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CHSE Odisha Elective Biotechnology study materials
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CHSE Odisha Elective Biotechnology practice questions

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Question 1Introduction to Biotechnology and Historical Development

Which discovery is most widely credited as the starting point of modern biotechnology, because it showed that a specific microorganism causes a specific disease?

  1. James Watson and Francis Crick's model of the DNA double helix
  2. Károly Ereky's coining of the term biotechnology
  3. Robert Koch's identification of Bacillus anthracis as the cause of anthrax Correct answer.
  4. Alexander Fleming's observation of penicillin mould inhibiting bacteria

Explanation

Koch's work in the 1870s established that a defined bacterium, Bacillus anthracis, causes a defined disease, anthrax, and he formalised the postulates linking microbe to disease. This germ-theory proof gave biotechnology its first reliable tool for controlling infectious disease. The other events are important but came later or concern different fields.

Question 2Introduction to Biotechnology and Historical Development

In the traditional history of biotechnology, which process is placed in the 'ancient' phase rather than the modern phase?

  1. Production of human insulin in genetically engineered bacteria
  2. Large-scale culture of hybridoma cells for monoclonal antibodies
  3. Use of selected yeasts and lactic acid bacteria to make bread, curd and fermented beverages Correct answer.
  4. Development of recombinant DNA vaccines against hepatitis B

Explanation

The ancient phase of biotechnology covers empirical food fermentations practised long before microbes were understood, such as leavening dough, souring milk and brewing. These rely on naturally occurring yeasts and bacteria rather than deliberate genetic manipulation. Recombinant insulin, subunit vaccines and hybridoma technology all belong to the modern phase.

Question 3Introduction to Biotechnology and Historical Development

Károly Ereky is remembered in the history of biotechnology chiefly because in 1919 he did which of the following?

  1. Discovered the enzyme reverse transcriptase in retroviruses
  2. First cloned a vertebrate animal from a somatic cell nucleus
  3. Coined the term biotechnology to describe technologies based on converting raw materials using living organisms Correct answer.
  4. Established the first industrial penicillin fermentation plant

Explanation

Ereky, a Hungarian agricultural engineer, introduced the word biotechnology in a 1919 book to describe the use of living organisms to convert raw materials into useful products. The word later broadened to cover genetic engineering and industrial biology. Cloning, reverse transcriptase and penicillin fermentation all belong to later decades.

Question 4Introduction to Biotechnology and Historical Development

Which statement best describes the contribution of Louis Pasteur that made later biotechnological work possible?

  1. He demonstrated that DNA is the transforming principle in bacteria
  2. He showed that microorganisms arise from pre-existing microbes and that heat treatment can destroy unwanted ones in liquids Correct answer.
  3. He constructed the first recombinant plasmid vector for gene cloning
  4. He proved that enzymes are proteins with catalytic activity

Explanation

Pasteur's experiments refuted spontaneous generation and established that microbes come from existing microbes, which underpins sterile technique. His work on heating liquids to kill spoilage organisms, pasteurisation, gave a practical method for controlling microbial contamination. These ideas made pure-culture fermentation and later industrial microbiology feasible.

Question 5Introduction to Biotechnology and Historical Development

A student argues that the 1953 DNA double-helix model was the single event that created genetic engineering. Which correction best reflects the historical record?

  1. The model supplied the structural basis for heredity, but gene transfer required later tools such as restriction enzymes and ligase Correct answer.
  2. The model was irrelevant, because genetic engineering depends only on fermentation technology
  3. The model was published after the first recombinant DNA experiment and merely explained it
  4. The model directly produced the first recombinant bacterium without any further discoveries

Explanation

Watson and Crick's model explained how genetic information could be stored and copied, but it did not by itself permit the cutting and joining of DNA. Recombinant DNA technology emerged in the 1970s after restriction endonucleases, DNA ligase and plasmid vectors were characterised. The double helix was therefore a necessary conceptual foundation, not the whole technology.

Question 6Introduction to Biotechnology and Historical Development

Which pair correctly matches a milestone with the achievement associated with it?

