Pra U STPM Biology Term1 CC039142a

Biology Y.K. Richard David Tan PRE-U TERM 1 © Penerbitan Pelangi Sdn. Bhd. 2022 All rights reserved. No part of this book may be reproduced, stored in a retrieval system, or transmitted in any form or by any means, electronic, photocopying, mechanical, recording or otherwise, without the prior permission of Penerbitan Pelangi Sdn. Bhd. ISBN: 978-967-2878-77-3 eISBN: 978-967-2878-82-7 (eBook) First Published 2022 Tel: 03-8922 3993 Fax: 03-8926 1223 / 8920 2366 E-mail: [email protected] Enquiry: [email protected] Printed in Malaysia by Percetakan Jiwabaru Sdn. Bhd. No. 2, Jalan P/8, Kawasan Miel Fasa 2, Bandar Baru Bangi, 43650 Bangi, Selangor Darul Ehsan. Please log on to https://plus.pelangibooks.com/errata/ for up-to-date adjustments to the contents of the book (where applicable). STPMText

ii Pre-U STPM Text Biology Term 1 is written based on the new syllabus made by the Malaysian Examinations Council (MEC). The book is well designed and organised with the following features to help students to understand the concepts taught. vi Term of Study Paper Code and Name Theme / Title Type of Test Mark (Weighting) Duration Administration First Term 964/1 Biology Paper 1 Biological Molecules and Metabolism Written test Section A 15 compulsory multiple-choice questions to be answered. Section B 2 compulsory short structured questions to be answered. Section C 2 out of 3 essay questions to be answered. All questions are based on topics 1 to 6. 60 (26.67%) 15 15 30 1—1 2 hours Central assessment Second Term 964/2 Biology Paper 2 Physiology Written test Section A 15 compulsory multiple-choice questions to be answered. Section B 2 compulsory short structured questions to be answered. Section C 2 out of 3 essay questions to be answered. All questions are based on topics 7 to 13. 60 (26.67%) 15 15 30 1—1 2 hours Central assessment vi STPM Scheme of Assessment CHAPTER BIOLOGICAL 1 MOLECULES Concept Map Biological Molecules Water Carbohydrates Monosaccharide, disaccharide, polysaccharide Functions Lipids Proteins Nucleic acids Analytical techniques Physiological roles in organisms Paper chromatography Electrophoresis Chemical properties Physical properties Triglycerides Polar, non-polar, acidic, basic Phospholipids (lecithin) Structures of nucleotides DNA & RNA (mRNA, tRNA and rRNA) Structure of DNA Steroids (cholesterol) Bilingual Keywords Constituent – Juzuk Carbohydrate – Karbohidrat Monosaccharide – Monosakarida Triglyceride – Trigliserida Phospholipid – Fosfolipid Bond – Ikatan Nucleotide – Nukleotida Electrophoresis – Elektroforesis Diffraction – Pembelauan Concept Map Provides an overall view of the concepts learnt in the chapter Bilingual Keywords A list of bilingual terms is provided Biology Term 1 STPM Chapter 1 Biological Molecules 1CHAPTER 15 Microfibril O CH2OH O β O O O O O O O -Glucose molecules Hydrogen bonds forming cross bridges CH2OH CH2OH CH2OH CH2OH Figure 1.12 The structure of cellulose 13. Cellulose performs a structural role in the form of cell wall of plants. In the cell wall, it protects the cell from both physical injuries and haemolysis. It helps in the support of plants through turgor pressure and mechanically when it is mixed with lignin. 2010 Quick Check 2 1. Why are all monosaccharides reducing sugars but not all disaccharides? 2. What are the differences between D-glucose and L-glucose? 3. Why do starch and glycogen make good food reserves? 4. How do cellulose form macrofibrils as in the cell wall of some plant cells? 1.3 Lipids 1. Lipids are organic compounds that are made up of carbon, hydrogen and oxygen. Unlike carbohydrate, the ratio of oxygen atoms to hydrogen atoms in a lipid molecule is much lower and have a common physical property of being soluble in non-polar organic solvent such as chloroform, ether or benzene. 2. Other physical properties are as follows: (a) They are insoluble in water and polar organic solvents. (b) Their densities are always less than water. (c ) They have high viscosity. (d) They are greasy and leave behind grease spots on paper. 3. Their chemical properties are as follows: (a) Their chemical structures differ between different categories. (b) Generally, they are esters formed from fatty acids and alcohol. Students should be able to: (a) describe the structures, properties and distribution of triglycerides, phospholipids (lecithin) and steroid (cholesterol); (b) state the functions of triglycerides, phospholipids (lecithin) and steroids (cholesterol); (c) differentiate between saturated and unsaturated fatty acids. Learning Outcomes Learning Outcomes A list of subtopics that students will learn in each chapter STPM Scheme of Assessment Latest STPM Scheme of Assessment starting 2012 Quick Check Provides short question for students to test their understanding of the concepts learnt in the subtopics PREFACE

Biology Term 1 STPM Chapter 1 Biological Molecules 1CHAPTER 30 (v) For example, haemoglobin has four subunits of two different kinds i.e. 2 α-units and 2 β-units. There are two genes involved, α gene and β gene forming two types of polypeptides, α and β respectively. Each polypeptide has to be linked to a haem group before four of them can form a subunit as in figure 1.26. Polypeptide chain 3 ( 2) Polypeptide chain 2 ( 1) Prosthetic group (haem) Polypeptide chain 1 (β1 ) Polypeptide chain 4 (β2 ) Figure 1.26 The quaternary structure of haemoglobin Denaturation and Renaturation of Protein 1. Proteins can be denatured by heat, extreme pH, chemicals and radiation. This is because the three dimensional structure or conformation of the proteins can be altered. The weak bonds that control the shape can easily be broken. The globular shape can be changed into a more fibrous insoluble state as shown in figure 1.27. Denaturation Globular structure Chain is loosen + + – + – + + – – Reacts with other Precipitate is formed Figure 1.27 Denaturation of protein 2. Factors that cause denaturation include: (a) Heat. Heat increases the kinetic energy of the protein molecules and causes the weak bonds to break. For example, enzymes cannot function at high temperature. (b) pH. Too high concentration of H+ or OH– hydrogen bonds. breaks ionic and (c) Heavy metallic ions easily attach to negative charged groups and break weak bonds. . Examples like mercury and silver ions can Exam Tips Remember the four types of proteins: the fibrous and globular proteins, the simple and conjugated proteins; and their functions. Denaturation and renaturation of protein 1. Denaturation of protein means less function due to change in its shape. 2. Heat supplies kinetic energy and breaks hydrogen and ionic bonds of protein. 3. pH changes the H+ concentration also breaks hydrogen and ionic bonds. 4. Organic solvent breaks hydrophobic interactions. 5. Renaturation means that the denatured protein can be reversed to become functional. 6. Renaturation is only possible if the weak broken bonds can be reformed. Summary Biology Term 1 STPM Chapter 2 Structure of Cells and Organelles 2 52 Cell Theory 1. A cell is the basic unit of the structure and function of all living organisms. This is according to the cell theory, the result of studies by Schleiden, a botanist, and Schwann, a zoologist. 2. According to Rudolp Virchow, all cells arise from pre-existing cells by cell division. 3. A cell is made up of protoplasm surrounded by a selectively permeable lipoprotein membrane. Within the cell, the basic chemistry of life exists. 4. Cells are divided into two types, the prokaryotic cells and the eukaryotic cells. VIDEO Cell Theory Comparison Between Structures of Prokaryotic and Eukaryotic Cells 1. Prokaryotic cells are primitive cells. They are not true cells. They do not have nucleus and are found in bacteria and cyanobacteria, which belong to the kingdom Prokaryotae. 2. A generalised prokaryotic cell is as shown in Figure 2.1. Bacterial flagellum Nucleoid (circular DNA) Pili Plasmid Ribosomes Cytoplasm Plasma membrane Cell wall Capsule Figure 2.1 Structure of a generalised prokaryotic cell Learning Outcomes Students should be able to: (a) state the cell theory; (b) compare the structures of prokaryotic and eukaryotic cells; (c) compare typical animal and plant cells as seen under electron microscopes; (d) describe the basic principles of light and electron microscopy. Cell theory 1. A cell is a basic unit of structure and function of all living things. 2. Cells arise from preexisting cells by cell division. Summary • Bacterium – singular Bacteria – plural • Flagellum – singular Flagella – plural Language Check Language Check 2013, 2015 2.1 Prokaryotic and Eukaryotic Cells VIDEO Prokaryotic and Eukaryotic cell Biology Term 1 STPM Chapter 6 Photosynthesis 6 209 8. CAM plants such as cacti have PEP carboxylase like C4 plants. They open their stomata at night to fix carbon dioxide. They are well adapted to live in arid land. 9. Cacti play an important role in hot arid areas where the night temperature is cool. They close their stomata to cut down transpiration during they day. Systematic planting of economic important cacti to cut down carbon emission is possible. 10. As the sea level increases, C3 mangrove plants can help to reduce carbon dioxide concentration. C4 crop plants adapted in increasing salt concentration are also important to produce more food. Both help to reduce global temperature through more transpiration and photosynthesis of absorbing more carbon dioxide. Exam Tips Remember that plants with compensation point in the presence of very low light intensity, temperature and carbon dioxide concentration is very efficient in photosynthesis. Quick Check 3 1. Explain factors that affect photosynthesis other than light, temperature and carbon dioxide concentration. 2. Explain the significance of compensation point. Objective Questions 1. What are the phases required during photosynthesis? A Photosystem I and II B Photosystem and photolysis of water C Dark and the carbon fixation reactions D Light and the carbon fixation reactions 2. The diagram shows the light dependent stage of photosynthesis. Primary acceptor Primary acceptor P700 P680 NADP+ + 2H+ 2H+ + —O2 1 2 2e– H2O ATP Light NADPH + H+ NADP+ reductase Light 2e– Cytochrome complex P Q R 2e– 2e– Which combination is true of P, Q and R? P Q R A Ferredoxin Plastocyanin Plastoquinone B Ferredoxin Plastoquinone Plastocyanin C Plastocyanin Plastoquinone Ferredoxin D Plastoquinone Plastocyanin Ferredoxin 3. Which statement is not true in the Z diagram of light dependent reaction below? 4e– 4e– Q P e– e– e– Quinone Ferredoxin 2 NADP+ 2 NDPH2 Sunlight Sunlight PS II PS I P680 P700 I NADP+ is oxidised in non-cyclic photophosphorylation II P680 and P700 are oxidised after their electrons are raised to higher energy levels 6 STPM PRACTICE iii Past Year Questions Tagging Enable students to familiarise the pattern of exam questions STPM Practice A variety of examinationtyped questions to check students’ understanding of the chapter learnt Exam Tips Provides helpful tips for students in answering exam questions Summary Summarises key concepts learnt in the section Language Check Additional language information. Eg. plural and singular Biology Term 1 STPM Chapter 2 Structure of Cells and Organelles 2 83 Animal tissues Homogenate Homogenisation Centrifugation at 600 g for 10 minutes Centrifugation at 10,000 g for 20 minutes Centrifugation at 100,000 g for 60 minutes Nuclei and unbroken cells Mitochondria, ER and Golgi bodies Ribosomes, microtubules and microfilaments Supernatant Supernatant Nucleic acids and proteins Figure 2.25 Step by step cell fractionation 3. Further differential centrifuge is ultra-centrifugation with more than using force 100,000 times gravity. 4. This technique is to separate mixture of macromolecules of different molecular weights or S values. S value is a scale of sedimentation units for molecule to move down in gel used in ultra-centrifugation. Higher S values are heavier and more stream-lined. For examples, DNA can be separated from RNA, nucleic acids can be separated from proteins and radioactive DNA can be separated from normal ones. 5. The method and precautions are as follows: (a) The space within the ultracentrifuge should be vacuumed to avoid any friction between the tubes and the air. (b) The temperature has to be lowered. (c) Gel is added to stop the molecule at certain levels in the tube. (d) Dye is added to the mixture to detect separation. Exam Tips Remember the basic principles of centrifugation. Remember the examples of uses in the isolation of cellular components. Quick Check 4 1. How can radioactive DNA and normal DNA be separated by ultra centrifugation? 2. How can the background of different refractive indexes with the object be lightened or darkened? 3. Why can electron beam and not light beam be used to resolve smaller objects? Differential centrifugation 1. Fractionation by homogenisation 2. Centrifuge 600 g for 10 min to obtain nuclei 3. Centrifuge 10, 000 g for 20 min to obtain mitochondria and chloroplasts 4. Centrifuge 100,000 g for 60 min to obtain ribosomes 5. Ultra-centrifuge with gel in vacuum to separate component of ribosome proteins, RNA and DNA of different S values Summary 80S ribosomes can be separated into 60s and 40s sub-units. Why 60s + 40s not equal to 100s? This is due to 80S ribosomes with one big and one small units are not so stream-lined. So, they have lower s units. Info Bio Info Bio Provides extra information that relates to the subtopics learnt

