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Microbiology सूक्ष्मजीवशास्त्र

BSc Semester 3 · 10 chapters · 10 with notes

Chapters

  1. 1EnzymesNotes
  2. 2Microbial MetabolismNotes
  3. 3Bioanalytical Techniques: Chromatography, Electrophoresis and CentrifugationNotes
  4. 4DNA Structure and ReplicationNotes
  5. 5Transcription in ProkaryotesNotes
  6. 6Genetic Code and TranslationNotes
  7. 7Regulation of Gene Expression: The Lac OperonNotes
  8. 8MutationNotes
  9. 9DNA Repair MechanismsNotes
  10. 10Gene Transfer in BacteriaNotes

BSc Semester 3 · Microbiology · Chapter 1

Enzymes

Quick summary

Enzymes (विकरे) are biological catalysts, mostly proteins, that speed up chemical reactions in cells without being used up. This topic covers enzyme structure, nomenclature and classification, mechanism of action, enzyme kinetics (Michaelis–Menten), factors affecting activity, inhibition and regulation.

Key points

  • Enzymes lower the activation energy of a reaction; they do not change the equilibrium.
  • They are highly specific for their substrate.
  • The active site is the small region where the substrate binds and the reaction occurs.
  • Lock and key model (Emil Fischer, 1894): active site has a fixed shape fitting the substrate. Induced fit model (Daniel Koshland, 1958): the active site changes shape when the substrate binds.
  • Holoenzyme = apoenzyme (protein part) + cofactor. Cofactors may be metal ions (Mg²⁺, Zn²⁺, Fe²⁺) or organic coenzymes (NAD⁺, FAD, coenzyme A — many derived from B vitamins). A tightly bound cofactor is a prosthetic group.
  • IUB/EC classification — six classes: 1 Oxidoreductases, 2 Transferases, 3 Hydrolases, 4 Lyases, 5 Isomerases, 6 Ligases. (A seventh class, Translocases, was added in 2018.)
  • Factors affecting activity: temperature (optimum; denatured at high temperature), pH (optimum, e.g. pepsin about 2, trypsin about 8), substrate concentration, enzyme concentration, inhibitors, activators.
  • Competitive inhibition: inhibitor resembles substrate and binds the active site; Km increases, Vmax unchanged; overcome by more substrate. Example: malonate inhibits succinate dehydrogenase; sulphonamides compete with PABA.
  • Non-competitive inhibition: inhibitor binds elsewhere; Vmax decreases, Km unchanged.
  • Allosteric enzymes have regulatory sites; feedback inhibition – the end product inhibits the first enzyme of the pathway (e.g. isoleucine inhibits threonine deaminase).
  • Isoenzymes (isozymes): different forms of an enzyme catalysing the same reaction (e.g. lactate dehydrogenase has five).
  • Ribozymes are catalytic RNA molecules (not proteins).
  • Microbes make constitutive enzymes (always) and inducible enzymes (only when needed, e.g. β-galactosidase in E. coli with lactose).

Important terms

  • Substrate – molecule acted on by the enzyme.
  • Km (Michaelis constant) – substrate concentration at which velocity is half of Vmax; low Km = high affinity.
  • Vmax – maximum reaction velocity when the enzyme is saturated.
  • Turnover number (kcat) – substrate molecules converted per enzyme molecule per second.
  • Zymogen – inactive precursor of an enzyme (e.g. trypsinogen).

Formulas

  • v = initial velocity, [S] = substrate concentration
  • Michaelis–Menten equation: v = Vmax [S] / (Km + [S])
  • When [S] = Km, v = Vmax / 2.
  • y-intercept = 1/Vmax; x-intercept = −1/Km; slope = Km/Vmax
  • Lineweaver–Burk (double reciprocal): 1/v = (Km/Vmax)(1/[S]) + 1/Vmax

5 important questions

Q1. Define Km and give its significance.

Km is the substrate concentration at which an enzyme works at half its maximum velocity. A low Km shows high affinity of the enzyme for its substrate.

Q2. An enzyme has Vmax = 100 units and Km = 2 mM. Find v when [S] = 2 mM and 6 mM.

  • [S] = 2 mM: v = 100 × 2 / (2 + 2) = 200/4 = 50 units (= Vmax/2).
  • [S] = 6 mM: v = 100 × 6 / (2 + 6) = 600/8 = 75 units.

Q3. Explain the IUB classification of enzymes with one example each.

  • Oxidoreductases: oxidation–reduction (alcohol dehydrogenase).
  • Transferases: transfer groups (hexokinase transfers phosphate).
  • Hydrolases: break bonds with water (amylase, lipase, protease).
  • Lyases: remove groups without water, forming double bonds (aldolase, decarboxylases).
  • Isomerases: rearrange within a molecule (phosphoglucose isomerase).
  • Ligases: join molecules using ATP (DNA ligase).

Q4. Compare competitive and non-competitive inhibition.

  • Binding site: active site vs another site.
  • Structural similarity to substrate: yes vs no.
  • Km: increases vs unchanged.
  • Vmax: unchanged vs decreases.
  • Reversed by excess substrate: yes vs no.
  • Examples: malonate / sulphonamides vs heavy metal ions like Hg²⁺ (often cited).
  • Lineweaver–Burk plots: lines meet on y-axis vs on x-axis.

Q5. Describe the factors affecting enzyme activity.

  • Temperature: rate rises to optimum, then falls due to denaturation.
  • pH: each enzyme has an optimum; extremes change ionization of active site.
  • Substrate concentration: hyperbolic curve, saturation at Vmax.
  • Enzyme concentration: rate proportional when substrate is in excess.
  • Inhibitors and activators.
  • Draw graphs for temperature, pH and [S].

Exam tip

Learn the six EC classes in their official order — O, T, H, L, I, L (mnemonic: "Our Tiny Hamster Likes Ice Lollies"). Draw the Michaelis–Menten curve and mark Vmax and Km.

BSc Semester 3 · Microbiology · Chapter 2

Microbial Metabolism

Quick summary

Metabolism is the sum of all chemical reactions in a cell: catabolism breaks down molecules to release energy, and anabolism builds cell materials using that energy. This topic covers nutritional types of microbes, glycolysis, the TCA cycle, electron transport, aerobic and anaerobic respiration, fermentation and ATP yield.

