Chapter 1 Biological Molecules PDF

Chapter 1 Chapter One- biological macromolecules Classes of carbohydrates Functional group- Carbonyl group Biological Macromolecules= polymers Aldose ALL...

Chapter 1 Chapter One- biological macromolecules Classes of carbohydrates Functional group- Carbonyl group Biological Macromolecules= polymers Aldose ALL key components of every living cell made from macromolecules. Carbonyl group is located at the end of carbon chain (e.g. glucose) Poly= many Ketose Mer= pieces Carbonyl group is located in the middle of carbon chain (e.g. fructose) polymers= molecules that consist of multiple monomers Numbering of carbon chain begins with carbon nearest to carbonyl group Dehydration= condensation—> two monomers bond together through loss of water molecules (form bonds) Hydrolysis= break bonds through gain of water molecules Carbohydrates= Carbon, hydrogen, oxygen Ratio of C,H, & O 1:2:1 (CH2O)n n= number of C atoms Monosaccharides 1 sugar molecule, simple sugar (3-7 atoms) Pentose and hexose can be converted into ring structure but triose can’t form Characteristics a ring structure because it is unstable. small sweet Function of monosaccharides: soluble in water energy source can be crystallized glucose- major respiratory substrate, primary energy source white/colourless basic building units or monomers for disaccharide and polysaccharide. reducing sugar- (Small carbohydrates usually containing one or two sugar units) Capable of acting as reducing agents such as Ag+ or Cu2+. Capable of acting as reducing agent. Disaccharides Starch 2 or more sugar moleules, Energy storage in plants or two monosaccharides Plants convert excess glucose into starch for storage in chloroplast joined by condensation (removing of water) Monomers- alpha glucose molecules formed by a strong covalent bond- glycosidic bond broken down by hydrolysis process with the help of amylase enzyme can be broken into monomers by hydrolysis (adding water) (found in saliva) Characteristics: Starch is a mixture of: sweet Amylose soluble in water Amylopectin can be crystallized white/colourless Amylose Monomers- alpha glucose Maltose Joined by alpha 1,4 glycosidic bonds made of alpha glucose+ alpha glucose Helical due to formation of hydrogen bonds Joined by alpha 1,4 glycosidic bond (carbon 1 and carbon 4) unbranched chain (variable length of monomers) folded- very compact (ideal for storage) Polysaccharides: Insoluble, coiled, compact Many sugar molecules, polymer that varies in length Joined together by condensation ( removing water Amylopectin molecules) Monomers- alpha 1,4 glycosidic, alpha 1,6 glycosidic bonds form covalent glycosidic bond Helical, branched chain (variable length of monomers) formaytion of polymer (large molecules made up of Branches occur within 30 units repeating units of monomers (polymerization) Folded & branched- very compact (ideal for storage) can be broken down into monomers by hydrolysis (addition Insoluble, coiled, compact of water molecules) Characteristics: Glycogen Large molecule Energy storage in animals (liver & muscles) Not sweet Structure- similar to amylopectin Insoluble in water (form colloid) More branched than amylopectin Cellulose Fatty Acids Major component of plant cell wall Has two regions Monomers- beta glucose Hydrophlic head (carboxyl group, COOH at 1 end) beta 1,4 glycosidic bond Hydrophobic tail, long unbranched hydrovarbon chain (known as side chain. R) Broken down by hydrolysis with the help of cellulase enzyme Hydrophobic tail varies in length unbranched, long straight chain Different fatty acids has different number of C atoms (usable 16 or 18) Straight chains are parallel to one another linked together by hydrogen bonds (form microfibril) Classification of fatty acids: Very stable and touch structure that provides support to the plant cells based on presence of double bonds within hydrocarbon chain Saturated fatty acids- no double bond between C atoms Functions of polysaccharides; Has maximum number of H atoms, straight chain, carry acids are closely packed Energy storage together. Solid at room temperature major component of cell walls, provide structural support to plant cells Mostly animal fats (butter, lard) Tests for starch- iodine Unsaturated fatty acids- has double bonds between C atoms presence of starch is indicated in blue black colouration Double bonding causes bending cannot be closely packed Benedict’s test- reducing sugar liquid at room temperature blue colour, then will become brick red when there is reducing sugar. Mostly plant and fishes fat (oil) Biuret test- When protein reacts with copper (II) sulphate positive test forms violet coloured complex. classification of fatty acids based on ability to synthesise: Emulsion test Essential fatty acids: Positive if there is cloudy white emulsion, negative is if no emulsion. cannot be synthesised in body must be present in diet Lipids E.g. linoleic, linolenic, arachidonic acids Consists of mainly carbon and hydrogen atoms, few oxygen atoms Group of hydrophobic molecules- insoluble in water Non-essential fatty acids can be synthesised in body Fats/triglyceride/triacylglycerol e.g. stearic acids and others most abundant Phospholipids made up of 1 glycerol and 3 fatty acids major component of cell membrane e.g. lecithin amphipathic molecule (comprising of hydrophobic and hydrophilic parts) Glycerol/Glycerine made of 1 glycerol, 2 fatty acids, phosphate group 3 C alcohol with 3 OH groups (soluble in water) Triglyceride Group of amino acids composed of 3 fatty acids and 1 glycerol backbone Essential amino acids Formed when 3 condensation reaction occurs which is known as esterification Cannot be syntehisied in sufficient amount, needs to be obtained through food Water removed by extracting OH from carbonyl group (fatty acid) and H from or supplement intake glycerol’s hydroxyl group isoleucine, leucine, lysine, methionine, phenylalanine, theoronine, tryptophan, Triglyceride can be broken down by hydrolysis valine infants: histidine Lipids Main energy storage in animals (due to higher number of hydrogen atom) Non-essential amino acids 1 g fat > twice energy than same weight of starch can be synthesised by body Lighter, hydrophobic so doesnt associate with water alanine, arginine, asparagine, aspartic acid, cysteine, glutamic acid, proline, Improve buoyancy in aquatic animals serine, tyrosine, glutamine provide better thermal insulation in mamals acts as padding for internal organs Formation of dipeptide Phospholipid is major component of plasma membrane 2 amino acids joined by condensation Steroid- some are hormones that regulate metabolism reomval of hydroxyl group from carboxyl end of one amino acid and hydrogen from amino group of another amino acid Protein form a covalent peptide bond. Building blocks are amino acids Breakdown of dipeptide Joined together to form long chain of polypeptides hydrolysed into 2 amino acids by hydrolysis Protein consist of 1 or more polypeptide chain that is folded or coiled into peptide bond is broken and produces two amino acids specific conformation Mostly composed of C,H,O,N, sometimes S Formation of polypeptide many amino acids joined by condensation process to form polypeptide Amino acids peptide bonds consists of 4 components attached to central carbon- hydrogen atom, carboxyl polypeptide backbone- repeated sequeance of (-N-C-C-) group(acidic) , amino group (basic), variable R group (or side chain) Amphoteric, have both hydrophilic and hydrophobic side All have basic strucutre but different side chain amino aicds grouped based on properties of R group Primary structure Maintained by 4 interactions between R groups linear sequence of aminno acids joined by polypeptide bonds within a disulfide bonds(covalent bonds( between R gropu and sulfhydrl groups polypeptide chain Ionic bonds between postively and negatiely charged R groups unique sequence of amino acids determined by specific DNA code Hydrogen bonds between sigma positive and negative R groups Different sequecne, different formation of bonds Hydrophobic/Vander waals interaction between non-polar R group causes polypeptide chain to fold and coil into unique 3 dimensional shape of protein Sickle cell anemia is a mutation of one of the sequence( supposed to be hydrophlic but its hydrophobic Secondary structure Liner chain of amino acieds coil or fold spontaneously due to formation of H bonds between backbones. (between H of NH groups and O of C=O groups) two types double helix beta pleated sheet Alpha helix helical coil H bonds formed between 1st amino and 4th amino stretches under tension, provides elascticity hydrogen bonds are reformed e.g. keratin, nails, wool E.g.