D1.2 Protein synthesis · Educerie Biology

6 higher-level sections hidden.

Educerie · IB Diploma · Biology

Theme D Continuity and change · D1.2 Protein synthesis

Level: SL and HL. Sections 10 to 15 are HL only. If you are SL, skip them; nothing in your papers tests them.
Themes (key concepts): continuity and change, at the level of molecules. The genetic code is the great continuity of life, the same codons read the same way in bacteria, plants and people; a change of one base in that code is enough to change a protein, and with it an organism.
The question this unit answers: how does a cell turn a sequence of DNA bases into a sequence of amino acids, and how is that done reliably?
Where it is examined: Paper 1A multiple choice, often a decoding item with a codon table; Paper 1B data on sequences or mutations; Paper 2 Section A, for "deduce the amino acid sequence" (2–3 marks) or "outline the role of tRNA" (2 marks); Paper 2 Section B, where "explain the process of transcription" or "explain how translation…" is a regular 7 or 8 mark part.

What you must be able to do

You must be able to	Level	What it looks like in the exam
Describe transcription and the roles of RNA polymerase	SL, HL	"Outline the process of transcription" (4 marks)
Explain the role of hydrogen bonding and complementary pairing in transcription, including A with U	SL, HL	Paper 1A; "state the mRNA sequence" (1 mark)
Explain why DNA templates stay unchanged, and why transcription is where genes are switched on and off	SL, HL	Short answer, 2 marks
Describe translation and the roles of mRNA, ribosomes and tRNA, codons and anticodons	SL, HL	"Explain the roles of mRNA and tRNA in translation" (4 marks)
Explain why the code is a triplet code, and what degeneracy and universality mean	SL, HL	"Explain why the genetic code is a triplet code" (2 marks)
Deduce an amino acid sequence from mRNA using a codon table	SL, HL	Paper 1A or 2, 2–3 marks
Describe elongation: the ribosome moving codon by codon and peptide bonds forming	SL, HL	Part of a 4–8 mark answer
Explain how a point mutation, such as the sickle-cell mutation, changes protein structure	SL, HL	"Explain how a base substitution can change a protein" (3 marks)
Explain 5′ to 3′ transcription and translation, and initiation of transcription at the promoter	HL only	Short answer, 2–3 marks
Outline non-coding DNA sequences, post-transcriptional modification and alternative splicing	HL only	"Outline the modifications to eukaryotic mRNA" (3–4 marks)
Describe initiation of translation and the roles of the A, P and E sites	HL only	Paper 2 Section B, 4–6 marks
Outline modification of pre-proinsulin to insulin, and the recycling of amino acids by proteasomes	HL only	Short answer, 2–3 marks

Before you start

You need A1.2: DNA and RNA as polymers of nucleotides, the bases A, C, G and T in DNA and A, C, G and U in RNA, and complementary pairing held by hydrogen bonds. From B1.2 you need proteins as chains of amino acids joined by peptide bonds, folded into a shape that depends on the sequence. D1.1 shows how a DNA strand acts as a template; transcription uses the same idea.

1The idea in one paragraph

A gene is a sequence of DNA bases that codes for a polypeptide. The cell cannot use the DNA directly, so it copies the gene into messenger RNA (mRNA): this is transcription, done by RNA polymerase in the nucleus, with RNA nucleotides pairing to one DNA strand by complementary base pairing. The mRNA travels to a ribosome, where it is read three bases at a time. Each three-base codon is matched, again by base pairing, to the anticodon of a transfer RNA (tRNA) that carries one particular amino acid, and the ribosome joins the amino acids with peptide bonds in the order the codons dictate: this is translation. The rules linking codons to amino acids are the genetic code, shared by almost all life. Change one base and you may change one amino acid, which is all it takes to cause sickle-cell disease.

2The overview: two stages, two places

Protein synthesis happens in two stages, shown in Figure 1 for a eukaryotic cell.

Figure 1 · From gene to polypeptide in a eukaryotic cell

	Transcription	Translation
What is made	RNA, copied from a DNA template	a polypeptide, built from the mRNA's instructions
Where (eukaryote)	nucleus	cytoplasm, at ribosomes
Main machinery	RNA polymerase	ribosome, tRNA
What pairs with what	RNA nucleotides with the DNA template strand	tRNA anticodons with mRNA codons

In a prokaryote there is no nucleus, so both stages happen in the cytoplasm.