  1. Ian Wilmut - construction of the first recombinant plasmid
  2. Kary Mullis - development of the polymerase chain reaction Correct answer.
  3. Khorana - chemical synthesis of a defined gene, later used in a functional bacterial gene
  4. Frederick Sanger - discovery of restriction endonucleases

Explanation

Kary Mullis conceived the polymerase chain reaction in 1983, a method for amplifying specific DNA sequences through repeated cycles of denaturation, annealing and extension. It became one of the most widely used techniques in biotechnology and diagnostics. The other pairings confuse the researchers with different achievements.

Question 7Introduction to Biotechnology and Historical Development

In 1973, Cohen and Boyer reported a landmark experiment in which they did which of the following?

  1. Sequenced the complete genome of Escherichia coli
  2. Inserted foreign DNA into a plasmid vector and propagated it in Escherichia coli Correct answer.
  3. Produced the first transgenic mouse expressing a foreign growth hormone gene
  4. Isolated the first restriction enzyme from Haemophilus influenzae

Explanation

Cohen and Boyer combined plasmid vectors with restriction enzymes and ligase to construct recombinant DNA molecules, then introduced them into E. coli where they replicated. This demonstrated that genes could be cloned in bacteria, the founding experiment of recombinant DNA technology. The other achievements belong to different researchers and later or earlier dates.

Question 8Introduction to Biotechnology and Historical Development

A biotechnology student is asked to explain why the development of restriction endonucleases was essential for the birth of recombinant DNA technology. Which reasoning is most accurate?

  1. They join DNA fragments by forming phosphodiester bonds between any two ends
  2. They synthesise DNA from an RNA template, allowing genes to be recovered from mRNA
  3. They cut DNA at specific recognition sequences, allowing defined fragments to be joined to vectors reproducibly Correct answer.
  4. They amplify DNA exponentially without the need for a host cell

Explanation

Restriction endonucleases recognise specific short sequences and cleave DNA at or near them, producing predictable fragments with defined ends. This specificity let researchers cut both vector and insert in a controlled way before ligation. Without such reproducible cutting, gene cloning would have been random and impractical.

Question 9Introduction to Biotechnology and Historical Development

Which early industrial biotechnology product is correctly paired with the organism traditionally used for its large-scale production?

  1. Penicillin - Saccharomyces cerevisiae
  2. Ethanol - Penicillium notatum
  3. Citric acid - Aspergillus niger Correct answer.
  4. Glutamic acid - Escherichia coli

Explanation

Citric acid has long been produced industrially by fermentation using the mould Aspergillus niger, a classic example of an early microbial industrial process. Penicillin comes from Penicillium moulds, ethanol from yeasts, and glutamic acid historically from Corynebacterium species. Matching product to organism is a standard historical fact in industrial biotechnology.

Question 10Introduction to Biotechnology and Historical Development

Which statement about the historical shift from empirical fermentation to rational biotechnology is most defensible?

  1. Empirical fermentation and rational biotechnology differ only in the scale of the vessels used
  2. Rational biotechnology abandoned microorganisms entirely and relies only on plant tissue culture
  3. Empirical fermentation relied on mixed natural cultures, while rational biotechnology depends on defined cultures and controlled genetic traits Correct answer.
  4. Empirical fermentation always used pure cultures, while rational biotechnology uses only mixed cultures

Explanation

Traditional fermentation depended on naturally occurring or mixed microbial populations whose behaviour was not fully understood. Modern biotechnology uses characterised pure cultures, defined media and, in many cases, deliberately engineered strains. This shift from empirical practice to controlled biological design is the defining historical change.

Question 11Introduction to Biotechnology and Historical Development

Which sequence orders these biotechnology milestones correctly from earliest to latest?

  1. Polymerase chain reaction → Cohen–Boyer recombinant DNA → discovery of penicillin → Pasteur's germ theory work
  2. Cohen–Boyer recombinant DNA → Pasteur's germ theory work → discovery of penicillin → polymerase chain reaction
  3. Discovery of penicillin → Pasteur's germ theory work → polymerase chain reaction → Cohen–Boyer recombinant DNA
  4. Pasteur's germ theory work → discovery of penicillin → Cohen–Boyer recombinant DNA → polymerase chain reaction Correct answer.

