Carbohydrates
Introduced| English | Meaning |
|---|---|
| glycogen/ˈɡlaɪkədʒn/ | A highly branched glucose polysaccharide used for energy storage in animals and fungi. |
| cellulose/ˈseljuːləʊs/ | A structural polysaccharide made of glucose units, forming a major component of plant cell walls. |
| monomer/ˈmɒnəʊmə/ | A molecule that can join with other molecules to form a polymer. |
| polymer/ˈpɒlɪmə/ | A large molecule made from many linked repeating units derived from smaller molecules. |
| macromolecule/ˈmækrəmɒlɪkjuːl/ | A very large molecule, such as a protein, polysaccharide or nucleic acid. |
| monosaccharide/ˈmɒnəʊsækɑːraɪd/ | A single sugar unit that cannot be hydrolysed into smaller carbohydrates. |
| disaccharide/ˌdɪsəkˈhɑːraɪd/ | A carbohydrate made of two monosaccharide units joined by a glycosidic bond. |
| polysaccharide/ˌpɒlɪˈsækɑːraɪd/ | A carbohydrate polymer composed of many sugar units linked together. |
| condensation/kɒndenˈseɪʃn/ | A chemical reaction in which molecules join and a small molecule, commonly water, is eliminated. |
| glycosidic bond/ˌɡlaɪkəˈsɪdɪk bɒnd/ | A covalent linkage joining a sugar to another group, including the links between sugar units in carbohydrates. |
| hydrolysis/haɪˈdrɒləsɪs/ | Breaking chemical bonds by reaction with water. |
| amylose/ˈæmɪləʊs/ | An unbranched alpha-glucose polymer in starch, with alpha-1,4 links and a coiled shape. |
| amylopectin/ˈæmɪləʊpktɪn/ | A branched alpha-glucose polymer in starch, with alpha-1,4 chains and alpha-1,6 branch points. |
| microfibril/ˈmaɪkrəʊfɪbrəl/ | A strong bundle of cellulose chains that reinforces the plant cell wall. |
One sugar can build a store or a wall
- A liver cell stores glucose in glycogen 糖原; a plant cell wall contains cellulose 纤维素. Both are built from glucose units, but their bonds and arrangement give them different jobs.
- Explain each link: monomer 单体 → bond and shape → arrangement → function. “It contains glucose” does not explain why a structure stores fuel or resists stretching.
Small units and large molecules
- A monomer is a small unit that can join to others; a polymer 聚合物 contains many repeating units joined by covalent bonds. A macromolecule 大分子 is a large molecule: size alone does not make it a polymer. A triglyceride is large, but is not a long chain of repeating monomers.
- A monosaccharide 单糖 is one sugar unit (glucose or fructose); a disaccharide 二糖 contains two joined units (maltose or sucrose); a polysaccharide 多糖 contains many (starch, glycogen or cellulose). Glucose, fructose and maltose are reducing sugars; sucrose is non-reducing.
Which is a monosaccharide?
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Glucose is a single sugar unit (monosaccharide); sucrose is a disaccharide; starch and glycogen are polysaccharides.
Match each carbohydrate to what it is.
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Mono = one unit, di = two, poly = many. Cellulose is β-glucose; glycogen is α-glucose.
Sucrose is a non-reducing sugar, so it gives a negative Benedict's test unless it is first hydrolysed.
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Glucose and maltose are reducing sugars (positive Benedict's); sucrose is non-reducing, so you must hydrolyse it (acid + heat) first.
A large molecule R contains many repeating glucose units. Another large molecule S consists of one glycerol joined to three fatty acids. Explain why both can be macromolecules but only R is a polymer.
Both are large molecules, so both are macromolecules. R is a polymer because it contains many repeating glucose units joined together. S is a triglyceride, with a glycerol and three fatty acids rather than a long chain of repeating monomers; being large does not by itself make it a polymer.
