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with oxidation causes this error. Hydrolysis specifically uses water to split the chemical bond."},{"text":"Hydrogen.","correct":false,"explanation":"Hydrogen alone cannot hydrolyse a bond. The reaction requires water, whose components are added across the bond as it breaks."},{"text":"Carbon dioxide.","correct":false,"explanation":"Carbon dioxide is not involved in hydrolysis. Water reacts with the bond, adding its components and allowing the original molecule to split."}]},{"id":"4a276cfe-8554-44fa-a6e9-6bd7921f8864","cardType":"basic","points":2,"front":"Why are large, insoluble food molecules digested?","back":"To form small, soluble molecules that cross membranes.","mcqOptions":[{"text":"To form small, soluble molecules that cross membranes.","correct":true,"explanation":"Large food molecules are usually too large to cross cell membranes. Digestion hydrolyses them into small, soluble molecules that can be absorbed."},{"text":"To form large, soluble molecules that cross membranes.","correct":false,"explanation":"Digestion does not make molecules larger; that would make membrane crossing more difficult. It produces small, soluble molecules that can be absorbed."},{"text":"To form small, insoluble molecules that cross membranes.","correct":false,"explanation":"Insoluble products cannot dissolve in the gut contents or pass effectively into solution. Digestion forms small, soluble molecules that can cross membranes."},{"text":"To form small, soluble molecules that release energy.","correct":false,"explanation":"Digestion makes molecules suitable for absorption, rather than releasing their energy directly. The products are small and soluble so they can cross membranes."}]},{"id":"6d203a59-6284-46f1-b344-2d57e182c513","cardType":"basic","points":1,"front":"Where does most absorption of digested molecules occur?","back":"The ileum.","mcqOptions":[{"text":"The ileum.","correct":true,"explanation":"Most absorption of digested nutrients occurs in the ileum, whose villi and microvilli provide a large surface area and a short diffusion distance."},{"text":"The duodenum.","correct":false,"explanation":"The duodenum receives bile and pancreatic enzymes, so digestion is completed there, but most absorption occurs later in the ileum."},{"text":"The jejunum.","correct":false,"explanation":"The jejunum can absorb some nutrients, but the A-level description identifies the ileum as the main site of absorption of digested molecules."},{"text":"The colon.","correct":false,"explanation":"The colon mainly absorbs water and mineral ions, not most digested food molecules. The principal site for nutrient absorption is the ileum."}]},{"id":"1e0f9d66-575b-491b-b778-12960fb2e4f1","cardType":"cloze","points":null,"front":"[ ] increase surface area; a thin epithelium shortens the [ ].","back":"Villi increase surface area; a thin epithelium shortens the diffusion distance.","mcqOptions":[{"text":"Villi increase surface area; a thin epithelium shortens the diffusion distance.","correct":true,"explanation":"Villi greatly increase the ileum’s surface area, while the thin epithelium reduces the distance nutrients must diffuse before entering blood or lymph."},{"text":"Goblet cells increase surface area; a thin epithelium shortens the diffusion distance.","correct":false,"explanation":"Goblet cells secrete mucus and do not create the main absorptive surface. Villi increase surface area, while a thin epithelium shortens diffusion distance."},{"text":"Muscle fibres increase surface area; a thin epithelium shortens the diffusion distance.","correct":false,"explanation":"Muscle fibres produce movement rather than forming the absorptive surface. Villi provide increased surface area, and the thin epithelium reduces diffusion distance."},{"text":"Capillaries increase surface area; a thick epithelium shortens the diffusion distance.","correct":false,"explanation":"Capillaries transport absorbed substances, but villi provide the increased surface area. A thick epithelium would increase, not shorten, the diffusion distance."