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form the central part of chlorophyll, allowing the pigment to absorb light energy during photosynthesis."},{"text":"$\\text{Fe}^{2+}$.","correct":false,"explanation":"Iron(II) is associated with haemoglobin and oxygen transport, not chlorophyll. The central ion in chlorophyll is magnesium(II), which helps the pigment absorb light."},{"text":"$\\text{Ca}^{2+}$.","correct":false,"explanation":"Calcium ions are important in bones, muscle contraction and blood clotting, but they are not found at chlorophyll’s centre. Magnesium(II) occupies that position."},{"text":"$\\text{PO}_4^{3-}$.","correct":false,"explanation":"Phosphate ions occur in ATP, nucleic acids and phospholipids, but they do not form chlorophyll’s central ion. Chlorophyll contains magnesium(II) at its centre."}]},{"id":"eb783a13-06c3-442a-b930-d41b3be6b17c","cardType":"basic","points":1,"front":"Which inorganic ion in haemoglobin helps oxygen transport?","back":"$\\text{Fe}^{2+}$ in haem groups.","mcqOptions":[{"text":"$\\text{Fe}^{2+}$ in haem groups.","correct":true,"explanation":"Iron(II) ions in haem groups bind oxygen reversibly, allowing haemoglobin to transport oxygen in the blood."},{"text":"$\\text{Mg}^{2+}$ in haem groups.","correct":false,"explanation":"Magnesium is the central ion in chlorophyll and is linked with photosynthesis, not oxygen transport. Haemoglobin’s haem groups contain iron(II), which binds oxygen."},{"text":"$\\text{Ca}^{2+}$ in haem groups.","correct":false,"explanation":"Calcium supports bones, clotting and muscle contraction, but it is not the ion in haem groups. Iron(II) binds oxygen reversibly within haemoglobin."},{"text":"$\\text{PO}_4^{3-}$ in haem groups.","correct":false,"explanation":"Phosphate is a component of ATP, DNA, RNA and phospholipids, not haem groups. Oxygen transport depends on iron(II) ions in haemoglobin."}]},{"id":"c354c1cc-c4a2-4d45-a312-a9c711158229","cardType":"basic","points":1,"front":"Which ion is important in bones, blood clotting, muscle contraction and synapses?","back":"$\\text{Ca}^{2+}$.","mcqOptions":[{"text":"$\\text{Ca}^{2+}$.","correct":true,"explanation":"Calcium ions provide strength in bones and also function in blood clotting, muscle contraction and synaptic transmission."},{"text":"$\\text{Mg}^{2+}$.","correct":false,"explanation":"Magnesium is commonly associated with chlorophyll, so it is a tempting substitute. The ion involved in bones, clotting, muscle contraction and synapses is calcium(II)."},{"text":"$\\text{Na}^{+}$.","correct":false,"explanation":"Sodium ions are important for nerve impulses and membrane potentials, but they do not have all these roles. Calcium(II) is needed for bones, clotting, contraction and synapses."},{"text":"$\\text{K}^{+}$.","correct":false,"explanation":"Potassium ions contribute to nerve impulses and cell function, but they are not the main ion in bones or blood clotting. Those combined roles identify calcium(II)."}]},{"id":"d4cbdc2e-9e3f-4adb-bfd6-9b8bd26ec861","cardType":"basic","points":1,"front":"Which ion forms part of ATP, DNA, RNA and phospholipids?","back":"$\\text{PO}_4^{3-}$.","mcqOptions":[{"text":"$\\text{PO}_4^{3-}$.","correct":true,"explanation":"Phosphate ions are present in ATP, the phosphate backbone of DNA and RNA, and the phosphate groups of phospholipids."},{"text":"$\\text{NO}_3^{-}$.","correct":false,"explanation":"Nitrate ions provide nitrogen for making amino acids and nucleotides in plants, but they do not form part of all these molecules. Phosphate does."},{"text":"$\\text{SO}_4^{2-}$.","correct":false,"explanation":"Sulphate ions provide sulphur for certain amino acids and proteins, but they are not structural components of ATP, nucleic acids and phospholipids. Phosphate is."},{"text":"$\\text{CO}_3^{2-}$.","correct":false,"explanation":"Carbonate ions are linked mainly with carbon dioxide buffering and some mineral structures, not these biological molecules. Phosphate forms their characteristic phosphate groups."}]},{"id":"53fbe533-5634-4db7-b4e0-cb5112a5c912","cardType":"cloze","points":null,"front":"Water is polar because oxygen attracts electrons more strongly than hydrogen, making oxygen [ ] and hydrogen [ ].","back":"Water is polar because oxygen attracts electrons more strongly than hydrogen, making oxygen slightly negative and hydrogen slightly positive.","mcqOptions":[{"text":"Water is polar because oxygen attracts electrons more strongly than hydrogen, making oxygen slightly negative and hydrogen slightly positive.","correct":true,"explanation":"Oxygen attracts the shared electrons more strongly, gaining a slight negative charge, while each hydrogen becomes slightly positive. This unequal charge distribution makes water polar."