Draw the two glucose rings
- The ring has five carbon atoms and one oxygen; the sixth carbon is in the $\text{CH}_2\text{OH}$ group. In the orientation below, that group points up: the OH on carbon 1 points down in α-glucose and up in β-glucose. The other groups remain in the same positions.
- For a ring drawing, check every group, not only carbon 1. In this orientation the OH groups on carbons 2, 3 and 4 point down, up and down. Both forms have the formula $\text{C}_6\text{H}_{12}\text{O}_6$; their different arrangement affects the polymers they form.
Original ring reference. Orange marks the groups at carbon 1; the other groups are unchanged.
α-glucose and β-glucose differ in:
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They are the same atoms — only the –OH on carbon 1 differs: down in α, up in β. This decides which polysaccharide forms.
On paper, draw the ring forms of α-glucose and β-glucose without copying the reference. Include the ring oxygen, CH₂OH, H and OH groups. Compare with the reference only after drawing, correct any misplaced group, and describe the carbon-1 difference here. Keep the drawing for teacher review.
Both drawings should have one oxygen and five carbons in the ring and a CH₂OH group attached to carbon 5. In the shown orientation CH₂OH points up; α has OH down on carbon 1 and β has OH up there, with H in the opposite position. The OH groups on carbons 2, 3 and 4 remain down, up and down. Check the complete structures and valences, not just an α/β label. The reflection alone does not verify the paper drawing.
Join with condensation; split with water
- A glycosidic bond 糖苷键 is a covalent link between sugar units. In condensation 缩合, an H and an OH are removed as one water molecule for each bond formed; hydrolysis 水解 uses water to break that bond and restore groups on the products.
- Two glucose units can form maltose; glucose and fructose form sucrose. A straight, unbranched chain of five units has four links: forming it releases four water molecules; completely hydrolysing those links uses four water molecules.
For maltose, write glucose + glucose → maltose + water. Reverse the arrow for hydrolysis and put water on the reactant side. In a structural drawing, break the inter-unit glycosidic link and restore the missing H and OH; do not break every bond within the glucose rings.
Condensation and hydrolysis
Step through how two sugars join. Condensation removes one water to make the bond; hydrolysis is the reverse — adding water splits it again.
Two monosaccharides join by a ______ bond, removing one water molecule.
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A glycosidic bond forms by condensation; one water molecule is removed for each bond.
Put the steps of condensation in order.
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Condensation removes one water to form the bond; hydrolysis (adding water) is the reverse.
Nine glucose units join into one straight, unbranched chain. Count the glycosidic bonds formed and water molecules released. How many water molecules are needed to break all those glycosidic bonds by hydrolysis? Explain the counting rule.
Nine units need eight links to make one unbranched chain. Eight glycosidic bonds form, releasing eight water molecules. Complete hydrolysis of those eight links uses eight water molecules. For one connected, unbranched chain of n units, the number of inter-unit links is n − 1; do not count the bonds inside each glucose ring.
Maltase completely hydrolyses one maltose molecule by breaking its alpha-1,4 glycosidic link. On paper, draw the two products in their alpha-glucose ring forms, include water on the correct side of the reaction arrow and mark the groups restored where the link breaks. State the bond and reaction names here, then compare your drawings with the ring reference.
Hydrolysis uses one water molecule to break the inter-unit glycosidic link and produces two glucose molecules. Draw both complete alpha-glucose rings in the reference orientation, with OH down at carbon 1. The groups restored at the broken link give a free OH at carbon 1 of the first unit and a free OH at carbon 4 of the second. Keep the other groups unchanged, put water on the reactant side and retain the structural drawing for teacher review.