3Transcription: RNA made on a DNA template

Transcription is the synthesis of RNA using a DNA template. Figure 2 shows it under way.

Figure 2 · Transcription: RNA polymerase builds mRNA on the template strand

RNA polymerase does all the work, and it has three roles.

It binds to the DNA at the start of a gene and separates the two strands by breaking the hydrogen bonds between their bases, over a short stretch at a time.
It pairs free RNA nucleotides with the exposed bases of one strand, the template strand, by complementary base pairing.
It links the RNA nucleotides together with covalent bonds to form a strand of RNA.

As the polymerase moves on, the RNA peels away from the template and the two DNA strands re-form their hydrogen bonds behind it. At the end of the gene the RNA is released.

Hydrogen bonding and complementary pairing. Each RNA nucleotide is held against its template base by hydrogen bonds, and only the complementary base forms them properly, so the RNA sequence is fixed by the template. The pairing rules are those of DNA with one change: RNA has uracil instead of thymine.

In transcription, A on the DNA template pairs with U in the RNA. T pairs with A, C with G, G with C.

Only one DNA strand is transcribed for a given gene. The other, the coding strand, has the same sequence as the mRNA except that it has T where the mRNA has U. In Figure 2 the template reads TACGGATTCACG, so the mRNA reads AUGCCUAAGUGC, and the coding strand reads ATGCCTAAGTGC.

4Stable templates and switching genes on and off

Why the template does not change. Transcription only reads the template. The hydrogen bonds between the DNA strands are broken briefly and then re-formed; the covalent backbone and the base sequence are untouched. So a single DNA strand can be used as a template again and again, for thousands of mRNA copies, without its sequence changing. That matters most in cells that never divide again, such as nerve cells and heart muscle cells. They cannot replace their DNA by replication, so the same DNA must serve as a template, unchanged, for the whole life of the cell, which in a human neuron can be many decades.

Transcription and gene expression. A gene is expressed when its information is used to make its product. Not every gene in a cell is expressed at any one time: a liver cell and a nerve cell have the same genes but use different selections of them, and a single cell changes which genes it uses as conditions change. Transcription is the first stage of gene expression, and it is a key point of control: if a gene is not transcribed, no mRNA is made and no protein can follow. Cells switch genes on and off largely by controlling whether they are transcribed.

5Translation: the roles of mRNA, ribosomes and tRNA

Translation is the synthesis of a polypeptide from the information in mRNA: the base sequence of the mRNA is translated into the amino acid sequence of a polypeptide. Three players take part, as Figure 3 shows.

Figure 3 · Translation at the ribosome

mRNA carries the information, as a sequence of codons. A codon is a group of three bases on the mRNA that codes for one amino acid (or for "stop").
Ribosomes are the site of translation. Each has two subunits. The mRNA binds to the small subunit. Two tRNA molecules can bind to the large subunit at the same time, side by side, and the ribosome catalyses the formation of a peptide bond between their amino acids.
tRNA molecules bring the amino acids. Each tRNA carries one specific amino acid, and has a group of three bases called the anticodon that is complementary to one codon.

Codon meets anticodon. A tRNA can only bind at the ribosome if its anticodon pairs with the codon exposed there, by complementary base pairing held by hydrogen bonds. In Figure 3 the codon CAU is matched by the anticodon GUA, and that tRNA carries histidine; the next codon, GCA, is matched by anticodon CGU, carrying alanine. Because each tRNA carries only the amino acid that matches its anticodon, the order of codons fixes the order of amino acids.

6The genetic code

Why a triplet code? Twenty different amino acids are used to build proteins, and there are only four bases. Count how many different "words" each code length allows.

1 base per codon: 4¹ = 4 combinations → not enough for 20 amino acids

2 bases per codon: 4² = 16 combinations → still not enough

3 bases per codon: 4³ = 64 combinations → enough, with some to spare

Three is the smallest codon length that can give every amino acid a codon of its own, so the code is a triplet code.

