as the fork opens.

Enzymes in the prokaryotic system (HL only).

Enzyme	Function
Primase	Makes short RNA primers
DNA polymerase III	Main enzyme extending primers by adding DNA nucleotides
DNA polymerase I	Removes RNA primers and replaces them with DNA
DNA ligase	Seals the gaps between fragments, joining sugar-phosphate backbone

Proofreading (HL only). DNA polymerase III can detect a mismatched base at the 3’ end it has just added, remove that nucleotide, then add the correct one. This greatly reduces errors.

Worked example (HL only). A bacterial chromosome of 4.6 × 10⁶ base pairs has one origin and two forks moving apart at 1000 nucleotides per second each (data given).

1. Each fork copies half: 2.3 × 10⁶ bp.
2. Time = 2.3 × 10⁶ ÷ 1000 = 2300 s, about 38 min.
3. Lagging strand per fork with fragments of 1500 nucleotides: 2.3 × 10⁶ ÷ 1500 ≈ 1530 fragments, so about 1530 primers per fork on the lagging side, plus one on each leading strand.

Proofreading: if polymerase makes 1 error per 10⁵ bases without proofreading and 1 per 10⁷ with it (data given), errors per replication fall from 4.6 × 10⁶ ÷ 10⁵ = 46 to 0.46, 100-fold fewer.

D1.3.1 to D1.3.3 Gene mutations

A gene mutation is a change in the base sequence of a gene. Types: substitution (one base swapped), insertion (extra base or bases) and deletion (loss of bases). A single-nucleotide polymorphism (SNP) is a substitution found at a particular position in a population.

Because the genetic code is degenerate, a substitution may be:

• silent: the new codon codes for the same amino acid;
• missense: a different amino acid is inserted, which may or may not alter function;
• nonsense: a stop codon appears, ending the polypeptide early.

Insertions and deletions not in multiples of three cause a frameshift: every codon after the change is read differently, so the polypeptide is usually non-functional. Large insertions or deletions, even in multiples of three, can also stop the protein working.

Worked example. Wild-type mRNA 5’ AUG CAU GGU ACU UAA 3’ codes Met-His-Gly-Thr-stop.

• Substitution CAU to CAC: His still, so silent.
• Substitution GGU to UGU: Gly becomes Cys, a missense change.
• Substitution UGG to UGA (a Trp codon elsewhere): becomes stop, a nonsense change.
• Insert G after CAU: AUG CAU GGG UAC UUA A reads Met-His-Gly-Tyr-Leu and onwards, a frameshift; the original stop codon is no longer read in frame.
• Delete the whole codon CAU: Met-Gly-Thr-stop, one amino acid lost, rest in frame.

D1.3.4 to D1.3.7 Causes, randomness, consequences, variation

Causes. Errors in replication or repair that are not corrected, and mutagens. Chemical mutagens include nitrous acid and the compounds in tobacco smoke. Mutagenic radiation includes ultraviolet light, X-rays and gamma rays.

Randomness. A mutation can occur at any base in a genome, though some bases mutate more often than others. No natural mechanism is known that makes a chosen change to a particular base in order to alter a trait.

Germ and somatic cells. A mutation in germ cells (gamete-producing cells) can be inherited by offspring and then appear in every cell. A mutation in a somatic cell is not passed on, but it can lead to uncontrolled division, which is cancer.

Worked example. Take a new mutation rate of 1.2 × 10⁻⁸ per base pair per generation and a diploid genome of 6.4 × 10⁹ bp (data given). Expected new mutations per child = 1.2 × 10⁻⁸ × 6.4 × 10⁹ = about 77. Many fall in non-coding DNA or are silent.

Variation. All genetic variation ultimately begins with mutation. Most are harmful or neutral for an individual, but over a long time new alleles supply the variation that natural selection acts on.

Nature of science. Commercial genetic tests can reveal disease risk, but without expert interpretation the results can be misunderstood or cause needless worry.

D1.3.8 to D1.3.10 Knockout, CRISPR and conserved sequences (HL only)

Gene knockout (HL only). A gene is altered so it no longer works, then the organism is compared with normal individuals to infer what the gene does. Libraries of knockout organisms exist for some model species.

Worked example (HL only). Mean body mass at 8 weeks: normal mice 24.5 g (n = 6, range 23.8 to 25.3); knockout mice 19.4 g (n = 6, range 18.7 to 20.1). Percentage difference = (19.4 − 24.5) ÷ 24.5 × 100 = −21%. The ranges do not overlap, supporting a role for the gene in growth, but a statistical test and other variables are needed before concluding.

CRISPR and Cas9 (HL only). A guide RNA matching a target sequence directs the enzyme Cas9 to cut the DNA at that site. Repair of the cut can disable the gene or, with a supplied template, change it to a chosen sequence. An example of use: a CRISPR-based therapy for sickle cell disease and transfusion-dependent beta-thalassaemia was authorised in the UK in 2023. Ethical issues, such as editing embryos, must be resolved first, and because countries regulate differently there is an international effort to harmonise rules.

Conserved sequences (HL only). These are identical or similar across species. Hypothesis one: the gene product has a function that tolerates little change, so natural selection removes variants. Hypothesis two: the sequence mutates more slowly.

Common errors

• Writing that replication is “conservative” or that the old strand is broken down.
• Forgetting that primers set the start and end of the amplified region.
• Saying bigger fragments travel further on a gel.
• Calling every substitution harmful.
• Forgetting that a frameshift needs a change not divisible by three.
• Saying mutations occur because the organism needs them.

Official syllabus

IB Diploma Programme Biology guide, first assessment 2025, published by the International Baccalaureate Organization: sections D1.1 (DNA replication) and D1.3 (Mutation and gene editing).