Holiday Assignment - Essay
HOLIDAY ASSIGNMENT: ESSAY
1. 2010 P2 Q9 (a) Describe the structure of an amino acid and how a peptide bond is formed with another amino acid.  Structure of amino acids 1. Each amino acid consists of a central α-carbon atom bonded covalently to four groups;; 2. A basic amino group (–NH2), and an acidic carboxyl group (–COOH);; 3. A hydrogen atom, and a variable group known as the R group ( or side chain);;
Structure of an amino acid (fully labelled);; How a peptide bond is formed with another amino acid 4. The peptide bond is formed between the amino end of an amino acid with the carboxyl end of another amino acid;; 5. The two amino acids react together in a condensation reaction with the loss of a water molecule to form a dipeptide;; 6. An enzyme, peptidyl transferase, which resides in the large ribosomal subunit catalyses the formation of the peptide bond;;
HOLIDAY ASSIGNMENT: ESSAY (b) Explain what is meant by primary, secondary, tertiary and quaternary structure of a named protein.  1. The primary structure of a protein molecule refers to its number, type and sequence of amino acids held together by peptide bonds in the linear strand of polypeptide chain;; 2. Haemoglobin consists of 2 α chains with each chain having a specific sequence of 141 amino acids and 2 β chains with each chain having a specific sequence of 146 amino acids;; 3. Secondary structures are geometrically regular repeating structures, including α-helix and the β-pleated sheet, stabilised by hydrogen bonds between groups in the main chain of the polypeptide;; 4. Tertiary structure is the three-dimensional conformation of a polypeptide maintained by ionic bonds, hydrogen bonds, disulphide bonds and hydrophobic interactions;; 5. Each polypeptide chain of haemoglobin coils into an α helix which is then folded upon itself into a rough spherical globular protein ;; 6. The shape of the haemoglobin is held by interactions between the R-groups of its amino acids - hydrogen bonds, ionic bonds, disulphide bonds and hydrophobic interactions;; 7. Quaternary structure consists of an aggregation of 2 or more polypeptide chains held together by hydrophobic interactions, disulphide linkages, hydrogen bonds and ionic bonds;; 8. Haemoglobin’s four polypeptide chains pack closely together, held together by hydrophobic interactions, disulphide linkages, hydrogen bonds and ionic bonds resulting in a nearly spherical haemoglobin molecule;; (c) Outline how the structure of a named globular protein is related to its specific function.  1. Haemoglobin is a globular protein found in red blood cells that functions to transport oxygen;; 2. Haemoglobin consists of 2 α chains and 2 β chains and each polypeptide chain is first coiled into an α helix which is then folded upon itself into a rough spherical shape;; 3. Each polypeptide chain contains a (non-proteinaceous) prosthetic haem group and each haem group contains an iron, Fe2+, which binds a molecule of oxygen;; 4. The four polypeptide chains pack closely together, held together by hydrophobic interactions, disulphide linkages, hydrogen bonds and ionic bonds resulting in a nearly spherical haemoglobin molecule;; 5. A complete haemoglobin molecule, with four haem groups, can carry four oxygen molecules at a time;; 6. Since haemoglobin is in contact with solution, their hydrophobic R-groups point into the centre of the molecule and their hydrophilic R-groups point outwards to form hydrogen bonds with water, hence maintaining its solubility in water;; 7. Due to its allosteric nature, the haemoglobin molecule can undergo drastic changes in shape as it binds O2 to make it easier for further O2 to bind;;
HOLIDAY ASSIGNMENT: ESSAY 2. 2007 P2 Q8 (a) Describe the main structural features of cellulose and collagen.  Cellulose 1. A polysaccharide made up of β-glucose residues joined by β-1, 4 glycosidic bonds;; 2. Successive β-glucose residues are rotated 180o with respect to its adjacent residue, results in the –OH groups projecting outwards from each chain in all directions;; 3. Many cellulose chains run parallel to each other, forming hydrogen bonds between the –OH groups of neighbouring chains, resulting in cross-linking;; 4. Cellulose is made of microfibrils held together in bundles called macrofibrils;; Collagen 5. A fibrous protein made up of a chain of amino acids held together by peptide bonds;; 6. Three helical polypeptide chains are twisted into a tropocollagen/ triple helix and bound to each other by intermolecular hydrogen bonds;; 7. Almost every third amino acid in each chain is glycine, and glycine’s small size allows the three strands of polypeptide to form a tight coil;; 8. Tropocollagen runs parallel to each other and the ends of the tropocollagens are staggered;; 9. Covalent cross links form between the carboxyl end of one tropocollagen and the amino end of another tropocollagen;; 10. Many tropocollagens lie parallel to form fibrils, which unite to form fibres;; [max 5] 3. 