iv A absorption spectrum The graph of percent of light absorbed by a pigment plotted against wavelengths of visible light. action spectrum The graph of photosynthesis rate by a pigment plotted against wavelengths of visible light. activation energy The initial or minimum energy that must be possessed by atoms or molecules in order to react. active site The specific portion of an enzyme that attaches to the substrate by means of weak chemical bonds. active transport The movement of a substance across a biological membrane against its concentration or electrochemical gradient, with the help of ATP and specific transport proteins. adenosine triphosphate (ATP) An adenine-containing nucleoside triphosphate that releases free energy when its phosphate bonds are hydrolysed. adenylyl cyclase An enzyme that converts ATP to cyclic AMP in response to a chemical signal. aldehyde An organic molecule with a carbonyl group located at the end of the carbon skeleton. allosteric site A specific receptor site on an enzyme molecule that binds to chemical resulting a change in the shape of the enzyme, making it either more or less receptive to the substrate on the active site. alpha helix A spiral shape constituting one form of the secondary structure of proteins, arising from a specific hydrogenbonding structure. amino acid An organic molecule possessing both carboxyl and amino groups. amino group A functional group that consists of a nitrogen atom bonded to two hydrogen atoms that can act as a base in solution. aminoacyl—tRNA synthetases (synthases – USA) A family of enzymes, at least one for each amino acid, that catalyses the attachment of an amino acid to its specific tRNA molecule. amphipathic molecule A molecule that has both hydrophilic region and hydrophobic region. anabolism Within a cell or organism, the sum of all biosynthetic reactions in which larger molecules are formed from smaller ones. anaerobic Lacking oxygen; referring to an organism, environment, or cellular process that lacks oxygen. anticodon The triplet on one end of a tRNA molecule that recognises a particular complementary codon on an mRNA molecule. apical meristem Undifferentiated tissue in the tips of roots and shoots that where the cells can continue to divide. B basement membrane The layer of fine protein fibres beneath epithelial cells. buffer A substance that consists of acid and base forms in a solution and that minimises changes in pH when extraneous acids or bases are added to the solution. C C4 plant A plant that prefaces the Calvin cycle with reactions that incorporate CO2 into four-carbon compounds, the end-product of which supplies CO2 for the Calvin cycle. Calvin cycle The second of two major stages in photosynthesis involving atmospheric CO2 fixation and reduction of the fixed carbon into carbohydrate in a cyclic series of reactions. 225 223 Biology Term 1 STPM Model Paper (964/1) Objective Questions 1. D 2. A 3. D 4. D 5. C 6. C 7. C 8. C 9. A 10. D 11. A 12. C 13. A 14. B 15. B Structured Questions 16. (a) Enzyme immobilisation is the technique of binding enzyme molecules to solid medium. Thus, the enzyme can be used continuously to carry out industrial process. (b) Adsorption (c) Entrapment (d) The enzyme molecules vibrate less as they are bound to solid. They are protected when the temperature is higher. They do not move about to be affected by other factors 17. (a) • It is a nucleotide. • It has adenine. • It has three phosphate groups. • It has a ribose. (b) • It is synthesised from ADP and inorganic phosphate (Pi ). • It is a soluble molecule and diffuses rapidly. • On hydrolysis of the third phosphate energy released. • It acts as an intermediary between energy yielding and energy requiring reactions. [maximum 3] (c) • It is synthesised by oxidative phosphorylation. • NADH moves to crista and transfers electrons to the electron transport chain. • H+ are pumped into the inter-membrane space to create H+ gradient. • Chemiosmosis resulting ADP and Pi forming ATP. Essay Questions 18. (a) • The structure of collagen consists of fibrous protein forming fibre to hold different tissues together as under the skin. • Each molecule of collagen is made up of only secondary structure twisted together with two similar molecules to have greater strength . • The tripolypeptide unit is further lengthened by disulfide bonds with similar units to form longer fibre for stronger binding of tissues. • Such collagen fibres can cross-linked together to form even larger fibres as found in the tendon and ligaments to produce high tensile strength. • Collagen fibres also cross-linked with other fibrous proteins for attachment firmly between two bones at the joints. (b) • The procedure to separate organelles is by differential centrifugation after the tissue such as liver is homogenised. • Homogenisation is by sophisticated ultrasound homogeniser to break up the cells without breaking organelles. • Chilled buffer is added before homogenisation to maintain the structures and functions of organelles and enzymes. • The homogenate obtained is centrifuged for 10 minutes at 600 g force so nucleic will be spun down. • The supernatant is then centrifuged at 10,000 g force for 20 minutes to spin down mitochondria, ER and Golgi bodies. • Then, the supernatant is further spun with 100,000 g force for 60 minutes to obtain ribosomes, microtubules and microfilaments. • The nuclei obtained may be mixed with unbroken cells that need further centrifugation to obtain higher concentration. • Similarly, mitochondria need to be separated from ER and Golgi bodies. • Ribosomes are most difficult to obtain as they exist as separated subunits or combined • To separate the two subunits of ribosomes, magnesium free buffer has to be used by ultracentrifugation using gel. 19. (a) • At body temperature, the barrier of activation energy prevents reaction e.g. the breakdown of glycogen to occur. • Enzyme e.g. phosphorylase lowers the barrier and permits the reaction to occur at that temperature. • Enzymes allow substrate to bind to their active sites to form complex. • Then the enzymes act directly or indirectly on the substrate molecules. • They break and reform certain bonds of the substrate to form the product. • An enzyme may bring two reactive substrate molecules together in its active site and a larger product is formed. • The enzyme molecule orientates two reacting substrate molecules so that the product can be easily formed. ANSWERS Answers Complete answers are provided 217 Section A [15 marks] Bahagian A [15 markah] Answer all questions in this section. Jawab semua soalan dalam bahagian ini. 1. Which chemical produces variation in DNA molecules? Bahan kimia manakah menghasilkan variasi dalam molekul-molekul DNA? I Phosphate group Kumpulan fosfat II Pentose sugar Kumpulan pentose III Pyrimidine Pirimidina IV Purine Purina A I and II C II and III I dan II II dan III B I and IV D III and IV I dan IV III dan IV 2. Guanine makes up 8% of the nucleotides in a sample of DNA. What is the percentage of thymine in the sample? Guanina membentuk 8% nukleotida dalam suatu sampel DNA. Apakah peratus timina dalam sampel tersebut? A 42 B 28 C 44 D 56 3. The diagram below shows part of a polymer. Gambar rajah di bawah menunjukkan sebahagian daripada sejenis polimer. C C N H H H2N CH3 CH3 C C C H O OH Which molecule is used to break the bond indicated by the arrow? Molekul yang manakah digunakan untuk memutuskan ikatan yang ditunjukkan dengan anak panah? A Amino acid C Peptide Asid amino Peptida B Amylase D Water Amilase Air 4. What differentiates prokaryotic from eukaryotic cells? Apakah yang membezakan antara sel prokariotik dengan sel eukariotik? A The presence or absence of ribosomes Kehadiran atau ketiadaan ribosom B The ability of the cell to carry out cellular metabolism Kebolehan sel untuk menjalankan metabolisme sel C The presence or absence of a rigid cell wall Kehadiran atau ketiadaan dinding sel yang tegar D Whether or not the cell is partitioned by internal membranes Sama ada sel dibahagikan mengikut membran internal atau tidak 5. The pH of the intermembrane space of mitochondria is lower than the matrix of mitochondria due to the pH di ruang antara membran mitokondria adalah lebih rendah daripada di matriks mitokondria kerana A Accumulation of lactic acid during anaerobic respiration Pengumpulan asid laktik semasa respirasi anerobik STPM Model Paper (964/1) STPM Model Paper (964/1) A model paper follows the latest STPM exam format is provided for practice Glosarry Gives a list of important terms of ease the students’ understanding of their meanings

Chapter 2015 2016 2017 2018 A B C A B C A B C A B C 1. Biological Molecules 4 1 – 4 – 1 4 – 0.33 4 – 0.5 2. Structure of Cells and Organelles 3 – – 3 – 1 3 1 0.67 3 – 0.5 3. Membrane Structure and Transport 1 – 0.67 1 1 – 1 – – 1 1 1 4. Enzymes 3 – 1 3 – 1 3 – 0.67 3 – – 5. Cellular Respiration 2 – 1 2 1 – 2 – 1 2 1 – 6. Photosynthesis 2 1 0.33 2 – – 2 1 0.33 2 – 1 Total 15 2 3 15 2 3 15 2 3 15 2 3 Analysis of STPM Papers (2015 – 2018) v

vi Term of Study Paper Code and Name Theme / Title Type of Test Mark (Weighting) Duration Administration First Term 964/1 Biology Paper 1 Biological Molecules and Metabolism Written test Section A 15 compulsory multiple-choice questions to be answered. Section B 2 compulsory short structured questions to be answered. Section C 2 out of 3 essay questions to be answered. All questions are based on topics 1 to 6. 60 (26.67%) 15 15 30 1—1 2 hours Central assessment Second Term 964/2 Biology Paper 2 Physiology Written test Section A 15 compulsory multiple-choice questions to be answered. Section B 2 compulsory short structured questions to be answered. Section C 2 out of 3 essay questions to be answered. All questions are based on topics 7 to 13. 60 (26.67%) 15 15 30 1—1 2 hours Central assessment vi STPM Scheme of Assessment

vii Term of Study Paper Code and Name Theme / Title Type of Test Mark (Weighting) Duration Administration Third Term 964/3 Biology Paper 3 Ecology and Genetics Written test Section A 15 compulsory multiple-choice questions to be answered. Section B 2 compulsory short structured questions to be answered. Section C 2 out of 3 essay questions to be answered. All questions are based on topics 14 to 19. 60 (26.67%) 15 15 30 1—1 2 hours Central assessment 964/5 Biology Paper 5 Written Practical Test 3 structured questions with diagram/graph/ table to be answered. 45 (20%) 1—1 2 hours Central assessment First, Second and Third Terms 964/4 Biology Paper 4 Biology Practical School-based Assessment of Practical 15 compulsory experiments to be carried out. 25 (20%) Throughout the three terms School-based assessment vii

Chapter • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • 1 BIOLOGICAL MOLECULES 1 1.1 Water 2 1.2 Carbohydrates 6 1.3 Lipids 15 1.4 Proteins 22 1.5 Nucleic Acids 33 1.6 Analytical Techniques 39 STPM Practice 1 42 QQ 47 Answers 48 Chapter • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • 2 STRUCTURE OF CELLS AND ORGANELLES 51 2.1 Prokaryotic and Eukaryotic Cells 52 2.2 Cellular Components 59 2.3 Specialised Cells 84 STPM Practice 2 102 QQ 107 Answers 107 Chapter • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • 3 MEMBRANE STRUCTURE AND TRANSPORT 111 3.1 Fluid Mosaic Model 112 3.2 Movement of Substances across Membrane 116 STPM Practice 3 128 QQ 131 Answers 131 Chapter • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • 4 ENZYMES 133 4.1 Catalysis and Activation Energy 134 4.2 Mechanism of Action and Kinetics 136 4.3 Cofactors 144 4.4 Inhibitors 146 4.5 Classification of Enzymes 150 4.6 Enzyme Technology 152 STPM Practice 4 156 QQ 159 Answers 160 Chapter • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • 5 CELLULAR RESPIRATION 163 5.1 The Need for Energy in Living Organisms 164 5.2 Aerobic Respiration 168 5.3 Anerobic Respiration 177 STPM Practice 5 180 QQ 184 Answers 185 Chapter • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • 6 PHOTOSYNTHESIS 187 6.1 Autotrophs 188 6.2 Light-dependent Reactions 193 6.3 Light-independent Reactions 197 6.4 Limiting Factors 206 STPM Practice 6 209 QQ 214 Answers 215 STPM Model Paper (964/1) 217 Answers 223 Glossary 225 viii CONTENTS

CHAPTER BIOLOGICAL 1 MOLECULES Concept Map Biological Molecules Water Carbohydrates Monosaccharide, disaccharide, polysaccharide Lipids Functions Proteins Nucleic acids Analytical techniques Physiological roles in organisms Paper chromatography Electrophoresis Chemical properties Physical properties Triglycerides Polar, non-polar, acidic, basic Phospholipids (lecithin) Structures of nucleotides DNA & RNA (mRNA, tRNA and rRNA) Structure of DNA Steroids (cholesterol) Bilingual Keywords Constituent – Juzuk Carbohydrate – Karbohidrat Monosaccharide – Monosakarida Triglyceride – Trigliserida Phospholipid – Fosfolipid Bond – Ikatan Nucleotide – Nukleotida Electrophoresis – Elektroforesis Diffraction – Pembelauan