Key points

  • ATP is the energy currency. It is made by substrate-level phosphorylation (in glycolysis and TCA cycle) and oxidative phosphorylation (via the electron transport chain and ATP synthase).
  • Nutritional types by energy and carbon source: photoautotrophs (light, CO₂ — cyanobacteria), photoheterotrophs (light, organic carbon — purple non-sulphur bacteria), chemoautotrophs/chemolithotrophs (inorganic chemicals, CO₂ — Nitrosomonas, Thiobacillus), chemoheterotrophs (organic compounds for both — most bacteria, fungi, all pathogens).
  • Glycolysis (EMP pathway) occurs in the cytoplasm: glucose → 2 pyruvate; net gain 2 ATP and 2 NADH.
  • Other glucose pathways in microbes: Entner–Doudoroff (ED) pathway (e.g. Pseudomonas; net 1 ATP, 1 NADH, 1 NADPH per glucose) and pentose phosphate pathway (gives NADPH and 5-carbon sugars).
  • Link reaction: pyruvate → acetyl-CoA + CO₂ + NADH (pyruvate dehydrogenase).
  • TCA / Krebs cycle: acetyl-CoA + oxaloacetate → citrate → … → oxaloacetate. Per acetyl-CoA: 3 NADH, 1 FADH₂, 1 GTP/ATP, 2 CO₂.
  • Electron transport chain (ETC): in bacteria it is in the cell membrane. Electrons pass from NADH/FADH₂ to a final acceptor; protons are pumped out, making a proton motive force that drives ATP synthase (chemiosmotic theory, Peter Mitchell).
  • Aerobic respiration: final acceptor is O₂ → H₂O.
  • Anaerobic respiration: final acceptor is an inorganic molecule other than O₂ — nitrate (→ nitrite, N₂), sulphate (→ H₂S), CO₂ (→ methane by methanogens).
  • Fermentation: no ETC; an organic molecule (like pyruvate) accepts electrons from NADH. Only 2 ATP per glucose (from glycolysis).
  • Fermentation types: homolactic (pyruvate → lactic acid; Lactobacillus, Streptococcus), alcoholic (pyruvate → ethanol + CO₂; yeast Saccharomyces cerevisiae), mixed acid (E. coli), butanediol (Enterobacter).

Important terms

  • Catabolism / anabolism – breakdown / synthesis reactions.
  • Oxidation – loss of electrons (or hydrogen); reduction – gain of electrons.
  • Proton motive force – energy stored as a proton gradient across a membrane.
  • Amphibolic pathway – a pathway used for both breakdown and synthesis (TCA cycle).
  • Chemolithotroph – organism getting energy by oxidizing inorganic compounds.

Formulas

  • Aerobic respiration: C₆H₁₂O₆ + 6O₂ → 6CO₂ + 6H₂O + energy (ATP)
  • Alcoholic fermentation: C₆H₁₂O₆ → 2C₂H₅OH + 2CO₂ (+ 2 ATP)
  • Homolactic fermentation: C₆H₁₂O₆ → 2CH₃CH(OH)COOH (+ 2 ATP)
  • ATP yield (aerobic, prokaryote, theoretical): about 38 ATP per glucose using older values (NADH = 3 ATP, FADH₂ = 2 ATP); modern estimates are lower (about 30–32).

ATP count (older textbook values):

StageATP directNADHFADH₂ATP total
Glycolysis22–8
Pyruvate → acetyl-CoA (×2)–2–6
TCA cycle (×2)26224
Total38

5 important questions

Q1. Differentiate fermentation and respiration.

In respiration, electrons go through an ETC to an external acceptor (O₂ or another inorganic molecule) and much ATP is made. In fermentation there is no ETC; an organic molecule accepts electrons and only 2 ATP per glucose are made.

Q2. What is anaerobic respiration? Give an example.

Respiration using an inorganic final electron acceptor other than oxygen. Example: denitrifying bacteria like Pseudomonas use nitrate, reducing it to nitrogen gas.

Q3. Describe glycolysis.

  • Site: cytoplasm; no oxygen needed.
  • Preparatory phase: glucose phosphorylated twice using 2 ATP (hexokinase, phosphofructokinase), split into two 3-carbon molecules.
  • Pay-off phase: each G3P oxidized (NADH formed), 4 ATP made by substrate-level phosphorylation.
  • End product: 2 pyruvate.
  • Net: 2 ATP + 2 NADH per glucose.

Q4. Explain the TCA cycle.

  • Acetyl-CoA (2C) + oxaloacetate (4C) → citrate (6C).
  • Citrate → isocitrate → α-ketoglutarate (CO₂, NADH) → succinyl-CoA (CO₂, NADH) → succinate (GTP/ATP) → fumarate (FADH₂) → malate → oxaloacetate (NADH).
  • Per turn: 2 CO₂, 3 NADH, 1 FADH₂, 1 ATP/GTP.
  • Two turns per glucose.
  • Amphibolic: gives precursors for amino acids.

Q5. Classify microorganisms on the basis of nutrition.

  • Energy source: phototrophs vs chemotrophs.
  • Carbon source: autotrophs vs heterotrophs.
  • Electron source: lithotrophs vs organotrophs.
  • Four main groups with examples (see Key points).

Exam tip

Write the ATP table clearly and state which values (older 38 or newer 30–32) you used.

BSc Semester 3 · Microbiology · Chapter 3

Bioanalytical Techniques: Chromatography, Electrophoresis and Centrifugation

Quick summary

To study enzymes, proteins, nucleic acids and metabolites, a microbiologist must separate and identify them. Three basic tools do most of this work: chromatography (separation by distribution between two phases), electrophoresis (separation by movement in an electric field) and centrifugation (separation by sedimentation). This topic explains their principles, types and uses.