2009 P2 Q8 (a) Describe the mode of action of enzymes.  1. Enzymes speed up biochemical reactions by lowering the activation energy of the reaction;; 2. and remain chemically unchanged at the end of the reactions they catalysed and can be reused;; 3. Enzymes reduce the activation energy by holding substrates close together at the correct angle and orientation for successful interaction and collision OR straining the chemical bonds within the substrates until they break;; 4. Enzymes are highly specific in the reactions they catalyse;; 5. Enzyme is specific because only substrates of a complementary shape will fit into the active site with distinctive configuration –spatial fit;; 6. According to Lock & Key hypothesis, the shape of the substrate is complementary to the shape of the active site of the enzyme;; 7. According to Induced Fit hypothesis, as the substrate enters and binds to the active site, it induces a conformational change in the shape of the enzyme which enables the substrate to fit even more snugly into the active site;; 8. Enzyme and substrate must have charge and hydrophobic/hydrophilic complementarity – chemical fit;; 9. Upon successful collisions between substrate and enzyme, enzyme-substrate complexes are formed;; 10. Temporary bonds such as ionic bonds, hydrogen bonds and hydrophobic interactions between the substrate and active site are formed;; 11. Eventually leading to the formation of products which then leaves the active site as the products no longer fit into the active site;;
HOLIDAY ASSIGNMENT: ESSAY
(b) Explain how pH affects the rate of an enzyme catalysed reaction. 
1. well-annotated graph;; 2. At optimum pH, maximum rate of reaction occurs;; 3. as bonds maintaining the secondary and tertiary structures of the enzyme are intact, enabling highest frequency of successful collisions between the substrate and enzyme;; 4. This increases the rate of formation of E-S complexes, increasing the rate of formation of products;; 5. At pH above or below this optimum pH/other than optimum pH, the rate of reaction decreases;; 6. as the change in pH alters the ionic charge of the acidic and basic R groups on the amino acids at the active site of the enzyme;; 7. the ionic bonds and hydrogen bonds that help to maintain the specific shape of the active site of the enzyme are disrupted;; 8. causing a loss of 3D conformation of enzymes’ active site, loss of 3D conformation of enzymes and the enzymes are denatured;; 9. substrate can no longer bind to the active site of the enzyme to form E-S complexes, decreasing the rate of formation of E-S complexes and hence decreasing the rate of formation of products;; (c) Explain the effect of non-competitive inhibitors on enzyme activity. 
1. well-annotated graph;; 2. non-competitive inhibitor has no structural similarity to the substrate and binds to the enzyme at site other than the active site (allosteric site);; 3. upon binding of the non-competitive inhibitor to the enzyme, the enzyme 3D conformation is changed such that its active site's conformation is altered and substrate can no longer bind to the active site;; 4. formation of such enzyme-inhibitor complexes prevents the formation of E-S complexes and formation of products and decreasing the rate of reaction;; 5. an increase in substrate concentration will not reverse the inhibition/reduce the effect of inhibition, even at very high substrate concentration, the maximum rate of reaction in the presence of non-competitive inhibitor is lower than that of reaction in the absence of noncompetitive inhibitor;; 6. allosteric inhibition can be reversible or irreversible and some non-competitive inhibitors bind to active site irreversibly;
HOLIDAY ASSIGNMENT: ESSAY 4. 2005 P2 Q8 (a) Describe the effect of inhibitors on the rate of enzyme activity.  1.
