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GCE A/L Bio Systems Technology study materials
GCE A/L Bio Systems Technology study resources

GCE A/L Bio Systems Technology practice questions

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Question 1Introduction to Bio Systems Technology and Laboratory Safety

In a bio systems technology laboratory, which practice best describes the principle of aseptic technique when transferring a microbial culture?

  1. Flame the neck of the culture vessel and work close to a lit burner Correct answer.
  2. Keep the petri dish lid fully removed to speed up the transfer
  3. Sterilise the loop only before the first transfer of the session
  4. Hold the vessel horizontally to reduce contamination from the air

Explanation

Aseptic technique aims to keep unwanted microorganisms out of the culture and the environment. Passing the vessel neck through a flame creates an upward air current that discourages airborne contaminants from entering, and working near the flame keeps the working zone sterile. Every transfer should use a freshly sterilised loop, and lids should be lifted only minimally.

Question 2Introduction to Bio Systems Technology and Laboratory Safety

A student reads a chemical label showing the hazard symbol for a substance that is corrosive. Which precaution is most directly required when handling this reagent in the laboratory?

  1. Neutralise any spill with concentrated acid before cleaning
  2. Handle the reagent only when a second student is present
  3. Store the reagent in a glass bottle inside a fume cupboard only
  4. Wear splash-resistant goggles and gloves and avoid skin contact Correct answer.

Explanation

Corrosive substances damage living tissue on contact, so the primary control is preventing contact with skin and eyes through appropriate personal protective equipment such as splash goggles and gloves. Storage location and supervision do not substitute for direct protection of the handler. Spills are dealt with using the material safety data sheet guidance, never by adding concentrated acid.

Question 3Introduction to Bio Systems Technology and Laboratory Safety

Which piece of equipment is designed specifically to sterilise heat-sensitive liquids such as growth media by filtration?

  1. Bunsen burner with a wire loop
  2. Membrane filter assembly with a suitable pore size Correct answer.
  3. Hot air oven
  4. Autoclave

Explanation

Filtration physically removes microorganisms from a liquid without applying heat, so it is suitable for heat-labile media and solutions. A membrane with a pore size around 0.22 micrometres retains bacteria while allowing the liquid to pass. Autoclaves, hot air ovens and flaming rely on heat or combustion and would damage heat-sensitive components.

Question 4Introduction to Bio Systems Technology and Laboratory Safety

In a laboratory risk assessment, a chemical is described as having an LD50 value that is very low. What does this indicate about the substance?

  1. It is safe to handle without gloves because the dose is small
  2. It is corrosive rather than toxic to living tissue
  3. It is only harmful after repeated exposure over many years
  4. It is highly toxic because a small dose can be lethal Correct answer.

Explanation

LD50 is the dose that kills half of a test population, so a very low value means only a small amount is needed to cause death. This identifies the substance as highly toxic and demands strict handling precautions. LD50 describes acute toxicity and does not by itself indicate corrosivity or long-term effects.

Question 5Introduction to Bio Systems Technology and Laboratory Safety

A biosystems technologist is setting up a small-scale fermenter to produce a microbial product. Which combination of conditions must be controlled to maintain reproducible growth?

  1. Colour of the vessel, light intensity and humidity only
  2. Temperature, pH and dissolved oxygen concentration Correct answer.
  3. Mass of the vessel, stirring speed and glass thickness
  4. Pressure of the room, day length and soil moisture

Explanation

Microbial growth in a fermenter is reproducible only when the key environmental variables affecting metabolism are monitored and controlled. Temperature, pH and dissolved oxygen directly influence enzyme activity, growth rate and product formation. Other listed factors do not govern microbial metabolism in a controlled fermentation.

Question 6Introduction to Bio Systems Technology and Laboratory Safety

A student must prepare a 1 in 10 dilution series of a bacterial culture for viable counting. Starting with 1.0 cm3 of culture, which procedure produces a correct first dilution?