Key points

  • Chromatography (Mikhail Tswett, 1906, plant pigments) separates molecules by their different distribution between a stationary phase and a mobile phase.
  • Paper chromatography: stationary phase is water held in cellulose paper (partition); used for amino acids and sugars. Spots are identified by Rf value.
  • Thin layer chromatography (TLC): a thin layer of silica gel or alumina on glass/plastic; faster and sharper than paper.
  • Gel filtration (size exclusion): porous beads (Sephadex); large molecules come out first, small ones enter pores and are delayed.
  • Ion exchange: charged resin (DEAE-cellulose = anion exchanger, CM-cellulose = cation exchanger); proteins separated by charge, eluted by salt or pH gradient.
  • Affinity chromatography: a specific ligand (antibody, substrate, Ni²⁺ for His-tagged proteins) binds only the target; most specific method.
  • HPLC (high performance liquid chromatography): high pressure, fine particles, fast and sensitive; GC (gas chromatography) for volatile compounds.
  • Column chromatography types:
  • Electrophoresis: charged molecules move toward the electrode of opposite charge; speed depends on charge, size, shape, voltage and gel concentration.
  • Agarose gel electrophoresis: for DNA; DNA is negative (phosphate groups) and moves to the anode (+); smaller fragments move faster; stained with ethidium bromide (or safer dyes) and seen under UV; size read against a DNA ladder.
  • SDS-PAGE: polyacrylamide gel; SDS (sodium dodecyl sulfate) denatures proteins and gives them uniform negative charge, so they separate by molecular weight only; stained with Coomassie blue.
  • Centrifugation separates particles by sedimentation in a strong centrifugal field. RCF (relative centrifugal force, ×g) = 1.118 × 10⁻⁵ × r × (rpm)², where r is radius in cm.
  • Sedimentation coefficient is measured in Svedberg units (S); 1 S = 10⁻¹³ second. S values are not additive (30S + 50S = 70S ribosome).
  • Differential centrifugation separates by size in steps (nuclei, then mitochondria, then microsomes). Density gradient centrifugation: rate-zonal (sucrose gradient, by size) and isopycnic (CsCl, by density; used by Meselson and Stahl, 1958).
  • Ultracentrifuges reach above about 60,000 rpm and need refrigeration and vacuum.

Important terms

  • Rf value – distance moved by the solute ÷ distance moved by the solvent front (always less than 1).
  • Eluent / elution – liquid used to wash a bound substance off a column / the washing process.
  • Ampholyte – molecule with both positive and negative groups (e.g. protein).
  • Isoelectric point (pI) – pH at which a molecule has no net charge and does not move in an electric field.
  • Svedberg unit (S) – unit of sedimentation coefficient, 10⁻¹³ s.

Formulas

  • Rf = distance travelled by compound ÷ distance travelled by solvent front.
  • RCF (×g) = 1.118 × 10⁻⁵ × r × N², where r = radius in cm and N = speed in rpm.
  • Example: r = 10 cm, N = 10,000 rpm → RCF = 1.118 × 10⁻⁵ × 10 × 10⁸ = 11,180 × g.
  • Sedimentation coefficient s = v ÷ (ω² r), where v = sedimentation velocity, ω = angular velocity (radian/s).

5 important questions

Q1. Define Rf value. A spot moves 3.2 cm when the solvent moves 8.0 cm. Find Rf.

Rf = distance moved by spot ÷ distance moved by solvent = 3.2 ÷ 8.0 = 0.4.

Q2. Why do large molecules come out first in gel filtration?

Large molecules cannot enter the pores of the gel beads, so they travel only through the spaces between beads and elute quickly. Small molecules enter the pores, take a longer path and come out later.

Q3. Explain the principle, types and applications of chromatography.

  • Principle: distribution between stationary and mobile phase.
  • Paper and TLC (partition/adsorption), column types: gel filtration, ion exchange, affinity, HPLC, GC.
  • Uses: amino acid and sugar identification, protein purification, antibiotic analysis, pigment separation.

Q4. Describe agarose gel electrophoresis and SDS-PAGE.

  • Apparatus: gel, buffer tank, wells, power supply.
  • Agarose for DNA: movement to anode, size separation, ethidium bromide, ladder.
  • SDS-PAGE for proteins: SDS gives uniform charge; stacking and resolving gel; Coomassie stain.
  • Applications: plasmid analysis, PCR product check, protein molecular weight.

Q5. Write a note on types of centrifugation.

  • Principle and RCF formula.
  • Differential centrifugation for cell fractionation.
  • Density gradient: rate-zonal (sucrose) and isopycnic (CsCl).
  • Analytical vs preparative ultracentrifuge; Svedberg unit.
  • Example: Meselson–Stahl experiment.

Exam tip

Solve one Rf and one RCF numerical in your answer sheet to show understanding. Remember: DNA runs to the positive electrode.

BSc Semester 3 · Microbiology · Chapter 4

DNA Structure and Replication

Quick summary

DNA (deoxyribonucleic acid) is the genetic material of all cells. This topic covers the experiments that proved DNA is genetic material, the Watson–Crick double helix model, and how DNA copies itself by semi-conservative replication with the help of many enzymes.

Key points

  • Griffith (1928): transformation in Streptococcus pneumoniae — heat-killed smooth (S, virulent) cells plus live rough (R) cells killed mice; something changed R into S.
  • Avery, MacLeod and McCarty (1944): the transforming principle is DNA (destroyed by DNase, not by protease or RNase).
  • Hershey and Chase (1952): with T2 phage labelled with ³²P (DNA) and ³⁵S (protein), only ³²P entered bacteria — DNA is the genetic material.
  • Watson and Crick (1953) proposed the double helix, using X-ray pictures by Rosalind Franklin and Maurice Wilkins.
  • Nucleotide = deoxyribose sugar + phosphate + nitrogen base. Purines: adenine (A), guanine (G). Pyrimidines: cytosine (C), thymine (T).
  • Chargaff's rule: A = T and G = C; so purines = pyrimidines.
  • Double helix: two antiparallel strands (5′→3′ and 3′→5′), right-handed (B-DNA), sugar-phosphate backbone outside, bases inside. A pairs with T by 2 hydrogen bonds, G with C by 3 hydrogen bonds.
  • B-DNA: about 10 base pairs per turn, 3.4 nm per turn, 0.34 nm between base pairs, diameter about 2 nm.
  • Semi-conservative replication: each new DNA has one old and one new strand — proved by Meselson and Stahl (1958) using ¹⁵N and ¹⁴N and CsCl density gradient centrifugation.
  • Bacterial replication starts at one origin (oriC) and moves in both directions (bidirectional) — a theta (θ) structure in circular DNA.
  • Enzymes: helicase (unwinds), single-strand binding proteins (keep strands apart), topoisomerase/DNA gyrase (removes supercoils), primase (makes RNA primer), DNA polymerase III (main synthesis in E. coli), DNA polymerase I (removes primer, fills gap), DNA ligase (joins fragments).
  • DNA polymerase adds nucleotides only in the 5′→3′ direction, so the leading strand is made continuously and the lagging strand discontinuously as Okazaki fragments.
  • DNA polymerases have 3′→5′ exonuclease (proofreading) activity, which keeps errors very low.