Explanation

Pasteur's germ-theory and fermentation studies date from the 1860s–1870s, Fleming observed penicillin in 1928, Cohen and Boyer cloned recombinant DNA in 1973, and the polymerase chain reaction was conceived in 1983. This ordering follows the documented historical record. Any other sequence places a later milestone before an earlier one.

Question 12Introduction to Biotechnology and Historical Development

In the history of biotechnology, the term 'second-generation biotechnology' is most often used to refer to which development?

  1. Products of recombinant DNA technology and genetic engineering, such as engineered human insulin Correct answer.
  2. The application of tissue culture to ornamental plant propagation
  3. The use of antibiotics discovered from moulds without genetic modification
  4. Traditional fermented foods such as curd, bread and wine

Explanation

Second-generation biotechnology is generally used for the era of deliberate genetic manipulation, including recombinant DNA products such as human insulin produced in bacteria. Traditional fermentation is placed in the first or ancient generation. The term highlights the transition from empirical microbial processes to engineered biological systems.

Question 13Cell Biology, Microbiology and Microbial Techniques

Which structure is the primary site of ATP synthesis by oxidative phosphorylation in a eukaryotic cell?

  1. Golgi apparatus
  2. Inner mitochondrial membrane Correct answer.
  3. Outer mitochondrial membrane
  4. Rough endoplasmic reticulum

Explanation

The electron transport chain and ATP synthase are embedded in the inner mitochondrial membrane, whose folded cristae increase surface area for chemiosmotic ATP production. The outer membrane is freely permeable and lacks these complexes, while the ER and Golgi handle protein processing and sorting.

Question 14Cell Biology, Microbiology and Microbial Techniques

A bacterial culture is grown in a medium containing only glucose, ammonium chloride and inorganic salts. Which nutritional category does the organism belong to if it grows well?

  1. Photoautotroph
  2. Chemoheterotroph
  3. Chemolithoautotroph Correct answer.
  4. Fastidious heterotroph

Explanation

Growth on a single organic carbon source plus ammonium and inorganic salts is not the key here; the decisive point is that the description matches a minimal medium, which supports organisms able to use inorganic nitrogen and salts. Among the choices, an autotroph synthesising its own organic carbon from inorganic sources fits, whereas heterotrophs need preformed organic carbon such as vitamins or amino acids.

Question 15Cell Biology, Microbiology and Microbial Techniques

Which statement correctly describes the outcome of the Gram stain for a Gram-positive bacterium?

  1. It retains crystal violet-iodine and appears purple after decolourisation. Correct answer.
  2. It loses crystal violet and takes up safranin, appearing pink.
  3. It retains safranin and appears red because of a thick peptidoglycan layer.
  4. It appears colourless because the cell wall excludes all dyes.

Explanation

In the Gram stain, crystal violet forms a complex with iodine inside the cell. The thick peptidoglycan layer of Gram-positive cells retains this complex during alcohol decolourisation, so the cells remain purple. Gram-negative cells lose the complex and counterstain pink with safranin.

Question 16Cell Biology, Microbiology and Microbial Techniques

During the exponential phase of a batch culture, a bacterial population increases from 1.0 × 10^4 to 6.4 × 10^5 cells. How many generations have elapsed?

  1. 6 Correct answer.
  2. 4
  3. 5
  4. 7

Explanation

Each generation doubles the population, so the number of generations n satisfies 2^n = final/initial = 64. Since 2^6 = 64, six generations have elapsed. This follows directly from the relation N = N0 × 2^n.

Question 17Cell Biology, Microbiology and Microbial Techniques

Which feature distinguishes archaeal cell membranes from those of bacteria?

  1. Absence of any lipid bilayer
  2. Ether linkages between glycerol and isoprenoid chains Correct answer.
  3. Presence of lipopolysaccharide in the outer leaflet
  4. Presence of sterols in the lipid bilayer

Explanation

Archaeal membrane lipids use ether bonds linking glycerol to branched isoprenoid chains, which confers chemical stability at extremes of temperature and pH. Bacterial and eukaryotic membranes typically use ester-linked fatty acids, and lipopolysaccharide is characteristic of Gram-negative bacterial outer membranes.

Question 18Cell Biology, Microbiology and Microbial Techniques

A culture is suspected of containing a heat-resistant endospore former. Which treatment most directly supports this diagnosis before staining?