  1. Add 9.0 cm3 of culture to 1.0 cm3 of sterile diluent and mix thoroughly
  2. Add 1.0 cm3 of culture to 1.0 cm3 of sterile diluent and mix thoroughly
  3. Add 1.0 cm3 of culture to 9.0 cm3 of sterile diluent and mix thoroughly Correct answer.
  4. Add 1.0 cm3 of culture to 10.0 cm3 of sterile diluent and mix thoroughly

Explanation

A 1 in 10 dilution means the sample is one part in ten parts total, so 1.0 cm3 of culture is added to 9.0 cm3 of diluent, giving a total of 10.0 cm3. Thorough mixing ensures the dilution is uniform before the next step. Adding the wrong volume of diluent changes the dilution factor and invalidates the viable count.

Question 7Introduction to Bio Systems Technology and Laboratory Safety

A culture is spread on an agar plate and incubated. After 24 hours, 60 colonies are counted on a plate prepared from a 10^-4 dilution, using 0.1 cm3 of that dilution. What is the estimated concentration of viable cells in the original culture?

  1. 6.0 x 10^3 cells per cm3
  2. 6.0 x 10^5 cells per cm3
  3. 6.0 x 10^6 cells per cm3 Correct answer.
  4. 6.0 x 10^4 cells per cm3

Explanation

The concentration in the diluted sample is 60 colonies divided by 0.1 cm3, which is 600 cells per cm3. Multiplying by the dilution factor 10^4 gives 600 x 10^4, equal to 6.0 x 10^6 cells per cm3 in the original culture. Each counted colony is assumed to have arisen from one viable cell.

Question 8Introduction to Bio Systems Technology and Laboratory Safety

During a practical, a small volume of bacterial culture is spilled on the bench. What is the most appropriate immediate action?

  1. Cover the spill with absorbent material and apply a suitable disinfectant, allowing contact time Correct answer.
  2. Leave the spill to dry in the air before cleaning at the end of the lesson
  3. Wipe the spill towards the edge of the bench with a dry paper towel
  4. Rinse the bench with water and continue working without further treatment

Explanation

A biological spill must be contained and decontaminated to prevent spread of microorganisms. Absorbing the liquid and applying a validated disinfectant with adequate contact time inactivates the organisms before removal. Spreading the spill or simply rinsing with water can enlarge the contaminated area without killing the microorganisms.

Question 9Introduction to Bio Systems Technology and Laboratory Safety

A technologist needs to sterilise glass Petri dishes and metal instruments that can withstand high temperatures. Which method is most appropriate?

  1. Pasteurisation at 63 degrees Celsius for 30 minutes
  2. Exposure to ultraviolet light for 20 minutes
  3. Boiling in water for 10 minutes
  4. Hot air oven at 160 degrees Celsius for about two hours Correct answer.

Explanation

Dry heat sterilisation in a hot air oven at around 160 degrees Celsius for about two hours reliably kills microorganisms, including endospores, on heat-stable glassware and metal items. The method requires dry items and good air circulation. Methods that rely on lower temperatures or on UV light do not reliably sterilise these items.

Question 10Introduction to Bio Systems Technology and Laboratory Safety

A student is asked to calibrate a micropipette before use. Why is this step important in a bio systems technology laboratory?

  1. It sterilises the pipette tip before each use
  2. It increases the maximum volume the pipette can deliver
  3. It reduces the need to wear gloves when handling samples
  4. It ensures accurate and reproducible volumes so that dilution factors and results are reliable Correct answer.

Explanation

Micropipettes deliver small volumes, so an error in volume directly changes dilution factors and reagent concentrations. Calibration checks that the delivered volume matches the set volume, ensuring results are accurate and reproducible. Calibration does not sterilise the pipette, change its capacity, or remove the need for personal protective equipment.

Question 11Introduction to Bio Systems Technology and Laboratory Safety

A growth curve of a bacterial batch culture shows a lag phase followed by a steep exponential phase. Which explanation best accounts for the lag phase?

