Chapter 11 with our Microbiology MCQs and explanations! Test your knowledge and understanding of key concepts with our complete set of multiple choice questions with detailed explanations for each answer. Increase your confidence and understanding of the fascinating world of microorganisms!

Microbiology MCQs 501 to 550
- a) Change from aerobic to anaerobic
- b) Providing oxygen to anaerobically respiring structures
- c) Rapid utilization of ATP
- d) Nonsynthesis of ATP
Why other options are incorrect:
- • Option a: Switching from aerobic to anaerobic is the opposite process and relates more to the onset of fermentation.
- • Option c: The Pasteur effect actually results in more efficient ATP production, meaning less glucose is needed to meet ATP demands.
- • Option d: ATP is synthesized in both aerobic and anaerobic respiration; it is never “not synthesized” in living cells.
- a) Translocation
- b) Inversion
- c) Crossing over
- d) Duplication
Why other options are incorrect:
- • Inversion: A segment of the chromosome breaks and reinserts in reverse order on the same chromosome; the linkage group does not change.
- • Crossing over: This is the exchange of genetic material between homologous chromosomes. Since homologous chromosomes belong to the same linkage group, genes are not moving to a “different” group.
- • Duplication: This involves the production of extra copies of a gene or DNA segment, but it does not inherently involve moving that gene to a different linkage group.
- a) Recon
- b) Cistron
- c) Muton
- d) Operon
Why other options are incorrect:
- • Recon: This is the smallest unit of DNA capable of undergoing genetic recombination.
- • Muton: This is the smallest unit of DNA that can undergo a mutation (usually a single nucleotide pair).
- • Operon: This is a cluster of genes under the control of a single promoter and operator, often found in prokaryotes, which may produce multiple polypeptides.
- a) Replicated chromosomes to be separated at anaphase
- b) Homologous chromosomes of a diploid set
- c) Non-homologous chromosomes joined at the centromere
- d) Maternal and paternal chromosomes joined at the centromere
Why other options are incorrect:
- • Option b: Homologous chromosomes are pairs (one from each parent) that carry the same genes but are not physically attached as chromatids.
- • Option c: Non-homologous chromosomes belong to different pairs and never join at a single centromere.
- • Option d: Maternal and paternal chromosomes make up a homologous pair; they do not fuse into a single metaphase chromosome with two chromatids.
- a) Transferase
- b) Hydrolase
- c) Isomerase
- d) Oxido reductase
Why other options are incorrect:
- • Transferase: These enzymes transfer a functional group (like a methyl or phosphate group) from one molecule to another.
- • Hydrolase: These enzymes catalyze the cleavage of chemical bonds by the addition of water (hydrolysis).
- • Isomerase: These enzymes catalyze the structural rearrangement of isomers (converting a molecule from one isomer to another).
- a) Gal and biogenes
- b) Bio and niacin genes
- c) Gal and B genes
- d) None of these
Why other options are incorrect:
- • Option b: Niacin genes are not the flanking markers for the λ attachment site in the E. coli genome.
- • Option c: While “Gal” is correct, “B genes” is an imprecise term. The correct neighbor is specifically the biotin (bio) operon.
- • Option d: Option (a) is the scientifically accepted location for the attB site.
- a) Structural gene
- b) Regulator gene
- c) Operator gene
- d) Promoter gene
Why other options are incorrect:
- • Regulator gene: It codes for a repressor or activator protein that controls the expression of other genes, rather than the metabolic polypeptide itself.
- • Operator gene: This is a segment of DNA to which a transcription factor (repressor) binds; it does not code for a protein.
- • Promoter gene: This is the binding site for RNA polymerase to initiate transcription; it is a regulatory sequence, not a coding sequence for a polypeptide.
- a) G1 – phase
- b) S – phase
- c) G2 – phase
- d) M – phase
Why other options are incorrect:
- • G1 – phase: This is the first gap phase where the cell grows and prepares for DNA replication, but the actual synthesis hasn’t started yet.
- • G2 – phase: This is the second gap phase where the cell continues to grow and produces proteins necessary for cell division after DNA replication is complete.
- • M – phase: This is the Mitotic phase where the cell actually divides its copied DNA and cytoplasm to make two new cells.
- a) Prophase I
- b) Prophase II
- c) Anaphase I
- d) Telophase II
Why other options are incorrect:
- • Prophase II: This phase occurs in haploid cells where homologous pairs are no longer present together; therefore, crossing-over cannot occur.