Degeneracy. There are 64 codons and only 20 amino acids (plus stop signals), so most amino acids are coded for by more than one codon. The code is therefore described as degenerate. Glycine, for example, has four codons: GGU, GGC, GGA and GGG. Leucine has six. Only methionine and tryptophan have a single codon each.

Universality. With minor exceptions, the same codons code for the same amino acids in all organisms, from bacteria to humans. The code is universal. That is why a human gene placed in a bacterium can be translated into the human protein, and it is strong evidence that all life shares a common ancestor.

Three codons, UAA, UAG and UGA, do not code for any amino acid: they are stop codons, which end translation. AUG codes for methionine and is also the start codon, where translation begins.

7Using the table of mRNA codons

Figure 4 is the genetic code as the exam presents it: a table of mRNA codons. You do not memorise it; you must be able to use it.

Figure 4 · The genetic code as a table of mRNA codons

To look up a codon, find its first base in the left-hand column, its second base along the top, and its third base down the right-hand side of that box. So GCA is row G, column C, line A: alanine.

Worked example. A section of a DNA template strand reads, from its 3′ end: TAC AAG GTA CGT ATT. Deduce the amino acid sequence.

DNA template (3′ → 5′): TAC AAG GTA CGT ATT

mRNA (5′ → 3′): AUG UUC CAU GCA UAApair each base: T->A, A->U, C->G, G->C

codons read from the table: AUG = Met, UUC = Phe, CAU = His, GCA = Ala, UAA = stop

polypeptide: Met – Phe – His – Ala

Two traps. Always convert the template DNA to mRNA before using the table, because the table is written in mRNA codons (with U, never T). And a stop codon is not an amino acid: the chain simply ends there.

8Elongation: the ribosome moves and the chain grows

The guide asks you to focus on how the polypeptide gets longer, a stage called elongation. It is a repeated cycle.

A tRNA whose anticodon matches the next codon binds at the ribosome, next to the tRNA holding the growing chain.
The ribosome catalyses a peptide bond between the amino acid on the new tRNA and the end of the growing chain, so the chain is now attached to the new tRNA.
The ribosome moves along the mRNA by one codon, a step of three bases.
The tRNA that has given up its chain leaves the ribosome, to be reloaded with another molecule of its amino acid in the cytoplasm, and a new codon is exposed for the next tRNA.

The cycle repeats, one amino acid per step, until a stop codon is reached and the finished polypeptide is released. The ribosome moves stepwise along the mRNA; the chain grows by one amino acid at every step.

9A mutation that changes protein structure

Because the base sequence dictates the amino acid sequence, a change to a gene's base sequence, a mutation, can change the protein. A point mutation changes a single base. Figure 5 shows the most famous example.

Figure 5 · A point mutation that changes a protein: sickle-cell haemoglobin

The gene for the β-globin chain of haemoglobin has GAG as the codon for the chain's sixth amino acid, glutamic acid. In the sickle-cell allele, one base is substituted: the A becomes a T in the DNA, so the mRNA codon becomes GUG, which codes for valine. One amino acid out of 146 in the chain is changed.

That is enough to change the protein's behaviour. Glutamic acid is charged and hydrophilic; valine is non-polar and hydrophobic. At low oxygen concentrations the hydrophobic valine on one haemoglobin molecule binds to a hydrophobic region on another, and the molecules stick together into long fibres. The fibres distort red blood cells into rigid sickle shapes, which carry less oxygen and block small blood vessels. D1.3 goes further into mutations; here the point is the chain of cause: base → codon → amino acid → protein shape → function.

10HLDirectionality of transcription and translation

SL students can skip to section 16.

Like DNA polymerase, RNA polymerase can only add nucleotides to the 3′ end of a growing strand, so RNA is always built 5′ to 3′. Because the RNA is antiparallel to its template, RNA polymerase moves along the template strand from its 3′ end towards its 5′ end. That is 5′ to 3′ transcription.

Translation is also 5′ to 3′: the ribosome starts near the 5′ end of the mRNA and moves towards the 3′ end, codon by codon. This fixes the reading frame: the codons are read in consecutive groups of three from the start codon, never from the other end. In Figure 3, reading from the 3′ end would give a completely different, meaningless sequence.

11HLInitiation of transcription at the promoter

Transcription of a gene begins at the promoter, a DNA sequence just before the start of the gene where RNA polymerase binds. The promoter is not transcribed itself; it marks where and on which strand transcription starts.

