IB Biology · Theme D Continuity and change · Molecules
D1.2 Protein synthesis
A gene is transcribed into mRNA, which a ribosome translates into a polypeptide. Complementary base pairing carries the sequence at each step: DNA to RNA, codon to anticodon. The triplet code is degenerate and near-universal; a point mutation can change one amino acid.
Compiled from the IB Biology guide (first assessment 2025, updated May 2026 for 2028) and our question bank ·
Specialist review in progress
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Assessed in Paper 1A (multiple choice), Paper 1B (data-based) and Paper 2 (short and extended response). IB Biology guide (first assessment 2025, updated May 2026 for 2028).
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D1.2.1 Transcription: RNA polymerase makes an RNA copy of one gene
Transcription builds RNA using one strand of a gene as a template.
RNA polymerase separates the DNA strands and moves along the template strand.
It links RNA nucleotides with covalent bonds, sugar to phosphate, into one strand.
Only the gene is copied, not the whole molecule; the DNA is left unchanged.
Students often think transcription needs helicase and a primer, as replication does. In fact RNA polymerase opens the DNA and starts the strand by itself.
Students often swap the two terms. In fact transcription makes RNA from DNA; translation makes a polypeptide from mRNA.
D1.2.2 Base pairing sets the RNA sequence; hydrogen bonds let it go
Each RNA nucleotide pairs with the DNA base opposite it on the template strand.
C pairs with G, G with C, T on DNA with A on RNA.
Adenine on the template pairs with uracil on the RNA; RNA has no thymine.
The pairs are held by hydrogen bonds, which break so the RNA can leave.
The finished RNA matches the non-template strand of the gene, with U in place of T.
Students often think the RNA copies the template strand like for like. In fact it is complementary to it, base by base.
Students often think uracil sits opposite thymine. In fact uracil pairs with adenine: wherever the template has A, the RNA has U.
D1.2.3 A DNA template can be read again and again without changing
The template is read by base pairing, not consumed or rewritten.
After the RNA leaves, the two DNA strands re-pair and the gene is as before.
So the same sequence can be transcribed for the whole life of a cell.
A non-dividing somatic cell, such as a neuron, must conserve its sequences for decades.
Students often think a gene wears out with use and needs replacing by replication. In fact the template is unchanged by any number of transcriptions.
Students often think a cell that will never divide no longer needs its DNA intact. In fact it transcribes that DNA for the rest of its life.
D1.2.4 Transcription is where gene expression is switched on or off
Gene expression uses a gene's information to make a product, via transcription and translation.
Not all genes are expressed at once; each cell type transcribes only some.
Transcription is the first stage, so expression is switched on or off there.
A gene that is not transcribed gives no mRNA and no polypeptide.
Students often think every cell makes every protein its DNA codes for. In fact each cell type expresses only some of its genes.
Students often think a liver cell has lost its haemoglobin genes. In fact the genes are present but not transcribed.
D1.2.5 Translation: a ribosome turns mRNA codons into a polypeptide
Translation makes a polypeptide on a ribosome, following the base sequence of mRNA.
The mRNA is read in codons of three bases, each specifying one amino acid.
Amino acids arrive on tRNA and are joined by peptide bonds in codon order.
It happens in the cytoplasm, on free ribosomes or on rough endoplasmic reticulum.
Students often think the ribosome converts bases into amino acids. In fact the amino acids come from the cytoplasm, each carried by a tRNA.
Students often think translation happens in the nucleus, next to the gene. In fact the mRNA leaves the nucleus first; translation is in the cytoplasm.
D1.2.6 What mRNA, the ribosome and tRNA each do
mRNA carries the gene's codons and binds to the small subunit of the ribosome.
The large subunit holds two tRNAs at once, so a peptide bond can form.
Each tRNA carries an anticodon at one end and its amino acid at the other.
After giving up its amino acid, a tRNA leaves and can be reloaded.
Students often think mRNA binds the large subunit because it is the larger molecule. In fact mRNA binds the small subunit; tRNAs bind the large one.
Students often think only one tRNA is on the ribosome at a time. In fact two are bound together, so the chain can pass from one to the other.
D1.2.7 Codon and anticodon pair by complementary bases
A codon is three mRNA bases that specify one amino acid, or start or stop.
An anticodon is three tRNA bases complementary to a codon (A with U, C with G).
Pairing is by hydrogen bonds, so only the matching tRNA can bind.
The anticodon is complementary, not identical: codon AUG pairs with anticodon UAC.
Students often think the anticodon has the same sequence as its codon. In fact each anticodon base is complementary to the codon base opposite it.
Students often think a tRNA binds the codon, then picks up an amino acid. In fact the tRNA is loaded before it arrives; the codon selects which loaded tRNA binds.
D1.2.8 Why the code uses triplets, and what degenerate and universal mean
Two-base codons give only 16 combinations; three-base codons give 64, enough for 20 amino acids.
The code is degenerate: most amino acids have more than one codon.
It is not ambiguous: each codon specifies only one amino acid.
It is universal: the same codons mean the same amino acids in almost all organisms.
Universality is evidence of common ancestry, and it lets a gene from one species be translated in another.
Students often think degenerate means one codon can give different amino acids. In fact it means several codons give the same amino acid; each codon has one meaning.
Students often think universal means all organisms share the same genes. In fact it means they share the same codon meanings.
D1.2.9 Reading an mRNA sequence with the codon table
From 2028 this is supplied in the Biology data booklet — you need to recognise and interpret it, not reproduce it from memory.
The table lists all 64 mRNA codons, so convert a DNA template sequence first.
Split the mRNA into codons from the start codon AUG, in a fixed reading frame.
AUG also codes for methionine, the first amino acid of every new polypeptide.
Stop codons UAA, UAG and UGA code for nothing; translation ends there.
