Microbial Genetics (I)
2 sub-topics · Pages 502–530
1. Introduction
Microbial genetics explores how genetic information is stored, replicated, expressed, and varied in microorganisms. Bacteria replicate their single circular chromosome from a defined origin of replication (oriC) with remarkable accuracy — one error per 10⁹ base pairs after repair. Yet their enormous population sizes and short generation times mean that even rare mutations are rapidly expressed and selected, driving the rapid evolutionary adaptation that underlies antibiotic resistance, host range expansion, and metabolic specialisation.
✏️ Fill in the Blank
1. The segment of DNA that encodes a functional product (protein or structural RNA) is called a _______.
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Gene2. The sigma (σ) factor of prokaryotic RNA polymerase recognises specific _______ sequences upstream of genes, directing transcription initiation.
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Promoter3. In DNA replication, the enzyme that synthesises short RNA primers needed for DNA polymerase III to start elongation is called _______.
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Primase4. A segment of DNA that can move from one location in the genome to another is called a _______.
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Transposon🔘 Multiple Choice
1. The enzyme responsible for synthesising a new DNA strand using an existing strand as a template is called:
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Correct: C) DNA polymerase2. In prokaryotic DNA replication, which enzyme removes RNA primers and replaces them with DNA?
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Correct: B) DNA polymerase I3. In prokaryotic transcription, the -10 and -35 consensus sequences of a promoter are recognised by:
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Correct: B) The σ (sigma) factor of RNA polymerase4. The Shine-Dalgarno sequence in prokaryotic mRNA is important for:
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Correct: B) 30S ribosomal subunit binding to position the start codon for translation initiation5. The genetic code is described as 'degenerate'. This means:
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Correct: B) Most amino acids can be encoded by more than one codon6. Which of the following describes the Ames test?
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Correct: B) A bacterial reverse mutation assay to screen chemicals for mutagenic potential7. SOS repair in bacteria is induced by:
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Correct: B) Single-stranded DNA accumulation due to extensive DNA damage, which activates RecA8. Restriction-modification systems in bacteria function to:
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Correct: B) Cleave foreign DNA at specific recognition sequences while protecting self-DNA by methylation💬 Open-Ended Questions
1. A single base-pair substitution in a gene results in a premature stop codon. Explain the molecular consequences for the protein product and predict how this might affect the bacterium's phenotype if the gene codes for a key metabolic enzyme.
Hint / Guidance
Nonsense mutation → premature stop codon → ribosome terminates prematurely → truncated polypeptide lacking C-terminal domain → loss of enzyme active site or structural integrity → enzyme non-functional → metabolic pathway blocked → auxotrophy or inability to catabolise substrate → bacterium outcompeted in mixed culture; if gene essential → lethal mutation. Nonsense-mediated mRNA decay may also degrade the truncated transcript.2. Explain the SOS response in E. coli. What triggers it, what genes are induced, and what are the consequences for bacterial mutation rate?
Hint / Guidance
Trigger: DNA damage (UV, alkylating agents) → single-stranded DNA exposed → RecA binds ssDNA → activated RecA cleaves LexA repressor. LexA normally represses >40 SOS genes including: recA, uvrA-C (nucleotide excision repair), sulA (cell division inhibitor), umuC/umuD (error-prone polymerase V). Consequence: transient cell division arrest (sulA inhibits FtsZ); error-prone repair (DNA pol V inserts random nucleotides across damaged sites — translesion synthesis). Mutation rate 100–1000× elevated during SOS. Adaptive significance: generates diversity for selection, allowing some cells to survive novel challenges; cost: increased mutation rate may generate deleterious mutations.3. Describe the structural basis of DNA double helix and explain how Watson-Crick base pairing rules ensure accurate replication.