In eukaryotes, RNA polymerase cannot bind to a promoter on its own. Proteins called transcription factors must first bind to the promoter, and RNA polymerase then binds to them and begins transcription. Which transcription factors are present in a cell therefore decides which genes are transcribed. This is one of the main ways genes are switched on and off (section 4). You do not need to name any transcription factors.

12HLNon-coding sequences in DNA

Most of the DNA in a eukaryotic genome does not code for polypeptides. The guide limits you to four kinds of non-coding sequence.

Regulators of gene expression. Promoters and other sequences that proteins bind to in order to switch transcription on or off. They control genes rather than coding for polypeptides.
Introns. Sequences within a gene that are transcribed but removed from the mRNA before translation (section 13).
Telomeres. Repetitive sequences at the ends of chromosomes. They protect the ends of the chromosome, because a little DNA is lost from the ends each time DNA is replicated, and it is better to lose telomere repeats than genes.
Genes for rRNA and tRNA. These are transcribed, but their RNA is the final product. Ribosomal RNA forms much of the ribosome and tRNA carries amino acids; neither is translated into a polypeptide.

13HLPost-transcriptional modification and alternative splicing

In eukaryotic cells the RNA made by transcription, the pre-mRNA (or primary transcript), is not ready to be translated. It is modified in the nucleus first, as Figure 6 shows.

Figure 6 · Turning pre-mRNA into mature mRNA, and alternative splicing (HL)

Removal of introns and splicing of exons. Eukaryotic genes are interrupted by introns, non-coding sequences, between the exons, the sequences that will be translated. Both are transcribed. The introns are then cut out and the exons are joined together, spliced, to make a continuous coding sequence: the mature mRNA.
A 5′ cap. A modified guanine nucleotide is added to the 5′ end.
A 3′ poly-A tail. A long chain of adenine nucleotides is added to the 3′ end.

The cap and the tail stabilise the mRNA, protecting it from being broken down by enzymes in the cytoplasm, so it lasts long enough to be translated many times. (The cap also helps the ribosome bind, section 14.)

Alternative splicing. The exons of one gene do not have to be spliced together in only one way. By including some exons and leaving others out, the same pre-mRNA can give different mature mRNAs, as panel (c) of Figure 6 shows. Each codes for a different polypeptide, so one gene can code for several variants of a protein. This is one reason humans make far more kinds of protein than they have genes.

14HLInitiation of translation, and the A, P and E sites

Translation starts in a set sequence.

The small ribosomal subunit attaches to the 5′ end of the mRNA (in eukaryotes, at the 5′ cap).
It moves along the mRNA to the start codon, AUG.
An initiator tRNA, with anticodon UAC and carrying methionine, pairs with the start codon.
The large subunit attaches, completing the ribosome. The initiator tRNA sits in the P site, and a second tRNA, matching the next codon, binds in the A site.

The large subunit has three binding sites for tRNA, each with a role in elongation, as Figure 7 shows.

Figure 7 · One step of elongation: the E, P and A sites (HL)

Site	Role during elongation
A site	where a tRNA carrying the next amino acid binds, its anticodon paired with the codon there
P site	holds the tRNA carrying the growing polypeptide chain
E site	where the tRNA that has given up its chain sits before it exits the ribosome

In one step: a tRNA enters the A site; a peptide bond forms and the chain passes from the tRNA in the P site to the amino acid on the tRNA in the A site; the ribosome moves one codon towards the 3′ end, so the tRNA carrying the chain moves into the P site and the empty tRNA into the E site, from which it leaves; the A site is free again for the next tRNA.

15HLMaking proteins functional, and breaking them down

Modification of polypeptides. Many polypeptides cannot work as first made; they must be modified into their functional state. Insulin is the guide's example, and it is modified in two stages, shown in Figure 8.

Figure 8 · Pre-proinsulin is cut twice to make insulin (HL)

Insulin is translated as pre-proinsulin, by ribosomes on the rough endoplasmic reticulum. Its first section is a signal peptide, which directs the chain into the ER. Inside the ER the signal peptide is cut off, leaving proinsulin.
Proinsulin folds and disulfide bonds form between what will become the two chains. In the Golgi apparatus and secretory vesicles, a middle section, the C peptide, is cut out. What remains is active insulin: an A chain of 21 amino acids and a B chain of 30, held together by disulfide bonds.