- enzymes, hormones, antibodies Beta pleated sheet zig zag pattern- provides strength and felxibiltiy e.g. silk, spider web Tertiary structure 1 polypeptide chain further coiled into globular shape, maintained by bonds and interactions among R groups Globular shape unique for polypeptide chain Quaternary structure single protein formed when two or more polypeptide chain joins Conjugated proteins bonds same as tertiary structure protein that can function well if join with other compound e.g. haemoglobin (consists of 4 polypeptide chains, two alpha non protein compound- prosthetic group chains, and two beta chains Effect of pH and temperature Structure of protein influenced by Classification of protein 1. pH Fibrous protein 2. temperature mostly secondary structure long parallel filaments/strands Globular protein, relatively unstable because maintained by weak ionic bonds, hydrogen insoluble and stable bonds, and hydrophobic interactions mechanical and structural support Protein is heated and exposed to extreme pH changes then bonds are broken e.g. keratin, myosin (basic structure of muscles), collagen, silk Causes the protein to uncoil and change its conformation, it then loess its biological function and is denatured (protein may also coagulate) Globular protein usually irreversible (sometimes renaturation occurs) mostly tertiary.quaternary strucutre folded to form compact spherical shaped relatively unstable relatively soluble (colloid) metabolic and chemical process e.g. enzyme, hormone, antibodies. haemoglobin Enzyme Enzyme activity is maximum at optimum temperature and pH. Biological catalyst produced by living cells that speeds up the rate of Enzyme is destroyed at extreme temperature and pH. chemical reaction by lowering the actication energy Acts faster, one enzyme molecule can catalyse thousand or more Two sites reactions in a second. active site- specific site on enzyme that binds to specific substrate allosteric site- other site other than active site, place to bind for non Amylase (saliva)- optimum pH 7 competitive inhibitors Pepsin (stomach lining)- Optimum pH 3 Activation energy minimum amount of energy required by reactant (substrate) to start a Specificity of enzyme is duye to complementary shape of active site and chemical reaction, in order for reaction to occur, requires energy substrate How does it work?? When enzyme is denatured by heat or pH change, shape of active site Substrate collides with enzyme in active site is changed. substrate no longer fits into active site. enzyme facilitates breaking/forming of bond in substrate to form product if without enzyme, reaction can still occur but will need alot more energy Enzyme action Lock and key model Substrate is key, fits exactly in lock (enzyme) Active site completely complementary to substrate. Steps: An enzyme collides with its substrate molecule Substrate binds to active site of enzyme Active site of enzyme exactly complementary substrate enzyme- substrate complex formed reaction occur where the substrate reacts within complex and product is released enzyme is not changed or damaged and can be reused. Properties of an enzyme: further coiled and compacted globular protein acts as biological catalust required in small amounts to catalyse reactions Enzyme does not change after reaction (reversible catalyst if can be used again, irreversible if can’t form bond again due to inhibitors) Enzyme is specific to substrate Induced fit model Factors affecting enzyme reactions: Attachemnt of substrate induces conformation change in active site of enzyme molecule temperature until its active site fits with substrate. (e.g. hydrogen peroxide) inhibitors active site is not exactly complementary to substrate pH attachment of substrate to active site of enzyme induces conformational change in substrate concentration enzyme’s active site. enzyme concentration Steps: active site of enzyme is not fully complementary to shape of substrate Nucleic acids Enzyme collides with substrate molecule polyneucleotide (polymer of nucleotides) form enzyme- substrate complex made of nucleotide binding induce a slight change in shape of enzyme composed of C,H,O, P, and N allows substrate to fit enzyme precisely active site changes shape and becomes fully complementary with substrate Structure of nucleotide: enables enzyme to carry out catalytic function pentose sugar, nitrogeneous base, phosphate group product is formed and enzyme changes or reconverts back to original conformation single ring myst always be complementary with double rings Single ring Pyramidine 1. Cytosine (C) 2. Uracil (U) (RNA) 3. Thymine (T) (DNA) Purine 1. Guanine (G) 2. Adenine (A) DNA- Guanine, adenine, cytosine, thymine RNA- Guanine, adenine, uracil , cytosine Dinucleotide 2 nucleotides combined by condensation process between OH group in phosephate (1 nucleotide) and other OH group at carbon 3 in pentose sugar of other nucelotide joined by phosphodiester bond  Formation of polypeptide Both chains held together by hydrogen bonds between complementary base pairs many nucleotides joined together Adenine with thymine elongation from 5’3 five and three prime Cytosine with guanine 5' carbon has a phosphate group attached to it and the 3' carbon a hydroxyl Between adenine and thymine (2 hydrogen bonds) (-OH) group Between cytosine and guanine (3 hydrogen bonds) form a backbone with repeating sugar phosphate units specific base parining rule, A=T, G=C breakdown by hydrolysis process DNA- carries genetic information that controls activites of a cell DNA structure based on Watson & Crick model RNA consists of 2 polynucleotide chains single-sstranded polynucleotide both chains twisted to form double helix pentose sugar- ribose each polynucleotide chain is made of nucelotide guanine, adenine. cytosine. uracil nucleotides joined by phosphodiester bond each ful turn of double helix has 10 base pairs Three types of RNA 2 polynucleotides are arragned in oppostite direction so antiparallel Messenger RNA (mRNA) one strand ends with 3 prime and other strand ends with 5 prime Ribosomal RNA (rRNA) sugar-phosphate forms backbone Transfer RNA (tRNA) 2 backbones are outside, nitrogenous bases are paired inside the helix. mRNA carries genetic information copied form DNA which acts as template for protein synthesis rRNA forms ribosomal subunits (together with proteins) tRNA Phosphodiester bond Transfer specific amino acids to ribosome during protein synthesis  Function of ATP Chemical- supplies energy to make macromolecules Transport- supplesi