  1. Heating the smear to 80 °C for 10 minutes before applying malachite green Correct answer.
  2. Treating the smear with dilute acid for 30 seconds
  3. Boiling the smear in water for 1 minute
  4. Washing the smear with ethanol for 5 minutes

Explanation

Endospores resist ordinary staining because of their tough coat, so heat or prolonged steaming is used to drive malachite green into the spore during the Schaeffer-Fulton method. Brief boiling or acid treatment does not reliably permeabilise the spore coat. A subsequent water wash removes dye from vegetative cells, which are then counterstained with safranin.

Question 19Cell Biology, Microbiology and Microbial Techniques

Which enzyme is responsible for relieving supercoiling ahead of the replication fork in bacteria?

  1. Primase
  2. DNA gyrase Correct answer.
  3. DNA ligase
  4. Helicase

Explanation

DNA gyrase is a type II topoisomerase that introduces negative supercoils and removes positive supercoiling generated ahead of the replication fork. It is the target of quinolone antibiotics. Ligase seals nicks, primase lays RNA primers and helicase unwinds the duplex.

Question 20Cell Biology, Microbiology and Microbial Techniques

A suspension of bacteria is serially diluted and 0.1 mL of a 10^-5 dilution is spread on an agar plate. After incubation, 68 colonies appear. What is the estimated viable count per mL of the original suspension?

  1. 6.8 × 10^5
  2. 6.8 × 10^4
  3. 6.8 × 10^7
  4. 6.8 × 10^6 Correct answer.

Explanation

The viable count is calculated as colony count divided by the plated volume and multiplied by the dilution factor: 68 ÷ 0.1 × 10^5 = 6.8 × 10^6 CFU per mL. Plating only 0.1 mL means the count must be scaled up tenfold before applying the dilution factor.

Question 21Cell Biology, Microbiology and Microbial Techniques

Which statement best explains why a bacterial cell maintains its chromosome in a negatively supercoiled state?

  1. It prevents the binding of RNA polymerase to promoters.
  2. It converts the chromosome into a linear molecule.
  3. It allows easier separation of strands during replication and transcription. Correct answer.
  4. It makes the DNA resistant to all nucleases.

Explanation

Negative supercoiling stores torsional energy that favours local unwinding of the duplex, lowering the energy barrier for strand separation by helicase and RNA polymerase. It also compacts the chromosome within the nucleoid. It does not make DNA nuclease-proof or linearise the chromosome.

Question 22Cell Biology, Microbiology and Microbial Techniques

Which component of the bacterial cell wall is the target of penicillin?

  1. The cytoplasmic membrane lipid bilayer
  2. Transpeptidase enzymes that cross-link peptidoglycan Correct answer.
  3. The lipopolysaccharide layer of the outer membrane
  4. The 30S ribosomal subunit

Explanation

Beta-lactam antibiotics such as penicillin acylate the active-site serine of transpeptidases (penicillin-binding proteins), blocking cross-linking of peptidoglycan strands. This weakens the wall and leads to lysis of actively growing cells. Ribosomes and membranes are targets of other antibiotic classes.

Question 23Cell Biology, Microbiology and Microbial Techniques

In a chemostat, which parameter directly sets the specific growth rate of the culture at steady state?

  1. The concentration of the limiting nutrient in the feed
  2. The dilution rate Correct answer.
  3. The initial inoculum size
  4. The volume of the vessel

Explanation

At steady state in a chemostat, the specific growth rate equals the dilution rate, because biomass is removed at the same rate it is produced. The feed concentration of the limiting nutrient determines the steady-state biomass concentration rather than the growth rate.

Question 24Cell Biology, Microbiology and Microbial Techniques

Which staining method is specifically used to detect the presence of a capsule around bacterial cells?

  1. Albert staining
  2. Capsule staining with India ink Correct answer.
  3. Gram staining
  4. Ziehl-Neelsen staining

Explanation

Capsule staining uses a negatively charged dye such as India ink or nigrosin, which is excluded by the capsule and stains the background, leaving the capsule as a clear halo around the cell. This negative staining principle makes the capsule visible against a dark background.