},{"text":"Water is polar because oxygen attracts electrons more strongly than hydrogen, making oxygen slightly positive and hydrogen slightly negative.","correct":false,"explanation":"The charges have been reversed: stronger attraction gives oxygen more electron density and a slight negative charge. Hydrogen therefore becomes slightly positive, producing water’s polarity."},{"text":"Water is polar because oxygen attracts electrons more strongly than hydrogen, making oxygen neutral and hydrogen slightly positive.","correct":false,"explanation":"Oxygen is not neutral because the shared electrons are distributed unevenly towards it. Oxygen becomes slightly negative, while hydrogen becomes slightly positive."},{"text":"Water is polar because oxygen attracts electrons more strongly than hydrogen, making oxygen slightly negative and hydrogen neutral.","correct":false,"explanation":"Hydrogen also has an unequal share of the bonding electrons, so it becomes slightly positive rather than neutral. Opposite partial charges across the molecule create polarity."}]},{"id":"3796809c-340d-46da-a341-9be9012ad8d0","cardType":"basic","points":2,"front":"Why is water useful as a solvent in cells and transport systems?","back":"It dissolves ionic and polar substances for reactions and transport.","mcqOptions":[{"text":"It dissolves ionic and polar substances for reactions and transport.","correct":true,"explanation":"Water’s polarity allows it to surround ions and interact with other polar molecules, keeping them dissolved so they can react and be transported."},{"text":"It dissolves non-polar and lipid-soluble substances for reactions and transport.","correct":false,"explanation":"Non-polar substances do not interact favourably with water and are generally insoluble in it. Water mainly dissolves ionic and polar substances for reactions and transport."},{"text":"It dissolves ionic and non-polar substances for reactions and transport.","correct":false,"explanation":"Water dissolves ions but not most non-polar substances, because its partial charges cannot form strong interactions with them. The correct combination is ionic and polar substances."},{"text":"It dissolves polar and lipid-soluble substances for reactions and transport.","correct":false,"explanation":"Lipid-soluble substances are usually non-polar and do not dissolve readily in water. Water is an effective solvent for ionic and polar substances instead."}]},{"id":"60c764ef-41b7-4f50-9fc0-9ee7a2cf9a9e","cardType":"cloze","points":null,"front":"Water resists temperature change due to high [ ], and evaporation removes lots of energy due to high [ ].","back":"Water resists temperature change due to high specific heat capacity, and evaporation removes lots of energy due to high latent heat of vaporisation.","mcqOptions":[{"text":"Water resists temperature change due to high specific heat capacity, and evaporation removes lots of energy due to high latent heat of vaporisation.","correct":true,"explanation":"A high specific heat capacity means much energy is needed to raise water’s temperature, while a high latent heat of vaporisation makes evaporation remove substantial energy."},{"text":"Water resists temperature change due to high latent heat of vaporisation, and evaporation removes lots of energy due to high specific heat capacity.","correct":false,"explanation":"These properties have been swapped. Specific heat capacity explains resistance to temperature change, whereas latent heat of vaporisation explains the energy removed during evaporation."},{"text":"Water resists temperature change due to high thermal conductivity, and evaporation removes lots of energy due to high latent heat of vaporisation.","correct":false,"explanation":"Thermal conductivity describes heat transfer through a substance, not the energy required to change its temperature. Specific heat capacity gives the first property; latent heat of vaporisation gives the second."},{"text":"Water resists temperature change due to high specific heat capacity, and evaporation removes lots of energy due to high surface tension.","correct":false,"explanation":"Surface tension results from cohesion at the surface and does not measure energy needed for evaporation. The relevant evaporation property is latent heat of vaporisation."}]},{"id":"26ca4ab3-a41d-48fb-a6e8-f020b6c2f7d7","cardType":"cloze","points":null,"front":"Condensation reactions [ ]; hydrolysis reactions [ ] to break bonds.","back":"Condensation reactions release water; hydrolysis reactions add water to break bonds.","mcqOptions":[{"text":"Condensation reactions release water; hydrolysis reactions add water to break bonds.","correct":true,"explanation":"Condensation joins molecules by forming a bond and releasing water. Hydrolysis breaks that bond by adding water, with its components contributing to the separated molecules."},{"text":"Condensation reactions add water; hydrolysis reactions release water to break bonds.","correct":false,"explanation":"The processes have been reversed. Condensation forms bonds while releasing water, whereas hydrolysis uses added water to split bonds between molecules."