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membrane and cytoplasm.","correct":false,"explanation":"All bacterial cells possess cytoplasm enclosed by a plasma membrane. As prokaryotes, they instead lack a nucleus and membrane-bound organelles."},{"text":"A cell wall and circular chromosome.","correct":false,"explanation":"Most bacteria have a cell wall and their main genetic material is usually a circular DNA molecule. They lack a nucleus and membrane-bound organelles."},{"text":"A capsule and slime layer.","correct":false,"explanation":"Capsules and slime layers occur in some bacteria but are not defining prokaryotic structures. Prokaryotes specifically lack a nucleus and membrane-bound organelles."}]},{"id":"68672f40-2a2e-48ec-b826-2c243a18d699","cardType":"cloze","points":null,"front":"Bacteria usually reproduce by [ ], where one cell forms two [ ].","back":"Bacteria usually reproduce by binary fission, where one cell forms two genetically identical cells.","mcqOptions":[{"text":"Bacteria usually reproduce by binary fission, where one cell forms two genetically identical cells.","correct":true,"explanation":"During binary fission, bacterial DNA is replicated and the original cell divides into two daughter cells that are genetically identical, apart from possible mutations."},{"text":"Bacteria usually reproduce by mitosis, where one cell forms two genetically identical cells.","correct":false,"explanation":"Mitosis involves a nucleus and spindle apparatus, which bacteria lack. Bacteria normally divide by binary fission to produce two genetically identical cells."},{"text":"Bacteria usually reproduce by binary fission, where one cell forms two genetically different cells.","correct":false,"explanation":"Binary fission copies the bacterial chromosome before division, so it normally produces two genetically identical cells rather than genetically different ones, except where mutations occur."},{"text":"Bacteria usually reproduce by budding, where one cell forms two genetically different cells.","correct":false,"explanation":"Budding is not the usual reproductive process in bacteria. Most reproduce by binary fission, producing two genetically identical daughter cells after DNA replication."}]},{"id":"d947c0bc-ae39-49a1-9864-952d89f2523d","cardType":"cloze","points":null,"front":"Cocci are [ ] bacteria; bacilli are [ ] bacteria.","back":"Cocci are spherical bacteria; bacilli are rod-shaped bacteria.","mcqOptions":[{"text":"Cocci are spherical bacteria; bacilli are rod-shaped bacteria.","correct":true,"explanation":"Bacterial shape terminology identifies cocci as spherical cells and bacilli as rod-shaped cells. These names are commonly used when classifying bacterial morphology."},{"text":"Cocci are spherical bacteria; bacilli are spiral-shaped bacteria.","correct":false,"explanation":"Cocci are correctly identified as spherical, but spiral-shaped bacteria are spirilla. Bacilli are rod-shaped bacteria."},{"text":"Cocci are rod-shaped bacteria; bacilli are spherical bacteria.","correct":false,"explanation":"This reverses the two morphological terms. Cocci are spherical bacteria, whereas bacilli are rod-shaped bacteria."},{"text":"Cocci are comma-shaped bacteria; bacilli are rod-shaped bacteria.","correct":false,"explanation":"Bacilli are correctly described as rod-shaped, but comma-shaped bacteria are vibrios. Cocci are spherical bacteria."}]},{"id":"c11f247a-d2cb-400c-abad-e0749c9a3ab4","cardType":"cloze","points":null,"front":"Spirilla are [ ] bacteria; vibrios are [ ] bacteria.","back":"Spirilla are spiral-shaped bacteria; vibrios are comma-shaped bacteria.","mcqOptions":[{"text":"Spirilla are spiral-shaped bacteria; vibrios are comma-shaped bacteria.","correct":true,"explanation":"Spirilla have a twisted or spiral morphology, while vibrios are curved rods commonly described as comma-shaped. These terms distinguish their characteristic cell shapes."},{"text":"Spirilla are spiral-shaped bacteria; vibrios are rod-shaped bacteria.","correct":false,"explanation":"Spirilla are correctly described as spiral-shaped, but ordinary rod-shaped bacteria are bacilli. Vibrios are curved, comma-shaped bacteria."