Genetic technology: A Level Biology questions with answers
Free Genetic technology questions for A Level Biology (9700), part of Genetics & evolution. Each question has the answer and the marking points an examiner would look for.
denaturation at about 95 °C separates strands B1 annealing of primers at about 55 °C B1 extension by Taq polymerase at about 72 °C B1
3
Question 2Structured · 11 marksA* challenge
(a)
Describe how the polymerase chain reaction (PCR) amplifies a DNA sequence.
[4]
(b)
Calculate the number of copies after 25 cycles starting from one molecule, and explain why the real number is lower.
[2]
(c)
Describe how gel electrophoresis separates DNA fragments.
[3]
(d)
Discuss one benefit and one concern of genetically modified crops.
[2]
Show answers and mark scheme
Part
Answer and marking points
Marks
(a)
denature at 95 °C separating strands; anneal primers at ~55 °C; extension by Taq polymerase at 72 °C; cycle repeated, doubling each cycle
denaturation ~95 °C B1 primers anneal ~55 °C B1 Taq (thermostable) polymerase extends at ~72 °C B1 repeated cycles; exponential amplification B1
4
(b)
225 ≈ 3.4 × 107; reagents (primers, nucleotides) run out / enzyme efficiency falls
3.4 × 107B1 limited reagents/efficiency B1
2
(c)
DNA negatively charged (phosphate); moves towards anode through agarose gel; smaller fragments move further
DNA negatively charged B1 moves towards positive electrode B1 smaller fragments travel further B1
3
(d)
benefit: herbicide/pest resistance, higher yield, vitamin A (Golden Rice); concern: gene transfer to wild relatives, reduced biodiversity, superweeds
valid benefit B1 valid concern B1
2
Question 3Structured · 10 marks
DNA samples from a mother, a child and two possible fathers (F1 and F2) are cut with the same restriction enzyme, and the fragments for one locus are separated by gel electrophoresis. Fig. 3.1 shows the results (M = mother, C = child); fragment sizes in base pairs are marked on the left.
Fig. 3.1
(a)
Use Fig. 3.1 to identify the likely father. Explain your answer.
[2]
(b)
Explain why DNA fragments move through the gel towards the anode.
[1]
(c)
Explain why the smallest fragments travel furthest.
[1]
(d)
Describe how PCR could be used if only a tiny sample of DNA were available.
[3]
(e)
Explain why several different loci are usually tested, not just one.
[2]
(f)
State one other use of DNA profiling.
[1]
Show answers and mark scheme
Part
Answer and marking points
Marks
(a)
F1
F1 B1 the child’s band at 1200 bp is not in the mother’s lane and is present in F1 but not in F2 B1
2
(b)
DNA negatively charged (phosphate)
DNA is negatively charged because of its phosphate groups B1
1
(c)
less resistance from gel mesh
smaller fragments move more easily through the pores / mesh of the gel B1
Any 3 of: heat to about 95 °C to separate the strands; cool to about 55–65 °C so primers anneal; about 72 °C for Taq polymerase to extend the primers; repeat for many cycles, doubling the DNA each cycle B3
3
(e)
many people share fragments at one locus; probability of match by chance smaller
many unrelated people could share the same fragments at one locus by chance B1 testing many loci makes a chance match extremely unlikely B1
2
(f)
forensic identification / species identification / conservation breeding
forensic identification of suspects / identifying remains / checking relatedness in captive breeding B1
1
Question 4Structured · 9 marks
This question is about genetic screening and gene therapy.
(a)
Describe how DNA sequencing data can be used in genetic screening for an inherited disease.
[2]
(b)
Explain how a DNA probe locates a particular allele.
[3]
(c)
Discuss two ethical issues raised by genetic screening of embryos.
[2]
(d)
State one vector used to deliver genes in gene therapy and one risk associated with it.