},{"text":"Condensation reactions release water; hydrolysis reactions remove water to break bonds.","correct":false,"explanation":"Hydrolysis requires water rather than removing it: adding water helps break the bond. Condensation releases water as a new bond forms."},{"text":"Condensation reactions remove water; hydrolysis reactions add water to break bonds.","correct":false,"explanation":"Removing water is not the usual description of condensation in this context; condensation releases water as molecules join. Hydrolysis correctly involves adding water to break bonds."}]},{"id":"70961bcc-b1af-4d59-9c14-1cc973a4cc6a","cardType":"basic","points":2,"front":"How many water molecules completely hydrolyse a polypeptide of 6 amino acids?","back":"5 - a 6-amino-acid chain has 5 peptide bonds.","mcqOptions":[{"text":"5 - a 6-amino-acid chain has 5 peptide bonds.","correct":true,"explanation":"A chain of 6 amino acids contains 5 peptide bonds because each bond joins two neighbouring amino acids. Hydrolysis requires one water molecule per peptide bond, so 5 water molecules are needed."},{"text":"4 - a 6-amino-acid chain has 4 peptide bonds.","correct":false,"explanation":"Using 4 water molecules misses one peptide bond. A chain of 6 amino acids has one fewer peptide bond than amino acids, giving 5 bonds and requiring 5 water molecules for complete hydrolysis."},{"text":"6 - a 6-amino-acid chain has 6 peptide bonds.","correct":false,"explanation":"The number of peptide bonds is not equal to the number of amino acids. Six amino acids form 5 bonds in a chain, so complete hydrolysis requires 5 water molecules."},{"text":"7 - a 6-amino-acid chain has 7 peptide bonds.","correct":false,"explanation":"Seven would be one too many because a linear chain has one fewer peptide bond than amino acids. Six amino acids therefore contain 5 peptide bonds and need 5 water molecules."}]},{"id":"30ba40ec-04a2-4091-8522-75fb0388bde5","cardType":"basic","points":1,"front":"Which bond joins monosaccharides in carbohydrates?","back":"Glycosidic bond.","mcqOptions":[{"text":"Glycosidic bond.","correct":true,"explanation":"A glycosidic bond forms when two monosaccharides join in a condensation reaction, with a molecule of water removed. These bonds link sugars into disaccharides and polysaccharides."},{"text":"Peptide bond.","correct":false,"explanation":"Peptide bonds join amino acids in polypeptides, not monosaccharides in carbohydrates. Monosaccharides are joined by glycosidic bonds formed during condensation reactions."},{"text":"Ester bond.","correct":false,"explanation":"Ester bonds join glycerol to fatty acids in triglycerides. They do not link monosaccharides; carbohydrate monomers are joined by glycosidic bonds."},{"text":"Hydrogen bond.","correct":false,"explanation":"Hydrogen bonds help stabilise structures such as cellulose and DNA, but they do not covalently join monosaccharides. The linking bond in carbohydrates is a glycosidic bond."}]},{"id":"9cf95305-73c5-4ed7-b41f-3fba1e4f5ecd","cardType":"cloze","points":null,"front":"Alpha and beta glucose are [ ]; in ring diagrams they differ at the hydroxyl group on [ ].","back":"Alpha and beta glucose are isomers; in ring diagrams they differ at the hydroxyl group on carbon-1.","mcqOptions":[{"text":"Alpha and beta glucose are isomers; in ring diagrams they differ at the hydroxyl group on carbon-1.","correct":true,"explanation":"Alpha and beta glucose have the same molecular formula and structural arrangement but differ in the orientation of the hydroxyl group attached to carbon-1, making them isomers."},{"text":"Alpha and beta glucose are polymers; in ring diagrams they differ at the hydroxyl group on carbon-1.","correct":false,"explanation":"They are not polymers, because each is a single glucose molecule. They are isomers, and the differing hydroxyl orientation occurs at carbon-1 in the ring structure."},{"text":"Alpha and beta glucose are isomers; in ring diagrams they differ at the hydroxyl group on carbon-2.","correct":false,"explanation":"The term isomers is correct, but carbon-2 is not the anomeric carbon where alpha and beta forms differ. Their hydroxyl groups have different orientations at carbon-1."},{"text":"Alpha and beta glucose are polysaccharides; in ring diagrams they differ at the hydroxyl group on carbon-6.","correct":false,"explanation":"Neither description is correct: glucose is not a polysaccharide, and carbon-6 does not determine the alpha or beta form. The molecules are isomers differing at carbon-1."}]},{"id":"3634ea0c-323a-4b8e-9cf4-2f1035a93f85","cardType":"basic","points":2,"front":"Why can amylopectin be hydrolysed rapidly by enzymes?","back":"Its branches provide many enzyme-action ends.","mcqOptions":[{"text":"Its branches provide many enzyme-action ends.","correct":true,"explanation":"Branches create many terminal ends where enzymes can attach and remove glucose units simultaneously. This gives amylopectin more enzyme-action sites and allows rapid hydrolysis."