Important terms

  • Replication fork – Y-shaped region where DNA is opened and copied.
  • Primer – short RNA piece giving a free 3′-OH to start synthesis.
  • Okazaki fragments – short DNA pieces on the lagging strand.
  • Antiparallel – strands run in opposite directions.
  • Replicon – unit of DNA replicated from one origin.

Formulas

  • Chargaff: %A = %T, %G = %C, and %A + %G = 50% (= %T + %C).
  • Length of B-DNA = number of base pairs × 0.34 nm.
  • Number of turns = base pairs ÷ 10.

5 important questions

Q1. A DNA sample has 30% adenine. Find the percentage of G, C and T.

T = A = 30%. A + T = 60%, so G + C = 40%. G = C = 20%.

Q2. Find the length of a DNA molecule with 5000 base pairs.

Length = 5000 × 0.34 nm = 1700 nm = 1.7 µm. Number of turns = 5000/10 = 500.

Q3. Describe the Watson and Crick model of DNA.

  • Two polynucleotide strands coiled in a right-handed double helix.
  • Antiparallel strands.
  • Sugar-phosphate backbone outside, bases inside.
  • Complementary pairing A=T (2 H-bonds), G≡C (3 H-bonds).
  • Dimensions: 2 nm diameter, 3.4 nm per turn, 10 bp per turn.
  • Major and minor grooves.

Q4. Explain the Meselson–Stahl experiment.

  • E. coli grown for many generations in ¹⁵N medium (heavy DNA).
  • Shifted to ¹⁴N medium.
  • After one generation: all DNA of intermediate (hybrid) density.
  • After two generations: half hybrid, half light.
  • Separated by CsCl density gradient centrifugation.
  • Proves semi-conservative replication.

Q5. Describe DNA replication in prokaryotes.

  • Initiation at oriC; helicase unwinds, SSB proteins stabilise, gyrase relieves tension.
  • Primase makes RNA primers.
  • DNA pol III extends 5′→3′: leading strand continuous, lagging strand as Okazaki fragments.
  • DNA pol I removes primers and fills gaps; ligase seals nicks.
  • Proofreading by 3′→5′ exonuclease.
  • Termination at ter sites.

Exam tip

Draw a replication fork with leading and lagging strands, Okazaki fragments and enzymes labelled.

BSc Semester 3 · Microbiology · Chapter 5

Transcription in Prokaryotes

Quick summary

Transcription is the copying of the information in a DNA gene into an RNA molecule. It is the first step of gene expression (DNA → RNA → protein, the central dogma stated by Francis Crick). In bacteria a single RNA polymerase makes all types of RNA, and transcription takes place in three stages: initiation, elongation and termination. Because there is no nucleus, translation can start on the mRNA even before transcription ends.

Key points

  • Only one DNA strand is copied: the template (antisense) strand. The other, the coding (sense) strand, has the same sequence as the RNA (with T instead of U).
  • RNA is made in the 5′ → 3′ direction; the template is read 3′ → 5′. No primer is needed.
  • E. coli RNA polymerase holoenzyme = core enzyme (α₂ββ′ω) + sigma (σ) factor. The core does the synthesis; σ recognizes the promoter. The main sigma factor is σ⁷⁰.
  • Promoter (upstream of the start, +1) has two conserved regions: the −10 box (Pribnow box, TATAAT) and the −35 box (TTGACA).
  • Initiation: holoenzyme binds the promoter (closed complex) → DNA unwinds about 12–14 bp near −10 (open complex) → first nucleotides join → after a short stretch σ is released and the core moves on.
  • Elongation: core enzyme adds ribonucleotides (ATP, GTP, CTP, UTP) complementary to the template; a moving transcription bubble of unwound DNA travels with it; rate is about 40–50 nucleotides per second.
  • Rho-independent (intrinsic): a GC-rich inverted repeat forms a hairpin (stem-loop) in the RNA, followed by a run of U's; the polymerase stalls and the RNA falls off.
  • Rho-dependent: the rho (ρ) protein, an ATP-dependent helicase, binds the RNA at a rut site, moves along it and pulls the RNA from the polymerase.
  • Termination happens in two ways:
  • Bacterial mRNA is often polycistronic (one mRNA carries several genes, e.g. lac operon); eukaryotic mRNA is usually monocistronic.
  • Prokaryotic mRNA has no 5′ cap, no poly-A tail and no introns, and it is short-lived (half-life of a few minutes).
  • Rifampicin blocks the β subunit of bacterial RNA polymerase (used in TB treatment); actinomycin D binds DNA and blocks elongation.
  • Types of RNA made: mRNA (message), tRNA (adapter) and rRNA (ribosome; 16S, 23S, 5S in bacteria).

Important terms

  • Promoter – DNA sequence where RNA polymerase binds to start transcription.
  • Sigma factor – protein that guides RNA polymerase to the promoter.
  • Transcription unit – stretch of DNA from promoter to terminator.
  • Polycistronic mRNA – mRNA coding for more than one protein.
  • Terminator – DNA sequence that signals the end of transcription.

Key facts

FeatureDNA replicationTranscription
ProductDNARNA
Strands copiedBothOne (template)
EnzymeDNA polymerase IIIRNA polymerase
Primer neededYes (RNA primer)No
Sugar / baseDeoxyribose / TRibose / U
ProofreadingYesVery limited
ExtentWhole genomeSelected genes

5 important questions

Q1. What is the Pribnow box?

A conserved six-base sequence, TATAAT, found about 10 bases upstream of the transcription start site in bacterial promoters. Here the DNA first opens for transcription.

Q2. What is the role of the sigma factor?

It makes RNA polymerase recognize and bind specifically to promoter sequences (−10 and −35). After initiation it leaves the core enzyme.

Q3. Describe the structure of E. coli RNA polymerase and the process of initiation.