Question 25Cell Biology, Microbiology and Microbial Techniques

Which process allows bacteria to transfer DNA between cells via a bacteriophage?

  1. Conjugation
  2. Binary fission
  3. Transformation
  4. Transduction Correct answer.

Explanation

Transduction is phage-mediated transfer of bacterial DNA, either by accidental packaging of host DNA or through integration into the host chromosome. Transformation involves uptake of free DNA, conjugation requires direct cell contact and binary fission is asexual reproduction.

Question 26Cell Biology, Microbiology and Microbial Techniques

A mutant bacterium cannot grow on lactose but grows normally on glucose. Which explanation is most consistent with the observation?

  1. A defect in the lac operon prevents lactose utilisation. Correct answer.
  2. A defect in the cell wall prevents nutrient uptake.
  3. A defect in glycolysis prevents glucose utilisation.
  4. A defect in the electron transport chain blocks all growth.

Explanation

The ability to grow on glucose but not lactose points to a specific lesion in lactose metabolism, such as loss of beta-galactosidase or the lac permease. Glucose utilisation remains intact, so glycolysis and general energy metabolism are functional. The phenotype is substrate-specific rather than global.

Question 27Cell Biology, Microbiology and Microbial Techniques

Which characteristic is typical of a bacterial species classified as psychrophilic?

  1. Optimal growth near 15 °C with membranes rich in unsaturated fatty acids Correct answer.
  2. Optimal growth near 80 °C with ether-linked lipids
  3. Optimal growth near 37 °C with a narrow tolerance range
  4. Optimal growth near 55 °C with heat-stable proteins

Explanation

Psychrophiles grow best at low temperatures, typically around 15 °C or below, and maintain membrane fluidity by increasing the proportion of unsaturated and short-chain fatty acids. Their enzymes are cold-adapted and often heat-labile. Thermophiles and hyperthermophiles occupy the higher temperature ranges.

Question 28Cell Biology, Microbiology and Microbial Techniques

Which observation would support the conclusion that a particular bacterium is motile?

  1. Growth spreading away from a stab line in semi-solid agar Correct answer.
  2. Formation of a pellicle on a broth surface
  3. Production of gas bubbles in a Durham tube
  4. Appearance of pink cells after Gram staining

Explanation

In semi-solid motility medium, motile bacteria swim away from the stab line and produce diffuse turbidity throughout the tube, whereas non-motile strains grow only along the stab. Pellicle formation, gas production and Gram reaction do not specifically demonstrate motility.

Question 29Cell Biology, Microbiology and Microbial Techniques

Which statement about the bacterial nucleoid is correct?

  1. It is a region of DNA not enclosed by a nuclear membrane. Correct answer.
  2. It is a membrane-bound organelle containing histone-packaged DNA.
  3. It is present only during sporulation.
  4. It contains RNA and protein but no DNA.

Explanation

The bacterial nucleoid is a discrete region of the cytoplasm where the circular chromosome is compacted by nucleoid-associated proteins, but it is not bounded by a membrane. This absence of a nuclear envelope is a defining feature distinguishing prokaryotes from eukaryotes.

Question 30Cell Biology, Microbiology and Microbial Techniques

In a culture, the number of viable cells remains constant while total cell count rises. Which explanation fits best?

  1. The culture is in the lag phase with no cell division.
  2. The viable count method is insensitive to cell wall type.
  3. The cells are forming spores and losing viability.
  4. Cells are dying at the same rate they divide. Correct answer.

Explanation

During the stationary phase, growth and death rates balance, so viable numbers plateau while total counts may still include dead cells and debris. A rising total count with constant viability indicates accumulation of non-viable cells rather than active net growth.

Question 31Cell Biology, Microbiology and Microbial Techniques

Which molecule serves as the terminal electron acceptor in anaerobic respiration by denitrifying bacteria?

  1. Sulphate
  2. Oxygen
  3. Carbon dioxide
  4. Nitrate Correct answer.

Explanation

Denitrifying bacteria use nitrate as the terminal electron acceptor, reducing it through nitrite, nitric oxide and nitrous oxide to nitrogen gas. This allows respiration to proceed in the absence of oxygen. Sulphate is used by a different group, the sulphate reducers.