  1. Cells have reached the stationary phase and stopped dividing
  2. Cells are being washed out of the vessel faster than they grow
  3. Cells are adapting metabolically and synthesising enzymes before rapid division Correct answer.
  4. Cells are dying faster than they divide because nutrients are exhausted

Explanation

During the lag phase, inoculated cells adjust to the new environment and build up the enzymes and metabolites needed for growth, so cell numbers change little. Once this adaptation is complete, the culture enters exponential growth. Nutrient exhaustion and washout describe later or continuous-culture situations, not the lag phase.

Question 12Introduction to Bio Systems Technology and Laboratory Safety

A laboratory protocol requires a Class II biological safety cabinet for handling a sample. What is the main function of this equipment?

  1. To protect the operator, the sample and the environment by controlling airflow and filtering it Correct answer.
  2. To store cultures at low temperature for long periods
  3. To measure the concentration of microorganisms in a sample
  4. To provide a sterile environment by heating the working area to high temperature

Explanation

A Class II biological safety cabinet uses directional airflow and high-efficiency particulate air filtration to protect the operator and the environment as well as the sample. It does not sterilise by heat, store cultures, or measure microbial numbers. These cabinets are used when handling hazardous biological material.

Question 13Introduction to Bio Systems Technology and Laboratory Safety

A student observes that a culture grown at 25 degrees Celsius reaches a lower final cell density than the same organism grown at 37 degrees Celsius with identical medium. Which conclusion is most reasonable?

  1. The medium at 25 degrees Celsius contained more nutrients
  2. The organism grows better at 37 degrees Celsius, suggesting this is closer to its optimum Correct answer.
  3. Temperature has no effect on the growth of the organism
  4. The 37 degrees Celsius culture reached the lag phase earlier

Explanation

Temperature affects enzyme activity and therefore growth rate and final yield. A higher final density at 37 degrees Celsius suggests that this temperature is nearer the organism's optimum for growth. The identical medium rules out nutrient differences, and the lag phase precedes growth rather than explaining a higher final density.

Question 14Introduction to Bio Systems Technology and Laboratory Safety

A culture is grown in a chemostat, where fresh medium is supplied continuously at a fixed dilution rate. What happens if the dilution rate is increased above the maximum specific growth rate of the organism?

  1. The organism is washed out of the vessel faster than it can grow Correct answer.
  2. The growth rate increases to match any chosen dilution rate
  3. The biomass concentration increases steadily without limit
  4. The organism switches permanently to the stationary phase

Explanation

In a chemostat, steady state requires the organism's growth rate to match the dilution rate. If the dilution rate exceeds the maximum specific growth rate, cells are removed faster than they can divide, and the population declines until it is washed out. Growth rate cannot increase indefinitely because it is limited by the organism's maximum capacity.

Question 15Introduction to Bio Systems Technology and Laboratory Safety

A technologist plans to use an enzyme immobilised on a solid support in a continuous packed-bed reactor. Which advantage is most directly associated with immobilising the enzyme rather than using it free in solution?

  1. The enzyme can be retained in the reactor and reused, and the product is easier to separate Correct answer.
  2. The enzyme becomes permanently inactive once immobilised
  3. The enzyme can tolerate any temperature without loss of activity
  4. The reaction no longer requires any substrate to be supplied

Explanation

Immobilising an enzyme on a support allows it to be retained within the reactor, so it can be reused over many cycles and separated easily from the product stream. This improves process economy and product purity. Immobilisation does not remove the need for substrate, and it does not make the enzyme immune to thermal denaturation.

Question 16Introduction to Bio Systems Technology and Laboratory Safety

A laboratory is designing a quality control scheme for a bio systems process. Which approach best ensures that measurements are trustworthy over time?

  1. Using control samples and regular calibration to monitor accuracy and precision Correct answer.
  2. Increasing the sample size without checking instrument performance
  3. Relying on a single measurement taken at the start of the process
  4. Recording results only when they agree with the expected value

Explanation

Trustworthy measurements require ongoing evidence of accuracy and precision, which is provided by analysing control samples of known value and by regular calibration of instruments. This detects drift and bias before they affect results. Selecting only agreeable results or taking a single measurement does not demonstrate reliability.