energy to transport substances across cell membrane Mechanical- Supplies energy to allow muscle contraction =, sepration of chromosome Regeneration of ATP Continual breakdown and regenration of ATP is ATP cycle ATP broken apart, energy released and left with ADP ATP charged battery, ADP and phosphates are 2 parts of dead battery Carbohydrates and fats are high energy storage moelcues that are burned to generate ATP. Cells need suppy of fat and carbohydrates to make enough ATP to keep cell alive. DNA replication to produce two identical copies of DNA molecule essential for cell division during growth and repair of damaged tissues DNA replication ensures each new cell receives its own copy of DNA DNA is semi conservative, original DNA strand are split in two and each half is used as template to make complementary strand (other half) ATP (Adenosine Triphosphate) energy used by all cells organic molecules containing high energy phosphate bonds Chemical structure of ATP Composed of three components center is sugar, ribose attached to one side is base other side is attached to a string of phosphate groups phosphates are key to activity of ATP  H atom of one atom attracted to O atom of nearby water molecules. Water water molecules held together held together by hydrogen bonds 1 oxygen atom & 2 hydrogen atoms (H2O) H bond is weak, represented by dotted lines Joined by sharing electrons (covalent bonds) 1 water molecule can form maximum 4 hydrogen bonds with 4 3 atoms that form a triangle water molecules 104.5 degrees Properties of water: 1. Universal solvent 2. High specific heat capacity 3. High latent heat of vaporization 4. High surface Tension 5. Maximum density at 4°C 6. Low viscosity Universal solvent Sharing of electrons between O& H atoms are not equal powerful solvent for ionic and polar substances (e.g. NaCl) O atoms slightly negative NaCl held together by ionic bonds between Na and Cl H atoms slightly positive Why? Unequal charge distribution within a molecule (polar) water is polar molecule O atoms are slightly negative and H atoms are slightly positive O atoms attracted to positively changed sodium ions H atoms attracted to negatively charged chloride ions ionic bonds between NaCl become weaker water molecules gather around sodium and chloride ions forming a hydration shell that separates the ions from one another these ions pull away from the salt crystal thus the ions separate and dissolve Importance? acts as transport medium for solutes (e.g. Dissolved nutrients carried throughout body in blood plasma facilitates chemical reactions ( solute are more reactive when dissolved.  High surface tension Hydrogen bonds make water molecules stick to each othe—> cohesion High specific heat capacity on surface, water has greater attractio for each other than for molecules Definition —> Amount of energy needed to increases in air temperature of 1 g of a substance by 1 degrees C water is pulled downward by other water molecules beneath them Large amount of energy is needed to increase the They produce a strong layer on the surface which is hard to break--> temperature of 1g of water by 1 degrees C creates surface tension Heat absorbed to break Hydrogen bonds between water Water moelcules stick to other substances--> adhesion molecules Importance? Value= 1cal/g (most substances have lower value) Allow some organism to move on water (e.g. water strider) Large body of water (e.g. lakes/ocean) have relatively constant temperature(temperature rises and falls slowly) Maximum density at 4 degrees C although their environment temperature changes. as water cools, molecules get closer together Importance? reach maximum density of 4 degrees C Organisms can maintain their normal body temperature at temperature below 4 degrees, water molecules expand until it freezes Aquatic organisms can live in relatively stable temperatures and solidifies at 0 degrees (protected from rapid temperature changes) Hydrogen bonds in ice space the water molecules far apart and become stable High Latent heat of vaporization ice is less dense than cold water-> floats Defination—> Amount of energy needed to change 1 g of Importance? substance rom liquid to vapor phase allows aquatic life to live under frozen surface of water in cold climate A lot of energy needed to change 1 g water from liquid to region vapor floating layer of ice insulates the water below, preventing them from Heat absorved to break Hydrogen bonds between water losing heat and freezing molecules so they can move faster and be vaporised Value: 540 cal/g Low viscosity Importance? Weak hydrogen bonds between water moelcules are constantly breaking cooling effect in organisms due to sweating (animals)or and reorming transpiration (plants) water molecules can slide easily over each other As water absrobs heat, energy increases flows with less friction through narrow vessels when water evaporates. takes a lot of heat energy away—> acts as medium of transportation in living organisms (e.g. blood easily lowering temperature of organism flows in the circulatory system) acts as a good lubricant to reduce friction within the body (e.g. mucus facilitates movement of feces through the bowel  : CHAPTER 1: BIOLOGICAL MOLECULES Ainul Farhana 1  1.0 Biological Molecules a) Monomers & Polymers (1/2) b) Carbohydrates (1/2) c) Lipids (1/2) d) Proteins & Enzymes (1) e) Nucleic Acids (1/2) f) ATP (1/2) g) Water (1/2) AINUL FARHANA 2  Learning Outcomes By the end of this chapter you should be able to: a) Explain what a monomer and polymer are. b) Identify some biological polymers, and the monomer from which they are made c) Explain the concept of condensation and hydrolysis reactions in forming/breaking down polymers. AINUL FARHANA 3 Learning Outcomes: a) Explain what a monomer and polymer are. Macromolecules are polymers All key components of every living cell are made of macromolecules. Biological Macromolecules are also known as POLYMERS. Poly = many and mer = pieces. Polymers= molecules that consists of multiple monomers. There are four major types of biological macromolecules: Proteins Carbohydrates Lipids Nucleic acids AINUL FARHANA 4 Learning Outcomes: b) Identify some biological polymers and the monomer from which they are made (Carbohydrate, proteins, nucleic acid) Functions of macromolecules Energy storage Structural Movement Support Functions Growth Protections Transport AINUL FARHANA 5 Learning Outcomes: c) Explain the concept of condensation and hydrolysis reactions in forming/breaking down polymers Monomers In living cells, a small set of monomers is used to create a variety of polymers. Each polymer is unique in the number and type of monomers used to build it. Macromolecule Monomers Carbohydrates monosaccharides Lipids glycerol, fatty acids Proteins amino acids Nucleic acid nucleotides AINUL FARHANA 6 Learning Outcomes: c) Explain the concept of condensation and hydrolysis reactions in forming/breaking down polymers The Synthesis and Breakdown of Polymers Dehydration=condensation A dehydration reaction occurs when two monomers bond together through the loss of a water molecule Purpose? = to form bonds Polymers are disassembled to monomers by hydrolysis, a reaction that is essentially the reverse of the dehydration reaction To break bonds AINUL FARHANA 7 Learning Outcomes: c) Explain the concept of condensation and hydrolysis reactions in forming/breaking down polymers AINUL FARHANA 8 Learning Outcomes: c) Explain the concept of condensation and hydrolysis reactions in forming/breaking down polymers AINUL FARHANA 9 OMBEA RESPONSE 1. What is the name of reaction process