Replication — AP Biology Study Guide

For: AP Biology candidates sitting AP Biology.

Covers: the semiconservative replication model, enzyme functions at the replication fork, leading/lagging strand synthesis, prokaryotic vs eukaryotic replication, telomere replication, and DNA proofreading aligned to the AP Biology CED.

You should already know: DNA nucleotide structure and complementary base pairing rules; directionality of nucleic acid strands; basic prokaryotic vs eukaryotic genome structure.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the AP Biology style for educational use. They are not reproductions of past College Board / Cambridge / IB papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official mark schemes for grading conventions.

1. What Is Replication?

DNA replication is the process by which a cell duplicates its entire genome before cell division, ensuring each daughter cell inherits a complete, largely identical set of genetic information. In the AP Biology CED, this topic is part of Unit 6: Gene Expression and Regulation, accounting for ~2-4% of total exam weight, and it regularly appears in both multiple-choice (MCQ) and free-response (FRQ) sections. Common related terms include DNA replication, genomic replication, and semiconservative replication.

Notation conventions for replication follow standard nucleic acid directionality: all strands are always described by their 5' → 3' orientation because DNA polymerase can only add new nucleotides to the free 3' hydroxyl group of an existing strand—a foundational constraint that shapes all features of replication. Replication occurs during S phase of the cell cycle, and its accuracy is critical for preventing lethal mutations or uncontrolled cell growth. AP exam questions frequently test understanding of replication concepts by asking for analysis of experimental data or comparison of replication mechanisms between prokaryotes and eukaryotes.

2. Semiconservative Replication and Experimental Evidence

Before the landmark Meselson-Stahl experiment, three competing models for DNA replication were proposed: conservative (the parent double helix remains intact, and the new daughter helix is entirely new), semiconservative (each daughter double helix has one original parent strand and one newly synthesized strand), and dispersive (parent strands are cut into fragments, so each daughter strand has a mix of old and new DNA along its length).

Meselson and Stahl tested these models by growing E. coli in medium containing heavy nitrogen (${}^{15}\text{N}$) for multiple generations, so all parental DNA was uniformly labeled with ${}^{15}\text{N}$. They then transferred the bacteria to medium containing only light nitrogen (${}^{14}\text{N}$) and allowed replication to proceed for one, then two generations. They separated DNA by density using centrifugation, and observed the position of DNA bands after each replication. After one replication, they saw a single band of intermediate density, which ruled out conservative replication (which would produce one heavy and one light band). After two replications, they saw two bands: one intermediate and one light, which ruled out dispersive replication (which would only produce one band of gradually lightening intermediate density) and confirmed semiconservative replication as the correct model. The AP exam often asks you to predict band positions for different models and different numbers of replications.

Worked Example

Predict the position and number of DNA bands after 3 rounds of semiconservative replication starting from fully ${}^{15}\text{N}$-labeled DNA grown in ${}^{14}\text{N}$ medium.

1. Start with Generation 0 (parental): all DNA has two ${}^{15}\text{N}$ strands, so one band of heavy density.
2. After 1 replication (Generation 1): every double helix has one ${}^{15}\text{N}$ strand and one ${}^{14}\text{N}$ strand, so all DNA is intermediate density, one band.
3. After 2 replications (Generation 2): half of all double helices are ${}^{15}\text{N}{/}^{14}\text{N}$ (intermediate), half are ${}^{14}\text{N}{/}^{14}\text{N}$ (light), so two distinct bands.
4. After 3 replications: we have 8 total double helices. Only 2 of the 8 still contain the original ${}^{15}\text{N}$ strand, paired with a new ${}^{14}\text{N}$ strand. The other 6 helices are fully ${}^{14}\text{N}$. The result is two distinct bands: a faint intermediate density band and a bright light density band.

Exam tip: When counting bands for semiconservative replication starting from one fully labeled population, you will never have more than two bands, regardless of how many generations pass. Only two original labeled strands exist, so only one intermediate band forms.

3. Replication Fork Mechanism and Enzyme Functions

The replication fork is the Y-shaped region where DNA is unwound and new strands are synthesized. The core constraint of replication is that all DNA polymerases can only add nucleotides to the free 3' hydroxyl end of a pre-existing strand, so all new strands are always synthesized 5' → 3'. Because the two parent strands are antiparallel, this produces two distinct types of new strands at the fork:

• Leading strand: synthesized continuously toward the opening replication fork, since its 3' end points toward the fork.
• Lagging strand: synthesized in short, discontinuous segments called Okazaki fragments, because its 5' end points toward the fork, so new 3' template ends are exposed as the fork opens.