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Structure: two antiparallel polynucleotide strands; deoxyribose-phosphate backbone outside; nitrogenous bases inside stacking (van der Waals) and hydrogen bonding. Watson-Crick pairing: A-T (2 H-bonds); G-C (3 H-bonds); complementarity ensures each strand is template for the other. Accuracy: DNA pol III base-selects complementary nucleotide via H-bonding geometry (wrong base doesn't fit correctly); 3'→5' exonuclease proofreading removes mismatches (error rate 10⁻⁶→10⁻⁹ with mismatch repair). Higher G-C content → higher Tm (more H-bonds) → thermophile DNA stability partially conferred by elevated G-C content (also reverse gyrase/supercoiling).4. What is horizontal gene transfer (HGT) and why is it ecologically and medically important? How does it challenge traditional notions of the phylogenetic tree of life?
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HGT: transfer of genetic material between organisms other than by parent-to-offspring (vertical) transmission. Mechanisms: transformation, transduction, conjugation. Medical importance: spread of antibiotic resistance (R-plasmids), virulence factors (pathogenicity islands), metabolic capabilities. Ecological: Prochlorococcus receives ~20% of genes from HGT; metabolic flexibility of soil bacteria; syntrophic partnerships. Challenge to phylogenetics: different genes in same organism may have different evolutionary histories; genome is mosaic; no single universal 'species tree'; network rather than bifurcating tree of life. 16S rRNA relatively resistant to HGT (due to complex function requiring co-evolution with many partners) so still useful as phylogenetic marker.5. Explain what a silent (synonymous) mutation is. Why are silent mutations generally neutral, but can sometimes have functional consequences?
Hint / Guidance
Silent mutation: nucleotide change in a codon that does not change the amino acid (e.g., GCA→GCG, both Ala; genetic code degeneracy). Generally neutral: protein sequence unchanged; no selective pressure. Sometimes functional: (1) Codon usage bias — rare codons slow ribosome translation; if rate-limiting for protein production, rare codon introduction reduces protein output; (2) mRNA secondary structure — synonymous change alters mRNA folding → affects translation efficiency, stability, co-translational folding; (3) Splicing signals — in eukaryotes, synonymous changes near exon-splicing enhancers can disrupt splicing; (4) Transcription factor binding sites within coding regions can be disrupted. Examples: silent mutation in MDR1 gene changes P-glycoprotein folding; silent SMN1 mutations affect splicing and spinal muscular atrophy severity.6. Explain the mechanism of CRISPR-Cas9 gene editing. How was this system discovered and what are its key molecular components?
Hint / Guidance
Discovery: Ishino (1987) noticed repeats; Mojica/Barrangou (2000s) linked to adaptive immunity against phage; Doudna/Charpentier (2012) programmed Cas9 for genome editing (Nobel Prize 2020). Components: Cas9 nuclease (RuvC + HNH domains cut each DNA strand); single guide RNA (sgRNA = crRNA + tracrRNA fusion); 20-nt spacer guides Cas9 to target; PAM sequence (NGG for SpCas9) required adjacent to target. Mechanism: sgRNA:Cas9 complex scans DNA; matches 20-nt target → R-loop formed → HNH cuts strand complementary to sgRNA; RuvC cuts other strand → DSB; repaired by NHEJ (indels, gene KO) or HDR (precise edit with donor template).7. How does recombinational DNA repair restore double-strand breaks in bacteria? What is the role of RecA in this process?
Hint / Guidance
RecBCD pathway (for DSB from collapsed replication forks or DNA damage): (1) RecBCD helicase/nuclease loads at DSB end; unwinds and degrades DNA until it encounters Chi (5'-GCTGGTGG-3') site; (2) At Chi: RecBCD switches from nuclease to recombinase loader; loads RecA onto 3' ssDNA. (3) RecA-ssDNA nucleofilament invades homologous duplex (intact sister chromosome); (4) Branch migration extends heteroduplex; (5) Holliday junction resolved by RuvABC (RuvA stabilises, RuvB drives migration, RuvC cleaves). RecA also activates SOS response (protease activity cleaves LexA repressor). In the absence of a sister chromosome (non-replicating cells), RecA uses template switching.8. What is the CRISPR-Cas adaptive immune system's natural role in bacteria? How do bacteria acquire new 'memories' (spacers) against phages?