},{"text":"Spirilla are comma-shaped bacteria; vibrios are comma-shaped bacteria.","correct":false,"explanation":"Comma-shaped correctly describes vibrios, but not spirilla. Spirilla have a spiral-shaped or twisted morphology."},{"text":"Spirilla are spherical bacteria; vibrios are rod-shaped bacteria.","correct":false,"explanation":"Spherical bacteria are cocci and straight rod-shaped bacteria are bacilli. Spirilla are spiral-shaped, while vibrios are comma-shaped."}]},{"id":"a4646dad-2a69-43ed-b700-dfdc87caea70","cardType":"basic","points":2,"front":"Why do Gram-positive bacteria remain purple after decolourisation?","back":"Their thick peptidoglycan wall retains crystal violet-iodine.","mcqOptions":[{"text":"Their thick peptidoglycan wall retains crystal violet-iodine.","correct":true,"explanation":"Alcohol dehydrates the thick peptidoglycan wall, trapping the large crystal violet-iodine complexes. Gram-positive cells therefore remain purple after decolourisation."},{"text":"Their outer membrane retains crystal violet-iodine.","correct":false,"explanation":"An outer membrane is characteristic of Gram-negative bacteria and is disrupted by the decolouriser. Gram-positive bacteria remain purple because their thick peptidoglycan wall retains the complex."},{"text":"Their thin peptidoglycan wall prevents decolouriser entering.","correct":false,"explanation":"A thin peptidoglycan layer is characteristic of Gram-negative bacteria and does not retain the dye complex during decolourisation. Gram-positive bacteria have thick peptidoglycan."},{"text":"Their lipopolysaccharide layer binds the safranin counterstain.","correct":false,"explanation":"Lipopolysaccharide occurs in the Gram-negative outer membrane, and safranin produces a pink or red result. Gram-positive cells remain purple because thick peptidoglycan retains crystal violet-iodine."}]},{"id":"99aa2993-f40a-4dc0-99e9-2eab7d358d15","cardType":"cloze","points":null,"front":"Gram-negative bacteria have a [ ] plus an [ ].","back":"Gram-negative bacteria have a thin peptidoglycan layer plus an outer membrane.","mcqOptions":[{"text":"Gram-negative bacteria have a thin peptidoglycan layer plus an outer membrane.","correct":true,"explanation":"The Gram-negative cell envelope contains a thin peptidoglycan layer between the plasma membrane and an additional outer membrane containing lipopolysaccharide."},{"text":"Gram-negative bacteria have a thick peptidoglycan layer plus an outer membrane.","correct":false,"explanation":"The outer membrane is correctly associated with Gram-negative bacteria, but their peptidoglycan is thin. A thick peptidoglycan wall is characteristic of Gram-positive bacteria."},{"text":"Gram-negative bacteria have a thin peptidoglycan layer plus an inner teichoic acid layer.","correct":false,"explanation":"The thin peptidoglycan layer is correct, but teichoic acids are associated with Gram-positive cell walls. Gram-negative bacteria instead possess an outer membrane."},{"text":"Gram-negative bacteria have a thick peptidoglycan layer plus an inner teichoic acid layer.","correct":false,"explanation":"Thick peptidoglycan and teichoic acids are features associated with Gram-positive bacteria. Gram-negative bacteria have thin peptidoglycan plus an outer membrane."}]},{"id":"f90a74d1-9771-4503-b9f1-ba59de5ff4a3","cardType":"basic","points":1,"front":"After Gram staining, what colour do Gram-negative bacteria appear?","back":"Pink/red.","mcqOptions":[{"text":"Pink/red.","correct":true,"explanation":"Decolourisation removes the crystal violet-iodine complex from Gram-negative cells, which then take up the safranin counterstain and appear pink or red."},{"text":"Purple.","correct":false,"explanation":"Purple indicates retention of the crystal violet-iodine complex by Gram-positive bacteria. Gram-negative bacteria are decolourised and then appear pink or red after counterstaining."},{"text":"Colourless.","correct":false,"explanation":"Gram-negative cells are temporarily colourless immediately after decolourisation, but the completed Gram-staining procedure includes safranin, making them appear pink or red."},{"text":"Blue/black.","correct":false,"explanation":"Blue or black is not the final colour produced by the standard Gram stain. Gram-negative bacteria absorb the counterstain and appear pink or red."