Biology Term 1 STPM Chapter 1 Biological Molecules 1CHAPTER 2 1.1 Water Water (H2 O) is a clear, colourless, odourless and tasteless liquid essential for all organisms. There are other properties of water that enable it to play its roles in organisms. Chemical Properties of Water 1. Each water molecule is composed of two hydrogen atoms and an oxygen atom as shown in Figure 1.1. H+ 104.5° H+ O– H Covalent bond Electron H + – – + O O — Oxygen H — Hydrogen δ+ — Small positive charge δ– — Small negative charge Figure 1.1 Structure of water molecule, a polar molecule 2. Two hydrogen atoms are attached to an oxygen atom at an angle of 104.5°. Each hydrogen atom shares a pair of electrons with the oxygen atom, so there are two covalent bonds. 3. Water is polar, the oxygen atom is slightly negative and the two hydrogen atoms are slightly positive. This is because the oxygen atom draws the electrons in the bond towards itself, giving a partial negative charge to the oxygen atom. The water molecule, as a whole is neutral. 4. Water is an excellent solvent to dissolve polar substances like electrolytes and non-electrolytes such as hydrophilic organic compounds, which have –OH, –COOH, –NH2 , –CO– and –PO4 groups. Water molecules can surround polar groups after weakening and separating inter-molecular or inter-ionic bonds within a substance such as in sodium chloride in Figure 1.2. – Cl– Na+ Water molecules + Figure 1.2 Shell of orientated water molecules surround ions INFO Why is Water a Polar Molecule? Students should be able to: (a) describe the chemical properties (solvent, bond angles and hydrogen bond) of water and relate its physiological roles in the organisms; (b) describe the physical properties (polarity, cohesiveness, density, surface tension, specific heat capacity, and latent heat of vaporisation) of water and relate its physiological roles in organisms. Learning Outcomes

Biology Term 1 STPM Chapter 1 Biological Molecules 1CHAPTER 3 5. Similarly, organic substances such as sucrose is soluble in water as water molecules interact and surround the sugar molecules, breaking down their inter-molecular bonds. Water molecules are linked through hydrogen bonds. The hydrogen bond is an electrostatic attraction between the positively charged hydrogen atom of one molecule and the negatively charged oxygen atom of another as shown in Figure 1.3. 6. Water is a liquid at room temperature. Each molecule of water can form a maximum of four hydrogen bonds with different molecules of water. Under room temperature, about 20% hydrogen bonds exist in water. The lower the temperature, the more hydrogen bonds are formed. At 0 °C, it freezes where all the molecules are involved in forming the three-dimensional structure of ice. This explains why ice expands and has a lower density. i.e. ice floats on water. These bonds are shown in Figure 1.4 below. Exam Tips Remember that water acts as reactant in hydrolysis and is chemically inert. Chemical properties of water 1. Water contains two hydrogen atoms and one oxygen atom in a bent molecule of 104.5° bond angle. 2. Water molecules exhibit polarity and form hydrogen bonds with one another. 3. Water is inert and a good solvent for reactions and transport. Summary Hydrogen bonds O H O H H O O H H H + + – + + H+ H+ H+ H+ – – – + + O– Figure 1.3 Polarisation of water molecules causes them to be linked together by hydrogen bonds Water molecules Weak hydrogen bonds Oxygen Hydrogen Water Water molecules Stable hydrogen bonds Oxygen Hydrogen Ice Figure 1.4 Hydrogen bonds in water and ice 7. Because of its chemical properties, the physiological roles of water are as follows: (a) Water provides a medium for reactions to take place. This can be seen in seeds that absorb water and start germinating almost immediately. Enzyme and substrate molecules in the cytoplasm of cells within the seeds are activated and will start reacting with each other to bring about faster respiration and to start cell divisions. (b) Water acts as a solvent to transport substances in the blood of animals, or in xylem and phloem of plants. Inorganic and organic nutrients dissolve in water and are transported in soluble forms in the blood of animals. In plants, inorganic ions such as nitrate are transported in the xylem. Organic substances such as sucrose, amino acids and some plant hormones are transported in the sieve tubes of the phloem.

Biology Term 1 STPM Chapter 1 Biological Molecules 1CHAPTER 4 (c) Water ionises and acts as substrates for photolysis during photosynthesis and hydrolysis reactions during digestion of food. The products of its ionisation i.e. hydrogen and hydroxyl ions are then used to form products during the light-dependent stage of photosynthesis. During hydrolysis, water is split into hydrogen ions and hydroxyl ions too to break bonds between complex biopolymers into monomers such as starch into glucose. (d) Water interacts with macromolecules such as proteins, nucleic acids and molecules in the lipoprotein membrane structure. The water molecules surround these macromolecules, making their structures more stable and thus maintaining their three dimensional structures to perform their functions. For example, a protein can act as an enzyme (catalyst) to speed up reactions when a substrate molecule binds to the enzyme molecule. Physical Properties of Water 1. Water has a high specific heat capacity of 4.2 kJ/K/kg, which means 4.2 kJ of heat is required to raise the temperature by 1 K for 1 kg of water. (a) This means that a lot of energy is required to raise the temperature of water. This is because the breaking of the hydrogen bonds within water molecules requires a large amount of energy before the temperature can be raised. (b) Temperature is raised when the individual molecules or atoms vibrate vigorously. (c) The physiological roles of a high specific heat capacity are as follows: • Water in our body can absorb a lot of heat from the muscles or the environment before our body temperature rises. This prevents our body temperature from rising too fast. • Water provides a more constant environment in the sea for aquatic organisms to live in. 2. Water has a high latent heat of vaporisation of 2260 kJ/kg, that is 2260 kJ of heat is required to vaporise 1 kg of water. (a) This means that a lot of heat is required to change water to vapour as the water molecules require a lot of heat energy to break all the hydrogen bonds between them to escape as gas. (b) Vaporisation always results in the cooling of the surroundings. There are two sorts of vaporisation which are evaporation, the slow vaporisation and boiling, the fast vaporisation. Evaporation occurs at the water surface which provides maximum cooling effect with minimum loss of water. Boiling occurs under the surface of water at the boiling point of water. (c) Water’s physiological roles of high heat of vaporisation are as follows:

Biology Term 1 STPM Chapter 1 Biological Molecules 1CHAPTER 5 • This enables many land invertebrates such as earthworms and snails to survive. These invertebrates are nocturnal. They come out at night to forage for food in order to cut down water loss through evaporation during the day. • This also helps to lower our body temperature when we sweat. When sweat evaporates, it absorbs heat from our skin and cools down our body. • Panting helps to get rid of the excessive heats in dogs and birds. Water evaporating from the lungs is more effective to cool body temperature than the skin. Heat is absorbed within the body rather than the body surface. • Similarly, water evaporating during transpiration in plants helps to lower the temperature of leaves. This cuts down overheating when they are exposed to bright sunlight during photosynthesis. 3. Density or specific gravity of water is highest at 4 °C. (a) This means that when water is cooled to 4 °C, it sinks. Further cooling will cause it to float. At 4 °C, the water molecules are most compact with the shortest intermolecular bonds. (b) At 0 °C, water freezes. Ice expands during freezing as hydrogen bonds occupy more space in solid medium. Thus, this results in a lower density than water causing it to float. (c) Freezing starts from the top of the water surface downwards as the surface is subjected to low temperature first. (d) After a certain thickness is formed, ice insulates the water below. Heat cannot escape from the water preventing further freezing as shown in Figure 1.5. Water = 4 °C Lake Ice = 0 °C Air = –20 °C Figure 1.5 Water in a lake would not be totally frozen (e) The physiological roles of water by having the highest density at 4 °C are as follows: • Water at the bottom of lakes or rivers will not freeze. This enables aquatic organisms to survive during winter or in the tundra ocean. This happens especially when the depth of water is more than a meter. • Nutrients can circulate in the lake, helping in the colonisation of organisms into a greater depth. When the atmospheric temperature decreases, the surface water sinks at 4 °C. When the temperature of water at the bottom increases after winter, water moves up bringing soluble salts along. Exam Tips Water in the new syllabus has chemical properties of 104.5°C bond angle, with polarity and solvent properties. It has physical properties of high specific capacity; its highest density at 4°C; it has high latent heat of vaporisation; high cohesive forces and high surface tension. (2006 essay question) Physical properties of water 1. Water has high specific heat capacity to prevent quick change of temperature in organisms or in the earth. 2. Water has high latent heat of vaporisation to lower temperature during transpiration or sweating. 3. Water’s highest density at 4°C allows it to freeze from top to bottom permitting organisms to live below and circulate nutrients during winter. 4. Water has high cohesive force allowing transport of water in xylem vessels. 5. Water has high surface tension allowing insects to carry out activities at the surface. Summary

Biology Term 1 STPM Chapter 1 Biological Molecules 1CHAPTER 6 4. Water has a high cohesive force. (a) This means that there is a high attractive force that exists among water molecules due to the hydrogen bonds. (b) Its physiological roles of high cohesive force are as follows: • Together with adhesive force, water creates a transpirational pull in the xylem vessels when evaporation occurs in the leaves. This helps to transport water and mineral ions up the leaves for photosynthesis. • Cohesion and adhesion of water molecules enable water to stay in the upper layer of soil called topsoil. Good quality soil of fine grain sizes together with humus retain the right amount of water so that plants and other soil organisms can live well in it. 5. Water has a high surface tension. (a) High surface tension means that there is a strong inward pull of water forming a skin-like layer at the surface of water. This is caused by the high cohesive forces of water molecules as a result of hydrogen bonding. (b) Surface tension creates a habitat on the water surface. It allows insects to stay on the surface where they can gather food or catch their preys like water-skaters do. (c) Surface tension allows female mosquitoes to stand on water to lay eggs. The tiny eggs can float on the surface of water and the larvae can attach to the surface so that their syphons can breathe in air. Quick Check 1 1. What are the other properties of water that are important to organisms? 1.2 Carbohydrates 1. Carbohydrates (hydrates of carbon) are a large group of biochemicals which consist of carbon, hydrogen and oxygen in the proportion, 1 : 2 : 1. 2. Carbohydrates are sugar-containing compounds. These are the most common organic compounds found on earth and in the largest amount as cellulose in plants. 3. Their general formula is Cx (H2 O)y , such as glucose (C6 H12O6 ) and sucrose (C12H22O11). Students should be able to: (a) classify carbohydrates into monosaccharides, disaccharides and polysaccharides with respect to their physical and chemical properties; (b) classify monosaccharides according to the number of carbon atoms and the functional groups: (i) triose e.g. glyceraldehydes, (ii) pentose e.g. ribose and deoxyribose, (iii) hexose e.g. glucose and fructose; Learning Outcomes

Biology Term 1 STPM Chapter 1 Biological Molecules 1CHAPTER 7 Classification of Carbohydrates Carbohydrates are divided into three groups with respect to their physical and chemical properties: (a) Monosaccharides (one sugar unit) 1. Monosaccharides are simple sugars, containing only one basic unit which can be used to form disaccharides and polysaccharides. They are the smallest carbohydrate molecules. 2. Monosaccharides have the following physical properties: (a) They are sweet in taste. (b) They dissolve in water. (c) They can be crystallised. 3. Monosaccharides have the following chemical properties: (a) They are either aldehydes which contain –CHO groups or ketones which contain C=O groups. Both ketones and aldehydes have carbonyl groups, C=O. (b) They are all reducing sugars. They can carry out reduction since carbonyl groups donate electrons. (c) They react with Benedict’s or Fehling solution to give a brick-red precipitate. They reduce alkaline copper(II) sulphate (CuSO4 ) into insoluble copper(I) oxide (Cu2 O). Cu2+ + e– → Cu+ Blue solution Brick-red precipitate (d) During reduction, they themselves become oxidised. H | –C=O or –CHO +O → Oxidation –COOH Aldehyde Carboxylic acid (e) They can be reduced in the cell, forming alcohols. H | –C=O +2H+ → Reduction –CH2 OH Aldehyde Primary alcohol | | –C=O +2H+ → Reduction –CHOH Ketone Secondary alcohol Classification of carbohydrates Carbohydrates Monosaccharides (Single sugar unit) Disaccharides (Two sugar units) Polysaccharides (Many sugar units) Summary (c) illustrate the molecular structure of a monosaccharide and differentiate between the reducing and nonreducing ends; (d) describe the formation of glycosidic bond in disaccharides (maltose and sucrose) and polysaccharides (starch, glycogen and cellulose); (e) relate the structure of disaccharides and polysaccharides to their functions in living organisms. INFO Carbohydrates: Definition, Classification and Functions

Biology Term 1 STPM Chapter 1 Biological Molecules 1CHAPTER 8 4. Monosaccharides have a general formula of (CH2 O)n . They can be classified into trioses, pentoses and hexoses according to the number of carbon atoms and the type of functional groups. Functional groups can be either aldehyde group (–CHO) or keto group (–CO). 5. Trioses (3C) – C3 H6 O3 . (a) Trioses are monosaccharides that contain three atoms of carbon e.g. glyceraldehyde and dihydroxyacetone. (b) Their molecular structures are shown below: H H C OH C OH C O H H C H C OH H C OH O H H Glyceraldehyde Dihydroxyacetone (Aldose) (Ketose) (c) Both trioses are the intermediate products of metabolism during the breakdown of glucose (glycolysis) or synthesis (photosynthesis) of it. Glyceraldehyde exists as phosphoglyceraldehyde and dihydroxyacetone exists as phosphodihydroxyacetone during glycolysis and carbon fixation during photosynthesis. They seldom exist long in the free state in cells. 6. Pentoses (5C) – C5 H10O5 . (a) They are monosaccharides that contain five atoms of carbon. Examples include ribose, deoxyribose and ribulose. H C1 C2 C3 C4 C5 O OH OH OH OH H H H H H Ribose (Aldose) H C1 C2 C3 C4 C5 O OH OH OH H H H H H OH Ribulose (Ketose) Exam Tips The molecular structure of glyceraldehydes, ribose, deoxyribose and glucose should be remembered to show that you know the presence of carbonyl group for reducing. Classification of monosaccharides Trioses (C3 sugars) – glyceraldehyde (aldose) Pentoses (C5 sugars) – ribose (aldose) and deoxyribose (aldose) Hexoses (C6 sugars) – glucose (aldose) and fructose (ketose) Summary

