Microbiology - MCQ Practice Questions
Microbiology sits behind a lot of applied biology, so the fundamentals here carry into medicine, food technology and biotechnology alike. Practice covers bacterial structure and growth, viruses, fungi and parasites, sterilisation and culture techniques, immunology, and microbial genetics. Technique based questions explain the purpose of each step, because that is usually what separates a memorised protocol from a usable one.
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Which of the following best describes the process of transformation in bacteria?
Understanding:
We need to identify the correct definition of bacterial transformation among four mechanisms of genetic exchange.
Step 1: Define transformation
Transformation is the process by which a competent bacterial cell takes up free (naked) DNA released into the environment — typically from lysed donor cells — and stably incorporates it into its own genome. This was first demonstrated by Frederick Griffith in 1928 using Streptococcus pneumoniae and later confirmed as a DNA-mediated process by Avery, MacLeod, and McCarty in 1944.
Step 2: Eliminate incorrect options
Transfer of genetic material via a bacteriophage describes transduction. Transfer through direct cell-to-cell contact via a pilus describes conjugation. Exchange of chromosomal segments during direct physical contact resembles conjugation or recombination, not transformation.
Step 3: Confirm the correct concept
Transformation specifically involves the direct uptake of exogenous naked DNA by a competent bacterium. Competence may be natural (e.g., Bacillus subtilis, Streptococcus pneumoniae) or artificially induced (e.g., by calcium chloride treatment in the laboratory).
Answer:
Transformation is defined as the direct uptake and incorporation of naked exogenous DNA from the surrounding environment by a competent bacterial cell.
Quick Tip:
Remember the triad — Transformation (naked DNA), Transduction (phage), Conjugation (pilus/cell contact). This distinction is a frequent exam trap.
In bacterial conjugation, the fertility factor (F plasmid) in an Hfr (High frequency recombination) strain is integrated into the bacterial chromosome. What is the primary consequence of mating between an Hfr strain and an F⁻ strain?
Understanding:
We need to understand the outcome of Hfr × F⁻ conjugation with respect to gene transfer and F factor acquisition by the recipient.
Step 1: Recall Hfr structure
In an Hfr strain, the F plasmid is integrated into the bacterial chromosome. During conjugation, DNA transfer begins at the origin of transfer (oriT) within the integrated F factor and proceeds linearly into the F⁻ recipient.
Step 2: Analyse the transfer sequence
The F factor is split upon integration: part of it (oriT end) leads the transfer, and the remaining part of F trails at the very end of the chromosome. Because the entire bacterial chromosome (~100 minutes in E. coli) rarely transfers completely before conjugal contact breaks, the trailing F sequences almost never reach the recipient.
Step 3: Determine the outcome
As a result, chromosomal genes are transferred at high frequency (hence the name Hfr), but the recipient cell does not receive the complete F factor. Therefore, the recipient remains F⁻ in the vast majority of matings. Only if the entire chromosome is transferred — a rare event — would the recipient become Hfr.
Answer:
In Hfr × F⁻ matings, chromosomal genes are transferred at high frequency but the recipient almost always remains F⁻ because the complete F factor is rarely transferred.
Quick Tip:
Contrast with F⁺ × F⁻ matings, where the F plasmid transfers efficiently but chromosomal gene transfer is rare. The key difference is the position of oriT relative to chromosomal genes.
A mutation that changes a codon from UAC (Tyrosine) to UAA is best classified as which type of mutation?
Understanding:
We need to classify a point mutation where codon UAC (Tyrosine) changes to UAA.
Step 1: Identify the original and mutant codons
UAC codes for the amino acid Tyrosine. UAA is one of the three stop codons (UAA, UAG, UGA) and does not code for any amino acid.
Step 2: Define mutation types
A missense mutation changes a codon so that it codes for a different amino acid. A silent (synonymous) mutation changes a codon but it still codes for the same amino acid. A frameshift mutation involves insertion or deletion of nucleotides (not a multiple of 3), shifting the reading frame. A nonsense mutation changes a sense codon (one coding for an amino acid) into a stop codon, causing premature termination of translation.