}]},{"id":"e0e42865-6ad6-4821-be2c-64be5eb109a6","cardType":"basic","points":1,"front":"What is iodine's role in Gram staining?","back":"It is a mordant forming the crystal violet-iodine complex.","mcqOptions":[{"text":"It is a mordant forming the crystal violet-iodine complex.","correct":true,"explanation":"Iodine acts as a mordant by combining with crystal violet to form a larger, less soluble complex that becomes trapped within thick Gram-positive peptidoglycan."},{"text":"It is a counterstain colouring decolourised cells pink.","correct":false,"explanation":"Safranin, not iodine, is the counterstain that colours decolourised Gram-negative cells pink. Iodine is a mordant that forms the crystal violet-iodine complex."},{"text":"It is a decolouriser removing crystal violet from cells.","correct":false,"explanation":"Alcohol or acetone is used as the decolouriser. Iodine instead acts as a mordant, combining with crystal violet to form a larger dye complex."},{"text":"It is a primary stain colouring all cells purple.","correct":false,"explanation":"Crystal violet is the primary stain that initially colours all cells purple. Iodine is the mordant that forms and stabilises the crystal violet-iodine complex."}]},{"id":"6adc21a7-cbbe-4390-9895-a7c8139a5f27","cardType":"basic","points":2,"front":"What can leaving alcohol on a Gram-stained smear too long cause?","back":"Gram-positive cells may falsely appear Gram-negative.","mcqOptions":[{"text":"Gram-positive cells may falsely appear Gram-negative.","correct":true,"explanation":"Excess alcohol removes the crystal violet-iodine complex from thick peptidoglycan walls, so Gram-positive cells can lose the purple stain and be falsely classified as Gram-negative."},{"text":"Gram-negative cells may falsely appear Gram-positive.","correct":false,"explanation":"Over-decolourisation does not make Gram-negative cells retain purple stain. The key error is that excessive alcohol removes stain from Gram-positive cells, making them falsely appear Gram-negative."},{"text":"Gram-positive cells may falsely appear unstained.","correct":false,"explanation":"Alcohol is not expected to remove all staining from Gram-positive cells in this specific false-result pattern. Excessive decolourisation makes Gram-positive cells appear Gram-negative."},{"text":"Gram-positive cells may falsely appear enlarged.","correct":false,"explanation":"Alcohol does not cause Gram-positive cells to become larger or change their shape. The relevant staining error is that Gram-positive cells may falsely appear Gram-negative after excessive decolourisation."}]},{"id":"6bde7171-7cb8-45f2-966d-453d057201e6","cardType":"basic","points":1,"front":"A Gram-stained smear shows pink/red rods. What are the bacteria?","back":"Gram-negative bacilli.","mcqOptions":[{"text":"Gram-negative bacilli.","correct":true,"explanation":"Gram-negative bacteria retain the counterstain and appear pink or red. Rod-shaped bacteria are called bacilli, so the smear shows Gram-negative bacilli."},{"text":"Gram-positive bacilli.","correct":false,"explanation":"Bacilli correctly identifies the rod shape, but Gram-positive cells retain the crystal violet and appear purple. Pink or red rods are therefore Gram-negative bacilli."},{"text":"Gram-negative cocci.","correct":false,"explanation":"Gram-negative correctly identifies the staining reaction, but cocci are spherical rather than rod-shaped. Pink or red rods are specifically Gram-negative bacilli."},{"text":"Gram-positive cocci.","correct":false,"explanation":"This option reverses both observations: the cells are pink or red, indicating Gram-negative bacteria, and rods are bacilli rather than cocci. The answer is Gram-negative bacilli."}]},{"id":"ed4275a7-3c73-4519-beff-7578548fc324","cardType":"basic","points":1,"front":"Which plating method spreads a measured volume evenly over agar for viable counts?","back":"Spread plate.","mcqOptions":[{"text":"Spread plate.","correct":true,"explanation":"A spread plate distributes a measured sample evenly across the agar surface, allowing colonies to develop separately and be counted to estimate viable cell numbers."