Biology Term 1 STPM Chapter 1 Biological Molecules 1CHAPTER 9 (b) Pentose can either exist in an open chain form or a ring structure e.g. ribose and deoxyribose are shown in Figure 1.6 below. (c) The structure of deoxyribose is the same as ribose except that the oxygen in carbon number 2 is missing. (d) The physiological roles of pentoses are as follows: • Pentoses are used for the synthesis of nucleotides in which the ribose is for the formation of ribonucleotides in RNA and the deoxyribose is for the formation of deoxyribonucleotides in DNA. • They are used for the synthesis of coenzymes such as NAD, NADP, FAD and coenzyme A. • They can be used for the synthesis of polysaccharides. • Ribulose biphosphate is used as a receptor for the fixation of CO2 at the beginning of the light independent stage of photosynthesis. 7. Hexoses (6C) – C6 H12O6 . (a) Hexoses such as glucose, galactose and fructose are monosaccharides that contain six atoms of carbon. H C1 C2 C3 C4 C5 O OH H OH OH H HO H H H C6 OH H Glucose (Aldose) H C1 C2 C3 C4 C5 O OH OH H H H OH HO H H C6 OH H Fructose (Ketose) 2013 H 1C 2 C 3 C 4 C 5CH2OH OH HO OH Pentose ring form Ribose CH2OH Open chain form O O H OH H OH H OH H 1C 2C 3C 4C 5CH2OH OH HO Pentose ring form CH2OH Open chain form Deoxyribose O O H H H OH H OH Figure 1.6 The straight-chain and ring form of ribose and deoxyribose

Biology Term 1 STPM Chapter 1 Biological Molecules 1CHAPTER 10 (b) Fructose and other ketoses have ketone (–CO) group as functional groups at one end. Both aldehyde and ketone groups have carbonyl (–CO) group as donor of electrons for reducing, at the reducing end. (c) All of them are isomers, which have the same empirical formula of C6 H12O6 . The glucose molecule can either exist in an open chain form or a ring structure as shown in Figure 1.7. Exam Tips Remember examples of monosaccharides (STPM 2013) H 1CHO 2C D-glucose α-glucose reducing end OH HO 3C 3C 2C H H 4C OH H 5C OH 6CH2OH 6CH2OH H 5C O 4C OH OH OH H H 1 C OH H non-reducing end H β-glucose 3C 2C 6CH2OH H 5C O 4C OH OH OH H H 1 C H OH H Figure 1.7 Straight chain structure and ring structure of glucose As shown in the open chain structure, glucose like all aldoses have aldehyde (–CHO) as a functional group at one end. (d) As shown in the straight chain structure, the glucose is polyhydroxyaldehyde or aldohexose. D-glucose and L-glucose are optical isomers or stereoisomers. (e) Hexose or pyranose ring structure of glucose can exist in two forms i.e. α-glucose and β-glucose depending on whether the OH group at the carbon 1 atom is below or above the plane respectively. (f) The physiological roles of hexoses are as follows: • Hexoses act as a source of ATP in respiration especially glucose when completely broken down into water and carbon dioxide. • They can balance water potential between the inside and outside of cells, so as to maintain the shape of cells e.g. red blood cells. • They are used to synthesise disaccharides such as sucrose and lactose. • They are used to synthesise polysaccharides such as starch, glycogen and cellulose. • They are used to synthesise other substances such as fats and amino acids in the body when required. Disaccharides (two sugar units) 1. Disaccharides are double-sugars formed from two units of monosaccharides through condensation reaction with the removal of a water molecule. Molecular structure of monosaccharide Reducing end – has carbonyl group. Non-reducing end – has no carbonyl group. Summary

Biology Term 1 STPM Chapter 1 Biological Molecules 1CHAPTER 11 2. The bond between the monosaccharides is called glycosidic bond. The glycosidic bond can be α or β depending on the disaccharide. 3. Their physical properties are as follows: (a) They are all sweet. (b) They are all soluble in water. (c) They are all crystalline substances. 4. Their chemical properties are as follows: (a) They are reducing sugars except sucrose. (b) They can be hydrolysed to form monosaccharides by heating in dilute acid or by specific enzymes. Sucrase can hydrolyse sucrose, maltase can hydrolyse maltose and lactase can hydrolyse lactose. 5. Examples of disaccharides are maltose, sucrose and lactose. 6. Maltose is formed from two molecules of glucose, which are bonded by α-1,4-glycosidic bond as shown in Figure 1.8. 6CH2OH 5 4 3 2 1 OH OH O –Glucose HO OH HO H2O 6CH2OH 5 4 3 2 1 OH OH O α –Glucose Condensation Hydrolysis α 6CH2OH 5 4 3 2 1 OH OH O Maltose HO O 1,4-glycosidic bond 6CH2OH 5 4 3 2 1 OH OH OH O Figure 1.8 The formation and hydrolysis of maltose 7. The reaction is condensation when a molecule of water is removed between two hydroxyl groups from C1 of one glucose and C4 of another glucose unit. The reaction is catalysed by synthetase (ligase) enzyme. 8. This occurs in the stroma of chloroplast, in amyloplast of parenchyma cells and cytoplastin of liver and muscle cells. 9. Sucrose is formed from the bonding of glucose and fructose by an α-1,2-glycosidic bond as shown below. 6CH2OH 5 4 3 2 1 OH OH O Glucose HO OH HO H2O HOH2C CH2OH OH OH O Fructose CH2OH OH OH O HOH2C CH2OH OH OH O Sucrose Condensation Hydrolysis O 1,2-glycosidic bond Figure 1.9 The formation and hydrolysis of sucrose

Biology Term 1 STPM Chapter 1 Biological Molecules 1CHAPTER 12 10. This is also a condensation reaction that occurs in the stroma chloroplast and cytoplasm of plant cells catalysed by synthase. This occurs in the mesophyll cells and the sucrose is transported away by phloem. This is a common reaction in sugar cane, sugar beet, flower and fruit cells, where sucrose is stored. Polysaccharides (many sugar units) 1. Polysaccharides are complex carbohydrates, a type of biopolymer i.e. long chains of molecules which are made of repeating units of monosaccharides. 2. The monosaccharides in polysaccharides may be all hexoses hence they are called hexosan, and their common formula is (C6 H10O5 )n . They can be all pentoses which then are called pentosan (C5 H8 O4 )n or a mixture of both. 3. Their physical properties are as follows: (a) Polysaccharides are not sweet. (b) They are insoluble in water. If dissolved in hot water, they form colloids because they are large molecules. (c) They cannot be crystallised but form amorphous mass if desiccated. (d) They are compact and not osmotically active in cells. (e) They can be extracted and purified to form white powder. 4. Their chemical properties are as follows: (a) All of them have no reducing power. (b) Polysaccharides like starch and glycogen are easily hydrolysed to maltose by amylase. (c) Polysaccharides like cellulose can be hydrolysed by cellulase, which is only produced by microbes and snails. 5. Examples of polysaccharides are starch, glycogen and cellulose. 6. Starch is formed by a-1,4-glycosidic bonds in amylose, an unbranched component. So, one amylose synthetase is involved. The other branched component, amylopectin is formed by both a-1,4-glycosidic and a-1,6-glycosidic bonds. So, two amylopectin synthetases are involved. 7. The reaction is polymerisation by adding one glucose monomer to maltose at a time to elongate in chloroplast and amyloplast. 8. Glycogen is a branched biopolymer with both α-1,4-glycosidic and α-1,6-glycosidic bonds. It is formed in the smooth endoplasmic reticulum of both liver and muscle cells, also by polymerisation catalysed by two synthetases. After forming a straight chain of more than 12 units, a short chain is cut by a hydrolase. Then, the chain is transferred to another chain to form a branch forming a-1,6-glycosidic bond catalysed by another synthetase.

Biology Term 1 STPM Chapter 1 Biological Molecules 1CHAPTER 13 9. Cellulose is formed by β-1,4-glycosidic bonds into a straight biopolymer. The β-glucose units are catalysed by synthetase to form long chains at the plasma membrane. 10. The enzymes involved are synthesised in the rough endoplasmic reticulum packaged into vesicles. The enzymes are released at the plasma membrane to form cellulose molecules, and then microfilaments to form new layer of cell wall. 11. After plant cells divide, vesicles gather in the central plate and fuse. The enzymes in the vesicles form cellulose molecules and microfilaments that form the new cell wall. The membrane of vesicles form new plasma membrane at both sides of the new cross wall. Functions of Disaccharides and Polysaccharides in Living Organisms 1. Disaccharides like maltose consists of two glucose units and is an intermediate in the synthesis of polysaccharides. It also acts as the intermediate when starch or glycogen is broken down. Thus, it acts as a control in both directions i.e. to form polysaccharides or to release glucose. 2. Disaccharides like sucrose acts as a medium to transport carbohydrate in phloem. It can lower water potential and create hydrostatic flow in the sieve tube. It can be stored in plant cells especially sugar cane and sugar beet as food reserve. Sucrose is found in fruits and nectar to attract animals and it is not easily destroyed like glucose or fructose as it needs sucrase to digest into glucose and fructose. 3. Another disaccharide, lactose that is found in milk acts as food. Like all disaccharides, lactose needs to be hydrolysed before it can be used by babies. Thus, the control of glucose and galactose release is determined by the rate of hydrolysis of the disaccharide. 4. Starch is a polysaccharide stored as food reserve that is formed from α-glucose in the form of amylose and amylopectin. 5. Amylose is unbranched and coiled to form an a-helix but is easily hydrolysed by amylase to form maltose then by maltase to release glucose for entry into cells for respiration. Formation of glycosidic bonds in disaccharides and polysaccharides Disaccharides (One synthetase) 1. Maltose – α-1,4- glycosidic bond 2. Sucrose – α-1,2- glycosidic bond Polysaccharides (More than one synthetase) 1. Starch – amylose – α-1,4-glycosidic bond 2. Starch – amylopectin – α-1,4 and α-1,6- glycosidic bonds 3. Glycogen – α-1,4 & α-1,6-glycosidic bonds 4. Cellulose – β-1,4- glycosidic bond Summary Remember that reducing sugars (monosaccharides and disaccharides) have free carbonyl (aldehyde or ketone) groups. Remember the bonds in maltose and sucrose with 2 monosaccharide units. They can be used only when the glycosidic bond is hydrolysed. So, there is a control in its usage. Exam Tips CH2OH OH OH O O O O CH2OH OH Amylose OH O CH2OH OH OH O O O CH2OH OH OH O Figure 1.10 The structure of amylose 2015

Biology Term 1 STPM Chapter 1 Biological Molecules 1CHAPTER 14 Exam Tips Remember the differences in structures related to the functions of starch and cellulose. (STPM 2009) 6. Amylopectin is branched, consisting of a-1,4 and a-1,6-glycosidic bonds. Thus, it needs isomaltase to hydrolyse a-1,6-glycosidic bonds causing the release dependent on the control of another enzyme. CH2OH OH OH O O O Amylopectin CH2OH HO OH O O CH2OH OH OH OH O O O O CH2OH OH O CH2 OH OH OH O O O CH2OH OH O Figure 1.11 The structure of amylopectin 7. Starch is insoluble, compact and not osmotically active making it a good food reserve in the cell. This is especially so in seed and storage organs where glucose can be released for growth when conditions are suitable. 8. Inter-conversion of starch and maltose maintains water potential in the cells especially in the guard cells for the opening and closure of stomata. 9. Glycogen is an ‘animal starch’ which is found in the liver and muscles as food reserve. Glycogen is formed from α-glucose and its molecular structure is the same as amylopectin but more branched. 10. In the liver and the muscles, glycogen can be broken down to glucose phosphate by phosphorylase. Glucose phosphate can be used directly for respiration. 11. Glycogen in the liver can be converted to free glucose and released into the blood in between meals when blood glucose falls. After meal, glucose is absorbed and stored as glycogen. Thus, glycogen acts to control blood sugar level. 12. Cellulose is a polysaccharide, which is used to make the cell wall of plants. The molecules are cross-linked by hydrogen bonds to form microfibril as shown in Figure 1.12. Functions of disaccharides and polysaccharides Disaccharides (a) Maltose – two unit stage to form starch and glycogen – two unit stage to release glucose for respiration or change to other compounds (b) Sucrose – to transport carbohydrate in phloem so can generate hydrostatic pressure – to store food that can release glucose and fructose under the control of sucrase Polysaccharides (a) Starch – to store food in compact forms in plants cells (b) Glycogen – to store glucose in compact forms and easily release glucose for respiration in animals (c) Cellulose – to form cell walls for protection and support in plants Summary