to form bond between two monomers? - - - - A) Hydrolysis B) Condensation C) Hydrogen bond D) Water A B C D AINUL FARHANA 10  1.0 Biological Molecules a) Monomers & Polymers (1/2) b) Carbohydrates (1/2) c) Lipids (1/2) d) Proteins & Enzymes (1) e) Nucleic Acids (1/2) f) ATP (1/2) g) Water (1/2) AINUL FARHANA 11  Learning outcomes a) Identify common monosaccharides, disaccharides and polysaccharides and explain their basic structure b) Represent the structure of α -glucose and β –glucose diagrammatically. c) Explain what is meant by a glycosidic bond and how polymerisation of α-glucose can form starch or glycogen, whilst polymerisation of β-glucose can form cellulose. d) Describe and perform the tests for starch, a reducing and non-reducing sugar in detail (Benedict’s test). AINUL FARHANA 12 Learning Outcomes : (a) Identify common monosaccharides, disaccharides and polysaccharides and explain their basic structure Introduction  Composed of carbon, hydrogen & oxygen atoms  Ratio of C, H & O ~ approximately 1:2:1 (CH2O)n  n = number of C atoms AINUL FARHANA 13 Learning Outcomes : (a) Identify common monosaccharides, disaccharides and polysaccharides and explain their basic structure Classes of carbohydrates: Classification Monosaccharide Disaccharide Polysaccharide 1 sugar unit 2 sugar units Many sugar units AINUL FARHANA 14 Learning Outcomes : (a) Identify common monosaccharides, disaccharides and polysaccharides and explain their basic structure MONOSACCHARIDES  1 sugar molecule; simple sugar (3-7 carbon atoms)  Characteristics: 1. Small 2. Sweet 3. Soluble in water 4. Can crystallized 5. White / colourless 6. Reducing sugar Reducing sugars are small carbohydrates (usually containing one or two sugar units) that are capable of acting as reducing agents towards metal salts such as Ag+ or Cu2+. Capable of acting as a reducing agent. AINUL FARHANA 15 Learning Outcomes : (a) Identify common monosaccharides, disaccharides and polysaccharides and explain their basic structure Classes of monosaccharide: Classification Based on the position Based on no. of C atom of functional group Aldose Ketose Triose 3C Pentose 5C Hexose 6C AINUL FARHANA 16 Learning Outcomes : (a) Identify Learning common Outcomes : monosaccharides, disaccharides and polysaccharides and explain their basic (a) structure Describe various forms and classes of carbohydrates Classes of monosaccharide: based on the position of functional group  Functional group ~ carbonyl group (C=O) 1. Aldose sugar Carbonyl group is located at the end of carbon chain Eg: glucose 2. Ketose sugar Carbonyl group is located in the middle of carbon chain Eg: fructose  Numbering of carbon chain begins with the carbon nearest to carbonyl group AINUL FARHANA 17 Learning Outcomes : (a) Identify common monosaccharides, disaccharides and polysaccharides and explain their basic structure Glucose, galactose,fructose is same molecular formula. but position of atom n functional group is different so different AINUL FARHANA 18 Learning Outcomes : (a) Identify common monosaccharides, disaccharides and polysaccharides and explain their basic structure Classes of monosaccharide: based on the number of carbon atom  3 classes: 1. Triose sugar (3 C atoms) Eg: glyceraldehyde & dihidroxyacetone; C3H6O3 2. Pentose sugar (5 C atoms) – will be discussed in nucleic acid Eg: ribose, ribulose; C5H10O5, & deoxyribose; C5H10O4 3. Hexose sugar (6 C atoms) Eg: glucose, galactose, fructose; C6H12O6  When dissolved in water, pentose & hexose form ring structure; which is more stable AINUL FARHANA 19 Learning Outcomes : (a) Identify common monosaccharides, disaccharides and polysaccharides and explain their basic structure Hexose sugar (6 C atoms)  Same molecular formulas but different arrangements of atoms ~ isomers  Glucose ~ blood sugar, moderate sweetness, found in fruits & vegetables  Galactose ~ less sweet, found in milk & yoghurt  Fructose ~ fruit sugar, sweetest, found in fruits & honey AINUL FARHANA 20 Learning Outcomes : (a) Identify common monosaccharides, disaccharides and polysaccharides and explain their basic structure Function of Monosaccharides  Energy source  Glucose ~ major respiratory substrate, primary energy source  Basic building units or monomers for disaccharide & polysaccharide AINUL FARHANA 21 Learning Outcomes : (a) Identify common monosaccharides, disaccharides and polysaccharides and explain their basic structure DISACCHARIDES  2 sugar molecules or two monosaccharides  Joined together by a condensation process (removal of a water molecule)  Form covalent bond → glycosidic bond  Can be broken down into monomers by hydrolysis process (addition of water molecule)  Characteristics: Disaccharide 1. Sweet 2. Soluble in water 3. Can crystallized 4. White / colourless Maltose Sucrose Lactose AINUL FARHANA 22 Learning Outcomes : (a) Identify common monosaccharides, disaccharides and polysaccharides and explain their basic structure CONDENSATION PROCESS Condensation HYDROLYSIS PROCESS Hydrolysis AINUL FARHANA 23 Learning Outcomes : (a) Identify common monosaccharides, disaccharides and polysaccharides and explain their basic structure MALTOSE  Malt sugar  Monomers : -glucose + -glucose  Joined together by -1,4 glycosidic bond (between C1 of one glucose molecule and C4 of another molecule) + condensation condensation hydrolysis hydrolysis -1,4 glycosidic bond -glucose -glucose Maltose (C6H12O6) (C6H12O6) (C12H22O11) 24 AINUL FARHANA Learning Outcomes : (b) Represent the structure of α -glucose and β –glucose diagrammatically. Glucose Each of these structure is glucose (C6H12O6) a. All carbon atoms are clearly shown b. The carbon atoms are omitted c. The hydrogen atoms are omitted AINUL FARHANA 25 Learning Outcomes : (b) Represent the structure of α -glucose and β –glucose diagrammatically. Glucose  2 isomeric forms, according to the position of OH group at C1 above the ring plane = -glucose below the ring plane = -glucose AINUL FARHANA 26 Learning Outcomes : (c) Explain what is meant by a glycosidic bond and how polymerisation of α-glucose can form starch or glycogen, whilst polymerisation of β-glucose can form cellulose. POLYSACCHARIDES  Many sugar molecules; polymer (vary in length)  Joined together by condensation process (removal of a water molecules)  Form covalent bond → glycosidic bond  Formation of polymer (large molecules made up of repeating units of monomers) → polymerization  Can be broken down into monomers by hydrolysis process (addition of water molecules) Polysaccharide  Characteristics: 1. Large molecule 2. Not sweet Starch Glycogen Cellulose 3. Insoluble in water (form colloid) Amylose Amylopectin 27 AINUL FARHANA Learning Outcomes : (c) Explain what is meant by a glycosidic bond and how polymerisation of α-glucose can form starch or glycogen, whilst polymerisation of β-glucose can form cellulose. STARCH  Energy storage in plants  Plants convert excess glucose into starch for storage in chloroplast  Monomers: -glucose molecules  Broken down by hydrolysis process with the help of amylase enzyme  Starch is a mixture of: Amylose Amylopectin 28 AINUL FARHANA Learning Outcomes : (c) Explain what is meant by a glycosidic bond and how polymerisation of α-glucose can form starch or glycogen, whilst polymerisation of β-glucose can form cellulose. AMYLOSE  Monomers: -glucose molecules  Joined together by -1,4 glycosidic bond  Helical, due to the formation of H bonds  Unbranched chain (variable length of monomers)  Folded ~ very compact (ideal for storage) 6 6 6 6 CH2OH CH2OH CH2OH CH2OH 5 5 5 5 4 OH 1 4 OH 1 4 OH 1 4 OH 1 3 2 3 2 3 2 3 2 O O O O O OH OH OH OH -glucose -glucose -glucose -glucose -1,4 glycosidic bond AINUL FARHANA 29 Learning Outcomes : (c) Explain what is meant by a glycosidic bond and how polymerisation of α-glucose can form starch or glycogen, whilst polymerisation of β-glucose can form cellulose. AMYLOPECTIN  Monomers: -glucose molecules  Joined together by -1,4 glycosidic bond & -1,6 glycosidic bond  Helical, branched chain (variable length of monomers)  Branches occur within 30 units  Folded & branched ~ very compact (ideal for storage) 30 AINUL FARHANA Learning Outcomes : (c) Explain what is meant by a glycosidic bond and how polymerisation of α-glucose can form starch or glycogen, whilst polymerisation of β-glucose can form cellulose. AMYLOPECTIN 6 6 CH2OH CH2OH 5 5 4 OH 1 4 OH 1 3 2 O 3 2 O O OH OH -1,6 glycosidic bond 6 6 6 6 CH2OH H C H CH2OH CH2OH 5 5 5 5 4 OH 1 4 OH 1 4 OH 1 4 OH 1 3 2 O 3 2 O 3 2 O 3 2 O O OH OH OH OH -glucose -glucose -glucose -glucose -1,4 glycosidic bond 31 AINUL FARHANA Learning Outcomes : (c) Explain what is meant by a glycosidic bond and how polymerisation of α-glucose can form starch or glycogen, whilst polymerisation of β-glucose can form cellulose. DIFFERENCES BETWEEN AMYLOSE & AMYLOPECTIN Features Amylose Amylopectin Monomers -glucose -glucose Bond -1,4 glycosidic bond -1,4 glycosidic bond -1,6 glycosidic bond Shape Helical, unbranched Helical, branched (within 30 units) Function Ideal for energy storage (insoluble, coiled, compact) AINUL FARHANA 32 Learning Outcomes : (c) Explain what is meant by a glycosidic bond and how polymerisation of α-glucose can form starch or glycogen, whilst polymerisation of β-glucose can form cellulose. GLYCOGEN  Energy storage in animals (in liver & muscle)  Structure ~ similar to amylopectin  More branched than amylopectin AINUL FARHANA 33 Learning Outcomes : (c) Explain what is meant by a glycosidic bond and how polymerisation of α-glucose can form starch or glycogen, whilst polymerisation of β-glucose can form cellulose. CELLULOSE  Major component of plant cell wall  Monomers: -glucose  Bond: -1,4 glycosidic bond  Broken down by hydrolysis with the help of cellulase enzyme  Unbranched, long straight chain 6 6 6 6 CH2OH CH2OH CH2OH CH2OH 5 5 5 5 4 OH 1 O 4 OH 1 O 4 OH 1 O 4 OH 1 3 2 3 2 3 2 3 2 OH OH OH OH -glucose -glucose -glucose -glucose -1,4 glycosidic bond AINUL FARHANA 34 Learning Outcomes : (c) Explain what is meant by a glycosidic bond and how polymerisation of α-glucose can form starch or glycogen, whilst polymerisation of β-glucose can form cellulose. CELLULOSE -1,4 glycosidic bond AINUL FARHANA 35 Learning Outcomes : (c) Explain what is meant by a glycosidic bond and how polymerisation of α-glucose can form starch or glycogen, whilst polymerisation of β-glucose can form cellulose. CELLULOSE 3D model structure: http://www.pslc.ws/modelhtms/cellpdb.htm AINUL FARHANA 36 Learning Outcomes : (c) Explain what is meant by a glycosidic bond and how polymerisation of α-glucose can form starch or glycogen, whilst polymerisation of β-glucose can form cellulose. CELLULOSE  Straight chains are arranged parallel to one another  Linked together by hydrogen bonds (form microfibril)  Very stable & tough structure ~ provide support to plant cell 37 AINUL FARHANA Learning Outcomes : (c) Explain what is meant by a glycosidic bond and how polymerisation of α-glucose can form starch or glycogen, whilst polymerisation of β-glucose can form cellulose. Functions of Polysaccharides  Energy Storage ~ eg: starch (in plant) & glycogen (in animals)  Major component of cell walls ~ provide structural support to plant cell ~ eg: cellulose AINUL FARHANA 38 Learning Outcomes : (d) Describe and perform the tests for starch, a reducing and non-reducing sugar in detail (Benedict’s test). Tests for Starch a) Add two drops of iodine solution and shake or stir a) The presence of starch is indicated by a blue-black colouration AINUL FARHANA 39 Learning Outcomes : (d) Describe and perform the tests for starch, a reducing and non-reducing sugar in detail (Benedict’s test). Benedict’s Test a) Add an equal volume of Benedict's reagent (which is an alkaline solution of copper (II) sulphate) b) Heat the mixture in a gently boiling water bath for 5 minutes AINUL FARHANA 40  Question FIGURE 2 shows the structure of a polysaccharides. a) Name the polysaccharides in FIGURE 2 b) State the monomer for (a) AINUL FARHANA 41  1.0 Biological Molecules a) Monomers & Polymers (1/2) b) Carbohydrates (1/2) c) Lipids (1/2) d) Proteins & Enzymes (1) e) Nucleic Acids (1/2) f) ATP (1/2) g) Water (1/2) AINUL FARHANA 42  Learning outcomes a) Perform the emulsion test, explain the purpose of each stage, and interpret the results of the emulsion test. b) Recognise, and explain the difference between, saturated and unsaturated fatty acids. c) Describe and explain the structure of phospholipids in relation to their function. d) Describe the structure of triglycerides and explain how triglycerides form e) Contrast the different properties of triglycerides and phospholipids 43 AINUL FARHANA Learning Outcomes : (a) Perform the emulsion test, explain the purpose of each stage, and interpret the results of the emulsion test. EMULSION TEST Positive result Negative result a) Since lipids do not dissolve in water, when the ethanol is diluted, it falls out of solution to give a cloudy white emulsion. b) Add ethanol to the solution you want to test, and add water. c) A colour change to cloudy white shows that lipids are present AINUL FARHANA 44 Learning Outcomes : (b) Recognise, and explain the difference between, saturated and unsaturated fatty acids. LIPIDS  Consist of mainly carbon & hydrogen atoms; few oxygen atom  Group of hydrophobic molecules ~ insoluble in water  3 major types: LIPIDS Fats Phospholipids Steroids Eg: oil, butter Eg: lecithin Eg: cholesterol 45 AINUL FARHANA Learning Outcomes : (b) Recognise, and explain the difference between, saturated and unsaturated fatty acids. FATS  Most abundant  Also known as (a.k.a) : Triglyceride / Triacylglycerol Building block 1 Glycerol 3 Fatty acids Saturated Unsaturated 46 AINUL FARHANA Learning Outcomes : (b) Recognise, and explain the difference between, saturated and unsaturated fatty acids. Glycerol  A.k.a : glycerine.  3C alcohol with 3 OH groups (soluble in water) H H C OH H C OH H C OH H 47 AINUL FARHANA Learning Outcomes : (b) Recognise, and explain the difference between, saturated and unsaturated fatty acids. Fatty Acids  Has 2 regions:. Hydrophilic head (a carboxyl group, COOH at 1 end ) ~ acid Hydrophobic tail ~ long unbranched hydrocarbon chain (known as side chain, R) O H H H H H H H H H H H H H HO C1 C2 C3 C4 C5 C6 C7 C8 C9 C10 C11 C12 C13 C14 H H H H H H H H H H H H H H Carboxyl group Hydrocarbon chain R 48 AINUL FARHANA Learning Outcomes : (b) State the types of lipid: fats, phospholipids and steroids Fatty Acids  Hydrophobic tail vary in length.  