Question 17Cell Biology, Microscopy and Biological Techniques

A student views a slide using a compound light microscope. Which combination of objective and eyepiece lenses gives the highest total magnification?

  1. 100× objective with 10× eyepiece Correct answer.
  2. 40× objective with 10× eyepiece
  3. 40× objective with 15× eyepiece
  4. 10× objective with 10× eyepiece

Explanation

Total magnification of a compound microscope is the product of the objective magnification and the eyepiece magnification. Multiplying 100 by 10 gives 1000×, which is larger than the products of the other combinations. Therefore the 100× objective with a 10× eyepiece yields the greatest total magnification.

Question 18Cell Biology, Microscopy and Biological Techniques

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

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

Explanation

The electron transport chain and ATP synthase are embedded in the inner mitochondrial membrane, where the proton gradient drives oxidative phosphorylation. Other organelles listed perform different functions such as protein modification, packaging, or hydrolysis.

Question 19Cell Biology, Microscopy and Biological Techniques

A suspension of cells is centrifuged so that organelles sediment in order of decreasing size and density. Which organelle would be expected to pellet first at low centrifugal force?

  1. Nucleus Correct answer.
  2. Ribosome
  3. Lysosome
  4. Mitochondrion

Explanation

Differential centrifugation separates organelles according to size and density, with the largest and densest structures pelleting at the lowest centrifugal force. The nucleus is much larger and denser than mitochondria, lysosomes and ribosomes, so it sediments first.

Question 20Cell Biology, Microscopy and Biological Techniques

Which statement correctly describes the behaviour of a plant cell placed in a hypotonic solution?

  1. The cell gains water and becomes turgid as the cell wall resists further expansion. Correct answer.
  2. The cell shows no net water movement because the wall is impermeable.
  3. The cell loses water and undergoes plasmolysis.
  4. The cell bursts because it lacks a rigid wall.

Explanation

In a hypotonic solution the external water potential is higher than that inside the cell, so water enters by osmosis. The rigid cellulose wall limits expansion, generating turgor pressure that eventually balances further water uptake, making the cell turgid rather than lysed.

Question 21Cell Biology, Microscopy and Biological Techniques

Which feature distinguishes prokaryotic cells from eukaryotic cells?

  1. Presence of a plasma membrane
  2. Presence of a membrane-bound nucleus Correct answer.
  3. Presence of ribosomes
  4. Presence of DNA

Explanation

Prokaryotes lack a membrane-bound nucleus and other membrane-bound organelles; their genetic material lies in a nucleoid region. Ribosomes, a plasma membrane and DNA are present in both prokaryotic and eukaryotic cells, so they do not distinguish the two groups.

Question 22Cell Biology, Microscopy and Biological Techniques

A student needs to observe the detailed internal structure of a mitochondrion at very high resolution. Which instrument is most appropriate?

  1. Transmission electron microscope Correct answer.
  2. Stereomicroscope
  3. Light microscope with 100× objective
  4. Hand lens

Explanation

The transmission electron microscope has a far shorter effective wavelength than visible light, giving resolution sufficient to reveal internal membranes such as mitochondrial cristae. Light microscopes cannot resolve structures at that scale, and stereomicroscopes and hand lenses give even lower magnification.

Question 23Cell Biology, Microscopy and Biological Techniques

Which molecule is the main structural component of plant cell walls?

  1. Chitin
  2. Peptidoglycan
  3. Phospholipid
  4. Cellulose Correct answer.

Explanation

Plant cell walls are composed largely of cellulose microfibrils, which are strong, rigid polysaccharide chains that provide tensile strength. Chitin forms fungal walls and arthropod exoskeletons, peptidoglycan forms bacterial walls, and phospholipids form membranes.