Biology Term 1 STPM Chapter 1 Biological Molecules 1CHAPTER 15 Microfibril O CH2OH O β O O O O O O O -Glucose molecules Hydrogen bonds forming cross bridges CH2OH CH2OH CH2OH CH2OH Figure 1.12 The structure of cellulose 13. Cellulose performs a structural role in the form of cell wall of plants. In the cell wall, it protects the cell from both physical injuries and haemolysis. It helps in the support of plants through turgor pressure and mechanically when it is mixed with lignin. 2010 Quick Check 2 1. Why are all monosaccharides reducing sugars but not all disaccharides? 2. What are the differences between D-glucose and L-glucose? 3. Why do starch and glycogen make good food reserves? 4. How do cellulose form macrofibrils as in the cell wall of some plant cells? 1.3 Lipids 1. Lipids are organic compounds that are made up of carbon, hydrogen and oxygen. Unlike carbohydrate, the ratio of oxygen atoms to hydrogen atoms in a lipid molecule is much lower and have a common physical property of being soluble in non-polar organic solvent such as chloroform, ether or benzene. 2. Other physical properties are as follows: (a) They are insoluble in water and polar organic solvents. (b) Their densities are always less than water. (c ) They have high viscosity. (d) They are greasy and leave behind grease spots on paper. 3. Their chemical properties are as follows: (a) Their chemical structures differ between different categories. (b) Generally, they are esters formed from fatty acids and alcohol. Students should be able to: (a) describe the structures, properties and distribution of triglycerides, phospholipids (lecithin) and steroid (cholesterol); (b) state the functions of triglycerides, phospholipids (lecithin) and steroids (cholesterol); (c) differentiate between saturated and unsaturated fatty acids. Learning Outcomes

Biology Term 1 STPM Chapter 1 Biological Molecules 1CHAPTER 16 4. Examples of lipids are triglycerides (fats), phospholipids, steroids, waxes and terpenes. Structure, Properties and Distribution of Triglycerides, Phospholipids and Steroids (a) Triglycerides 1. Triglycerides are fats or oils formed from one molecule of glycerol and three molecules of fatty acids. The process of their formations is called esterification, which is a condensation between an alcohol group and a carboxylic group. 2. Esterification is also called lipogenesis in which one molecule of glycerol reacts with three molecules of fatty acids as shown in Figure 1.13. H C H C OH Glycerol O H + + + H H C O H H O C Fatty acids Triglyceride Water O R O H C R O H O C R O H O C R H C H H O H C O H C O O C R O C R + H O H + H + H O H O H Figure 1.13  Esterification (a) The process involves condensation of the three hydrogen atoms from the hydroxyl (OH) groups of the glycerol, each forming an ester bond with a hydroxyl group from three fatty acids. It is catalysed by a ligase. (b) This process takes place in the adipose tissue and can happen also in the liver. In plants, the process can occur in the chloroplast and seeds. (c) The three fatty acids in a triglyceride are usually the same type forming products such as tristearin, tripalmitin and triolein. (d) Tristearin, therefore, is made up of one glycerol and three stearic acids. 3. The physical properties of triglycerides are as follows: (a) Triglycerides are insoluble in water. (b) Their specific gravity is less than water, therefore they float on water. (c) They are soluble in non-polar organic solvents like acetone. (d) They form emulsion if shaken with alcohol. (e) They leave behind grease spots on paper. Structures, properties and distribution of triglycerides (TG), phospholipids (PL) and steroid (cholesterol) 1. Structures: (a) TG: Ester of 1 glycerol and 3 fatty acids (b) PL: Ester of 1 glycerol, 2 fatty acids, 1 phosphoric acid & 1 choline (c) Cholesterol: 4 fused rings with 1 hydroxyl group and 1 hydrocarbon tail 2. Properties: (a) All soluble in nonpolar organic solvent. (b) Lower density than water. (c) PL more soluble than cholesterol with TG as the least soluble in water. (d) TG – hydrophobic, PL – amphipathic & cholesterol little amphipathic 3. Distribution (a) TG – in adipose tissue under skin or around internal organs – in seeds and cuticle (b) PL – in membrane and blood (c) cholesterol – in membrane & fatty tissues Summary

Biology Term 1 STPM Chapter 1 Biological Molecules 1CHAPTER 17 4. Their chemical properties are as follows: (a) Triglycerides react with atmospheric oxygen and become rancid if kept for too long a period. (b) They react with Sudan III reagent to form a dark red complex. (c) They can be hydrolysed by lipase or boiling with dilute alkali to form glycerol and fatty acid. 5. In animal, triglycerides are stored as oil droplets in the cytoplasm of adipose cells. The adipose cells are found underneath the skin, in the mesentery or surrounding the intestines, kidney and even the heart. It can be found in between muscle fibres. 6. In plants, they are found in seeds and fruit walls such as in oil palm fruit, seeds of groundnuts, rapeseeds, sunflower seeds and kernels of all cereals. (b) Phospholipids 1. Phospholipids are lipids that contain phosphorus atoms usually in the phosphate form that can be removed by hydrolysing with phosphatase. They are the membrane lipids forming the basic bimolecular layers within all cell membranes. 2. There are two major types, the derivatives of phosphatidic acid and sphingolipid, for which the former is more common. 3. An example of the derivatives of phosphotidic acid is lecithin, which is phosphatidylcholine. 4. The molecular structure of lecithin is shown in Figure 1.14. Exam Tips Remember the structure of lecithin and its importance in cell membrane structure. Hydrophilic head Hydrophobic tail Hydrophilic head Hydrophobic tail Hydrophilic head Choline Phosphate ↑  Glycerol → Fatty acid chains → →  Hydrophobic tail H O | || H –C–O–C–C17H35 O || H –C – O – C – C17H35 CH3H H O– | | | | CH3– N+ – C – C – O – P ——O – CH | | | || | CH3 H H O H ↑ Figure 1.14 The structure of lecithin

Biology Term 1 STPM Chapter 1 Biological Molecules 1CHAPTER 18 5. As shown in Figure 1.14, phospholipids are derived from one molecule of glycerol bonded with two molecules of fatty acids, one molecule of phosphoric acid forming phosphatidic acid through ester bonds. Choline is an alcohol complex bonded to the phosphatidic acid forming lecithin. 6. The physical properties of phospholipids are as follows: (a) They are amphipathic, containing hydrophobic and hydrophilic groups. (b) They form a layer of micelle at the water surface if poured slowly on water as shown in Figure 1.15. Water Surface Hydrophilic head Hydrophobic tail Spherical micelle "Detergent" effect of micelle Lipid droplet Figure 1.15 Micelle of phospholipid (c) They are soluble in water forming emulsion of spherical micelles if shaken in water. (d) They act as ‘detergent’, helping to clear droplets of lipid in aqueous medium especially in the blood. (e) They can combine with protein, forming lipoprotein units to transport cholesterol. Low density lipoproteins transport cholesterol from the liver to tissues including arterial walls, forming blockage in arteries. High density lipoproteins transport cholesterol from tissues to the liver to be excreted. Remember the structure of phospholipid (STPM 2007 structure question) and its functions compared to those of triglyserides. Remember the structure of cholesterol. Exam Tips High density lipoprotein (HDL) Cholesterol Phospholipid Protein Less cholesterol More protein Phospholipid Low density lipoprotein (LDL) Figure 1.16 Phospholipids combined with proteins forming lipoprotein unit 7. Phospholipid is found in the cell membrane as bilayer. The membrane surrounds the cell and organelles and also within some organelles in both plant and animal cells.

Biology Term 1 STPM Chapter 1 Biological Molecules 1CHAPTER 19 (c) Steroids 1. Steroids are lipids that are made of four hydrocarbon rings as shown in Figure 1.17. Cholesterol HO 3 4 5 6 7 8 14 15 16 17 13 12 11 9 10 1 2 Hydrophilic part of molecule Hydrophobic part of molecule Figure 1.17 The structure of steroid and cholesterol 2. There are 17 atoms of carbon within the fused rings. They are labelled according to standard practice. Different steroids differ from one another from the different organic groups attached to the different carbon atoms. 3. Examples of steroids include cholesterol, testosterone, estrogen, progesterone, aldosterone, vitamin D, bile salts i.e. sodium glycocholate dan taurocholate and those found in plants known as phytosterol. 4. Properties of steroids: (a) They are insoluble in water as they are hydrophobic in nature. (b) Most steroids are soluble in organic solvents such as petroleum, ether and acetone. (c) They can be solified at low temperature. (d) They can form micelles with phospholipids and can be soluble in water or alcohol as emulsion. Functions of Triglycerides, Phospholipids and Steroids Function of Triglycerides 1. They can act as energy sources, in which they yield 38 kJ g –1 of energy compared to that of 17 kJ g –1 for carbohydrates. 2. They act as energy reserves in the adipose tissue of animals and seeds of many plants. They offer advantages of more quantity of energy, less space involved, insolubility in water and easily respire. 3. They can be used for the formation of other chemicals including glucose to maintain blood glucose level and amino acids to make proteins. They can be used to form all other chemicals especially in plants and herbivores after dormancy during winter.

Biology Term 1 STPM Chapter 1 Biological Molecules 1CHAPTER 20 4. They insulate the animal body from cold in the form of subcutaneous fat layer, which is a bad conductor of heat. This is to ensure survival in cold temperature especially aquatic mammals like seals as they are not covered with fur. 5. They protect organs such as the kidney from physical damage. 6. They waterproof both animals and plants. Our skin is covered with oil from sebaceous glands. Fur and feather of animals are covered with oil to make better insulators. Plants and other animals have cuticles of fat to reduce excessive evaporation. Functions of phospholipid 1. Phospholipid is the main component and forms the basic structure of the membrane with hydrophilic heads facing outside, hydrophobic tails opposite each other as in Figure 1.18. Outside of cell Inside of cell Cholesterol Lipid tail Phosphate head Protein Glycocalyx Figure 1.18 The membrane structure with phospholipid as basic component 2. It forms the bimolecular layer in the membrane in which other components are bonded to it. 3. The phospholipid molecules in the membrane are dynamic which means each of the molecules can move freely or they are fluid-like. 4. The molecules ensure stability in the membrane. The outer surfaces interact with aqueous medium on both sides whereas the inner layers are locked in non-polar bonding. The bilayer can easily reform even if the structure is temporarily broken. 5. Its lipid nature permits hydrophobic substances especially those of lower molecular size like methanol to diffuse easily across the membrane. It acts as a barrier for polar molecules and ions to cross the membrane. 6. The bimolecular nature of the phospholipid in the membrane allows reorientation of their molecules to form vesicles or the vesicles fuse with membrane become part of it. Thus, endocytosis and exocytosis can easily take place. Functions of Triglycerides (TG), Phospholipids (PL) and cholesterol 1. TG – food reserve to release glycerol + fatty acids for respiration or conversion to other compounds – insulation, protection and water proofing 2. FL – Basic structure of membrane and transport lipids 3. Cholesterol – regulatory function in membrane and precursor for steroids (e.g. steroid hormones) Summary

Biology Term 1 STPM Chapter 1 Biological Molecules 1CHAPTER 21 7. The phospholipid layer allows globular proteins called intrinsic proteins, with corresponding hydrophobic and hydrophilic surfaces, to span across the membrane. 8. Thus, the bilayer with such intrinsic proteins such as enzymes and transport proteins can respectively perform enzymatic functions and allow the passage of hydrophilic substances to cross the membrane. 9. Similarly, some extrinsic proteins that are attached to the outer or inner surface of the phospholipid layer can perform structural and enzymatic functions. 10. Short polysaccharides can bind with phospholipids or proteins. They serve as receptors or cementing substances. Such carbohydrates can also stabilise the membrane structure by forming hydrogen bonds with water molecules both outside and within the cell. Functions of steroids 1. Steroids such as sex hormones like testosterone, oestrogen and progesterone are required for maintaining sexual health. 2. Other steroid hormones from adrenal cortex, for example, the corticoids are required for glucose and mineral metabolism. Steroid hormones being lipids can pass through the membrane into the nucleus to activate genes. 3. Cholesterol is important in the synthesis of membrane structures as such it is for the general well being of the skin cells and body. 4. Steroids in the form of vitamin D in the skin is needed for the metabolism of calcium and phosphate. 5. Steroids such as bile salts, emulsify lipids in our food in the small intestine. This is due to the solubility of the hydrocarbon rings in lipid and the ionic heads in water. Emulsified lipids can be easily digested and absorbed. Differences between Saturated and Unsaturated Fatty Acids 1. Saturated fatty acids are those that contain only single bonds with no double bonds within their carbon linkages. Examples of saturated fatty acids are as follows: (a) Palmitic acid, CH3 (CH2 )14COOH (b) Stearic acid, CH3 (CH2 )16COOH, as shown below: Remember an example of saturated fatty acid is stearic acid and unsaturated fatty acid is oleic acid. Ester bond is formed between alcohol and acid. Esterification is the process of forming the ester bond. Exam Tips