Students often apply the table straight to the DNA template, reading T as U. In fact the table is for mRNA, which is complementary to the template.
Students often think a stop codon adds an amino acid, or the ribosome carries on past it. In fact translation ends and the polypeptide is released.
D1.2.10 Elongation: one codon, one peptide bond, one step along
With two tRNAs bound, a peptide bond joins the chain to the newly arrived amino acid.
The chain is transferred to the new tRNA; the old tRNA leaves.
The ribosome moves along the mRNA by exactly one codon and the cycle repeats.
A peptide bond is covalent, formed by condensation between carboxyl and amino groups.
Students often think amino acids are held by hydrogen bonds, like codon and anticodon. In fact they are joined by covalent peptide bonds.
Students often think each tRNA stays attached to its amino acid in the finished chain. In fact the tRNA leaves empty and is reused.
D1.2.11 A single base change can change a protein
A point mutation alters one base pair: a substitution, an insertion or a deletion.
A substitution changes one codon, so it may change one amino acid, or none.
In sickle cell anaemia, GAG becomes GUG, so valine replaces glutamic acid in beta haemoglobin.
That one change makes haemoglobin form fibres at low oxygen, sickling the red cells.
An insertion or deletion of one base shifts the reading frame, so every codon after it is misread.
Students often think every substitution damages the protein. In fact a degenerate code means some substitutions leave the amino acid unchanged.
Students often think any point mutation shifts the reading frame. In fact only insertions or deletions do; a substitution alters just one codon.
D1.2.12 Both transcription and translation run 5' to 3' HL
A strand's 5' end carries a phosphate; its 3' end carries a hydroxyl group.
RNA polymerase adds nucleotides only at the 3' end, so RNA grows 5' to 3'.
The antiparallel template strand is therefore read 3' to 5'.
The ribosome moves along mRNA from the 5' end; that codon is translated first.
Students often think the template strand is read 5' to 3'. In fact the RNA grows 5' to 3', so the template is read 3' to 5'.
Students often name direction after the end being extended. In fact adding at the 3' end means growing 5' to 3'.
D1.2.13 Transcription starts at the promoter, once transcription factors bind HL
The promoter is non-coding DNA next to a gene's start, where RNA polymerase binds.
It positions the enzyme and sets which strand is the template.
In eukaryotes, transcription factors must bind the promoter before RNA polymerase can.
Cell types differ in transcription factors, so the same gene can be on or off.
Students often think the promoter is the start codon or the first part of the mRNA. In fact it is DNA beside the gene and is not translated.
Students often think RNA polymerase binds the promoter alone. In fact in eukaryotes transcription factors must bind first.
D1.2.14 Most eukaryotic DNA does not code for polypeptides, but has functions HL
Non-coding DNA includes sequences that regulate expression, such as promoters and enhancers.
Introns lie inside genes; telomeres protect the ends of chromosomes.
Genes for rRNA and tRNA are transcribed but their RNA is never translated.
Students often think non-coding DNA is junk and never transcribed. In fact much of it has a function, and rRNA and tRNA genes are transcribed.
Students often think tRNA and rRNA are made by translation. In fact they are transcribed; the RNA itself is the product.
D1.2.15 Pre-mRNA is spliced and capped before it leaves the nucleus HL
The first transcript, pre-mRNA, contains introns and exons.
Introns are removed and exons are spliced together to form mature mRNA.
A 5' cap and a 3' poly-A tail are added to stabilise the transcript.
Only mature mRNA leaves the nucleus; the DNA itself is unchanged.
Students often think introns are cut out of the DNA, or skipped by RNA polymerase. In fact they are transcribed, then removed from the RNA.
Students often think the cap and tail are translated into extra amino acids. In fact they protect the mRNA and add no amino acids.
D1.2.16 One gene can give several polypeptides by splicing exons differently HL
Alternative splicing joins different combinations of exons from one pre-mRNA.
Each combination gives a different mature mRNA and a different polypeptide.
So an organism makes far more polypeptides than it has genes.
The gene's own base sequence is not changed by splicing.
Students often think each gene codes for exactly one polypeptide. In fact alternative splicing lets one gene code for several.
Students often think alternative splicing is a kind of mutation. In fact it recombines exons of the RNA; the gene is unchanged.
D1.2.17 How translation starts, and the A, P and E sites HL
The small subunit binds the 5' end of mRNA and moves to the start codon.
The initiator tRNA, carrying methionine, pairs with AUG; the large subunit then attaches.
A loaded tRNA enters the A site; the growing chain sits in the P site.
After bonding and one step along, the empty tRNA exits from the E site.
Students often think the whole ribosome binds the start codon as one unit. In fact the small subunit binds first and the large subunit joins later.
Students often think E means entry. In fact tRNAs enter at the A site and exit from the E site.
D1.2.18 Many polypeptides are cut or altered before they work HL
Post-translational modification may remove sections, form disulfide bonds or add carbohydrate groups.
Translation gives pre-proinsulin; removing its signal sequence gives proinsulin.
Proinsulin folds and forms disulfide bonds; cutting out the C-peptide leaves insulin.
Insulin is two chains, A and B, from one gene and one polypeptide.
Students often think a polypeptide works as soon as it leaves the ribosome. In fact many must be modified first.
Students often think insulin's two chains come from two genes. In fact both are pieces of one cut polypeptide, held by disulfide bonds.
D1.2.19 Proteins are constantly broken down and rebuilt HL
A proteasome breaks down damaged or unneeded proteins into amino acids.
The amino acids are recycled into new proteins, not excreted.
The proteome is all the proteins in a cell at one time, and it changes.
Keeping a working proteome needs constant breakdown as well as constant synthesis.
Students often think a protein lasts until it is damaged or the cell dies. In fact proteins are continually replaced.
Students often think breakdown happens only when a cell is starved. In fact it is a normal, continuous activity.
