IB Biology · Theme D Continuity and change · Molecules
D1.1 DNA replication
DNA is copied exactly before every cell division, so each new cell gets a full set. Each old strand is a template; complementary base pairing keeps the copy accurate. PCR copies chosen DNA sections in a tube, and gel electrophoresis sorts the fragments by length.
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
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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D1.1.1 Replication makes exact copies, and every dividing cell needs it
DNA replication makes two DNA molecules with identical base sequences to the original.
It happens before every cell division, not only before reproduction.
Multicellular organisms need it for growth and for tissue replacement.
An exact copy means the same sequence, not just the same length or composition.
Students often think replication happens only when an organism reproduces. In fact it happens before every cell division, so growth and repair need it too.
Students often think dividing cells share out the DNA, half each. In fact the DNA is replicated first and each daughter cell gets a complete copy.
D1.1.2 Each copy keeps one old strand; base pairing keeps the copy accurate
Replication is semi-conservative: each new molecule has one original strand and one new strand.
The two parent strands separate and each acts as a template.
Complementary base pairing (A with T, G with C) fixes which nucleotide fits each position.
The new strand is complementary to its template, so it matches the template's old partner.
Students often think the old molecule stays intact and a wholly new one is built beside it. In fact each daughter molecule has one old strand and one new strand.
Students often think the new strand is a copy of its template. In fact it is complementary to it, which makes it identical to the strand the template used to pair with.
D1.1.3 Helicase opens the helix; DNA polymerase builds the new strands
Helicase unwinds the double helix and breaks the hydrogen bonds between bases.
Each sugar–phosphate backbone stays intact, so each strand is a whole template.
DNA polymerase links free nucleotides, once paired with the template, into a growing strand.
It forms covalent bonds between one nucleotide's sugar and the next nucleotide's phosphate.
Students often think helicase cuts the DNA backbone. In fact it breaks only the hydrogen bonds between bases; the covalent backbone is untouched.
Students often think the enzymes make the hydrogen bonds between new nucleotides and the template. In fact those form on their own; polymerase makes the covalent links along the strand.
D1.1.4 PCR copies a chosen DNA section; electrophoresis sorts fragments by length
Taq polymerase is heat-stable, so it survives the strand-separation step of every cycle.
Copies double each cycle; primers set which section is copied and give polymerase a start.
In gel electrophoresis DNA moves to the positive electrode; shorter fragments travel further.
Primers are short single-stranded DNA pieces binding either side of the target; polymerase can only extend an existing strand. DNA moves because its phosphates are negatively charged, and small fragments slip through the gel pores more easily.
Students often think helicase is added to PCR to separate the strands. In fact heat does it, by breaking the hydrogen bonds; no enzyme is needed.
Students often think the largest fragments travel furthest because they carry more charge. In fact the shortest travel furthest, because they slip through the pores most easily.
D1.1.5 Using PCR and electrophoresis: DNA profiling in forensics and paternity
A DNA profile is the band pattern from variable markers, amplified and run on a gel.
In forensics, a crime-scene sample is compared with samples from suspects.
In paternity tests, each band in the child must come from mother or father.
Many people share any one marker; more markers make a false match far less likely.
Each extra independent marker multiplies the chance of a coincidental match by a further small fraction, an example of reliability rising with the number of measurements.
Students often think a single-marker match proves a sample came from someone. In fact many unrelated people share any one fragment length; the chance of a false match depends on how many markers are used.
Students often think a child's profile must match the father's completely. In fact only the child's bands not from the mother must be present in the father.
D1.1.6 Strands have 5' and 3' ends, and polymerase builds one way only HL
The 5' terminal carries a free phosphate on the sugar's fifth carbon.
The 3' terminal carries a free hydroxyl on the sugar's third carbon.
The two strands are antiparallel: one strand's 5' end lies beside the other's 3' end.
DNA polymerase joins the incoming nucleotide's 5' phosphate to the strand's 3' hydroxyl.
