OCR A Level Biology A: The genetic code, transcription and translation (H420) – 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 – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

These questions are for OCR A Level Biology A (H420), written to the OCR specification Version 4.1 (April 2026), for first assessment in 2025. They cover module 2.1.3 learning outcomes (f) and (g). Codons needed for each question are given in the question. Learn the content first in the study guide or the revision notes. Course pages: OCR A Level Biology hub, the printable checklist and the free 10-minute diagnostics.

Questions

1. State four features of the genetic code. [4]

2. The genetic code is a triplet code.

Calculate how many different codes are possible with a group of two bases, and explain why a code based on two bases could not code for all 20 amino acids used in proteins. State how many codes a triplet gives. [3]

3. A section of the template strand of a gene reads, 3’ to 5’: TAC GTT CTT ACT

Use these codons: AUG = methionine, CAA = glutamine, GAA = glutamic acid, UGA = stop.

(a) Write the mRNA sequence transcribed from this section. [2] (b) State the amino acid sequence of the polypeptide and explain why it has fewer amino acids than the mRNA has codons. [2] (c) Give the anticodon of the tRNA that pairs with the second codon. [1]

4. Describe the role of RNA polymerase in transcription. [3]

5. (a) Describe the structure of a tRNA molecule. [3] (b) State the role of mRNA and the role of rRNA in making a polypeptide. [2]

6. Describe how a polypeptide is made at a ribosome once mRNA has reached the cytoplasm. [6]

7. The coding region of an mRNA, including the stop codon, is 1203 bases long.

(a) Calculate the number of amino acids in the polypeptide. [2] (b) A ribosome adds 16 amino acids per second (data supplied). Calculate the time taken to make one polypeptide. [2]

8. A section of mRNA is 15 bases long.

(a) State how many codons it contains if the code is non-overlapping, and how many it would contain if the code were overlapping with a new codon starting at every base. [2] (b) Explain why a base in an overlapping code would affect more than one amino acid. [2]

9. The codons GGU, GGC, GGA and GGG all code for glycine. The codons AAA and AAG both code for lysine.

(a) Explain what this shows about the genetic code. [2] (b) A cell makes an mRNA with GGC at a position where GGU was expected. Predict the effect on the polypeptide and explain your answer. [2]

10. A synthetic mRNA made of the repeating unit GAU, so it reads GAUGAUGAU and so on, is added to a cell-free system containing ribosomes, tRNA and amino acids. Two kinds of polypeptide are made, one containing only aspartic acid and one containing only methionine. No third kind is made.

Use these codons: GAU = aspartic acid, AUG = methionine, UGA = stop.

(a) Show the three ways the repeating sequence can be read in groups of three, and the result of each. [3] (b) Explain why only two kinds of polypeptide are made. [2] (c) Explain which features of the genetic code this result supports. [3]

11. Explain how the base sequence of a gene determines the structure and function of a protein. [6]

12. A gene from a human cell is inserted into a bacterium, and the bacterium makes the human polypeptide.

Explain, with reference to the genetic code, why this is possible. [3]

Answers

1. Any four from: it is a triplet code, with three bases coding for one amino acid [1]; it is non-overlapping, so each base is part of only one codon [1]; it is degenerate, so most amino acids have several codons [1]; it is universal, so the same codon codes for the same amino acid in virtually all organisms [1]; a stop codon codes for no amino acid, and AUG is the start codon (extra points beyond four features are not needed). [4] Examiner insight: Each feature needs a short explanation as well as its name. Writing only “degenerate” is a label, and a clear meaning in the same line shows you know the term.

2. Two bases give 4 × 4 = 16 combinations [1]. 16 is fewer than 20, so not every amino acid could have its own code [1]. Three bases give 4 × 4 × 4 = 64 codes, enough for 20 amino acids [1]. [3] Examiner insight: Show the multiplication as well as the totals. A correct comparison between 16 and 20 is the reasoning step that gains the explanation point.

3. (a) Each template base is paired with its complement [1]: T-A, A-U, C-G, G-C, with uracil in place of thymine, giving AUG CAA GAA UGA [1]. (b) Methionine-glutamine-glutamic acid [1]. The last codon, UGA, is a stop codon that codes for no amino acid, so there are four codons but three amino acids [1]. (c) Second codon CAA, so anticodon GUU [1]. [5] Examiner insight: A sequence containing T in mRNA is wrong, however neat the rest. Check each line for uracil before you move on.

4. RNA polymerase binds to the DNA at the start of the gene and separates the two strands by breaking the hydrogen bonds between bases [1]. It lines up free RNA nucleotides with the exposed template strand by complementary base pairing [1]. It joins the RNA nucleotides together by forming phosphodiester bonds, making mRNA [1]. [3] Examiner insight: Naming the bonds broken and the bonds formed separates a precise answer from a vague “unzips the DNA and builds RNA”. Describe three distinct actions to match three marks.