Diagnostic a bearings check, not a test
10 questions, one per part of the topic where we can. Answer them, then see which statements you own and which to read.
1 Which statement describes the roles of RNA polymerase in transcription?
Answer and reasoning
It separates the DNA strands and links RNA nucleotides, positioned by pairing with the template strand, into one RNA strand. — RNA polymerase binds to the gene, separates the DNA strands, positions free RNA nucleotides by complementary base pairing with the template strand and forms the covalent bonds that join them into a single RNA strand.
It links RNA nucleotides into a strand once helicase has opened the DNA and a primer has been laid down. — A student who transfers the replication machinery to transcription picks this. RNA polymerase separates the DNA strands itself and needs no primer; helicase and primase are not involved in transcription.
It copies both DNA strands of the gene at the same time, producing a double-stranded RNA molecule from the gene. — A student who models transcription on replication picks this. Only one strand, the template strand, is transcribed, and the product is a single strand of RNA.
It joins amino acids together, one after another, in the order specified by the base sequence of the gene. — A student who has swapped transcription and translation picks this. Amino acids are joined during translation on a ribosome; RNA polymerase joins RNA nucleotides, not amino acids.
2 During transcription, what holds each RNA nucleotide in position against the DNA template strand while it is being added to the growing RNA?
Answer and reasoning
Covalent bonds made by RNA polymerase between the RNA base and the DNA base opposite it. — A student who thinks the enzyme bonds the RNA to the DNA picks this. RNA polymerase forms covalent bonds between successive RNA nucleotides along the backbone; base pairs are held by hydrogen bonds.
Hydrogen bonds between identical bases, so the RNA is an exact copy of the template. — A student who expects a like-for-like copy picks this. Bases pair with their complements (A with U, C with G), so the RNA is complementary to the template, not identical to it.
Hydrogen bonds between complementary bases, which break again when the RNA is released. — Complementary base pairing between the RNA nucleotide and the template base is held by hydrogen bonds. These are weak and temporary, so the completed RNA separates from the DNA and the DNA strands re-pair.
Hydrogen bonds between adenine and thymine, since RNA contains the same four bases as DNA. — A student who forgets that RNA contains uracil picks this. Adenine on the DNA template strand pairs with uracil on the RNA strand; there is no thymine in RNA.
3 Why is transcription regarded as a key stage at which the expression of a gene can be switched on or off?
Answer and reasoning
Every gene in a cell is transcribed and translated all the time, so expression is not switched on or off at any stage. — A student who thinks every gene is always expressed picks this. Not all genes in a cell are expressed at any given time; a gene that is not transcribed produces no mRNA and no polypeptide, so transcription is the stage that is switched on and off.
It is the first stage of gene expression, so a gene that is not transcribed produces no mRNA and no polypeptide. — Transcription is the first stage of gene expression. If a gene is not transcribed there is no mRNA, so nothing further in the expression of that gene can happen; switching transcription on or off switches the gene on or off.
All genes in a cell are transcribed at all times, so expression is switched on and off only at translation. — A student who locates the control of expression at translation picks this. Not all genes are transcribed at any given time; because transcription is the first stage, switching it off stops the mRNA being made at all, so it is the key stage at which a gene is switched on or off.
It is the stage at which amino acids are assembled into a chain, so it decides which protein is made. — A student who has swapped transcription and translation picks this. Amino acids are assembled in translation; transcription produces the mRNA and is the key control stage because it comes first.
4 Which statement about the binding of mRNA and tRNA to a ribosome is correct?
Answer and reasoning
The mRNA binds to the large subunit, because it is the larger molecule, and the tRNAs bind to the small subunit. — A student who matches molecule size to subunit size picks this. It is the small subunit that binds the mRNA; the tRNA binding sites are on the large subunit.
The mRNA binds to the small subunit, and only one tRNA can bind to the large subunit at any moment. — A student who pictures one tRNA delivering at a time picks this. Two tRNAs bind simultaneously, so the amino acid on the new tRNA can be joined to the chain held by the other.
The mRNA binds to the small subunit, and every tRNA needed for the polypeptide binds along it at once. — A student who takes diagrams of several tRNAs literally picks this. Only two tRNAs are bound at a time, and the ribosome works along the mRNA stepwise.
The mRNA binds to the small subunit, and two tRNAs can bind to the large subunit at the same time. — mRNA binds to the small subunit of the ribosome, and two tRNAs can bind simultaneously to the large subunit, allowing a peptide bond to form between the amino acid on one and the polypeptide on the other.
5 Why does the genetic code use codons of three bases rather than two?
Answer and reasoning
Two bases would give 16 codons, which is close enough to 20 for nearly all the amino acids to be coded. — A student who has not seen the constraint picks this. Sixteen codons cannot specify 20 amino acids; every amino acid needs at least one codon of its own.
Three bases allow each codon to code for up to three different amino acids whenever these are needed. — A student who thinks the code is ambiguous picks this. Each codon specifies only one amino acid; the number of bases sets how many different codons exist, not how many amino acids one codon can mean.
Three bases give 64 possible codons, enough for 20 amino acids, whereas two bases give only 16. — With four bases, two-base codons give 4 × 4 = 16 combinations, fewer than the 20 amino acids used in polypeptides. Three-base codons give 4 × 4 × 4 = 64, more than enough, so a triplet code is the shortest that works.
Three bases let the ribosome start reading at any point, since any group of three is a codon. — A student without a fixed reading frame picks this. Codons are read in a fixed frame from the start codon, and reading from another base gives a wrong sequence; the triplet length is about having enough codons.
6 How is the polypeptide chain lengthened during translation?
Answer and reasoning
All the tRNAs first line up along the whole length of the mRNA, and their amino acids are then joined together in one step. — A student who takes a diagram of many tRNAs literally picks this. Only two tRNAs are on the ribosome at once, and amino acids are added one at a time as the ribosome moves.
