Biology • Year 12 • Module 5 • Lesson 11
Translation — From mRNA to Polypeptide
Apply codon-anticodon matching, the genetic code table and ribosome behaviour to data, scenarios and a predict-and-justify problem.
1. Sequence the steps of translation
The seven events below have been shuffled. They describe the elongation of a polypeptide from an mRNA that has already been exported from the nucleus. In the "Order" column write 1–7 in the correct sequence. 7 marks (1 per correctly placed event)
| Order (1–7) | Event |
|---|---|
| The ribosome catalyses formation of a peptide bond between the two adjacent amino acids. | |
| mRNA leaves the nucleus and enters the cytoplasm. | |
| The ribosome moves one codon along the mRNA so that the next codon is ready to be read. | |
| A second tRNA arrives, its anticodon pairing with the next codon on the mRNA. | |
| The ribosome attaches to the mRNA and begins reading codons three bases at a time. | |
| A first tRNA, carrying its specific amino acid, pairs its anticodon with the first codon being read. | |
| The polypeptide chain has elongated and the cycle of codon-anticodon matching and peptide bond formation repeats. |
2. Codon-table interpretation — predict the amino acid sequence
The fragment of the genetic code below shows which amino acid is specified by selected mRNA codons. Use it together with the rules of codon-anticodon pairing to answer 2.1–2.4. 9 marks
| mRNA codon | Amino acid (3-letter) | mRNA codon | Amino acid (3-letter) |
|---|---|---|---|
| AUG | Met (start) | CCU | Pro |
| GAA | Glu | UAC | Tyr |
| GUG | Val | UAA | STOP |
| CGA | Arg | GGC | Gly |
Reminder: tRNA anticodons are written 3′→5′ but in this lesson we read both strands left-to-right and require the anticodon to be complementary to the codon, base-for-base (A↔U, G↔C).
2.1 The mRNA fragment AUG GAA CCU UAC is translated. Write out the polypeptide sequence (4 amino acids, 3-letter codes) produced. 2 marks
2.2 Write the complementary tRNA anticodon that pairs with each of the four mRNA codons in 2.1. 2 marks
2.3 A point mutation changes the second codon from GAA to GUG. Identify the new polypeptide sequence and state which amino acid has been substituted. 3 marks
2.4 A different mutation changes the third codon from CCU to UAA. Use the table to predict and justify what happens to the polypeptide. 2 marks
3. Graph interpretation — ribosome translation rate vs cycloheximide
The graph below shows the average rate of polypeptide synthesis (amino acids added per ribosome per second) in cultured human cells as the concentration of cycloheximide — a drug that binds the large ribosomal subunit — is increased. 7 marks
Figure 3.1. Translation rate vs cycloheximide concentration in cultured HeLa cells. Adapted from Schneider-Poetsch et al. (2010), Nature Chemical Biology 6: 209-217.
3.1 Describe the overall trend shown by the data from 0 µM to 100 µM cycloheximide. 2 marks
3.2 Estimate the translation rate at 20 µM and at 60 µM cycloheximide, and calculate the % reduction at each compared with the 0 µM baseline. 2 marks
3.3 Cycloheximide binds the large ribosomal subunit and blocks the movement of the ribosome along the mRNA. Use lesson content to explain why blocking ribosome movement reduces the rate of polypeptide synthesis. 3 marks
4. Predict and justify — a translation error in proinsulin
A point mutation in the INS gene changes a single codon in the mRNA from UGC (Cys) to UGG (Trp). Cysteine residues at this position normally form a disulfide bridge that holds the two chains of mature insulin together. 5 marks
4.1 Identify the amino acid that will now be inserted into the polypeptide at this position, and predict how this could affect the final insulin protein. 2 marks
4.2 A classmate claims: "Because translation is so accurate, this mutation should not affect insulin — the ribosome will fix the codon before adding the wrong amino acid." Predict whether the classmate is correct and justify using lesson content. 3 marks
Q1 — Sequence the steps (7 marks)
Correct order: 1 mRNA leaves the nucleus and enters the cytoplasm. 2 The ribosome attaches to the mRNA and begins reading codons three bases at a time. 3 A first tRNA, carrying its specific amino acid, pairs its anticodon with the first codon being read. 4 A second tRNA arrives, its anticodon pairing with the next codon on the mRNA. 5 The ribosome catalyses formation of a peptide bond between the two adjacent amino acids. 6 The ribosome moves one codon along the mRNA so that the next codon is ready to be read. 7 The polypeptide chain has elongated and the cycle of codon-anticodon matching and peptide bond formation repeats.
Marking notes. 1 mark per event placed in the correct position. Accept swapping steps 3 and 4 only if students explicitly state both tRNAs arrive in parallel at the ribosome's two binding sites — penalise reversal of 5 and 6.
Q2 — Codon-table interpretation (9 marks)
2.1 (2) Polypeptide: Met – Glu – Pro – Tyr. (1 mark for correct decoding of AUG–GAA; 1 mark for correct decoding of CCU–UAC.)
2.2 (2) Complementary anticodons: UAC · CUU · GGA · AUG. (1 mark for first two correct; 1 mark for last two correct. Penalise the use of T in place of U.)
2.3 (3) New codon GUG → Val, so the new polypeptide is Met – Val – Pro – Tyr. Glu has been substituted by Val. (1 mark for new amino acid identified; 1 mark for full new sequence; 1 mark for naming the substitution Glu→Val.) Accept students noting that this is the same nucleotide change responsible for sickle-cell haemoglobin.
2.4 (2) CCU → UAA is a STOP codon, so translation terminates at this codon. The polypeptide produced is only Met – Glu — the ribosome releases an incomplete chain. (1 mark for identifying STOP / premature termination; 1 mark for stating the truncated product.)
Q3 — Cycloheximide graph (7 marks)
3.1 (2) As cycloheximide concentration increases, the translation rate falls steeply at first and then levels off near zero [1]. The rate drops from about 5.6 amino acids/s at 0 µM to about 0.4 amino acids/s at 100 µM — a negative, non-linear relationship that plateaus by ~80 µM [1].
3.2 (2) At 20 µM: ≈ 2.8 aa/s, a reduction of ≈ 50% from baseline. At 60 µM: ≈ 0.9 aa/s, a reduction of ≈ 84% from baseline. (1 mark per correct estimate-with-% pair. Accept ±0.3 aa/s and ±5% on each.)
3.3 (3) Translation depends on the ribosome moving along the mRNA so the next codon can be read and the next tRNA can pair its anticodon [1]. If cycloheximide blocks that translocation, codon-anticodon matching cannot bring in the next amino acid, so the next peptide bond cannot form [1]. As more ribosomes are stalled, fewer amino acids are added per second across the cell — exactly the drop the graph shows [1].
Q4 — Insulin point mutation (5 marks)
4.1 (2) The new amino acid is tryptophan (Trp) — the tRNA with anticodon ACC pairs with UGG and brings Trp instead of Cys [1]. Tryptophan cannot form the disulfide bridge that the original cysteine residue provided, so the two chains of the mature insulin protein may not be held together correctly; the protein may misfold or fail to function as the insulin hormone [1].
4.2 (3) The classmate is incorrect [1]. The ribosome does not "fix" or proofread the codon during translation — its role is to read codons and coordinate amino acid joining; codon-anticodon matching simply pairs whichever tRNA is complementary to the codon presented [1]. If the codon now reads UGG, a Trp-charged tRNA will pair with it and Trp will be inserted, so the mutation will be translated faithfully into a faulty insulin polypeptide [1].