}]},{"id":"9791247d-b704-487f-a66a-b617c366cd1e","cardType":"basic","points":1,"front":"Which ileum structures carry absorbed lipid products away?","back":"Lacteals.","mcqOptions":[{"text":"Lacteals.","correct":true,"explanation":"Lacteals are lymphatic vessels inside villi. They take up absorbed fatty acids and monoglycerides, which are transported away as chylomicrons in the lymph."},{"text":"Capillaries.","correct":false,"explanation":"Capillaries carry absorbed glucose and amino acids, but most lipid products enter lacteals because chylomicrons are too large to enter blood capillaries directly."},{"text":"Venules.","correct":false,"explanation":"Venules drain blood from capillaries and do not specifically transport absorbed lipid products. Lipids are carried away from villi through lacteals in the lymphatic system."},{"text":"Arterioles.","correct":false,"explanation":"Arterioles deliver blood to capillary networks rather than removing absorbed lipids. Lipid products leave the villi through lacteals before entering the bloodstream."}]},{"id":"3bc9e0c2-95d9-4f8f-93b8-4ae3be9a57d2","cardType":"basic","points":1,"front":"What product does amylase form from starch?","back":"Maltose.","mcqOptions":[{"text":"Maltose.","correct":true,"explanation":"Amylase hydrolyses starch into the disaccharide maltose. Maltose is then hydrolysed by maltase at the brush border to form glucose."},{"text":"Glucose.","correct":false,"explanation":"Glucose is the final monosaccharide product of starch digestion, but amylase first produces maltose. Maltase subsequently breaks maltose into glucose."},{"text":"Sucrose.","correct":false,"explanation":"Sucrose is a different disaccharide made from glucose and fructose. It is not formed when amylase hydrolyses starch; the product is maltose."},{"text":"Glycogen.","correct":false,"explanation":"Glycogen is a storage polysaccharide, not a product of starch hydrolysis. Amylase breaks starch down into the disaccharide maltose."}]},{"id":"50ccf813-e04a-4566-84cc-bb3447ba87dc","cardType":"cloze","points":null,"front":"Amylase hydrolyses [ ] in starch to form [ ].","back":"Amylase hydrolyses glycosidic bonds in starch to form maltose.","mcqOptions":[{"text":"Amylase hydrolyses glycosidic bonds in starch to form maltose.","correct":true,"explanation":"Starch contains glycosidic bonds between glucose units. Amylase hydrolyses some of these bonds, producing the disaccharide maltose."},{"text":"Amylase hydrolyses peptide bonds in starch to form maltose.","correct":false,"explanation":"Peptide bonds join amino acids in proteins, not glucose units in starch. Starch contains glycosidic bonds, whose hydrolysis produces maltose."},{"text":"Amylase hydrolyses glycosidic bonds in starch to form glucose.","correct":false,"explanation":"Glucose is produced after maltose is hydrolysed by maltase. Amylase’s immediate product from starch is maltose, formed by breaking glycosidic bonds."},{"text":"Amylase hydrolyses hydrogen bonds in starch to form glucose.","correct":false,"explanation":"Hydrogen bonds help maintain molecular structure but are not the covalent links hydrolysed in starch digestion. Amylase breaks glycosidic bonds to form maltose."}]},{"id":"fd9334bd-b84e-459a-b839-b52590ab9a9c","cardType":"cloze","points":null,"front":"Membrane-bound disaccharidases on the [ ] hydrolyse disaccharides into [ ].","back":"Membrane-bound disaccharidases on the brush border hydrolyse disaccharides into monosaccharides.","mcqOptions":[{"text":"Membrane-bound disaccharidases on the brush border hydrolyse disaccharides into monosaccharides.","correct":true,"explanation":"The brush border contains membrane-bound disaccharidases such as maltase. They hydrolyse disaccharides into absorbable monosaccharides."},{"text":"Membrane-bound disaccharidases on the stomach lining hydrolyse disaccharides into polysaccharides.","correct":false,"explanation":"These enzymes are located on the ileum’s brush border, not the stomach lining, and hydrolysis reduces disaccharides to monosaccharides rather than larger polysaccharides."},{"text":"Membrane-bound disaccharidases on the cell wall hydrolyse disaccharides into disaccharides.","correct":false,"explanation":"Animal intestinal cells have cell membranes, not cell walls, and hydrolysis must split disaccharides into smaller monosaccharides rather than leaving them unchanged."