So a new strand grows only 5' to 3'; its template is read 3' to 5'.
Students often think polymerase adds nucleotides to the 5' end. In fact it adds to the free 3' hydroxyl, so the strand grows 5' to 3'.
Students often think both new strands are built the same way towards the fork. In fact the templates are antiparallel, so one strand is built towards the fork and the other away from it.
D1.1.7 Leading strand is built continuously, lagging strand in fragments HL
The leading strand is made continuously, following the fork as it opens.
Its template runs 3' to 5' towards the fork, so polymerase never has to stop.
The lagging strand is made discontinuously, in short Okazaki fragments built away from the fork.
The leading strand needs one RNA primer; the lagging strand needs one per fragment.
Students often think the lagging strand is built 3' to 5'. In fact every fragment is built 5' to 3', just in the direction away from the fork.
Students often think one primer per fork is enough. In fact the leading strand needs one, but the lagging strand needs a new one for every Okazaki fragment.
D1.1.8 Four prokaryotic enzymes: primase, polymerase III, polymerase I, ligase HL
DNA primase makes a short RNA primer wherever DNA synthesis must start.
DNA polymerase III extends each primer 5' to 3': the leading strand and every Okazaki fragment.
DNA polymerase I removes the RNA primer and replaces it with DNA nucleotides.
DNA ligase seals the remaining nick with a covalent bond, joining fragments into one strand.
Students often think ligase builds the new strand nucleotide by nucleotide. In fact it only joins adjacent fragments once the primer between them has been replaced with DNA.
Students often think the primer is DNA because the enzyme is called DNA primase. In fact the primer is RNA; the name says what is being primed.
D1.1.9 Polymerase III checks each nucleotide as it adds it HL
Proofreading happens during replication, one nucleotide at a time.
If the base just added is mismatched, polymerase III removes it from the 3' terminal.
It then adds a correctly paired nucleotide and carries on.
This cuts the error rate sharply but does not remove every error.
A mismatch that escapes becomes a permanent change in sequence, a mutation, the next time that strand is copied.
Students often think proofreading is a separate scan of the finished molecule. In fact polymerase III checks each nucleotide as it is added and removes a mismatch at once.
Students often think base pairing alone makes replication perfect. In fact pairing plus proofreading make errors rare, not impossible; a few slip through as mutations.
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 A cut in the skin heals as new skin cells are produced by cell division. Why must DNA replication take place before each of these divisions?
Answer and reasoning
So that each new cell receives a complete copy of the DNA with an identical base sequence. — DNA replication produces exact copies of DNA with identical base sequences. It is required not only for reproduction but for growth and tissue replacement, because every cell produced by division must receive a complete copy.
So that the existing DNA can be divided equally, half going to each of the new cells. — A student who thinks division shares out the DNA picks this. The DNA is copied before division so that each new cell receives a complete set, not half of the original.
It does not need to take place, because replication is only required for reproduction. — A student who links replication only with reproduction picks this. Replication is also required for growth and tissue replacement in multicellular organisms, since each new cell needs a full copy of the DNA.
So that the original DNA can stay intact in one cell and the other cell gets a new copy. — A student who imagines conservative replication picks this. Each daughter molecule contains one original strand and one new strand, so neither cell receives the intact original molecule.
2 During replication, a section of a template strand with the base sequence TACGGA is copied. Which statement describes the new strand built on this section?
Answer and reasoning
Its sequence is TACGGA, an exact copy of the template strand it was assembled on. — A student who expects the new strand to copy its template directly picks this. The new strand is complementary to the template; it is identical to the strand that was originally paired with the template.
Its sequence is ATGCCT, the complement of the template, held to it by hydrogen bonds. — Complementary base pairing places A opposite T, T opposite A, G opposite C and C opposite G, so the new strand reads ATGCCT and is held to the template by hydrogen bonds between the paired bases.
