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IB Biology B1.1: Carbohydrates and lipids worked solutions

Worked solutions to every practice question in B1.1: Carbohydrates and lipids of Photon Biology: Study Guide for the IB Diploma (SL & HL). Try each question, then compare your working with the full solution.

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Basic practice

Question A1

[1 mark]
Which feature of carbon atoms explains why so many different biological molecules are built on carbon? (A) Four covalent bonds per atom, including C to C bonds (B) Carbon is the most abundant element in cells (C) Carbon forms ionic bonds with oxygen (D) Carbon atoms form hydrogen bonds to each other
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Four covalent bonds per atom, including carbon-to-carbon bonds, allow chains, branches and rings (A). Abundance (B) does not explain variety, and carbon forms covalent rather than ionic or hydrogen bonds with other carbons (C, D).

Final answer

A

Question A2

[1 mark]
Two glucose molecules react to form maltose. Which statement is correct? (A) A water molecule is released and a glycosidic bond forms. (B) A water molecule is used and a glycosidic bond forms. (C) A water molecule is released and a peptide bond forms. (D) A water molecule is used and an ester bond forms.
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Joining two sugars is condensation: water is a product and the new link is a glycosidic bond (A). B has the water in the wrong direction, and peptide and ester bonds link other molecules (C, D).

Final answer

A

Question A3

[1 mark]
Which statement describes the role of water in the digestion of starch to glucose? (A) H and OH from water are added at each broken link (B) Water supplies the energy to break the bonds (C) Water is released as each bond breaks (D) Water only dissolves the starch
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Hydrolysis uses water as a reactant: H goes to one end and OH to the other end of each broken link (A). Water does not supply energy (B), is used not released (C), and does more than dissolve (D).

Final answer

A

Question A4

[1 mark]
A ring-form monosaccharide has one oxygen atom and five carbon atoms in its ring, and a \ce{CH2OH} group attached to the ring. How many carbon atoms does the molecule have, and what is it? (A) 5 carbons, a pentose (B) 6 carbons, a hexose (C) 6 carbons, a pentose (D) 7 carbons, a hexose
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The ring contributes five carbons (the oxygen is not a carbon) and the \ce{CH2OH} group adds one outside the ring, so there are 6 carbons and the sugar is a hexose (B).

Final answer

B

Question A5

[1 mark]
Glucose is soluble in water and chemically stable. What do these two properties allow? (A) Carried dissolved in blood without reacting (B) Stored in cells as insoluble granules (C) Built into cell walls (D) Oxidised to give more energy than fat
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Solubility lets it dissolve in plasma and stability lets it be carried unchanged (A). Insoluble stores, cell walls and a higher energy yield than fat describe other molecules or are untrue (B, C, D).

Final answer

A

Question A6

[1 mark]
Which feature makes glycogen suitable as an energy store? (A) Compact and branched, with many free ends (B) Made of beta-glucose, so it is straight (C) Soluble, so carried to other tissues (D) Contains many C–H bonds
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Branching makes glycogen compact and gives many ends for condensation and hydrolysis (A). Glycogen is made of alpha-glucose (B), is relatively insoluble (C), and C–H bond richness describes lipids (D).

Final answer

A

Question A7

[1 mark]
Which feature gives cellulose fibres their strength? (A) Hydrogen bonds between parallel straight chains (B) Branched chains of alpha-glucose (C) Coiled chains with alpha-1,6 links (D) Peptide bonds between chains
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Straight chains of beta-glucose are held side by side by many hydrogen bonds, bundling them into microfibrils (A). B and C describe starch and glycogen, and cellulose has no peptide bonds (D).

Final answer

A

Question A8

[1 mark]
Which description fits a fatty acid? (A) A long chain of carbons ending in COOH (B) Three carbons each bearing an OH group (C) Four fused carbon rings (D) A ring of carbons with one oxygen
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A fatty acid is a hydrocarbon chain ending in a carboxyl group (A). B describes glycerol, C a steroid and D a sugar ring.