Compare the storage polysaccharides
| Polysaccharide | Monomer and glycosidic links | Shape |
|---|---|---|
| Amylose 直链淀粉, part of starch | α-glucose; α-1,4 links | Unbranched chain that coils |
| Amylopectin 支链淀粉, part of starch | α-glucose; α-1,4 chains and α-1,6 branch points | Branched |
| Glycogen | α-glucose; α-1,4 chains and α-1,6 branch points | More frequently branched than amylopectin |
- Starch stores glucose in plants; glycogen stores it in animals and fungi. Their compact form allows much glucose to be stored in a small space, and their insolubility avoids the large osmotic effect of the same amount stored as free glucose.
- Branches provide many ends where enzymes can add or remove glucose units. This supports rapid mobilisation when glucose is needed; glycogen stores fuel rather than producing energy itself.
The diagram compares chain arrangements; the table names their monomers and links.
Why is glycogen a good energy store in animals?
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Compact glycogen stores many glucose units in a small space. Its insolubility avoids the osmotic effect of storing the same glucose as separate dissolved molecules, and many branch ends allow rapid enzyme action.
Molecule A contains only α-1,4 links and is coiled. B contains α-1,4 links with occasional α-1,6 branches. C has the same link types as B but much more frequent branches. Identify A, B and C and explain one functional benefit of C’s branches.
A is amylose, B is amylopectin and C is glycogen. More frequent branching gives glycogen many ends where enzymes can act, supporting rapid release or addition of glucose units. The branch link is α-1,6; branching does not mean the units are β-glucose.
Build a wall from cellulose
- Cellulose has β-glucose units joined by β-1,4 glycosidic bonds, with alternate units rotated. The chains are straight and unbranched; many chains lie parallel with hydrogen bonds between their OH groups.
- Many chains form microfibrils 微纤丝 with high tensile strength: they resist stretching. Their arrangement strengthens the wall and helps it resist the pressure as water enters the cell. Glycosidic bonds run within a chain; hydrogen bonds link neighbouring chains.
“Strong bonds” is too vague. Name the bond and its position: covalent glycosidic links within each cellulose chain; many hydrogen bonds between chains in a microfibril. Cellulose is a structural polymer, not the plant's starch store.
Cellulose is strong and structural because it is made of:
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β-glucose chains are straight; many lie side by side and hydrogen-bond into strong microfibrils — the cell wall.
A cellulose chain alone is not a complete cell wall. Explain how the arrangement of many cellulose molecules helps a plant cell resist pressure when water enters.
Straight unbranched cellulose chains lie parallel. Many hydrogen bonds form between neighbouring chains, grouping them into microfibrils with high tensile strength. The wall’s microfibrils resist stretching, helping the wall withstand turgor pressure as water enters. This is different from glycosidic bonds linking glucose units within a chain.
Turn a feature into an explanation
- Worked comparison: glycogen's branching gives many ends for enzyme action, supporting rapid glucose release. Cellulose's parallel unbranched chains form hydrogen-bonded microfibrils, supporting resistance to stretching. Each explanation links a visible structural difference to its effect.
- Do not explain every function with “large molecule”. A large molecule may be a polymer or a non-polymer; use repeating units to decide. To distinguish a storage polymer from a wall polymer, use its monomer, links and chain arrangement.
You've got it
- Macromolecule describes size; polymer describes many repeating linked units.
- Draw α- and β-glucose with the same groups except at carbon 1; follow the shown ring orientation.
- Condensation forms glycosidic links and releases water; hydrolysis breaks them using water.
- Amylose coils; amylopectin and glycogen branch at α-1,6 links. Glycogen has more frequent branches.
- Cellulose has straight β-1,4 chains and hydrogen-bonded microfibrils; explain how that arrangement strengthens walls.
Polymer P has α-glucose units and many branches. Polymer Q has β-glucose units in parallel unbranched chains. Compare their likely roles and explain how the different arrangements support those roles.
Use the monomer and arrangement to support a function. A name alone is weaker than a structure-to-function explanation.
Which statement correctly locates the bonds in a cellulose microfibril?
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Cellulose has β-1,4 glycosidic links within straight chains. Many hydrogen bonds between parallel chains contribute to microfibril tensile strength.