StudyStore

Free MCAT-style practice, 1B

MCAT DNA Replication, Mutation and Repair Practice Questions

By StudyStore · Published 2026-10-20 · 8 original questions · about 12 min

Eight free MCAT-style practice questions on semiconservative replication, replication enzymes, Okazaki fragments, telomeres, mutations and DNA repair pathways, with full explanations.

Quick concept primer

DNA replication is semiconservative: each daughter duplex contains one old strand and one new strand. Helicase unwinds the fork, single-strand binding proteins keep the template open, primase lays down short RNA primers, and DNA polymerase extends them in the 5' to 3' direction only. The leading strand is made continuously; the lagging strand is made in Okazaki fragments, whose primers are later removed (DNA polymerase I in bacteria) and whose nicks are sealed by DNA ligase. Eukaryotic chromosome ends are protected and extended by telomerase, a reverse transcriptase carrying its own RNA template.

Mutations that add or delete a number of nucleotides not divisible by three cause frameshifts. Cells repair damage with specialized pathways: base excision repair for single damaged bases, nucleotide excision repair for bulky distortions such as UV-induced thymine dimers, mismatch repair for replication errors, and end-joining or homologous recombination for double-strand breaks.

Traps: polymerase cannot start a chain without a primer, and a tumor suppressor loss typically requires both alleles to be hit, while a proto-oncogene gain of function is dominant.

8 practice questions

Try each question before opening the explanation. Click "Show answer and explanation" to check yourself. All questions are original and written for study practice.

Question 1

E. coli grown for many generations in medium containing 15N are transferred to medium containing only 14N and allowed to divide exactly twice. What fraction of the DNA duplexes is expected to be hybrid (one 15N strand and one 14N strand)?

  1. One quarter
  2. One half
  3. All of them
  4. None of them
Show answer and explanation

Answer: B. One half

Replication is semiconservative. After one division, every duplex is 15N/14N. Each of those separates into one 15N strand and one 14N strand; each strand gets a new 14N partner in the second division. The result is four duplexes from one original: two hybrids (15N/14N) and two fully light (14N/14N), so half are hybrid.

Question 2

Three double-stranded DNA samples of equal length and concentration have melting temperatures (Tm) of 78 degrees C, 85 degrees C and 91 degrees C under identical salt conditions. Which conclusion is best supported?

  1. The three samples must have identical sequences
  2. The sample with Tm 91 degrees C has the highest proportion of A-T base pairs
  3. The sample with Tm 91 degrees C has the highest proportion of G-C base pairs
  4. Tm depends only on length
Show answer and explanation

Answer: C. The sample with Tm 91 degrees C has the highest proportion of G-C base pairs

A G-C pair has three hydrogen bonds and stronger base stacking than an A-T pair with two hydrogen bonds, so a GC-rich duplex needs more thermal energy to separate. For samples of equal length, a higher Tm indicates higher GC content. Salt concentration and length also influence Tm, but both are held constant here.

Question 3

During E. coli replication, which enzyme removes the RNA primers from the lagging strand and fills the gap with DNA?

  1. DNA gyrase
  2. DNA polymerase I
  3. Primase
  4. DNA polymerase III
Show answer and explanation

Answer: B. DNA polymerase I

DNA polymerase I has a 5' to 3' exonuclease activity that removes the RNA primer ahead of it, while its polymerase activity fills the gap with DNA. DNA polymerase III is the main replicative enzyme that extends primers, primase synthesizes the RNA primers, and gyrase (a topoisomerase) relieves supercoiling ahead of the fork. DNA ligase then seals the final nick.

Question 4

A bacterial strain has a temperature-sensitive mutation in DNA ligase. At the restrictive temperature, which pattern is expected in newly synthesized DNA?

