Transcription, Translation and the Genetic Code
General Biology · Section 2.2 · 18 study cards
From gene to protein: making RNA, processing it, reading codons and building a polypeptide on the ribosome.
Practice this set → Spaced repetition, card by card. No account needed.
Method
Hold the directions steady and everything else follows
Most lost marks on this unit are direction errors, not knowledge gaps. Three rules, memorised as a block, prevent nearly all of them: the template is read 3ʹ to 5ʹ, RNA is built 5ʹ to 3ʹ, and mRNA is translated 5ʹ to 3ʹ starting from the first AUG. The polypeptide correspondingly grows from its amino end to its carboxyl end.
A procedure for any gene-to-protein problem
- Identify which strand is the template. If the question gives the coding strand instead, the mRNA is that sequence with U in place of T.
- Transcribe. Write the mRNA with its 5ʹ and 3ʹ labels, then check that it is antiparallel to the template.
- Find the first AUG. That fixes the reading frame for everything after it. Do not assume the frame starts at base one of the sequence you were given.
- Split into triplets from AUG onwards, in a single pass, and only then look up amino acids. Splitting and translating at the same time is how frames get lost.
- Stop at the first stop codon and say so explicitly. Anything after it is untranslated.
- If a mutation is applied, redo the sequence from the mutated base onwards rather than editing your first answer in place.
Processing is the eukaryotic difference
Group the eukaryote-only steps together: cap, tail, splice, export. Each exists because the nucleus separates transcription from translation in time and space, and each is also a place where expression can be regulated. Bacteria, having no nucleus, translate a transcript while it is still being made, and that single fact answers most comparison questions.
Know who reads what
Assign every reading job to its molecule: RNA polymerase reads the DNA template, tRNA anticodons read the codons, aminoacyl-tRNA synthetase reads both the amino acid and the tRNA, and release factors read stop codons. If you can say which molecule is doing the recognising at each step, you can reason out the effect of almost any lesion the exam invents.
Definitions and theorems
Worked example
A short bacterial gene has the template strand 3ʹ-TAC CGG TTA GCA ATC-5ʹ. Transcribe it, translate the mRNA, then apply a mutation in which the seventh base of the template strand, a T, is deleted. Give the new mRNA and the new polypeptide, and name the type of mutation.
Transcribe the original. RNA is antiparallel and complementary to the template, with U replacing T, so reading the template 3ʹ to 5ʹ gives mRNA 5ʹ-AUG GCC AAU CGU UAG-3ʹ. Check the first three bases: template TAC gives AUG, which is the expected start.
Set the frame and translate. Start at AUG. AUG is methionine, GCC is alanine, AAU is asparagine, CGU is arginine, and UAG is a stop codon. The polypeptide is Met-Ala-Asn-Arg, four residues, and translation stops there.
Apply the deletion. Counting from the 3ʹ end as written, the template is T A C C G G T T A G C A A T C and the seventh base is the first T of the TTA triplet. Removing it leaves the template 3ʹ-TACCGG TAGCAATC-5ʹ, now fourteen bases long.
Transcribe the mutant. The new mRNA is 5ʹ-AUG GCC AUC GUU AG-3ʹ. Notice that the first six bases are unchanged, because the deletion lies downstream of them.
Translate the mutant. AUG is methionine, GCC is alanine, AUC is isoleucine, GUU is valine, and the trailing AG is an incomplete codon, so the ribosome would carry on into whatever sequence follows in the real gene. The polypeptide begins Met-Ala-Ile-Val and no longer stops where it did.
Classify and explain. Deleting one base is a frameshift mutation, because one is not a multiple of three. Everything before the deletion is unaffected, everything after it is read in a new frame, and the original UAG stop codon has been destroyed, so the protein is both wrong from residue three onwards and abnormally long.
Contrast with a substitution. Had the same base merely been changed rather than removed, at most one codon would have altered, giving a silent, missense or nonsense mutation. The severity of a frameshift comes entirely from the loss of the reading frame, not from the number of bases involved.
Common mistakes
- Using the coding strand as the template. The mRNA sequence matches the coding strand, so if you transcribe the coding strand by complementing it you get the template strand written in RNA, which is the exact opposite of the answer. Always name which strand you have been given before writing anything.
- Splitting into codons from the wrong place. The frame starts at the first AUG, not at the first base of whatever sequence was printed on the exam. Students who chop the whole sequence into triplets before finding AUG usually get every amino acid wrong while making no other mistake.
- Writing thymine in RNA or using DNA triplets in the codon table. Codon tables are RNA tables. Looking up TAC instead of the mRNA codon AUG is a common and costly slip, and any RNA answer containing T is wrong by inspection.
- Saying a tRNA checks its own amino acid. It does not. The aminoacyl-tRNA synthetase is the only step that matches amino acid to tRNA, and the ribosome checks only codon to anticodon pairing. This is the whole point of questions about mischarged tRNAs.
- Claiming eukaryotic mRNA is translated as transcribed. Only prokaryotes couple the two. A eukaryotic transcript must be capped, tailed, spliced and exported first, and forgetting this sinks the standard compare-and-contrast question.
Practice it
Reading the method is not the same as being able to recall it under pressure. This set drills 18 cards one at a time and schedules each card separately, so the ones you keep missing come back sooner.