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Worked solutions: Chapter 2 Human biochemistry Section 2.9 Exercises 1 Aerobic respiration of glucose releases 2860 kJ of energy per mole of glucose whereas anaerobic respiration in humans releases 120 kJ of energy per mole of glucose. In yeast, only 69 kJ of energy is released per mole of glucose. 2 In aerobic respiration glucose is oxidized and oxygen is reduced. 3 Electron transport in the oxidation of food involves the metal ions Cu2+ and Fe2+ in cytochromes. 4 The bonding between the four nitrogen atoms and the iron is coordinate (dative) bonding. 5 Iron plays a principle role in the transport and storage of oxygen in mammalian blood and tissues. The transport of oxygen in the blood is carried out by hemoglobin. This is made up of four myoglobin-like units in which a hexacoordinate Fe2+ ion is coordinated to four nitrogen atoms of the porphyrin ring and to a nitrogen atom of the protein chain. The sixth coordination position is occupied by oxygen. The hemoglobin carries oxygen in the blood to the cells of the body where it is needed (in short supply) and releases it to the cells. Chapter 2 Review questions Standard level questions 1 .T = 22.72 – 20.10 = 2.62 0C q = (m ) c ) .T)calorimeter + (m ) c ) .T)water = (500 ) 2.62) + (100 ) 4.18 ) 2.62) = 1310 + 1095.16 J = 2.405 kJ E= 2.405 = 0.481 kJ g–1 5.00 The heat of combustion of the walnuts is 0.481 kJ g–1. Page 14 Copyright © Pearson Australia 2009 (a division of Pearson Australia Group Pty Ltd) ISBN 9780733993794 This page from the Chemistry: For use with the IB Diploma Programme Options SL and HL Teacher’s Resource may be reproduced for classroom use. Worked solutions: Chapter 2 Human biochemistry 2 a b c 3 The molecular formula of maltose is C12H22O11 (2 ) C6H12O6 – H2O). a b c Page 15 Copyright © Pearson Australia 2009 (a division of Pearson Australia Group Pty Ltd) ISBN 9780733993794 This page from the Chemistry: For use with the IB Diploma Programme Options SL and HL Teacher’s Resource may be reproduced for classroom use. Worked solutions: Chapter 2 Human biochemistry 4 Proteins have a secondary, tertiary and quaternary structure due to intermolecular forces between functional groups in the polymer chain, in addition to the initial primary structure, whereas poly(ethene), a non-polar polymer, only has a primary structure. Also, there are 20 possible monomers for protein synthesis, but only a single monomer in poly(ethene) synthesis. 5 The egg white is mostly protein. The protein has denatured in the acid, with the tertiary structure breaking down. The egg white has then coagulated, to give a white appearance, and become hard. 6 a Hydrogen bonding holds the protein strand together in its secondary structure, to form an alpha helix or a beta-pleated sheet. It also contributes to the tertiary structure, together with other intermolecular forces and even ionic and covalent bonding. b Hydrogen bonding holds the two strands of DNA together in the double helix. It binds a purine base to its matching pyrimidine base of the second strand. In addition, hydrogen bonding would help to hold the DNA molecule in its helical shape. a Amylose and amylopectin are both forms of starch and as such are used as a source of energy by animals. They are food storage molecules in plants. 7 Cellulose is a structural polysaccharide that is used for the walls of plant cells. Glycogen is a food/energy storage molecule used by animals. b All four polymers are formed by condensation reactions in which a water molecule is made each time two monomers react. All four polymers are polymers of glucose. c For example: The linkages between monomers makes the four polymers different. In amylose the 1-1,4 linkage occurs between two 1-glucose monomers, whereas in amylopectin both 1-1,4 and 1-1,6 linkages form between 1-glucose monomers. Cellulose has a 2-1,4 linkage; that is, carbon 1 of a 2-glucose molecule forms a glycoside (ether) linkage with carbon 4 of an 1-glucose. Glycogen is a highly branched polymer with a similar structure to amylopectin, but more branching. 8 9 a Water is a product formed when two glucose monomers join by a condensation reaction in the polymerization of glucose to starch. b Water is a product formed when two amino acids join by a condensation reaction in the formation of a dipeptide. c Water needs to be added in order for starch to be broken down by enzymes to glucose molecules. a Two amino acids are joined by an amide or peptide linkage. b Two monosaccharides are joined by an ether or glycosidic linkage. c Glycerol is joined to a fatty acid in a triglyceride by an ester linkage. d Guanine is joined to cytosine by hydrogen bonding. 10 Amino acids and monosaccharides are soluble in water because they both have functional groups—carboxyl groups in amino acids, and amino groups in monosaccharides—that can hydrogen bond with water molecules. Page 16 Copyright © Pearson Australia 2009 (a division of Pearson Australia Group Pty Ltd) ISBN 9780733993794 This page from the Chemistry: For use with the IB Diploma Programme Options SL and HL Teacher’s Resource may be reproduced for classroom use. Worked solutions: Chapter 2 Human biochemistry 11 12 a b A saturated fatty acid will have the general formula CnH2n + 1COOH. Docosahexaenoic acid has the formula C22H32O2 or C21H31COOH. If n = 21, then 31 = 2n – 11. This represents 12 Hs less than a saturated fatty acid, so there must be 6 carbon-carbon double bonds in docosahexaenoic acid. M(C22H32O2) = 328.54 Considering 100 g of the lipid: n(C22H32O2) = 100 = 0.304 mol 328.54 C20H30O2 : I2 1 : 6 n(I2) = 6 ) 0.304 = 1.826 mol m(I2) = 1.826 ) 253.8 = 463.5 g The iodine number of docosahexaenoic acid is 464. 