},{"text":"Membrane-bound disaccharidases on the pancreatic duct hydrolyse disaccharides into glucose.","correct":false,"explanation":"Disaccharidases are attached to the ileum’s brush border, not pancreatic ducts. Their products are monosaccharides; glucose is only one possible monosaccharide product."}]},{"id":"aec776bd-08a3-43ff-9338-80004e14f376","cardType":"basic","points":2,"front":"What does an endopeptidase do to a polypeptide?","back":"Hydrolyses peptide bonds within the chain, forming shorter peptides.","mcqOptions":[{"text":"Hydrolyses peptide bonds within the chain, forming shorter peptides.","correct":true,"explanation":"Endopeptidases act at peptide bonds within a polypeptide rather than at its ends, producing several shorter peptide chains for further hydrolysis."},{"text":"Hydrolyses peptide bonds at the ends of the chain, forming amino acids.","correct":false,"explanation":"Removing terminal amino acids is the action of exopeptidases. Endopeptidases instead hydrolyse peptide bonds within the chain, forming shorter peptides."},{"text":"Hydrolyses disulfide bonds within the chain, forming shorter peptides.","correct":false,"explanation":"Disulfide bonds stabilise a protein’s tertiary structure but are not the substrate bonds hydrolysed by endopeptidases. Endopeptidases split internal peptide bonds to form shorter peptides."},{"text":"Forms peptide bonds within the chain, producing longer peptides.","correct":false,"explanation":"Proteases catalyse hydrolysis rather than condensation, so they do not create peptide bonds. Endopeptidases hydrolyse internal peptide bonds and thereby shorten polypeptide chains."}]},{"id":"89f8a413-62cb-4aef-8761-59043d859917","cardType":"basic","points":1,"front":"What does an exopeptidase remove from peptide chains?","back":"Single amino acids from the ends.","mcqOptions":[{"text":"Single amino acids from the ends.","correct":true,"explanation":"Exopeptidases hydrolyse terminal peptide bonds, releasing one amino acid at a time from either end of a peptide chain."},{"text":"Single amino acids from the middle.","correct":false,"explanation":"Removing amino acids from within a chain would require cleavage of internal peptide bonds, which is characteristic of endopeptidases. Exopeptidases release single amino acids from the ends."},{"text":"Whole dipeptides from the ends.","correct":false,"explanation":"Dipeptides are subsequently hydrolysed by dipeptidases rather than being the characteristic products removed by exopeptidases. Exopeptidases release single amino acids from chain ends."},{"text":"Short peptide chains from the ends.","correct":false,"explanation":"Exopeptidases remove individual amino acids rather than detaching long peptide sections. Their terminal action releases single amino acids from the ends of peptide chains."}]},{"id":"5fac19da-6391-4af4-beb1-b42040163360","cardType":"cloze","points":null,"front":"Dipeptidases hydrolyse [ ] into individual [ ].","back":"Dipeptidases hydrolyse dipeptides into individual amino acids.","mcqOptions":[{"text":"Dipeptidases hydrolyse dipeptides into individual amino acids.","correct":true,"explanation":"A dipeptide consists of two amino acids joined by one peptide bond. Dipeptidases hydrolyse this bond, releasing the two individual amino acids."},{"text":"Dipeptidases hydrolyse polypeptides into individual amino acids.","correct":false,"explanation":"Polypeptides contain many peptide bonds and are initially shortened by endopeptidases and exopeptidases. Dipeptidases specifically hydrolyse dipeptides into individual amino acids."},{"text":"Dipeptidases hydrolyse dipeptides into individual short peptides.","correct":false,"explanation":"Dipeptidase activity breaks the peptide bond rather than producing smaller peptides. It hydrolyses dipeptides into individual amino acids, the absorbable monomers of proteins."},{"text":"Dipeptidases hydrolyse triglycerides into individual fatty acids.","correct":false,"explanation":"Triglycerides and fatty acids belong to lipid digestion and are acted on by lipase. Dipeptidases act on dipeptides and release individual amino acids."