Final answer

A

Question A9

[1 mark]
A fatty acid has 18 carbons, a carboxyl group and two C=C double bonds in the chain. How is it classified? (A) saturated (B) monounsaturated (C) polyunsaturated (D) a steroid
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Two C=C bonds in the chain make it polyunsaturated (C). The C=O of the carboxyl group is not counted.

Final answer

C

Question A10

[1 mark]
Why can testosterone cross the plasma membrane without a transport protein? (A) Non-polar: it crosses the hydrophobic centre (B) Hydrophilic: it uses water-filled channels (C) A glycoprotein recognised by the cell (D) Made of glycerol and fatty acids
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Testosterone is a non-polar steroid, so it dissolves in the hydrophobic centre of the bilayer and crosses directly (A). It is not hydrophilic (B), a glycoprotein (C) or a glycerol-fatty acid lipid (D).

Final answer

A

Tutorial practice

Question B1

[4 marks]
Starch, glycogen and cellulose are all polysaccharides of glucose. (a) Outline two structural differences between starch and cellulose. [2] (b) State one structural similarity between starch and glycogen. [1] (c) Explain why glycogen can release glucose more quickly than starch. [1]
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(a) Starch is alpha-glucose, cellulose is beta-glucose (1); starch is coiled or branched, cellulose is straight and unbranched (1). (b) Both are chains of alpha-glucose joined by alpha-1,4 links with alpha-1,6 links at branches (1). (c) Glycogen has more branches, so more ends are available at which glucose is removed by hydrolysis at the same time (1).

Final answer

(a) Alpha against beta glucose; coiled or branched against straight and unbranched. (b) Both are alpha-glucose joined by alpha-1,4 links, with alpha-1,6 branches. (c) More branches mean more ends for hydrolysis.

Question B2

[5 marks]
A triglyceride forms from glycerol and fatty acids. (a) State the type of reaction and the name of the link that forms between glycerol and a fatty acid. [2] (b) Calculate the number of water molecules released when 40 molecules of triglyceride are formed. [1] (c) Explain why a triglyceride is not a polymer, and state how a phospholipid differs from it. [2]
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(a) Condensation (1); ester link (1). (b) Each triglyceride releases three waters, so 40\times3=120 (1). (c) It is made of one glycerol and three fatty acids, not a repeated monomer (1); a phospholipid has a phosphate group in place of one of the fatty acids (1).

Final answer

(a) Condensation; ester link. (b) 120. (c) No repeating unit; a phospholipid has phosphate in place of one fatty acid.

Question B3

[6 marks]
Three triglycerides were each made from three identical fatty acid chains. All the chains have 18 carbons.\begin{array}{lccc}\hline \text{Triglyceride}&\text{S}&\text{O}&\text{L}\\ \hline \text{C=C bonds in each chain}&\text{0}&\text{1}&\text{2}\\ \hline \text{Melting point / }^\circ\text{C}&\text{72}&-4&-13\\ \hline\end{array} (a) Calculate the number of C=C bonds in one molecule of triglyceride O. [1] (b) Describe the relationship between the number of C=C bonds per chain and the melting point. [1] (c) Explain this relationship. [3] (d) State whether S is a solid or a liquid at body temperature, 37\,{}^\circC. [1]
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(a) Three chains with one C=C each give 3\times1=3 (1). (b) As the number of C=C bonds per chain rises, the melting point falls (1). (c) Each cis C=C bond puts a kink in the chain (1); kinked chains cannot pack closely (1); attractions between molecules are weaker, so less energy is needed to separate them (1). (d) Solid: 72^\circC is above 37^\circC (1).

Final answer

(a) 3. (b) More C=C bonds, lower melting point. (c) Kinks; poorer packing; weaker attractions; less energy to separate. (d) Solid.

