Genomics
Level 4 (Application: Novel Problems)
Time limit: 60 minutes Total marks: 50
Question 1 (10 marks)
A research team sequences a 100 bp fragment using the Sanger method. They set up four reaction tubes and run capillary electrophoresis producing a chromatogram.
(a) In the reaction, the ratio of normal dNTP to ddATP is 100:1 in the "A" tube. Explain qualitatively what happens to fragment length distribution if this ratio is increased to 1000:1, and what problem this would cause for reading a 100 bp fragment. (4)
(b) A ddNTP lacks the 3′-OH group. Using this fact, explain precisely why chain elongation stops after a ddNTP is incorporated. (3)
(c) The team wishes to sequence 3 billion base pairs (a whole human genome). Explain two specific reasons why Sanger sequencing is impractical here and why NGS is preferred. (3)
Question 2 (12 marks)
A clinician is investigating a child with an undiagnosed inherited disorder. Sequencing options: (i) whole-genome sequencing (WGS), (ii) whole-exome sequencing (WES).
(a) The exome represents roughly 1.5% of the genome but contains ~85% of known disease-causing mutations. Calculate how many megabases (Mb) the exome represents, given a genome of 3200 Mb. Show working. (3)
(b) Using the data in (a), justify two advantages and one disadvantage of choosing WES over WGS for this case. (5)
(c) The child's variant is later found in a deep intronic regulatory region. Explain why WES would have missed this variant, and what ENCODE findings tell us about the importance of such non-coding regions. (4)
Question 3 (10 marks)
A GWAS is conducted on 20,000 individuals to study Type 2 diabetes. Researchers test 1,000,000 SNPs for association.
(a) Using a standard significance threshold of , calculate how many SNPs would be expected to appear "significant" by chance alone if there were no true associations. (2)
(b) Explain why researchers instead use a genome-wide significance threshold of . Relate your answer to your result in (a). (4)
(c) A SNP shows strong association with the disease but lies in a non-coding region and is not itself causal. Explain the concept that allows this SNP to still be a useful marker. (4)
Question 4 (10 marks)
Two patients are prescribed the same drug, which is metabolised by the enzyme CYP2C9. Patient A carries two normal-function alleles; Patient B carries two reduced-function alleles.
(a) Predict and explain the difference in drug metabolism rate and blood drug concentration between the two patients at a standard dose. (4)
(b) Explain, using the term pharmacogenomics, how genotyping before prescribing could improve outcomes for Patient B. (3)
(c) Describe how this scenario illustrates the shift from "one-size-fits-all" medicine to precision/personalized medicine. (3)
Question 5 (8 marks)
Comparative genomics reveals that humans and mice share ~85% of protein-coding sequences, and that a particular gene region is nearly identical ("conserved") across humans, mice, and fish.
(a) Explain what strong sequence conservation across distantly related species suggests about the functional importance of that region. (3)
(b) Genome annotation must identify genes in a newly sequenced species. Describe two methods/lines of evidence used to identify a protein-coding gene during annotation. (3)
(c) Distinguish clearly between the terms genome, transcriptome, and proteome in a single cell. (2)
Answer keyMark scheme & solutions
Question 1 (10 marks)
(a) (4 marks)
- Higher dNTP:ddNTP ratio (1000:1) means ddATP is incorporated less frequently → chains terminate less often early → produces predominantly longer fragments (1)
- Fewer short fragments generated (1)
- Problem: gaps in the "ladder" — some positions near the start won't be represented, so early bases can't be read reliably (1)
- Reading accuracy at short lengths / short-read resolution is lost, making the 100 bp read incomplete at the beginning (1)
(b) (3 marks)
- Chain elongation requires formation of a phosphodiester bond between the 3′-OH of the last nucleotide and the 5′-phosphate of the incoming nucleotide (1)
- ddNTP lacks the 3′-OH group (1)
- Therefore no bond can form with the next nucleotide → polymerisation terminates at that point (1)
