MCAT content library / Biology & biochemistry / DNA replication and repair: follow strands, fragments and errors
Use strand polarity and the measured product to identify what a replication or repair experiment actually tests.
About 15 minutes, including the worked examples and checks.
Useful foundations
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One template, one complementary product
Replication copies DNA; repair corrects particular errors or damaged structures. A useful experiment can separate those jobs by measuring how much DNA is synthesized, how it is joined, and whether its bases are correct. The AAMC outline includes these mechanisms, their enzymes, replication origins and the challenge of linear chromosome ends.1
Replication is semiconservative: each completed daughter duplex contains one parental strand and one newly synthesized strand. The parental strands separate and provide complementary templates. “Half old” refers to the strand composition of each newly made duplex after a complete replication round. It does not mean that half of every daughter strand is newly synthesized.2
For a simple label-tracing thought experiment, begin with one duplex whose two strands carry label X. Supply only unlabeled precursors for two complete rounds, with no degradation or repair exchange. Round one produces two duplexes, each with one labeled strand. Round two produces four duplexes: two retain one original labeled strand and two contain no label. The original two strands persist, but they are distributed among a growing number of DNA molecules. The fraction of duplexes containing an original strand is therefore 2/4 after round two.
Direction comes from the growing end
Replicative DNA polymerase extends an existing strand by adding to its free 3′ hydroxyl. The new strand therefore grows 5′→3′ while its template is read 3′→5′. Primase makes a primer that supplies the initial extendable end. Helicase separates the parental strands; proteins stabilize the exposed single strands. Topoisomerases relieve the topological stress generated as the helix is unwound.34
Imagine a replication fork moving to the right. Write both parental templates as they appear from the left of the page to the right, without silently reversing either sequence:
| Template, left to right | New synthesis direction on the page | Relationship to the fork |
|---|---|---|
| 3′ → 5′ | Left to right, always 5′ → 3′ | Leading synthesis follows the opening fork |
| 5′ → 3′ | Right to left within each new fragment | Lagging synthesis starts repeatedly near the opening fork |
The leading strand can be extended continuously in the fork’s direction. The lagging strand is made as separate Okazaki fragments because its template has the opposite orientation. Every fragment still grows 5′→3′. “Discontinuous” describes how synthesis is organized along the chromosome, not a reversal of the polymerase’s chemistry.3
After a lagging fragment is extended, its RNA primer must be removed and the region replaced with DNA. Ligase seals a remaining nick between adjacent DNA segments. Filling a gap and sealing a nick are different tasks: a gap lacks one or more nucleotides, whereas a nick interrupts backbone continuity between neighboring nucleotides already present. Merely making longer fragments does not guarantee that the complete daughter strand is continuous.
Worked experiment: follow labeled material through a chase
In this original illustrative experiment, a short pulse labels newly synthesized bacterial DNA. Excess unlabeled precursor is then supplied for a two-minute chase. Extracted nascent DNA is separated by size under denaturing conditions, so a break in backbone continuity can reveal a short strand. Each percentage is the fraction of recovered pulse label in the indicated size class.
| Preparation and sampling time | Label in short strands | Label in long strands |
|---|---|---|
| Control: end of pulse | 40% | 60% |
| Control: after chase | 5% | 95% |
| Variant: end of pulse | 43% | 57% |
| Variant: after chase | 41% | 59% |
| Variant: chase sample after functional ligase is supplied in an extract | 4% | 96% |
Initial total incorporation and total recovered pulse label are matched. Independent tests find normal primer removal and gap filling in the variant; degradation is negligible. Supplying ligase is the only difference in the final extract treatment. Those controls are part of the experiment, not facts to assume from the table.
The control shifts 35 percentage points of label out of the short-strand class during the chase. The variant shifts only 2 points. Short labeled material remains even though synthesis occurred. Rescue changes the variant’s short fraction from 41% to 4%, with a matching rise in long material. Together with normal primer removal and gap filling, that pattern supports failure of nick sealing rather than failure to incorporate nucleotides.
The label itself does not “move” from one nucleotide into another. A labeled nucleotide becomes part of a longer continuous molecule when preexisting DNA segments are joined. This distinction makes pulse-chase experiments powerful: they follow the fate of material made during a defined interval, rather than merely comparing the total amount present at two times.
Without the functional controls, persistent short products would not uniquely identify ligase. Failure to remove primers, incomplete gap filling or DNA breakage could also prevent long continuous strands from accumulating. A rescue result narrows the mechanism only to the extent that the added activity and other conditions are well specified.
