E1. Since the amount of polyunsaturated fats (PUFA) in your diet may affect your health, it is worthwhile to look closely at the contents of several different foods that you might use to prepare sandwiches. Use the Internet and food labels to investigate the PUFA content of butter, several different brands of margarine, and avocados. a Which have the highest total fat content? b What proportion of the fat content is unsaturated? c Does the labelling distinguish between the amount of saturated, mono-
unsaturated and polyunsaturated fats? d Is information about the source of these fats provided on the label?
EA1. Answer will depend on student’s research.
Q1. Which two functional groups are directly involved in the formation of a lipid?
A1. hydroxyl and carboxyl
Q2. a Write an equation showing the formation of the triglyceride glyceryl trioleate
from the reaction of glycerol and oleic acid, CH3(CH2)7CH=CH(CH2)7COOH. b Draw the structure of glyceryl trioleate, showing the ester bonds.
A2.
a
b
Worked solutions to student book questions Chapter 12 Pathways to biomolecules
Q3. A fat present in a vegetable oil has the structure shown in Figure 12.8.
Figure 12.8 A fat found in a vegetable oil.
a Circle an ester functional group. b Is this fat saturated or unsaturated? Explain. c The fat is broken down into two different substances in the stomach and small
intestine. i Write the semi-structural formulas of the reaction products. ii What chemical, apart from the fat, is a reactant in this process? iii What name is given to this type of reaction?
A3.
a
b The fat is saturated because the alkyl groups in the molecules (CH3(CH2)14
–) contain only single carbon–carbon bonds. Alkyl groups with the general formula CnH2n + 1 are saturated.
c i HOCH2CHOHCH2OH and CH3(CH2)14COOH ii water iii hydrolysis
Q4. Describe the types of fatty acids that combine with glycerol to form: a saturated fats b mono-unsaturated fats c polyunsaturated fats
A4. a Saturated fats have single bonds between carbon atoms. b Monounsaturated fats have a single carbon to carbon double bond. c Polyunsaturated fats have multiple double bonds.
Worked solutions to student book questions Chapter 12 Pathways to biomolecules
Q5. Sunflower oil is a liquid and can be used to make margarine. How are polyunsaturated oils, such as sunflower oil, converted to a solid form?
A5. Some of the double bonds in sunflower oil are converted to single bonds via a catalysed addition reaction with hydrogen.
Q6. Glucose is the most abundant monosaccharide. Draw a structural formula for glucose and use it to explain why it is highly soluble in water.
A6.
The glucose molecule has five polar OH groups which form hydrogen bonds with water. As a result glucose is highly soluble in water.
Q7. Glucose is described as a monosaccharide, maltose as a disaccharide and glycogen as a polysaccharide. a What is the difference between these three types of carbohydrate? b What type of reaction is involved in converting glucose to maltose and then to
glycogen? c Which functional group in the molecules is involved in these reactions?
A7.
a Whereas glucose is a monosaccharide, maltose is made up of two monosaccharide molecules bonded together and starch is a polymer of monosaccharide molecules.
b condensation c hydroxyl functional groups
Worked solutions to student book questions Chapter 12 Pathways to biomolecules
Q8. The structure of lactose is shown in Figure 12.21. Lactose undergoes hydrolysis in the digestion process.
Figure 12.21 The structure of lactose.
a What is meant by ‘hydrolysis’? b Circle a hydroxyl and a glycosidic functional group in the molecule. c Draw the structures of the hydrolysis products. d Lactose is a carbohydrate. Why does 10 g of lactose provide more energy for our
bodies than 10 g of the main carbohydrate found in celery?
A8. a A hydrolysis reaction is one in which water is a reactant. b
c
d Our digestive system secretes enzymes that can rapidly hydrolyse lactose to
glucose, which is used in respiration. We do not produce enzymes that can break down cellulose, the main carbohydrate in celery.
