Translation & the Genetic Code

You’ll be able to

Explain the roles of the ribosome, mRNA, and tRNA in building a polypeptide
Read the genetic code to translate mRNA codons into an amino acid sequence
Identify the start codon and stop codons that frame every reading frame

The genetic code: reading in threes

mRNA is read three bases at a time, and each triplet is a codon that specifies one amino acid. With four possible bases, there are 4³ = 64 codons for only 20 amino acids, so the code is redundant (degenerate) — several codons can call for the same amino acid. The code is nearly universal across all life. AUG does double duty: it is the start codon and also codes for methionine (Met), setting the reading frame. Three codons — UAA, UAG, and UGA — are stop codons that code for no amino acid and end translation.

The machinery: ribosome and tRNA

Translation happens on the ribosome, built of ribosomal RNA (rRNA) and protein in two subunits. The ribosome has three sites — A (arrival), P (peptide), and E (exit) — through which transfer RNAs move. Each tRNA is an adapter: at one end it carries a specific amino acid, and at the other it displays a three-base anticodon that base-pairs with a complementary mRNA codon. When the anticodon matches the codon in the A site, that tRNA’s amino acid is added to the growing chain, and a peptide bond links it to the previous one.

Initiation, elongation, termination

Translation mirrors transcription’s three stages. In initiation, the small ribosomal subunit binds the mRNA and finds the AUG start codon, where the first tRNA (carrying Met) settles into the P site; the large subunit then joins. In elongation, tRNAs deliver amino acids to the A site codon by codon, peptide bonds form, and the ribosome ratchets along the mRNA 5′→3′. In termination, a stop codon enters the A site; no tRNA matches it, a release factor binds instead, and the finished polypeptide is set free.

Codon → anticodon pairing

mRNA codon 5′–A U G–3′ pairs with tRNA anticodon 3′–U A C–5′

The anticodon is antiparallel and complementary to the codon (A–U, G–C), just like the two strands of DNA.

Worked example

A gene’s template strand reads 3′–T A C G G A T T C A C T–5′. Transcribe it to mRNA, then translate the mRNA into the amino acid sequence. (Codons: AUG = Met/Start, CCU = Pro, AAG = Lys, UGA = Stop.)

1.Transcribe: pair each template base for RNA (T→A, A→U, C→G, G→C). The template 3′–TAC GGA TTC ACT–5′ gives mRNA 5′–A U G C C U A A G U G A–3′.
2.Split the mRNA into codons starting at the 5′ AUG: AUG · CCU · AAG · UGA.
3.Translate each codon: AUG = Met (start), CCU = Pro, AAG = Lys, UGA = Stop (no amino acid — translation ends here).

Answer: mRNA is 5′–AUG CCU AAG UGA–3′, which translates to the polypeptide Met – Pro – Lys (the UGA stop codon ends the chain).

Checkpoint

An mRNA reads 5′–A U G U U C U A A–3′. What polypeptide does it produce? (AUG = Met, UUC = Phe, UAA = Stop.)

Watch out

Stop codons (UAA, UAG, UGA) are not amino acids — never write "Stop" as a residue in the chain. And always start reading at the first AUG; beginning one base off shifts the entire reading frame and gives a completely different protein.

Checkpoint

A tRNA carries the anticodon 3′–U A C–5′. Which mRNA codon does it read, and what does that codon do?

On the exam

Practice the full pipeline until it is automatic: DNA template → (transcribe, A–U) → mRNA codons → (read code 5′→3′) → amino acids. On the exam, show the mRNA and the codon splits explicitly — partial credit often lives in those middle steps.

Answer the 2 checkpoints as you read.