Protein synthesis links DNA information to functional proteins through transcription, RNA processing, translation, genetic-code reading and post-translational modification in cells.
mRNA Carries a Working Copy of Selected DNA Information
Chromosomal DNA is the long-term information store. Protein synthesis instead uses a short-lived messenger RNA (mRNA) copy of a selected gene.
selected gene in DNA → complementary mRNA copy → mRNA reaches a ribosome → nucleotide sequence directs amino-acid order
A working copy solves two problems:
The original DNA sequence remains protected and reusable.
Many mRNA copies can be made, so one gene can support repeated protein production.
mRNA carries information; it is not converted into protein matter. The ribosome reads its sequence while amino acids are joined into a separate molecule.
RNA Polymerase Builds mRNA from One DNA Template Strand
1
RNA polymerase binds near the gene and opens a short region of the DNA double helix, exposing the template strand.
2
Free RNA nucleotides align by complementary base pairing: DNA A pairs with RNA U, DNA T with RNA A, and C with G.
3
RNA polymerase joins the RNA nucleotides into a sugar–phosphate backbone. The mRNA separates, and the DNA strands pair again.
Template DNA Determines the mRNA Sequence
Strand
Relationship to the new mRNA
DNA template
antiparallel and complementary
DNA coding strand
same base order as mRNA, except DNA has T where RNA has U
DNA template: 3′–TAC CTT GCG–5′
mRNA: 5′–AUG GAA CGC–3′
Codon tables are read from mRNA 5′→3′. Do not look up a DNA template triplet directly.
Transcription Opens DNA Locally and Leaves Its Sequence Unchanged
Hydrogen bonds between complementary bases can break and re-form. This lets a short DNA region open temporarily without breaking the covalent sugar–phosphate backbones.
After the mRNA leaves, the original complementary DNA strands pair again. The gene's base sequence is conserved and can be transcribed repeatedly.
Transcription copies one gene region into RNA; it does not duplicate the whole DNA molecule and does not consume the template strand.
Selective Transcription Controls Which Proteins a Cell Can Make
Gene expression is the use of gene information to produce a functional RNA or polypeptide. Transcription is a major control point because an untranslated gene cannot supply mRNA to ribosomes.
Transcription state
Immediate consequence
Possible protein outcome
gene active
mRNA is produced
translation can occur
gene inactive
little or no mRNA is produced
little or no corresponding polypeptide is made
Most cells in one organism contain the same genome, but different cell types transcribe different gene sets. Their different protein mixtures produce different structures and functions.
Turning transcription on permits expression; it does not guarantee a fixed protein amount. RNA processing, translation and protein breakdown can also be regulated.
Reconstruct the Route from a Selected Gene to mRNA
gene selected → DNA opens locally → RNA polymerase reads one template strand → complementary RNA nucleotides are joined → mRNA leaves → DNA re-forms unchanged
Question
Correct check
Which DNA strand determines mRNA?
the template strand
Which base replaces thymine in RNA?
uracil
Which direction is mRNA written?
5′→3′
Why can the gene be reused?
DNA sequence and covalent backbones remain intact
The mRNA now carries codons in an order that a ribosome can translate into an amino-acid sequence.
Translation Converts Codon Order into Amino-Acid Order
Translation is the synthesis of a polypeptide whose amino-acid sequence is determined by the codon sequence of an mRNA.
mRNA codons read 5′→3′ → matching tRNAs deliver amino acids → ribosome forms peptide bonds → polypeptide grows from its amino end toward its carboxyl end
The immediate product is a polypeptide: a specific linear amino-acid sequence. Folding and later processing are needed before many polypeptides become functional proteins.
The genetic code determines amino-acid order, not the final three-dimensional shape directly. Shape emerges from interactions within the amino-acid sequence and its environment.
mRNA, tRNA and the Ribosome Divide Translation Work
Component
Information or material carried
Translation job
mRNA
ordered codons
supplies the sequence to be read
tRNA
one anticodon and its attached amino acid
matches a codon and delivers the corresponding amino acid
small ribosomal subunit
mRNA-binding site
positions the message for decoding
large ribosomal subunit
A, P and E sites
positions tRNAs and catalyses peptide-bond formation
Translation is accurate only when all three information links agree: the mRNA codon, the tRNA anticodon and the amino acid attached to that tRNA.