},{"text":"Its double bonds provide many enzyme-action ends.","correct":false,"explanation":"Double bonds are features of some unsaturated fatty acids, not the reason amylopectin is hydrolysed quickly. Its branches provide many accessible enzyme-action ends."},{"text":"Its hydroxyl groups provide many enzyme-action ends.","correct":false,"explanation":"Hydroxyl groups occur throughout glucose and do not create the multiple terminal sites responsible for rapid hydrolysis. Branching provides many enzyme-action ends."},{"text":"Its rings provide many enzyme-action ends.","correct":false,"explanation":"Amylopectin is not rapidly hydrolysed because of rings in its glucose units. The polymer’s branches create numerous ends for enzymes to act on."}]},{"id":"cae68d66-7b70-47e5-910f-81daaf47d05d","cardType":"basic","points":2,"front":"Why do starch and glycogen not greatly affect cell water potential?","back":"They are insoluble.","mcqOptions":[{"text":"They are insoluble.","correct":true,"explanation":"Starch and glycogen are insoluble, so they do not dissolve into many solute particles. Consequently, they have little effect on the water potential of the cell."},{"text":"They have few glucose units.","correct":false,"explanation":"Starch and glycogen contain many glucose units, rather than few. Their limited effect on water potential is mainly because they are insoluble and therefore do not contribute many dissolved particles."},{"text":"They are large.","correct":false,"explanation":"Their large size can reduce movement, but size alone does not explain the small water-potential effect. Starch and glycogen are insoluble, so few solute particles are present in solution."},{"text":"They are hydrophobic.","correct":false,"explanation":"Starch and glycogen are not hydrophobic substances like lipids. Their small effect on water potential results from insolubility, which prevents them forming many dissolved particles."}]},{"id":"b97ecc6d-5513-47dd-b0bd-49ccb73b69a2","cardType":"basic","points":2,"front":"Why are cellulose microfibrils strong?","back":"Straight beta-glucose chains form many hydrogen bonds together.","mcqOptions":[{"text":"Straight beta-glucose chains form many hydrogen bonds together.","correct":true,"explanation":"Hydrogen bonds form between hydroxyl groups on neighbouring straight beta-glucose chains. Many such bonds hold the chains together into strong cellulose microfibrils."},{"text":"Straight beta-glucose chains form many ionic bonds together.","correct":false,"explanation":"Ionic bonds require oppositely charged ions and are not the main links between cellulose chains. Their strength comes from many hydrogen bonds between neighbouring beta-glucose chains."},{"text":"Straight beta-glucose chains form many peptide bonds together.","correct":false,"explanation":"Peptide bonds join amino acids in proteins, not glucose chains in cellulose. Cellulose microfibrils are strengthened by numerous hydrogen bonds between adjacent chains."},{"text":"Straight beta-glucose chains form many disulfide bonds together.","correct":false,"explanation":"Disulfide bonds form between sulfur-containing amino acids in some proteins. Cellulose contains glucose and gains microfibril strength from hydrogen bonds between its chains."}]},{"id":"148c2bb9-0964-4e6b-be67-0427e1136a3a","cardType":"basic","points":1,"front":"Which bond joins glycerol to fatty acids in triglycerides?","back":"Ester bond.","mcqOptions":[{"text":"Ester bond.","correct":true,"explanation":"An ester bond forms in a condensation reaction between a hydroxyl group of glycerol and a carboxyl group of a fatty acid. Three ester bonds link the components of a triglyceride."},{"text":"Peptide bond.","correct":false,"explanation":"Peptide bonds join amino acids to form polypeptides, not glycerol and fatty acids. Triglycerides contain ester bonds formed between these two types of molecule."},{"text":"Glycosidic bond.","correct":false,"explanation":"Glycosidic bonds join monosaccharides in carbohydrates. Glycerol and fatty acids in triglycerides are joined by ester bonds."},{"text":"Hydrogen bond.","correct":false,"explanation":"Hydrogen bonds are weak attractions and do not covalently join glycerol to fatty acids. The covalent links in triglycerides are ester bonds."}]},{"id":"b1286bcd-b4ab-461d-bdc4-75f8a0ce52be","cardType":"basic","points":2,"front":"Why do triglycerides store lots of chemical energy?","back":"They contain many carbon-hydrogen bonds.","mcqOptions":[{"text":"They contain many carbon-hydrogen bonds.","correct":true,"explanation":"Carbon-hydrogen bonds contain electrons at a relatively high energy level. Triglycerides have many such bonds, so their oxidation releases a large amount of chemical energy."