Question B4

[6 marks]
A mammal that lives in a cold climate stores 1.2\,\mathrm{kg} of fat. Use 37\,\mathrm{kJ\,g^{-1}} for lipid and 17\,\mathrm{kJ\,g^{-1}} for carbohydrate. (a) Calculate the energy stored in the fat and the mass of carbohydrate that would store the same energy. [2] (b) Explain two reasons why triglyceride is a suitable long-term energy store. [2] (c) Explain how adipose tissue beneath the skin helps the mammal to survive in the cold. [2]
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(a) 1200\times37=44\,400\,\mathrm{kJ} (1); 44\,400/17\approx2600\,\mathrm{g} (1). (b) Any two: long chains rich in C–H bonds release a lot of energy per gram (1); it is hydrophobic and insoluble, so it does not alter the osmotic balance of cells (1); it is compact, so less mass is carried (1). (c) Fat is a poor conductor of heat (1), so heat is lost more slowly to the surroundings (1).

Final answer

(a) 44 400 kJ; about 2 600 g of carbohydrate. (b) Much energy per gram; insoluble, so no osmotic effect. (c) Poor conductor of heat; slows heat loss.

Question B5

[6 marks]
Phospholipids and steroids are both found in cell membranes. (a) Draw a simple labelled diagram of one phospholipid molecule. [3] (b) Define the term amphipathic. [1] (c) Explain why oestradiol can pass through a phospholipid bilayer but glucose cannot without help. [2]
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(a) A head drawn as a circle attached to two tails (1); the head labelled hydrophilic or phosphate (1); the tails labelled hydrophobic or fatty acid (1). (b) A molecule with a hydrophilic part and a hydrophobic part (1). (c) Oestradiol is a non-polar steroid, so it dissolves in the hydrophobic centre of the bilayer and crosses (1); glucose is polar and hydrophilic, so it cannot enter the hydrophobic centre (1).

Final answer

(a) One round head labelled hydrophilic (phosphate); two tails labelled hydrophobic (fatty acids). (b) Having both a hydrophilic and a hydrophobic region. (c) Oestradiol is non-polar so dissolves in the hydrophobic centre; glucose is polar and is repelled.

Question B6

[6 marks]
An enzyme hydrolyses maltose in a closed tube.\begin{array}{lcc}\hline \text{Time / min}&\text{0}&\text{10}\\ \hline \text{Maltose molecules present}&\text{100}&\text{60}\\ \hline \text{Free glucose molecules measured}&\text{0}&\text{70}\\ \hline\end{array} (a) Calculate the number of maltose molecules hydrolysed in 10 minutes and the number of water molecules used. [2] (b) Calculate the number of glucose molecules formed and the number missing from the measurement. [2] (c) Show that the glucose units in the tube have not been conserved, and suggest why. [2]
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(a) 100-60=40 maltose (1); one water per link, so 40 (1). (b) Each maltose gives two glucose, 40\times2=80 (1); 80-70=10 missing (1). (c) Units at 10 min: free 70 plus 60\times2=120 in maltose =190, but the start had 100\times2=200 (1); ten glucose units have been removed or used, for example absorbed or respired (1).

Final answer

(a) 40 maltose; 40 water. (b) 80 formed; 10 missing. (c) 70+120=190, not 200; 10 units removed or used.

Question B7

[5 marks]
A student shook 200\,\mathrm{mg} of each of three solids with 20\,\mathrm{cm^3} of water and with 20\,\mathrm{cm^3} of hexane, and measured the mass that dissolved.\begin{array}{lcc}\hline \text{Solid}&\text{In water / mg}&\text{In hexane / mg}\\ \hline \text{X}&\text{6}&\text{176}\\ \hline \text{Y}&\text{190}&\text{4}\\ \hline \text{Z}&\text{2}&\text{3}\\ \hline\end{array} (a) Calculate the percentage of X that dissolved in hexane and deduce which solid behaves as a lipid. [2] (b) State why the test cannot classify Z. [1] (c) State one variable that must be controlled. [1] (d) State why the results do not show that X is a triglyceride. [1]
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(a) 176/200\times100=88\% (1); X is the lipid: it dissolves in the non-polar solvent and hardly in water, so it is hydrophobic (1). (b) Z is almost insoluble in both solvents, so the test places it in neither group (1). (c) Any one: volume of each solvent, temperature, shaking time or mass of solid (1). (d) Other lipids, such as waxes or steroids, give the same pattern (1).