Worked solutions to student book questions Chapter 12 Pathways to biomolecules
Q9. a i Write a chemical equation for the condensation reactions between two
glucose molecules to form a maltose molecule. ii Draw structural formulas for all molecules involved in this reaction and
circle the glycosidic linkage in maltose. b i Write a chemical equation for the condensation reactions between a glucose
molecule and a fructose molecule to form a sucrose molecule. ii Draw structural formulas for all molecules involved in this reaction and
circle the glycosidic linkage in sucrose.
A9.
a i, ii
b i, ii
Q10. Starch, glycogen and cellulose are all polymers that contain glucose monomers. Explain why the human body can completely digest the starch and glycogen but cannot digest cellulose to any great extent.
A10. Humans lack the enzyme (cellulase) that catalyses the hydrolysis of cellulose to form glucose.
Q11. Write a chemical equation for: a the hydrolysis of sucrose b the hydrolysis of tristearin (see Figure 12.3)
A11.
a
Worked solutions to student book questions Chapter 12 Pathways to biomolecules
E2. Why must vegetarians ensure that their diet contains complementary proteins?
AE2. The human body only manufactures some of the required amino acids. The other amino acids, called essential amino acids are provided in our food. Animal protein provides all these essential amino acids. Vegetable protein is often deficient in one or more essential amino acids.
E3. A healthy diet includes proteins that include essential amino acids. Why should essential amino acids be included in a healthy diet?
AE3. Nine out the 20 amino acids required by the body cannot be manufactured in the body. They must be provided by the food we eat. These are called essential amino acids.
E4. Kwashiorkor or protein–energy malnutrition is a disease found in some traditional African societies. Find out more about this disease and its causes.
AE4. Lack of protein in the diet of young children causes a form of starvation known as protein–energy malnutrition. It is the most common form of death of young children in the developing world.
Q12.
What two functional groups are present in all amino acids?
A12. amino and carboxyl
Q13. Draw the structural formula of glycine as it is likely to exist in: a an acidic solution b an alkaline solution c neutral solution.
Worked solutions to student book questions Chapter 12 Pathways to biomolecules
Q14. With the aid of Table 12.7 on page 189 of the student book: a draw structural formulas of serine and cysteine b write an equation to show the formation of a dipeptide from these amino acids c name the type of reaction in part b. A14. a
b H2NCH(CH2OH)COOH + H2NCH(CH2SH)COOH →
H2NCH(CH2OH)CONHCH(CH2SH)COOH + H2O c condensation
Worked solutions to student book questions Chapter 12 Pathways to biomolecules
Q15. Why is protein more important in your diet as a source of raw materials for building other molecules, rather than as a source of energy?
A15. There are no substitutes for the essential amino acids required by your body. The other nutrients, fats and carbohydrates, can supply your energy needs.
Q16. Explain what is meant by the primary, secondary and tertiary structures of a protein. Clearly outline the bonding type involved at each level of protein structure. (Include labelled diagrams.)
A16.
Primary structure refers to the sequence of amino acids in a protein.
The folding of a section due to, for example, hydrogen bonding between peptide links is known as the secondary protein structure.
The overall three-dimensional shape of a protein is called the tertiary structure.
Worked solutions to student book questions Chapter 12 Pathways to biomolecules
Q17. Draw the structure of the two amino acids produced when the dipeptide in Figure 12.31 undergoes hydrolysis.
Figure 12.31
A17.
Q18. Use the information in Table 12.7 on page 189 of the student book to sketch the structural formula of the tripeptide that would be represented as ala-ser-val.
A18.
Q19. a Find the names and describe the functions of two enzymes in the body. b Explain why the action of enzymes justifies the statement ‘Enzymes make life
possible’. c Why is the action of an enzyme often described as operating like a ‘lock and
key’?