},{"text":"They contain many carbon-carbon bonds.","correct":false,"explanation":"Triglycerides do contain carbon-carbon bonds, but these are not the key explanation for their high energy yield. Their many carbon-hydrogen bonds release substantial energy when oxidised."},{"text":"They contain many oxygen-hydrogen bonds.","correct":false,"explanation":"Oxygen-hydrogen bonds are not abundant in triglycerides and are not responsible for their high energy content. The important feature is their many carbon-hydrogen bonds."},{"text":"They contain many phosphate bonds.","correct":false,"explanation":"Phosphate bonds are characteristic of molecules such as ATP and phospholipids, not the main energy-rich feature of triglycerides. Their high energy yield comes from many carbon-hydrogen bonds."}]},{"id":"cc783576-14e6-45de-b974-862c091abefa","cardType":"cloze","points":null,"front":"A phospholipid has a [ ] phosphate-containing head and [ ] fatty acid tails.","back":"A phospholipid has a hydrophilic phosphate-containing head and hydrophobic fatty acid tails.","mcqOptions":[{"text":"A phospholipid has a hydrophilic phosphate-containing head and hydrophobic fatty acid tails.","correct":true,"explanation":"The phosphate-containing head is attracted to water, making it hydrophilic, while the fatty acid tails avoid water, making them hydrophobic."},{"text":"A phospholipid has a hydrophobic phosphate-containing head and hydrophilic fatty acid tails.","correct":false,"explanation":"This reverses the properties of the two regions. The phosphate-containing head interacts with water and is hydrophilic, whereas the fatty acid tails are hydrophobic."},{"text":"A phospholipid has a hydrophilic phosphate-containing head and hydrophilic fatty acid tails.","correct":false,"explanation":"Only the phosphate-containing head is hydrophilic. The fatty acid tails are non-polar and hydrophobic, so they avoid contact with water."},{"text":"A phospholipid has a hydrophobic phosphate-containing head and hydrophobic fatty acid tails.","correct":false,"explanation":"The tails are correctly identified as hydrophobic, but the phosphate-containing head is hydrophilic because it is attracted to water."}]},{"id":"ed7a2247-e766-45ce-9d4c-7350a871e6e6","cardType":"basic","points":2,"front":"How does amphipathic structure help phospholipids form membranes?","back":"They form bilayers with heads toward water and tails away.","mcqOptions":[{"text":"They form bilayers with heads toward water and tails away.","correct":true,"explanation":"The hydrophilic heads face the surrounding water, while the hydrophobic tails point inward away from water, producing a stable phospholipid bilayer."},{"text":"They form bilayers with tails toward water and heads away.","correct":false,"explanation":"This reverses the arrangement. Hydrophilic heads face water, while hydrophobic tails avoid it and point towards the centre of the bilayer."},{"text":"They form monolayers with heads toward water and tails away.","correct":false,"explanation":"A membrane normally contains two opposing phospholipid layers, not one. The heads face the aqueous environments and the tails meet in the middle."},{"text":"They form bilayers with heads and tails mixed together.","correct":false,"explanation":"Random mixing would expose hydrophobic tails to water and destabilise the membrane. Bilayers arrange heads towards water and tails away from it."}]},{"id":"7e9830e6-b47f-420d-88ac-32335b720b23","cardType":"cloze","points":null,"front":"Saturated fatty acids have [ ]; unsaturated fatty acids have [ ].","back":"Saturated fatty acids have no carbon-carbon double bonds; unsaturated fatty acids have at least one carbon-carbon double bond.","mcqOptions":[{"text":"Saturated fatty acids have no carbon-carbon double bonds; unsaturated fatty acids have at least one carbon-carbon double bond.","correct":true,"explanation":"Saturated fatty acids contain only single carbon-carbon bonds. Unsaturated fatty acids contain one or more carbon-carbon double bonds."},{"text":"Saturated fatty acids have at least one carbon-carbon double bond; unsaturated fatty acids have no carbon-carbon double bonds.","correct":false,"explanation":"The two definitions have been reversed. Saturated fatty acids have no carbon-carbon double bonds, while unsaturated fatty acids have at least one."},{"text":"Saturated fatty acids have no carbon-carbon single bonds; unsaturated fatty acids have at least one carbon-carbon double bond.","correct":false,"explanation":"Fatty acids necessarily contain carbon-carbon single bonds, so their absence does not define saturation. Saturated molecules specifically lack carbon-carbon double bonds."},{"text":"Saturated fatty acids have no carbon-carbon double bonds; unsaturated fatty acids have at least one carbon-oxygen double bond.","correct":false,"explanation":"Unsaturation refers to carbon-carbon double bonds, not carbon-oxygen bonds. Saturated fatty acids lack carbon-carbon double bonds, while unsaturated ones contain at least one."