Key enzymes act in sequence at the fork: 1) Helicase unwinds the double helix and separates parent strands; 2) Single-strand binding proteins (SSBs) keep separated strands from reannealing; 3) Topoisomerase relieves supercoiling ahead of the fork; 4) Primase synthesizes a short RNA primer to provide the 3' OH DNA polymerase needs to start; 5) DNA polymerase extends the new strand; 6) DNA ligase joins Okazaki fragments on the lagging strand.

Worked Example

A researcher adds an inhibitor that blocks DNA ligase activity to replicating E. coli. Predict the state of the new DNA after one complete round of replication.

1. First recall DNA ligase's core function: it seals the nicks between adjacent Okazaki fragments on the lagging strand after RNA primers are removed and replaced with DNA.
2. Leading strand synthesis is continuous, so only one RNA primer is needed at the start, and no fragments form. All nicks are resolved without ligase activity on the leading strand.
3. The lagging strand is made of many separate Okazaki fragments. After primer removal, there is no covalent phosphodiester bond between the 3' end of one fragment and the 5' end of the next, and ligase cannot form this bond.
4. Final result: The leading strand is a complete, covalently continuous new strand, while the lagging strand remains as short, disconnected Okazaki fragments with nicks between them.

Exam tip: If a question gives you an unlabeled replication fork diagram, first mark the 5' and 3' ends of the parent strands, then match the 3' end of the new strand to the direction of fork movement to label leading and lagging strands in 10 seconds.

4. Prokaryotic vs Eukaryotic Replication and Telomeres

Prokaryotes have a single circular chromosome, so replication starts at one origin of replication and proceeds bidirectionally around the circle until completion. Eukaryotes have long linear chromosomes, so they use multiple origins of replication along each chromosome to replicate the entire genome in a reasonable timeframe during S phase.

A key unique problem for eukaryotes is the end-replication problem: because RNA primers at the 5' end of the new strand cannot be replaced with DNA (there is no upstream 3' OH end to extend from), each round of replication shortens the chromosome by the length of the terminal primer. To protect coding DNA from being lost, eukaryotic chromosomes have non-coding repetitive sequences at their ends called telomeres. In germ cells, stem cells, and most cancer cells, the enzyme telomerase adds new telomere repeats to the ends of chromosomes, preventing shortening. Most somatic cells do not have active telomerase, so their chromosomes shorten with each division, a process linked to aging and cellular senescence.

Worked Example

A eukaryotic somatic cell has a coding region located 600 base pairs (bp) from the end of its chromosome, and telomeres are 2000 bp long. If each round of replication removes 15 bp from the 5' end of the chromosome, how much telomere remains after 40 rounds of replication, and is the coding region affected?

1. Calculate total base pairs lost after 40 rounds: $40\times 15=600$ bp total lost.
2. Initial telomere length is 2000 bp, so remaining telomere length = $2000-600=1400$ bp.
3. The coding region starts 600 bp inward from the original chromosome end, so removing 600 bp of telomere brings the new end exactly to the start of the coding region.
4. After 40 rounds, all remaining sequence after the end is non-coding telomere, so the coding region has not yet been affected. Additional rounds of replication would begin to remove coding sequence.

Exam tip: On FRQ, when asked to explain the end-replication problem, always connect it to two key points: the 5'→3' synthesis constraint of DNA polymerase, and the requirement for an RNA primer to start synthesis. AP graders require both to award full credit.

5. Common Pitfalls (and how to avoid them)

• Wrong move: Labeling the leading strand as synthesized 3' → 5' because it follows the direction of the replication fork. Why: Students confuse the orientation of the parent template strand with the direction of synthesis of the new strand. Correct move: Always remember all new DNA synthesis is 5' → 3', then match the 3' end of the new strand to fork movement to identify leading/lagging.
• Wrong move: Claiming three bands (heavy, intermediate, light) form after two rounds of semiconservative replication in the Meselson-Stahl experiment. Why: Students incorrectly expect the original heavy double helix to remain intact. Correct move: Remember after the first generation, all original heavy strands are separated and each paired with a light strand, so only one intermediate band forms, no heavy band remains.
• Wrong move: Stating that topoisomerase unwinds the double helix at the replication fork. Why: Students mix up the functions of helicase and topoisomerase. Correct move: Memorize the distinct jobs: helicase unwinds the helix, topoisomerase relieves supercoiling ahead of the fork.
• Wrong move: Claiming that all human cells have active telomerase. Why: Students generalize telomerase function from germ cells to all cell types. Correct move: Remember only germ cells, stem cells, and most cancer cells have active telomerase; most somatic cells do not.
• Wrong move: Stating that DNA polymerase can start DNA synthesis from scratch without a primer. Why: Students confuse DNA polymerase with RNA polymerase (which can start from scratch, learned later in transcription). Correct move: Always note DNA polymerase requires a free 3' OH from an RNA primer to start synthesis.

