Microbiology

Question 1

(a) (4 marks) From $N = N_0 2^{t/g}$: $\dfrac{N}{N_0} = 2^{t/g}$. Take $\log_2$: $\dfrac{t}{g} = \log_2\!\dfrac{N}{N_0}$, so

$g = \frac{t}{\log_2(N/N_0)} = \frac{t\ln 2}{\ln(N/N_0)}.$ (2 for derivation)

$N/N_0 = 8.0\times10^4 / 5.0\times10^3 = 16 = 2^4$, so $\log_2 16 = 4$. $g = 80/4 = \mathbf{20\ \text{min}}$. (2 for value)

(b) (4 marks) $\dfrac{dN}{dt}=\mu N \Rightarrow N = N_0 e^{\mu t}$ (solution of the linear ODE). (1) Also $N = N_0 2^{t/g} = N_0 e^{(t/g)\ln 2}$. (1) Equating exponents: $\mu t = \dfrac{t\ln2}{g} \Rightarrow \mu = \dfrac{\ln 2}{g}$. (1) $\mu = \ln 2 / 20 = 0.6931/20 = \mathbf{0.0347\ \text{min}^{-1}}$. (1)

(c) (5 marks) Curve of $\log N$ vs $t$ showing: Lag (flat), Log/exponential (steep straight line on log axis), Stationary (plateau), Death/decline (downward). (1 each phase, up to 4) The equation describes the exponential (log) phase — cells divide at constant maximal rate with unlimited nutrients. It fails because nutrient depletion, waste/toxin accumulation and space limitation slow division, so growth becomes density-dependent (stationary phase). (1)

(d) (5 marks) For the logistic model $\dfrac{dN}{dt}=\mu N(1-N/K)$, treat RHS as function of $N$: $f(N)=\mu N - \mu N^2/K$. $\dfrac{d}{dN}f = \mu - 2\mu N/K = 0 \Rightarrow N = K/2$. (2) $N_{max\ rate} = K/2 = 1.0\times10^{9}\ \text{CFU mL}^{-1}$. (1) Maximum absolute rate: $\dfrac{dN}{dt}\Big|_{N=K/2} = \mu \cdot \dfrac{K}{2}\cdot\dfrac12 = \dfrac{\mu K}{4}$ $= 0.03466 \times 2.0\times10^{9}/4 = 1.7\times10^{7}\ \text{CFU mL}^{-1}\,\text{min}^{-1}$. (2)

(e) (4 marks) Any two, each 2 marks (precaution + prevented event):

• Flaming the inoculating loop / neck of the culture tube → kills contaminating microbes on the metal/glass, preventing introduction of airborne or surface contaminants.
• Working near a Bunsen flame (updraught) or in a laminar-flow hood → convection current/filtered air carries airborne spores/bacteria away from the culture, preventing airborne contamination.
• Keeping plates lidded / inverting → prevents settling dust and condensation-borne contaminants.

Question 2

(a) (8 marks) Architecture (2): Gram-positive = thick multilayer peptidoglycan (with teichoic acids), no outer membrane. Gram-negative = thin peptidoglycan layer plus an outer lipopolysaccharide (LPS) membrane and periplasmic space. Steps (½ colour + ½ reason each ≈ 6 marks):

1. Crystal violet — stains all cells purple (enters both).
2. Iodine mordant — forms large crystal-violet–iodine (CV-I) complex trapped inside cells.
3. Alcohol decolouriser — dehydrates and shrinks the thick peptidoglycan of Gram-positives, closing pores and retaining CV-I (stay purple); in Gram-negatives alcohol dissolves the outer lipid membrane and the thin peptidoglycan cannot retain CV-I, so they lose colour. (key discriminating step)
4. Safranin counterstain — decolourised Gram-negatives take up safranin → pink/red; Gram-positives remain purple.

(b) (4 marks) (i) $\beta$-lactams inhibit transpeptidase, blocking cross-linking of new peptidoglycan. Only dividing cells actively synthesise wall material; without cross-links the growing wall weakens and the cell lyses under turgor. Dormant cells make no new wall, so nothing is disrupted → bacteriostatic tolerance, not death. (2) (ii) The Gram-negative outer LPS membrane is a permeability barrier; the drug must cross via porins, and periplasmic $\beta$-lactamases can degrade it, so less drug reaches the PBP targets → reduced efficacy. (2)