}]},{"id":"3ef0f826-81df-469a-b1c8-9ded4360470b","cardType":"basic","points":2,"front":"Why do unsaturated lipids usually have lower melting points?","back":"Double bonds bend chains, so molecules pack less closely.","mcqOptions":[{"text":"Double bonds bend chains, so molecules pack less closely.","correct":true,"explanation":"Carbon-carbon double bonds, especially in cis configurations, introduce bends into fatty acid chains. The molecules therefore pack less closely and require less energy to melt."},{"text":"Single bonds bend chains, so molecules pack less closely.","correct":false,"explanation":"Single bonds allow rotation but do not usually create the fixed kinks responsible for lower melting points. Double bonds bend chains and reduce close packing."},{"text":"Double bonds straighten chains, so molecules pack more closely.","correct":false,"explanation":"Double bonds commonly introduce bends rather than straightening chains. These bends disrupt packing, which lowers the energy needed to melt the lipid."},{"text":"Hydrogen bonds bend chains, so molecules pack less closely.","correct":false,"explanation":"Fatty acid chain shape is mainly affected by carbon-carbon double bonds, not hydrogen bonds. Double bonds create bends that prevent molecules packing closely."}]},{"id":"9e24f048-660d-45a6-be85-cb702051512b","cardType":"basic","points":2,"front":"Which two groups react to form a peptide bond?","back":"Carboxyl group and amino group.","mcqOptions":[{"text":"Carboxyl group and amino group.","correct":true,"explanation":"A peptide bond forms by condensation between the carboxyl group of one amino acid and the amino group of another, releasing a molecule of water."},{"text":"Hydroxyl group and amino group.","correct":false,"explanation":"An amino group is involved, but peptide bonds form specifically with a carboxyl group. A hydroxyl group is not the reacting group in this bond."},{"text":"Carboxyl group and phosphate group.","correct":false,"explanation":"The carboxyl group is involved, but the second reacting group is an amino group, not a phosphate group. Their condensation creates the peptide linkage."},{"text":"Methyl group and carboxyl group.","correct":false,"explanation":"Methyl groups do not form peptide bonds. The required reaction is between a carboxyl group on one amino acid and an amino group on another."}]},{"id":"11ce33e1-50ff-4c97-a2c0-e82b4c5264ba","cardType":"cloze","points":null,"front":"Primary structure is the [ ]; secondary structure is held by [ ].","back":"Primary structure is the amino acid sequence; secondary structure is held by hydrogen bonds between the peptide backbone.","mcqOptions":[{"text":"Primary structure is the amino acid sequence; secondary structure is held by hydrogen bonds between the peptide backbone.","correct":true,"explanation":"Primary structure records the order of amino acids. Secondary structure, such as alpha helices and beta sheets, is stabilised by hydrogen bonds between peptide backbone groups."},{"text":"Primary structure is the three-dimensional shape; secondary structure is held by disulfide bonds between R groups.","correct":false,"explanation":"Three-dimensional shape describes higher-level folding, while disulfide bonds between R groups help stabilise tertiary structure. Primary structure is the amino acid sequence."},{"text":"Primary structure is the amino acid sequence; secondary structure is held by ionic bonds between R groups.","correct":false,"explanation":"The sequence correctly describes primary structure, but secondary structure is stabilised by hydrogen bonds between backbone groups, not ionic bonds between R groups."},{"text":"Primary structure is the number of peptide bonds; secondary structure is held by hydrogen bonds between the peptide backbone.","correct":false,"explanation":"Primary structure is the specific order of amino acids, not simply the number of peptide bonds. Backbone hydrogen bonds do stabilise secondary structure."}]},{"id":"f46037ac-ec90-41fb-bada-9d77e588e665","cardType":"basic","points":1,"front":"Which protein structure level is held by interactions between R groups?","back":"Tertiary structure.","mcqOptions":[{"text":"Tertiary structure.","correct":true,"explanation":"Tertiary structure is the overall three-dimensional shape of one polypeptide chain, stabilised by interactions between R groups, including ionic bonds and disulfide bridges."},{"text":"Primary structure.","correct":false,"explanation":"Primary structure is only the amino acid sequence joined by peptide bonds. Interactions between R groups determine the higher-level folding of the polypeptide."},{"text":"Secondary structure.","correct":false,"explanation":"Secondary structure consists mainly of alpha helices and beta sheets held by hydrogen bonds between backbone groups, rather than interactions between R groups."