Genomics

Question 1

(a) (6 marks)

• Genome = the complete set of DNA (all genes + non-coding sequence) of an organism. (2)
• Transcriptome = the complete set of RNA transcripts (mRNA etc.) present in a cell/tissue at a given time/condition. (2)
• Proteome = the complete set of proteins expressed. Transcriptome maps back to genomic loci (subset of transcribed genome), but proteome is not a simple subset because alternative splicing, RNA editing, and post-translational modifications mean one gene/transcript can yield many protein forms — the mapping is one-to-many, not one-to-one. (2)

(b) (6 marks) Poisson: expected reads covering a base $=c=NL/G$. Number of reads whose start places them over a given base follows Poisson mean $c$; probability a base is covered by zero reads $=e^{-c}$. (2)

• Expected uncovered bases $=G\,e^{-c}$; expected fraction of gaps $=e^{-c}$. (2)
• $c = NL/G = (2.0\times10^{5})(150)/(5.0\times10^{6}) = 3.0\times10^{7}/5.0\times10^{6} = 6.0$. (1)
• Uncovered fraction $=e^{-6}\approx 2.48\times10^{-3}$ (~0.25%, ≈12 400 bp). (1)

(c) (5 marks) Geometric: $P(X=k)=(1-p)^{k-1}p$ (first termination at position $k$). (1)

• $P(X=30)=(0.98)^{29}(0.02)$. $(0.98)^{29}=e^{29\ln0.98}=e^{-0.5857}=0.5568$; $\times0.02=0.01114$. (2)
• Mean $=1/p = 1/0.02 = 50$ bases. (1)
• Random termination at each position generates a population of fragments of every possible length; separated by size (electrophoresis), each length reveals the terminating ddNTP base → sequence read as a ladder. (1)

(d) (3 marks)

• Read length: Sanger long (~500–1000 bp) > individual NGS reads (~100–300 bp). (1)
• Throughput/cost: NGS massively parallel (millions–billions of reads simultaneously), far lower per-base cost. (1)
• WGS needs enormous throughput → NGS; confirming one known point mutation needs high accuracy on a short region only → Sanger remains gold-standard/validation. (1)

Question 2

(a) (3 marks) SNP = a single base-pair position in the genome that varies among individuals in a population (2). Conventionally called a polymorphism when the minor allele frequency (MAF) ≥1%; below that it is regarded as a rare variant/mutation (1).

(b) (9 marks) Genotype frequencies under HWE: $AA=p^{2}$, $Aa=2pq$, $aa=q^{2}$. (1) Allele counting (2N = 2000 alleles):

• $A$ count $=2(640)+320 = 1600$; $\hat p = 1600/2000 = 0.80$. (2)
• $\hat q = 0.20$. (1) Expected counts (×1000):
• $AA=p^2\cdot1000=0.64\cdot1000=640$
• $Aa=2pq\cdot1000=0.32\cdot1000=320$
• $aa=q^2\cdot1000=0.04\cdot1000=40$ (2) Chi-square: observed = expected exactly, so $\chi^2=\sum (O-E)^2/E = 0$. (2) $0<3.84$ → HWE not rejected; population is in equilibrium at this locus. (1)

(c) (5 marks) With $10^6$ independent tests at $\alpha=0.05$, expected false positives $=0.05\times10^6=5\times10^4$ — far too many. (2) Bonferroni: threshold $=\alpha/m = 0.05/10^{6}=5\times10^{-8}$ (the standard genome-wide significance line). (2) Expected false positives using this threshold among null SNPs $=m\times(\alpha/m)=\alpha=0.05$ (i.e. controls family-wise error at 0.05, <1 expected). (1)

(d) (5 marks) Causal chain (award up to 5):

1. GWAS/pharmacogenomic association identifies a SNP in/near a gene affecting drug response (e.g. CYP450 metabolizer variant). (1)
2. Genotype alters enzyme activity → poor vs ultra-rapid metabolizer. (1)
3. This changes plasma drug concentration/half-life for a standard dose. (1)
4. Clinician genotypes the patient before prescribing. (1)
5. Dose (or drug choice) is tailored to genotype to maximise efficacy and avoid toxicity — the essence of precision/personalized medicine. (1)

Question 3

(a) (6 marks) — any 4 stages, 1.5 each:

1. Repeat masking (RepeatMasker) — removes repetitive/transposon DNA using repeat libraries.
2. Structural annotation / gene prediction — ab initio HMM gene finders + alignment of RNA-seq/EST (transcript) and protein evidence to define exon–intron structure.
3. Non-coding feature annotation — tRNA/rRNA/ncRNA via specialized tools (evidence: covariance models/homology).
4. Functional annotation — BLAST/HMMER against protein databases (UniProt, Pfam), assign GO terms/enzyme function (evidence: homology to characterized proteins).

(b) (7 marks)

def gc_content(seq):
    seq = seq.upper()
    gc = sum(1 for b in seq if b in "GC")
    return gc / len(seq) if seq else 0.0        # (2)
 
def find_orfs(seq):
    seq = seq.upper()
    stops = {"TAA", "TAG", "TGA"}
    orfs = []
    n = len(seq)
    for frame in range(3):                       # 3 forward frames
        i = frame
        while i < n - 2:
            if seq[i:i+3] == "ATG":              # start
                j = i
                while j < n - 2:
                    codon = seq[j:j+3]
                    if codon in stops:
                        orfs.append((i, j+3))    # start,stop coords
                        break
                    j += 3
            i += 3
    return orfs                                  # (4)

Complexity: each frame scanned once; inner scan does not revisit already-passed positions more than a constant number of times per codon → overall O(n) (linear in sequence length). (1)

(c) (5 marks) Non-coding DNA is not "junk": functions include promoters, enhancers/silencers, insulators, origins of replication, telomeres/centromeres, and genes for functional RNAs (tRNA, rRNA, miRNA, lncRNA); ENCODE reported the majority of the genome is transcribed or shows biochemical activity (DNase hypersensitivity, TF binding, histone marks). (3) Comparative genomics gives independent evidence: sequences conserved across evolutionarily distant species (evolutionary constraint / low substitution rate) are inferred to be functional because purifying selection preserves them — conserved non-coding elements corroborate biochemical activity without relying on the same assays. (2)

[
  {"claim":"Coverage c = NL/G = 6 and uncovered fraction = e^-6",
   "code":"G=5e6; N=2e5; L=150; c=N*L/G; frac=exp(-c); result = (abs(c-6)<1e-9) and (abs(frac-exp(-6))<1e-12)"},
  {"claim":"Sanger geometric: P(X=30)=0.98**29*0.02 and mean=50",
   "code":"p=Rational(2,100); Pk=(1-p)**29*p; mean=1/p; result = (abs(float(Pk)-0.011135)<1e-4) and (mean==50)"},
  {"claim":"HWE chi-square is 0 for AA640 Aa320 aa40 with p=0.8",
   "code":"pA=Rational(1600,2000); qa=1-pA; obs=[640,320,40]; exp=[pA**2*1000,2*pA*qa*1000,qa**2*1000]; chi=sum((o-e)**2/e for o,e in zip(obs,exp)); result = (chi==0)"},
  {"claim":"Bonferroni threshold 0.05/1e6 = 5e-8 and expected false positives = 0.05",
   "code":"m=10**6; alpha=Rational(5,100); thr=alpha/m; fp=m*thr; result = (thr==Rational(5,100000000)) and (fp==alpha)"}
]