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DNA replication: one old strand in every new molecule questions

Semi-conservative replication, helicase unwinding the helix and breaking hydrogen bonds, DNA polymerase and the 5 prime to 3 prime constraint that produces leading and lagging strands, the Meselson-Stahl experiment and why its results ruled out the conservative and dispersive models, and mutations arising from copying errors.

5 original questions · 17 marks · the dna replication: one old strand in every new molecule notes · Nucleic acids, genomes and protein synthesis

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• 5 marks · Describe

  Describe how a molecule of DNA is copied semi-conservatively, naming the enzymes involved and the bonds that are broken and formed.

  Mark scheme

  1. B1 DNA helicase breaks the hydrogen bonds between complementary bases, so the two strands separate and the helix unwinds
  2. B1 each of the two separated strands acts as a template
  3. B1 free DNA nucleotides align against the exposed bases by complementary base pairing, adenine to thymine and cytosine to guanine
  4. B1 DNA polymerase catalyses the condensation reactions that join the aligned nucleotides, forming phosphodiester bonds along the new backbone
  5. A1 each daughter molecule ends up with one original strand and one newly synthesised strand, which is what semi-conservative means

• 4 marks · Explain

  Explain why one of the two new DNA strands is built continuously while the other is built as a series of short fragments.

  Mark scheme

  1. B1 DNA polymerase can only attach a nucleotide to a free 3′ hydroxyl group, so it can only build a strand in the 5′ to 3′ direction
  2. B1 the two template strands are antiparallel, so the enzyme cannot travel the same way along both of them
  3. B1 on one template it moves in the direction the replication fork is opening, producing one continuous leading strand
  4. B1 on the other it must work away from the fork, so it waits for template to be exposed and copies a short piece backwards each time, giving the Okazaki fragments of the lagging strand

• 3 marks · Calculate

  Bacteria whose DNA contains only ¹⁵N are transferred to a medium containing only ¹⁴N and allowed to divide four times. Calculate the percentage of the DNA molecules that still contain a ¹⁵N strand.

  Mark scheme

  1. M1 the two original heavy strands are never destroyed, so exactly two molecules contain one
  2. M1 after four divisions there are 2⁴ = 16 molecules altogether, so the proportion is 2 ÷ 16
  3. A1 12.5 per cent of the molecules contain a ¹⁵N strand, the remaining 87.5 per cent being entirely light

• 3 marks · Suggest

  DNA polymerase corrects most of its own copying errors, but a few substitutions survive in every cell division. Suggest three reasons why such a substitution often has no effect on the organism.

• 2 marks · State

  State the role of DNA helicase and state the role of DNA ligase in the replication of a DNA molecule.

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