Question 32Cell Biology, Microbiology and Microbial Techniques

A researcher needs to sterilise a heat-labile antibiotic solution without denaturing it. Which method is most appropriate?

  1. Filtration through a 0.22 µm membrane Correct answer.
  2. Dry heat at 160 °C for 2 hours
  3. Boiling for 30 minutes
  4. Autoclaving at 121 °C for 15 minutes

Explanation

Membrane filtration physically removes microorganisms without heating, so heat-labile solutions such as antibiotics and vitamins retain activity. Autoclaving, dry heat and boiling all rely on heat and would degrade such compounds. A 0.22 µm filter retains most bacteria.

Question 33Molecular Biology: DNA, RNA, Replication, Transcription and Translation

In the Watson–Crick model of B-DNA, which pair of hydrogen-bonding partners is held together by three hydrogen bonds?

  1. Adenine and uracil
  2. Guanine and thymine
  3. Adenine and thymine
  4. Guanine and cytosine Correct answer.

Explanation

Guanine and cytosine form three hydrogen bonds because the G–C pair has an extra donor and acceptor site compared with A–T. Adenine pairs with thymine in DNA using only two hydrogen bonds, and adenine pairs with uracil in RNA using two. This is why GC-rich regions of DNA require higher temperature to separate.

Question 34Molecular Biology: DNA, RNA, Replication, Transcription and Translation

Chargaff's rules applied to a double-stranded DNA sample show that the mole percentage of adenine equals that of thymine. Which structural feature of the double helix directly explains this equality?

  1. The major and minor grooves along the helix
  2. The phosphodiester backbone linking the nucleotides
  3. The antiparallel orientation of the two strands
  4. The specific A–T base pairing between the strands Correct answer.

Explanation

Because every adenine in one strand is paired with a thymine in the opposite strand, the total number of A residues equals the total number of T residues in the duplex. The antiparallel arrangement, the sugar-phosphate backbone and the grooves do not impose any base composition equality. Chargaff's observation therefore reflects complementary base pairing.

Question 35Molecular Biology: DNA, RNA, Replication, Transcription and Translation

During DNA replication, DNA polymerase III in bacteria extends a new strand only when which condition is satisfied?

  1. The two parental strands are still covalently joined
  2. A free 5'-phosphate end of the growing strand is exposed
  3. A short RNA primer is removed before any addition
  4. A free 3'-OH end of a primer is available for nucleotide addition Correct answer.

Explanation

DNA polymerases add deoxynucleotides to the 3'-hydroxyl group of an existing primer, so synthesis always proceeds in the 5' to 3' direction. A free 5'-phosphate cannot serve as the site of addition. Primer removal happens later during maturation of Okazaki fragments, after extension.

Question 36Molecular Biology: DNA, RNA, Replication, Transcription and Translation

In the Meselson–Stahl experiment, bacteria grown in 15N medium were shifted to 14N medium and DNA was analysed after one generation. What banding pattern in the density gradient confirmed semiconservative replication?

  1. A single hybrid band of intermediate density Correct answer.
  2. A single light band
  3. Two bands, one heavy and one light
  4. A single heavy band

Explanation

After one round of replication in 14N, each daughter duplex contains one old 15N strand and one new 14N strand, giving a single band of intermediate density. A single heavy band would indicate conservative retention of the original duplex, and a single light band would require no incorporation of old strands. The intermediate band is the key evidence for semiconservative replication.

Question 37Molecular Biology: DNA, RNA, Replication, Transcription and Translation

Which enzyme relieves the supercoiling ahead of the replication fork by transiently breaking and rejoining DNA strands?

  1. Topoisomerase Correct answer.
  2. DNA ligase
  3. Helicase
  4. Primase

Explanation

Topoisomerases nick the DNA, allow the strands to swivel and then reseal the break, removing torsional stress generated ahead of the advancing fork. Helicase unwinds the duplex but does not relieve supercoiling, and ligase joins nicks rather than creating them. Primase synthesises short RNA primers.

Question 38Molecular Biology: DNA, RNA, Replication, Transcription and Translation

Why is one of the two new strands at a replication fork synthesised discontinuously as Okazaki fragments?

