D1.2 Protein synthesis

Protein synthesis links DNA information to functional proteins through transcription, RNA processing, translation, genetic-code reading and post-translational modification in cells.

Syllabus
First assessment 2025
Topic
D1.2
Level
HL

Learning objectives

D1.2.1Transcription• Transcription makes mRNA as a mobile copy of gene information• RNA polymerase synthesizes RNA complementary to the DNA template strandD1.2.2Hydrogen bonding in transcription• Free RNA nucleotides align by complementary base pairing and hydrogen bonding• DNA adenine pairs with RNA uracil, while cytosine pairs with guanineD1.2.3DNA template stability• DNA template strands are transcribed without altering the base sequence• Sugar-phosphate backbone and base pairing preserve genetic informationD1.2.4Transcription for gene expression• Transcription is the first stage of gene expression• Cells regulate which genes are transcribed according to tissue, stage, and signalsD1.2.5Translation• Translation decodes mRNA at ribosomes to synthesize polypeptides• mRNA codon order determines amino acid sequenceD1.2.6Roles in translation• mRNA provides codons; tRNA carries activated amino acids with anticodons• Ribosomes hold mRNA and tRNAs so peptide bonds can formD1.2.7Complementary base pairing• tRNA anticodons pair with complementary mRNA codons by hydrogen bonding• Specific tRNA-amino acid attachment helps ensure correct amino acid additionD1.2.8Genetic code features• The genetic code is triplet, degenerate, and almost universal• Codons specify amino acids, a start signal, or stop signalsD1.2.9Using genetic code table• Genetic code tables use mRNA codons, not DNA triplets• Convert template DNA to mRNA first, then read codons 5' to 3'D1.2.10Elongation of polypeptide• Ribosomes move along mRNA one codon at a time from start to stop• Peptide bonds join amino acids; multiple ribosomes can form a polysomeD1.2.11Mutations changing protein structure• Mutations can change codons and therefore amino acid sequence• Changed primary structure may alter folding and function, such as sickle-cell haemoglobinD1.2.12(HL)—Directionality• RNA polymerase reads template DNA 3' to 5' and synthesizes RNA 5' to 3'• Ribosomes translate mRNA codons in the 5' to 3' directionD1.2.13(HL)—Initiation of transcription at promoter• Promoters mark transcription start regions and orientation• Transcription factors help RNA polymerase bind and initiate in eukaryotesD1.2.14(HL)—Non-coding sequences in DNA• Non-coding DNA does not code for polypeptide amino acid sequences• Includes introns, regulatory sequences, telomeres, rRNA genes, and tRNA genesD1.2.15(HL)—Post-transcriptional modification• Eukaryotic pre-mRNA is modified before export and translation• Processing adds a 5' cap and poly-A tail and removes introns by splicingD1.2.16(HL)—Alternative splicing• Alternative splicing joins different exon combinations from one pre-mRNA• One gene can produce multiple protein variants in different cells or stagesD1.2.17(HL)—Translation initiation• Translation initiation assembles ribosomal subunits at the start codon AUG• Initiator tRNA enters the P site; A, P, and E sites organize tRNA movementD1.2.18(HL)—Polypeptide modification• Newly made polypeptides may be folded, cleaved, or chemically modified• Preproinsulin processing to active insulin is a key exampleD1.2.19(HL)—Amino acid recycling by proteasomes• Proteasomes degrade tagged, damaged, or unneeded proteins• Amino acid recycling supports new protein synthesis and proteome quality control

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.

RNA polymerase opens a short DNA region, reads the template strand and synthesizes an mRNA strand that leaves the transcription bubble.
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.

DNA template bases pair with RNA nucleotides according to A–U and C–G complementarity.

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.

DNA opens locally for mRNA synthesis, the mRNA leaves and the original DNA double helix reforms.

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.

A DNA template is transcribed into mRNA codons read 5′ to 3′, and a genetic-code table maps those codons to amino acids.
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.

A ribosome positions mRNA codons and complementary tRNA anticodons in E, P and A sites while peptide bonds extend a polypeptide.
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.

Several ribosomes move in the same 5′ to 3′ direction along one mRNA, each producing a polypeptide.

