← Unit 4: Heredity and continuity of life
Topic 1: DNA, genes and the continuity of life
Explain gene expression through transcription and translation, including the role of mRNA, tRNA, rRNA, the codon table and ribosomes, and compare prokaryotic and eukaryotic gene expression
A focused answer to the QCE Biology Unit 4 dot point on gene expression. Covers transcription (RNA polymerase, template strand, mRNA), the codon to amino acid code (universal, degenerate, non-overlapping), translation at the ribosome (initiation, elongation, termination) and the main differences between prokaryotic and eukaryotic expression.
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What this dot point is asking
QCAA wants you to explain how the information in a gene becomes a protein, name the molecular players, use the codon table to translate sequences and identify the main differences between prokaryotic and eukaryotic gene expression. Genetic code questions where you decode a short sequence are routine.
The answer
A gene is a stretch of DNA that codes for a functional product, usually a polypeptide. Gene expression is the two-step process that turns that sequence into protein: transcription copies DNA into RNA, and translation decodes RNA into a chain of amino acids.
Transcription
Location. Nucleus in eukaryotes, cytoplasm in prokaryotes.
Enzyme. RNA polymerase. It binds to a region of DNA called the promoter just upstream of the gene, unwinds a short stretch of the double helix and reads the template strand in the 3 prime to 5 prime direction.
Process.
- Initiation. RNA polymerase recognises the promoter (TATA box in eukaryotes; specific sequences such as the to 10 box in prokaryotes). Transcription factors help recruit the enzyme in eukaryotes.
- Elongation. RNA polymerase synthesises mRNA 5 prime to 3 prime, adding ribonucleotides complementary to the template strand. Uracil replaces thymine. The non-template (sense) strand has the same sequence as the mRNA except for the T to U change.
- Termination. RNA polymerase reaches a termination signal (hairpin loops in prokaryotes; a polyadenylation signal in eukaryotes) and releases the new RNA molecule.
Eukaryotic processing. The primary transcript (pre-mRNA) is modified before leaving the nucleus.
- 5 prime cap. A modified guanine (7-methylguanosine) is added to the 5 prime end. Helps the ribosome find the start codon and protects the mRNA from degradation.
- 3 prime poly-A tail. A string of about 200 adenines is added to the 3 prime end. Increases mRNA stability and export.
- Splicing. Introns (non-coding) are removed and exons (coding) are joined together by the spliceosome. Alternative splicing of the same pre-mRNA can produce different mature mRNAs and therefore different proteins.
The genetic code
The genetic code links sequences of nucleotides in mRNA to sequences of amino acids in protein. Its key features:
- Triplet. Three nucleotides (one codon) code for one amino acid.
- 64 codons. AUG is the start codon (also codes for methionine) and UAA, UAG and UGA are the three stop codons.
- Degenerate. Most amino acids are coded for by more than one codon (for example leucine has six codons). This buffers against some mutations.
- Non-overlapping. Codons are read consecutively, not in overlapping windows.
- Universal. The same code is used in almost every organism, which is why a human gene can be expressed correctly in a bacterium.
A standard codon table is read in three steps: find the first base on the left axis, the second across the top and the third on the right axis, then read off the amino acid.
Translation
Location. Cytoplasm, on ribosomes (free or membrane-bound).
Components.
- mRNA. The message, read 5 prime to 3 prime.
- Ribosome. Two subunits made of rRNA and protein. Has three tRNA binding sites: A (aminoacyl, incoming), P (peptidyl, holds the growing chain) and E (exit). The peptidyl transferase activity that joins amino acids is catalysed by rRNA, making the ribosome a ribozyme.
- tRNA. A small RNA molecule folded into a cloverleaf shape. Carries a specific amino acid at the 3 prime end and a three-base anticodon at the other end that base pairs with the mRNA codon.
- Aminoacyl tRNA synthetases. A family of 20 enzymes that load each tRNA with the correct amino acid, using ATP.
Process.
- Initiation. The small ribosomal subunit binds the 5 prime cap (eukaryotes) or the Shine-Dalgarno sequence (prokaryotes) and scans to the start codon AUG. The initiator tRNA carrying methionine enters the P site. The large subunit then joins.
- Elongation. The next codon is exposed in the A site. The tRNA with the matching anticodon enters, carrying its amino acid. The ribosome catalyses a peptide bond between the amino acid in the A site and the growing chain in the P site. The ribosome translocates: the tRNA in the A site moves to the P site, the tRNA in the P site moves to the E site and exits.
- Termination. When a stop codon (UAA, UAG or UGA) enters the A site, release factors bind, the peptide chain is hydrolysed off the final tRNA and the ribosome dissociates.
The newly made polypeptide folds (sometimes assisted by chaperone proteins) into its functional three-dimensional shape. It may be modified further (phosphorylation, glycosylation, cleavage) before becoming an active protein.
Prokaryotic versus eukaryotic expression
| Feature | Prokaryotes | Eukaryotes |
|---|---|---|
| Location | Cytoplasm | Transcription in nucleus, translation in cytoplasm |
| Coupling | Coupled (translation begins while still transcribing) | Separated in time and space |
| Gene structure | Often continuous, often in operons | Split into exons and introns |
| mRNA processing | Minimal | 5 prime cap, 3 prime poly-A tail, splicing |
| Regulation | Mostly transcriptional (operons) | Multiple layers (chromatin, transcription factors, splicing, RNA stability, translation) |
The lac operon in E. coli is the classic prokaryotic example: three genes for lactose metabolism transcribed from one promoter, with a repressor that releases when lactose is present.
