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Quiz: The Central Dogma — DNA Replication and Protein Synthesis

Test your understanding of DNA replication, transcription, mRNA processing, and translation with these review questions.

1. What is the primary function of DNA ligase during DNA replication?

  1. Unwinds the double helix ahead of the replication fork
  2. Synthesizes short RNA primers to initiate new strands
  3. Removes RNA primers and replaces them with DNA
  4. Seals nicks in the DNA backbone by joining adjacent Okazaki fragments with phosphodiester bonds
Show Answer

The correct answer is D. After DNA polymerase I removes the RNA primers from Okazaki fragments and fills the gaps with DNA, small nicks remain in the sugar-phosphate backbone between adjacent fragments. DNA ligase seals these nicks by forming phosphodiester bonds, joining the fragments into one continuous lagging strand.

Concept Tested: DNA replication enzymes (ligase)

2. Which three-nucleotide sequence on mRNA signals the start of translation and codes for methionine?

  1. UAA
  2. AUG
  3. UGA
  4. UAG
Show Answer

The correct answer is B. AUG is the universal start codon that both signals the beginning of translation and codes for methionine, the first amino acid incorporated into nearly every polypeptide. The first AUG encountered by the ribosome also establishes the reading frame for the rest of the message. UAA, UGA, and UAG are stop codons that do not code for any amino acid.

Concept Tested: Start and stop codons

3. Why is DNA replication described as semiconservative?

  1. Each daughter DNA molecule contains one original parental strand and one newly synthesized strand
  2. Both strands of each daughter molecule are entirely new
  3. Both strands of each daughter molecule are entirely original
  4. Replication produces one fully parental molecule and one fully new molecule
Show Answer

The correct answer is A. The Meselson-Stahl experiment demonstrated that each daughter DNA molecule retains one original parental strand paired with one newly synthesized strand, rather than conserving both original strands together (conservative model) or mixing old and new segments within both strands (dispersive model).

Concept Tested: Semiconservative replication

4. Why must the lagging strand be synthesized discontinuously as Okazaki fragments, rather than continuously like the leading strand?

  1. Because the lagging strand template contains no origins of replication
  2. Because ligase cannot act on the lagging strand until replication is complete
  3. Because DNA polymerase synthesizes only in the 5'→3' direction, so on the lagging strand (whose template runs 5'→3' relative to the fork) new DNA must be made in short segments moving away from the fork, each requiring its own primer
  4. Because the lagging strand is transcribed rather than replicated
Show Answer

The correct answer is C. DNA polymerase can only add nucleotides in the 5'→3' direction. On the leading strand, this direction matches the direction the replication fork opens, so synthesis is continuous. On the lagging strand, the template's orientation forces new DNA to be synthesized in short segments moving away from the fork, each requiring a new RNA primer, producing the discontinuous Okazaki fragments.

Concept Tested: Leading and lagging strands

5. Why does the redundancy (degeneracy) of the genetic code help protect against the effects of some DNA mutations?

  1. Because redundancy allows DNA polymerase to skip damaged bases entirely
  2. Because most amino acids are specified by more than one codon, a mutation changing a codon's third position often still specifies the same amino acid, producing a silent mutation with no effect on the protein
  3. Because redundant codons are never actually translated by ribosomes
  4. Because redundancy eliminates the need for stop codons
Show Answer

The correct answer is B. Because 64 possible codons encode only 20 amino acids, many amino acids are specified by multiple synonymous codons that often differ only in their third nucleotide position (sometimes called the "wobble" position). A mutation at this position frequently still codes for the identical amino acid, resulting in a silent mutation that does not alter the resulting protein sequence.

Concept Tested: Genetic code redundancy

6. If the template strand of DNA reads 3'-TACGGATCCAATT-5', what is the sequence of the resulting mRNA?

  1. 5'-TACGGATCCAATT-3'
  2. 3'-AUGCCUAGGUUAA-5'
  3. 5'-UACCCUAGGUUAA-3'
  4. 5'-AUGCCUAGGUUAA-3'
Show Answer

The correct answer is D. RNA polymerase reads the template strand 3'→5' and synthesizes a complementary mRNA strand 5'→3', substituting uracil for thymine. Pairing each template base with its complement (T→A, A→U, C→G, G→C) in the correct 5'→3' orientation for the new strand yields 5'-AUGCCUAGGUUAA-3', which conveniently begins with the start codon AUG and ends with the stop codon UAA.