6. Practice Questions (AP Biology Style)

Question 1 (Multiple Choice)

A biologist repeats the Meselson-Stahl experiment starting with fully ${}^{15}\text{N}$-labeled E. coli, then grows the culture for 3 generations in ${}^{14}\text{N}$ medium. After extracting DNA and centrifuging to separate by density, how many bands will she observe for semiconservative replication, and what are their densities? A. One band of intermediate density B. Two bands: one intermediate, one light C. Three bands: one heavy, one intermediate, one light D. Two bands: one heavy, one light

Worked Solution: We track strand density generation by generation. Starting with 1 double helix of fully ${}^{15}\text{N}$ DNA. After 1 generation, all 2 double helices are ${}^{15}\text{N}{/}^{14}\text{N}$ (intermediate). After 2 generations, 4 double helices: 2 intermediate, 2 light. After 3 generations, 8 total double helices: only 2 still contain the original ${}^{15}\text{N}$ strand (paired with ${}^{14}\text{N}$), and 6 are fully ${}^{14}\text{N}$ (light). This produces two distinct bands: intermediate and light. The correct answer is B.

Question 2 (Free Response)

The antibiotic ciprofloxacin inhibits bacterial topoisomerase II (DNA gyrase), the enzyme that relieves supercoiling during bacterial replication. (a) Explain why inhibiting topoisomerase stops bacterial replication and kills the cell. (b) Predict the effect of topoisomerase inhibition on leading vs lagging strand synthesis, and justify your prediction. (c) Explain why ciprofloxacin does not harm human DNA replication at therapeutic doses.

Worked Solution: (a) As helicase unwinds the DNA double helix at the replication fork, it creates increasing torsional stress and supercoiling in the DNA ahead of the fork. Topoisomerase cuts DNA strands to release this supercoiling, preventing DNA breakage and allowing helicase to continue unwinding. If topoisomerase is inhibited, supercoiling builds up until helicase can no longer unwind the DNA, replication forks stall, and replication stops, killing the bacterial cell. (b) Both leading and lagging strand synthesis will stop completely. Supercoiling builds up ahead of the entire replication fork, regardless of which strand is being synthesized. Both strands require the fork to continue opening to expose new template DNA, so stalling of the fork stops synthesis of both strands. (c) Human (eukaryotic) topoisomerase has a different protein structure than bacterial topoisomerase II. Ciprofloxacin binds specifically to the active site of the bacterial enzyme, and does not interact with eukaryotic topoisomerase at therapeutic concentrations, so human replication is unaffected.

Question 3 (Application / Real-World Style)

Telomerase activity is measured in three human cell types: fibroblasts (somatic connective tissue cells) have 0 units of telomerase activity, testicular germ cells have 120 units, and melanoma (skin cancer) cells have 145 units. Fibroblasts have an average telomere length of 8000 base pairs after 40 divisions. Assuming each division without active telomerase shortens telomeres by 40 base pairs, and active telomerase replaces all lost sequence in germ cells and cancer cells, what is the expected telomere length in germ cells after 40 divisions? What does the high telomerase activity in melanoma suggest about their ability to divide indefinitely?

Worked Solution: First calculate total telomere shortening over 40 divisions without telomerase: $40\text{ divisions}\times 40\text{ bp/division}=1600\text{ bp lost}$. The starting telomere length (before any divisions) is the fibroblast end length plus total lost: $8000+1600=9600\text{ bp}$. In germ cells, telomerase replaces all lost sequence, so expected telomere length after 40 divisions is 9600 base pairs. The high telomerase activity in melanoma cells means they can replace all lost telomere sequence each division, so they will never lose coding DNA from chromosome shortening, allowing them to continue dividing indefinitely to support tumor growth.

7. Quick Reference Cheatsheet

8. What's Next

Replication is the foundational process that underlies all gene expression and inheritance, and it is a required prerequisite for the next core topics in Unit 6: transcription and translation. Without mastering the directionality of nucleic acids and base pairing rules from replication, you will struggle to correctly predict the sequence of RNA and protein products from a given DNA template, a common skill tested on both MCQ and FRQ. Beyond Unit 6, replication is core to understanding cell division, mutation, and the inheritance of genetic traits in Unit 5: Heredity. Errors during replication are also the primary source of new genetic variation that fuels evolution, connecting this topic to Big Idea 1. Follow-on topics to study next: Transcription, Translation, Mutations, Meiosis and Inheritance