Chromosome ends create a different problem
Eukaryotic nuclear chromosomes usually initiate replication at multiple origins. Each origin can establish forks that extend into neighboring regions. The end of a linear chromosome presents a special problem: after removal of a terminal primer, ordinary polymerase cannot replace that segment without an upstream extendable 3′ end. Telomerase uses an internal RNA template to extend a chromosome’s 3′ end, providing additional template for complementary-strand synthesis. Telomerase activity is not equivalent to sealing every nick in a lagging strand.4
Proofreading and repair act on different substrates
Many replicative polymerases proofread by removing an incorrectly incorporated terminal nucleotide using 3′→5′ exonuclease activity. This direction describes removal from the new strand, not DNA synthesis running backward. Mismatch repair can remove and replace a newly synthesized segment containing a mismatch that escaped proofreading. Nucleotide excision repair instead removes a short segment containing a bulky DNA lesion, such as a UV-induced pyrimidine dimer, followed by resynthesis and sealing.5
A mismatch is an abnormal base pairing; a damaged base or covalent lesion is a chemical alteration; a mutation is a persistent sequence change. These categories can be connected without being interchangeable. If an uncorrected mismatch is copied again, daughter molecules can acquire different stable sequences. Whether a particular persistent change alters a protein or phenotype requires additional information about its position and function.
For a new passage, first identify the measured product: nucleotide incorporation, strand length, sequence accuracy or lesion removal. Then follow the altered substrate to the step that normally processes it. An increase in copying errors, an accumulation of short strands and defective removal of UV lesions are different patterns, even if all can compromise genome integrity.
ACTIVE PRACTICE · DNA REPLICATION
Follow the Labeled DNA, Not Just the Total Label
The original pulse–chase measures the fraction of recovered labeled DNA in short versus long strands. Label recovery is matched, degradation is negligible, and primer removal and gap filling are normal in the variant.
Distribution of recovered pulse label:
Short 40%; long 60% (control, end of pulse).
Gold: label in short strands. Teal: label in long strands.
The assay measures size distribution after denaturing extraction. A change in the percentage does not mean a labeled nucleotide migrated between molecules; covalent joining changes which size class contains it.
Check your reasoning
Choose an answer before opening its feedback. Each option explains the reasoning behind that choice.
Check 1
A replication fork moves to the right. Its lower template segment is written left to right as 3′-TACCGT-5′. A primer is already present immediately to the left, and polymerase copies all six displayed bases. Which new DNA segment is produced left to right on this leading strand?
A. 3′-ATGGCA-5′
Review this choice. The bases are complementary, but the new strand’s direction must be opposite to the displayed template, giving 5′ at the left.
B. 3′-TACCGT-5′
Review this choice. This retains both the template sequence and orientation rather than constructing its antiparallel complement.
C. 5′-ATGGCA-3′
Correct. Base pairing gives A-T, T-A, G-C, G-C, C-G and A-T. The new strand grows to the right from its 3′ end, so it is 5′-ATGGCA-3′.
D. 5′-TACCGT-3′
Review this choice. The direction is appropriate, but copying is complementary rather than identical to the template.
The principle to carry forward
Write both the bases and the strand ends; correct complementarity with reversed polarity is still an incorrect product.
Check 2
A reconstituted replication system is pulsed with labeled nucleotides and then chased. A treatment selectively blocks nick sealing after normal primer removal and DNA gap filling. Total nucleotide incorporation is initially unchanged. Which result is most directly predicted in a denaturing size assay?
A. A smaller total amount of label is incorporated during the pulse.
Review this choice. The stem specifies unchanged initial incorporation. Nick sealing can fail after nucleotides have already been incorporated.
B. Labeled short DNA segments persist instead of becoming continuous long strands.
Correct. DNA synthesis and primer processing can occur, but unsealed backbone nicks keep adjacent segments from behaving as one continuous strand after denaturation.
C. Only DNA synthesized in the 3′→5′ direction accumulates.
Review this choice. DNA polymerization remains 5′→3′; a joining defect does not reverse it.
D. The old parental strands acquire all newly added label.
Review this choice. In a replication pulse, new nucleotides are incorporated into daughter strands; the stem does not describe repair labeling of the parental template.
The principle to carry forward
A pulse-chase size shift reports continuity of labeled products, not simply whether synthesis occurred.
Check 3
A polymerase variant retains normal 5′→3′ elongation but lacks its usual 3′→5′ proofreading exonuclease activity. Separate mismatch-repair and UV-lesion excision systems are functional. Which immediate replication result is most consistent with this isolated defect?
A. More mismatched terminal nucleotides survive copying, without requiring impaired UV-lesion removal.
Correct. Proofreading removes incorrectly added nucleotides during copying. Losing that activity raises immediate error survival; later repair can still act, and a separate UV-lesion pathway need not be defective.
B. New DNA must be synthesized 3′→5′ to replace the missing exonuclease.
Review this choice. Elongation direction is fixed by addition to the growing 3′ hydroxyl and does not reverse to compensate.
C. RNA primers cannot be made even when primase is normal.
Review this choice. Primase is a separate activity and is stated to be normal. Loss of proofreading does not itself prevent primer production.
D. UV lesions must persist even if nucleotide excision repair is intact.
Review this choice. A proofreading defect does not imply failure of the separate excision pathway specified as intact.
The principle to carry forward
Match the defect to its substrate and timing; proofreading loss and damage-excision loss need not produce the same experimental pattern.
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Related lessons
- Nucleotides, Base Pairing, and DNA Stability
- Meiosis and recombination: track chromosomes, DNA and alleles
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Sources
1. AAMC: B/B Content Category 1B
2. OpenStax Biology 2e: Basics of DNA Replication
3. OpenStax Biology 2e: DNA Replication in Prokaryotes