The ribosome does not choose amino acids by recognizing their chemical identity. It accepts charged tRNAs whose anticodons pair with the exposed codons.
Two Recognition Events Protect Amino-Acid Identity
At the ribosome, a tRNA anticodon pairs antiparallel with a complementary mRNA codon. This selects which tRNA can occupy the decoding site.
Before translation, a specific aminoacyl-tRNA synthetase attaches the correct amino acid to its matching tRNA. This charging step links anticodon identity to amino-acid identity.
Recognition
Example
What it secures
codon–anticodon pairing
mRNA 5′–AUG–3′ pairs with tRNA 3′–UAC–5′
the correct tRNA is selected
enzyme–tRNA/amino-acid recognition
methionine is attached to the initiator tRNA
the selected tRNA carries the correct amino acid
Complementary base pairing alone cannot check which amino acid is attached. Translation fidelity depends on both correct pairing and correct tRNA charging.
The Genetic Code Is Triplet, Degenerate and Nearly Universal
The genetic code has several linked properties:
Triplet: three mRNA bases form one codon.
Degenerate: most amino acids are specified by more than one codon.
Unambiguous: each codon specifies only one amino acid or stop signal.
Nearly universal: the same codons usually have the same meanings across organisms, with limited exceptions.
Punctuated: a start codon establishes the reading frame; stop codons end translation.
Degeneracy means some base substitutions are silent, but it does not mean one codon can represent several different amino acids.
Codon type
Function
AUG
usually codes for methionine and can establish the start of translation
UAA, UAG, UGA
stop signals recognized by release factors; they do not add an amino acid
Read the mRNA 5′→3′ before Using a Codon Table
1
If a DNA template is supplied, write the antiparallel complementary mRNA, replacing DNA A with RNA U.
2
Write the mRNA 5′→3′, locate the stated or biologically relevant start, and divide the sequence into consecutive triplets from that frame.
3
Look up each mRNA codon, record amino acids in order and stop when a stop codon is reached.
4
Template 3′–TAC CTT GCG ACT–5′ gives mRNA 5′–AUG GAA CGC UGA–3′, which translates as Met–Glu–Arg–Stop.
Elongation Repeats Until a Stop Codon Ends Translation
1
A charged tRNA with a complementary anticodon enters the A site beside the tRNA carrying the growing chain in the P site.
2
The ribosome catalyses a peptide bond, transferring the growing polypeptide to the amino acid on the A-site tRNA.
3
The ribosome moves one codon along the mRNA. The peptidyl-tRNA shifts to P, the empty tRNA shifts to E and exits, and the A site becomes available again.
4
The entry–bond–movement cycle repeats for successive codons. When a stop codon reaches the A site, no tRNA enters; a release factor releases the completed polypeptide and the ribosome dissociates.
A Polysome Makes Several Copies from One mRNA at Once
A polysome is one mRNA being translated simultaneously by several ribosomes. Each ribosome starts near the 5′ end and moves toward the 3′ end.
All ribosomes read the same codon sequence, so they produce polypeptides with the same amino-acid order. Simultaneous translation increases the rate of protein production from that mRNA.
The ribosomes do not cooperate to make one long polypeptide. Each ribosome makes a separate copy.
A DNA Mutation Does Not Always Change the Protein
A mutation can alter a DNA base sequence. If the changed region is transcribed, it can alter an mRNA codon and therefore the amino-acid sequence or the timing of translation.
Codon outcome
Possible protein consequence
codon still specifies the same amino acid
no change to primary structure
codon specifies a different amino acid
altered side-chain interactions, folding or function
codon becomes a stop signal
shortened polypeptide
insertion or deletion shifts the reading frame
many downstream codons may change
Effect size depends on where the mutation occurs and what the affected amino acid does. A substitution at an active site or interaction surface can matter more than one in a tolerant region.
Mutation creates a possibility of changed protein structure; it does not guarantee harm. Some mutations are silent, neutral or beneficial in a particular environment.
One β-Globin Substitution Can Reshape a Red Blood Cell
1
In the sickle-cell allele, a β-globin DNA substitution changes an mRNA codon from GAG to GUG, replacing glutamate with valine in the polypeptide.