Worked solutions to student book questions Chapter 12 Pathways to biomolecules
A19. a Enzymes in the body include pepsin, which hydrolyses peptide bonds of certain
amino aids: DNA polymerase – which replicates and repairs DNA lactase – which breaks down the sugar lactose in the small intestine salivary amylase – which breaks down polysaccharides in the mouth.
b Almost all the chemical reactions occurring in living creatures are controlled by enzymes. Enzymes speed up the reactions that are essential for life processes by as much as 1010 times; reactions that do not contribute to the functioning of a creature are not catalysed and occur at much slower rates.
c The shape and functional groups in the active site of the enzyme allow it to bind only with certain substrates so that only a specific reaction is catalysed. In a similar way, a lock will only open using a key of a certain shape.
Q20. What are the main differences between enzymes and inorganic catalysts?
A20. Compared to inorganic catalysts, enzymes can produce much faster reaction rates, operate under milder conditions and are highly selective. Whereas an inorganic catalyst might increase the rates of many different reactions, enzymes are usually effective for just one reaction.
Q21. The enzyme carbonic anhydrase catalyses the decomposition of carbonic acid molecules to carbon dioxide and water in the lungs. When heated to more than 60°C, the enzyme becomes denatured. a What is meant by the term ‘denatured’? b Describe the events that usually occur to the structure of an enzyme when it is
denatured. c Does the primary structure of the carbonic anhydrase enzyme change during this
process? d Why is the functioning of the enzyme closely related to its tertiary structure?
A21. a Denaturation occurs when there is a change to the structure and shape of an
enzyme that prevents it from functioning. b Denaturation usually involves disruption of the bonds that hold the enzyme’s
protein chain in a particular shape, followed by unfolding of the chain. The unfolded chains will often clump together, in a process called coagulation.
c no d The overall three-dimensional shape of an enzyme is known as its tertiary
structure. Since enzyme action depends upon the existence of a suitable hollow or cavity within the molecule, the tertiary structure is crucial to the enzyme’s operations.
Worked solutions to student book questions Chapter 12 Pathways to biomolecules
Q22. Explain why LDH is not used as a specific marker for heart attack.
A22. Increase in the level of the protein marker LDH can be caused by other diseases such as anaemia or pneumonia.
Q23. A competitor collapses during a race. A blood test showed a raised level of total CPK enzyme. a Does the test indicate that the competitor has had a heart attack? b What other enzyme tests would confirm a heart attack?
A23. a CPK levels rise after vigorous exercise. b Troponin and a specific form of CPK, CPK-MB
Q24. Explain why doctors sometimes rely on information from a number of tests and observations to make a diagnosis rather than relying solely on the presence of a protein marker.
A24. A number of factors other than the disease under investigation may lead to an increase in the level of a specific protein marker.
Worked solutions to student book questions Chapter 12 Pathways to biomolecules
Q25. Explain the difference between: a a fatty acid and a fat b a polyunsaturated fat and a saturated fat.
A25. a A fatty acid molecule consists of a hydrocarbon chain with a carboxyl (–COOH)
group at one end. A fat, or triglyceride, is the product of a condensation reaction between three fatty acid molecules and a glycerol molecule.
b The fatty acid groups that make up a saturated fat contain only single carbon–carbon bonds. A polyunsaturated fat contains fatty acid groups with more than one carbon–carbon double bond.
Q26. By referring to Figure 12.3, write reactions that show: a the hydrolysis of the fat tristearin b the complete oxidation of the products from part a
Q27. The structure of a fatty acid, linoleic acid, is shown in Figure 12.39. a Name the chemical needed to convert linoleic acid into a lipid. b Draw the structure of the lipid made from linoleic acid. c Name and circle the functional group found in the lipid molecule. d Classify the lipid as saturated, mono-unsaturated or polyunsaturated. e Is this fat likely to be a solid or liquid at room temperature? Explain.
Figure 12.39 The structure of linoleic acid.