Final answer

(a) 88 %; X. (b) Insoluble in both. (c) Volume, temperature or time shaken. (d) Fats, waxes and steroids all behave this way.

Question B8

[4 marks]
Cells carry a glycoprotein chain. Probe a binds an A end sugar. Probe b binds a B end sugar. Four samples were tested; yes means binding.\begin{array}{lcccc}\hline \text{Probe}&\text{Sample 1}&\text{Sample 2}&\text{Sample 3}&\text{Sample 4}\\ \hline \text{a}&\text{yes}&\text{no}&\text{yes}&\text{no}\\ \hline \text{b}&\text{no}&\text{yes}&\text{yes}&\text{no}\\ \hline\end{array} (a) Deduce which end sugar or sugars each sample carries. [2] (b) Sample 1 cells are treated with an enzyme that removes only the end sugar. Predict the result with probe a. [1] (c) Explain why this result shows that recognition depends on the carbohydrate chain. [1]
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(a) Sample 1 carries A, sample 2 carries B (1); sample 3 carries both and sample 4 carries neither (1). (b) Probe a no longer binds (1). (c) The enzyme removes only the sugar and leaves the protein, yet binding is lost, so the sugar is what the probe recognises (1).

Final answer

(a) 1: A; 2: B; 3: A and B; 4: neither. (b) No binding. (c) The protein is unchanged but binding is lost.

Advanced practice

Question C1

[7 marks]
Two groups of 20 rats were fed diets with the same energy content for 12 weeks. Diet 1 was rich in saturated fat. Diet 2 was rich in unsaturated fat. Values are mean \pm standard deviation.\begin{array}{lcc}\hline \text{Measurement}&\text{Diet 1}&\text{Diet 2}\\ \hline \text{Body fat / \% of body mass}&24.1\pm2.0&18.6\pm1.8\\ \hline \text{Food eaten / g per day}&22&25\\ \hline\end{array} (a) Calculate the percentage by which the body fat of Diet 1 exceeds that of Diet 2. [1] (b) Evaluate whether the difference in body fat is likely to be real, using the standard deviations. [2] (c) A newspaper says that saturated fat causes obesity in all mammals. Evaluate this claim using the data. [2] (d) Suggest two further pieces of information needed before the claim could be accepted. [2]
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(a) (24.1-18.6)/18.6\times100=29.6\approx30\% (1). (b) The ranges are 22.1 to 26.1 and 16.8 to 20.4 (1) and do not overlap, so the difference is likely to be real (1). (c) The data come from one species of rat over twelve weeks, so they cannot support a claim about all mammals (1); rats on Diet 2 ate more, so body fat is not explained by food intake and the data show an association with fat type, not a proven cause (1). (d) Any two: results from other species (1); activity levels or other variables controlled (1); a longer study or repeat trials (1).

Final answer

(a) 30 %. (b) Ranges 22.1-26.1 and 16.8-20.4 do not overlap, so likely real. (c) Only rats, one duration; food intake differs; not all mammals. (d) Other species, measured activity, longer time.

Question C2

[6 marks]
A glycogen molecule contains 4000 glucose units. It is a branched chain with 160 branch points. (a) Calculate the number of glycosidic links in the molecule and state the number of water molecules needed to hydrolyse it completely to glucose. [2] (b) Explain why the branches allow glucose to be released more quickly. [2] (c) Explain why storing 4000 glucose units as one glycogen molecule has less effect on the osmotic balance of a cell than storing 4000 free glucose molecules. [2]
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(a) A branched molecule of n units still has n-1 links, so 4000-1=3999 (1); one water per link, so 3999 (1). (b) Each branch creates another end (1); glucose is removed from every end at the same time, so release is faster (1). (c) One large molecule is one particle in solution, whereas 4000 free glucose molecules are 4000 particles (1); the solute concentration is far lower with glycogen, so less water enters by osmosis (1).