2
Glutamate is charged and hydrophilic; valine is non-polar. The replacement creates a hydrophobic patch on deoxygenated haemoglobin S.
3
Under low oxygen conditions, haemoglobin S molecules associate into long fibres. The fibres distort the red blood cell into a rigid sickle shape.
4
Rigid cells can block small blood vessels and are removed rapidly from circulation, reducing oxygen delivery and causing anaemia and painful crises.
The mutation does not directly bend the whole cell. It first changes one amino acid, which changes intermolecular interactions, which then changes cell shape under particular oxygen conditions.
Trace Information—and the Places Where It Can Change
Is the mRNA complementary to the template and written 5′→3′?
code table
Are mRNA codons grouped from the correct start?
tRNA selection
Does the anticodon pair and is the tRNA correctly charged?
elongation
Has the ribosome advanced one codon while the chain remains on the peptidyl-tRNA?
mutation
Did the codon, amino acid or reading frame actually change?
A sequence change matters biologically only through a causal path: altered primary structure → altered interactions or folding → altered protein activity → possible cellular or organism effect.
Transcription exam focus
8 marks
Explain the process of transcription in prokaryotes.
Hydrogen bonding in transcription
1 mark
The sequence of bases on a short section of the antisense strand of a gene undergoing transcription is shown:
5′ CATG 3′
What is the sequence of bases on the resulting mRNA?
Transcription for gene expression
2 marks
The scientists concluded that auxin activates the transcription of the GH 3 gene. Using the information on the auxin concentration in the stem base in the graph on page 4 and the Northern blot, evaluate whether this conclusion is supported.
Translation exam focus
8 marks
Explain how polypeptides are produced by the process of translation.
Roles in translation
7 marks
Explain the role of RNA in translation, resulting in the formation of polypeptide chains.
Complementary base pairing
3 marks
Outline how translation depends on complementary base pairing.
Genetic code features
5 marks
Describe the genetic code and its relationship to polypeptides and proteins.
Using genetic code table
1 mark
Which sequence of mRNA bases and amino acids could be produced by transcription and translation of the DNA molecule shown?
3' AAAGTGGCACGTATATTT 5' 5' TTTCACCGTGCATATAAA 3′
\begin{tabular}{|l|l|l|l|l|l|l|l|} \hline \multirow{6}{*}{} & \multicolumn{6}{|c|}{2nd base in codon} & \multirow{14}{*}{3rd base in codon} \\ \hline & & U & C & A & G & & \\ \hline & \multirow{4}{*}{U} & Phe & Ser & Tyr & Cys & U & \\ \hline & & Phe & Ser & Tyr & Cys & C & \\ \hline & & Leu & Ser & STOP & STOP & A & \\ \hline & & Leu & Ser & STOP & Trp & G & \\ \hline \multirow[t]{8}{*}{} & C & Leu & Pro & His & Arg & U & \\ \hline & \multirow{3}{*}{A} & lle & Thr & Asn & Ser & U & \\ \hline & & lle & Thr & Lys & Arg & A & \\ \hline & & Met & Thr & Lys & Arg & G & \\ \hline & \multirow{4}{*}{G} & Val & Ala & Asp & Gly & U & \\ \hline & & Val & Ala & Asp & Gly & C & \\ \hline & & Val & Ala & Glu & Gly & A & \\ \hline & & Val & Ala & Glu & Gly & G & \\ \hline \end{tabular}
Sequence of mRNA bases
Sequence of amino acids
UUU-GAG-GCU-CGA-UAU-UUU
Phe-Glu-Ala-Arg-Tyr-Phe
AAA-CUC-CGA-GCU-AUA-UUU
Lys-Leu-Arg-Ala-lle-Phe
UUU-CAC-CGU-GCA-UAU-AAA
Phe-His-Arg-Ala-Tyr-Lys
AAA-GUG-GCA-CGU-AUA-UUU
Lys-Val-Ser-Arg-Ile-Phe
Elongation of polypeptide
2 marks
Six polypeptides are shown in the diagram. Explain the different lengths of these polypeptides.
Mutations changing protein structure
8 marks
Explain the cause of sickle cell anemia and how this disease affects humans.