}]},{"id":"150e76fa-6f49-40a0-b243-956410caace3","cardType":"basic","points":2,"front":"Why do bile salts increase lipid digestion rate?","back":"They emulsify lipids, increasing surface area for lipase.","mcqOptions":[{"text":"They emulsify lipids, increasing surface area for lipase.","correct":true,"explanation":"Bile salts break large lipid droplets into many smaller droplets. This emulsification increases the total surface area available for lipase action."},{"text":"They hydrolyse lipids, increasing substrate concentration for lipase.","correct":false,"explanation":"Bile salts do not enzymatically hydrolyse triglycerides; lipase performs that reaction. They emulsify lipids, increasing the surface area on which lipase can act."},{"text":"They acidify lipids, providing the optimum pH for lipase.","correct":false,"explanation":"Bile helps neutralise acidic chyme, but lowering pH would inhibit pancreatic lipase. Bile salts principally increase digestion by emulsifying lipids and increasing their surface area."},{"text":"They combine lipid droplets, increasing contact time with lipase.","correct":false,"explanation":"Combining small droplets into larger ones would reduce the surface-area-to-volume ratio and slow digestion. Bile salts instead emulsify lipids into smaller droplets."}]},{"id":"c874e501-e069-4b5c-9d72-f9cb57225ba1","cardType":"basic","points":1,"front":"What bond does lipase hydrolyse in triglycerides?","back":"Ester bonds.","mcqOptions":[{"text":"Ester bonds.","correct":true,"explanation":"Each fatty acid is attached to glycerol by an ester bond. Lipase hydrolyses these bonds during the digestion of triglycerides."},{"text":"Peptide bonds.","correct":false,"explanation":"Peptide bonds join amino acids in proteins and are hydrolysed by proteases. Lipase digests triglycerides by hydrolysing their ester bonds."},{"text":"Glycosidic bonds.","correct":false,"explanation":"Glycosidic bonds join monosaccharides in carbohydrates and are targeted by carbohydrases. The fatty acids in triglycerides are linked to glycerol by ester bonds."},{"text":"Phosphodiester bonds.","correct":false,"explanation":"Phosphodiester bonds form the sugar-phosphate backbone of nucleic acids and are not present between glycerol and fatty acids. Lipase hydrolyses ester bonds."}]},{"id":"0cef0c57-f168-41a4-9ef1-7ea4fc97394e","cardType":"cloze","points":null,"front":"Lipase hydrolyses triglycerides to [ ] and [ ].","back":"Lipase hydrolyses triglycerides to fatty acids and monoglycerides.","mcqOptions":[{"text":"Lipase hydrolyses triglycerides to fatty acids and monoglycerides.","correct":true,"explanation":"In the small intestine, lipase hydrolyses ester bonds in triglycerides, producing fatty acids and monoglycerides that can associate with bile salts in micelles."},{"text":"Lipase hydrolyses triglycerides to fatty acids and disaccharides.","correct":false,"explanation":"Disaccharides are carbohydrate digestion products and cannot be formed from triglycerides. Intestinal lipase produces fatty acids and monoglycerides from triglycerides."},{"text":"Lipase hydrolyses triglycerides to amino acids and monoglycerides.","correct":false,"explanation":"Amino acids result from protein digestion, not triglyceride digestion. Lipase releases fatty acids while leaving monoglycerides as the other principal products."},{"text":"Lipase hydrolyses triglycerides to glucose and maltose.","correct":false,"explanation":"Glucose and maltose are products associated with carbohydrate digestion, not lipid digestion. Lipase digests triglycerides into fatty acids and monoglycerides."}]},{"id":"52bcd5f1-363e-408c-b14d-4a1d4551152f","cardType":"basic","points":2,"front":"How do micelles help lipid products enter epithelial cells?","back":"They bring products close to the epithelial surface, maintaining diffusion.","mcqOptions":[{"text":"They bring products close to the epithelial surface, maintaining diffusion.","correct":true,"explanation":"Micelles transport poorly soluble fatty acids and monoglycerides through the watery intestinal contents to the epithelial membrane, helping maintain a steep diffusion gradient."