Biology Term 1 STPM Chapter 1 Biological Molecules 1CHAPTER 22 2. Unsaturated fatty acids contain at least one double bond within their hydrocarbon chains. An example of unsaturated fatty acid is Oleic acid as shown below: 3. Saturated fatty acids have higher ratio of H:C compared to unsaturated fatty acids. 4. Saturated fatty acids solidified more readily when cooled. 5. Saturated fatty acids are for energy storage and they easily associate with cholesterol. 6. Saturated fatty acids are found in animal fats except fish oil. 7. Unsaturated fatty acids are found in plant oils such as corn and rapeseed. Differences between saturated and unsaturated fatty acids Saturated Unsaturated Single –C–C– bonds Double –C=C– bonds More H:C ratio Less H:C ratio Easier to solidify Less tendency to solidify Less reactive Easier to metabolise Found more in animals More in plants Summary Quick Check 3 1. Differentiate fat and oil. 2. What are the functions of phospholipids other than for the formation of membrane? 3. What are the advantages and disadvantages of having low levels of cholesterol in the diet? 1.4 Proteins 1. Proteins are polypeptides, biopolymers that are formed from amino acids. 2. Amino acids are carboxylic acids, which have amino groups. There are 20 amino acids used as basic units for protein synthesis. However there are over a hundred others. Structure and Classification of Amino Acids 1. The basic structure of amino acid is as shown below: H H O | | || H – N – C – C – O – H | R 2. Each amino acid is different from the other by having different R-side chains. Students should be able to: (a) classify amino acids into four main classes based on their side chains: polar, non-polar, acidic and basic; (b) describe the structure of an amino acid and the formation of peptide bonds in polypeptides; (c) explain the properties of protein (amphoteric, isoelectric point, buffer and colloid); (d) differentiate the various levels of organisation of protein structure (primary, secondary, tertiary and quaternary) and relate the functions of each structure to the organisation of proteins; Learning Outcomes

Biology Term 1 STPM Chapter 1 Biological Molecules 1CHAPTER 23 (e) explain the denaturation and renaturation of protein; (f) classify proteins according to their structures, compositions (simple and conjugated) and functions. Structure of amino acid H | NH2– C – COOH | R Carboxylic acid with amine Summary 3. The 20 amino acids are divided into four main classes depending on their R-side chains. 2010 (a) Non-polar / aliphatic group: glycine, alanine, valine, leucine, isoleucine, methionine and phenylalanine. (b) Polar group: serine, threonine, cysteine, asparagine, tyrosine and glutamine. (c) Acidic group: aspartic acid and glutamic acid. (d) Basic group: lysine, arginine, histidine, tryptophan and proline. 4. The four classes of amino acids are as shown below. (a) non-polar class Glycine (Gly) C H H C O O– H3N+ Alanine (Ala) C H CH3 C O O– H3N+ Valine (Val) C H CH CH3 CH3 C O O– H3N+ C H CH2 Leucine (Leu) CH CH3 CH3 C O O– H3N+ C H CH2 Isoleucine (Ile) CH CH3 C O O– H3N+ H3C Phenylalanine (Phe) CH2 C H C O O– H3N+ C H CH2 CH2 Methionine (Met) S C O O– H3N+ CH3 C H CH2 CH2 H2C Proline (Pro) C O O– H2N+ C H CH2 NH Tryptophan (Trp) C O O– H3N+ (b) Polar class Serine (Ser) C H CH2 OH C O O– H3N+ Threonine (Thr) C H CH OH CH3 C O O– H3N+ Tyrosine (Tyr) CH2 OH C H C O O– H3N+ C H SH CH2 Cysteine (Cys) C O O– H3N C – H CH2 NH2 Asparagine (Asa) C O C O O– H3N+ C H CH2 CH2 NH2 Glutamine (Glu) C O C O O– H3N+ C H CH2 CH2 NH2 Glutamine (Glu) C O C O O– H3N+ 2013 Classification of amino acids 1. Polar – e.g. serine with hydroxyl side chain 2. Non-polar – e.g. glycine with hydrophobic side chain 3. Acidic – e.g. glutamic acid with carboxylic acid side chain 4. Basic – e.g. Lysine with basic amine side chain Summary

Biology Term 1 STPM Chapter 1 Biological Molecules 1CHAPTER 24 (c) Acidic class C H CH2 Aspartic acid (Asp) C – O O C O O– H3N+ Glutamic acid (Glu) C H CH2 CH2 C – O O C O O– H3N+ (d) Basic class C H CH2 CH2 Lysine (Lys) CH2 CH2 NH3 + C O O– H3N+ C H CH2 CH2 Arginine (Arg) CH2 NH C = NH2 + NH2 C O O– H3N+ C H CH2 NH NH+ Histidine (His) C O O– H3N+ Formation of Peptide Bond in Polypeptides 1. Two amino acids can be linked by a peptide bond by condensation catalysed by enzyme to form dipeptide as shown in Figure 1.19. H NH H C R1 Amino acid 1 Amino acid 2 O C O H NH H C R2 O H + C O Dipeptide Water Peptide bond H NH H C R1 O C N H H C R2 O C O H + H O H Figure 1.19 Formation of a dipeptide Since there are 20 different types of amino acids, the number of different combinations of dipeptides is 20 × 20 = 400. Tripeptides and polypeptides can be formed from dipeptides. 2. Formation of polypeptide: (a) This is the protein synthesis in which amino acids are linked together to form a long chain of polypeptide or protein. (b) The process happens in cells. A gene controls the synthesis of each type of protein. (c) Human cells contain about 30, 000 such genes, which correspondingly control the synthesis of such large number of different proteins. Exam Tips Remember how amino acids form peptide bond and polymerisation.

Biology Term 1 STPM Chapter 1 Biological Molecules 1CHAPTER 25 (d) The sequence of amino acids in each type of protein is determined by the base sequence within a gene, a part of DNA. (e) The part of DNA has to be transcribed or copied into another form called mRNA before it is translated into protein by the ribosomes. (f) The transcription occurs in nucleus catalysed by RNA polymerase and translation occurs in the cytoplasm involving tRNA catalysed by a ligase enzyme. Properties of Proteins 1. Globular protein are soluble in water. However, as macromolecules, these proteins exhibit a hydrophillic surface preventing them to dissolve entirely. Instead, they disperse evenly throughout water forming colloid. Colloid exhibits the following effects: (a) Colloid if added to water form sols. When the amount of water is reduced by heating, it forms gel. (b) Colloid cannot be crystallised but form amorphous mass when dried and the protein is denatured by heating. (c) Colloid exhibits Tyndall effect i.e. if shone with light, cloudiness is observed. This is due to the reflection of light by the macromolecules. (d) Colloid exhibits Brownian movement i.e. if observed under the microscope, vibrations can be seen as the protein molecules obtain kinetic energy. 2. Proteins are amphoteric in water. This means protein molecules can behave like acid and alkali. Proteins contain carboxylic side chains that can behave like an acid. –COOH → –COO– + H+ They also contain amino side chains that can behave like an alkali. –NH2 + + H+ → NH3 + Therefore, the protein molecules are zwitterions in water, exhibiting both positive and negative charges. 3. At isoelectric point, which is when the total amount of positive charges is equal to the total amount of negative charges at a certain pH, proteins are least soluble in water or become insoluble precipitate. This is caused by the intermolecular bonding of the two charges. 4. Protein molecules can act as buffers i.e. they can resist pH changes when a small amount of acid or alkali is added. The presence of amino side chains neutralise acid. –NH2 + H+ → –NH3 + Formation of peptide bond in polypeptide protein synthesis → transcription → translation → amino acids are condendesed in fix sequence on ribosome and linked by peptide bond. Summary Properties of proteins 1. Colloid – solution of large molecules 2. Amphoteric – with both positive and negative charges due to side chains of component amino acids 3. Isoelectric point – least soluble due to intermolecular bonding of positive and negative charges 4. Buffer – can resist pH changes due to side chains that bind with H+ or OH– ions. Summary INFO Brownian Motion

Biology Term 1 STPM Chapter 1 Biological Molecules 1CHAPTER 26 The carboxylic side chains, on the other hand, neutralise alkali. –COOH + OH– → –COO– + H2 O Levels of Protein Structure 1. This refers to the four levels of organisation of the protein structures. The protein molecules are large, complex and can be divided into primary, secondary, tertiary and quaternary levels. (a) Primary structure (i) This refers to the sequence of amino acids in the protein linked together by peptide bonds. This also includes the number of amino acids, the number of chains and the position of the disulfide bonds that link the chains as in insulin. 2011 H s ala–ser s s s H – N+ – gly–ile–val–glu–gln–cys–s–s–cys–ser–leu–tyr–gln–leu–glu–asn–tyr–cys–asn–C Chain A: 21 amino acids Chain B: 30 amino acids H O O– H H–N–phe–val–asn–gln–his–leu–cys–gly–ser–his–leu–val–glu–ala–leu–tyr–leu–val–cys–gly–glu–arg–gly–phe–phe–tyr–thr–pro–lys–ala–C H O O– cys val Figure 1.20 The primary structure of bovine insulin, a protein (ii) The primary structure of each type of protein is different and unique. This is controlled by the base sequence of the gene. (iii) The primary structure determines the function of a protein as it controls the other higher levels of the structure i.e. it determines the shape of the protein molecule. So, the primary structure also determines the properties of a protein. (iv) Any changes in the primary structure can cause changes in the other levels. (v) For example, the primary structural change of the 6th amino acid in the β gene of haemoglobin from glutamic acid to valine, which is caused by a mutation leads to a genetic disease called sickle-celled anaemia i.e. the secondary, tertiary and quarternary levels are changed. Thus, the properties of the protein is also affected. (vi) The haemoglobin formed cannot fit in the valine, which is non-polar in nature. This results in the linking of haemoglobin molecules by non-polar interactions and elongation of red blood cells to become sickle-shaped.

Biology Term 1 STPM Chapter 1 Biological Molecules 1CHAPTER 27 Sickle-celled haemoglobin Sickle-shape red blood cell Chain of haemoglobin Valine Normal haemoglobin Normal red blood cell Figure 1.21 Formation of sickle-celled haemoglobin as a result of the replacement of one amino acid (b) Secondary structure (i) This refers to the α-helix coiling and the β-pleated sheet of the primary structure. (ii) The a-helix structure is formed by hydrogen bonding between the carbonyl group ( C=O) in one amino acid and the amide group ( N–H) of another amino acid four positions away. α-helix C C N C C C H O O O C O H N C C N C C H O N C N C β - pleated sheet O H N H O C O H C O N H C C C O C Figure 1.22 The secondary structure of proteins (iii) Such regularity may exist in the entire length of the primary structure as in fibrous proteins or just certain portions as in globular proteins like enzymes. Therefore, the proportion of a-helix structure in different protein varies. (iv) Fibrous protein like keratin has very high proportion of a-helix structures. The a-helix structures in keratin found in tendon, horn and hoof are cross-linked by hydrogen bonds or disulfide bonds to form hard and tough structures. (v) Likewise, the β-pleated sheet structure is formed by regular hydrogen bonding of adjacent primary structure that lies close to each other in an anti-parallel way. It also happens in intra-chain bonding after the primary structure is folded in a ‘z’ shape. Exam Tips Remember the primary, secondary, tertiary and quaternary structures of proteins with examples and bonds involved. Relate the protein structures to their functions. (STPM 2005 essay question)

Biology Term 1 STPM Chapter 1 Biological Molecules 1CHAPTER 28 (vi) β-pleated sheet may exist in small portions in globular proteins like enzymes. It may exist in its entire primary structure as in silk protein called fibrion. (vii) The higher the proportion of secondary structure, the more fibrous the proteins become. Fibrous proteins are insoluble in water except fibrinogen. They are tough and hard. (c) Tertiary structure (i) This refers to the folding and coiling of the already coiled primary structure of the polypeptide to form a globule, which has a three dimensional shape. Hence, tertiary structure is found in all globular proteins. With – pleated sheet With –helix coil Figure 1.23 The tertiary structure of protein (ii) Each type of protein has a unique three dimensional shape called conformation, which has a specific function. For examples, the functions of enzyme as a catalyst, insulin as a hormone and myoglobin as storage of oxygen are determined by the tertiary structure. (iii) Tertiary structure determines the globular shape of proteins. This also results in the solubility of the proteins. The outer surface has hydrophilic groups to interact with water and the inner part usually has hydrophobic groups. (iv) The bonds that are involved in the folding include the followings: • Disulphide bond, which involves two side chains of amino acid cysteine. The two SH groups form a disulfide (–S–S–) covalent bond with the removal of the two hydrogen atoms as shown in Figure 1.24. • Ionic or polar bond, which involves the side chains that are of opposite charges. For example, ionised –NH3 + bonds with –COO– . • Hydrogen bond, which involves the sharing of an electron of hydrogen atom between two side chains of different polar groups. For example, NH (amide) group with C=O (carbonyl) group. 2014