},{"text":"Pour plate.","correct":false,"explanation":"A pour plate mixes the measured sample with molten agar rather than spreading it across the surface. The described surface method is the Spread plate."},{"text":"Streak plate.","correct":false,"explanation":"A streak plate separates organisms to obtain isolated colonies using an inoculating loop, but it is not normally used to spread a measured volume for viable counts. The answer is Spread plate."},{"text":"Membrane filtration.","correct":false,"explanation":"Membrane filtration concentrates cells on a filter before placing it on agar; it does not spread a measured volume directly over the agar surface. That method is the Spread plate."}]},{"id":"481010e9-f774-4a8b-b204-12722fa5280e","cardType":"basic","points":2,"front":"Why are school agar plates commonly incubated at about 25 °C, not 37 °C?","back":"37 °C favours many human pathogens.","mcqOptions":[{"text":"37 °C favours many human pathogens.","correct":true,"explanation":"Incubating at 37 °C can encourage growth of microorganisms adapted to the human body, including potential pathogens. Using about 25 °C reduces this safety risk in school cultures."},{"text":"25 °C favours many human pathogens.","correct":false,"explanation":"Many environmental microorganisms can grow at 25 °C, but 37 °C is the temperature that more strongly favours organisms adapted to humans. The correct reason is that 37 °C favours many human pathogens."},{"text":"37 °C prevents many human pathogens.","correct":false,"explanation":"A temperature of 37 °C generally promotes, rather than prevents, growth of organisms adapted to humans. It is avoided because 37 °C favours many human pathogens."},{"text":"37 °C kills many human pathogens.","correct":false,"explanation":"Incubation temperature does not sterilise plates or eliminate pathogens. The safety concern is that 37 °C favours many human pathogens, so lower temperatures are used."}]},{"id":"e3e96444-5976-408d-97d7-97b81b429bf9","cardType":"basic","points":1,"front":"Which growth phase has rapid binary fission and regular population doubling?","back":"Log/exponential phase.","mcqOptions":[{"text":"Log/exponential phase.","correct":true,"explanation":"During the log or exponential phase, cells divide rapidly at a regular rate, producing repeated population doublings while nutrients remain plentiful and conditions are favourable."},{"text":"Lag phase","correct":false,"explanation":"The lag phase involves adjustment to the new environment and little population increase, not regular rapid doubling. Rapid binary fission occurs in the Log/exponential phase."},{"text":"Stationary phase","correct":false,"explanation":"The stationary phase occurs when nutrient limitation and waste accumulation balance cell division with cell death. Regular rapid doubling occurs during the Log/exponential phase."},{"text":"Death/decline phase","correct":false,"explanation":"The death or decline phase is characterised by more cells dying than being produced, so population size falls. Rapid regular doubling belongs to the Log/exponential phase."}]},{"id":"d4382589-2d52-4993-a048-6b66272dd624","cardType":"cloze","points":null,"front":"In the [ ], the division rate equals the [ ].","back":"In the stationary phase, the division rate equals the death rate.","mcqOptions":[{"text":"In the stationary phase, the division rate equals the death rate.","correct":true,"explanation":"In the stationary phase, limiting nutrients and accumulating waste reduce net population growth. Cells still divide, but the division rate equals the death rate."},{"text":"In the lag phase, the division rate equals the death rate.","correct":false,"explanation":"The lag phase is mainly a period of cellular adjustment, not the point where division and death balance. That equality occurs in the stationary phase, where the division rate equals the death rate."},{"text":"In the exponential phase, the division rate equals the death rate.","correct":false,"explanation":"During exponential growth, division greatly exceeds death, producing rapid population doubling. Equal rates occur in the stationary phase, where the division rate equals the death rate."},{"text":"In the death phase, the division rate equals the death rate.","correct":false,"explanation":"In the death phase, cells die faster than new cells are produced, so the rates are not equal. Equality occurs in the stationary phase, when the division rate equals the death rate."