},{"text":"They carry products directly into the lacteals, maintaining absorption.","correct":false,"explanation":"Micelles do not transport lipids directly into lacteals; chylomicrons perform that role after processing inside epithelial cells. Micelles deliver digestion products close to the epithelial surface."},{"text":"They pump products through epithelial membranes, using active transport.","correct":false,"explanation":"Lipid products enter epithelial cells mainly by diffusion rather than active transport. Micelles bring these products close to the epithelial surface, maintaining the diffusion gradient."},{"text":"They hold products away from the epithelial surface, preventing aggregation.","correct":false,"explanation":"Keeping lipid products away from epithelial cells would increase their diffusion distance and slow absorption. Micelles instead carry them close to the epithelial surface."}]},{"id":"09027e2f-44a1-4246-92d6-f5ad34732345","cardType":"basic","points":1,"front":"Which particles carry re-formed triglycerides out into lacteals?","back":"Chylomicrons.","mcqOptions":[{"text":"Chylomicrons.","correct":true,"explanation":"Inside epithelial cells, absorbed fatty acids and monoglycerides are re-formed into triglycerides and packaged with proteins into chylomicrons, which enter the lacteals."},{"text":"Micelles.","correct":false,"explanation":"Micelles deliver fatty acids and monoglycerides to epithelial membranes but do not enter lacteals carrying re-formed triglycerides. Those triglycerides are packaged into chylomicrons."},{"text":"HDLs.","correct":false,"explanation":"High-density lipoproteins mainly transport cholesterol in the bloodstream and are not the particles assembled by intestinal epithelial cells for lacteal uptake. Chylomicrons perform this role."},{"text":"Lysosomes.","correct":false,"explanation":"Lysosomes contain hydrolytic enzymes for intracellular digestion and do not transport dietary triglycerides. Re-formed triglycerides leave epithelial cells in chylomicrons."}]},{"id":"74b8e71d-a090-4839-8c1f-00057c5424ff","cardType":"cloze","points":null,"front":"The [ ] uses [ ] to move sodium ions into the blood.","back":"The sodium-potassium pump uses ATP to move sodium ions into the blood.","mcqOptions":[{"text":"The sodium-potassium pump uses ATP to move sodium ions into the blood.","correct":true,"explanation":"The sodium-potassium pump hydrolyses ATP to actively transport sodium ions from epithelial cells into the blood, maintaining the sodium ion gradient needed for glucose absorption."},{"text":"The sodium-glucose co-transporter uses ATP to move sodium ions into the blood.","correct":false,"explanation":"The co-transporter moves sodium and glucose into epithelial cells, not sodium into the blood. The sodium-potassium pump uses ATP to export sodium and maintain the gradient."},{"text":"The sodium-potassium pump uses glucose to move sodium ions into the blood.","correct":false,"explanation":"Glucose provides no energy for this active transport mechanism. ATP supplies the energy, while the sodium-potassium pump exports sodium to maintain the concentration gradient."},{"text":"The sodium-glucose co-transporter uses glucose to move sodium ions into the blood.","correct":false,"explanation":"This reverses both stages of absorption: the co-transporter uses the sodium gradient to bring glucose into cells, whereas ATP powers the sodium-potassium pump moving sodium towards the blood."}]},{"id":"377952de-9b4a-458b-b8f3-b05a25113bb9","cardType":"basic","points":2,"front":"How does the sodium ion gradient drive glucose absorption?","back":"Sodium ions enter via a co-transporter, carrying glucose in.","mcqOptions":[{"text":"Sodium ions enter via a co-transporter, carrying glucose in.","correct":true,"explanation":"The sodium ion gradient provides the energy for co-transport. Sodium moves into the epithelial cell down its gradient and carries glucose against its concentration gradient."},{"text":"Sodium ions enter via a channel protein, carrying glucose in.","correct":false,"explanation":"A channel protein would allow ions to diffuse but would not couple sodium movement to glucose transport. The correct mechanism is a co-transporter using the sodium gradient."