Biology Term 1 STPM Chapter 1 Biological Molecules 1CHAPTER 29 • Hydrophobic interactions which involve two side chains that are hydrophobic (non-polar). • Van der Waal’s forces which involve two side chains that are of very weak opposite charges or of very weak hydrophobic nature. CH H Side chain SH of cysteine Disulphide link SH C CH H S S H H H C H C O Ionic bond O H+ H H N Disulphide bond Ionic or polar bond Figure 1.24 Disulphide bond and ionic or polar bond involved in tertiary and quaternary structures Hydrogen bonds Hδ+ C Oδ– N Oδ– Oδ– H Hδ+ Oδ– H H δ+ N Hδ+ H Van der Waal’s force of attraction C Water excluded in hydrophobic interaction CH3 CH3 CH3 Hydrogen bond Hydrophobic interaction and Van der Waal’s forces Figure 1.25 Hydrogen bond, Van der Waal’s forces and hydrophobic interactions involved in tertiary and quaternary structures (v) The tertiary structure is determined by the primary structure which determines the position of amino acid. Therefore, any replacement of amino acid in the primary structure can affect the tertiary structure and its function. (d) Quarternary structure (i) This refers to protein with more than one polypeptide, each with its own tertiary structure, to form a functional macrounit. The overall structure is usually globular in shape. (ii) This structure is found in protein with more than one subunit. They can be of the same or different primary structures. (iii) Such primary structures may involve one or more genes. (iv) The bonds involved are the same as those involved in tertiary structure, which include disulfide bond, ionic bond, hydrogen bond, hydrophobic interactions and van der Waals forces. Levels of protein structure 1. Primary – sequence of amino acids, determines overall function. 2. Secondary – α-helix or β-pleated sheet, determines regular coiling or folding especially in fibrous proteins. 3. Tertiary – 3-D shape, determines function of globular proteins likes enzymes. 4. Quarternary – 2 or more polypeptides, determine the 3-D shape of some globular proteins. Summary

Biology Term 1 STPM Chapter 1 Biological Molecules 1CHAPTER 30 (v) For example, haemoglobin has four subunits of two different kinds i.e. 2 α-units and 2 β-units. There are two genes involved, α gene and β gene forming two types of polypeptides, α and β respectively. Each polypeptide has to be linked to a haem group before each of them can form a subunit as in Figure 1.26. 2018 Polypeptide chain 3 ( 2) Polypeptide chain 2 ( 1) Prosthetic group (haem) Polypeptide chain 1 (β1 ) Polypeptide chain 4 (β2 ) Figure 1.26 The quaternary structure of haemoglobin Denaturation and Renaturation of Protein 1. Proteins can be denatured by heat, extreme pH, chemicals and radiation. This is because the three dimensional structure or conformation of the proteins can be altered. The weak bonds that control the shape can easily be broken. The globular shape can unwind and each bond with one another and becomes more insoluble or precipitate as shown in Figure 1.27. Denaturation Globular structure Chain is loosen + + – + – + + – – Reacts with other Precipitate is formed Figure 1.27 Denaturation of protein 2. Factors that cause denaturation include: (a) Heat. Heat increases the kinetic energy of the protein molecules and causes the weak bonds to break. For example, enzymes cannot function at high temperature. (b) pH. Too high concentration of H+ or OH– breaks ionic and hydrogen bonds. (c) Heavy metallic ions. Examples like mercury and silver ions can easily attach to negative charged groups and break weak bonds. (d) Organic solvents. Examples like benzene and petroleum either form bonds with non-polar groups in the protein or disrupt hydrophobic interactions. Exam Tips Remember the four types of proteins: the fibrous and globular proteins, the simple and conjugated proteins; and their functions. Denaturation and renaturation of protein 1. Denaturation of protein means less function due to change in its shape. 2. Heat supplies kinetic energy and breaks hydrogen and ionic bonds of protein. 3. pH changes the H+ concentration also breaks hydrogen and ionic bonds. 4. Organic solvent breaks hydrophobic interactions. 5. Renaturation means that the denatured protein can be reversed to become functional. 6. Renaturation is only possible if the weak broken bonds can be reformed. Summary

Biology Term 1 STPM Chapter 1 Biological Molecules 1CHAPTER 31 Exam Tips Remember the amphoteric, buffering and colloidal properties of proteins. Factors causing denaturation of proteins include heat, extreme pH, chemicals and radiations. Exam Tips Collagen is an example of a fibrous protein, so it is a non-globular protein. (e) Radiation. X-ray and radioactive radiations can break weak and covalent bonds in the proteins. 3. Some proteins including globular proteins can withstand temperatures up to 80°C like those found in hot spring bacteria. Their tertiary structure contains more disulfide bonds and would not change shape and denature so easily. 4. Renaturation of protein is possible when the denaturation does not occur beyond the critical stage. This is especially so when the temperature is not too high or when the pH changes are minimal by lowering the temperature or reversing the pH changes respectively. 5. Protein molecules and amino acids have their own isoelectric point and could be charged. Hence, protein molecules and amino acids can be separated and identified through electrophoresis. 2018 Classification of Proteins 1. Proteins can be classified according to their structures. (a) Globular proteins. They are spherical in shape and soluble in water. Examples include enzymes, insulin, antibodies and haemoglobin. Their hydrophobic side chains are found within the ball-like structure. Their hydrophilic side chains are found on the surface interacting with water. They have tertiary or quaternary structure. (b) Fibrous proteins. They are hair-like in shape and some are in sheet or block forms as found in nail, hoof and horn. Another example includes collagen found in skin, tendons, cartilage, bones, teeth and the walls of blood vessels. Other examples include fibrion in silk, keratin found in hair, actin and myosin found in muscles. They are insoluble in water and most of them cannot be digested by proteinase. 2017 (c) Fibrous proteins that are soluble. An example is fibrinogen, a plasma protein used for blood clotting. 2. Proteins can also be classified according to their composition. (a) Simple protein.They are formed only from amino acids and no other non-protein component. Examples include insulin, enzymes like pepsin, fibrous proteins like myosin and collagen. (b) Conjugated proteins. They are formed not just from amino acids but a non-protein component called prosthetic group. Conjugated proteins can be divided into several categories depending on the types of prosthetic groups. Most of them form globular proteins and are soluble. The prosthetic groups include: (i) phosphate, forming phosphoproteins such as casein in milk. (ii) carbohydrate, forming glycoproteins such as mucus. (iii) lipid, forming lipoproteins in the membrane. (iv) haem, forming haemoproteins such as haemoglobin and myoglobin. Classification of proteins. Based on: 1. Structure • Globular – globule shape – soluble in water e.g. enzymes • Fibrous – fibre-like – insoluble in water e.g. nail 2. Composition • Simple – Only consists of amino acid e.g. insulin • Conjugated – + non-protein prosthetic group like haem e.g. haemoglobin 3. Function • Enzymic – Proteins catalysts e.g. amylase • Non-enzymic – Non-catalysts include hormones and antibodies Summary

Biology Term 1 STPM Chapter 1 Biological Molecules 1CHAPTER 32 (v) flavin, forming flavoproteins such as dehydrogenase with FAD. (vi) nucleic acid, forming nucleoproteins in the chromosomes and ribosomes. 3. Proteins can be classified according to their functions. (a) Enzymic proteins. All enzymes are protein catalysts with function to speed up biochemical reactions. Each enzyme has an active site to bind to a substrate which is then converted into a product. (b) Hormonal proteins. Some hormones like insulin, glucagon and growth hormone are proteins. They are required in small amounts to activate enzymes and control metabolic processes. (c) Transport proteins. They are found in the membrane like channel proteins and carrier proteins for transport of substances across the membrane. (d) Complement proteins. These are proteins involved in immune response. These include antibodies, interleukins, histamine and perforin. They are produced by white blood cells. (e) Structural proteins. They are part of the structure of cells or tissues. Examples of such proteins include soluble globular proteins in the cytoplasm. More noticeable ones are fibrous proteins such as collagen found in skin; and keratin found in hair. (f) Contractile proteins. These fibrous proteins are found in muscles. Examples are actin and myosin that enable muscles to contract. (g) Storage proteins. An example is casein found in milk, albumin found in eggs and gluten in aleurone found in cereal grains. Quick Check 4 1. What are the major differences among amino acids? 2. Why do globular proteins have hydrophilic amino acids on the outside and hydrophobic amino acids on the inside? 3. What makes fibrous proteins so hard? 4. Why does albumen turn white on boiling?

Biology Term 1 STPM Chapter 1 Biological Molecules 1CHAPTER 33 1.5 Nucleic Acids Nucleic acids are polynucleotides, biopolymers of nucleotides that include DNA and RNA. Nucleic acid is the important genetic material of all living organisms. Nucleotides and Formation of Phosphodiester Bonds 1. Nucleotides are biochemical compounds that are made of a base, a pentose sugar and phosphate (Figure 1.28). Sugar Base N-glycosidic bond P 4ʹ 5ʹ 1ʹ 3ʹ 2ʹ Figure 1.28 Bacis structure of a nucleotide 2. The bases are nitrogenous biochemicals that are grouped as purines and prymidines as shown in Figure 1.29. (a) Purine bases N N H Adenine N NH2 N N N H Guanine N O N H2N (b) Pyrimidine bases H Thymine (DNA only) N O NH CH3 O H Uracil (RNA only) N O NH O H Cytosine N NH2 N O Figure 1.29 The nitrogenous bases of DNA and RNA 3. Purines, i.e. adenine (A) and guanine (G), are bases with two fused hydrocarbon rings. Exam Tips Remember the structure of nucleotide in general (consisting a base, a pentose and three phosphates), specific examples (ATP, CTP, GTP, TTP, UTP) and other forms (ADP and AMP). Nucleotides Structure – Bases (A, C, G, T/U) – Pentose (ribose and deoxyribose) – Phosphate (1, 2 or 3) Types – Ribonucleotides (A, C, G and U) – Units in RNA Deoxyribonucleotides – units in DNA (A, C, G & T) Bases – Purines (A & G) – Pyrimidines (C & T/U) Summary Students should be able to: (a) describe the structures of nucleotides and the formation of phosphodiester bonds in a polynucleotide; (b) distinguish between DNA and RNA and the three types of RNAs (mRNA, tRNA and rRNA); (c) describe the structure of DNA based on Watson and Crick model. Learning Outcomes

Biology Term 1 STPM Chapter 1 Biological Molecules 1CHAPTER 34 Exam Tips Remember that the phosphodiester bond links nucleotides within DNA and RNA. Phosphodiester bond is between a pentose and phosphate or a nucleoside and phosphate within a mono-phosphate nucleotide. 4. Pyrimidines, i.e. cytosine (C), thymine (T) and uracil (U), are bases with one hydrocarbon ring. 5. The pentose sugars include ribose and deoxyribose. 6. There are two types of nucleotides i.e. ribonucleotides and deoxyribonucleotides. Ribonucleotides comprise of bases of A, C, G and U with ribose and phosphate while deoxyribonucleotides comprise of A, C, G and T with deoxyribose and phosphate. 7. There may be 1, 2 or 3 phosphate groups bonded to it. Therefore, nucleotides can exist as nucleotide monophosphates e.g. AMP; nucleotide diphosphates e.g. ADP or nucleotide triphosphates e.g. ATP. They are all soluble in water. 8. The bond between the base and the sugar is called the N-glycosidic bond, while the bonds between pentose and phosphates are called the phosphoester bonds. 9. Nucleotides can be hydrolysed to form phosphates and nucleoside by phosphatase or nucleotidase. 10. Nucleosides can be hydrolysed by nucleosidase to form bases and pentoses. Formation of phosphodiester bond in a polynucleotide 1. This is a process of polymerisation in which nucleotides are linked together to form polynucleotides by condensation with the formation of phophodiester bonds. 2. Figure 1.30 shows a condensation reaction between the hydroxyl (–OH) group of C3 of one nucleotide and the phosphoric acid group of another nucleotide forming a dinucleotide. The repeated condensation process later forms a polynucleotide. P P P P Sugar T Base Phosphate Sugar and phosphate join here by α-phosphodiester bond Sugar-phosphate backbone Polynucleotide chain C A G OH PO OH O CH2 O OH OH P O CH2 O Base OH H 2 Nucleotides O OH Base H H H H H H H OH PO OH O CH2 H2O Phospho diester bond O O P O CH2 O Base OH H Dinucleotide O OH Base H H H H H H H Figure 1.30 Formation of polynucleotide

