Question 1
Paper 1A style
What is meant by semi-conservative replication of DNA?
- Half of the DNA in a cell is copied before each division.
- Each new DNA molecule consists of one original strand and one new strand.
- Each new strand is made of sections of old and new nucleotides.
- One new DNA molecule is entirely original and the other is entirely new.
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Answer: B ;
Question 2
Paper 1A style
Why do DNA fragments move towards the positive electrode during gel electrophoresis?
- The bases are positively charged.
- The deoxyribose sugars are attracted to the positive electrode.
- The phosphate groups give DNA a negative charge.
- Hydrogen bonds between bases carry a negative charge.
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Answer: C ;
Question 3
Paper 1B style
A student set up a PCR to amplify a 400 base pair region of a plant gene. The thermal cycler was programmed as shown in the table (practice data). (a) State what happens to the DNA during stage 1. [1] (b) Explain why the temperature is lowered in stage 2. [1] (c) Explain why Taq polymerase, rather than a DNA polymerase from a human cell, is used in this reaction. [2] (d) The reaction started with 20 copies of the target region. Calculate the number of copies after the full programme, assuming each cycle doubles the number of copies. [1]
| Stage | Temperature / °C | Time / s | Number of cycles |
|---|---|---|---|
| 1 | 95 | 30 | 25 |
| 2 | 55 | 30 | 25 |
| 3 | 72 | 45 | 25 |
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- (a) hydrogen bonds (between bases) break / strands separate / DNA denatures;
- (b) allows primers to bind/anneal to (complementary sequences at the ends of) the target region / single strands;
- (c) Taq polymerase is heat stable / not denatured at 95 °C;
- (c) human DNA polymerase would be denatured (in stage 1 of the first cycle) / Taq has optimum temperature of about 72 °C so works in stage 3;
- (c) Taq does not need to be added again each cycle; Accept from Thermus aquaticus, a hot-spring bacterium, as context only if linked to heat stability
- (d) 20 × 2²⁵ = 671 088 640 / 6.7 × 10⁸ (copies);
- max 5
Question 4
Paper 1B style
DNA profiling was used in a paternity case. Three STR regions (loci) were amplified by PCR and the fragment lengths were found by gel electrophoresis. The table shows the fragment lengths, in base pairs, for a mother, her child and three men (practice data). (a) Identify which man could be the father. [1] (b) Using the data, explain why the other two men are excluded. [2] (c) Explain why several STR loci, rather than one, are used in DNA profiling. [2]
| Person | Locus A | Locus B | Locus C |
|---|---|---|---|
| Mother | 120, 150 | 210, 230 | 300, 330 |
| Child | 120, 180 | 230, 250 | 300, 310 |
| Man 1 | 150, 200 | 250, 260 | 310, 340 |
| Man 2 | 140, 180 | 220, 250 | 310, 320 |
| Man 3 | 160, 190 | 240, 270 | 330, 350 |
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- (a) Man 2;
- (b) child's non-maternal fragments are 180 (A), 250 (B) and 310 (C) / fragments not present in the mother must come from the father;
- (b) Man 1 lacks the 180 bp fragment at locus A;
- (b) Man 3 lacks the non-maternal fragments at all three loci / has none of 180, 250, 310;
- (c) one locus may be shared by chance by unrelated people / many people have the same allele at one locus;
- (c) more loci reduce the probability of a false/random match / increase reliability;
- (c) a match at all loci makes the conclusion much more certain; OWTTE
- max 5
Question 5
Paper 2A style
Outline the roles of helicase and DNA polymerase in DNA replication.
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- helicase unwinds the double helix;
- helicase breaks hydrogen bonds between (complementary) bases / separates the strands;
- (each) separated strand acts as a template;
- DNA polymerase adds/links free nucleotides to form a new strand;
- DNA polymerase follows complementary base pairing (A–T, C–G) with the template;
- max 3