Final answer

(a) 3999 links; 3999 waters. (b) More ends, so many hydrolysis reactions at once. (c) One particle instead of 4000, so the solute concentration is far lower and little water enters.

Question C3

[5 marks]
Fish that live at 4\,{}^\circC have more unsaturated fatty acid tails in the phospholipids of their membranes than fish that live at 25\,{}^\circC.\begin{array}{lcc}\hline \text{Fish at}&4\,{}^\circ\text{C}&25\,{}^\circ\text{C}\\ \hline \text{Tails with C=C / \%}&\text{62}&\text{38}\\ \hline\end{array} (a) Calculate the difference between the two percentages. [1] (b) Explain why more unsaturated tails help the cold-water fish to keep their membranes fluid. [3] (c) Predict the effect on a membrane of the 25\,{}^\circC fish if it were cooled to 4\,{}^\circC. [1]
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(a) 62-38=24 percentage points (1). (b) Each C=C puts a kink in a tail (1); kinked tails cannot pack closely, so the attractions between phospholipids are weaker (1); the bilayer therefore stays fluid at a temperature where saturated tails would pack tightly and solidify (1). (c) The membrane would become less fluid and more rigid, because mostly saturated tails pack closely (1).

Final answer

(a) 24 percentage points. (b) C=C gives kinks; poor packing; weaker attractions; stays fluid at low temperature. (c) Less fluid, more rigid.

Question C4

[6 marks]
Artificial vesicles were made from a phospholipid bilayer only, with no proteins. Four labelled substances were added outside and the percentage that had entered after 10 minutes was measured.\begin{array}{lc}\hline \text{Substance}&\text{Entered / \%}\\ \hline \text{Testosterone}&\text{92}\\ \hline \text{Oestradiol}&\text{88}\\ \hline \text{Glucose}&\text{1}\\ \hline \text{Sodium ions}&\text{0}\\ \hline\end{array} (a) Compare the entry of the steroids with that of glucose. [2] (b) Explain the pattern. [3] (c) Suggest why cells take up glucose much faster than these vesicles do. [1]
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(a) Both steroids entered at 88\% or more (1) whereas glucose entered at only 1\%, so entry of the steroids was far greater (1). (b) The steroids are non-polar and hydrophobic, so they dissolve in the hydrophobic tails at the centre of the bilayer and cross (1); glucose is polar and sodium ions are charged (1), so they are repelled by the hydrophobic centre and cannot cross without proteins (1). (c) Real membranes contain transport proteins that move glucose, and the vesicles have none (1).

Final answer

(a) Steroids over 88 %, glucose 1 %. (b) Non-polar steroid dissolves in hydrophobic centre; polar glucose and charged ions cannot. (c) Transport proteins.

Question C5

[5 marks]
Strips of plant cell wall were treated and their tensile strength measured. Strength is relative to untreated wall, which is 100.\begin{array}{lc}\hline \text{Treatment}&\text{Relative strength}\\ \hline \text{None}&\text{100}\\ \hline \text{Cellulase (cuts the chains)}&\text{12}\\ \hline \text{Hydrogen-bond breaker}&\text{35}\\ \hline\end{array} (a) Calculate the percentage decrease in strength for each treatment. [2] (b) Use the data to evaluate the roles of the covalent links within chains and the hydrogen bonds between chains. [3]
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(a) Cellulase: (100-12)/100\times100=88\% (1); hydrogen bond breaker: (100-35)/100\times100=65\% (1). (b) Breaking the links in the chains removes most of the strength, because without intact chains there is nothing to bundle (1); breaking only the hydrogen bonds still removes about two thirds of the strength, so the bundling of chains into microfibrils also contributes a lot (1); the strength depends on both: strong chains and many hydrogen bonds between them (1).

Final answer

(a) 88 % and 65 %. (b) Both matter; losing chain links is the greater loss; but hydrogen bonds alone give a large share.