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

DNA template → complementary mRNA → codons read 5′→3′ → charged tRNAs selected → peptide bonds join amino acids → polypeptide primary structure

Checkpoint Question to ask
transcription 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.

Information Is Read in Opposite Strand Directions but Built 5′→3′

HL only
Process Molecule being read Reading direction Product growth or movement
transcription DNA template 3′→5′ RNA is synthesized 5′→3′
translation mRNA 5′→3′ ribosome moves toward the mRNA 3′ end; polypeptide grows N-terminus→C-terminus

Nucleotides are added to a free 3′ hydroxyl group, so every new nucleic-acid strand grows 5′→3′. The template must therefore be read antiparallel to the new strand.

A diagram arrow showing polymerase movement and an arrow showing new-strand growth refer to opposite strands; identify which molecule each arrow belongs to.

RNA polymerase reads a DNA template 3′ to 5′ while synthesizing RNA 5′ to 3′; a ribosome then moves along mRNA 5′ to 3′.

Promoters and Transcription Factors Assemble the Start Complex

HL only

A promoter is a DNA region that positions the transcription machinery near a gene's transcription start site. General transcription factors bind first and help recruit RNA polymerase.

Regulatory input Binding relationship Typical effect on initiation
activator transcription factor binds a regulatory DNA sequence such as an enhancer stabilizes or recruits the initiation machinery
repressor transcription factor binds a regulatory sequence or interferes with the complex reduces assembly or activity of the initiation machinery

Regulatory DNA may be far from the promoter in the linear sequence. DNA looping can bring bound factors into contact with promoter-associated proteins.

The combined transcription-factor state changes how often RNA polymerase initiates, altering the amount of pre-mRNA produced from the gene.

Non-Coding DNA Includes Several Functional Categories

HL only

Non-coding DNA is DNA that is not translated into a polypeptide sequence. Non-coding does not mean untranscribed, unused or without function.

Important non-coding categories include:

  • Regulatory sequences such as promoters and enhancers that affect transcription.
  • Introns that are transcribed into pre-mRNA and removed during splicing.
  • Genes for functional RNAs, including rRNA and tRNA, whose products are not translated.
  • Structural regions such as telomeres and centromeric DNA.
  • Repeated sequences, some functional and some with no known function.

An exon can include untranslated regions, so exon is not a synonym for protein-coding sequence. Likewise, an intron can influence regulation or splicing even though it is removed from mature mRNA.

Capping, Polyadenylation and Splicing Produce Mature mRNA

HL only
Processing change What happens Main consequence
5′ cap a modified nucleotide is added to the 5′ end protects RNA and supports export and translation initiation
poly-A tail many adenine nucleotides are added to the 3′ end increases stability and supports export and translation
splicing introns are removed and exons joined creates a continuous mature RNA sequence for translation

pre-mRNA in nucleus → processing completed → mature mRNA exported through a nuclear pore → translation in cytoplasm

The poly-A tail is added after transcription; it is not copied from a long template-DNA sequence of thymine bases.

A eukaryotic pre-mRNA receives a 5′ cap and poly-A tail, has introns removed and exons joined, then is exported from the nucleus as mature mRNA.

Alternative Splicing Changes Which Exons Reach the Ribosome

HL only

In alternative splicing, the same pre-mRNA is spliced in different patterns, so different exon combinations are retained in mature mRNA.

Different mature mRNAs contain different codon sequences and can therefore produce related polypeptides with different amino-acid segments, properties or tissue distributions.

Alternative splicing does not change the DNA sequence. It changes which parts of one RNA transcript remain in the message sent to the ribosome.

One troponin T pre-mRNA is spliced into two mature mRNAs containing different combinations of exons.

Translation Initiation Places the Start Codon in the P Site

HL only
1

The small ribosomal subunit binds the mRNA and positions the start region for scanning or recognition.

A translation initiation complex has AUG paired with an initiator methionine tRNA in the ribosomal P site, with A and E sites labelled.
2

The initiator tRNA carrying methionine pairs with AUG. This tRNA is placed directly in the P site, establishing the reading frame.

3

The large subunit joins to complete the ribosome. The A site is now open for the tRNA matching the next codon.

AUG inside an already established reading frame can code for methionine without serving as a new start. Initiation depends on context and assembly, not the triplet alone.