},{"text":"Quaternary structure.","correct":false,"explanation":"Quaternary structure describes how multiple polypeptide chains associate. Interactions between R groups within one chain primarily determine tertiary structure."}]},{"id":"24f334a1-8d20-4f5f-9119-495eb76ea2f6","cardType":"basic","points":2,"front":"How can a change in amino acid sequence alter protein function?","back":"It changes bonding and folding, altering the final shape.","mcqOptions":[{"text":"It changes bonding and folding, altering the final shape.","correct":true,"explanation":"Changing the amino acid sequence can alter R-group interactions and folding, producing a different three-dimensional shape that may change the protein’s function."},{"text":"It changes bonding and folding, preserving the final shape.","correct":false,"explanation":"A sequence change can alter interactions between amino acids, so the folding and final shape may change. Preserving the shape would not explain altered function."},{"text":"It changes the number of amino acids, altering the final shape.","correct":false,"explanation":"A mutation may change an amino acid or its position rather than simply changing the total number. Altered bonding and folding then affect the final shape."},{"text":"It changes bonding and folding, altering the amino acid sequence.","correct":false,"explanation":"The amino acid sequence is the cause of the altered folding, not the result. Changed bonding and folding alter the protein’s final shape and potentially its function."}]},{"id":"2a107ea8-3365-4e52-bffa-e42ccbc2bec0","cardType":"cloze","points":null,"front":"Iodine-potassium iodide changes orange-brown to [ ] if starch is present; Biuret changes blue to [ ] if protein is present.","back":"Iodine-potassium iodide changes orange-brown to blue-black if starch is present; Biuret changes blue to lilac/purple if protein is present.","mcqOptions":[{"text":"Iodine-potassium iodide changes orange-brown to blue-black if starch is present; Biuret changes blue to lilac/purple if protein is present.","correct":true,"explanation":"Starch forms a blue-black complex with iodine, while peptide bonds in proteins produce a lilac or purple colour with Biuret reagent."},{"text":"Iodine-potassium iodide changes orange-brown to blue-green if starch is present; Biuret changes blue to lilac/purple if protein is present.","correct":false,"explanation":"A positive starch test is blue-black, not blue-green, because iodine fits into the starch helix; Biuret gives lilac or purple when peptide bonds are present."},{"text":"Iodine-potassium iodide changes orange-brown to blue-black if starch is present; Biuret changes blue to orange-red if protein is present.","correct":false,"explanation":"Biuret changes to lilac or purple for protein, whereas a starch-positive sample becomes blue-black; confusing these colours commonly leads to assigning the Biuret result to iodine."},{"text":"Iodine-potassium iodide changes orange-brown to red-brown if starch is present; Biuret changes blue to green if protein is present.","correct":false,"explanation":"Neither test gives this pair of colours: starch produces blue-black with iodine and protein produces lilac or purple with Biuret, not the colours associated with reducing sugar tests."}]},{"id":"20e583ba-e232-424c-b7ab-f651b01c52d9","cardType":"basic","points":2,"front":"What condition is needed for Benedict’s test for reducing sugars?","back":"Heat with Benedict’s in a water bath at about 80 °C.","mcqOptions":[{"text":"Heat with Benedict’s in a water bath at about 80 °C.","correct":true,"explanation":"Benedict’s reagent is heated with the sample in a water bath at about 80 °C so reducing sugars can reduce the copper(II) ions and produce the characteristic colour change."},{"text":"Heat with Benedict’s in a water bath at about 100 °C.","correct":false,"explanation":"A boiling water bath is hotter than the specified condition and can cause excessive heating or evaporation; the standard test uses a water bath at about 80 °C."},{"text":"Mix with Benedict’s and leave at room temperature for about 80 minutes.","correct":false,"explanation":"Benedict’s testing requires heat to promote the reduction reaction, so room temperature is insufficient; the sample should be heated in a water bath at about 80 °C."},{"text":"Cool with Benedict’s in an ice bath at about 0 °C.","correct":false,"explanation":"An ice bath prevents the reduction reaction from proceeding effectively; Benedict’s reagent must be heated with the sample in a water bath at about 80 °C."}]},{"id":"5169d180-d130-4c71-b552-df7275502835","cardType":"basic","points":1,"front":"What positive Benedict’s result indicates reducing sugar?","back":"Blue to green, yellow, orange or brick-red.","mcqOptions":[{"text":"Blue to green, yellow, orange or brick-red.","correct":true,"explanation":"Reducing sugars reduce blue copper(II) ions during heating, producing a progression through green, yellow and orange to a brick-red precipitate as concentration increases."