Its sequence is AUGCCU, because uracil replaces thymine in each newly made strand. — A student who confuses replication with transcription picks this. Replication uses DNA nucleotides, so the new strand contains thymine, not uracil, and has the sequence ATGCCT.
Its sequence is ATGCCT, held to the template by covalent bonds made by DNA polymerase. — A student who thinks an enzyme bonds the bases together picks this. The bases pair by hydrogen bonds that form spontaneously; DNA polymerase forms covalent bonds along the new strand's backbone, not between bases.
3 What is the role of helicase in DNA replication?
Answer and reasoning
It cuts through the sugar–phosphate backbone so that the two strands can come apart. — A student who thinks helicase cuts the DNA picks this. Helicase breaks only the hydrogen bonds between bases; the covalent backbone of each strand stays intact so that it can act as a template.
It adds free nucleotides to each exposed template strand to assemble the two new strands. — A student who has swapped the two enzymes picks this. Assembling the new strands is the general role of DNA polymerase, not of helicase.
It forms the hydrogen bonds that hold each new nucleotide to its complementary base on the template. — A student who believes an enzyme must make the base pairs picks this. Hydrogen bonds between complementary bases form spontaneously; helicase breaks hydrogen bonds rather than forming them.
It unwinds the double helix and breaks the hydrogen bonds between the two strands. — Helicase unwinds the helix and breaks the hydrogen bonds between complementary bases, separating the strands so that each can serve as a template for DNA polymerase.
4 Why is Taq polymerase, rather than a DNA polymerase from human cells, used in the polymerase chain reaction?
Answer and reasoning
It is a bacterial enzyme that copies DNA much faster than a human polymerase, so fewer cycles are needed. — A student who assumes the special enzyme is chosen for speed picks this. Taq polymerase is used because it is heat-stable; the number of cycles is set by how many copies are wanted, not by enzyme speed.
It is not denatured at the high temperature used to separate the strands, so it lasts through every cycle. — Each cycle heats the mixture to about 95°C to separate the strands. Taq polymerase, from the hot-spring bacterium Thermus aquaticus, keeps its shape at this temperature, so it does not have to be replaced after each cycle.
It separates the two strands of the template by itself, so no helicase has to be added to the reaction mixture. — A student who expects an enzyme to separate the strands picks this. In PCR the strands are separated by heat, not by any enzyme; the point of Taq polymerase is that it survives this heating.
It can start a new strand anywhere on a bare template, so no primers are needed in the reaction mixture. — A student who thinks primers are optional picks this. Like every DNA polymerase, Taq polymerase can only extend an existing strand, so primers are essential in PCR.
5 A forensic laboratory compares a sample from a crime scene with a suspect's DNA using 20 different markers rather than one. Why does this make the result more reliable?
Answer and reasoning
One marker is already conclusive, because everyone's DNA is unique; extra markers only make the bands easier to read. — A student who transfers the uniqueness of a whole genome to a single marker picks this. Each marker is shared by a proportion of the population, so one marker gives many chance matches.
Testing 20 markers is equivalent to repeating a single marker 20 times, and repeating a measurement makes it more accurate. — A student who reads 'more measurements' as 'repeat the same measurement' picks this. Repeating one marker gives the same lengths each time and does not change the chance that another person shares them.
Each extra independent marker multiplies down the chance that an unrelated person matches by coincidence. — The chance that an unrelated person matches at one marker might be a few per cent; the chances at independent markers multiply together, so with 20 markers a false match becomes extremely improbable. This is the NOS principle that reliability is enhanced by increasing the number of measurements.
With 20 markers the whole base sequence of the suspect's genome is read, so the identification is made complete. — A student who thinks profiling reads DNA sequences picks this. A profile compares fragment lengths at a small number of variable regions; it does not read the base sequence of the genome.
6 During replication, how does DNA polymerase add each new nucleotide to the strand it is building? HL
Answer and reasoning
It joins the 3' hydroxyl of the incoming nucleotide to the 5' terminal of the growing strand, so that the strand grows in the 3' to 5' direction. — A student who takes the direction the template is read as the direction of synthesis picks this. DNA polymerase can only add to the 3' terminal, so the new strand grows 5' to 3'.
