Skip to content
MarkSchemeMarkScheme
Sign up — no card

9700 · 6.2

Protein synthesis — practice questions

Practice and worked examples for 9700 Protein synthesis. Short previews only — attempt the full question in MarkScheme against the official scheme.

Worked example 1

A segment of a non-template strand of DNA has the sequence 5'-ATG CTC GAT TTA-3'.

  1. Deduce the sequence of the mRNA molecule synthesised from the template strand of this DNA segment.
  2. Using the genetic code table (assume standard table where AUG is Methionine), identify the amino acid sequence of the polypeptide produced from this mRNA.

Standard Genetic Code Table Excerpt:

CodonAmino AcidCodonAmino Acid
AUGMetUUALeu
------------
CUCLeuGAUAsp
CGAArgUAAStop
UAGStopUGAStop
Show solution outline

Part 1: Deduce mRNA sequence

  1. Identify the template strand: The given non-template (sense) strand is 5'-ATG CTC GAT TTA-3'. The template (antisense) strand runs in the opposite direction (3' to 5') and is complementary. Non-template DNA: 5'-A T G C T C G A T T T A-3' Template DNA: 3'-T A C G A G C T A A A T-5'

  2. Transcribe the template strand to mRNA: mRNA is synthesised 5' to 3', complementary to the template strand, with Uracil (U) replacing Thymine (T). Template DNA: 3'-T A C G A G C T A A A T-5' mRNA: 5'-A U G C U C G A U U U A-3'

    Shortcut: The mRNA sequence is identical to the non-template (sense) strand, with T replaced by U.

Part 2: Identify the amino acid sequence Read the mRNA codons and use the provided genetic code table:

  • AUG codes for Methionine (Met)
  • CUC codes for Leucine (Leu)
  • GAU codes for Aspartic Acid (Asp)
  • UUA codes for Leucine (Leu)

Therefore, the amino acid sequence is Met-Leu-Asp-Leu.

Worked example 2

A specific enzyme in humans has a single polypeptide chain consisting of 250 amino acids. The average relative molecular mass of an amino acid is 110.

  1. Calculate the minimum length (in number of bases) of the coding region of the mature mRNA molecule required to synthesise this enzyme.
  2. Calculate the relative molecular mass of the final enzyme polypeptide. (The relative molecular mass of water is 18).
Show solution outline

Part 1: Calculate mRNA length

  1. Bases for amino acids: Each amino acid is coded by a 3-base codon. Number of bases for the amino acids = 250 amino acids × 3 bases/amino acid = 750 bases.
  2. Bases for stop codon: Translation must be terminated by a stop codon, which does not code for an amino acid but is essential for the process. A stop codon is also 3 bases long.
  3. Total minimum length: The coding region includes the codons for all amino acids plus one stop codon. Minimum length of coding region = Bases for amino acids + Bases for stop codon = 750 + 3 = 753 bases.

Therefore, the minimum length of the coding region of the mRNA is 753 bases.

Part 2: Calculate polypeptide molecular mass

  1. Number of peptide bonds: When 'n' amino acids are joined to form a polypeptide, 'n-1' peptide bonds are formed. In this case, 250 amino acids are joined by 250 - 1 = 249 peptide bonds.
  2. Mass of water removed: Each peptide bond is formed by a condensation reaction, releasing one molecule of water. Total mass of water removed = 249 bonds × 18 (mass of water) = 4482.
  3. Initial mass of amino acids: Calculate the theoretical mass if no water was removed. Total initial mass of all amino acids = 250 amino acids × 110 = 27,500.
  4. Final polypeptide mass: Subtract the mass of the water molecules that were removed. Final molecular mass of polypeptide = Total initial mass - Total mass of water removed = 27,500 - 4,482 = 23,018.

Therefore, the relative molecular mass of the final enzyme polypeptide is 23,018.

Mark a 9700 question on this topicFull lessonOpen in course studio