Question 24Cell Biology, Microscopy and Biological Techniques

Which statement best explains why a specimen must be thin when viewed with a transmission electron microscope?

  1. Thin specimens are needed so that the specimen can be stained with coloured dyes.
  2. Thin specimens reduce the magnification required for viewing.
  3. Thin specimens prevent the electron beam from being scattered too much, improving image resolution.
  4. Thin specimens allow the electron beam to pass through, forming a clear image. Correct answer.

Explanation

In TEM, electrons must pass through the specimen to form an image, so specimens are cut into ultrathin sections. If the specimen is too thick, electrons are absorbed or scattered excessively and no useful image is transmitted.

Question 25Cell Biology, Microscopy and Biological Techniques

Which process allows a large polar molecule to enter a cell without direct expenditure of ATP by the cell?

  1. Active transport by a pump
  2. Phagocytosis
  3. Facilitated diffusion through a channel protein Correct answer.
  4. Simple diffusion

Explanation

Facilitated diffusion uses membrane proteins to allow polar or charged substances to cross the lipid bilayer down their concentration gradient, without ATP. Simple diffusion is restricted to small nonpolar molecules, active transport requires ATP, and phagocytosis is a form of bulk uptake rather than passive transport of a single molecule.

Question 26Cell Biology, Microscopy and Biological Techniques

Which part of a light microscope controls the amount of light reaching the specimen?

  1. The stage clips
  2. The coarse focus knob
  3. The iris diaphragm Correct answer.
  4. The objective lens

Explanation

The iris diaphragm is an adjustable aperture beneath the stage that regulates the intensity and cone of light passing through the specimen. The coarse focus knob adjusts focus, stage clips secure the slide, and the objective lens magnifies the image.

Question 27Cell Biology, Microscopy and Biological Techniques

A student calculates the actual size of a cell that appears 40 mm long under a microscope using a 10× eyepiece and a 40× objective. What is the actual length of the cell?

  1. 4.0 mm
  2. 0.4 mm
  3. 0.1 mm Correct answer.
  4. 1.0 mm

Explanation

Total magnification is the product of eyepiece and objective magnifications, that is, 10 × 40 = 400×. Actual size equals measured image size divided by magnification: 40 mm ÷ 400 = 0.1 mm.

Question 28Cell Biology, Microscopy and Biological Techniques

Which technique would be most suitable for separating proteins in a cell extract according to their molecular mass?

  1. Simple staining
  2. Differential centrifugation
  3. Paper chromatography
  4. SDS-polyacrylamide gel electrophoresis Correct answer.

Explanation

SDS-PAGE denatures proteins and gives them a uniform negative charge, so migration through the gel depends mainly on molecular mass, with smaller proteins moving faster. Chromatography separates by solubility and affinity, centrifugation separates by size and density of particles, and staining visualises rather than separates.

Question 29Cell Biology, Microscopy and Biological Techniques

A cell is treated with a drug that blocks the function of the Golgi apparatus. Which process would be most directly affected?

  1. Replication of chromosomal DNA
  2. Modification and sorting of proteins for secretion Correct answer.
  3. Translation of mRNA on ribosomes
  4. Generation of the proton gradient across the inner mitochondrial membrane

Explanation

The Golgi apparatus modifies proteins and lipids and packages them into vesicles for transport to their destinations. Blocking it therefore impairs post-translational modification and sorting of secretory proteins, whereas translation, DNA replication and the mitochondrial proton gradient occur elsewhere.

Question 30Cell Biology, Microscopy and Biological Techniques

Which of the following correctly describes the fluid mosaic model of the plasma membrane?

  1. A double layer of protein surrounding a lipid core
  2. A phospholipid bilayer with proteins and other molecules able to move within it Correct answer.
  3. A single layer of phospholipids coated on both sides by carbohydrate
  4. A rigid layer of protein with lipid channels embedded in it

Explanation

The fluid mosaic model describes a bilayer of phospholipids with proteins and other components able to move laterally within the membrane. This mobility gives the membrane its fluid character, while the distribution of proteins gives the mosaic appearance.