Worked solutions to student book questions Chapter 12 Pathways to biomolecules
Q30. The human brain relies almost exclusively on glucose as an energy source. Normal functioning of the brain depends on a constant supply of glucose from the blood. Explain how your body maintains a fairly constant concentration of glucose in the blood even though you don’t eat foods that supply glucose constantly throughout the day.
A30. When the body digests food and absorbs glucose, the glucose is transported by the blood to the liver where it is converted to glycogen, a storage polysaccharide. The glycogen is slowly hydrolysed between meals to maintain a fairly constant concentration of glucose in the blood for use by tissues such as those in the brain.
Q31. Explain why body builders and weightlifters eat a large amount of protein.
A31. Body builders and weight-lifters wish to develop muscle tissue. Muscle tissue comprises a substantial amount of protein; to build muscle tissue the diet must include a rich source of amino acids—protein.
Q32. When alanine and glycine react, two different dipeptides can be formed. a Write the formula for each dipeptide. b How many tripeptides can be formed from three different amino acids? c Proteins are formed using 20 different amino acids and can be hundreds of amino
acid units in length. On the basis of your answers to parts a and b, what can you say about the number of different proteins possible?
A32. a H2NCHCH3CONHCH2COOH
H2NCH2CONHCHCH3COOH b 6 (using each amino acid once in each peptide) c very large numbers
Q33. Aspartic acid is one of the non-essential amino acids (Figure 12.40). It exists as a zwitterion at pH 2.8.
Figure 12.40 The structure of aspartic acid.
a Name: i the acidic functional group ii the basic functional group.
b Draw the structure of the molecule as it is most likely to exist in a solution at: i pH much more than 2.8
Worked solutions to student book questions Chapter 12 Pathways to biomolecules
Q34. a Circle the peptide linkages in the part of the protein chain shown in Figure 12.41. b Write the formulas of the amino acids that made up this part of the chain. c Where in the body does hydrolysis of proteins to form amino acids take place?
Figure 12.41 A section of a protein molecule.
A34.
a
b H2NCHCH3COOH; H2NCH(CH2SH)COOH
H2NCH2COOH H2NCH(CH2OH)COOH
c stomach and small intestine
Worked solutions to student book questions Chapter 12 Pathways to biomolecules
Q35. Children need proportionately more protein in their diets than adults. Suggest why.
A35. Humans and other animals use protein to build bones, tendons, ligaments and muscle. Since these processes occur more rapidly during childhood and adolescence, children and teenagers need proportionately more protein in their diets to provide the amino acids required for the formation of protein.
Q36. Draw a structure for an amino acid that could not be classified as a 2-amino acid (α-amino acid).
A36.
(i.e. –NH2 and –COOH are not joined to the same carbon)
Q37. Scientists have spent many years trying to develop a form of the protein insulin that can be given to diabetics orally rather than by injection. Suggest an aspect of the chemistry of proteins that has so far prevented insulin from being taken orally.
A37. Proteins are readily denatured by acids. Stomach acid has prevented the ingestion of the protein insulin. The protein is hydrolysed into its component amino acids during digestion.
Q38. Proteins are often described in terms of their primary, secondary and tertiary structures. What is the difference between these levels of structure and how does each contribute to the role of the protein?
A38. Primary structure: the sequence of the amino acids that make up the polymer chain. This determines the entire shape of the protein. Secondary structure: the coiling, folding or pleating of the protein chain caused by hydrogen-bonding interactions of adjacent amino acids. These regular structures extend along one dimension like coils in a spring and are important in fibres such as hair and wool. Tertiary structure: the bending and folding of the protein chain to form a compact three-dimensional shape. These globular shapes with a low surface to volume ratio are important for proteins in solution such as enzymes in blood.
Worked solutions to student book questions Chapter 12 Pathways to biomolecules
Q39. Mutations can cause the wrong amino acid to be inserted into a protein chain when it is being assembled. In some instances such a mistake has no effect on the function of the protein, but in others it disables the protein. Explain how a single change in a protein’s primary structure could cause such dramatically different results.