Each tRNA stays attached to its own amino acid, so the chain is a series of tRNAs joined through their amino acids. — A student who leaves the tRNAs in the chain picks this. Each tRNA is released once its amino acid has been joined by a peptide bond, and the chain consists of amino acids only.
Each new amino acid is held to the previous one by hydrogen bonds, like those between codon and anticodon. — A student who generalizes hydrogen bonding to the whole topic picks this. Amino acids are linked by peptide bonds, which are covalent; hydrogen bonds hold codon to anticodon only temporarily.
The ribosome moves along the mRNA one codon at a time, and a peptide bond links each new amino acid to the chain. — Elongation is stepwise: the ribosome moves along the mRNA one codon at a time, a tRNA pairs with each codon in turn, and the amino acid it carries is linked to the growing polypeptide by a peptide bond.
7 Which statement about the direction of translation is correct? HL
Answer and reasoning
The ribosome can begin at either end of the mRNA, since the start codon is recognized wherever it is met. — A student who gives the mRNA no direction picks this. The ribosome binds at the 5' end and moves towards the 3' end; it does not begin from the 3' end.
The ribosome moves along the mRNA from its 5' end towards its 3' end, so the codon nearest the 5' end is translated first. — Translation is 5' to 3': the small subunit binds near the 5' end of the mRNA, finds the start codon and the ribosome moves towards the 3' end, so the polypeptide is built from the 5'-most codon onward.
The ribosome moves along the mRNA from its 3' end towards its 5' end, matching the direction the template is read. — A student who transfers the 3' to 5' reading of the DNA template to the mRNA picks this. The mRNA is read in its own 5' to 3' direction, the same direction in which it was synthesized.
The ribosome begins at the 5' cap and translates every base up to and including the poly-A tail at the 3' end. — A student who treats the whole mRNA as code picks this. Translation runs 5' to 3' but only from the start codon to the stop codon; the cap and tail are not translated.
8 Which statement about non-coding sequences in eukaryotic DNA is correct? HL
Answer and reasoning
Non-coding sequences such as introns and telomeres have no function at all and so are never transcribed into RNA. — A student who equates non-coding with junk picks this. Telomeres protect chromosome ends, promoters regulate expression, and introns are transcribed before being spliced out.
Every gene codes for a polypeptide, so tRNA and rRNA are both produced by translation of their own mRNA molecules. — A student who defines a gene by its protein product picks this. tRNA and rRNA are made directly by transcription of their genes and are never translated.
The promoter of a gene is transcribed and forms the first codons of the mRNA for that gene. — A student who confuses the promoter with the start of the mRNA picks this. The promoter is where RNA polymerase binds; it is a regulator of expression, not part of the coding sequence.
The genes for tRNA and rRNA are transcribed, but their RNA products are never translated into polypeptides. — Genes for tRNAs and rRNAs are non-coding in the sense that their products are functional RNA molecules, not polypeptides: they are transcribed but never translated. Introns are also transcribed but are removed before translation, while promoters and telomeres are neither translated nor part of any mRNA.
9 The human genome contains about 20 000 protein-coding genes, yet human cells produce well over 20 000 different polypeptides. Which process accounts for this? HL
Answer and reasoning
Mutation of the gene during transcription, so that each transcript carries a slightly different base sequence. — A student who treats mutation as a routine copying error picks this. Transcription copies the gene faithfully; the variants arise from combining exons differently, not from changing the sequence.
Degeneracy of the genetic code, which allows a codon to be translated as different amino acids in different cells. — A student who thinks the code is ambiguous picks this. Each codon specifies one amino acid in every cell; degeneracy means several codons per amino acid, which cannot create new polypeptides.
Alternative splicing, in which different combinations of exons from one pre-mRNA are joined to give different mature mRNAs. — Splicing together different combinations of exons allows a single gene to code for several different polypeptides, so the number of polypeptides an organism makes can far exceed its number of genes.
Duplication of each gene during differentiation, so that every different polypeptide is coded for by a separate gene of its own. — A student holding 'one gene, one polypeptide' picks this. Somatic cells all have the same genes; one gene gives several polypeptides through alternative splicing.
10 Insulin consists of two polypeptide chains, A and B, held together by disulfide bonds, yet it is coded for by a single gene. How is functional insulin produced? HL
Answer and reasoning
Two separate polypeptides, A and B, are translated from the gene and are then joined to each other by disulfide bonds in the Golgi apparatus. — A student who explains two chains by two polypeptides picks this. One polypeptide is translated; the two chains are what remains of it after a central section is removed.
Pre-proinsulin loses its signal sequence to give proinsulin; a central section is then cut out, leaving the two chains. — This is the two-stage modification of pre-proinsulin: removal of the signal sequence gives proinsulin, which folds and forms disulfide bonds, and removal of the central section then leaves the A and B chains joined as active insulin.
The polypeptide made by translation is already insulin, and the two chains form as it folds when it leaves the ribosome. — A student who thinks translation makes the finished protein picks this. Pre-proinsulin is not functional; two sections must be removed before it becomes insulin.
Alternative splicing of the insulin pre-mRNA produces two different mature mRNAs, one of which codes for each of the two chains. — A student who over-applies alternative splicing picks this. The two chains of insulin are not splice variants; they are two parts of a single polypeptide cut after translation.
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21 more questions. Every wrong answer here is a real misconception, and you see why it is wrong straight away.
1 A section of the DNA template strand of a gene has the base sequence TACGCAT. What is the base sequence of the RNA transcribed from this section?
Answer and reasoning
UACGCAU — A student who copies the template like-for-like, only swapping T for U, picks this. The RNA is complementary to the template strand, not identical to it.