}]},{"id":"d873e8b5-9ef0-42d0-ac88-2181d11a26f4","cardType":"basic","points":2,"front":"Why open Petri dish lids for the shortest possible time?","back":"To reduce contamination by unwanted microorganisms.","mcqOptions":[{"text":"To reduce contamination by unwanted microorganisms.","correct":true,"explanation":"Keeping the lid open briefly limits the time during which airborne particles and microbes can enter the dish, reducing contamination by unwanted microorganisms."},{"text":"To increase contamination by unwanted microorganisms.","correct":false,"explanation":"Leaving the lid open longer increases exposure to airborne microbes rather than reducing it. The correct practice is to minimise opening time to reduce contamination by unwanted microorganisms."},{"text":"To increase nutrient availability for microorganisms.","correct":false,"explanation":"Opening the lid briefly does not increase the nutrient supply in the agar. Its purpose is to reduce exposure and therefore reduce contamination by unwanted microorganisms."},{"text":"To accelerate growth of the inoculated microorganism.","correct":false,"explanation":"A short opening does not make the culture grow faster. It limits airborne exposure, helping to reduce contamination by unwanted microorganisms."}]},{"id":"92720d6e-2d25-4d33-b199-60d3d6c309a0","cardType":"basic","points":2,"front":"Why incubate agar plates inverted?","back":"To stop condensation dripping onto colonies.","mcqOptions":[{"text":"To stop condensation dripping onto colonies.","correct":true,"explanation":"Inverting the plate keeps condensation on the lid rather than allowing water droplets to fall onto the agar. This prevents colonies from spreading or becoming mixed."},{"text":"To stop condensation forming on the lid.","correct":false,"explanation":"Inverting plates does not stop condensation forming; it stops droplets on the lid from falling onto the agar. The purpose is to stop condensation dripping onto colonies."},{"text":"To increase oxygen entering the agar.","correct":false,"explanation":"The lid still allows some gas exchange around the plate, and inversion is not primarily for increasing oxygen supply. It prevents condensation dripping onto colonies."},{"text":"To kill microorganisms on the agar surface.","correct":false,"explanation":"Inversion does not sterilise the agar or kill microorganisms. It prevents liquid droplets from disturbing colonies by stopping condensation dripping onto colonies."}]},{"id":"e034a74a-bba4-4d6a-9e2d-1e110de706ed","cardType":"basic","points":1,"front":"What is a viable count reported as?","back":"CFU cm^-3.","mcqOptions":[{"text":"CFU cm^-3.","correct":true,"explanation":"A viable count is reported as colony-forming units per cubic centimetre because each visible colony represents one viable unit or clump that was able to reproduce."},{"text":"CFU cm^-1.","correct":false,"explanation":"The exponent must describe volume, not length. Viable counts are reported as CFU cm^-3 because the original sample volume is measured in cubic centimetres."},{"text":"cells cm^-3.","correct":false,"explanation":"A colony does not necessarily arise from one individual cell, so the result is expressed as colony-forming units rather than cells. The correct unit is CFU cm^-3."},{"text":"CFU cm^-2.","correct":false,"explanation":"The sample concentration is based on volume, not the surface area of the agar plate. Therefore the correct reporting unit is CFU cm^-3, not CFU cm^-2."}]},{"id":"fe31f976-d926-459e-b179-114f60db438d","cardType":"cloze","points":null,"front":"Transferring 1 cm³ into 9 cm³ sterile diluent makes a [ ] dilution, written [ ].","back":"Transferring 1 cm³ into 9 cm³ sterile diluent makes a tenfold dilution, written 10^-1.","mcqOptions":[{"text":"Transferring 1 cm³ into 9 cm³ sterile diluent makes a tenfold dilution, written 10^-1.","correct":true,"explanation":"The final volume is 10 cm³, with 1 cm³ of the original sample, so the sample is diluted to one tenth of its original concentration: a tenfold or 10^-1 dilution."