Hint / Guidance
Natural function: defence against phage and plasmid invasion. Memory acquisition (adaptation): phage injects DNA; Cas1-Cas2 complex recognises PAM on phage DNA; excises a 30 bp protospacer; integrates it into CRISPR array between first repeat and leader sequence as a new spacer — creates immunological memory. Expression: CRISPR locus transcribed → pre-crRNA → processed into individual crRNAs by Cas6/RNase III. Interference: crRNA loaded into Cas-effector complex (Cas9/Cas12/Cas13); guides complex to matching phage DNA/RNA → cleavage → phage neutralised. Bacterial population evolves rapid new spacers during phage epidemic; phage escapes by mutating protospacer/PAM = co-evolutionary arms race.2. Flow of Genetic Information
The central dogma (DNA → RNA → Protein) describes the directional flow of genetic information. In prokaryotes, this flow is tightly coupled in space and time: ribosomes begin translating mRNA while the RNA polymerase is still transcribing the gene, a process called co-transcriptional translation. This coupling enables gene expression to respond within seconds to environmental changes — a critical advantage for organisms competing in dynamic environments. It also allows regulatory mechanisms such as attenuation, which monitors translation to control transcription termination.
✏️ Fill in the Blank
1. In DNA, adenine pairs with _______ via two hydrogen bonds, while guanine pairs with cytosine via three hydrogen bonds.
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Thymine2. A mutation that changes an amino acid codon to a stop codon, producing a truncated protein, is called a _______ mutation.
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Nonsense3. The enzyme that synthesises a new DNA strand using an existing strand as a template is called DNA _______.
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Polymerase4. The process by which a bacteriophage integrates its DNA into the host chromosome is called _______.
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Lysogeny🔘 Multiple Choice
1. The central dogma of molecular biology describes the flow of genetic information as:
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Correct: B) DNA → RNA → Protein2. An organism's genotype differs from its phenotype in that:
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Correct: B) The genotype is the complete genetic information; the phenotype is the observable traits expressed under specific conditions3. A frameshift mutation is caused by:
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Correct: B) Insertion or deletion of a number of nucleotides not divisible by 3, disrupting the reading frame4. Which of the following describes a missense mutation?
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Correct: B) A mutation that changes one amino acid codon to a codon for a different amino acid5. In E. coli, which repair mechanism removes thymine dimers caused by UV irradiation?
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Correct: C) Nucleotide excision repair (UvrABC system)6. During bacterial conjugation, which DNA is transferred from donor to recipient?
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Correct: B) Primarily the F plasmid or mobilisable plasmid DNA via a mating bridge (pilus)7. A point mutation that changes a codon to a stop codon is called a:
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Correct: C) Nonsense mutation8. A Hfr (high frequency recombination) strain differs from an F⁺ strain because:
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Correct: B) The F factor is integrated into the chromosome, transferring chromosomal genes at high frequency💬 Open-Ended Questions
1. Describe the process of DNA replication in prokaryotes. Include the roles of helicase, primase, DNA polymerase III, DNA polymerase I, and ligase, and explain why one strand is synthesised continuously while the other is synthesised in fragments.
Hint / Guidance
Origin of replication (oriC) unwound by DnaA + helicase (DnaB); SSB proteins stabilise single strands; primase synthesises short RNA primers; DNA pol III (holoenzyme) extends 5'→3': leading strand continuous; lagging strand as Okazaki fragments (100–200 nt). DNA pol I (5'→3' exonuclease activity) removes primers + replaces with DNA. Ligase seals nicks between Okazaki fragments. Anti-parallel template strands force discontinuous synthesis because DNA pol can only extend 3' end.2. Compare transcription in prokaryotes with eukaryotes. Include at least three differences relevant to gene regulation and mRNA processing.
