OCR GCSE Biology A: DNA structure, protein synthesis and enzyme action (J247)

This guide is written for OCR GCSE (9-1) Gateway Science Biology A (J247), using the specification Version 4.0 (August 2026), for first assessment in 2018. It covers module B1.2, “What happens in cells (and what do cells need)?”, statements B1.2a to B1.2g. Statements B1.2d and B1.2e are in bold in the specification and are tested only in the Higher Tier papers; they are labelled “Higher Tier only” below. Everything else is for both tiers.

Use it with the revision notes and the practice questions. For the wider topic see Cell level systems, the OCR GCSE Biology hub and the printable checklist. The free diagnostics are a quick way to see which statements need work.

What this unit covers

Statement	What you must be able to do	Tier
B1.2a	Describe DNA as a polymer	Both
B1.2b	Describe DNA as two strands forming a double helix	Both
B1.2c	Describe four different nucleotides (common sugar and phosphate, one of four bases); pairs A-T and G-C	Both
B1.2d	Recall a simple description of protein synthesis: unzipping around the gene, copying to mRNA in the nucleus (transcription), translation in the cytoplasm, tRNA carrying amino acids	Higher Tier only
B1.2e	Explain simply how DNA structure affects the proteins made: the triplet code sets amino acid order	Higher Tier only
B1.2f	Describe experiments to investigate enzymatic reactions	Both
B1.2g	Explain enzyme action: role in metabolism, active site, specificity (lock and key), and the effects of pH, temperature, substrate and enzyme concentration	Both

The specification’s maths outcomes for this module are rate calculations for chemical reactions and simple compound measures such as rate. Both appear below.

DNA as a polymer (B1.2a, B1.2b)

A polymer is a long molecule made of many repeating units joined in a chain. The units are monomers. DNA is a polymer whose monomers are nucleotides.

DNA is made of two strands. They wind around each other to form a double helix, like a twisted ladder. Each strand is a chain of nucleotides, and the two strands are held side by side by the bases in the middle.

A common misconception, noted in the specification, is that DNA is made of protein or sugar. It is made of nucleotides. It does contain a sugar and a phosphate group in every nucleotide, but the molecule is a nucleic acid, not a sugar and not a protein.

Nucleotides and base pairing (B1.2c)

Every nucleotide has three parts:

• a sugar, the same in every nucleotide (a common sugar)
• a phosphate group, the same in every nucleotide
• one of four bases attached to the sugar: A (adenine), T (thymine), G (guanine) or C (cytosine)

Only the base differs between the four nucleotides. The sugar and phosphate groups link up along the outside of each strand. The bases point inwards and pair up across the helix.

The pairs are complementary: A pairs only with T, and G pairs only with C. So if you know one strand you can write the other.

A rough picture of a short section:

Strand 1	A	T	T	G	C
Strand 2	T	A	A	C	G

Worked example 1: complementary strands

One strand reads ATTGCCAGT. Write the other strand.

A pairs with T, T with A, T with A, G with C, C with G, C with G, A with T, G with C, T with A.

Other strand: TAACGGTCA.

Worked example 2: percentages of bases

A sample of double-stranded DNA is 28% adenine. Find the percentage of each base.

• A pairs with T, so T is also 28%.
• A + T = 56%, so G + C = 100 - 56 = 44%.
• G and C are equal in number, so each is 44 ÷ 2 = 22%.

A 28%, T 28%, G 22%, C 22%.

Protein synthesis (B1.2d) - Higher Tier only

A gene is a section of DNA that codes for one protein. Proteins are made of amino acids. The order of the amino acids is what makes one protein different from another.

DNA never leaves the nucleus, yet proteins are built in the cytoplasm. A working copy of the gene has to be made and sent out. The steps:

1. Unzipping. The DNA double helix unwinds and the two strands separate around the gene being used.
2. Transcription. A copy of the gene is made as a single strand of mRNA (messenger RNA) in the nucleus. The mRNA is built by matching nucleotides to one of the DNA strands, so its sequence is complementary to that strand.
3. Leaving the nucleus. The mRNA moves out into the cytoplasm.
4. Translation. The nucleotide sequence of the mRNA is read in the cytoplasm at a ribosome. tRNA (transfer RNA) molecules carry amino acids to the ribosome, and the amino acids are joined in the order the mRNA specifies.
5. The chain folds. The finished chain of amino acids folds into a protein with a specific 3D shape.

Extra detail, not required by the statement but useful: RNA contains the base uracil (U) where DNA has thymine, so an A in the DNA strand is matched by U in the mRNA.

How DNA structure affects proteins (B1.2e) - Higher Tier only

The bases are read in groups of three, called triplets. This is the triplet code. In mRNA each triplet specifies a single amino acid. The sequence of bases along the gene therefore fixes the sequence of amino acids in the protein.

A different base sequence means a different amino acid order. A different order folds differently and gives a different shape. Since shape decides function (an enzyme’s active site, for example), a change in the DNA can change what the protein does.

Worked example 3: from gene to amino acids

Data supplied: AUG = Met, CCA = Pro, GAA = Glu, GGC = Gly, AAA = Lys, AAG = Lys, AAC = Asn, GUA = Val.

The template strand of a gene reads TAC GGT CTT.

• Match each base: T→A, A→U, C→G gives AUG.
• G→C, G→C, T→A gives CCA.
• C→G, T→A, T→A gives GAA.

mRNA: AUG CCA GAA. Amino acids: Met - Pro - Glu.