},{"text":"Green to blue, yellow, orange or brick-red.","correct":false,"explanation":"The test begins blue because Benedict’s reagent contains copper(II) ions; the positive colour sequence moves away from blue towards green, yellow, orange or brick-red."},{"text":"Blue to purple, pink, orange or brick-red.","correct":false,"explanation":"A reducing sugar does not produce a purple sequence in Benedict’s test; blue copper(II) ions change through green, yellow and orange to brick-red after heating."},{"text":"Blue to green, yellow, orange or black.","correct":false,"explanation":"Black is not part of the Benedict’s colour scale; a positive result changes from blue through green, yellow and orange to brick-red as reducing sugar concentration rises."}]},{"id":"1a8b0538-b4c9-41f1-a910-9e79085ce50e","cardType":"basic","points":2,"front":"Why must a non-reducing sugar sample be neutralised after acid hydrolysis?","back":"Benedict’s reagent needs alkaline conditions.","mcqOptions":[{"text":"Benedict’s reagent needs alkaline conditions.","correct":true,"explanation":"Acid hydrolysis uses acid to break a non-reducing sugar into reducing sugars, but the remaining acid would interfere with Benedict’s test, which requires alkaline conditions."},{"text":"The hydrolysis reaction needs alkaline conditions.","correct":false,"explanation":"The sample is neutralised because Benedict’s reagent needs alkaline conditions, not because hydrolysis requires an alkaline solution; acid is used during the hydrolysis step."},{"text":"Benedict’s reagent needs acidic conditions.","correct":false,"explanation":"Neutralisation is not primarily to make the sample acidic; Benedict’s test works under alkaline conditions, so residual acid must be removed before adding the reagent."},{"text":"The hydrolysis reaction needs neutral conditions.","correct":false,"explanation":"Hydrolysis does not need to remain at a neutral pH for Benedict’s testing; the important reason is that Benedict’s reagent needs alkaline conditions after the acid has been used."}]},{"id":"12afe72e-d096-404a-82b8-04277173b661","cardType":"basic","points":1,"front":"What positive result shows lipid in the emulsion test?","back":"Cloudy white emulsion.","mcqOptions":[{"text":"Cloudy white emulsion.","correct":true,"explanation":"Lipids dissolve in ethanol but not in water, so adding water produces suspended lipid droplets that scatter light and form a cloudy white emulsion."},{"text":"Clear colourless solution.","correct":false,"explanation":"A clear solution means no visible emulsion has formed; a positive lipid test produces a cloudy white appearance when the ethanol extract is mixed with water."},{"text":"Blue-black solution.","correct":false,"explanation":"A blue-black colour is associated with iodine reacting with starch, not the lipid emulsion test; lipids produce a cloudy white emulsion after ethanol and water are mixed."},{"text":"Lilac-purple solution.","correct":false,"explanation":"A lilac or purple colour is the positive Biuret result for protein; the lipid emulsion test instead gives a cloudy white suspension of lipid droplets."}]},{"id":"1c6841c4-636e-492f-b7f8-6a31b20155a9","cardType":"basic","points":2,"front":"How can Benedict’s test be made semi-quantitative?","back":"Use known glucose concentrations to make a calibration curve.","mcqOptions":[{"text":"Use known glucose concentrations to make a calibration curve.","correct":true,"explanation":"Testing known glucose concentrations gives corresponding colour or absorbance values, allowing a calibration curve to estimate the glucose concentration in an unknown sample."},{"text":"Use unknown glucose concentrations to make a calibration curve.","correct":false,"explanation":"A calibration curve requires standards with known glucose concentrations; unknown samples alone cannot establish the relationship needed to estimate concentration."},{"text":"Use known protein concentrations to make a calibration curve.","correct":false,"explanation":"Protein concentrations are unrelated to Benedict’s response, which measures reducing sugars such as glucose; known glucose standards are needed for calibration."},{"text":"Use known lipid concentrations to make a calibration curve.","correct":false,"explanation":"Lipid concentrations do not provide standards for Benedict’s reaction; the curve must be produced using known glucose concentrations and their test responses."}]}]}],"examCountdown":{"examDate":"2027-06-07","paperLabel":"Component 1"},"heading":"Chemical elements are joined together to form biological compounds","title":"Chemical elements are joined together to form biological compounds Flashcards","topbar":{"backHref":"/a-level/biology/eduqas/flashcards"},"topicSidebar":"$L38"}]}],"topicId":"b44cdea8-ca5d-5d45-ae5e-040b1c0ad495","walled":"$L39"}],"$L3a","$L3b"]