Biology Term 1 STPM Chapter 1 Biological Molecules 1CHAPTER 35 3. There are two types of polynucleotides or nucleic acids formed i.e. Deoxyribonucleic acid (DNA) and Ribonucleic acid (RNA). 4. The formation of DNA is known as replication catalysed by DNA polymerase while that of RNA is known as transcription catalysed by RNA polymerase. (They will be discussed in Chapter 3.) 5. This occurs during the synthesis of DNA before cell division or synthesis of RNA especially messenger RNA before protein or polypeptide can be formed. DNA Structure and RNA Structure DNA structure 1. The DNA structure is based on the Watson and Crick model of double helix (Figure 1.31). 2. It consists of two strands of polynucleotides, which coil around each other on the same axis and interlock. 3. The nucleotides are of deoxyribonucleotide type, which compose of bases A, C, G and T. They are bonded by phosphodiester bond with deoxyribose and phosphates act as the backbone. 4. The two strands of polynucleotides are complementary to each other i.e. A in one strand is bonded by hydrogen bonds with T of another strand; and C bonded with G. Thus, A is paired with T with two hydrogen bonds and C is paired with G with three hydrogen bonds. 5. The two strands are anti-parallel i.e. in one strand, the carbon in the pentose is arranged in a 5’ to 3’ pattern while the other in a 3’ to 5’ pattern. 6. Each strand is very long, and can contain up to millions of nucleotides. 7. One coil of the strand is 34 nm long with ten pairs of bases. The diameter of the coil is 2 nm. 8. The structure of the molecule is very stable as the whole of both strands are linked by hydrogen bonds. 9. There are hydrophobic interactions between the bases in the centre aided by water molecules surrounding it. These interactions contribute to the stability of the molecule. 2011 Exam Tips Remember that the DNA structure is based on the Watson and Crick’s model of double helix. You need to know how to draw it. Formation of phosphodiester bond. 1. Between –OH group of C3 of pentose of 1 nucleotide and phosphoric acid group of another nucleotide. 2. It is a condensation reaction catalysed by polymerase, DNA template is required to synthesis DNA (replication) and RNA (transcription) Summary DNA Structure 1. Two strands coil to form α-helix. 2. Bases are A, C, G and T. 3. Complementary pairing of A = T and C G by hydrogen bonds. 4. Backbone of deoxyribose and phosphate is antiparallel. 5. Each strand contains more than a million nucleotides Summary 2014

Biology Term 1 STPM Chapter 1 Biological Molecules 1CHAPTER 36 O HO O P O HCH O O S N N C C N HN HN C Cytosine Guanine Thymine Adenine C HO H H H H C N N C N S O O PHO O O H OH HCH C C CH O PHO O HCH O O S N 1 O C HC NH N HN H C C HO H C N N C N S O O PHO O O H OH HCH C C CH CH 5 4 3 1 6 2 4 3 2 1 1 2 3 P P P P C G Hydrogen bond Deoxyribose sugar 5’ end 3’ end 3’ end 5’ end Sugar - phosphate backbone of one DNA strand Phosphate Organic base (guanine) T A G C C A T G 5’ 3’ 3’ 5’ C G A Sugarphosphate backbones Base pairs T C G T A A A T T T A T T GC C G GC A C G Hydrogen bonding of Arrangement of α-helix structure nucleotide pairs components in DNA of DNA Figure 1.31 The structure of DNA RNA structure 1. RNA consists of one strand of polynucleotide and is made up of A, C, G and U bases. The pentose is ribose. 2. There are three types of RNA i.e. ribosomal RNA (rRNA), transfer RNA (tRNA) and messenger RNA (mRNA). (a) rRNA (i) rRNA is found within the ribosome and consists about 80% of the total RNA inside the cell. They have long lifespan of several days before they are hydrolysed. (ii) Molecules of rRNA are of various sizes, some with more than 1000 nucleotides. (iii) They consist of intra-chain single-coiled sections and also double-coiled sections that are bound to protein molecules, forming the subunits of ribosomes. (iv) There are two types of ribosomes, the bigger eukaryotic (80S) ribosome and the smaller prokaryotic (70S) ribosome. The smaller ones are found in bacteria and inside mitochondria and chloroplasts. (v) The ribosomes in turn consist of two subunits. Eukaryotes: 80S → 60S + 40S Prokaryotes: 70S → 50S + 30S

Biology Term 1 STPM Chapter 1 Biological Molecules 1CHAPTER 37 (vi) The bigger ribosomal subunits of eukaryotes contain about 3 types of rRNA and the smaller subunits contain only 1 type. In that of prokaryotes, the 50S also contain 3 types of rRNA and only one in the 30S subunits. (vii) The rRNAs are transcribed from specific genes. These genes become the organisers for the nucleolus. (viii)The ribosomes in the eukaryotic cells are formed in the nucleolus. (ix) The role of rRNA is to form the ribosome. It enables the ribosome to combine with mRNA, to ‘read’ the codons in the mRNA and ‘translate’ the codons into sequences of amino acids in the primary structure of protein with the help of tRNA. So, rRNA forms ribosomes that provide two sites for tRNA to bind. Each tRNA can carry one specific amino to the site. (xi) Recent findings indicate that the rRNA in the ribosome functions as enzyme (ribozyme) where it catalyses the formation of peptide bond between the two amino acids brought to the ribosome. (b) tRNA (i) tRNAs are transcribed from genes in the DNA in the nucleus and are moved to the cytoplasm. They make up 10-15% of the total RNA in the cell. They have a shorter life span of several hours. (ii) They have a specific shape like a clover leaf. This is formed by the intra-chain coiling to form double strands, aided by hydrogen bonds where A pairs with U and C pairs with G. (iii) Their molecules are the smallest with about 80 nucleotides. (iv) There are 61 types of tRNA, 3 less than the number of genetic codes. 3’ 3’ C U A Amino acid attaches here C C G OH Hydrogen bonds Anticodon Unpaired bases A Two dimensional structure Three dimensional structure Amino acids attaches here 5 3’ ’ Anticodon Figure 1.32 The structure of tRNA RNA Structure 3 types i.e. rRNA, tRNA & mRNA – all single strand, consists of A, C, G & U with ribose – All are “copied” (transcribed) using DNA as template. rRNA – Mixed with protein to form ribosome – No fixed shape – May contain 1,000 nucleotides – No codons or anticodon tRNA – Not mixed with protein – Fixed clover leaf shape – About 80 nucleotides – Anticodon is found in central loop mRNA – Not mixed with protein – No fixed shape – Few hundreds to thousand bases – Contains codons used for determining primary structure of protein 2014 Summary

Biology Term 1 STPM Chapter 1 Biological Molecules 1 38 (v) Each type of tRNA has three bases at one loop called anticodon which is complementary to the codon of the amino acid it carries. (vi) At its 3’ end, there is a triplet sequence of CCA that can be bonded with a specific amino acid as shown in Figure 1.32. (vii) A tRNA transfers a specific amino acid to the ribosome, according to the codon in the mRNA, complementary to its anticodon to form a protein. Each tRNA can be reused to bind to amino acid and transfer it to the ribosome. (c) mRNA (i) mRNA is also transcribed from DNA and make up about 5% of the total RNA in the cell. (ii) It “copies” the information from at least one gene in the nucleus. Its sequence is complementary to one of the strands of DNA (coding or sense strand) which is exactly the same as the other strand. (iii) It then moves across the nuclear pore into the cytoplasm where it combines with a ribosome. The ribosome will ‘read’ the information that the mRNA carries with the help of tRNA and translates it into a protein. (iv) mRNA can be the longest RNA as it can copy several genes, each with about 300 bases. (v) It does not have a specific shape; it may coil or fold. (vi) It has the shortest life span in the cell. After several minutes, when enough protein is synthesised, the mRNA is hydrolysed, although some may last longer. (vii) mRNA is a transcript of a gene base sequence and it acts as template in which amino acid are joined into a protein of specific kind. Differences between DNA and RNA Table 1.1 The differences between DNA and RNA DNA RNA 1. It consists of double strands of polynucleotide. It consists of a single strand of polynucleotide. 2. It is a bigger molecule with more than a million bases. It is a smaller molecule with less than 1,000 bases. 3. It consists of an entirely double helix strand. It contains single and/or double helix of intra-strand. 4. The pentose within is deoxyribose. The pentose within is ribose. 5. The bases consist of A, C, G and T. The bases consist of A, C, G and U. Exam Tips Remember that mRNA is a messenger to ‘copy’ information from a gene and passes it to the ribosome for translation into a protein. rRNA is used to make ribosome. tRNAs transfer amino acids to ribosome for making protein.

Biology Term 1 STPM Chapter 1 Biological Molecules 1 39 DNA RNA 6. The ratio of A + G:C + T = 1:1. The ratio of A+G:C+U≠1:1. 7. Almost all DNAs are found in the nucleus. RNAs are found in both nucleus and cytoplasm. 8. Its amount is constant in all cells while gametes have half the amount. Its amount varies among cells and can be altered. 9. It is chemically stable. It is not so stable. 10. It will not be broken down, in living cells. It can be broken down by enzymes. 11. There is only one type. There are three types i.e. rRNA, tRNA and mRNA. Quick Check 5 1. What are the differences between purine and pyrimidine bases? 2. How does a DNA behave inside the nucleus? 3. List the differences between the three types of RNA. 1.6 Analytical Technique Paper Chromatography 1. Chromatography is a technique used to separate mixtures of chemicals of similar nature (e.g. photosynthesis pigments) by allowing their common solvent to flow over them in a solid medium such as paper. 2. Chromatography can separate other mixtures which include proteins, amino acids, nucleic acids, nucleotides, fatty acids, monosaccharides and disaccharides. Principles of Paper Chromatography in Pigment Separation 1. The solid medium used is a piece of paper cut to fit into a boiling tube. The paper of cellulose fibers is porous and absorptive of liquid. 2014 Students should be able to: (a) describe the basic principles of paper chromatography in pigment separation, electrophoresis for protein and nucleic acid separation. Learning Outcomes

Biology Term 1 STPM Chapter 1 Biological Molecules 1 40 Exam Tips Remember that electrophoresis is used to analyse antibodies within the blood and to diagnose sickness. It is also used for DNA fingerprinting. 4.5 cm 1 cm 1 cm 2 cm 10 cm Extract is applied here Filter paper Extract Carotene (orange) Rf = 0.95 Xanthophyll (yellow) Rf = 0.71 Chlorophyll a (blue green) Rf = 0.65 Chlorophyll b (yellow green) Rf = 0.45 Phaeophytin (spoiled chlorophyll -grey) Rf = 0.83 Solvent (a) Paper chromatography for leaf extract x x x x x x x Figure 1.33 Paper chromatography for leaf extract 2. A small but concentrated amount of leaf extract is applied on one end of the paper. The paper is hung touching a common solvent such as petroleum ether. The solvent moves up, separating the solute molecules as indicated by the different coloured spots. 3. The pigment molecules separate due to their differential adsorptions on the medium used. The reasons why their adsorptions differ are as follows: (a) Solubilities. The more soluble the pigment in the solvent, the faster it can move. (b) Molecular size. The smaller the size of the pigment, the faster it can move. (c) Charges. If the pigment is of similar charge with the medium, the faster it can move. If the pigment is of opposite charge with the medium, there would be an attraction between them thus slowing down its movement in the medium. 4. Rf is used to identify a pigment when a certain solvent is used. It is the ratio of the distance travelled by the pigment to that travelled by the solvent. Rf = Distance travelled by the pigment Distance travelled by the solvent Rf is a constant and when Rf of an unknown pigment spot is compared with Rf of standard known pigment, its identity can be determined. 5. Paper chromatography is useful because of several reasons as follows: (a) The technique is simple and can be easily carried out. (b) Other physical separation is almost impossible to carry out. (c) It takes a short time. A simple separation of leaf pigments takes less than 30 minutes. (d) Paper chromatography requires only simple apparatus such as paper, dropper and boiling tube.

Biology Term 1 STPM Chapter 1 Biological Molecules 1 41 6. Paper chromatography has its limitations. (a) Only small amounts of pigments can be separated at one time. (b) Certain pigments are too similar like chlorophyll a and b that they tend to overlap and are not easily separable by this technique. Electrophoresis 1. Electrophoresis is a technique used to separate substances with different charges such as amino acids, proteins and fragments of DNA using an electric field (Figure 1.34). 2018 Electrodes – Cathode Wells for sample 80 A B C D 70 60 50 40 30 25 10 5 + Anode Agarose gel Unknown DNA sample size Chamber filled with buffer solution Lane A of known fragment size (kilobase pairs) Figure 1.34 Electrophoresis 2. Mixtures such as amino acids and nucleic acid fragments, especially DNA fragments are separated for DNA fingerprinting. 3. The medium used is usually agarose gel layer or gel in a column even though paper can be used (Figure 1.34). 4. Electrodes are placed on both ends of a box of cooled gel. Small wells are cut at one end and filled with mixtures to be separated. Acidic buffer is added to change the mixtures of proteins or nucleic acids to positive charge. Dye is added for identification. When the electricity is switched on, the charged molecules move to the electrode of the opposite charge. 5. Its uses are as follows: (a) It is very useful to separate proteins, as they are delicate. Enzymes separated by this technique are still active. (b) It is used to diagnose diseases, as when blood plasma proteins are separated, extra proteins found could be antibodies formed to combat certain pathogens. The antibodies are compared with the standard ones and extracted to determine the actual type. (c) In forensic science, it is used for DNA-fingerprinting to identify individuals. A long DNA molecule is cut into different lengths and is separated to form a specific pattern according to the enzyme used to cut it. Paper chromatography 1. Separate pigments based on adsorption on paper 2. Most soluble moves fastest 3. Bigger size moves slower 4. Same charge with paper moves faster Electrophoresis 1. Separate proteins and fragments of DNA using direct current 2. Agarose gel is used as medium 3. Mixtures are put at one end 4. Negative charged protein or DNA fragments move to positive end (anode) 5. Smallest molecules move further away Summary 2015