- • Anaphase I: In this stage, homologous chromosomes are pulled apart toward opposite poles; the exchange of DNA has already finished.
- • Telophase II: This is the final stage of meiosis where nuclear membranes reform around the four resulting haploid daughter cells.
- a) Conservative
- b) Semiconservative
- c) Dispersive
- d) None of the above
Why other options are incorrect:
- • Conservative: In this hypothetical model, the original parent DNA molecule would remain entirely intact, and an entirely new double-stranded DNA molecule would be formed.
- • Dispersive: This model suggested that the parent DNA strands would be broken into fragments and the resulting daughter strands would contain a mixture of old and new DNA interspersed throughout.
- • Option d: Semiconservative replication is the universally accepted biological mechanism for DNA duplication.
- a) Translation
- b) RNA splicing
- c) Transcription
- d) Transposition
Why other options are incorrect:
- • Translation: This is the process that follows transcription, where the mRNA sequence is used to assemble a chain of amino acids (protein).
- • RNA splicing: This is a modification process where non-coding introns are removed from the pre-mRNA transcript.
- • Transposition: This refers to “jumping genes” or the movement of DNA segments to different positions within the genome.
- a) Alanine
- b) Adenine
- c) Lysine
- d) Arginine
Why other options are incorrect:
- • Alanine: This is a non-essential amino acid used in the biosynthesis of proteins, not nucleic acids.
- • Lysine: This is an essential amino acid required for protein synthesis.
- • Arginine: This is an amino acid that plays a role in cell division and immune function, but is not a part of the nucleic acid structure.
- a) N-bases
- b) Nucleosides
- c) Nucleotides
- d) Histones
Why other options are incorrect:
- • N-bases: Nitrogenous bases (A, T, C, G, U) are only one component of a nucleotide, not the complete structural unit itself.
- • Nucleosides: A nucleoside is composed only of a nitrogenous base and a sugar. It becomes a nucleotide only after a phosphate group is added.
- • Histones: Histones are alkaline proteins that provide structural support to chromosomes, but they are not part of the chemical structure of nucleic acids.
- a) Enzymes
- b) Hormones
- c) RNA
- d) DNA
Why other options are incorrect:
- • Hormones: While some hormones are proteins coded by genes, many others (like steroids) are lipids synthesized by enzymatic pathways, not directly by a gene.
- • RNA: Although genes are transcribed into RNA, in most cases, RNA is simply an intermediate messenger used to reach the final goal of protein/enzyme production.
- • DNA: DNA synthesis (replication) is the process of copying the genetic material for cell division, rather than the “functional expression” of a gene.
- a) Baldness
- b) Red-green colour blindness
- c) Facial hair/moustache in males
- d) Night blindness
Why other options are incorrect:
- • Baldness: Pattern baldness is typically a sex-influenced trait, meaning the genes are on autosomes but their expression is influenced by male hormones (testosterone).
- • Facial hair/moustache: These are secondary sexual characteristics influenced by Y-chromosome linked genes or hormonal responses; they are not exclusive to X-chromosome inheritance.
- • Night blindness: Most commonly caused by Vitamin A deficiency or autosomal genetic disorders like Retinitis Pigmentosa, rather than being an exclusive X-linked trait.
- a) Nucleotides
- b) Amino acids
- c) Glucose molecules
- d) Sucrose
Why other options are incorrect:
- • Nucleotides: These are joined together by phosphodiester bonds to form nucleic acids (DNA/RNA).
- • Glucose molecules: These are linked by glycosidic bonds to form polysaccharides like starch or glycogen.
- • Sucrose: This is a disaccharide already held together by a glycosidic bond between glucose and fructose; it is not a building block for peptide linkages.
- a) DNA
- b) RNA – (+) type
- c) t-RNA
- d) m-RNA
Why other options are incorrect:
- • DNA: Poliovirus belongs to the Picornaviridae family, which are exclusively RNA viruses; they do not contain DNA.
- • t-RNA: Transfer RNA is a functional RNA molecule used in protein synthesis, not the genomic material of the virus.
- • m-RNA: While the (+) RNA behaves like mRNA, the nucleic acid itself is classified as the viral genome (+ssRNA) rather than cellular m-RNA.