(c) (3 marks) — any two, 1.5 each (round to 3):
- Sanger reads only ~500–1000 bp per reaction → 3 billion bp needs enormous numbers of separate reactions = slow/labour-intensive (1.5)
- Cost per base far higher; NGS is massively parallel (millions of reads simultaneously) making it faster and cheaper (1.5)
- (Accept: throughput of NGS enables whole genome in days, Sanger would take years)
Question 2 (12 marks)
(a) (3 marks)
- Exome = 1.5% of 3200 Mb (1)
- Mb (2 for correct calculation & answer)
- Answer: 48 Mb
(b) (5 marks) Advantages (2 × 2 = 4, capped appropriately):
- WES targets only ~1.5% (48 Mb) → far less data, lower cost, faster analysis (2)
- Exome contains ~85% of known disease-causing mutations → high diagnostic yield for coding-region disorders (2) Disadvantage (1):
- WES misses ~15% of disease variants located in non-coding/regulatory/intronic regions (1)
(c) (4 marks)
- WES only captures exons (coding regions); a deep intronic variant lies outside the captured/enriched regions so it is not sequenced (2)
- ENCODE showed a large fraction of non-coding DNA is biochemically functional — regulatory elements, enhancers, promoters, transcription-factor binding sites (1)
- Therefore intronic/regulatory variants can affect gene expression and cause disease, justifying WGS in such cases (1)
Question 3 (10 marks)
(a) (2 marks)
- Expected false positives = (1)
- SNPs by chance (1)
(b) (4 marks)
- Testing 1,000,000 SNPs = 1,000,000 statistical tests (multiple testing problem) (1)
- At , 50,000 appear significant purely by chance → unusable (1)
- A stricter threshold corrects for multiple comparisons (Bonferroni-type: ) (1)
- This controls the false-positive rate so that "significant" hits are very likely true associations (1)
(c) (4 marks)
- The concept is linkage disequilibrium (1)
- The non-causal SNP is physically close on the chromosome to the true causal variant and is inherited together with it (non-random association) (2)
- So the marker SNP "tags" the causal variant and can be used to locate the disease-associated region even though it isn't itself functional (1)
Question 4 (10 marks)
(a) (4 marks)
- Patient A (normal alleles): normal CYP2C9 activity → normal/faster metabolism → drug cleared at expected rate, normal blood concentration (2)
- Patient B (reduced-function alleles): reduced enzyme activity → slower metabolism → drug accumulates → higher blood concentration, greater risk of toxicity/side effects at standard dose (2)
(b) (3 marks)
- Pharmacogenomics = study of how an individual's genome affects drug response (1)
- Genotyping Patient B beforehand reveals reduced-function alleles (1)
- Allows dose reduction or alternative drug → avoids toxicity and improves safety/efficacy (1)
(c) (3 marks)
- Traditional medicine gives the same dose to all patients regardless of genotype (1)
- Precision medicine tailors drug/dose to the individual's genetic makeup (1)
- Improves efficacy and reduces adverse reactions — treatment matched to the patient (1)
Question 5 (8 marks)
(a) (3 marks)
- Conserved sequences change little over evolutionary time because mutations there are selected against (2)
- This implies the region has an essential function (e.g. vital protein/regulatory role) that must be preserved (1)
(b) (3 marks) — any two:
- Look for open reading frames (ORFs) — start codon, coding sequence, stop codon (1.5)
- Match to known genes/proteins via homology/database (BLAST) alignment (1.5)
- (Accept: presence of promoters/splice sites; evidence from transcriptome/cDNA/RNA-seq showing the region is transcribed)
(c) (2 marks)
- Genome = complete set of DNA/genes (same in all cells) (⅔)
- Transcriptome = all RNA molecules transcribed (varies by cell/condition) (⅔)
- Proteome = complete set of proteins expressed (varies by cell/condition) (⅔)
[
{"claim": "Exome size = 1.5% of 3200 Mb = 48 Mb", "code": "result = (0.015*3200 == 48)"},
{"claim": "Expected chance-significant SNPs = 0.05 * 1,000,000 = 50000", "code": "result = (0.05*1000000 == 50000)"},
{"claim": "Bonferroni threshold 0.05/1e6 = 5e-8", "code": "result = (Rational(5,100)/1000000 == Rational(5,100000000))"},
{"claim": "Genome-wide threshold equals 5e-8", "code": "result = (5e-8 == 0.05/1000000)"}
]