},{"text":"Sodium ions enter via active transport, carrying glucose in.","correct":false,"explanation":"The sodium ion itself enters down its electrochemical gradient rather than being actively pumped in at this membrane. A co-transporter couples that movement to glucose uptake."},{"text":"Sodium ions enter via a facilitated diffusion carrier, carrying glucose in.","correct":false,"explanation":"Facilitated diffusion does not provide the energy needed to move glucose against its gradient in this step. Sodium-glucose co-transport uses the sodium gradient instead."}]},{"id":"b8165b95-c03e-4544-9db0-90f5b792d577","cardType":"basic","points":1,"front":"How do glucose and amino acids leave epithelial cells for blood?","back":"By facilitated diffusion down a concentration gradient.","mcqOptions":[{"text":"By facilitated diffusion down a concentration gradient.","correct":true,"explanation":"Glucose and amino acids move from a higher concentration inside epithelial cells to a lower concentration in the blood through specific carrier proteins, requiring no ATP."},{"text":"By active transport against a concentration gradient.","correct":false,"explanation":"Export into the blood normally occurs down the concentration gradient, so ATP is not required for this step. Facilitated diffusion through carrier proteins is correct."},{"text":"By simple diffusion through the phospholipid bilayer.","correct":false,"explanation":"These polar molecules cannot readily cross the hydrophobic phospholipid bilayer unaided. They leave through membrane proteins by facilitated diffusion down their gradients."},{"text":"By osmosis down a water potential gradient.","correct":false,"explanation":"Osmosis describes water movement, not the export of glucose and amino acids. These solutes cross via specific carrier proteins by facilitated diffusion."}]},{"id":"d7d085a3-6037-46b5-aa6c-212eca1eda96","cardType":"basic","points":2,"front":"What happens to glucose absorption if ATP production falls?","back":"It decreases because the sodium ion gradient is less steep.","mcqOptions":[{"text":"It decreases because the sodium ion gradient is less steep.","correct":true,"explanation":"Lower ATP production reduces sodium-potassium pump activity, so less sodium is exported into the blood. The weakened gradient provides less driving force for glucose co-transport."},{"text":"It increases because the sodium ion gradient is less steep.","correct":false,"explanation":"A less steep sodium gradient provides less energy for co-transport, so absorption decreases rather than increases. The sodium-potassium pump depends on ATP to maintain that gradient."},{"text":"It decreases because glucose cannot be hydrolysed.","correct":false,"explanation":"Glucose is absorbed as a monosaccharide and does not need hydrolysis during uptake. Reduced ATP weakens the sodium gradient, which is why absorption decreases."},{"text":"It stays constant because glucose absorption uses no ATP.","correct":false,"explanation":"Although the co-transporter does not directly hydrolyse ATP, it depends indirectly on the ATP-powered sodium-potassium pump. Less ATP therefore reduces the sodium gradient and absorption."}]},{"id":"03e98945-2269-4eca-9daa-d208d230a7a0","cardType":"basic","points":1,"front":"In the starch-amylase practical, what does blue-black iodine show?","back":"Starch is present.","mcqOptions":[{"text":"Starch is present.","correct":true,"explanation":"Iodine forms a blue-black complex with starch, so this colour indicates that starch remains in the sample and has not been completely digested by amylase."},{"text":"Reducing sugar is present.","correct":false,"explanation":"Reducing sugars are detected using Benedict’s reagent after heating, producing a coloured precipitate. Blue-black iodine specifically indicates the presence of starch."},{"text":"Amylase is present.","correct":false,"explanation":"Iodine does not test for the enzyme itself. A blue-black result means starch remains; amylase activity would be inferred from starch disappearing over time."