Worked example 4: effect of a change

The same template, TAC GGT CTT, has its last triplet changed to CAT.

• C→G, A→U, T→A gives mRNA triplet GUA, which is Val.
• The protein is now Met - Pro - Val instead of Met - Pro - Glu.

One amino acid has changed. This could alter the folded shape. Not every change matters: if CTT became CTC, the mRNA triplet would be GAG, which codes for the same amino acid (Glu) in the real genetic code, so the protein would be unchanged. A deleted base is more serious, because every triplet after it shifts and the amino acid order changes along the rest of the chain.

Worked example 5: how many bases?

A protein has 120 amino acids. Each amino acid needs one triplet, so the coding sequence is 120 × 3 = 360 bases. On a double-stranded gene those 360 bases sit opposite 360 partners.

Enzymes: what they do (B1.2g)

Metabolism is the sum of all the chemical reactions in a cell or organism. Enzymes are biological catalysts: they speed up reactions without being used up. Without them most reactions in cells would be far too slow.

Enzymes are proteins. Each has an active site, a region with a specific shape where the substrate binds. The substrate fits the active site and the reaction takes place, making the product.

Enzymes are specific: the active site shape matches only one substrate (or a very small group). This is the lock and key hypothesis: the substrate is the key and the active site is the lock.

Denaturation

If the enzyme’s structure is disrupted, the active site changes shape, the substrate no longer fits and the enzyme cannot work. The enzyme is denatured. This is permanent: lowering the temperature again does not restore it. High temperatures and extreme pH values both denature enzymes.

Factors affecting enzyme-controlled reactions (B1.2g)

Temperature. Warmer particles move faster, so enzyme and substrate meet more often and each meeting is more energetic. The rate rises up to the optimum. Above it the enzyme starts to denature and the rate falls steeply to zero. Cooling slows the reaction but does not denature the enzyme.

The specification warns against assuming that every enzyme works best at 37 °C. Optimum temperature varies between enzymes and organisms.

pH. Each enzyme has an optimum pH. Away from it the shape of the active site changes and the enzyme denatures. Pepsin in the stomach works best in acid conditions, while enzymes in the small intestine work best in alkaline conditions, so optima are not all near pH 7.

Substrate concentration. With a fixed amount of enzyme, more substrate means more collisions, so the rate rises. Eventually every active site is busy, and adding more substrate has no effect. The graph levels off. Something else, such as enzyme concentration, is now limiting.

Enzyme concentration. If substrate is in excess, adding enzyme adds active sites and the rate climbs. The rise is in proportion, and stops only when substrate becomes the limiting factor.

Investigating enzymes (B1.2f) and rate calculations

You are expected to describe experiments and analyse the numbers. Two common set-ups:

• Catalase and hydrogen peroxide. Measure the volume of oxygen collected in a fixed time (gas syringe or inverted measuring cylinder).
• Amylase and starch. Sample the mixture at intervals on a spotting tile of iodine solution. Record the time until iodine stays orange-brown, which shows that starch has gone.

To test a factor, change one variable only (the independent variable), measure the dependent variable, and control the rest (volumes, concentrations, temperature or pH, enzyme source). Repeat and find a mean. Use a water bath or buffer solutions to control temperature or pH.

To show results on a graph, put the independent variable on the horizontal axis and the measured variable on the vertical axis, label both with units, plot every point and draw a smooth line or curve through the trend. Reading values and gradients from a graph is part of the analysis.

Rate = amount of product formed (or substrate used) ÷ time. Where only a time is measured, 1 ÷ time is a measure of rate.

Worked example 6: temperature data

Catalase breaks down hydrogen peroxide. The table gives oxygen collected in 60 s.

Temperature / °C	15	25	35	45	55	65
O₂ collected in 60 s / cm³	5	11	21	33	9	0

Rate at 45 °C = 33 ÷ 60 = 0.55 cm³/s. At 35 °C it is 21 ÷ 60 = 0.35 cm³/s.

Percentage increase from 35 °C to 45 °C = (0.55 - 0.35) ÷ 0.35 × 100 = 57% (2 s.f.).

The optimum lies between 35 and 55 °C, and 45 °C is the best of the values tested. To narrow it down, repeat at 40, 45 and 50 °C. At 65 °C no gas forms because the enzyme is denatured.

Worked example 7: rate from a graph

Oxygen collected over time:

Time / s	0	10	20	30	40	50	60
Volume / cm³	0	14	24	30	34	36	36

• 0 to 10 s: 14 ÷ 10 = 1.4 cm³/s
• 20 to 30 s: (30 - 24) ÷ 10 = 0.6 cm³/s
• 40 to 50 s: (36 - 34) ÷ 10 = 0.2 cm³/s
• 50 to 60 s: 0 cm³/s

The rate falls because substrate is used up, so there are fewer collisions with active sites. After 50 s the line is flat: the reaction has finished. On a graph, the steepness of the line at a point is the rate at that moment.

Common errors

• Writing “the enzyme is killed”. Enzymes are not alive; say denatured.
• Saying low temperature denatures an enzyme. It slows collisions only.
• Saying the substrate changes shape to fit. In the model you use, the active site matches the substrate.
• Pairing A with G or C with T.
• Forgetting units in rates, such as cm³/s or g/min.

Official syllabus

OCR GCSE (9-1) Gateway Science Biology A (J247), Version 4.0 (August 2026), for first assessment in 2018, published by OCR: module B1.2.