Concept Tested: Transcription and mRNA sequence prediction

7. A mutation inactivates the telomerase enzyme in a particular cell lineage. What is the most likely long-term consequence for that cell's chromosomes across successive divisions?

  1. Telomeres will progressively shorten with each cell division until the cell reaches senescence (the Hayflick limit)
  2. Telomeres will lengthen with each division due to compensatory DNA repair
  3. The cell's chromosomes will immediately fuse end-to-end
  4. DNA replication will halt entirely at the very first division
Show Answer

The correct answer is A. Telomerase normally counteracts the end-replication problem by adding repetitive TTAGGG sequences to chromosome ends. Without telomerase activity, telomeres shorten a little with each round of replication because the lagging strand template cannot be fully copied at the very end, eventually triggering replicative senescence once telomeres become critically short — the basis of the Hayflick limit.

Concept Tested: Telomeres and telomerase

8. A drug specifically inhibits topoisomerase in actively dividing cells. What is the most likely immediate effect on DNA replication?

  1. Okazaki fragments would no longer require primers
  2. The leading strand would be synthesized in the 3'→5' direction instead
  3. Torsional strain ahead of the replication fork would not be relieved, stalling or blocking fork progression
  4. RNA polymerase activity would increase to compensate
Show Answer

The correct answer is C. Topoisomerase relieves the torsional strain (supercoiling) that builds up in the DNA ahead of the replication fork as helicase unwinds the double helix. Without this enzyme, the accumulating strain would make it increasingly difficult for helicase to continue unwinding the DNA, stalling or halting the progression of the replication fork.

Concept Tested: Replication fork enzymes (topoisomerase)

9. Why could an error in a single aminoacyl-tRNA synthetase be more damaging to a cell than a single point mutation in one DNA gene?

  1. A point mutation in DNA affects every protein in the cell simultaneously, while a synthetase error affects only one gene's product
  2. A synthetase error affects every protein made using that particular amino acid across the entire proteome, since the same enzyme charges every copy of that tRNA, whereas a DNA point mutation typically affects only the product of the single gene involved
  3. Both types of errors have identical, gene-limited effects on the proteome
  4. A synthetase error only affects ribosomal proteins, not other cellular proteins
Show Answer

The correct answer is B. Each of the 20 aminoacyl-tRNA synthetases charges every molecule of its corresponding tRNA type throughout the cell. If a synthetase mischarges its tRNA, that error is repeated across every protein synthesized that incorporates that amino acid at that codon, potentially affecting thousands of different proteins. A DNA point mutation, in contrast, is generally limited to the single gene in which it occurs.

Concept Tested: Aminoacyl-tRNA synthetases and translation fidelity

10. What allows eukaryotic cells to produce far more distinct proteins than the number of genes in their genome, a capability prokaryotic cells lack?

  1. Prokaryotic gene expression produces more protein diversity per gene because prokaryotes have more introns
  2. Eukaryotic and prokaryotic gene expression produce identical numbers of protein variants per gene
  3. Alternative splicing has no effect on the total number of distinct proteins a genome can produce
  4. Alternative splicing allows a single pre-mRNA to be spliced in multiple ways, including or excluding different exons to generate multiple distinct mature mRNAs and protein variants from one gene, an expansion mechanism unavailable to prokaryotes, which lack introns and spliceosomes
Show Answer

The correct answer is D. Eukaryotic genes are split into exons and introns, and the spliceosome can join exons in different combinations through alternative splicing, generating multiple distinct mature mRNAs, and therefore multiple protein variants, from a single gene. Prokaryotic genes generally lack introns and the spliceosome machinery entirely, so their protein diversity is more directly tied to the number of genes they carry.

Concept Tested: Alternative splicing and protein diversity