It adds to whichever terminal faces the fork, so both new strands grow in the same direction as the replication fork moves. — A student who pictures the two strands as parallel picks this. The strands are antiparallel and DNA polymerase adds only to a 3' terminal, so only one new strand can grow towards the fork.
It forms hydrogen bonds between the incoming nucleotide and the template base, which then hold the nucleotide in the strand. — A student who thinks the enzyme makes the base pair picks this. The hydrogen bonds to the template form spontaneously; DNA polymerase forms the covalent bond that links the nucleotide's 5' phosphate to the 3' terminal of the strand.
It joins the 5' phosphate of the incoming nucleotide to the 3' terminal of the growing strand, so the strand grows in the 5' to 3' direction. — The phosphate on the 5' carbon of the incoming nucleotide bonds to the free hydroxyl on the 3' carbon of the last nucleotide in the strand. Growth is therefore only at the 3' terminal, in the 5' to 3' direction, while the template is read 3' to 5'.
7 Which statement correctly compares replication on the leading strand with replication on the lagging strand? HL
Answer and reasoning
The leading strand is made continuously towards the fork; the lagging strand discontinuously, as Okazaki fragments. — The leading strand's template runs 3' to 5' towards the fork, so its new strand is built continuously behind the fork. The lagging strand is built in short Okazaki fragments, each 5' to 3' away from the fork, and each needing its own RNA primer.
The leading strand is made in the 5' to 3' direction, while the lagging strand is made in the opposite, 3' to 5', direction. — A student who thinks the difference between the strands is the direction of synthesis picks this. Both strands are made 5' to 3'; the lagging strand is different because that direction points away from the fork.
Both new strands are made continuously towards the fork, because the two template strands run in the same direction. — A student who pictures parallel strands picks this. The templates are antiparallel, and since DNA polymerase adds only to a 3' terminal, one strand must be made discontinuously, away from the fork.
The lagging strand is made as Okazaki fragments because helicase cuts its template strand into short pieces. — A student who thinks helicase cuts DNA picks this. Helicase breaks only hydrogen bonds and both templates stay intact; the fragments arise because the lagging strand can only be built in short lengths away from the fork.
8 What is the function of DNA polymerase I in prokaryotic DNA replication? HL
Answer and reasoning
It synthesizes most of the new DNA on both strands, being the first polymerase to act at the fork. — A student who reads 'I' as the main enzyme picks this. The bulk of DNA synthesis on both strands is carried out by DNA polymerase III.
It joins the sugar–phosphate backbones of adjacent Okazaki fragments to make one continuous lagging strand. — A student who blurs the polymerases with ligase picks this. Sealing the nick between fragments is the function of DNA ligase, which acts after DNA polymerase I has replaced the primer.
It removes the RNA nucleotides of each primer and replaces them with DNA nucleotides, filling the gap. — DNA polymerase I removes each RNA primer and fills the gap with DNA nucleotides, extending the 3' terminal of the adjacent DNA. DNA ligase then seals the remaining nick.
It synthesizes the short DNA primer that DNA polymerase III needs at the start of each fragment. — A student who thinks the primer is DNA made by a DNA polymerase picks this. The primer is RNA, made by DNA primase; DNA polymerase I removes it rather than making it.
9 As DNA polymerase III builds a new strand it adds a nucleotide carrying G opposite a T on the template. What happens next? HL
Answer and reasoning
DNA polymerase III removes the G nucleotide from the 3' terminal and adds an A in its place. — Proofreading is the action of DNA polymerase III in removing any nucleotide with a mismatched base from the 3' terminal of the growing strand and replacing it with a correctly matched nucleotide, here A opposite T, before continuing.