Different fatty acids has different number of C atoms (usually 16 or 18) O H H H H H H H H H H H H H H H H H HO C1 C2 C3 C4 C5 C6 C7 C8 C9 C10 C11 C12 C13 C14 C15 C16 C17 C18 H H H H H H H H H H H H H H H H H H Eg: Stearic acid (stearate) C18H36O2 AINUL FARHANA 49 Learning Outcomes : (b) Recognise, and explain the difference between, saturated and unsaturated fatty acids. Classification of Fatty Acids  Based on the presence of double bonds within hydrocarbon. chain: Saturated fatty acids ~ no double bond between C atoms; eg: stearic acid Unsaturated fatty acids ~ has double bonds between C atoms; eg: oleic acid O H H H H H H H H H H H H H H H H H HO C1 C2 C3 C4 C5 C6 C7 C8 C9 C10 C11 C12 C13 C14 C15 C16 C17 C18 H H H H H H H H H H H H H H H H H H Eg: Stearic acid (stearate) C18H36O2 AINUL FARHANA 50 Learning Outcomes : (b) Recognise, and explain the difference between, saturated and unsaturated fatty acids. SATURATED FATTY ACID  Has maximum number. of H atoms  Straight chain, fatty acids can be closely packed  Solid at room temperature 51 AINUL FARHANA Learning Outcomes : (b) Recognise, and explain the difference between, saturated and unsaturated fatty acids. SATURATED FATTY ACID  Fats that consist of saturated fatty acids ~ saturated fat. Mostly animal fats (eg: butter, lard) 52 AINUL FARHANA Learning Outcomes : (b) Recognise, and explain the difference between, saturated and unsaturated fatty acids. SATURATED FATTY ACID 53 AINUL FARHANA Learning Outcomes : (b) Recognise, and explain the difference between, saturated and unsaturated fatty acids. UNSATURATED FATTY ACID  Double bond causes bending in unsaturated fatty acids  Fatty acids cannot be closely packed  Liquid at room temperature 54 AINUL FARHANA Learning Outcomes : (b) Recognise, and explain the difference between, saturated and unsaturated fatty acids. UNSATURATED FATTY ACID  Fats that consist of unsaturated fatty acids ~ unsaturated fat  Mostly plant & fishes fats (eg: oil) AINUL FARHANA 55 Learning Outcomes : (b) Recognise, and explain the difference between, saturated and unsaturated fatty acids. Classification of Fatty Acids  Based on the ability to be synthesized in the body:. Essential fatty acids ~ cannot be synthesized in the body ~ must be present in the diet ~ eg: linoleic, linolenic and arachidonic acids Non-essential fatty acids ~ can be synthesized in the body ~ eg: stearic acid & others AINUL FARHANA 56 Learning Outcomes : (c) Describe and explain the structure of phospholipids in relation to their function. PHOSPHOLIPIDS  Major component of cell membrane; eg: lecithin  Amphipathic molecule  i.e one end hydrophilic polar, another end hydrophobic non-polar Building block 1 Glycerol 2 Fatty acids Phosphate group AINUL FARHANA 57 Learning Outcomes : (c) Describe and explain the structure of phospholipids in relation to their function. PHOSPHOLIPIDS 58 AINUL FARHANA Learning Outcomes : (d) Describe the structure of triglycerides and explain how triglycerides form TRIGLYCERIDE  Composed of 3 fatty acids & 1 glycerol backbone  Fats are formed when 3 condensation reaction occurs ~ also known as esterification  Water is removed by extracting the OH from the carboxyl group (of fatty acid) and H from the glycerol’s hydroxyl group  Triglyceride can be broken down by hydrolysis 59 AINUL FARHANA Learning Outcomes : (d) Describe the structure of triglycerides and explain how triglycerides form Formation of Triglycerides Ester bond H O H O H C OH HO C R H C O C R O O Ester bond Condensation H C OH HO C R’ H C O C R’ Hydrolysis O O Ester bond H C OH HO C R’’ H C O C R’’ H H + 3H2O 1 glycerol 3 fatty acids Triglyceride 60 AINUL FARHANA Learning Outcomes : (d) Describe the structure of triglycerides and explain how triglycerides form Importance of Lipids  Main energy storage in animals (due to higher number of hydrogen atom) 1 g of fat has > twice energy than same weight of starch Lighter ~ hydrophobic nature, doesn’t associates with water  Improve buoyancy in aquatic animals  Provide better thermal insulation of mammals  Act as padding for internal organs  Phospholipid ~ major component of plasma membrane  Steroid ~ some are hormones that regulates metabolism 61 AINUL FARHANA Learning Outcomes : (e) Contrast the different properties of triglycerides and phospholipids Triglycerides VS Phospholipids Components Triglycerides Phospholipids Building block Location in human body Function Diagram Similarities 62 AINUL FARHANA  1.0 Biological Molecules a) Monomers & Polymers (1/2) b) Carbohydrates (1/2) c) Lipids (1/2) d) Proteins & Enzymes (1) e) Nucleic Acids (1/2) f) ATP (1/2) g) Water (1/2) AINUL FARHANA 63  Learning outcomes a) Describe the general structure of an amino acid b) Describe the Biuret test and how it can be interpreted c) Explain the variety of functions that proteins have and why they are so important to the body. d) Explain how dipeptides polypeptides form e) Explain the hierarchical organisation of protein structure: primary, secondary, tertiary & quaternary levels f) Describe the types of bond involved in protein structure and the relative properties of each g) Relate the structure of proteins to properties of proteins. AINUL FARHANA 64 Learning Outcomes : (a) Describe the general structure of an amino acid PROTEIN Polymer Building blocks ~ amino acids Joined together to form a long chain ~ polypeptide Protein consist of 1 or more polypeptide chains which folded & coiled into specific conformation Mostly composed of C, H, O, N & sometimes S AINUL FARHANA 65 Learning Outcomes : (a) Describe the general structure of an amino acid BASIC STRUCTURE OF AMINO ACIDS Amino acids consist of 4 components attached to a central carbon These components include: a. a hydrogen atom b. a carboxyl group c. an amino group d. a variable R group (or side chain). R amino group H2N C COOH carboxyl group H 66 AINUL FARHANA Learning Outcomes : (a) Describe the general structure of an amino acid BASIC STRUCTURE OF AMINO ACIDS 2 functional groups: i. COOH (carboxyl group) : acidic ii. NH2 (amino group) : basic Amino acids are amphoteric; have both acidic & basic properties R amino group H2N C COOH carboxyl group H 67 AINUL FARHANA Learning Outcomes : (a) Describe the general structure of an amino acid GROUP OF AMINO ACIDS 20 types of amino acids commonly found in proteins All have the same basic structure but differ in the side chain, R Amino acids are grouped based on the properties of R group R amino group H2N C COOH carboxyl group H AINUL FARHANA 68 Learning Outcomes : (a) Describe the general structure of an amino acid 4 groups Non-polar Polar Acidic Basic Hydrophobic Hydrophilic Has COOH group Has NH2 group Glycine (gly) Serine (ser) Aspartic acid (asp) AINUL FARHANA 69 Lysine (Lys) Learning Outcome: b) Describe the Biuret test and how it can be interpreted Test for Protein: Biuret Test Negative Positive test test Protein present Learning Outcome: b) Describe the Biuret test and how it can be interpreted When a protein reacts with copper(II) sulfate (blue), the positive test is the formation of a violet colored complex. Purple / Lilac: Positive test Learning Outcomes : (c) Explain the variety of functions that proteins have and why they are so important to the body. GROUP OF AMINO ACIDS i. Essential amino acids (for human) Cannot be synthesized in sufficient amount, must be obtained through food or supplement intake 8 amino acids; eg: isoleucine, leucine, lysine, methionine, phenylalanine, threonine, tryptophan, valine For infants : histidine ii. Non-essential amino acids Can be synthesized in the body Eg: alanine, arginine, asparagine, aspartic acid, cysteine, glutamic acid, glycine, proline, serine, tyrosine, glutamine AINUL FARHANA 72 Learning Outcomes : (d) Explain how dipeptides & polypeptides form Formation of the dipeptide 2 amino acids are joined by condensation process to form a dipeptide By removal of a hydroxyl group from the carboxyl end of one amino acid and a hydrogen from the amino group of another amino acid. Form a covalent bond ~ peptide bond AINUL FARHANA 73 Learning Outcomes : (c) describe Learning the formation Outcomes : and breakdown of dipeptide (d) Explain how dipeptides & polypeptides form Formation & breakdown of the dipeptide R1 O R2 O H N C C OH H N C C OH H H H H Condensation Hydrolysis H 2O H 2O R1 O R2 O N-terminal H N C C N C C OH C-terminal H H H H Peptide bond AINUL FARHANA 74 Learning Outcomes : (d) Explain how dipeptides & polypeptides form Breakdown of dipeptide Dipeptide is hydrolysed into 2 amino acids by hydrolysis process (addition of water). Peptide bond is broken down and produces 2 amino acids. 