Step 3: Classify the mutation
Since UAC (amino acid codon) → UAA (stop codon), this is a classic nonsense mutation. It causes premature termination of translation, typically producing a truncated, non-functional protein.
Answer:
A change from UAC (Tyrosine) to UAA (stop codon) is a nonsense mutation because a sense codon is converted into a stop codon.
Quick Tip:
The three stop codons — UAA (ochre), UAG (amber), UGA (opal/umber) — are worth memorising. Any point mutation that generates one of these from an amino acid codon is, by definition, a nonsense mutation.
Which of the following correctly describes the function of the lac repressor in the absence of lactose in the lac operon system of E. coli?
Understanding:
We need to describe the role of the lac repressor when lactose is absent from the medium in E. coli.
Step 1: Recall the lac operon architecture
The lac operon consists of the promoter (P), operator (O), and three structural genes — lacZ, lacY, and lacA — encoding β-galactosidase, lactose permease, and thiogalactoside transacetylase, respectively. The lacI gene, located upstream, constitutively produces the lac repressor protein.
Step 2: Repressor behaviour without lactose
In the absence of lactose, the lac repressor (a tetrameric protein encoded by lacI) is in its active conformation. It binds with high affinity to the operator sequence (O), which overlaps with the transcription start site. This binding physically blocks the movement of RNA polymerase from the promoter into the structural genes, preventing transcription.
Step 3: Eliminate incorrect options
The repressor does not bind the promoter — it binds the operator. Allolactose (the true inducer) binds to the repressor and causes it to dissociate from the operator — the repressor is not activated by allolactose. The repressor is not degraded; it remains stable and active in the absence of inducer.
Answer:
In the absence of lactose, the active lac repressor binds to the operator region and blocks transcription of the structural genes.
Quick Tip:
Remember: inducer (allolactose) binds the repressor → repressor leaves the operator → transcription occurs. This is the classic example of negative inducible regulation.
Transposons (transposable elements) are sometimes called 'jumping genes'. Which enzyme, encoded by the transposon itself, is essential for the transposition process?
Understanding:
We need to identify the specific enzyme encoded by transposons that catalyses the transposition reaction.
Step 1: Recall transposon structure
A transposon (e.g., Tn3 in bacteria) typically contains inverted terminal repeats (ITRs) at both ends and carries genes including antibiotic resistance markers and the gene encoding the enzyme required for its own movement.
Step 2: Identify the catalytic enzyme
Transposase is the enzyme encoded by the transposon that recognises the terminal inverted repeats, catalyses the excision (cut) of the transposon from the donor site, and mediates its insertion into a new target site in the genome. Without transposase, the transposon cannot move.
Step 3: Eliminate incorrect options
Ligase joins DNA strands but is a host enzyme in this context, not transposon-encoded for movement. Helicase unwinds DNA during replication and is unrelated to transposition. Primase synthesises RNA primers for DNA replication — it has no role in transposition.
Answer:
Transposase is the enzyme encoded by the transposon itself that is essential for recognising the terminal repeats and catalysing the cut-and-paste mechanism of transposition.
Quick Tip:
Some transposons also encode a resolvase enzyme (e.g., Tn3), which resolves the cointegrate intermediate in replicative transposition — a detail that distinguishes replicative from conservative (cut-and-paste) transposition.
In the Ames test used to assess the mutagenic potential of chemical compounds, which of the following organisms is used as the test system?
Understanding:
We need to identify the specific test organism used in the Ames test for mutagenicity screening.
Step 1: Recall the principle of the Ames test
Developed by Bruce Ames in the 1970s, the Ames test detects mutagenic chemicals by measuring their ability to cause reverse mutations (revertants) in specially designed bacterial strains. Revertants are bacteria that regain the ability to synthesise an essential nutrient they previously could not produce.