  • Subunits α₂, β, β′, ω and σ with roles.
  • Promoter elements −35 and −10.
  • Closed complex → open complex → abortive initiation → promoter clearance.
  • Release of σ.

Q4. Explain the mechanisms of transcription termination in bacteria.

  • Rho-independent: inverted repeat, hairpin, poly-U, weak A–U pairing releases RNA (draw hairpin).
  • Rho-dependent: rho binds rut site, ATP-driven movement, unwinding of RNA–DNA hybrid.
  • Compare the two.

Q5. Compare transcription in prokaryotes and eukaryotes.

  • One RNA polymerase vs three (Pol I, II, III).
  • Sigma factor vs general transcription factors (TATA box binding protein).
  • Polycistronic vs monocistronic mRNA.
  • No processing vs capping, poly-A tail and splicing.
  • Coupled with translation vs separated by the nuclear membrane.

Exam tip

Draw a promoter diagram marking −35, −10 and +1 and a hairpin terminator; examiners give marks for these labels. Remember "RNA grows 5′ to 3′ like DNA".

BSc Semester 3 · Microbiology · Chapter 6

Genetic Code and Translation

Quick summary

Translation is the making of a protein using the base sequence of mRNA as a guide. The rules that link three-base codons to amino acids form the genetic code. Translation takes place on ribosomes with the help of tRNA, enzymes and protein factors, in three stages: initiation, elongation and termination. In bacteria it is fast and directly coupled to transcription.

Key points

  • The genetic code is a triplet code: each codon of three bases codes for one amino acid. 4³ = 64 codons: 61 sense codons and 3 stop codons (UAA, UAG, UGA).
  • Properties: triplet, non-overlapping, comma-less, degenerate (most amino acids have more than one codon, e.g. leucine has six), unambiguous (one codon → one amino acid), nearly universal (exceptions in mitochondria, e.g. UGA codes tryptophan in human mitochondria), and has a defined start codon AUG (methionine; sometimes GUG in bacteria).
  • Nirenberg and Matthaei (1961) made poly-U RNA and got polyphenylalanine, so UUU = phenylalanine was the first codon decoded. Khorana and Holley also helped (Nobel 1968 shared with Nirenberg).
  • Wobble hypothesis (Crick, 1966): pairing between the third base of the codon and the first base of the anticodon is flexible, so one tRNA can read more than one codon (e.g. inosine in anticodon pairs with U, C or A).
  • tRNA has a cloverleaf shape (L-shaped in 3-D), an anticodon loop and a 3′ CCA end where the amino acid is attached.
  • Aminoacyl-tRNA synthetases (one for each amino acid) attach the correct amino acid to its tRNA using ATP ("charging").
  • Bacterial ribosome (70S) = 30S (16S rRNA + about 21 proteins) + 50S (23S + 5S rRNA + about 31–34 proteins). It has three tRNA sites: A (aminoacyl), P (peptidyl), E (exit).
  • Initiation: the Shine–Dalgarno sequence (AGGAGG) on mRNA pairs with the 3′ end of 16S rRNA; the initiator fMet-tRNA (N-formylmethionine) sits in the P site on AUG with IF1, IF2, IF3 and GTP; the 50S subunit joins to form the 70S initiation complex.
  • Elongation: EF-Tu–GTP brings the next aminoacyl-tRNA to the A site → peptidyl transferase (a ribozyme, the 23S rRNA) forms the peptide bond → EF-G–GTP moves the ribosome one codon (translocation).
  • Termination: a stop codon in the A site is recognized by release factors (RF1: UAA, UAG; RF2: UAA, UGA; RF3 helps); the polypeptide is released and the ribosome separates (with ribosome recycling factor).
  • Several ribosomes on one mRNA form a polysome.
  • Antibiotics act here: streptomycin and tetracycline (30S), chloramphenicol and erythromycin (50S).

Important terms

  • Codon – three-base sequence on mRNA for one amino acid or stop signal.
  • Anticodon – three-base sequence on tRNA complementary to the codon.
  • Shine–Dalgarno sequence – ribosome-binding site on bacterial mRNA, upstream of AUG.
  • Reading frame – the way mRNA is divided into codons starting from AUG.
  • Polysome – many ribosomes translating one mRNA at the same time.

Key facts

StageMain factorsEvent
InitiationIF1, IF2, IF3, fMet-tRNA, GTP70S complex forms on AUG
ElongationEF-Tu, EF-Ts, EF-G, GTPPeptide bonds form, ribosome moves 5′ → 3′
TerminationRF1, RF2, RF3, RRFPolypeptide released at stop codon

5 important questions

Q1. What are stop codons?

UAA (ochre), UAG (amber) and UGA (opal). They code for no amino acid and signal the end of translation; release factors recognize them.

Q2. What is meant by degeneracy of the genetic code?

More than one codon can code for the same amino acid. For example, glycine is coded by GGU, GGC, GGA and GGG. Only methionine (AUG) and tryptophan (UGG) have a single codon.

Q3. Describe the properties of the genetic code.

  • Triplet, non-overlapping, comma-less.
  • Degenerate, unambiguous, universal (with exceptions).
  • Start and stop codons.
  • Wobble at the third position.
  • Nirenberg's poly-U experiment as evidence.

Q4. Explain the process of translation in prokaryotes.

  • Charging of tRNA.
  • Initiation with Shine–Dalgarno, fMet-tRNA, IFs.
  • Elongation cycle: entry at A site, peptide bond, translocation, exit through E site.
  • Termination by release factors.
  • Draw the ribosome with A, P, E sites.

Q5. Write the structure and function of tRNA.

  • Cloverleaf: acceptor arm with CCA, D arm, anticodon arm, variable loop, TΨC arm.
  • L-shaped tertiary structure.
  • Carries amino acid; reads codon via anticodon.
  • Role of aminoacyl-tRNA synthetase.

Exam tip

Remember "AUG start; UAA, UAG, UGA stop" and the three ribosome sites A–P–E. A cloverleaf tRNA diagram earns easy marks.