A39. The activity of an enzyme, which is a type of protein, is dependent on its three-dimensional shape. The tertiary structure results from a combination of hydrogen bonding, electrostatic attraction between polar groups and disulfide linkages. If the wrong amino acid is introduced into the chain, these bonds may not form and the tertiary structure may be distorted. In some instances the enzyme becomes inactive, depending upon how the tertiary structure is altered.
Q40. Enzymes allow some reactions to proceed far more rapidly than they would if inorganic catalysts were used. Find out why enzymes are not used widely in industrial processes.
A40. Scientists are investigating the use of enzymes as catalysts in industrial reactions. At present, enzymes are not used widely in industry because they tend to be present in biological systems in small quantities, so supplies for industrial use are limited (in future, enzymes may be generated in larger quantities by using genetic engineering techniques); they function only within a narrow range of temperatures, pressures and pH, and outside these ranges they are easily broken down; and their catalytic action and structure are often poorly understood by chemists. Since enzymes function under mild conditions, their use in industry could lead to enormous cost savings. One area of current research involves the synthesis of molecules with shapes similar to natural enzymes and the capacity to catalyse specific reactions.
Q41. Jellied pineapple dessert cannot be made using gelatine and fresh pineapple because, instead of setting, an enzyme in the pineapple causes molecules in the gelatine to break down. Suggest how jellied pineapple might be prepared.
A41. The catalytic property of the enzyme can be destroyed by changing its shape—in a process called denaturation. This can be done by heating the pineapple. Alternatively, canned pineapple can be used to make this dessert since the fruit is heated during the canning process.
Q42. Digestion of proteins occurs in both the stomach and the small intestine. Trypsin is an enzyme that catalyses the hydrolysis of proteins. Trypsin is secreted into the small intestine, where the pH is approximately 8. Would you expect this same enzyme to be secreted into the stomach, where the pH is approximately 2? Explain your reasoning.
Worked solutions to student book questions Chapter 12 Pathways to biomolecules
A42. The functioning of enzymes is specific. A different enzyme catalyses the hydrolysis of protein in the stomach.
Q43. a The steps in the action of an enzyme involve, in particular, an active site and a
substrate. With the use of a diagram describe in detail the action of an enzyme. b The forces of attraction that enable a substrate to bind to an active site can vary.
Describe four such forces.
A43.
a
b hydrogen bonds, ion–dipole interaction, dipole–dipole interactions, ionic interactions.
Worked solutions to student book questions Chapter 12 Pathways to biomolecules
Q44. Read the magazine article and answer the following questions. a This article reflects current research in the development of enzymes for
commercial use. What are the advantages of enzymes compared to inorganic catalysts?
b What tests would you require the Novo-Nordisk company to perform on the enzyme Lipolase before it was released commercially?
c Why might governments be cautious about licensing the use of genetically engineered organisms to produce enzymes? Discuss the issues involved in this question with other class members.
Engineered enzyme washes whiter than white
Until now, no one has developed a commercially available enzyme that
will break down fats.
In its hunt for a suitable enzyme, Novo-Nordisk investigated a range of
enzymes called lipases. Molecular biologists at the company identified a
fungus which will produce a specific type of lipase that is good at
breaking down the fats found in human foods.
Although the fungi produce these useful lipases naturally, they do not
produce enough to be financially viable. So the geneticists identified the
sequence of nitrogen bases which make up the DNA of the fungus and
cloned that sequence. They then spliced the section of the DNA which
holds the gene that codes for the lipase into a new host organism, a
fungus. [They] found that it will produce the lipase in sufficient
quantities.
Novo complains bitterly at the length of time it has had to wait for a
licence to produce Lipolase in Denmark … The same process took only
four weeks in Japan.