- a) Naked RNA virus
- b) Naked DNA virus
- c) Enveloped RNA virus
- d) Enveloped DNA virus
Why other options are incorrect:
- • Naked RNA virus: Rabies possesses a lipid bilayer envelope; “naked” viruses (like Polio) lack this outer layer.
- • Naked DNA virus: Rabies is an RNA virus, not a DNA virus, and it is not naked.
- • Enveloped DNA virus: While it is enveloped, the genetic material is RNA, making this option incorrect.
- a) Polio virus
- b) Adeno virus
- c) Echo virus
- d) Poty virus
Why other options are incorrect:
- • Polio virus: A member of the Picornaviridae family, which uses single-stranded RNA (ssRNA).
- • Echo virus: Also part of the Picornaviridae family (Enterovirus genus), it is an RNA virus.
- • Poty virus: A large group of plant viruses that possess a single-stranded RNA genome.
- a) Restriction enzymes
- b) Ligases
- c) Nucleases
- d) Hydralases
Why other options are incorrect:
- • Ligases: These enzymes act as “molecular glue” that join DNA fragments together rather than breaking them.
- • Nucleases: While restriction enzymes are a type of nuclease, the term is too broad; in genetic engineering contexts, “Restriction enzymes” is the specific answer for targeted DNA cutting.
- • Hydralases: This is a general class of enzymes that catalyze the hydrolysis of various chemical bonds, but they are not the specific tools used for DNA manipulation in genetic engineering.
- a) Transduction
- b) Induction
- c) Transfection
- d) Infection
Why other options are incorrect:
- • Induction: This refers to the process where a prophage is stimulated to leave the bacterial chromosome and enter the lytic cycle.
- • Transfection: This is the process of deliberately introducing naked or purified nucleic acids into eukaryotic cells.
- • Infection: This is a general term for the invasion and multiplication of microorganisms in a host, but it does not specifically describe the mechanism of gene transfer.
- a) Acetic acid
- b) Pyruvic acid
- c) α-ketoglutaric acid
- d) Fumaric acid
Why other options are incorrect:
- • Acetic acid: While Acetyl-CoA is the activated form of acetate, microorganisms typically derive it from the breakdown of sugars via pyruvate, not by oxidizing acetic acid directly as a primary pathway.
- • α-ketoglutaric acid: This is a keto acid produced *within* the TCA cycle after Acetyl-CoA has already reacted with oxaloacetate.
- • Fumaric acid: This is an intermediate found late in the TCA cycle, resulting from the oxidation of succinate.
- a) Semi conservative
- b) Conservative
- c) Dispersive
- d) Rolling loop
Why other options are incorrect:
- • Conservative: This model suggested that the original DNA molecule stayed intact and an entirely new double-stranded molecule was synthesized.
- • Dispersive: This model suggested that the parental DNA was broken into fragments and the resulting DNA molecules were a patchwork of old and new segments.
- • Rolling loop: Also known as rolling circle replication, this is a specific mechanism used by some circular DNAs (like plasmids or viruses), but it was not the general model proposed for DNA by Watson and Crick.
- a) 20 Å
- b) 34 Å
- d) 42 Å
Why other options are incorrect:
- • 20 Å: This is the diameter (width) of the DNA double helix, not the length of a turn.
- • 28 Å: This value does not correspond to the standard pitch of the B-DNA helix.
- • 42 Å: This value is significantly larger than the standard dimensions of a B-DNA turn.
- a) RNA
- b) DNA
- c) Proteins
- d) All of these
Why other options are incorrect:
- • RNA: The detection and study of RNA sequences is performed using Northern blotting.
- • DNA: The detection of specific DNA sequences is performed using Southern blotting.
- • All of these: Blotting techniques are specific to the molecule being studied; “Western” specifically refers only to protein analysis.
- a) Transcription
- b) Transformation
- c) Translation
- d) Replication
Why other options are incorrect:
- • Transformation: This refers to the genetic alteration of a cell resulting from the direct uptake and incorporation of exogenous genetic material from its surroundings.
- • Translation: This is the subsequent step where the genetic code carried by mRNA is decoded to produce a specific sequence of amino acids in a polypeptide chain (protein).
- • Replication: This is the process of producing two identical replicas of DNA from one original DNA molecule, essentially copying the entire genome before cell division.
- a) DNA
- b) RNA
- c) Protein
- d) Polysaccharides
Why other options are incorrect:
- • DNA: DNA is detected using the Southern blotting technique.
- • RNA: RNA is detected using the Northern blotting technique.
