Key ideas
Gene expression turns the information in DNA into protein through transcription and translation. The questions test base pairing, the genetic code and the effect of mutations.
- Transcription: RNA polymerase reads the template strand 3′ → 5′ and builds mRNA 5′ → 3′, pairing A with U, T with A, G with C and C with G.
- Translation: ribosomes read mRNA in codons. Each tRNA carries an amino acid and an anticodon that pairs with the codon. AUG starts, and UAA, UAG and UGA stop.
- The code is degenerate (several codons per amino acid) but not ambiguous (each codon gives one amino acid).
- Point mutations can be silent, missense or nonsense. An insertion or deletion not in multiples of three shifts the reading frame.
Key equations
DNA template → mRNA (A→U, T→A, G→C, C→G)codon = 3 basesAUG = startUAA, UAG, UGA = stop
Worked example
A template strand reads 3′-TAC GGT ATT-5′. The mRNA is 5′-AUG CCA UAA-3′. AUG codes for Met, CCA for Pro and UAA is a stop codon, so the peptide is Met–Pro. Deleting a single base after the start shifts the frame and changes every codon downstream, while deleting three bases removes just one amino acid.
Common mistakes
- Reading the coding strand instead of the template strand when asked for the mRNA from the template.
- Writing T in RNA. RNA uses U in place of T.
- Calling a frameshift a point mutation of the same type as a silent change. A frameshift usually destroys the protein.
Practice
Test yourself with practice questions on this topic: Gene expression and protein synthesis