Common traps
Using the template strand directly as the protein code. The mRNA is complementary to the template strand. The mRNA reads the same as the non-template strand except U replaces T.
Forgetting the start codon codes for an amino acid. AUG codes for methionine in addition to being the start signal.
Saying stop codons code for an amino acid. Stop codons do not code for any amino acid; they signal termination.
Saying tRNAs read DNA. tRNAs read mRNA codons through their anticodons.
Treating introns as junk. Introns can contain regulatory sequences and enable alternative splicing.
In one sentence
Gene expression is transcription of DNA into mRNA by RNA polymerase (in eukaryotes followed by 5 prime capping, poly-A tailing and intron splicing) and translation of mRNA into a polypeptide at the ribosome, where tRNAs deliver the amino acid specified by each codon of the universal, degenerate, non-overlapping triplet code from start codon AUG to one of the three stop codons.
Past exam questions, worked
Real questions from past QCAA papers on this dot point, with our answer explainer.
2024 QCAA6 marksUsing the DNA template strand 3 prime TAC GGA TCA ACT 5 prime, determine the mRNA sequence and then the amino acid sequence of the resulting peptide. Outline the role of each major component of the translation machinery.Show worked answer →
A 6-mark answer needs the mRNA, the amino acid sequence and the named components with their roles.
mRNA sequence. Transcription is complementary and antiparallel. The template 3 prime TAC GGA TCA ACT 5 prime gives mRNA 5 prime AUG CCU AGU UGA 3 prime. Remember that thymine in DNA is replaced by uracil in RNA.
Amino acid sequence. Read codons 5 prime to 3 prime.
- AUG = methionine (start codon)
- CCU = proline
- AGU = serine
- UGA = stop codon (no amino acid added)
The peptide is Met to Pro to Ser, three amino acids long.
Translation machinery.
- mRNA. Carries the codon sequence from the nucleus to the ribosome.
- Ribosome. Made of rRNA and protein. Has a small subunit that reads the mRNA and a large subunit with the A, P and E binding sites for tRNAs and the catalytic peptidyl transferase activity.
- tRNA. Cloverleaf-shaped molecules with an anticodon at one end (which base pairs with the mRNA codon) and the corresponding amino acid attached at the 3 prime end.
- Aminoacyl tRNA synthetases. Load each tRNA with the correct amino acid using ATP.
- Initiation, elongation, termination factors. Protein factors that drive the cycle (not always required in detail but worth naming).
Markers reward the correct mRNA sequence including U for T, the correct codon table reads, identification of start and stop codons, and at least three named machinery components.
2022 QCAA4 marksCompare and contrast gene expression in prokaryotes and eukaryotes, referring to the location of transcription and translation, the structure of the gene, and post-transcriptional processing.Show worked answer →
A 4-mark answer needs at least three contrasted features.
Location. Prokaryotes have no nucleus, so transcription and translation occur in the cytoplasm and can happen simultaneously (coupled transcription to translation). Eukaryotes transcribe in the nucleus and translate in the cytoplasm, so the two processes are separated in space and time.
Gene structure. Prokaryotic genes are usually a single continuous coding sequence, often grouped into operons (clusters of genes transcribed from one promoter, for example the lac operon). Eukaryotic genes are split into exons (coding) and introns (non-coding) and are transcribed individually.
Post-transcriptional processing. Prokaryotic mRNA is generally used directly. Eukaryotic pre-mRNA is processed: a 5 prime guanine cap is added, a 3 prime poly-A tail is added, and introns are removed by the spliceosome so that only exons remain. Alternative splicing can produce more than one protein from a single gene.
Result. Eukaryotic gene expression offers more layers of regulation and protein diversity at the cost of speed. Prokaryotic expression is faster and more directly tied to environmental signals.
Markers reward at least three named differences and a stated consequence.
Related dot points
- Describe the structure of DNA, the process of semi-conservative replication and the role of key enzymes including helicase, DNA polymerase, primase and ligase
A focused answer to the QCE Biology Unit 4 dot point on DNA. Walks through the double-helix structure (sugar, phosphate, four bases, complementary pairing, antiparallel strands), the semi-conservative model demonstrated by Meselson and Stahl, and the roles of helicase, primase, DNA polymerase III, DNA polymerase I and ligase on the leading and lagging strands.
- Describe types of mutation (point, frameshift, chromosomal) and the sources of genetic variation including meiosis, fertilisation and mutation, and explain the consequences of mutations for phenotype and population polymorphism
A focused answer to the QCE Biology Unit 4 dot point on mutations and variation. Covers point mutations (silent, missense, nonsense), frameshift indels, chromosomal mutations (deletion, duplication, inversion, translocation, non-disjunction) and the three sources of variation (independent assortment, crossing over, random fertilisation) plus mutation as the ultimate source.
- Describe key biotechnology techniques including PCR, gel electrophoresis, recombinant DNA technology, transgenic organisms (GMOs) and CRISPR-Cas9, and evaluate their applications
A focused answer to the QCE Biology Unit 4 dot point on biotechnology. Covers PCR (denaturation, annealing, extension, Taq, primers), gel electrophoresis (charge, size, ladder), recombinant DNA (restriction enzymes, plasmids, ligase, transformation), transgenic organisms (Bt cotton, golden rice, recombinant insulin) and CRISPR-Cas9 (guide RNA, PAM, repair pathways).