- • Polysaccharides: These are typically analyzed through methods like periodic acid-Schiff (PAS) staining or chromatography, rather than standard blotting techniques.
Mnemonic: SNOW DROP (Southern-DNA, Northern-RNA, Western-Protein).
- a) Amino acids
- b) Nucleosides
- c) Nucleotides
- d) Nucleo proteins
Why other options are incorrect:
- • Amino acids: These are the building blocks of proteins, not nucleic acids.
- • Nucleosides: They contain a nitrogenous base and sugar but lack the phosphate group required for nucleotides.
- • Nucleo proteins: These are complexes of nucleic acids and proteins, not the basic building blocks of nucleic acids.
- a) Repetitive sequences
- b) Unique sequences
- c) Amplified sequences
- d) Non-coding sequences
Why other options are incorrect:
- • Unique sequences: These are generally the same in all cells of an individual and are not used for DNA profiling.
- • Amplified sequences: PCR amplification may be used during analysis, but fingerprinting is not based on amplification itself.
- • Non-coding sequences: Many fingerprint markers are located in non-coding regions, but the key principle is the presence of repetitive sequences.
- a) RNA polymerase
- b) Reverse transcriptase
- c) DNA polymerase
- d) Terminal transferase
Why other options are incorrect:
- • RNA polymerase: Synthesizes RNA from a DNA template, not DNA from RNA.
- • DNA polymerase: Synthesizes DNA from DNA template during replication.
- • Terminal transferase: Adds nucleotides to the 3′ end of DNA but does not use an RNA template.
- a) Reo virus
- b) Rhabdo virus
- c) Parvo virus
- d) Retro virus
Why other options are incorrect:
- • Rhabdo virus: Contains single-stranded RNA (negative sense), e.g., rabies virus.
- • Parvo virus: Has single-stranded DNA genome, not RNA.
- • Retro virus: Contains single-stranded RNA but uses reverse transcriptase to form DNA.
- a) Adeno virus
- b) Bacteriophage T1, T2, T3, T4, T5, T6
- c) Papova virus
- d) Herpes virus and cauliflower mosaic
- e) All of the above
Why other options are incorrect:
- Each individual option (a–d) is correct, so the best answer is “All of the above.”
- a) Reo viruses
- b) Retro viruses
- c) Bacteriophage Φ C
- d) TMV and Bacteriophages MS2, F2
- e) Dahlia mosaic virus and Bacteriophages ΦX174, M12, M13
Why other options are incorrect:
- a) Reo viruses: Double-stranded RNA viruses.
- b) Retro viruses: RNA viruses that use reverse transcriptase.
- c) Bacteriophage ΦC: Considered in RNA phage grouping in basic classifications.
- d) TMV, MS2, F2: All are RNA viruses.
- a) Ionic bonds
- b) Covalent bonds
- c) Hydrogen bonds between bases
- d) Polar charges
Why other options are incorrect:
- • Ionic bonds: Not responsible for base pairing between DNA strands.
- • Covalent bonds: Occur within each strand (phosphodiester bonds), not between strands.
- • Polar charges: Not a bonding type responsible for DNA strand pairing.
- a) Double hydrogen bonds
- b) Single hydrogen bonds
- c) Triple hydrogen bonds
- d) Both Single hydrogen bonds and Triple hydrogen bonds
Why other options are incorrect:
- • Single hydrogen bonds: Not sufficient for stable A–T pairing in DNA.
- • Triple hydrogen bonds: Found in Guanine–Cytosine pairing, not A–T.
- • Both Single and Triple hydrogen bonds: Incorrect combination; A–T specifically forms double bonds only.
- a) 15 Å
- b) 34 Å
- c) 30 Å
- d) 5 Å
Why other options are incorrect:
- • 15 Å: Not the standard measurement for one DNA helical turn.
- • 30 Å: Close but incorrect; actual value is ~34 Å.
- • 5 Å: Represents spacing between individual base pairs (~3.4 Å), not a full turn.
- a) There is phosphodiester bond between 5’- hydroxyl of one ribose and 3’–hydroxyl of next ribose
- b) They have positive and negative ends
- c) Nucleotides are charged structures
- d) Nitrogenous bases are highly ionized compounds
Why other options are incorrect:
- • Positive and negative ends: DNA has directionality (5’ and 3’ ends) but not opposite electrical charges.