BSc Semester 3 · Microbiology · Chapter 7

Regulation of Gene Expression: The Lac Operon

Quick summary

Bacteria do not make all their proteins all the time; they switch genes on and off to save energy. The operon model, proposed by François Jacob and Jacques Monod (1961) (Nobel Prize 1965 with André Lwoff), explains this. The lac operon of E. coli is an inducible system switched on by lactose, and the trp operon is a repressible system switched off by tryptophan. This topic also covers catabolite repression.

Key points

  • An operon is a group of structural genes transcribed together from one promoter under one operator, giving one polycistronic mRNA.
  • Constitutive genes are always expressed (e.g. glycolysis enzymes); regulated genes are expressed only when needed.
  • lacZ → β-galactosidase (splits lactose into glucose + galactose; also makes allolactose).
  • lacY → lactose permease (brings lactose into the cell).
  • lacA → thiogalactoside transacetylase.
  • Parts of the lac operon (in order): lacI (regulator gene, has its own promoter, makes the repressor) → promoter (P) → operator (O) → structural genes lacZ, lacY, lacA.
  • No lactose: the active lac repressor (a tetramer) binds the operator and blocks RNA polymerase → genes OFF. (A few molecules are still made — basal level.)
  • Lactose present: lactose is changed to allolactose, the true inducer; it binds the repressor, changes its shape (allosteric change), and the repressor leaves the operator → genes ON. This is negative control.
  • IPTG (isopropyl-β-D-thiogalactoside) is a synthetic inducer that is not broken down; used in labs and in blue–white screening with X-gal.
  • Catabolite repression (positive control): when glucose is low, cAMP levels rise; cAMP binds CAP (catabolite activator protein, also CRP); the CAP–cAMP complex binds near the promoter and helps RNA polymerase. When glucose is high, cAMP is low and the operon stays weak even if lactose is present.
  • So the lac operon is fully ON only when lactose is present and glucose is absent. This explains diauxic growth (two growth phases with a lag between them) when both sugars are given.
  • Trp operon (repressible): genes trpE, D, C, B, A make enzymes for tryptophan synthesis. The aporepressor is inactive alone; when tryptophan (the corepressor) is high, it binds the aporepressor, which then binds the operator → genes OFF.
  • The trp operon is also controlled by attenuation (a leader sequence forms a terminator hairpin when tryptophan is plentiful), discovered by Charles Yanofsky.
  • Mutants: lacI⁻ or O^c (operator constitutive) mutants make the enzymes all the time; I^s (super-repressor) cannot be induced.

Important terms

  • Inducer – small molecule that switches on an inducible operon (allolactose, IPTG).
  • Repressor – protein that binds the operator and blocks transcription.
  • Corepressor – small molecule that activates a repressor (tryptophan).
  • Operator – DNA site where the repressor binds.
  • Diauxie – two-phase growth when a bacterium uses glucose first and then lactose.

Key facts

ConditionRepressor on operator?CAP–cAMP bound?Lac operon
Glucose present, lactose absentYesNoOFF
Glucose present, lactose presentNoNoVery low
Glucose absent, lactose absentYesYesOFF
Glucose absent, lactose presentNoYesFully ON
FeatureLac operonTrp operon
TypeInducible (catabolic)Repressible (anabolic)
Repressor madeActiveInactive (aporepressor)
Small moleculeInducer: allolactoseCorepressor: tryptophan
Extra controlCatabolite repressionAttenuation

5 important questions

Q1. What is the true inducer of the lac operon?

Allolactose, an isomer of lactose made by β-galactosidase. It binds the lac repressor and removes it from the operator.

Q2. Define operon.

A unit of bacterial DNA consisting of a promoter, an operator and a group of structural genes that are transcribed together and controlled as one unit.

Q3. Explain the lac operon model with diagrams.

  • Components and gene products.
  • State in absence of lactose (repressor bound).
  • State in presence of lactose (inducer, repressor released, mRNA made).
  • Two labelled diagrams.

Q4. What is catabolite repression? Explain diauxic growth.

  • Glucose lowers cAMP; CAP–cAMP needed for strong lac transcription.
  • Glucose used first; when it is finished cAMP rises and lac genes switch on.
  • Growth curve with two log phases separated by a lag (draw).

Q5. Describe the trp operon and compare it with the lac operon.

  • Structural genes and aporepressor.
  • High tryptophan → corepressor complex → OFF; low tryptophan → ON.
  • Brief idea of attenuation.
  • Comparison table.

Exam tip

Use the four-row glucose/lactose table — it shows you understand both negative and positive control. Always name allolactose, not lactose, as the real inducer.

BSc Semester 3 · Microbiology · Chapter 8

Mutation

Quick summary

A mutation (उत्परिवर्तन) is a permanent, heritable change in the base sequence of DNA. This topic covers types of mutations, spontaneous and induced mutations, mutagens and how they act, DNA repair systems, methods to detect mutants (replica plating), and the Ames test.

Key points

  • Transition: purine → purine or pyrimidine → pyrimidine (A↔G, C↔T).
  • Transversion: purine ↔ pyrimidine (e.g. A → T).
  • Point mutations change one base pair:
  • Effects of base substitution: silent (same amino acid, due to code degeneracy), missense (different amino acid, e.g. sickle-cell anaemia: glutamic acid → valine), nonsense (creates a stop codon, protein cut short).
  • Frameshift mutations: insertion or deletion of bases (not in multiples of 3) shifts the reading frame and changes all later amino acids.
  • Large changes: deletions, duplications, inversions, translocations.
  • Spontaneous mutations occur naturally, e.g. by tautomeric shifts of bases, replication errors, deamination. Luria and Delbrück (1943) showed by the fluctuation test that mutations arise randomly, before exposure to the selecting agent (e.g. phage resistance).
  • Base analogues: 5-bromouracil (resembles thymine), 2-aminopurine (resembles adenine) — cause transitions.
  • Chemical modifiers: nitrous acid (deaminates bases), hydroxylamine, alkylating agents (EMS, nitrosoguanidine).
  • Intercalating agents: acridine orange, ethidium bromide, proflavin — cause frameshifts.
  • UV light: forms thymine (pyrimidine) dimers.
  • Ionizing radiation (X-rays, gamma rays): break DNA strands.
  • Induced mutations are caused by mutagens:
  • DNA repair: photoreactivation (photolyase uses visible light to split thymine dimers), excision repair (cut out damaged part, refill), mismatch repair, SOS repair (error-prone), recombination repair.
  • Replica plating (Joshua and Esther Lederberg, 1952): colonies on a master plate are copied with sterile velvet onto test plates to detect auxotrophs or resistant mutants. It also supported the idea that mutations are pre-existing.
  • Ames test (Bruce Ames): uses histidine-requiring (his⁻) mutants of Salmonella typhimurium; a chemical that causes many revertants (his⁺) is mutagenic and possibly carcinogenic. Liver extract (S9) is added to mimic body metabolism.