AUGCGUA — Each RNA base is complementary to the template base opposite it: T pairs with A, A on the template pairs with U, C with G and G with C, giving AUGCGUA.
UAGCGAU — A student who places uracil opposite thymine, and adenine opposite adenine, picks this. Uracil pairs with adenine on the template; thymine on the template pairs with adenine on the RNA.
ATGCGTA — A student who applies the DNA pairing rules and writes T opposite A picks this. RNA contains uracil, not thymine, so adenine on the template strand pairs with uracil on the RNA.
2 A neuron formed before birth does not divide again, yet it transcribes the same genes throughout a lifetime of many decades. Which statement explains how this is possible?
Answer and reasoning
Each round of transcription slightly alters the template, so the neuron periodically replicates its DNA to restore the gene. — A student who thinks a template wears out picks this. The template is read by base pairing and left unchanged, and a non-dividing neuron does not replicate its DNA at all.
Because the neuron does not replicate its DNA again, its base sequences need not be conserved and can change freely. — A student who thinks DNA matters only when copied picks this. The neuron's DNA is the template for all its transcription, so a changed sequence would change every transcript for the rest of its life.
Transcription copies both strands of the gene just as replication does, so every round of transcription renews the gene. — A student who models transcription on replication picks this. Only the template strand is transcribed, into single-stranded RNA; the gene persists because transcription does not change it, not because it is re-copied.
Transcription reads the template strand without altering its base sequence, so the sequence is conserved for life. — A single DNA strand can be used as a template for transcription without its base sequence changing. In a somatic cell that does not divide, the sequences are conserved and reused for the life of the cell.
3 Developing red blood cells produce large amounts of the beta chain of haemoglobin, but liver cells produce none, although both cell types contain the gene. Which statement explains the difference?
Answer and reasoning
The gene is transcribed in developing red blood cells but not in liver cells, so liver cells make no mRNA for the protein. — Not all genes in a cell are expressed at any given time. Transcription is the first stage of gene expression, so a gene that is not transcribed in a cell type gives no mRNA and no polypeptide there.
Liver cells lost the gene when they differentiated, so they no longer contain the DNA sequence for the beta chain. — A student who thinks differentiated cells discard genes picks this. The stem states that both cell types contain the gene; the difference is in whether the gene is transcribed.
Both cell types transcribe the gene, but only red blood cells translate the mRNA into the beta chain polypeptide. — A student who locates all control at translation picks this. Transcription is the key stage at which a gene is switched on or off; in liver cells the gene is not transcribed, so there is no mRNA to translate.
Both cell types make the beta chain, but liver cells break the protein down as soon as it has been produced. — A student who believes every gene is expressed in every cell picks this. A liver cell does not make the beta chain at all, because the gene is not transcribed in liver cells.
A base sequence of DNA is used as a template to make a complementary mRNA molecule inside the cell nucleus. — A student who has swapped the two terms picks this. Making mRNA on a DNA template is transcription; translation is the synthesis of a polypeptide from the mRNA.
Ribosomes convert the bases of the mRNA into amino acids, which are then joined into a chain. — A student who thinks amino acids are made from bases picks this. The amino acids already exist in the cytoplasm and are brought by tRNA; the mRNA only determines the order in which they are joined.
The base sequence of an mRNA molecule is used to determine the sequence of amino acids in a polypeptide. — Translation is the synthesis of a polypeptide from mRNA: the base sequence of the mRNA, read as codons, is translated into the amino acid sequence of the polypeptide on a ribosome.
Amino acids are joined into a polypeptide inside the nucleus, right beside the DNA of the gene that codes for it. — A student who keeps the whole process in the nucleus picks this. Translation takes place on ribosomes in the cytoplasm, after the mRNA has left the nucleus.
5 A cell contains many different kinds of tRNA molecule. What is the role of a tRNA during translation?
Answer and reasoning
It brings a specific amino acid to the ribosome and pairs its anticodon with the matching codon on the mRNA. — Each tRNA carries a specific amino acid and has an anticodon complementary to the codons for that amino acid, so complementary base pairing at the ribosome delivers the amino acid the codon specifies.
It binds to a codon first and then collects from the cytoplasm whichever amino acid that codon specifies. — A student who thinks the codon tells the tRNA what to fetch picks this. The amino acid is attached to the tRNA before it reaches the ribosome; the codon selects which loaded tRNA binds.
It delivers an amino acid and remains bonded to it, becoming part of the finished polypeptide chain. — A student who thinks tRNAs stay in the chain picks this. Once its amino acid has been joined to the polypeptide by a peptide bond, the tRNA leaves the ribosome and is reused.
It carries the mRNA bases that the ribosome converts into the amino acids of the polypeptide. — A student who thinks amino acids are made from bases picks this. tRNA carries an amino acid, not bases to be converted; the mRNA bases only specify which amino acid is added.
6 The start codon on an mRNA molecule is AUG. Which tRNA anticodon binds to this codon, and how is it held there?
Answer and reasoning
AUG, the same three bases as the codon — A student who reads 'matching' as 'identical' picks this. The anticodon pairs with the codon by complementary base pairing, so it is UAC, not AUG.
UAC, held to the codon by hydrogen bonds — The anticodon is complementary to the codon: A pairs with U, U with A and G with C, giving UAC. The codon and anticodon are held together by hydrogen bonds, which break when the tRNA leaves.
TAC, with thymine pairing to adenine — A student who applies the DNA pairing rules picks this. tRNA is RNA and contains uracil, not thymine, so the base that pairs with the A of the codon is U.
UAC, bonded to the codon by covalent bonds — A student who thinks the ribosome bonds the tRNA to the mRNA picks this. Codon and anticodon pair by hydrogen bonds, which must break so the tRNA can leave; covalent bonds form only between amino acids.
