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9700 · 19.1

Principles of genetic technology — practice questions

Practice and worked examples for 9700 Principles of genetic technology. Short previews only — attempt the full question in MarkScheme against the official scheme.

Worked example 1

Explain how sticky ends are produced by restriction enzymes and why they are important in forming recombinant DNA.

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  1. Restriction endonucleases recognise specific, short, palindromic nucleotide sequences (known as recognition sites) on a double-stranded DNA molecule.
  2. They catalyse the hydrolysis of the phosphodiester bonds within the DNA backbone. Often, this cut is made in an asymmetrical fashion across the two strands, leading to overhanging single-stranded sections at each end of the DNA fragment. These overhangs are called sticky ends.
  3. For example, the restriction enzyme EcoRI recognises the sequence GAATTC and cuts between G and A on both strands, leaving single-stranded AATT overhangs.
  4. Importance: Sticky ends are crucial in recombinant DNA technology because they are complementary to each other. A desired gene fragment and a vector (e.g., plasmid) that have both been cut with the same restriction enzyme will possess complementary sticky ends.
  5. This complementarity allows for hydrogen bonding to occur between the bases of the sticky ends of the gene fragment and the vector, temporarily annealing them together.
  6. Subsequently, DNA ligase forms permanent phosphodiester bonds in the sugar-phosphate backbone, covalently joining the gene into the vector to create a stable recombinant DNA molecule.

Worked example 2

A scientist starts a PCR reaction with 15 molecules of a target double-stranded DNA sequence. Assuming 100% efficiency, calculate the number of copies of the target sequence after 25 cycles.

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The number of DNA molecules produced by PCR increases exponentially. The formula to calculate the number of copies (N) after a certain number of cycles (n) is:

N=N0×2nN = N_0 \times 2^n

Where N0N_0 is the initial number of DNA molecules.

Step 1: Identify the given values.

  • Initial number of DNA molecules (N0N_0) = 15
  • Number of PCR cycles (n) = 25

Step 2: Substitute the values into the formula. N=15×225N = 15 \times 2^{25}

Step 3: Calculate the value of 2252^{25}. 225=33,554,4322^{25} = 33,554,432

Step 4: Calculate the final number of DNA copies. N=15×33,554,432=503,316,480N = 15 \times 33,554,432 = 503,316,480

Answer: After 25 cycles, a theoretical maximum of 503,316,480 copies of the target DNA sequence will be produced.