75 AINUL FARHANA Learning Outcomes : (d) Explain how dipeptides & polypeptides form Formation of polypeptide Many amino acids are joined by condensation process to form a polypeptide Bond ~ peptide bonds Polypeptide backbone ~ repeated sequence of (-N-C-C-) R1 O R2 O R3 O H N C C OH H N C C OH H N C C OH H H H H H H Peptide bond Peptide bond AINUL FARHANA 76 Learning Outcomes : (e) Explain the hierarchical organisation of protein structure Classification of Protein Two ways based on:- Level of protein structure i. Primary ii. Secondary iii. Tertiary iv. Quaternary Structure i. Fibrous ii. Globular iii. Conjugated AINUL FARHANA 77 Learning Outcomes : (e) Explain the hierarchical organisation of protein structure Primary structure Refers to the linear sequence of amino acids joined by peptide bonds within a polypeptide chain Each protein has a unique sequence of amino acids, determined by specific DNA code Different sequence results in the formation of different bonds between amino acids Causes a polypeptide chain to fold & coil into a unique 3 dimensional shape of protein Eg : lysozyme & all protein that has been sequenced AINUL FARHANA 78 Learning Outcomes : (e) Explain the hierarchical organisation of protein structure Primary structure AINUL FARHANA 79 Learning Outcomes : (e) Explain the hierarchical organisation of protein structure Sickle cell anemia is a mutation of one of the Sickle cell anemia sequence (from hydrophilic to hydrophobic) I’m But I’m hydrophilic! hydrophobic! Learning Outcomes : (e) Explain the hierarchical organisation of protein structure Secondary structure Linear chain of amino acids will coil or fold spontaneously Due to the formation of H bonds between the backbones (between H of the NH groups & O of the C=O groups of amino acids in the primary chain) 2 types -helix -pleated sheet AINUL FARHANA 81 Learning Outcomes : (e) Explain the hierarchical organisation of protein structure -helix Helical coil H bonds is formed between 1 amino acid & the 4th amino acid away from it Can stretch under tension ~ provide elasticity Because the hydrogen bonds can be reformed Eg: keratin in hair, nails & wool | | S S | | S S | | AINUL FARHANA 82 Learning Outcomes : (e) Explain the hierarchical organisation of protein structure -pleated sheet Zig-zag pattern ~ provide strength & flexibility Eg: silk, spider web AINUL FARHANA 83 Learning Outcomes : (e) Explain the hierarchical organisation of protein structure Tertiary structure 1 polypeptide chain in secondary structure may further coiled into a globular shape which is maintained by bonds and interactions among R groups Globular shape unique to each polypeptide chain Mantained by 4 interactions between R groups:- Disulfide bonds (covalent bonds) between R groups with sulfhydryl groups Ionic bonds between +ve & -ve charged R groups Hydrogen bonds between + & - charged R groups Hydrophobic / Van der Waals interaction between non- polar R groups AINUL FARHANA 84 Learning Outcomes : (e) Explain the hierarchical organisation of protein structure Tertiary structure AINUL FARHANA 85 Learning Outcomes : (f) Explain the hierarchical organisation of protein structure Tertiary structure Eg: enzymes, hormones, antibodies Type of bonds exist in different levels of protein: Features Primary Secondary Tertiary Type of i. Peptide i. Peptide bond i. Peptide bond bond exist bond ii. H bond (between ii. H bond (between C=O C=O & NH group of & NH group of backbone) backbone) iii. Disulfide bond iv. Ionic bond v. H bond (between R groups) vi. Hydrophobic (van der Waals) interaction AINUL FARHANA 86 Learning Outcomes : (e) Explain the hierarchical organisation of protein structure Quaternary structure Refers to a single protein that is formed when two or more polypeptide chain joins Bonds exist ~ same as tertiary structure Eg: hemoglobin (consist of 4 polypeptide chains, two α chains and two β chains) AINUL FARHANA 87 Learning Outcomes : (e) Explain the hierarchical organisation of protein structure 4 Levels of Protein AINUL FARHANA 88 Learning Outcomes : (g) Relate the structure of proteins to properties of proteins. Classification of Protein Based on Structure Fibrous Globular Conjugated FIBROUS Mostly secondary structure Form long parallel filaments or strands Insoluble & stable Function : mechanical & structural support Eg: Basic structure of muscle cells Keratin, actin, myosin, collagen, silk AINUL FARHANA 89 Learning Outcomes : (g) Relate the structure of proteins to properties of proteins. Globular Protein Mostly tertiary/quaternary structure Folded to form compact spherical shaped Relatively unstable Relatively soluble ~ colloid Function : metabolic & chemical process Eg: Enzyme Hormone Antibodies Hemoglobin AINUL FARHANA 90 Learning Outcomes : (g) Relate the structure of proteins to properties of proteins. Conjugated Proteins Protein that can function well if it joins with other compound Non-protein compound ~ prosthetic group Conjugated Protein Prosthetic group Location Glycoprotein Polysaccharide Component of cell membrane Mucin (in saliva) Lipoprotein Lipid Component of cell membrane Lipid transported in blood plasma Chromoprotein Pigment Myoglobin, hemoglobin Nucleoprotein Nucleic acid Chromosome, ribosome AINUL FARHANA 91 Learning Outcomes : (h) Explain the effect of pH and temperature on the structure of protein Effect of pH and Temperature Structure of protein is affected by:- 1. pH 2. temperature Globular protein is relatively unstable because it is maintained by weak ionic bonds, hydrogen bonds & hydrophobic interactions When protein is heated or exposed to an extreme pH changes, those bonds are broken Causes it to uncoiled & change its conformation Lose its biological function ~ denatured → protein coagulate Usually irreversible (sometimes, renaturation may occur) 92 AINUL FARHANA Learning Outcomes : (h) Explain the effect of pH and temperature on the structure of protein Effect of pH and Temperature Factor Effect to the protein structure Heat or radiation ✓ Increased kinetic energy ✓ Protein atoms vibrate ✓ H & ionic bonds, van der Waals interaction break Strong acids & ✓ H & ionic bonds, van der Waals interaction break alkali ✓ Breakdown of peptide bonds occur if exposed very long 93 AINUL FARHANA  Learning outcomes a) Interpret energy level diagrams and identify the activation energy b) Apply knowledge of tertiary structure to explain enzyme specificity and the formation of enzyme- substrate complexes. c) Explain, compare and evaluate the induced fit model and lock and key hypotheses of enzyme action d) Explain how temperature, pH, substrate concentration, enzyme concentration, and the presence of inhibitors affect enzyme catalysis AINUL FARHANA 94 Learning Outcome: a) Interpret energy level diagrams and identify the activation energy Biological catalyst produce by living cells that speed up the rate of chemical reaction by lowering the ACTIVATION ENERGY Dont copy and paste this exact defination, keep only the key words. Introduction Learning Outcome: a) Interpret energy level diagrams and identify the activation energy Has two sites : A pocket / groove into which the substrate fits ✓ ACTIVE SITE Specific site on enzyme that binds to specific substrate ✓ ALLOSTERIC SITE Other site than the Attachment for non competitive inhibitors active site