7 The genetic code is described as degenerate. What does this mean?
Answer and reasoning
A single codon can code for two or more different amino acids. — A student who reverses the many-to-one relationship picks this. The code is degenerate but not ambiguous: a codon always specifies the same single amino acid.
The same codons code for the same amino acids in all organisms. — A student who has interchanged the two terms picks this. This is universality; degeneracy is about several codons sharing one amino acid.
Some amino acids are coded for by two bases and some by three bases. — A student who takes 'degenerate' to mean irregular picks this. Every codon is exactly three bases long; degeneracy refers to the number of codons per amino acid, not the number of bases per codon.
Most amino acids are coded for by more than one codon. — Degeneracy means that several codons can specify the same amino acid: 61 codons code for 20 amino acids, so for example GGU, GGC, GGA and GGG all code for glycine.
8 The gene for human insulin has been inserted into bacteria, which then translate it into a polypeptide with the same amino acid sequence as the polypeptide made in human cells. Which feature of the genetic code makes this possible?
Answer and reasoning
Universality: the bacterium's tRNAs assign exactly the same amino acids to the codons as human cells do. — The genetic code is universal: with minor exceptions, the same codons specify the same amino acids in all organisms, so the bacterium translates the human mRNA into the same amino acid sequence.
Universality: bacteria and humans have the same genes, so the bacterium already had an insulin gene. — A student who takes universal to mean 'same genes' picks this. Universality is about the meaning of codons, not the genes; bacteria have no insulin gene, which is why one had to be inserted.
Degeneracy: each codon can be read as several amino acids, so the bacterium can make any protein. — A student who thinks degenerate means ambiguous picks this. Each codon specifies one amino acid; the bacterium makes the human polypeptide because it reads the codons the same way, not because it can read them differently.
Complementary pairing: bacterial tRNA anticodons pair directly with the bases of the inserted DNA. — A student who thinks tRNA pairs with DNA picks this. The inserted gene is transcribed into mRNA first, and it is the shared meaning of the mRNA codons that makes the product identical.
9 Part of an mRNA molecule, written in the order in which it is translated, has the base sequence AUGGGCAACAGA. The genetic code table gives: AUG methionine, GGC glycine, AAC asparagine, AGA arginine, UAC tyrosine, CCG proline, UUG leucine, UCU serine, UGG tryptophan, GCA alanine, ACA threonine. Which amino acid is the third in the polypeptide?
Answer and reasoning
arginine — A student who treats AUG as a start signal only, and begins counting from GGC, makes arginine (AGA) the third amino acid. AUG codes for methionine, which is the first amino acid, so the third is asparagine.
asparagine — Reading from the start codon in groups of three gives AUG GGC AAC AGA: methionine, glycine, asparagine, arginine. The third codon, AAC, codes for asparagine.
leucine — A student who first writes the complementary sequence (UAC CCG UUG UCU) and looks that up picks leucine. The table is for mRNA codons and the sequence given is already mRNA, so no complement should be taken.
threonine — A student who begins reading from the second base (UGG GCA ACA) picks threonine. The reading frame is fixed by the start codon AUG, so codons must be counted from the first base.
10 An mRNA molecule has the base sequence AUGUUUGCUUAAGGC, read from the first base. The genetic code table gives: AUG methionine (start), UUU phenylalanine, GCU alanine, UAA stop, GGC glycine. How many amino acids are in the polypeptide translated from it?
Answer and reasoning
two amino acids — A student who does not count the start codon picks this. AUG codes for methionine as well as marking the start, so methionine, phenylalanine and alanine make three.
four amino acids — A student who reads past the stop codon to include GGC picks this. UAA ends translation, so the glycine codon after it is not translated.
three amino acids — The codons are AUG UUU GCU UAA GGC. AUG gives methionine, UUU phenylalanine and GCU alanine; UAA is a stop codon, so translation ends there and GGC is never translated. The polypeptide has three amino acids.
fifteen amino acids — A student who thinks each base codes for one amino acid picks this. The code is a triplet code, so 15 bases form 5 codons, of which one is a stop codon and only three are translated.
11 An antibiotic stops the ribosome from moving along the mRNA but does not affect the binding of tRNAs or the formation of peptide bonds. What effect will this have on translation of an mRNA?
Answer and reasoning
Only the first peptide bond forms, since the next codon cannot be moved into place for a tRNA to bind. — Two tRNAs can bind and a peptide bond can form between their amino acids, but the chain then grows only if the ribosome moves one codon along so that the used tRNA leaves and the next codon is exposed. Without movement, elongation stops.
Translation is unaffected, because all the tRNAs bind along the mRNA before any peptide bonds are formed. — A student who pictures all the tRNAs lining up at once picks this. Only two tRNAs are bound at a time, so each further codon has to be brought into the ribosome by stepwise movement.
Translation is unaffected, because each tRNA leaves the ribosome before the next one binds to the mRNA. — A student who thinks one tRNA is bound at a time picks this. Even so, the next codon must be moved into the ribosome for the next tRNA to pair with it, and the ribosome cannot move.
The chain still grows, because each tRNA stays attached to its amino acid and the next tRNA simply joins onto it. — A student who thinks tRNAs chain together picks this. tRNAs leave after giving up their amino acid, and new tRNAs bind to codons in the ribosome, which the ribosome must move along the mRNA to reach.
12 In the allele that causes sickle cell anaemia, the sixth codon of the mRNA for the beta chain of haemoglobin is GUG instead of GAG. GAG codes for glutamic acid and GUG codes for valine. What is the effect of this mutation on the protein?
Answer and reasoning
One amino acid out of more than a hundred is changed, too small a difference to alter the structure or function of haemoglobin. — A student reasoning proportionally picks this. A protein's function depends on its exact sequence; this single substitution is enough to cause a serious disease.