(c) (4 marks) Expected mutants $= N \times \mu_m = 1.0\times10^{9} \times 1.0\times10^{-9} = \mathbf{1}$ resistant cell (order 1). (2) Selection argument: the antibiotic kills the ~$10^9$ susceptible cells but the pre-existing resistant mutant survives and, freed from competition, proliferates to regenerate the population — a resistant clone. Thus resistance is selected, not created, by the drug → relapse/treatment failure. (2)

(d) (4 marks) (condition + consequence, ~1⅓ each)

• Conjugation: direct cell–cell contact via pilus; requires a conjugative plasmid (F/R factor) and cell contact → efficiently spreads plasmid-borne resistance, even between species; main driver of clinical multi-drug resistance.
• Transformation: uptake of naked DNA from environment; requires competence and free DNA released by lysed cells → spreads resistance where cells die and release DNA (e.g. biofilms).
• Transduction: transfer of DNA packaged in a bacteriophage; requires a phage infecting both donor and recipient (host range) → spreads resistance within phage-susceptible strains; limited by host range.

Question 3

(a) (5 marks) Lytic: phage injects DNA → immediate replication → assembly → lysis releasing progeny. (1) Lysogenic: phage DNA integrates as a prophage into host genome, replicates passively with the host, no immediate lysis. (1) Decision point: the regulatory switch controlling the phage repressor (e.g. cI/lambda repressor vs Cro) determines commitment. (2) Trigger: DNA damage / stress (e.g. UV irradiation) inactivates the repressor (SOS response), inducing excision and entry into the lytic cycle. (1)

(b) (5 marks) Flow: $\text{ssRNA genome} \xrightarrow{\text{reverse transcriptase}} \text{cDNA / dsDNA} \xrightarrow{\text{integrase}} \text{provirus (host genome)} \xrightarrow{\text{host RNA pol}} \text{viral mRNA/genomic RNA} \xrightarrow{\text{translation}} \text{proteins}.$ (3, enzymes named) Eradication difficulty: reverse transcription + integrase permanently insert a provirus into the host cell genome (a latent reservoir), which is copied with host DNA and invisible to drugs/immune system when silent; error-prone reverse transcriptase also drives rapid mutation and drug/immune escape. (2)

(c)(i) (2 marks) $V_3 = 10 \times 200^3 = 10 \times 8.0\times10^6 = \mathbf{8.0\times10^{7}}$.

(c)(ii) (4 marks) Function (2):

import math
def cycles_to_reach(V0, b, target):
    n = 0
    V = V0
    while V < target:
        V *= b
        n += 1
    return n
# closed form check: n = ceil( log(target/V0) / log(b) )

By hand (2): need $10\times200^n \ge 10^9 \Rightarrow 200^n \ge 10^8$. $\log_{10}(200)=2.301$; $n \ge 8/2.301 = 3.477$ → smallest integer $n=\mathbf{4}$. Check: $V_4 = 10\times200^4 = 10\times1.6\times10^9 = 1.6\times10^{10} \ge 10^9$. ✓ ($V_3=8\times10^7<10^9$.)

(d) (2 marks)

• Prion: misfolded protein (no nucleic acid); propagates by templating conversion of normal PrP$^{C}$ to PrP$^{Sc}$; e.g. CJD / BSE in mammals.
• Viroid: small circular naked ssRNA (no protein coat); replicates via host RNA polymerase (rolling circle); e.g. potato spindle tuber viroid in plants.

[
  {"claim":"Generation time g = 20 min from N/N0=16 over 80 min","code":"import sympy as sp\nt=80\nratio=8.0e4/5.0e3\ng=t/ (sp.log(ratio,2))\nresult = abs(float(g)-20)<1e-9"},
  {"claim":"Specific growth rate mu = ln2/20 approx 0.03466 /min","code":"mu=sp.log(2)/20\nresult = abs(float(mu)-0.034657)<1e-4"},
  {"claim":"Max-rate population is K/2 = 1.0e9 and max dN/dt = mu*K/4","code":"K=2.0e9\nmu=float(sp.log(2)/20)\nNstar=K/2\nrate=mu*K/4\nresult = abs(Nstar-1.0e9)<1 and abs(rate-1.733e7)<1e5"},
  {"claim":"Expected pre-existing resistant mutants = N*mu_m = 1","code":"N=1.0e9\nmm=1.0e-9\nresult = abs(N*mm-1.0)<1e-9"},
  {"claim":"V3 = 8.0e7 and cycles to reach 1e9 is 4","code":"V0=10;b=200\nV3=V0*b**3\nimport math\nn=math.ceil(math.log(1e9/V0)/math.log(b))\nresult = V3==80000000 and n==4"}
]