13 There should be more HDL cholesterol than LDL cholesterol in the bloodstream. Higher concentrations of LDL (or inversely, low concentrations of HDL) are strongly associated with cardiovascular disease. LDL promotes the narrowing of arteries (atherosclerosis) by accumulating beneath the inner elastic wall of the artery and the smooth muscle surrounding it. This disease process leads to heart attack, stroke and other diseases caused by the blockage of large peripheral arteries. 14 a Lipids provide a source of energy for bodily processes, and when stored provide insulation and protection for vital organs. b Lipids have a high calorific value, so too great a proportion of lipids in the diet can lead to obesity. 15 Any four of: vitamins and trace minerals such as iron, copper, fluorine, zinc, iodine, selenium, manganese, molybdenum, chromium, cobalt and boron. Micronutrients mostly function as co-factors of enzymes, and so are vital to digestion and other important bodily processes. Page 17 Copyright © Pearson Australia 2009 (a division of Pearson Australia Group Pty Ltd) ISBN 9780733993794 This page from the Chemistry: For use with the IB Diploma Programme Options SL and HL Teacher’s Resource may be reproduced for classroom use. Worked solutions: Chapter 2 Human biochemistry 16 Fat-soluble vitamins usually have a long non-polar hydrocarbon chain and very few polar groups such as hydroxyl groups and amino groups. Water-soluble vitamins usually have several hydroxyl or amino groups that can form hydrogen bonds with water. 17 a Goitre is caused by a deficiency of iodine. b Anemia is caused by a deficiency of iron. c Beriberi is caused by a deficiency of thiamine (vitamin B1). d Scurvy is caused by a deficiency of vitamin C. e Night blindness is caused by a deficiency of vitamin A. f Rickets is caused by a deficiency of vitamin D. 18 a The steroid backbone – a series of four fused rings – three of which have six members and one with five members is common to cholesterol and to the sex hormones. b Testosterone has a carbonyl group where cholesterol has a hydroxyl group and a hydroxyl group where cholesterol has a hydrocarbon chain. Estradiol has a hydroxyl group where cholesterol has a hydrocarbon chain. Progesterone has a carbonyl group where cholesterol has a hydroxyl group and a methyl carbonyl group where cholesterol has a hydrocarbon chain. 19 Hormone a Progesterone Endocrine gland in which it is produced Ovaries Effect of the hormone in the body b Testosterone Testes, adrenal gland Development of male secondary sex characteristics c Adrenaline (epinephrine) Adrenal gland Fight or flight response to a perceived threat d Insulin Pancreas Causes glucose to be taken up from the blood and to be sorted in liver and muscle Prepares uterus for fertilized egg 20 Steroids may be used to gain an advantage in sport. This is because steroids increase muscle mass, making the user stronger than he/she would naturally be. Page 18 Copyright © Pearson Australia 2009 (a division of Pearson Australia Group Pty Ltd) ISBN 9780733993794 This page from the Chemistry: For use with the IB Diploma Programme Options SL and HL Teacher’s Resource may be reproduced for classroom use. Worked solutions: Chapter 2 Human biochemistry Higher level questions 21 a i An inorganic catalyst increases the rate of a reaction without being consumed in the reaction. ii An enzyme increases the rate of a biological reaction without being consumed in the reaction. iii Enzymes have faster reaction rates and operate under milder conditions than inorganic catalysts. b 22 a The rate of reaction reaches a peak around 400C because this is just a little higher than the usual temperature of the body. The rate falls off rapidly above 420C because the high temperature denatures the enzyme and it is no longer able to function. The rate also falls off rapidly at temperatures below 400C because only a small proportion of enzyme– substrate collisions have energy greater than the activation energy of the reaction. A substrate is the reactant in an enzyme-catalysed reaction. b The enzyme–substrate complex is formed when the enzyme and substrate have bonded together temporarily. c The active site is the part of the enzyme to which the substrate bonds. The substrate ‘fits’ into the active site. d The induced-fit model of enzyme action suggests that the enzyme changes shape as the substrate binds to the active site. Since enzymes are rather flexible structures, the active site is continually reshaped by interactions with the substrate. This enables a more precise fit to be achieved between the enzyme and the substrate. When the products leave the enzyme, it returns to its original form. 23 a In the cytosine–guanine base pair, the hydrogen bonds occur between the primary amine group of cytosine and the carbonyl group of guanine, the tertiary amine of cytosine and the secondary amine of guanine and the carbonyl of cytosine and the primary amine of guanine. b In the adenine–thymine base pair, the hydrogen bonds occur between the primary amine group of adenine and the carbonyl group of thymine, and the tertiary amine of adenine and the secondary amine of thymine. c In the adenine–uracil base pair, the hydrogen bonds occur between the primary amine group of adenine and the carbonyl group of uracil and the tertiary amine of adenine and the secondary amine of uracil. 24 P: phosphate group S: sugar (deoxyribose) A: adenine (a purine base) T: thymine (a pyrimidine base) C: cytosine (a pyrimidine base) Page 19 Copyright © Pearson Australia 2009 (a division of Pearson Australia Group Pty Ltd) ISBN 9780733993794 This page from the Chemistry: For use with the IB Diploma Programme Options SL and HL Teacher’s Resource may be reproduced for classroom use.