Valine replaces glutamic acid at one position, which alters haemoglobin enough for it to form fibres at low oxygen concentrations. — This base substitution changes one codon and so one amino acid. The valine on the surface of the beta chain makes deoxygenated haemoglobin molecules stick together into fibres, which distort the red blood cells into a sickle shape.
The reading frame is shifted from the sixth codon onward, so every amino acid after it in the beta chain is different. — A student who attaches the frameshift effect to every point mutation picks this. A substitution leaves the number of bases unchanged, so only the sixth codon is altered.
The codon GUG is degenerate, so the ribosome can still insert glutamic acid and the haemoglobin produced is normal. — A student who thinks a codon can be read as more than one amino acid picks this. GUG always codes for valine; degeneracy means several codons for one amino acid, not several amino acids for one codon.
13 A base substitution in a gene changes an mRNA codon from GGU to GGC. Both codons code for glycine. What is the effect on the polypeptide?
Answer and reasoning
Its structure changes, because a different codon is bound to give a different amino acid. — A student who expects every substitution to change the protein picks this. Degeneracy means a changed codon can still specify the same amino acid, as here.
Every amino acid after this point changes, because the reading frame has been shifted. — A student who thinks all point mutations shift the frame picks this. A substitution replaces one base with another and leaves the reading frame intact.
There is no effect, because the same amino acid, glycine, is inserted at that position. — Because the genetic code is degenerate, GGU and GGC both code for glycine. The polypeptide's amino acid sequence, and so its structure, is unchanged by this silent substitution.
Glycine may be replaced, because GGC can also code for a different amino acid. — A student who thinks the code is ambiguous picks this. Each codon specifies one amino acid only; GGC always codes for glycine.
14 Two mutations occur near the start of the coding sequence of the same gene, in different cells: in one cell a single base is substituted; in the other a single base is inserted. Which mutation is more likely to change the structure of the protein, and why?
Answer and reasoning
Neither is likely to, because each changes only a single base and so can affect at most one amino acid. — A student who assumes a one-base change means a one-amino-acid change picks this. An insertion shifts the reading frame and alters every subsequent codon.
The substitution, because a substitution is certain to change an amino acid, while an insertion adds only one. — A student who thinks every substitution changes an amino acid picks this. Many substitutions are silent, and an insertion does not add one amino acid but changes the reading of all that follow.
Neither is likely to, because the ribosome recognizes each codon wherever it starts reading. — A student without a fixed reading frame picks this. The ribosome reads consecutive triplets from the start codon, so an inserted base makes it read every later triplet in the wrong frame.
The insertion, because it shifts the reading frame so that every codon after it is read differently. — Adding one base changes how all the following bases are grouped into codons, so the amino acid sequence from that point onward is altered. A substitution changes at most one amino acid and may be silent.
15 During transcription, RNA polymerase adds each new nucleotide to the free 3' end of the growing RNA strand. What does this mean for the direction of transcription? HL
Answer and reasoning
The RNA is synthesized 5' to 3' while the template strand is read 3' to 5'. — Adding nucleotides to the 3' end means the RNA grows from its 5' end towards its 3' end, which is what 5' to 3' transcription means. The strands are antiparallel, so the template is read from its 3' end towards its 5' end.
The RNA is synthesized 5' to 3' and the template strand is read in the same 5' to 3' direction. — A student who forgets that the strands are antiparallel picks this. The RNA runs opposite to the template it pairs with, so the template is read 3' to 5'.
The RNA is synthesized 3' to 5', because the 3' end is where the chain is being extended. — A student who names the direction by the end being extended picks this. Direction of synthesis is from the first nucleotide (5' end) to the last, so extending the 3' end is 5' to 3' synthesis.
The RNA can be synthesized in either direction, since nucleotides can be joined at either end. — A student who gives strands no direction picks this. RNA polymerase can add only to the 3' hydroxyl, so synthesis is always 5' to 3'.
16 A gene is transcribed in muscle cells but not in skin cells, although the DNA of the two cell types is identical. Which statement explains this in terms of initiation of transcription at the promoter? HL
Answer and reasoning
RNA polymerase binds to the promoter by itself in both cell types, so the difference between the two cell types must arise at a stage after transcription. — A student who thinks RNA polymerase needs no help picks this. Eukaryotic RNA polymerase cannot initiate without transcription factors at the promoter, and the stem states that the gene is not transcribed in skin cells.
In muscle cells the promoter is transcribed as the first part of the mRNA, and this acts as a signal for the gene to be expressed. — A student who thinks the promoter is copied into the mRNA picks this. The promoter is the DNA sequence where the transcription machinery binds; transcription begins after it.
Muscle cells contain transcription factors that bind to the promoter, allowing RNA polymerase to bind and begin transcription; skin cells lack them. — In eukaryotes, transcription is initiated when transcription factors bind to the promoter and enable RNA polymerase to bind. Cell types differ in the transcription factors they contain, so the same gene is switched on in one and off in another.
All genes are transcribed in every cell, and the mRNA for this gene is simply broken down faster in skin cells than in muscle cells. — A student who thinks every gene is always transcribed picks this. Not all genes are transcribed in every cell, and here the stem states that skin cells do not transcribe the gene at all.
17 A eukaryotic gene contains three exons separated by two introns. Which statement describes the mature mRNA produced from this gene? HL
Answer and reasoning
It contains the three exons joined together, with a 5' cap and a 3' poly-A tail added on. — Post-transcriptional modification removes the introns from the pre-mRNA and splices the exons together, and adds a 5' cap and a 3' poly-A tail that stabilize the transcript. The result is the mature mRNA.
It contains the three exons only, because the introns were cut out of the DNA before the gene was transcribed. — A student who removes introns from the gene picks this. The whole gene, introns included, is transcribed into pre-mRNA; the introns are removed from the RNA, and the DNA is unchanged.