Connected questions

Question X1

Connected questions[6 marks]
B1.1 + A1.1. Glucose dissolves readily in water. Triglycerides do not. In a plant, sucrose is moved in solution and oil is kept in seeds. (a) Explain, with reference to the polarity of water, why glucose dissolves and a triglyceride does not. [2] (b) Outline why a plant moves sucrose in solution but keeps oil in the seeds where it is stored. [2] (c) State the two products when sucrose is hydrolysed and name the reaction that formed it. [2]
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(a) Glucose has polar \ce{-OH} groups that form hydrogen bonds with the polar water molecules (1); a triglyceride is mostly non-polar hydrocarbon chains that water does not attract, so it does not dissolve (1). (b) Sucrose is soluble and stable, so it can be carried in solution to where it is needed (1); oil is insoluble and compact, so it stays in the seed as a long-term store (1). (c) Glucose and fructose (1); condensation formed the glycosidic bond (1).

Final answer

(a) Glucose has polar OH groups and forms hydrogen bonds with water; triglyceride chains are non-polar. (b) Sucrose soluble so moves; oil insoluble so stays in place. (c) Glucose and fructose; condensation.

Question X2

Connected questions[8 marks]
B1.1 + B2.1. A plasma membrane has phospholipids, steroid cholesterol and glycoproteins. (a) Explain how phospholipids form a bilayer in water. [2] (b) Outline how the glycoproteins allow other cells to recognise this cell. [2] (c) Explain why steroid hormones need no channel to cross the membrane, while glucose does. [2] (d) Explain how kinks in the fatty acid tails affect the membrane. [2]
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(a) The head is hydrophilic and the tails hydrophobic (1); the heads face the water on both sides and the tails face each other in the middle (1). (b) The carbohydrate chains of the glycoproteins project from the outer surface (1); other cells recognise the particular shape of the chain (1). (c) The steroid is non-polar, so it dissolves in the hydrophobic centre of the bilayer (1); glucose is polar and cannot cross it without help from a protein (1). (d) Kinks stop the tails packing closely (1), so the bilayer stays fluid (1).

Final answer

(a) Amphipathic; heads out to water, tails in. (b) Carbohydrate chains project outward; shape recognised. (c) Non-polar steroid dissolves in the hydrophobic centre; glucose is polar. (d) Kinks prevent close packing; membrane stays fluid.

Question X3

Connected questions[8 marks]
B1.1 + C1.2. A castor bean seed stores 60\% of its dry mass as oil and a pea seed stores most of its energy as starch. Use 37\,\mathrm{kJ\,g^{-1}} for oil and 17\,\mathrm{kJ\,g^{-1}} for starch. (a) Calculate the energy in the oil of a castor bean seed of dry mass 0.5\,\mathrm{g}. [2] (b) Calculate the mass of starch that would store the same energy. [2] (c) Outline how the stored starch becomes available to the cells of the germinating pea seed. [2] (d) Explain why oxidising a gram of oil takes more oxygen than oxidising a gram of starch. [2]
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(a) Oil mass 0.6\times0.5=0.30\,\mathrm{g} (1); energy 0.30\times37=11.1\,\mathrm{kJ} (1). (b) 11.1/17=0.65\,\mathrm{g} (1); this is more than the 0.30\,\mathrm{g} of oil, so starch is the heavier store (1). (c) Starch is hydrolysed to glucose by water (1), and the glucose is oxidised in cell respiration to release energy for ATP (1). (d) Oil has many more C–H bonds per gram (1), so more hydrogen is oxidised and more oxygen is used as the final electron acceptor (1).

Final answer

(a) 11.1 kJ. (b) About 0.65 g. (c) Hydrolysis to glucose, used in respiration. (d) More C–H bonds; more hydrogen to oxidise.

From the book

Photon Biology: Study Guide for the IB Diploma (SL & HL)

Carbohydrates and lipids from the core ideas to full exam questions, with question types, worked examples and graded practice.

About the book