A, P and E Sites Track Each tRNA's State

HL only
Ribosomal site Typical tRNA state during elongation
A (aminoacyl) incoming charged tRNA paired with the next codon
P (peptidyl) tRNA carrying the growing polypeptide before peptide transfer
E (exit) uncharged tRNA leaving after translocation

Before peptide transfer: growing chain in P, new amino acid in A.

After peptide transfer and translocation: the chain-bearing tRNA occupies P, the empty tRNA moves to E, and A opens for the next charged tRNA.

The site letters describe functional positions in the ribosome, not permanent labels on a tRNA. A single tRNA can move A → P → E during one cycle.

A New Polypeptide May Need Folding, Cleavage or Chemical Modification

HL only

Translation produces a polypeptide primary structure. Many proteins become functional only after post-translational processing changes their shape, length, location or chemical properties.

Common processing routes include:

  • Folding into a specific three-dimensional conformation, sometimes assisted by chaperones.
  • Cleavage to remove a signal peptide or inactive segment.
  • Covalent modification, such as phosphorylation or glycosylation, which can alter activity, stability or location.
  • Assembly with other polypeptide subunits or prosthetic groups.

Not every protein undergoes every modification. The required pathway depends on the protein's sequence, destination and function.

Insulin Becomes Active through Ordered Cleavage and Folding

HL only
1

Preproinsulin contains an N-terminal signal peptide that directs the growing polypeptide into the endoplasmic reticulum. Removing the signal peptide produces proinsulin.

Preproinsulin loses its signal peptide, folds as proinsulin with disulfide bonds and loses the C-peptide to form mature insulin A and B chains.
2

Proinsulin folds, and disulfide bonds form between regions that will become the A and B chains.

3

The connecting C-peptide is removed. The A and B chains remain linked by disulfide bonds, producing mature insulin.

Ubiquitin Directs Selected Proteins to the Proteasome

HL only
1

Proteins selected for degradation are commonly marked with a chain of ubiquitin molecules.

A ubiquitin-tagged protein is recognized, unfolded and threaded into a proteasome, which releases short peptides while ubiquitin is recycled.
2

The proteasome recognizes the tag, removes and recycles ubiquitin, unfolds the target protein and threads it into a proteolytic core.

3

The protein is cut into short peptides, which can be hydrolysed to amino acids for new synthesis or metabolism.

Selective degradation removes damaged or misfolded proteins and can rapidly lower the abundance of regulatory proteins. Protein concentration therefore depends on both synthesis and breakdown.

Control Protein Output at Several Checkpoints

HL only
Checkpoint What can change Consequence for protein output
transcription initiation promoter and transcription-factor activity amount of pre-mRNA made
RNA processing cap, tail and exon combination mRNA stability, export and encoded sequence
translation initiation successful AUG recognition and ribosome assembly whether and where translation begins
post-translational processing folding, cleavage, modification and assembly protein activity or location
proteasomal degradation ubiquitin tagging and breakdown rate protein lifetime and abundance

Across these checkpoints, sequence direction stays consistent: RNA is made 5′→3′, the ribosome reads mRNA 5′→3′ and the polypeptide grows from N-terminus to C-terminus.

Gene expression is therefore not a single on/off event. A cell controls which message is made, how it is processed, whether it is translated, how the product is activated and how long the protein remains.

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′5^{\prime} \text { CATG } 3^{\prime}

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′3^{\prime}

\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.

Directionality exam focus

HL only

1 mark

Using the diagram, identify, with a reason, whether X or Y is the start codon.

Initiation of transcription at promoter

HL only

2 marks

Explain the function of a promoter in DNA.

Non-coding sequences in DNA

HL only

2 marks

DNA has regions that do not code for proteins. State two functions of these regions.
1.
2.

Post-transcriptional modification

HL only

1 mark

What happens to mRNA after transcription in eukaryotic cells?

Alternative splicing

HL only

1 mark

The number of protein-coding genes in the human genome is estimated to be about 20000 , which is much less than the size of the proteome. What is one reason for this?

Translation initiation

HL only

4 marks

Outline the roles of the different binding sites for tRNA on ribosomes during translation.

Polypeptide modification

HL only

2 marks

Insulin is produced by cutting C -peptide from the precursor molecule proinsulin. Suggest why group 1 has a greater level of C-peptide than group 2.