  1. A mixture of single-stranded DNA with no fragments
  2. Accumulation of RNA primers on the leading strand only
  3. Complete failure to initiate replication at the origin
  4. Accumulation of short, unjoined fragments on the lagging strand
Show answer and explanation

Answer: D. Accumulation of short, unjoined fragments on the lagging strand

Ligase seals the nicks between adjacent Okazaki fragments after primer removal and gap filling. Without ligase, the lagging strand remains as small pieces, while the continuously made leading strand has few breaks. Initiation does not require ligase (choice C is false), and the leading strand needs only a single primer near the origin.

Question 5

Cells from a culture lacking telomerase are passaged many times. Which change is expected?

  1. Chromosome ends gradually shorten, because the final RNA primer at the end of the lagging strand cannot be replaced with DNA
  2. Okazaki fragments disappear
  3. Chromosome ends lengthen, because primers cannot be removed
  4. The cells become immortal
Show answer and explanation

Answer: A. Chromosome ends gradually shorten, because the final RNA primer at the end of the lagging strand cannot be replaced with DNA

After the final RNA primer is removed at the end of a linear chromosome, there is no upstream 3' end for polymerase to extend, so a short single-stranded stretch is left unreplicated each cycle. Telomerase extends the 3' end using its internal RNA template, allowing the lagging strand to be completed. Without it, telomeres progressively shorten, eventually triggering senescence or apoptosis, so choice D is wrong.

Question 6

Which of the following mutations in the coding region of a gene is most likely to have the most severe effect on the protein?

  1. A point mutation that changes a codon to another codon for the same amino acid
  2. Deletion of three consecutive nucleotides in the middle of the coding sequence
  3. A point mutation that changes a leucine codon to an isoleucine codon
  4. Deletion of a single nucleotide near the start of the coding sequence
Show answer and explanation

Answer: D. Deletion of a single nucleotide near the start of the coding sequence

Deleting one nucleotide shifts the reading frame for every downstream codon, producing a different amino acid sequence and usually a premature stop codon. Choice A is silent, choice C is a conservative missense change, and choice B removes one amino acid without changing the frame (an in-frame deletion).

Question 7

Which pairing of DNA lesion and primary repair pathway is correct?

  1. A double-strand break is repaired by mismatch repair
  2. A replication mismatch is repaired by non-homologous end joining
  3. A thymine dimer caused by UV light is repaired by base excision repair
  4. A uracil arising from spontaneous deamination of cytosine is removed by a DNA glycosylase in base excision repair
Show answer and explanation

Answer: D. A uracil arising from spontaneous deamination of cytosine is removed by a DNA glycosylase in base excision repair

Uracil does not belong in DNA, and a uracil DNA glycosylase excises the base, leaving an abasic site that is processed and filled in by base excision repair. Thymine dimers distort the helix and are removed by nucleotide excision repair, and defects in that pathway cause xeroderma pigmentosum. Replication mismatches are corrected by mismatch repair, while double-strand breaks are handled by end joining or homologous recombination.

Question 8

Which statement about cancer-related genes is most accurate?

  1. Tumor suppressors promote cell cycle progression when mutated
  2. Both oncogenes and tumor suppressors require loss of function in both alleles
  3. Proto-oncogenes are only found in cancer cells
  4. One activating mutation in a single RAS allele can drive transformation, whereas a tumor suppressor typically has to be lost from both copies
Show answer and explanation

Answer: D. One activating mutation in a single RAS allele can drive transformation, whereas a tumor suppressor typically has to be lost from both copies

Oncogenes arise from proto-oncogenes through gain-of-function changes, which act dominantly at the cellular level. Tumor suppressors such as p53 and Rb restrain cell division or promote repair and apoptosis, and generally both copies must be inactivated before the brake is lost (the two-hit model). Proto-oncogenes are normal genes present in every cell.

Keep practicing

Written by StudyStore from standard scientific facts as a study aid. These are not official exam questions and are not affiliated with or endorsed by any testing organization. Always check values against the reference provided with your exam or course.