  1. Because the lagging strand is chemically different from the leading strand
  2. Because the lagging template strand contains more adenine residues
  3. Because RNA primers can only be laid down on the leading strand
  4. Because DNA polymerase can only add nucleotides in the 5' to 3' direction Correct answer.

Explanation

Both strands are synthesised 5' to 3', but the template strands are antiparallel. On the lagging template the polymerase must move away from the fork, so synthesis restarts repeatedly in short pieces. Base composition and chemical identity of the strands are identical; only the direction of template readout differs.

Question 39Molecular Biology: DNA, RNA, Replication, Transcription and Translation

Which statement correctly describes the role of the sigma factor in bacterial transcription?

  1. It proofreads the RNA transcript after synthesis
  2. It catalyses the formation of phosphodiester bonds in the RNA chain
  3. It directs RNA polymerase to specific promoter sequences for initiation Correct answer.
  4. It terminates transcription by recognising the terminator hairpin

Explanation

Sigma factor binds RNA polymerase and guides the holoenzyme to promoter elements such as the -10 and -35 regions, allowing specific initiation. Catalysis of phosphodiester bonds is carried out by the core enzyme, and termination depends on intrinsic terminators or rho factor. Sigma dissociates after initiation rather than proofreading the transcript.

Question 40Molecular Biology: DNA, RNA, Replication, Transcription and Translation

In eukaryotes, the primary transcript is processed before export to the cytoplasm. Which modification adds a 7-methylguanosine structure to the 5' end?

  1. Methylation of histone tails
  2. Capping Correct answer.
  3. Polyadenylation
  4. Splicing

Explanation

Capping adds a 7-methylguanosine residue to the 5' end of the nascent transcript through an unusual 5'-5' triphosphate linkage. Polyadenylation adds an adenine tail at the 3' end, whereas splicing removes introns. Histone methylation modifies chromatin proteins, not the RNA transcript.

Question 41Molecular Biology: DNA, RNA, Replication, Transcription and Translation

A template DNA strand reads 3'-TAC GGA TTC-5'. Applying standard base-pairing rules, what is the sequence of the RNA synthesised from it?

  1. 5'-ATG CCT AAG-3'
  2. 5'-UAC GGA UUC-3'
  3. 5'-AUG CCU AAG-3' Correct answer.
  4. 5'-TAC GGA TTC-3'

Explanation

RNA is synthesised antiparallel and complementary to the template, with uracil replacing thymine. Reading the template 3' to 5' gives AUG CCU AAG written 5' to 3'. The transcript contains uracil, so any option containing thymine cannot be RNA. The option that copies the template sequence directly ignores complementarity.

Question 42Molecular Biology: DNA, RNA, Replication, Transcription and Translation

Which feature distinguishes the genetic code as degenerate?

  1. More than one codon can specify the same amino acid Correct answer.
  2. The code differs between mitochondria and the nucleus
  3. A single codon can specify more than one amino acid
  4. Codons are read in overlapping triplets

Explanation

Degeneracy means that several different codons may encode the same amino acid, which is possible because there are 64 codons but only about 20 amino acids. A codon specifying multiple amino acids would make the code ambiguous, which it is not. Overlapping reading and mitochondrial variation are separate phenomena.

Question 43Molecular Biology: DNA, RNA, Replication, Transcription and Translation

During translation, which molecule delivers the initiator amino acid to the P site of the ribosome in bacteria?

  1. EF-Tu carrying a charged tRNA to the A site
  2. RF1 recognising a stop codon
  3. fMet-tRNA bound to IF2 Correct answer.
  4. Peptidyl transferase acting on the 50S subunit

Explanation

Bacterial initiation uses N-formylmethionyl-tRNA, which is delivered to the P site by initiation factor IF2 together with GTP. EF-Tu delivers elongator tRNAs to the A site during elongation, not initiation. Release factors and peptidyl transferase act later in the process.

Question 44Molecular Biology: DNA, RNA, Replication, Transcription and Translation

A frameshift mutation arises when a single base pair is inserted into a coding region. What is the direct consequence for the polypeptide?

  1. Only the codon at the mutation site changes
  2. Replication of the gene stops at the insertion
  3. The transcript is no longer capped at its 5' end
  4. All codons downstream of the insertion are read out of frame Correct answer.