A44. a Enzymes are much faster, operate under milder conditions and are more selective
than inorganic catalysts. b Tests could include a study of the effect of the enzyme on human health and on
the environment. c Issues to consider when answering this question include the potential hazards
associated with the development of new organisms and enzymes.
Worked solutions to student book questions Chapter 12 Pathways to biomolecules
Q45. What functional groups are required to react to form each of the following, and with which biological polymers are each of these linkages associated? a an ester linkage b a glycosidic linkage c a peptide linkage
A45. a carboxyl and hydroxyl groups; lipids b two hydroxyl groups; carbohydrates c carboxyl and amino groups; proteins
Q46. a Explain the difference between condensation reactions and hydrolysis reactions. b What is the importance of these reactions in the human body? c Write equations for condensation reactions between:
i two glucose molecules ii a glycerol molecule and three palmitic acid molecules (see Table 12.1 on
page 179 of the student book) iii two alanine molecules
A46. a Condensation reactions involve the linking together of two small molecules and
the elimination of a small molecule, usually water. Hydrolysis reactions consume water and they can often be regarded as the reverse of condensation reactions.
b Vital biochemical processes include condensation reactions (e.g. synthesis of lipids, proteins and polysaccharides) and hydrolysis reactions (e.g. digestion).
c i
ii
iii
Worked solutions to student book questions Chapter 12 Pathways to biomolecules
a By referring to Figure 12.33 on page 195, draw the structure of the dipeptide produced when the ACE enzyme catalyses the conversion of angiotensin I to its active form.
b Identify the amino acids formed when the amide functional group is hydrolysed. c List the types of bonds involved at the active site of ACE when it is catalysing
this reaction. d The other molecule produced is known to stimulate the constriction of blood
vessels which can lead to an increase in blood pressure. It is a small polypeptide called angiotensin II that contains the amino acid sequence asp-arg-val-tyr-ile-his-pro-phe. What is the number of each of the following types of functional groups in angiotensin II? i amide (peptide) ii carboxyl
A47. a
b histidine, leucine c ion–dipole interaction, hydrogen bonds, ionic interactions d i 7 amide
ii 1 terminal carboxyl group, 1 carboxyl group on the Z group of asp
Q48. A fatty acid that was extracted from sunflower oil is analysed in a laboratory in a number of ways. a Using IR spectroscopy, it was deduced that the fatty acid is unsaturated. Explain
what feature of the IR spectrum would lead to such a conclusion. b The unsaturated fatty acid was reacted quantitatively with hydrogen gas to
determine the number of double bonds per molecule. A 0.0102 mol sample of the fatty acid reacted with exactly 500 mL of hydrogen gas, under SLC conditions, for the complete conversion of the fatty acid sample to a saturated fatty acid. i What type of reaction occurred between the unsaturated fatty acid and
hydrogen gas? ii Calculate the amount, in moles, of hydrogen gas that reacted with the fatty
acid. c Determine the number of double bonds per molecule. d Use the data in Table 12.1 to identify the unsaturated fatty acid.
Worked solutions to student book questions Chapter 12 Pathways to biomolecules
A48. a Absorbance at 1600–1700 cm–1 indicative of C=C bond b i addition
ii
volumemolar volume0.50024.5
0.0204 mol
n =
=
=
c Each double bond reacts with one H2 molecule.
2(H ) 0.0204 2(acid) 0.0102
nn
= =
1 mol of fatty acid reacts with 2 mol H2. Acid contains 2 double bonds. d For example, linoleic acid
Q49. From many thousand proteins in a human tissue sample, a research scientist needed to separate and identify individual proteins that might be used as markers for particular diseases. Briefly describe how each of these analytical techniques could be used in such a process. a HPLC b proton NMR spectroscopy c IR spectroscopy d mass spectroscopy
A49. a HPLC—separation of proteins b proton NMR spectroscopy—structure c IR spectroscopy—functional groups d mass spectroscopy—molar mass