},{"text":"Transferring 1 cm³ into 9 cm³ sterile diluent makes a tenfold dilution, written 10^-2.","correct":false,"explanation":"The volume ratio is 1 in 10, not 1 in 100. This is a common exponent error; the correct notation is 10^-1 for a tenfold dilution."},{"text":"Transferring 1 cm³ into 9 cm³ sterile diluent makes a tenfold dilution, written 10¹.","correct":false,"explanation":"Dilution reduces concentration, so its factor is written with a negative exponent. One part in ten is 10^-1, not 10¹."},{"text":"Transferring 1 cm³ into 9 cm³ sterile diluent makes a hundredfold dilution, written 10^-2.","correct":false,"explanation":"A hundredfold dilution would require one part sample in 100 parts total volume. Here the total is 10 cm³, giving the correct dilution of tenfold, written 10^-1."}]},{"id":"27dd12af-6b81-4eea-b063-721022416fa7","cardType":"basic","points":1,"front":"Which colony count range should be used for plates?","back":"30-300 colonies.","mcqOptions":[{"text":"30-300 colonies.","correct":true,"explanation":"Plates with 30-300 colonies provide enough colonies for a reliable count while avoiding overcrowding, merged colonies and competition that make higher counts inaccurate."},{"text":"1-10 colonies.","correct":false,"explanation":"Very few colonies produce a large percentage counting uncertainty, so random variation has a strong effect. The recommended reliable range is 30-300 colonies."},{"text":"10-100 colonies.","correct":false,"explanation":"This range includes plates with too few colonies at its lower end, increasing sampling error. The standard usable range is 30-300 colonies."},{"text":"300-1000 colonies.","correct":false,"explanation":"Dense plates in this range may have overlapping colonies or exhausted nutrients, causing undercounting. Reliable estimates normally use plates containing 30-300 colonies."}]},{"id":"f872753e-06c3-4b2f-b7b2-06563582ef2b","cardType":"basic","points":1,"front":"What formula calculates viable count from plate counts?","back":"colonies counted ÷ (volume plated in cm³ × dilution).","mcqOptions":[{"text":"colonies counted ÷ (volume plated in cm³ × dilution).","correct":true,"explanation":"The colonies counted are divided by the plated volume and by the dilution used. This converts the observed count back to the viable concentration in the original sample."},{"text":"colonies counted × (volume plated in cm³ × dilution).","correct":false,"explanation":"Multiplying applies the dilution and plated volume in the wrong direction, making the calculated concentration much too small. Both factors belong in the denominator."},{"text":"colonies counted ÷ (volume plated in cm³ ÷ dilution).","correct":false,"explanation":"Dividing by the dilution reverses how the dilution factor is used inside the denominator. The correct calculation divides colonies counted by volume plated multiplied by dilution."},{"text":"(colonies counted ÷ volume plated in cm³) × dilution.","correct":false,"explanation":"Applying the dilution factor by multiplication reduces the result instead of correcting it back to the original concentration. The dilution must divide the colony concentration."}]},{"id":"1847f28a-ef06-4721-9e88-ffd9fe809ec8","cardType":"basic","points":2,"front":"Mean count = 87.3, volume = 0.1 cm³, dilution = 10^-5. What is the viable count?","back":"8.73 × 10⁷ CFU cm^-3.","mcqOptions":[{"text":"8.73 × 10⁷ CFU cm^-3.","correct":true,"explanation":"Using colonies ÷ (volume × dilution), 87.3 ÷ (0.1 × 10^-5) = 8.73 × 10⁷ CFU cm^-3."},{"text":"8.73 × 10⁶ CFU cm^-3.","correct":false,"explanation":"This results from forgetting to divide by the plated volume of 0.1 cm³. Including that factor increases the result tenfold to 8.73 × 10⁷ CFU cm^-3."},{"text":"8.73 × 10⁸ CFU cm^-3.","correct":false,"explanation":"This is ten times too large, usually caused by treating 0.1 cm³ as 0.01 cm³. The correct calculation gives 8.73 × 10⁷ CFU cm^-3."},{"text":"8.73 × 10² CFU cm^-3.","correct":false,"explanation":"This ignores the correction for the 10^-5 dilution and leaves the value close to the raw plate count. Dividing by both factors gives 8.73 × 10⁷ CFU cm^-3."