Step 2: Identify the test strain
The Ames test uses histidine auxotrophic mutants of Salmonella typhimurium (now reclassified as Salmonella enterica). These strains carry mutations in genes of the histidine biosynthesis pathway (his⁻) and cannot grow on histidine-deficient medium. When exposed to a mutagen, some cells undergo reverse mutation back to his⁺ (prototrophy) and form colonies on histidine-free agar. The number of revertant colonies indicates the mutagenic potency of the test compound.
Step 3: Note the role of rat liver extract
To simulate mammalian metabolism, rat liver S9 fraction (containing cytochrome P450 enzymes) is often added, as many compounds require metabolic activation to become mutagenic.
Answer:
The Ames test employs histidine auxotrophic mutants of Salmonella typhimurium to detect chemical mutagens through reversion to histidine prototrophy.
Quick Tip:
The Ames test has a sensitivity of about 90% for carcinogens, because most carcinogens are also mutagens. The addition of the S9 liver fraction is the key feature that bridges in vitro bacterial testing with in vivo mammalian metabolism.
Which of the following statements correctly describes the difference between generalised transduction and specialised transduction?
Understanding:
We need to distinguish between generalised and specialised (restricted) transduction based on their mechanisms and the genes they transfer.
Step 1: Generalised transduction
During generalised transduction, a lytic bacteriophage (e.g., P1 in E. coli, P22 in Salmonella) accidentally packages a fragment of host bacterial DNA instead of phage DNA during the assembly stage. Because this packaging error can occur at essentially any location in the chromosome, any bacterial gene can be transferred. The phage head is filled with host DNA by a 'headful' packaging mechanism that is not sequence-specific.
Step 2: Specialised (restricted) transduction
Specialised transduction is mediated by temperate phages (e.g., lambda phage in E. coli). When the integrated prophage excises imprecisely from the bacterial chromosome, it takes along adjacent bacterial genes. Lambda integrates at a specific att site between the gal and bio operons; imprecise excision therefore transfers only these specific flanking genes (gal or bio), not arbitrary genes.
Step 3: Eliminate incorrect options
Generalised transduction does NOT require phage integration; it occurs during lytic infection. Specialised transduction IS mediated by lysogenic (temperate) phages, not lytic phages — so option D has the relationship reversed.
Answer:
Generalised transduction can transfer any bacterial gene randomly, whereas specialised transduction transfers only specific bacterial genes located adjacent to the phage integration site.
Quick Tip:
A useful mnemonic: Generalised = General (any gene, lytic phage accident); Specialised = Specific (only flanking genes, lysogenic phage imprecise excision).
DNA repair mechanism that removes a damaged or incorrect base by breaking the N-glycosidic bond between the base and the deoxyribose sugar, leaving an apurinic/apyrimidinic (AP) site, is called:
Understanding:
We need to identify the DNA repair mechanism that specifically cleaves the N-glycosidic bond to remove a damaged base, generating an AP site.
Step 1: Define the mechanism described
The cleavage of the N-glycosidic bond between a damaged base and its deoxyribose sugar is the hallmark of base excision repair (BER). The enzyme that performs this cleavage is called a DNA glycosylase. Each glycosylase has specificity for particular types of damaged or abnormal bases (e.g., uracil-DNA glycosylase removes uracil from DNA; 8-oxoguanine DNA glycosylase removes oxidised guanine).
Step 2: Describe what follows
After glycosylase removes the base, an AP endonuclease cleaves the phosphodiester backbone at the AP site. DNA polymerase fills in the gap, and DNA ligase seals the nick.
Step 3: Eliminate incorrect options
Nucleotide excision repair removes a short oligonucleotide segment (12–13 nt in prokaryotes, 25–32 nt in eukaryotes) containing the bulky lesion — it does not specifically create AP sites. Mismatch repair corrects replication errors by recognising base-pair mismatches and removing the newly synthesised strand. Photoreactivation uses photolyase enzyme and visible light to directly reverse pyrimidine dimers without creating AP sites.