Question 31Cell Biology, Microscopy and Biological Techniques

Which statement best explains why a light microscope cannot resolve two points that are 50 nm apart?

  1. The eyepiece lens is of poor quality.
  2. The magnification of the objective lens is too low.
  3. The resolution is limited by the wavelength of visible light, which is much longer than 50 nm. Correct answer.
  4. The specimen cannot be stained adequately.

Explanation

Resolution is limited by the wavelength of the radiation used; visible light has a wavelength of roughly 400–700 nm, so two points 50 nm apart cannot be distinguished. Increasing magnification without improving resolution merely produces a larger but blurred image.

Question 32Cell Biology, Microscopy and Biological Techniques

Which process describes the movement of water molecules across a selectively permeable membrane from a region of higher water potential to a region of lower water potential?

  1. Endocytosis
  2. Active transport
  3. Diffusion of solutes
  4. Osmosis Correct answer.

Explanation

Osmosis is the passive movement of water across a selectively permeable membrane down a water potential gradient. It does not require ATP and is distinct from the movement of solutes or the bulk uptake of material by endocytosis.

Question 33Cell Biology, Microscopy and Biological Techniques

A student wishes to determine the mitotic index of a root tip. Which procedure is most appropriate?

  1. Counting the number of cells in mitosis and dividing by the total number of cells counted Correct answer.
  2. Measuring the length of the root tip with a ruler
  3. Counting the number of chromosomes in a single cell
  4. Counting the total number of cells in a field of view

Explanation

The mitotic index is the proportion of cells undergoing mitosis, calculated by dividing the number of mitotic cells by the total number of cells counted and often expressed as a percentage. This requires counting both dividing and non-dividing cells in the same field.

Question 34Cell Biology, Microscopy and Biological Techniques

Which reagent is used to test a food sample for the presence of reducing sugars?

  1. Biuret reagent
  2. Iodine solution
  3. Benedict's solution Correct answer.
  4. Ethanol emulsion test

Explanation

Benedict's solution contains copper(II) ions that are reduced to copper(I) oxide by reducing sugars on heating, giving a brick-red precipitate. Biuret tests for protein, iodine for starch, and the ethanol emulsion test for lipids.

Question 35Cell Biology, Microscopy and Biological Techniques

During an enzyme-controlled reaction, a student measures product formation over time. Which observation would indicate that the rate of reaction has decreased because the substrate is being used up?

  1. The gradient of the product-time graph decreases as the curve levels off. Correct answer.
  2. The initial rate of product formation increases with temperature.
  3. The enzyme concentration doubles.
  4. The pH of the buffer changes.

Explanation

As substrate is consumed, fewer substrate molecules are available to bind to enzyme active sites, so the rate of product formation falls and the product-time curve flattens. This declining gradient reflects substrate depletion rather than a change in enzyme activity per se.

Question 36Cell Biology, Microscopy and Biological Techniques

Which statement about the resolving power of an electron microscope compared with a light microscope is correct?

  1. The electron microscope has a higher resolving power because electrons have a much shorter wavelength. Correct answer.
  2. The electron microscope has a lower resolving power because electrons carry a charge.
  3. The light microscope has a higher resolving power because visible light is more energetic.
  4. Both instruments have the same resolving power because they use lenses.

Explanation

Resolution improves as the wavelength of the radiation decreases. Electrons used in electron microscopy have wavelengths far shorter than visible light, allowing the electron microscope to resolve much finer detail than a light microscope.

Question 37Cell Biology, Microscopy and Biological Techniques

A cell biologist wants to isolate intact chloroplasts from a leaf homogenate. Which precaution is most important during the procedure?