}]},{"id":"4d1ee4f0-f6e3-49fb-a40d-f77fc26d4764","cardType":"cloze","points":null,"front":"In the milk practical, the independent variable is [ ]; the dependent variable is [ ].","back":"In the milk practical, the independent variable is milk condition; the dependent variable is viable bacterial count.","mcqOptions":[{"text":"In the milk practical, the independent variable is milk condition; the dependent variable is viable bacterial count.","correct":true,"explanation":"Milk condition is deliberately changed between samples, making it the independent variable. The viable bacterial count is measured in response, so it is the dependent variable."},{"text":"In the milk practical, the independent variable is viable bacterial count; the dependent variable is milk condition.","correct":false,"explanation":"This reverses the roles of the variables. The milk condition is changed by the investigator, while the viable bacterial count is measured as the outcome."},{"text":"In the milk practical, the independent variable is incubation temperature; the dependent variable is milk condition.","correct":false,"explanation":"Unless temperature is the factor deliberately varied, it should be controlled rather than identified as the independent variable. The investigation changes milk condition and measures viable bacterial count."},{"text":"In the milk practical, the independent variable is viable bacterial count; the dependent variable is incubation time.","correct":false,"explanation":"A bacterial count is measured, not deliberately set, so it cannot be the independent variable. The practical compares milk condition and measures viable bacterial count."}]},{"id":"d7ebd98d-911a-406a-b2b1-496cfb8c9d1e","cardType":"basic","points":2,"front":"Why should stale milk usually have a higher viable count than fresh milk?","back":"Bacteria have had more time to divide.","mcqOptions":[{"text":"Bacteria have had more time to divide.","correct":true,"explanation":"During storage, bacteria repeatedly undergo cell division, so stale milk generally contains more viable bacteria than fresh milk, provided conditions allow growth."},{"text":"Bacteria have had less time to divide.","correct":false,"explanation":"Less time would generally allow fewer cell divisions and therefore a lower count. Stale milk usually has a higher viable count because bacteria have had more time to divide."},{"text":"Bacteria have had more nutrients added.","correct":false,"explanation":"The usual explanation is not that nutrients are added during storage. The population increases because existing bacteria have had more time to divide."},{"text":"Bacteria have had less competition for oxygen.","correct":false,"explanation":"Oxygen availability is not the main general explanation for the difference between fresh and stale milk. Longer storage usually permits more bacterial divisions and a higher viable count."}]},{"id":"1cc53920-a91c-4825-b887-178297b57dae","cardType":"basic","points":2,"front":"Why can a colony count overestimate the number of individual cells?","back":"One colony may come from a clump.","mcqOptions":[{"text":"One colony may come from a clump.","correct":true,"explanation":"Several cells stuck together can grow as one visible colony, so the count records one colony-forming unit rather than every individual cell present."},{"text":"One colony must come from one cell.","correct":false,"explanation":"Cells may remain attached in clumps, allowing several cells to produce one colony. Therefore the count can underestimate individual cells; the correct explanation is that a colony may come from a clump."},{"text":"One colony may come from a dead cell.","correct":false,"explanation":"Dead cells cannot reproduce to form colonies on the agar. The overestimate concern is instead that one colony-forming unit may contain several living cells in a clump."},{"text":"One colony may contain no bacterial cells.","correct":false,"explanation":"A visible bacterial colony requires viable cells to reproduce. The relevant limitation is that several cells can be grouped in one clump and produce only one counted colony."}]}]}],"examCountdown":{"examDate":"2027-06-07","paperLabel":"Component 1"},"heading":"Microbiology","title":"Microbiology Flashcards","topbar":{"backHref":"/a-level/biology/eduqas/flashcards"},"topicSidebar":"$L38"}]}],"topicId":"fa296282-1975-5284-8f9a-507fcf21f1da","walled":"$L39"}],"$L3a","$L3b"]