Synthesis continues, and the mismatch is found and corrected once the whole molecule has been copied. — A student who pictures proofreading as a scan of the finished molecule picks this. DNA polymerase III can only remove the nucleotide at the 3' terminal, so the mismatch is dealt with before synthesis continues.
DNA polymerase III removes the mismatched nucleotide from the 5' terminal of the new strand and replaces it. — A student who muddles the two terminals picks this. The nucleotide just added is at the 3' terminal, the growing end of the strand, and that is where DNA polymerase III removes it.
DNA polymerase I, the enzyme that removes nucleotides, cuts out the mismatched G and replaces it with an A. — A student who generalizes DNA polymerase I as the removal enzyme picks this. DNA polymerase I removes RNA primers; proofreading of a just-added nucleotide is done by DNA polymerase III itself.
10 Bacteria were grown for many generations with only heavy nitrogen (¹⁵N), so all their DNA was heavy. They were then moved to a medium with only light nitrogen (¹⁴N). After one round of replication, all the DNA had a density midway between heavy and light. After a second round, half the DNA was of this intermediate density and half was light. What do these results show about replication?
Answer and reasoning
The original double-stranded molecule stays intact and a completely new molecule is made beside it. — A student who pictures conservative replication picks this. Conservative replication would give heavy and light DNA, never intermediate, after the first round; the observed intermediate density rules it out.
Old and new nucleotides become mixed along both strands of every daughter molecule. — A student who pictures dispersive replication picks this. Mixing would fit the first round, but after the second round all molecules would be of a single, lighter intermediate density; no wholly light DNA would appear.
Each daughter molecule keeps one original strand and gains one strand that is newly made. — After one round every molecule is one heavy strand plus one light strand, giving intermediate density. After a second round, the two strands of each intermediate molecule separate again, giving half intermediate and half wholly light molecules. Only semi-conservative replication predicts both results.
The DNA is not copied but is shared out equally between the daughter cells at each division. — A student who thinks DNA is halved at division picks this. If no new DNA were made, no light nitrogen could be incorporated and all the DNA would have stayed heavy; the shift to intermediate and then to light density shows that new strands were synthesized.
Read the ones marked not yet in Learn, then Verify.
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8 more questions. Every wrong answer here is a real misconception, and you see why it is wrong straight away.
1 A drug inhibits helicase but has no effect on DNA polymerase. What will happen to DNA replication in a cell treated with the drug?
Answer and reasoning
Replication cannot begin, because the hydrogen bonds between the strands are not broken, so no template strand is exposed. — Helicase is needed to unwind the helix and break the hydrogen bonds between the strands. Without it the strands stay paired, no template is exposed and DNA polymerase has nothing to copy.
Replication continues normally, because DNA polymerase unwinds the double helix itself as it copies each strand. — A student who gives the unwinding job to DNA polymerase picks this. DNA polymerase only assembles new strands on templates that helicase has already exposed.
New strands are made but fall off the template, because helicase forms the hydrogen bonds that hold new bases in place. — A student who thinks an enzyme makes the base pairs picks this. Hydrogen bonds between complementary bases form spontaneously; helicase's role is to break them, and without it no strand is even started.
Cell division continues without replication, because the existing DNA is simply shared between the two daughter cells. — A student who thinks division halves the DNA picks this. Each daughter cell must receive a complete copy of the DNA, so without replication normal division cannot proceed.
2 What is the role of the primers in the polymerase chain reaction?
Answer and reasoning
They break the hydrogen bonds between the two strands so that the polymerase can reach the template. — A student who has not separated the roles of heat and primers picks this. The strands are separated by heating to about 95°C; primers bind to the separated strands only after the mixture has cooled.
They only speed the reaction up, because Taq polymerase could begin a new strand at any point along the bare template. — A student who thinks primers are optional picks this. No DNA polymerase can begin a strand on its own; without primers Taq polymerase makes no product at all.
They bind to sequences either side of the target, giving the polymerase a start point and defining the region copied. — The two primers anneal to complementary sequences flanking the target on opposite strands. Taq polymerase can only extend an existing strand, so it extends each primer, and only the DNA between the primers is amplified.