  1. Using distilled water to lyse all membranes
  2. Adding a strong detergent to dissolve membranes
  3. Keeping the homogenate cold and using an isotonic buffer Correct answer.
  4. Heating the homogenate to 60 °C to denature enzymes

Explanation

Chloroplasts must remain intact, so the homogenate is kept cold to reduce enzyme activity and in an isotonic buffer to prevent osmotic swelling and bursting. Heating, distilled water or detergent would damage the organelle membranes.

Question 38Cell Biology, Microscopy and Biological Techniques

Which observation would provide evidence that a molecule crosses a membrane by active transport rather than facilitated diffusion?

  1. The molecule is transported only when its concentration gradient is steep.
  2. The molecule moves from higher to lower concentration through a channel protein.
  3. The molecule moves through the phospholipid bilayer without a carrier.
  4. The molecule moves from a region of lower to higher concentration, and the process requires ATP. Correct answer.

Explanation

Active transport is defined by movement against a concentration gradient, which requires an energy source such as ATP. Facilitated diffusion moves substances down their gradient through proteins and does not require ATP.

Question 39Cell Biology, Microscopy and Biological Techniques

Which technique would allow a student to estimate the surface area of a leaf using a microscope image?

  1. Measuring the image area and dividing by the square of the total magnification Correct answer.
  2. Counting the number of stomata per field of view
  3. Counting the number of chloroplasts in a single cell
  4. Measuring the thickness of the leaf with a ruler

Explanation

The actual area of a specimen is found by measuring the area on the image and dividing by the square of the linear magnification, since area scales with the square of linear dimensions. The other procedures measure different features and do not yield leaf surface area.

Question 40Cell Biology, Microscopy and Biological Techniques

Which statement correctly describes the role of the nucleolus?

  1. It contains the enzymes for the Krebs cycle.
  2. It packages proteins into secretory vesicles.
  3. It synthesises ribosomal RNA and assembles ribosomal subunits. Correct answer.
  4. It controls the entry and exit of substances from the nucleus.

Explanation

The nucleolus is a dense region within the nucleus where ribosomal RNA is transcribed and combined with proteins to form ribosomal subunits. Protein packaging occurs in the Golgi, the Krebs cycle occurs in mitochondria, and nuclear pores regulate exchange with the cytoplasm.

Question 41Microbiology and Applied Microbiology

Which bacterial structure is primarily responsible for the selective permeability that controls what enters and leaves a prokaryotic cell?

  1. The cytoplasmic membrane Correct answer.
  2. The capsule
  3. The peptidoglycan wall
  4. The flagellum

Explanation

The cytoplasmic membrane is a phospholipid bilayer with embedded proteins that regulates the passage of solutes, so it acts as the selective permeability barrier. The wall, capsule and flagellum serve structural, protective and motility roles respectively and do not control molecular traffic.

Question 42Microbiology and Applied Microbiology

In the standard Gram stain, what is the essential role of the iodine step?

  1. It counterstains Gram-negative cells pink
  2. It decolourises Gram-negative cells by dissolving their outer membrane
  3. It forms a large crystal violet-iodine complex trapped in the thick peptidoglycan Correct answer.
  4. It fixes the smear to the slide before staining

Explanation

Iodine acts as a mordant, combining with crystal violet to form a large complex that is retained by the heavily cross-linked peptidoglycan of Gram-positive cells during decolourisation. Decolourisation is done with alcohol or acetone, and the counterstain is safranin applied afterwards.

Question 43Microbiology and Applied Microbiology

A culture of bacteria is inoculated into a closed flask of sterile broth and incubated. Which phase is characterised by the number of viable cells remaining constant while the rate of cell division equals the rate of cell death?

  1. Stationary phase Correct answer.
  2. Lag phase
  3. Death phase
  4. Log phase

Explanation

In the stationary phase, growth-limiting factors such as nutrient depletion and waste accumulation balance division and death, so viable counts plateau. The lag phase precedes active division and the log phase is exponential growth, while the death phase shows net decline.

Question 44Microbiology and Applied Microbiology

Which group of microorganisms is defined by lacking a membrane-bound nucleus and typically reproducing by binary fission?