It contains the two introns joined together, because the exons were the sections removed by splicing. — A student who has the names the wrong way round picks this. Exons are expressed and retained; introns are the intervening sequences that are removed.
It contains all five sections, with the cap and tail translated as extra amino acids at each end. — A student who treats everything on the mRNA as code picks this. The introns are removed, and the cap and tail are not translated; they stabilize the mRNA.
18 What is the role of the 5' cap and the 3' poly-A tail added to a eukaryotic mRNA transcript? HL
Answer and reasoning
They are translated, adding a short run of extra amino acids to each end of the polypeptide. — A student who assumes every part of the mRNA is code picks this. Translation runs from the start codon to the stop codon; the cap and tail lie outside this and add no amino acids.
They stabilize the mRNA, protecting it from being broken down before it is translated. — The 5' cap and 3' poly-A tail are added after transcription to stabilize mRNA transcripts, so that the mRNA survives to be exported from the nucleus and translated.
They mark the positions of the introns so that the splicing enzymes know which sections to remove. — A student who muddles the modifications picks this. The cap and tail are at the two ends of the transcript; introns are internal and are recognized by their own sequences.
They have no function, being left over from transcription of non-coding DNA at each end of the gene. — A student who equates non-coding with functionless picks this. The cap and tail are not transcribed from the gene at all; they are added afterwards and have the function of stabilizing the mRNA.
19 Which sequence of events initiates translation of an mRNA in a eukaryotic cell? HL
Answer and reasoning
The complete ribosome binds directly to the start codon as a single unit, and the initiator tRNA then enters the ribosome and pairs with the codon. — A student who pictures the ribosome as a single unit picks this. The small subunit binds first, at the 5' end, and the large subunit attaches only after the initiator tRNA is in place at the start codon.
The small subunit binds to the 3' end of the mRNA and moves along until it reaches the start codon, where the large subunit attaches. — A student who gives the mRNA no direction picks this. The small subunit attaches to the 5' terminal of the mRNA and moves towards the 3' end to find the start codon.
The small subunit binds at the start codon, and the initiator tRNA enters at the E site before the large subunit attaches to it. — A student who reads E as 'entry' picks this. The initiator tRNA occupies the P site once the large subunit has attached; the E site is the exit site from which used tRNAs leave.
The small subunit binds to the 5' end of the mRNA and moves to the start codon; the initiator tRNA binds there; then the large subunit attaches. — Initiation begins with attachment of the small ribosomal subunit to the 5' terminal of the mRNA. It moves to the start codon, the initiator tRNA carrying methionine pairs with AUG, and the large subunit then attaches so that a second tRNA can enter the A site.
20 During elongation, what are the roles of the A, P and E sites of the ribosome? HL
Answer and reasoning
A tRNA carrying the next amino acid enters the A site, the tRNA holding the growing polypeptide is in the P site, and a used tRNA leaves from the E site. — The A site accepts the incoming tRNA whose anticodon pairs with the codon, the P site holds the tRNA attached to the polypeptide, and after the ribosome moves one codon the tRNA that has given up the chain sits in the E site and exits.
A tRNA enters the ribosome at the E site, moves to the P site where its amino acid is added to the growing chain, and finally leaves the ribosome from the A site. — A student who reads E as 'entry' picks this. Entry is at the A site, and the E (exit) site is where a tRNA that has given up its amino acid sits before leaving.
Only one site is occupied at any one time: a tRNA enters the A site, passes through the P site and then the E site, and leaves before the next tRNA enters. — A student who thinks one tRNA is bound at a time picks this. Two tRNAs are bound during peptide bond formation, one in the A site and one in the P site.
The tRNAs in all three sites remain attached to their amino acids, so the polypeptide is held to the ribosome by three tRNAs at once. — A student who thinks tRNAs stay bonded to their amino acids picks this. Only the tRNA in the P site (or, just after a peptide bond forms, the A site) holds the polypeptide; the E-site tRNA has no amino acid.
21 Why does a cell continually break down some of its proteins in proteasomes while synthesizing new ones? HL
Answer and reasoning
Proteins are permanent once they have been made, so proteasomes act only when a cell has been badly damaged or is about to die. — A student who sees proteins as permanent parts picks this. Protein breakdown by proteasomes is continuous in healthy cells and is needed to keep the proteome functional.
Proteasomes break down proteins only when the cell is short of glucose and needs amino acids as a source of energy. — A student who knows protein breakdown only from starvation picks this. Proteasome activity is a normal, constant part of regulating which proteins are present, whatever the energy supply.
Sustaining a functional proteome needs constant turnover: proteins no longer needed are removed and their amino acids reused. — The proteome changes as the cell's needs change. Proteasomes break down proteins that are damaged or no longer required, and the amino acids released are reused in the synthesis of the proteins the cell now needs.
The amino acids released by proteasomes are waste products that the cell must excrete, so breakdown removes unwanted material. — A student who applies the GCSE idea of excreting excess amino acids picks this. The amino acids from proteasomes are recycled into new proteins; the point of breakdown is turnover, not excretion.
That was your twenty minutes. Real practice on D1.2 is past-paper questions marked against the mark scheme.
What the exam asks of D1.2
Paper 1A tests base pairing, uracil, the roles of mRNA, tRNA and the ribosome, and degeneracy versus universality. Paper 1B gives a codon table or a base sequence and asks you to deduce the polypeptide, or to trace the effect of a stated mutation. Paper 2 uses *outline* and *explain* for transcription and translation: name the molecule, say what it pairs with, then say what bond forms. HL questions use *describe* and *explain* for initiation, splicing, 5' to 3' direction and insulin processing; keep the order of events exact.
Compiled from the IB Biology guide (first assessment 2025, updated May 2026 for 2028) and our question bank · Specialist review in progress. How these pages are made ·
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