Recycling amino acids by proteasomes. Proteins do not last for ever. Damaged, misfolded and no-longer-needed proteins are broken down by proteasomes, large barrel-shaped protein complexes that digest proteins into short peptides and amino acids. The amino acids are reused to make new proteins. A cell's proteome, its full set of proteins, is kept functional by this constant balance of breakdown and synthesis: old proteins removed, new ones made to replace them.

16Where marks are lost

Converting the template to mRNA with T. mRNA never contains T. Opposite A on the template, write U.

Using the codon table with DNA. The table is for mRNA codons. Convert the template strand to mRNA first; if you are given the coding strand, just swap T for U.

Mixing up codon and anticodon. The codon is on the mRNA; the anticodon is on the tRNA. "The mRNA anticodon" scores zero.

Saying the ribosome "makes" amino acids, or tRNA "codes for" them. Amino acids are made elsewhere or taken in food; tRNA carries them; the ribosome joins them with peptide bonds.

Explaining the triplet code with "three bases fit best". The reason is arithmetic: 4² = 16 is fewer than 20 amino acids, 4³ = 64 is enough.

Confusing degenerate with universal. Degenerate: several codons for one amino acid. Universal: the same code in almost every organism.

Describing sickle cell as "a change in the protein" with no mechanism. Give the chain: A → T in the DNA, GAG → GUG, glutamic acid → valine, hydrophobic, molecules stick together in fibres.

HL: saying introns are "junk" that is never transcribed. Introns are transcribed and then removed from the pre-mRNA.

17Draw it right

Transcription: two DNA strands with one labelled template; RNA polymerase labelled; an RNA strand paired to the template with correct bases (U opposite A); the RNA's 5′ and 3′ ends labelled (HL).
Translation: a ribosome with small and large subunits; mRNA on the small subunit with codons marked in threes; two tRNAs in the large subunit, each with an anticodon complementary to its codon and an amino acid; a peptide bond between adjacent amino acids.
HL, the three sites: A, P and E labelled in order E–P–A from the 5′ side, the growing chain on the tRNA in the P site, the incoming tRNA in the A site.
Decoding: always show the mRNA sequence in codons before the amino acids, so method marks are available if you misread the table.

18Try it

Marks in brackets. Answers and marker's notes are at the end. Use Figure 4 for codons.

Q1. A section of the template strand of a gene reads, from its 3′ end: TAC CGA CTT TGG ATC. Deduce the mRNA sequence and the amino acid sequence it codes for. 3 marks

Q2. Outline the roles of RNA polymerase in transcription. 3 marks

Q3. Explain the roles of mRNA, tRNA and ribosomes in translation. 4 marks

Q4. Explain why the genetic code is described as a triplet code, and as degenerate. 3 marks

Q5. The codon GAG in the gene for β-globin is changed to GUG in people with the sickle-cell allele. Explain how this changes the structure and function of haemoglobin. 3 marks

Q6 (HL). Outline how pre-mRNA is modified to produce mature mRNA in a eukaryotic cell, and how one gene can code for more than one polypeptide. 4 marks

Q7 (HL). Describe the roles of the A, P and E sites during one step of elongation. 3 marks

19In one breath

Transcription makes RNA on a DNA template: RNA polymerase separates the strands, pairs RNA nucleotides with the template strand by hydrogen bonding and complementary pairing, A with U, and links them. The template is only read, never changed, so it lasts the life of a non-dividing cell, and transcription is where genes are switched on or off. Translation turns the mRNA's base sequence into an amino acid sequence: mRNA binds the ribosome's small subunit, two tRNAs bind the large subunit, each tRNA's anticodon pairs with a codon, and peptide bonds join the amino acids as the ribosome steps along one codon at a time. The code is a triplet because 4³ = 64 is the first power of four above 20; it is degenerate (several codons per amino acid) and universal; AUG starts, UAA, UAG and UGA stop. Decode by converting the template to mRNA, then reading the table. One base substitution, GAG → GUG, swaps glutamic acid for valine and causes sickle-cell disease. HL: RNA is made 5′ to 3′ and mRNA read 5′ to 3′; transcription factors bind the promoter; regulators, introns, telomeres and rRNA/tRNA genes do not code for polypeptides; pre-mRNA loses its introns, is capped and tailed, and can be spliced in alternative ways; translation starts with the small subunit at the 5′ end, the initiator tRNA at AUG, then the large subunit; A site incoming, P site chain, E site exit; pre-proinsulin loses a signal peptide then a C peptide; proteasomes recycle amino acids.