  1. Yeasts
  2. Bacteria Correct answer.
  3. Protozoa
  4. Moulds

Explanation

Bacteria are prokaryotes, meaning their genetic material is not enclosed in a nuclear membrane, and they commonly divide by binary fission. Yeasts, moulds and protozoa are all eukaryotes with a true nucleus.

Question 45Microbiology and Applied Microbiology

Which method achieves sterilisation by removing microorganisms physically rather than by killing them?

  1. Membrane filtration with a 0.22 µm filter Correct answer.
  2. Pasteurisation at 72 °C for 15 s
  3. Autoclaving at 121 °C
  4. Boiling at 100 °C for 10 min

Explanation

Filtration with pores of about 0.22 µm physically retains bacteria and many spores, so it removes rather than kills them, which suits heat-labile liquids. Autoclaving, pasteurisation and boiling all rely on heat to inactivate or kill microorganisms.

Question 46Microbiology and Applied Microbiology

What is the primary function of the bacterial capsule in relation to a host immune system?

  1. It provides the site of protein synthesis
  2. It resists phagocytosis and aids adherence to surfaces Correct answer.
  3. It transfers genetic material between cells
  4. It generates ATP for active transport

Explanation

The capsule is a polysaccharide or polypeptide layer that impedes phagocytosis and helps cells adhere to surfaces, contributing to virulence. ATP generation, gene transfer and protein synthesis are performed by the membrane, conjugative pili or plasmids, and ribosomes respectively.

Question 47Microbiology and Applied Microbiology

Which statement correctly describes the difference between a disinfectant and an antiseptic?

  1. A disinfectant is used on inanimate surfaces, whereas an antiseptic is applied to living tissue Correct answer.
  2. A disinfectant is a vaccine adjuvant, whereas an antiseptic is a sterilant
  3. A disinfectant is always applied to skin, whereas an antiseptic is used on instruments
  4. An antiseptic kills all spores, whereas a disinfectant does not

Explanation

Disinfectants are chemical agents applied to inanimate objects and surfaces, while antiseptics are formulated for safe application to living tissue such as skin. Both reduce microbial numbers but differ in the target surface and required tissue tolerance.

Question 48Microbiology and Applied Microbiology

A Gram-negative rod isolated from a food sample ferments lactose with gas production and is oxidase negative. Which organism is most consistent with these results?

  1. Staphylococcus aureus
  2. Escherichia coli Correct answer.
  3. Bacillus subtilis
  4. Pseudomonas aeruginosa

Explanation

Escherichia coli is a Gram-negative rod that ferments lactose with gas and is oxidase negative, matching the described profile. Pseudomonas is oxidase positive, while Staphylococcus and Bacillus are Gram-positive.

Question 49Microbiology and Applied Microbiology

In a viable plate count, a sample is diluted 10^-4 and 0.1 mL of the dilution is spread on an agar plate, giving 60 colonies after incubation. What is the estimated number of colony-forming units per mL of the original sample?

  1. 6.0 × 10^6
  2. 6.0 × 10^5 Correct answer.
  3. 6.0 × 10^3
  4. 6.0 × 10^4

Explanation

The count is divided by the plated volume and multiplied by the dilution factor: 60 ÷ 0.1 mL × 10^4 = 600 × 10^4 = 6.0 × 10^5 CFU per mL. Always express the count per mL of the original sample.

Question 50Microbiology and Applied Microbiology

Why is a candle jar or gas-generating sachet used when culturing microaerophilic bacteria?

  1. To remove all oxygen completely from the atmosphere
  2. To provide a pure oxygen atmosphere
  3. To maintain anaerobic conditions with hydrogen gas
  4. To reduce oxygen to a low but non-zero level while raising carbon dioxide Correct answer.

Explanation

Microaerophiles require oxygen at concentrations below atmospheric but cannot tolerate full air, so a candle jar or sachet lowers oxygen and enriches carbon dioxide. Complete oxygen removal would be used for strict anaerobes, not microaerophiles.