- • Nucleotides are charged structures: Individual nucleotides are not the main reason for overall polymer charge.
- • Nitrogenous bases: Bases are not highly ionized; the charge mainly comes from phosphate groups.
- a) Genome
- b) Gene map
- c) Gene-structure
- d) Chromatin
Why other options are incorrect:
- • Genome: Refers to the entire genetic material of an organism, not a diagram.
- • Gene-structure: Refers to internal organization of a gene, not mapping.
- • Chromatin: DNA-protein complex in chromosomes, not a representation of gene numbers.
- a) Large amount of DNA is transferred
- b) A few no. of genes are transferred
- c) Whole DNA is transferred
- d) None of these
Why other options are incorrect:
- • Large amount of DNA: Occurs in generalized transduction, not specialized.
- • Whole DNA is transferred: Not possible in transduction; only fragments are transferred.
- • None of these: Incorrect because option (b) is correct.
- a) Endogenate
- b) Exogenate
- c) Mesozygote
- d) Merosite
Why other options are incorrect:
- • Endogenate: Refers to the recipient or internal genetic material, not the donor.
- • Mesozygote: An intermediate state in partial diploidy (e.g., in conjugation), not transformation.
- • Merosite: Not related to bacterial DNA transfer mechanisms.
- a) Transcription (Transcriptase)
- b) Transduction
- c) Transformation
- d) Recombination
Why other options are incorrect:
- • Transduction: Transfer of DNA via bacteriophages.
- • Transformation: Uptake of naked DNA from environment.
- • Recombination: Rearrangement or exchange of genetic material.
- a) Transformation
- b) Transduction
- c) Conjugation
- d) Cell fusion
- e) All of the above
Why other options are incorrect:
- Each option individually represents a correct gene transfer mechanism, so the most complete answer is “All of the above.”
- a) HCV
- b) HAV
- c) HBV
- d) HIV
Why other options are incorrect:
- • HCV: Also blood-borne but historically not called serum hepatitis in classical classification.
- • HAV: Causes infectious (fecal-oral) hepatitis, not serum hepatitis.
- • HIV: Causes AIDS, not hepatitis.
- a) IgE
- b) IgA
- c) IgM
- d) IgG
Why other options are incorrect:
- • IgE: Involved in allergic reactions and parasitic infections.
- • IgA: Found in mucosal secretions like saliva and mucus.
- • IgG: Most abundant antibody but not primarily responsible for agglutination of RBCs.
- a) Immunogenic
- b) Non-immunogenic
- c) Antigenic
- d) None of these
Why other options are incorrect:
- • Immunogenic: Haptens alone cannot trigger immune response.
- • Antigenic: Partially true, but MCQ focus is on immunogenic property; correct classification is non-immunogenic alone.
- • None of these: Incorrect because option (b) is correct.
- a) IgG
- b) IgM
- c) IgD
- d) IgE
Why other options are incorrect:
- • IgG: Appears later and provides long-term immunity.
- • IgD: Mainly acts as a receptor on B cells.
- • IgE: Involved in allergic reactions and parasitic infections.
- a) B-lymphocytes
- b) T-lymphocytes
- c) Monocytes
- d) RBC’s
Why other options are incorrect:
- • T-lymphocytes: Involved in cell-mediated immunity, not antibody production.
- • Monocytes: Differentiate into macrophages, mainly involved in phagocytosis.
- • RBC’s: Function in oxygen transport, not immune response.
- a) Chiasmata
- b) Synaptonemal complex
- c) Centromeres
- d) Protein axes
Why other options are incorrect:
- • Synaptonemal complex: Protein structure that holds homologous chromosomes together, but is not the crossover point.
- • Centromeres: Regions where sister chromatids attach, not involved in crossing over.
- • Protein axes: Structural components of chromosomes, not the site of crossover.
- a) 8%
- b) 12%
- c) 16%
- d) 4%
Why other options are incorrect:
- • 8%: Too low for total globulin fraction in standard MCQ schemes.
- • 16%: Higher than commonly used simplified values in exam keys.
- • 4%: This value is closer to fibrinogen, not globulins.
- a) Activators
- b) Substrates
- c) Inhibitor
- d) Cofactor
Why other options are incorrect:
- • Activators: Increase enzyme activity, not inhibit metabolism.
- • Substrates: Natural molecules acted upon by enzymes, not inhibitory analogs.
- • Cofactor: Non-protein helper molecules required for enzyme activity.