[2]
Show answers and mark scheme
Part
Answer and marking points
Marks
(a)
compare person’s sequence with known disease alleles; identify mutation / carriers
the person’s DNA is sequenced or probed for known disease-causing alleles B1 identifying affected individuals or carriers B1
2
(b)
single-stranded labelled DNA complementary to target; hybridises; label detected (fluorescent/radioactive)
a short single-stranded DNA sequence complementary to the target allele B1 labelled with a fluorescent or radioactive marker B1 it hybridises with the target and is detected B1
3
(c)
selection of embryos (designer babies); discarding embryos; insurance/discrimination
embryos may be selected for non-medical traits / discarded embryos B1 information could be used for discrimination (insurance, employment) B1
2
(d)
viral vector (adenovirus/retrovirus); immune response / insertion causes cancer
viruses (e.g. adeno-associated virus / retrovirus) or liposomes B1 immune reaction / insertional mutagenesis causing cancer B1
2
Question 5Structured · 8 marks
This question is about the tools of genetic technology.
(a)
Explain the roles of restriction enzymes and DNA ligase.
[3]
(b)
Explain why reverse transcriptase is used to make the insulin gene for insertion into bacteria.
[2]
(c)
Explain why PCR is useful in forensic science.
[2]
(d)
State one use of bioinformatics in genetic technology.
[1]
Show answers and mark scheme
Part
Answer and marking points
Marks
(a)
cut at specific recognition sequences; sticky ends; ligase joins sugar–phosphate backbones
restriction enzymes cut DNA at specific base sequences B1 often producing complementary sticky ends B1 DNA ligase forms phosphodiester bonds to join fragments B1
3
(b)
mRNA from beta cells has no introns; bacteria cannot remove introns
mRNA from pancreatic β cells is used to make cDNA without introns B1 bacteria cannot splice introns out of a eukaryotic gene B1
2
(c)
amplifies tiny samples; enough DNA for profiling
PCR makes many copies from a tiny sample of DNA B1 providing enough for electrophoresis / DNA profiling B1
2
(d)
storing and comparing sequence databases
storing, searching and comparing DNA and protein sequence databases B1
1
Question 6Structured · 13 marks
This question is about gene technology.
(a)
Describe the polymerase chain reaction (PCR).
[4]
(b)
Explain why Taq polymerase is used.
[1]
(c)
Explain how gel electrophoresis separates DNA fragments.
[3]
(d)
Describe how genetic engineering produces human insulin using bacteria, including how transformed bacteria are identified.
[5]
Show answers and mark scheme
Part
Answer and marking points
Marks
(a)
denature at 95 °C; anneal primers 55 °C; extend at 72 °C with Taq polymerase; repeat, doubling DNA each cycle
heat to about 95 °C to separate the DNA strands B1 cool to about 55 °C so primers anneal B1 heat to about 72 °C so Taq polymerase adds nucleotides B1 repeat cycles, doubling the DNA each time B1
4
(b)
thermostable, not denatured at 95 °C
thermostable: not denatured at 95 °C (optimum about 72 °C), so it survives each heating cycle (ignore: from thermophilic bacteria on its own) B1
1
(c)
DNA negatively charged (phosphate); moves towards anode; smaller fragments move faster through gel
DNA is negatively charged due to phosphate groups B1 moves towards the positive electrode B1 smaller fragments move further through the gel in a given time B1
3
(d)
mRNA → cDNA by reverse transcriptase; sticky ends, restriction enzyme; ligase into plasmid; transformation; marker gene e.g. fluorescent / antibiotic resistance
insulin mRNA used to make cDNA with reverse transcriptase B1 plasmid cut with a restriction enzyme leaving sticky ends B1 DNA ligase inserts the gene into the plasmid B1 plasmids taken up by bacteria (transformation, e.g. heat shock) B1 bacteria containing the recombinant plasmid identified using a marker gene (e.g. fluorescence / antibiotic resistance) B1
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Original Cambridge-style practice questions written for mentorriyaz.in. These are not Cambridge past paper questions, and this site is not affiliated with or endorsed by Cambridge University Press & Assessment. Updated 2026-11-04. Paper generator · mentorriyaz.in