Important terms

  • Mutant / wild type – changed form / normal natural form.
  • Auxotroph – mutant that needs an extra nutrient (e.g. his⁻); prototroph – grows on minimal medium.
  • Reversion – mutation restoring the original phenotype.
  • Mutagen – agent that increases mutation rate.
  • Suppressor mutation – a second mutation that hides the effect of the first.

Key facts

MutagenTypeEffect
5-BromouracilBase analogueTransition (A-T → G-C)
Nitrous acidDeaminating agentTransition
Acridine orangeIntercalating dyeFrameshift
UV lightPhysicalThymine dimers
X-raysIonizing radiationStrand breaks

5 important questions

Q1. Differentiate transition and transversion.

Transition is replacement of a purine by another purine or a pyrimidine by another pyrimidine (A↔G, C↔T). Transversion is replacement of a purine by a pyrimidine or vice versa.

Q2. What is a frameshift mutation?

A mutation caused by adding or deleting one or more bases (not a multiple of three), which shifts the reading frame so that every codon after it is read wrongly. Acridine dyes cause it.

Q3. Describe the replica plating technique.

  • Master plate on complete medium with many colonies.
  • Sterile velvet pressed on master plate, then onto minimal medium and complete medium plates in the same orientation.
  • Colonies growing on complete but not on minimal medium are auxotrophs.
  • Can detect antibiotic- or phage-resistant mutants on selective plates.
  • Showed mutations exist before selection.

Q4. Explain the Ames test and its importance.

  • Uses his⁻ Salmonella typhimurium strains.
  • Test chemical + liver S9 extract + bacteria on medium lacking histidine (trace only).
  • Mutagens cause reversion to his⁺ → many colonies.
  • Compare with control plate (spontaneous revertants).
  • Importance: quick, cheap screening of chemicals, food additives, drugs for cancer risk.

Q5. Describe the mechanisms of DNA repair.

  • Photoreactivation: photolyase + visible light splits thymine dimers.
  • Nucleotide excision repair: UvrABC endonuclease cuts damaged stretch, DNA pol I fills, ligase seals.
  • Base excision repair: glycosylase removes wrong base.
  • Mismatch repair: corrects replication errors using methylation to identify the parent strand.
  • SOS response: emergency, error-prone repair.

Exam tip

For every mutagen, write its class, example and the type of mutation it causes — a table scores well.

BSc Semester 3 · Microbiology · Chapter 9

DNA Repair Mechanisms

Quick summary

DNA is damaged all the time by UV light, chemicals, radiation and copying errors. If the damage is not fixed it becomes a permanent mutation or kills the cell. Cells therefore have several DNA repair systems: some reverse the damage directly, some cut out the damaged part and fill the gap, and one emergency system (the SOS response) allows survival at the cost of new mutations. Much of this was first worked out in E. coli.

Key points

  • Types of DNA damage: base mismatches (replication errors), pyrimidine (thymine) dimers from UV light, deamination (cytosine → uracil), loss of bases (AP sites), alkylation (by EMS, nitrosoguanidine), oxidation (8-oxoguanine), and single- and double-strand breaks (ionizing radiation).
  • Photoreactivation (light repair): the enzyme photolyase uses visible (blue) light energy to split thymine dimers back into normal bases. Discovered by Albert Kelner (1949) — UV-treated bacteria survived better when kept in light.
  • O⁶-methylguanine DNA methyltransferase (Ada protein) removes a methyl group from guanine; it is used up in the process ("suicide enzyme").
  • 1. Direct reversal:
  • 2. Base excision repair (BER): a DNA glycosylase removes the single wrong base (e.g. uracil-DNA glycosylase) → AP endonuclease cuts the backbone → DNA polymerase I fills in → DNA ligase seals.
  • 3. Nucleotide excision repair (NER) (dark repair): for bulky damage such as thymine dimers. In E. coli the UvrABC endonuclease cuts the damaged strand on both sides (an oligonucleotide of about 12–13 nucleotides is removed), UvrD helicase removes it, DNA polymerase I fills the gap and ligase seals. In humans, defects in NER cause xeroderma pigmentosum (extreme sun sensitivity and skin cancer).
  • 4. Mismatch repair (MMR): corrects wrong bases left after replication proofreading. MutS recognizes the mismatch, MutL joins, MutH nicks the new (unmethylated) strand at a GATC site. The old strand is recognized because it is methylated by Dam methylase. Defects increase mutation rate about 100–1000 times (mutator strains).
  • 5. Recombination repair: gaps or breaks are repaired using the undamaged sister DNA as template, with the RecA protein (strand exchange). Double-strand breaks can also be joined by non-homologous end joining (mainly in eukaryotes).
  • 6. SOS response: heavy damage activates RecA (bound to single-stranded DNA), which helps the LexA repressor cut itself. More than 40 SOS genes switch on, including uvr genes and error-prone polymerases (Pol IV and Pol V, umuDC). These translesion polymerases copy across damage but make mistakes → SOS mutagenesis.
  • DNA polymerase III's 3′ → 5′ exonuclease proofreading is the first line of defence during replication.
  • The Ames test uses repair-deficient Salmonella strains to detect chemical mutagens (link with the Mutation topic).

Important terms

  • Thymine dimer – covalent link between two neighbouring thymines caused by UV light.
  • Photolyase – light-activated enzyme that splits pyrimidine dimers.
  • AP site – apurinic or apyrimidinic site, a spot where a base is missing.
  • Excision repair – removal of the damaged part followed by gap filling.
  • SOS response – inducible emergency repair system controlled by RecA and LexA.

