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A2 · Practice questions · PCR and gel electrophoresis

Denature, anneal, extend.

Six original Cambridge-style questions on PCR and gel electrophoresis, including copy-number calculations and reading a gel.

Original questions All questions on this page are original work, written in the Cambridge AS & A Level style. They are not from past papers. They test the same concepts and skills the syllabus rewards.
Keep these straight

Three temperatures, one direction.

01
Recall
[3 marks]

Describe what happens during each of the three stages of one cycle of the polymerase chain reaction.

  • Denaturation (about 95 °C): hydrogen bonds between complementary bases break and the two DNA strands separate. ✓
  • Annealing (about 55 °C): primers bind by complementary base pairing to the ends of the target sequence on each single strand. ✓
  • Extension (about 72 °C): Taq polymerase adds free DNA nucleotides to the primers, forming new complementary strands. ✓
02
Explain
[3 marks]

Explain why Taq polymerase is used in PCR and why the extension stage is carried out at 72 °C.

  • Taq polymerase comes from Thermus aquaticus, a bacterium living in hot springs, so it is thermostable. ✓
  • It is not denatured at 95 °C during denaturation, so it does not need to be added again every cycle (a human enzyme would be denatured). ✓
  • 72 °C is close to its optimum temperature, so it adds nucleotides quickly; this temperature is also high enough to stop primers binding to the wrong sequences. ✓
03
Calculation
[2 marks]

A forensic sample contains 50 copies of a target DNA sequence. Calculate how many copies there will be after 25 cycles of PCR, assuming every molecule is copied in every cycle. Give your answer in standard form.

  • Number of copies = 50 × 225 = 50 × 33 554 432. ✓
  • = 1 677 721 600 = 1.68 × 109 copies. ✓
04
Explain
[4 marks]

Explain how gel electrophoresis separates DNA fragments of different lengths.

  • DNA samples are placed in wells at the cathode (negative) end of an agarose gel in a buffer, and a voltage is applied. ✓
  • DNA is negatively charged because of its phosphate groups, so the fragments move towards the anode (positive electrode). ✓
  • The gel acts as a mesh or sieve; shorter fragments move through it more easily and so travel further in the same time. ✓
  • Fragments of the same length form a band, made visible by a stain or fluorescent / radioactive label; lengths are estimated by comparison with a ladder of known lengths. ✓
05
Data
[3 marks]

On a gel, ladder fragments of 1000, 500 and 250 base pairs travelled 12 mm, 22 mm and 32 mm from the wells. An unknown fragment travelled 26 mm. Estimate its length and explain your reasoning.

  • It travelled further than the 500 base pair fragment but not as far as the 250 base pair fragment, so its length is between 250 and 500 base pairs. ✓
  • Each halving of length adds 10 mm, so distance is proportional to the logarithm of length, not to length itself. ✓
  • 26 mm is 4/10 of the way from 22 to 32 mm, giving about 380 base pairs (accept 350 to 400). ✓
06
Analysis
[3 marks]

DNA from a blood stain at a crime scene and from three suspects was amplified by PCR and run on a gel. The blood stain gave bands at 900, 450 and 250 base pairs. Suspect A gave 800, 450 and 150; suspect B gave 900, 450 and 250; suspect C gave 900, 600 and 250. State and explain which suspect the evidence links to the blood stain, and why PCR was needed.

  • Suspect B: all three bands match the blood stain in length (position on the gel). ✓
  • Suspects A and C share only some bands, so the stain could not have come from them. ✓
  • The stain contained very little DNA, so PCR was used to amplify the regions to give enough DNA to see on the gel. ✓

Mark this once you have attempted all six and checked your working. It records a Practiced badge on the topic and adds a one-time bonus. Revealing the solutions alone does not count.