They are identical in sequence to the section being copied, so that the polymerase has an exact copy of it to work from. — A student who expects copies rather than complements picks this. A primer binds by complementary base pairing to the template, so its sequence is complementary to the template, and the polymerase copies the template, not the primer.
3 A PCR reaction starts with 10 double-stranded DNA molecules of the target section. How many double-stranded molecules of the target are present after 3 complete cycles?
Answer and reasoning
60 — A student who thinks each cycle produces a fixed two copies per starting molecule picks this (10 × 2 × 3). Each cycle doubles the whole population, so the growth is 2ⁿ, not 2n.
30 — A student who adds one new copy per starting molecule per cycle and forgets the originals, or who simply multiplies by the number of cycles, picks this. Doubling three times gives 8 times the start, not 3 times.
40 — A student who adds one copy per starting molecule in each cycle picks this (10 + 3 × 10). Copies are themselves copied in later cycles, so the number doubles each time.
80 — Every molecule is copied in every cycle, so the number doubles each cycle: 10 × 2 × 2 × 2 = 10 × 2³ = 80.
4 Three DNA fragments of 200, 500 and 1200 base pairs are loaded into the same well of a gel and a voltage is applied. Which statement describes the result?
Answer and reasoning
The 200 base pair fragment moves furthest through the gel, towards the positive electrode. — All DNA fragments are negatively charged and move towards the positive electrode. The gel acts as a sieve, so the shortest fragment passes through its pores most easily and travels furthest in the time available.
The 1200 base pair fragment moves furthest, because it carries the most negative charge and is pulled harder. — A student who reasons from total charge picks this. Charge per unit length is the same for every fragment; the gel holds back longer fragments, so the largest moves least.
The 200 base pair fragment moves furthest, towards the negative electrode at the far end of the gel. — A student who has the electrodes reversed picks this. DNA is negatively charged because of its phosphate groups, so all fragments move towards the positive electrode.
The three fragments separate according to their base sequences rather than by their lengths. — A student who thinks the gel reads the sequence picks this. Fragments of equal length travel equally far whatever their sequence; the gel separates by size only.
5 A DNA profile was prepared for a mother, her child and a man who may be the father. The fragment lengths (in base pairs) in each profile were: mother 100, 220, 260, 340; child 100, 150, 220, 300; man 150, 180, 300, 330. What can be concluded?
Answer and reasoning
The man is excluded as the father, because two of his bands (180 and 330) do not appear in the child's profile. — A student who expects the child to carry all of the father's bands picks this. A child inherits only half of each parent's DNA, so bands in the father that the child did not inherit are expected.
The man cannot be excluded as the father: the child's two bands not from the mother (150 and 300) are both in his profile. — The child's bands 100 and 220 came from the mother. The remaining bands, 150 and 300, must come from the biological father, and both appear in the man's profile, so he cannot be excluded. Further markers would be used to make the conclusion more reliable.
The man is excluded as the father, because the child's profile does not match his profile completely. — A student who applies forensic-style complete matching to paternity picks this. Only the child's bands that are not explained by the mother need to be present in the father.
The man is proved to be the father, because DNA profiles are unique and two of his bands match the child's. — A student who treats any match as proof picks this. Sharing bands at a few markers is consistent with paternity but does not prove it; other men could share the same fragment lengths by chance.
6 Starting from the origin, one replication fork in a bacterium copies a 6000 base pair section of the chromosome. Okazaki fragments in this bacterium are about 1000 nucleotides long. Approximately how many RNA primers are needed at this fork? HL
Answer and reasoning
One: replication has to be initiated with an RNA primer only once, at the origin, for the whole fork. — A student who thinks a primer only starts replication picks this. DNA polymerase III cannot start a strand, so every Okazaki fragment on the lagging strand needs its own primer.