Answer:
The repair mechanism that removes damaged bases by cleaving the N-glycosidic bond and creating an AP site is base excision repair.
Quick Tip:
Base excision repair handles small, non-helix-distorting lesions (deamination, oxidation, alkylation). Nucleotide excision repair handles bulky, helix-distorting lesions such as pyrimidine dimers caused by UV light — this distinction is frequently tested.
The SOS response in bacteria is a global regulatory response to DNA damage. Which protein acts as the co-protease that facilitates auto-cleavage of the LexA repressor to induce the SOS genes?
Understanding:
We need to identify the protein that acts as a co-protease to promote LexA repressor cleavage and thereby induce the bacterial SOS response.
Step 1: Recall LexA and SOS regulation
Under normal conditions, the LexA protein represses over 40 SOS genes (including recA, uvrA, uvrB, sulA) by binding to SOS boxes (operator sequences) in their promoters. The SOS genes encode proteins involved in DNA repair, damage tolerance, and mutagenesis.
Step 2: Activation of the SOS response
When DNA damage generates single-stranded DNA (ssDNA) — for example, at stalled replication forks — RecA protein binds cooperatively to the ssDNA and forms a nucleoprotein filament. This activated RecA filament (RecA*) acts as a co-protease: it stimulates the latent auto-cleavage activity of LexA. LexA cleaves itself at an Ala-Gly bond, inactivating the repressor and de-repressing all SOS genes simultaneously.
Step 3: Eliminate incorrect options
UvrA is itself an SOS-induced gene product involved in nucleotide excision repair, not a co-protease for LexA. DnaA is the initiator protein for chromosomal replication. MutL is a component of the mismatch repair system and has no role in LexA cleavage.
Answer:
RecA, upon binding to single-stranded DNA, forms an activated filament (RecA*) that acts as a co-protease to stimulate auto-cleavage of the LexA repressor, thereby inducing the SOS response.
Quick Tip:
RecA also facilitates the auto-cleavage of the UmuD protein (to UmuD') and the lambda phage CI repressor by the same co-protease mechanism — a detail that links SOS mutagenesis (UmuC/UmuD') and phage induction.
Which of the following plasmid properties ensures that copies of the plasmid are distributed to both daughter cells during bacterial cell division, preventing plasmid loss from the population?
Understanding:
We need to identify the plasmid property responsible for ensuring faithful distribution (segregation) of plasmid copies to daughter cells at cell division.
Step 1: Define the problem of plasmid maintenance
For a plasmid to be stably maintained in a bacterial population over many generations, it must be reliably partitioned into both daughter cells each time the host divides. If partitioning is random or absent, some daughter cells will not receive a copy — a phenomenon called plasmid curing.
Step 2: Role of the par locus
The partitioning system, encoded by the par locus (e.g., parABS system in low-copy plasmids like F and P1), actively mediates plasmid segregation. It consists of a centromere-like sequence (parS/incC), a DNA-binding protein (ParB/SopB), and an ATPase (ParA/SopA) that generates directed movement to push plasmid copies toward opposite poles of the dividing cell, analogous to the mitotic spindle in eukaryotes.
Step 3: Eliminate incorrect options
Conjugative ability (tra genes) and the origin of transfer (oriT) are required for plasmid transfer to other cells via conjugation — they have no role in segregation within a dividing cell. Antibiotic resistance genes provide a selective advantage for cells that retain the plasmid, but they do not actively ensure partitioning; they only allow plasmid-free cells to be eliminated under selection pressure.
Answer:
The partitioning system (par locus) actively ensures plasmid copies are distributed to both daughter cells, preventing plasmid loss during cell division.
Quick Tip:
High-copy-number plasmids can rely on random distribution (statistical certainty that each daughter receives at least one copy), but low-copy-number plasmids like F require an active par system for stable maintenance — this is a common exam distinction.