Explanation

Inserting one base shifts the reading frame so that every codon after the insertion site is grouped differently, usually producing a truncated or non-functional protein. A single codon change describes a missense substitution, not a frameshift. Capping and replication are unaffected by the reading frame of translation.

Question 45Molecular Biology: DNA, RNA, Replication, Transcription and Translation

Which property of the DNA double helix is most directly responsible for the fidelity of base pairing during replication?

  1. The negative charge of the phosphate backbone
  2. The hydrophobic interior formed by stacked bases
  3. The right-handed twist of the helix
  4. Complementary hydrogen-bonding surfaces of the bases Correct answer.

Explanation

Complementary donor and acceptor groups on the bases allow only A–T and G–C pairs to fit geometrically and hydrogen-bond correctly. Base stacking contributes to stability but does not select the correct partner. Backbone charge and helical twist are general structural features that do not distinguish among bases.

Question 46Molecular Biology: DNA, RNA, Replication, Transcription and Translation

Which enzyme is responsible for removing the RNA primer and replacing it with DNA during bacterial replication?

  1. DNA polymerase I Correct answer.
  2. Primase
  3. DNA ligase
  4. DNA polymerase III

Explanation

DNA polymerase I has 5' to 3' exonuclease activity that removes the RNA primer while extending the adjacent DNA, after which ligase seals the remaining nick. DNA polymerase III performs the bulk of chain elongation but lacks this primer-removal role. Ligase joins ends and primase synthesises primers.

Question 47Molecular Biology: DNA, RNA, Replication, Transcription and Translation

Why does a high concentration of guanine and cytosine in a DNA sample increase its melting temperature?

  1. GC pairs are held together by covalent bonds
  2. GC pairs have three hydrogen bonds and stronger stacking Correct answer.
  3. GC pairs carry a net positive charge
  4. GC pairs are larger and distort the helix

Explanation

Each G–C pair has three hydrogen bonds and favourable stacking interactions, so more energy is needed to separate the strands. A–T pairs have only two hydrogen bonds. Base pairs are not covalently bonded to each other across strands, and they do not carry a net positive charge.

Question 48Molecular Biology: DNA, RNA, Replication, Transcription and Translation

Which statement best describes the difference between the leading and lagging strands during replication?

  1. The leading strand is synthesised continuously toward the fork; the lagging strand is synthesised discontinuously Correct answer.
  2. The leading strand uses helicase; the lagging strand uses ligase
  3. The leading strand is synthesised 3' to 5'; the lagging strand is synthesised 5' to 3'
  4. The leading strand is made of RNA; the lagging strand is made of DNA

Explanation

The leading strand grows continuously in the same direction the fork moves, while the lagging strand is made in short Okazaki fragments as more template is exposed. Both strands are DNA and both are synthesised 5' to 3'. Enzymes such as helicase and ligase act on both strands, not one exclusively.

Question 49Molecular Biology: DNA, RNA, Replication, Transcription and Translation

In eukaryotic transcription, which enzyme transcribes protein-coding genes into mRNA precursors?

  1. RNA polymerase II Correct answer.
  2. DNA polymerase II
  3. RNA polymerase III
  4. RNA polymerase I

Explanation

RNA polymerase II transcribes protein-coding genes, producing pre-mRNA that is capped, spliced and polyadenylated. RNA polymerase I makes the large ribosomal RNAs, and RNA polymerase III makes tRNA and small RNAs. DNA polymerase II is not a transcription enzyme.

Question 50Molecular Biology: DNA, RNA, Replication, Transcription and Translation

The wobble hypothesis explains which feature of codon–anticodon interaction?

  1. Codons are read in pairs rather than triplets
  2. A single tRNA can recognise two different amino acids
  3. The first base of a codon can pair with any base of the anticodon
  4. The third base of a codon can pair non-standardly with the first base of the anticodon Correct answer.

Explanation

Wobble allows the third (3') base of the codon to form non-standard hydrogen bonds with the first (5') base of the anticodon, so one tRNA can read several synonymous codons. Pairing at the first two codon positions remains strict. A tRNA is always charged with a single specific amino acid.