},{"text":"Glycogen is present.","correct":false,"explanation":"Glycogen does not give the characteristic blue-black starch result in this test. Iodine turning blue-black indicates starch is present in the mixture."}]},{"id":"8f8fe1d2-e4de-4f89-ace0-6e23e02dfca8","cardType":"basic","points":1,"front":"How is digestion rate calculated from endpoint time?","back":"$\\text{rate}=\\frac{1}{\\text{time taken}}$.","mcqOptions":[{"text":"$\\text{rate}=\\frac{1}{\\text{time taken}}$.","correct":true,"explanation":"When the endpoint time is measured, a shorter time means faster digestion. Taking the reciprocal converts the measured time into a rate proportional to digestion speed."},{"text":"$\\text{rate}=\\text{time taken}$.","correct":false,"explanation":"Using time directly gives larger values for slower reactions, reversing the expected relationship. The digestion rate is calculated as the reciprocal of the endpoint time."},{"text":"$\\text{rate}=\\frac{\\text{time taken}}{1}$.","correct":false,"explanation":"Dividing time by one leaves the time unchanged, so it does not convert a shorter endpoint into a greater rate. The required calculation is one divided by time."},{"text":"$\\text{rate}=\\frac{1}{\\text{time taken}^2}$.","correct":false,"explanation":"Squaring the time introduces an unnecessary mathematical operation and changes the relationship between measurements. Rate is defined here as the reciprocal of the time taken."}]},{"id":"ec5e17b9-2cbe-444a-a1f8-7c55c7362416","cardType":"basic","points":2,"front":"Why does phenolphthalein turn colourless in the lipase practical?","back":"Lipase releases fatty acids, lowering pH.","mcqOptions":[{"text":"Lipase releases fatty acids, lowering pH.","correct":true,"explanation":"Lipase hydrolyses triglycerides into fatty acids and glycerol. The fatty acids increase hydrogen ion concentration, lowering pH and causing phenolphthalein to become colourless."},{"text":"Lipase releases glycerol, raising pH.","correct":false,"explanation":"Glycerol is produced, but it does not cause the pH change responsible for the indicator’s colour loss. Fatty acids lower pH, making phenolphthalein colourless."},{"text":"Bile releases fatty acids, lowering pH.","correct":false,"explanation":"Bile emulsifies fats into smaller droplets but does not hydrolyse them into fatty acids. Lipase performs the chemical digestion that lowers pH."},{"text":"Lipase releases fatty acids, raising pH.","correct":false,"explanation":"Fatty acids increase acidity rather than alkalinity, so pH falls. Phenolphthalein is colourless at the lower pH produced during lipid digestion."}]},{"id":"21c029ef-f240-40db-a290-0e82b6cdae06","cardType":"basic","points":2,"front":"Why is Visking tubing an incomplete ileum absorption model?","back":"It models diffusion only, not co-transport or active transport.","mcqOptions":[{"text":"It models diffusion only, not co-transport or active transport.","correct":true,"explanation":"Visking tubing is a selectively permeable membrane model that allows some substances to diffuse. It lacks living epithelial cells, transport proteins and ATP-powered pumps."},{"text":"It models active transport only, not diffusion or co-transport.","correct":false,"explanation":"Visking tubing has no ATP supply or membrane pumps, so it cannot model active transport. Its behaviour represents diffusion through a selectively permeable membrane."},{"text":"It models co-transport only, not diffusion or active transport.","correct":false,"explanation":"Co-transport requires specific proteins and an ion gradient maintained by ATP-dependent pumps, none of which Visking tubing contains. It can demonstrate diffusion instead."},{"text":"It models osmosis only, not diffusion or active transport.","correct":false,"explanation":"Although water may move by osmosis, the tubing model is not restricted to water movement and cannot represent epithelial transport mechanisms. Its main limitation is modelling diffusion only."}]}]}],"examCountdown":{"examDate":"2027-06-07","paperLabel":"Paper 1"},"heading":"Digestion and absorption","title":"Digestion and absorption Flashcards","topbar":{"backHref":"/a-level/biology/aqa/flashcards"},"topicSidebar":"$L36"}]}],"topicId":"dd29829e-a59f-59d7-ac05-7c6fccd26479","walled":"$L37"}],"$L38","$L39"]