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(1) Location: prokaryotes — nucleoid, no nuclear membrane, transcription and translation coupled; eukaryotes — nucleus, separated. (2) RNA processing: prokaryotes — no introns/no splicing, polycistronic mRNA; eukaryotes — pre-mRNA splicing, 5' cap, 3' poly-A tail, monocistronic. (3) RNA polymerase: prokaryotes — single RNA pol (core + σ); eukaryotes — RNA pol I (rRNA), II (mRNA), III (tRNA). (4) Promoters: prokaryotes — -10 and -35 Pribnow box recognised by σ; eukaryotes — TATA box, CAAT box, GC box recognised by TBP/transcription factors.3. What is the difference between constitutive and regulated gene expression? Describe how operons allow coordinated regulation of metabolic pathways in bacteria.
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Constitutive: genes expressed continuously regardless of conditions (housekeeping genes, e.g., ribosomal protein genes). Regulated: expression adjusted in response to signals. Operon: group of genes under single promoter + operator (e.g., lac operon): lacZ (β-galactosidase), lacY (permease), lacA (transacetylase). Negative regulation: lac repressor (lacI) binds operator → blocks transcription; allolactose (inducer) binds repressor → conformational change → dissociates from operator → genes transcribed. Positive regulation: catabolite repression (CAP-cAMP); high glucose → low cAMP → no CAP binding → reduced transcription even without repressor. Combined result: lac genes maximally expressed only when lactose present AND glucose absent.4. Explain how spontaneous mutations arise in bacteria. Describe three sources of spontaneous mutation and their molecular mechanisms.
Hint / Guidance
(1) Replication errors: DNA pol III mismatch (tautomeric shifts in bases → transient wrong base pairing) → transition or transversion; rate ~10⁻⁹ per bp per replication after repair. (2) Depurination: spontaneous loss of purine (A or G) from sugar-phosphate backbone → abasic site → if not repaired → random base insertion → transversion mutations; ~5,000 depurination events per cell per day. (3) Cytosine deamination: cytosine → uracil → pairs with adenine → C-G to T-A transition; repaired by uracil DNA glycosylase; evolutionary reason why thymine (methylated uracil) exists in DNA. (4) Reactive oxygen species: oxidative damage → 8-oxoguanine → pairs with adenine → G-C to T-A transversion.5. Describe the process of transcription in prokaryotes from promoter recognition to termination. Include the roles of sigma factor, core enzyme, and Rho factor.
Hint / Guidance
Initiation: σ factor associates with core enzyme (α₂ββ'ω) → holoenzyme; σ₇₀ recognises -35 (TTGACA) and -10 (TATAAT) sequences; DNA unwound forming open complex (~13 bp bubble). Promoter escape: after ~8–10 nt synthesised, σ released; elongation begins. Elongation: RNA pol moves 3'→5' on template, synthesises RNA 5'→3'; non-template strand = coding strand. Termination: (1) Intrinsic/Rho-independent: palindromic G-C rich sequence → hairpin loop in RNA → stalls RNA pol; A-U base pairs weak → transcript released. (2) Rho-dependent: Rho factor (hexameric helicase) binds C-rich rut site on mRNA; translocates 5'→3'; catches paused RNA pol; unwinds RNA:DNA hybrid; releases transcript.6. What are insertion sequences (IS elements) and transposons? How do they contribute to bacterial genome evolution and the spread of antibiotic resistance?
Hint / Guidance
IS elements: simplest transposons; ~1–2 kb; encode only transposase flanked by inverted terminal repeats (ITRs); transpose by cut-and-paste (conservative) or replicative mechanism. Composite transposons: antibiotic resistance genes flanked by two IS elements (e.g., Tn10 = IS10 flanking tetracycline resistance). Complex transposons (class II): Tn3, encode transposase + resolvase + passenger genes. Contribution to evolution: IS insertions disrupt/activate genes; genome rearrangements; new gene combinations. Antibiotic resistance spread: resistance genes captured between IS elements form transposons → inserted into plasmids → conjugation spreads to new hosts (transposon + plasmid = mobile element stack). Class 1 integrons + gene cassettes: VIM metallo-β-lactamase in Pseudomonas.7. Explain the difference between constitutive and inducible gene expression in bacteria, using the lac operon as a model. Include the roles of the repressor, inducer, and catabolite activator protein (CAP).