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Six original Cambridge-style questions on protein synthesis, from converting base sequences to explaining the roles of each kind of RNA.

Original questions All questions on this page are original work, written in the Cambridge AS & A Level style. They are not from past papers. They test the same concepts and skills the syllabus rewards.
Keep these straight

Template, codon, anticodon.

01
Recall
[2 marks]

State what is meant by a gene, and name the DNA strand that is used in transcription.

  • A gene is a sequence of nucleotides, forming part of a DNA molecule, that codes for a polypeptide. ✓
  • The template strand (also called the transcribed strand); the other strand is the non-transcribed strand. ✓
02
Explain
[4 marks]

Describe how an mRNA molecule is made from a gene in the nucleus of a eukaryotic cell.

  • RNA polymerase binds to the gene, breaks hydrogen bonds and separates the two DNA strands. ✓
  • Free RNA nucleotides pair with exposed bases on the template strand by complementary base pairing (A-U, T-A, C-G, G-C). ✓
  • RNA polymerase joins the nucleotides with phosphodiester bonds to form the primary transcript. ✓
  • Introns are removed and exons are joined together to form mRNA, which leaves through a nuclear pore. ✓
03
Data
[3 marks]

Part of a template DNA strand reads 3′ TAC AAA CCG ATT 5′. Using the information that AUG = Met, UUU = Phe, GGC = Gly and UAA = stop, give the mRNA sequence, the tRNA anticodons and the amino acid sequence.

  • mRNA: 5′ AUG UUU GGC UAA 3′. ✓
  • Anticodons: UAC, AAA, CCG (none for the stop codon). ✓
  • Amino acids: Met-Phe-Gly, then the polypeptide is released at UAA. ✓
04
Explain
[5 marks]

Explain how the sequence of codons in mRNA determines the sequence of amino acids in a polypeptide.

  • mRNA binds to a ribosome; translation starts at the start codon AUG. ✓
  • Each tRNA carries a specific amino acid and has an anticodon. ✓
  • The anticodon pairs with the complementary codon on the mRNA by hydrogen bonds, so the right amino acid is brought for each codon. ✓
  • A peptide bond forms between adjacent amino acids; the ribosome moves along one codon at a time. ✓
  • At a stop codon no tRNA binds and the polypeptide is released; so the codon order fixes the amino acid order. ✓
05
Calculation
[2 marks]

A eukaryotic gene contains 3 exons of 420, 270 and 231 nucleotides and 2 introns. The mRNA includes a start codon and a stop codon and no other non-coding bases. Calculate the number of amino acids in the polypeptide.

  • Coding mRNA = 420 + 270 + 231 = 921 nucleotides = 307 codons. ✓
  • The stop codon codes for no amino acid, so 306 amino acids. ✓
06
Explain
[2 marks]

Suggest why a human gene for insulin can be translated into human insulin inside a bacterium, but the bacterium could not use the gene directly as it is found in human chromosomes.

  • The genetic code is universal, so the bacterial ribosomes and tRNAs read the same codons as the same amino acids. ✓
  • The human gene contains introns, and bacteria cannot remove introns, so a version without introns (made from mRNA) must be used. ✓

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