Key facts

Repair systemDamage repairedKey enzymes (E. coli)
PhotoreactivationThymine dimersPhotolyase + visible light
Base excisionSingle altered baseDNA glycosylase, AP endonuclease, Pol I, ligase
Nucleotide excisionBulky lesions, dimersUvrA, UvrB, UvrC, UvrD, Pol I, ligase
Mismatch repairReplication mismatchesMutS, MutL, MutH, Dam methylation
Recombination repairGaps, double-strand breaksRecA, RecBCD
SOS responseHeavy damageRecA, LexA, Pol IV, Pol V

5 important questions

Q1. What is photoreactivation?

A direct repair method in which the enzyme photolyase, activated by visible light, breaks the bonds of UV-induced thymine dimers and restores normal DNA. No nucleotides are removed.

Q2. How does mismatch repair identify the newly made strand?

The parent strand is methylated at the A of GATC sequences by Dam methylase, but the new strand is not yet methylated. MutH nicks the unmethylated (new) strand, so the error is removed from the copy, not the original.

Q3. Explain nucleotide excision repair with a diagram.

  • Recognition of the lesion by UvrA–UvrB.
  • Dual incision by UvrC on both sides.
  • Removal of the oligonucleotide by UvrD.
  • Gap filling by Pol I and sealing by ligase.
  • Human disease: xeroderma pigmentosum.

Q4. Describe the SOS repair system.

  • Signal: single-stranded DNA from damage.
  • RecA activation and LexA self-cleavage.
  • Genes induced: recA, uvr, umuDC, sulA (blocks cell division).
  • Error-prone translesion synthesis and mutagenesis.
  • Return to normal when damage is repaired.

Q5. Classify DNA repair mechanisms and explain base excision repair.

  • Direct reversal, excision (BER, NER), mismatch, recombination, SOS.
  • BER steps with uracil-DNA glycosylase as example.
  • Importance of repair in preventing mutation and cancer.

Exam tip

Learn the names "Uvr" (UV repair) and "Mut" (mutator) proteins with their jobs; a flow diagram for NER or MMR is the safest way to score in long answers.

BSc Semester 3 · Microbiology · Chapter 10

Gene Transfer in Bacteria

Quick summary

Bacteria do not reproduce sexually, but they can exchange genes. This happens by three main methods: transformation (uptake of free DNA), conjugation (transfer through cell contact) and transduction (transfer by a bacteriophage). This topic explains each, with key experiments and importance.

Key points

  • Gene transfer in bacteria is horizontal (lateral), from donor to recipient, and usually one-way.
  • Transformation: a competent recipient cell takes up naked DNA from the surroundings and joins it into its chromosome by recombination. Discovered by Frederick Griffith (1928) with Streptococcus pneumoniae; DNA shown as the agent by Avery, MacLeod and McCarty (1944).
  • Competence: the state in which a cell can take up DNA. Natural in Streptococcus, Bacillus, Haemophilus; artificial competence in E. coli by CaCl₂ treatment and heat shock or by electroporation.
  • Conjugation: DNA transfer through direct contact by a sex pilus. Discovered by Joshua Lederberg and Edward Tatum (1946) in E. coli. Bernard Davis's U-tube experiment showed cell contact is needed.
  • F⁺ × F⁻: the F (fertility) plasmid is copied by rolling circle replication and transferred; the F⁻ cell becomes F⁺.
  • Hfr × F⁻: in Hfr (high frequency of recombination) cells, the F factor is integrated into the chromosome; chromosomal genes are transferred in a fixed order, but the recipient usually stays F⁻ because the whole chromosome rarely transfers. Interrupted mating (Wollman and Jacob) used this to map genes in minutes.
  • F′ (F-prime): an F plasmid that has picked up some chromosomal genes; transfer makes the recipient partly diploid (merodiploid) — called sexduction.
  • Generalized transduction: any bacterial gene may be packed by mistake into a phage head (e.g. P22, P1) during the lytic cycle.
  • Specialized (restricted) transduction: only genes next to the prophage site are carried, due to faulty excision of a temperate phage (e.g. lambda carries gal or bio genes).
  • Transduction: discovered by Zinder and Lederberg (1952) in Salmonella with phage P22.
  • Importance: spread of antibiotic resistance (R plasmids), gene mapping, genetic engineering.

Important terms

  • Recombination – joining of donor DNA into the recipient's DNA.
  • F factor – plasmid carrying genes for pilus formation and transfer.
  • Transducing particle – phage carrying bacterial DNA.
  • Abortive transduction – donor DNA enters but is not integrated or replicated.
  • Merozygote – partially diploid bacterium after gene transfer.

Key facts

FeatureTransformationConjugationTransduction
DNA formFree (naked) DNAPlasmid or chromosome via pilusDNA inside phage
Cell contactNot neededNeededNot needed
Affected by DNaseYesNoNo
Discovered byGriffith (1928)Lederberg & Tatum (1946)Zinder & Lederberg (1952)

5 important questions

Q1. What is competence?

The physiological state in which a bacterial cell can bind and take up free DNA from its surroundings. It may be natural or made artificially by CaCl₂ treatment or electroporation.

Q2. What is an Hfr cell?

A bacterial cell in which the F plasmid has integrated into the chromosome, so it transfers chromosomal genes to F⁻ cells at a high frequency.

Q3. Describe Griffith's experiment.

  • Two strains of S. pneumoniae: S (smooth, capsulated, virulent), R (rough, non-capsulated, non-virulent).
  • Live S → mouse dies; live R → lives; heat-killed S → lives.
  • Heat-killed S + live R → mouse dies; live S bacteria recovered.
  • Conclusion: a "transforming principle" from dead S changed R into S.
  • Later shown to be DNA.

Q4. Explain conjugation in E. coli.

  • F⁺ cell forms sex pilus, contacts F⁻ cell.
  • Rolling circle copy of F plasmid moves into F⁻.
  • Both cells become F⁺.
  • Hfr × F⁻: chromosomal genes transferred in order, recipient remains F⁻.
  • F′ transfer — sexduction.
  • Davis U-tube: no recombination without contact.

Q5. Compare generalized and specialized transduction.

  • Phage: virulent/temperate (P22, P1) vs temperate (lambda).
  • Genes carried: any gene vs only genes near attachment site.
  • Origin: wrong packaging during lytic cycle vs faulty excision of prophage.
  • Frequency and use in mapping.

Exam tip

A comparison table of the three methods is almost always asked; draw the Hfr transfer and F⁺ × F⁻ diagrams.