Answers

Q1. mRNA (5′ to 3′): AUG GCU GAA ACC UAG. Amino acids: Met – Ala – Glu – Thr, then stop. 1 for the mRNA with U and no T, 1 for Met–Ala–Glu–Thr, 1 for recognising UAG as stop rather than an amino acid. A correct amino acid sequence read from the DNA with T treated as U can still earn the second mark.

Q2. RNA polymerase binds to the DNA at the start of the gene and separates the two strands by breaking hydrogen bonds between bases. It pairs free RNA nucleotides with the bases of the template strand by complementary base pairing (U opposite A). It links the RNA nucleotides with covalent bonds to form a strand of RNA, moving along the gene. 1 for each role. "It copies DNA into RNA" alone scores 0.

Q3. mRNA carries the base sequence copied from the gene to the ribosome, as a series of codons of three bases, each coding for one amino acid. tRNA molecules each carry a specific amino acid and have an anticodon complementary to a codon, so they bring the correct amino acid to each codon by base pairing. Ribosomes are the site of translation: mRNA binds to the small subunit, two tRNAs bind to the large subunit at once, and the ribosome forms peptide bonds between their amino acids, moving along the mRNA one codon at a time. 1 for mRNA carrying codons, 1 for tRNA carrying a specific amino acid, 1 for anticodon–codon pairing, 1 for the ribosome's role, binding and peptide bonds. Confusing codon and anticodon caps the answer at 2.

Q4. There are 20 amino acids and four bases. Codons of one base give only 4 combinations and of two bases only 16, too few; three bases give 4³ = 64 combinations, enough for all 20, so the code is a triplet code. Because 64 codons code for only 20 amino acids, most amino acids have more than one codon (for example glycine: GGU, GGC, GGA, GGG), so the code is degenerate. 1 for 20 amino acids and four bases, 1 for 4² too few and 4³ enough, 1 for degeneracy explained with more than one codon per amino acid.

Q5. The codon change GAG to GUG replaces glutamic acid with valine at one position in the β-globin chain. Glutamic acid is hydrophilic and charged, whereas valine is hydrophobic, so the haemoglobin molecule has a hydrophobic patch on its surface. At low oxygen, haemoglobin molecules stick together through these patches, forming long fibres that distort red blood cells into a sickle shape, so they carry less oxygen and can block capillaries. 1 for glutamic acid replaced by valine, 1 for the change in polarity or hydrophobicity changing how the molecules interact, 1 for fibres and sickled cells with a consequence for function.

Q6 (HL). Introns, non-coding sequences within the gene, are removed from the pre-mRNA, and the exons are spliced together. A 5′ cap is added to the 5′ end and a poly-A tail to the 3′ end, which stabilise the mRNA. In alternative splicing, different combinations of exons are spliced together from the same pre-mRNA, so one gene produces different mature mRNAs, each translated into a different polypeptide. 1 for removal of introns and splicing of exons, 1 for the 5′ cap and poly-A tail, 1 for their stabilising role, 1 for alternative splicing leading to different polypeptides.

Q7 (HL). A tRNA carrying the next amino acid binds in the A site, its anticodon paired with the codon there. The P site holds the tRNA attached to the growing polypeptide chain; a peptide bond forms and the chain is transferred to the amino acid on the tRNA in the A site. The ribosome then moves one codon along the mRNA, so the tRNA with the chain moves into the P site and the empty tRNA moves into the E site, from which it exits. 1 for each site's role correctly described. Reversing A and P scores 0 for both.

Educerie · written from the published IB Diploma Programme Biology guide, first assessment 2025, section D1.2 Protein synthesis. Original text, examples and questions. Diagrams drawn by Educerie. Last reviewed 25 September 2026.

Mocks: in the future, hold tight!