on enzyme Introduction Learning Outcome: a) Interpret energy level diagrams and identify the activation energy E + S E S E + P Substrate binds at active site of enzyme Can form or break bond between two substrates Enzyme-Substrate Complex (transition state) E : enzyme S : substrate (reactant) ES : enzyme - substrate complex P : product Used for many processes, eg. Metabolic reaction Learning Outcome: a) Interpret energy level diagrams and identify the activation energy Minimum amount of energy required by a reactant (substrate) to start a chemical reaction. How do enzymes lower the activation energy of a reaction? Substrate collide with enzyme at its active site Enzyme facilitate breaking/formation of bond in the substrate to form product Without enzyme reaction can still occur but will need a lot more Energy Enzyme speed up the rate of chemical reactions by lowering EA Energy required almost double without enzyme Learning Outcome: b) Apply knowledge of tertiary structure to explain enzyme specificity and the formation of enzyme-substrate complexes. PROPERTIES OF ENZYME Further coiled and compacted 4 interactions: 1. Globular protein Ionic, sulfide, hydrophobic/ van der waals, hydrogen Has peptide & hydrogen bond : COOH+NH2. 2. Act as biological catalyst Hydrogen bond: R group 3. Required in small quantity to catalyse reaction 4. Enzyme do not change after the reaction @ reusable 5. Reversible catalyst ; Occur between enzymes & substrate - enzyme catalyse reversible or irreversible reaction 6. Enzyme is specific to its substrate Can't form bond again, may be due to inhibitors. 7. Enzyme activity is maximum at optimum temperature and pH. Amylase (saliva) optimum pH 7 Pepsin (stomach lining) optimum PH 3 - enzyme will be denatured @ destroyed at extreme temperature and pH. 8. Act faster ; one enzyme molecule can catalyze thousands or more reactions in a second Learning Outcome: b) Apply knowledge of tertiary structure to explain enzyme specificity and the formation of enzyme-substrate complexes. SPECIFICITY OF ENZYME The specificity of an enzyme is due to the complementary shape of the active site and the substrate. Amylase - pH 7 Pepsin-pH3 They will not work in their not optimum temperatures When an enzyme is denatured by heat or a change in pH, the shape of active site is changed and the substrate can no longer fit into active site. Learning Outcome: c) Explain, compare and evaluate the induced fit model and lock and key hypotheses of enzyme action HYPOTHESIS RELATED TO THE MECHANISM OF ENZYME ACTION 1) LOCK AND KEY MODEL 2) INDUCED FIT MODEL (widely accepted) Learning Outcome: c) Explain, compare and evaluate the induced fit model and lock and key hypotheses of enzyme action LOCK AND KEY MODEL The substrate is the ‘key’ that fits exactly the ‘lock’ (enzyme) Enzyme + substrate = enzyme-substrate complex Active site of enzyme exactly complementary to substrate Learning Outcome: c) Explain, compare and evaluate the induced fit model and lock and key hypotheses of enzyme action LOCK AND KEY MODEL (LOCK) (KEY) * Active site of enzyme exactly complementary to substrate LOCK : Enzyme KEY : Substrate Learning Outcome: c) Explain, compare and evaluate the induced fit model and lock and key hypotheses of enzyme action LOCK AND KEY MODEL ✓ An enzyme collides with its substrate molecule ✓ The substrate binds to the active site of enzyme ✓ Active site of enzyme exactly complementary to substrate ✓ Enzyme-substrate complex formed ✓ Reaction occur where the substrate reacts within the complex and product is released ✓ Enzyme is not change or damage and can be reused Learning Outcome: c) Explain, compare and evaluate the induced fit model and lock and key hypotheses of enzyme action LOCK AND KEY MODEL Learning Outcome: c) Explain, compare and evaluate the induced fit model and lock and key hypotheses of enzyme action LOCK AND KEY MODEL Learning Outcome: c) Explain, compare and evaluate the induced fit model and lock and key hypotheses of enzyme action INDUCED-FIT MODEL Attachment of substrates induces the conformation change in the active site of enzyme molecule until its active site fits with the substrate  * Active site of enzyme NOT exactly complementary to substrate * Attachment of substrate to active site of enzyme induce conformational change in enzyme’s active site Learning Outcome: c) Explain, compare and evaluate the induced fit model and lock and key hypotheses of enzyme action INDUCED-FIT MODEL ✓ The active site of an enzyme is not fully complementary to the shape of substrate ✓ Enzyme collides with the substrate molecule ✓ To form enzyme-substrate complex ✓ The binding induce a slight change in the shape of enzyme ✓Allowing the substrate fit to enzyme precisely Learning Outcome: c) Explain, compare and evaluate the induced fit model and lock and key hypotheses of enzyme action INDUCED-FIT MODEL ✓ The active site changes shape and becomes fully complementary with the substrate ✓ enable the enzyme to carry out their catalytic function ✓ product is formed and enzyme changes back or reconvert to its original conformation Learning Outcome: c) Explain, compare and evaluate the induced fit model and lock and key hypotheses of enzyme action Learning Outcome: c) Explain, compare and evaluate the induced fit model and lock and key hypotheses of enzyme action Differences between two models of enzyme actions LOCK & KEY MODEL INDUCED-FIT MODEL The active site is not flexible The active site is flexible and and exactly complementary not exactly complementary to substrate to substrate The active site is not The binding of substrate changed after binding with induce the conformational substrate changes to active site of enzyme Enzyme does not change Enzyme reverts/reconvert to and can be reused after its original conformation reaction when the product is released  QUESTION 1 Why are the products separate from enzyme at the end of reaction ? [ 1 mark ] Product are not complementary with the active site of enzyme Learning Outcome: d) Explain how temperature, pH, substrate concentration, enzyme concentration, and the presence of inhibitors affect enzyme catalysis Temperature Inhibitors pH Factors Enzyme Substrate concentration concentration 115  1.0 Biological Molecules a) Monomers & Polymers (1/2) b) Carbohydrates (1/2) c) Lipids (1/2) d) Proteins & Enzymes (1) e) Nucleic Acids (1/2) f) ATP (1/2) g) Water (1/2) AINUL FARHANA 116  Learning outcomes a) Explain the structure of DNA, and identify structural components from diagrams b) Apply knowledge of complementary base pairing rules to work out the frequency of certain bases, when provided with information about the frequency the other bases c) Explain the significance of DNA to organisms d) Explain the role of RNA in transferring genetic information and as a component of ribosomes e) Explain the structure of RNA, and identify structural components of an RNA nucleotide from diagrams f) Compare and contrast the similarities and differences between DNA and RNA. 117 AINUL FARHANA Learning Outcomes : (a) Explain the structure of DNA, and identify structural components from diagrams Introduction Nucleic acid is polynucleotide ~ polymer of nucleotides Building blocks ~ nucleotide Mostly composed of C, H, O, P & N 118 AINUL FARHANA Learning Outcomes : (a) Explain the structure of DNA, and identify structural components from diagrams NUCLEIC ACIDS 2 types Deoxyribonucleic Acid Ribonucleic Acid (DNA) (RNA) 119 AINUL FARHANA Learning Outcomes : (a) Explain the structure of DNA, and identify structural components from diagrams Structure of nucleotides Each nucleotide has 3