5. (a) A single polynucleotide strand [1] folded by hydrogen bonding between some bases into a clover-leaf shape [1]. It has an anticodon of three bases at one end and a site for a specific amino acid at the other [1]. (b) mRNA carries the code (the codons) from the gene to the ribosome [1]. rRNA forms part of the ribosome, which holds the mRNA and tRNA in position [1]. [5] Examiner insight: Give the role in each case, not a restatement of the name. “Messenger” tells you the label only, and a role involves what is carried and where.

6. Any six from: the small ribosomal subunit binds to the mRNA at the start codon AUG [1]; a tRNA with the complementary anticodon, carrying methionine, pairs with the codon [1]; a second tRNA pairs with the next codon, so the ribosome holds two tRNA at a time [1]; the two amino acids are joined by a peptide bond in a condensation reaction [1]; the first tRNA is released and the ribosome moves one codon along the mRNA [1]; the process repeats, building a chain in the order of the codons [1]; at a stop codon the polypeptide is released (a seventh point, beyond the six needed). [6] Examiner insight: A sequence question rewards order. Number your steps so that pairing, bond formation, movement and release appear in that sequence.

7. (a) 1203 ÷ 3 = 401 codons [1]. Subtract the stop codon: 401 - 1 = 400 amino acids [1]. (b) Time = 400 ÷ 16 [1] = 25 seconds [1]. [4] Examiner insight: Write down the division and the subtraction as separate lines. Forgetting the stop codon gives 401, so separate lines make a slip easy to spot. Include the unit in the final answer.

8. (a) Non-overlapping: 15 ÷ 3 = 5 codons [1]. Overlapping: a codon starts at each base except the last two, so 15 - 2 = 13 codons [1]. (b) In an overlapping code each base is part of up to three codons [1], so a change to one base would change up to three amino acids [1]. [4] Examiner insight: Explain the reasoning in terms of bases belonging to more than one codon. A bare statement that “it would be worse” gives nothing to credit.

9. (a) More than one codon can code for the same amino acid, so the code is degenerate [1]. The codons for one amino acid often differ only in the third base [1]. (b) GGC also codes for glycine, so the amino acid is the same [1]. The primary structure and the protein are unchanged [1]. [4] Examiner insight: A prediction needs a reason from the data. Quote the amino acid named in the question, rather than saying only “it makes no difference”.

10. (a) Reading from the first base: GAU GAU GAU, giving aspartic acid each time [1]. From the second base: AUG AUG AUG, giving methionine each time [1]. From the third base: UGA UGA UGA, giving stop each time [1]. (b) The reading frame depends on the base at which the ribosome starts reading [1]. The third frame begins with a stop codon, so no amino acids are added and no polypeptide forms [1]. (c) The sequence is read in groups of three (triplet) [1]. Each base belongs to one codon in a given frame, so the code is non-overlapping [1]. UGA acts as a stop signal and codes for no amino acid [1]. [8] Examiner insight: Use the data given in the question. Each frame should be written out in codons, because an unsupported statement that “three results are possible” earns no reading credit.

11. The gene’s base sequence is transcribed into an mRNA with a complementary sequence of codons [1]. Each codon (three bases) codes for one amino acid [1]. tRNA anticodons pair with the codons by complementary base pairing, bringing amino acids in order [1]. Amino acids are joined by peptide bonds at the ribosome [1]. The order of amino acids is the primary structure [1]. The primary structure decides how R group interactions fold the chain into a specific three-dimensional shape, which gives the protein its function [1]. [6] Examiner insight: Finish the chain from gene to function. Many answers stop at the amino acid sequence, but the last step (primary structure leads to shape, and shape leads to function) is the one that links the gene to the protein.

12. The genetic code is universal, so the same codon codes for the same amino acid in a human and in a bacterium [1]. The bacterial ribosomes and tRNA therefore read the human mRNA codons in the same way [1], and so the same amino acid sequence is built [1]. [3] Examiner insight: Link the universal code to what the bacterial cell does with the codons. A statement that the code is universal, with no consequence, is only the definition.

Where marks are usually lost

• Writing T in mRNA, or forgetting that tRNA anticodons are complementary to codons.
• Counting the stop codon as an amino acid when finding polypeptide length.
• Giving the definition of degenerate the wrong way round.
• Describing transcription without naming RNA polymerase, or without the bonds broken and formed.
• Placing translation in the nucleus, or saying that DNA leaves the nucleus.
• Giving the role of rRNA as “makes ribosomes” with no role in translation.
• Stopping an explanation at the amino acid sequence instead of linking it to protein shape and function.
• Not showing division by three in calculations.

Next steps

• Re-read the revision notes for anything you missed.
• Go back to the study guide for the explanations.
• Go to the OCR A Level Biology hub and the printable checklist.
• Take the free 10-minute diagnostics.
• Book a free trial class.

Official syllabus

OCR, A Level Biology A (H420) specification, Version 4.1 (April 2026), first assessment 2025: module 2.1.3 Nucleotides and nucleic acids, learning outcomes (f) and (g).