About seven: one on the leading strand plus one for each of the six or so Okazaki fragments on the lagging strand. — The leading strand is initiated with an RNA primer only once. The lagging strand is initiated repeatedly, once per Okazaki fragment: 6000 ÷ 1000 gives about six fragments, so about six primers. The total is about seven.
Two: one to start each of the two new strands, which are both built continuously to the end of the section. — A student who treats both strands alike picks this. Only the leading strand is continuous; the lagging strand is made in about six fragments, each beginning with a primer.
About twelve: one for each of about six Okazaki fragments on each of the two new strands being made at the fork. — A student who thinks both strands are made in fragments picks this. The leading strand is continuous and needs a single primer; only the lagging strand is made as Okazaki fragments.
7 On the lagging strand, DNA polymerase III has extended an Okazaki fragment until it reaches the RNA primer of the previous fragment. Which sequence of events completes this section of the strand? HL
Answer and reasoning
DNA ligase joins the new fragment to the RNA primer, and then DNA polymerase I replaces the primer with DNA nucleotides. — A student who thinks the fragments must be joined as soon as they meet picks this. DNA ligase acts last: it seals the nick only after the primer has been replaced with DNA.
DNA polymerase III removes the RNA primer and replaces it with DNA nucleotides, and then DNA ligase joins the two fragments. — A student who credits DNA polymerase III with every polymerase task picks this. Removing and replacing primers is the function of DNA polymerase I, not of DNA polymerase III.
DNA primase replaces its RNA primer with DNA nucleotides, and then DNA ligase joins the two fragments together. — A student who takes DNA primase to be a DNA-making enzyme picks this. Primase makes RNA primers only; it is DNA polymerase I that removes them and replaces them with DNA.
DNA polymerase I replaces the RNA primer with DNA nucleotides, and then DNA ligase joins the two fragments together. — DNA polymerase I removes the RNA primer and replaces it with DNA, leaving a nick in the backbone between the last nucleotide it added and the next fragment. DNA ligase then forms the covalent bond that seals the nick.
8 A mutant strain of Escherichia coli has a DNA polymerase III that can add nucleotides normally but cannot remove them. Which prediction about this strain is correct? HL
Answer and reasoning
Nothing changes, because complementary base pairing on its own ensures that every nucleotide added is correct. — A student who believes base pairing is infallible picks this. A wrong nucleotide is occasionally added; proofreading is what removes most of these, so losing it raises the error rate.
Its mutation rate rises, because mismatched nucleotides at the 3' terminal are left in place and copied in later replications. — Proofreading depends on DNA polymerase III removing a mismatched nucleotide from the 3' terminal. Without this, occasional mismatches from base pairing errors remain in the new strand and are copied as permanent changes in sequence.
Nothing changes, because DNA polymerase I proofreads the new strand once DNA polymerase III has finished with it. — A student who assigns the removal job to DNA polymerase I picks this. DNA polymerase I does not check the strand that polymerase III has built; its function in replication is to replace the RNA primers. Each added nucleotide is proofread by DNA polymerase III itself, and strains lacking this activity have greatly raised mutation rates.
Nothing changes, because mismatches are found and corrected by a separate check of the whole molecule after replication is complete. — A student who pictures proofreading as a post-replication scan picks this. In the scope of D1.1.9 the only correction is DNA polymerase III removing a just-added mismatched nucleotide from the 3' terminal; without that activity the mismatch remains.
That was your twenty minutes. Real practice on D1.1 is past-paper questions marked against the mark scheme.
What the exam asks of D1.1
Paper 1A tests the enzymes' roles, the semi-conservative model and the direction fragments move in a gel. Paper 1B may give a gel image or a PCR cycle graph and ask you to read band sizes or explain the doubling. Paper 2 uses *outline* and *explain*: name the enzyme, say what bond it acts on, then link it to the strand it builds. HL questions use *distinguish* and *explain* for leading versus lagging strand and for the four prokaryotic enzymes in sequence; keep 5' and 3' straight in every line.
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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