2. 3. 4. 5. 6. 7.
Enzyme inhibitors are molecules which prevent enzymes from catalysing the reactions, hence reducing the rate of reaction and inhibition can be competitive or noncompetitive;; Competitive inhibitor (max 4m) competitive inhibitor is structurally similar to the actual substrate and can fit into the active site of the enzyme;; when a competitive inhibitor is bound at the active site, it prevents the actual substrate from entering the site, preventing the formation of E-S complexes and formation of products;; enzyme-inhibitor complexes are formed instead as competitive inhibitor competes with the actual substrate for binding at the active site;; At low substrate concentration, the frequency of enzyme-substrate collisions is similar to the frequency of enzyme-inhibitor collisions, number of E-S complexes formed is about the same as E-I complexes formed, and this leads to a low rate of enzyme activity;; at high substrate concentrations, substrate competes more successfully for active site, resulting in higher frequency of successful collision between substrate and enzyme, hence more E-S complex are formed, and this leads to a higher rate of reaction;; competitive inhibition is reversible and can be overcome by high substrates concentration, as substrates can out-compete the inhibitors for binding to the active site and allow the maximum rate of reaction to be reached;;
Non-competitive inhibitor (max 4m) non-competitive inhibitor has no structural similarity to the substrate and binds to the enzyme at site other than the active site (allosteric site);; 9. upon binding of the non-competitive inhibitor to the allosteric site, the enzyme 3D conformation is changed such that its active site's conformation is altered and substrate can no longer bind to the active site;; 10. formation of such enzyme-inhibitor complexes prevents the formation of E-S complexes and formation of products, decreasing the rate of reaction;; 11. an increase in substrate concentration will not reverse the inhibition/reduce the effect of inhibitor, even at very high substrate concentration, the maximum rate of reaction in the presence of non-competitive inhibitor is lower than that of reaction in the absence of non-competitive inhibitor;; 12. allosteric inhibition can be reversible or irreversible and some non-competitive inhibitors bind to active site irreversibly;; 8.
HOLIDAY ASSIGNMENT: ESSAY (b) Cells have metabolic pathways commonly made up of sequences of enzyme catalysed reactions. Explain, with examples, the advantages of such sequences of enzyme catalysed reactions.  1. Cells have metabolic pathways commonly made up of sequences of enzyme catalysed reactions in which the product from one reaction acts as the substrate for the next reaction;; 2. Each reaction is catalysed by different enzymes and this allows intermediates to go down different pathways;; 3. The enzymes which catalyse such chain reactions often form a linear series bound to membranes within the cell, constituting a multi-enzyme complex, and such close proximity between enzymes is efficient since collisions between enzymes and their substrates are made more likely;; 4. Example of metabolic pathway;; a) Calvin cycle in stroma of chloroplast In Calvin cycle, carbon dioxide is carboxylated to RuBP, catalysed by Rubisco, to form a 6C unstable intermediate which splits immediately to 2 molecules of GP. OR b) Glycolysis in cytosol of cells In glycolysis, glucose is phosphorylated by ATP to glucose 6-phosphate, catalysed by hexokinase. Glucose-6-phosphate is isomerised by a different enzyme to become fructose-6-phosphate. OR c) Krebs cycle in matrix of mitochondrion In Krebs cycle, acetyl-coA combines with oxaloacetate to form citrate, catalysed by citrate synthase. Citrate undergoes oxidative decarboxylation to become αketoglutarate, catalysed by citrate dehydrogenase 5. For many metabolic pathways, the final product of the pathway is usually an allosteric inhibitor of one of the earlier enzymes in the pathway, and such inhibition of an earlier stage in a process by the final product is termed negative feedback inhibition;; 6. In this way, an accumulation of final product will thus slow down or stop its further production, preventing wastage of resources; when the product is used up, the inhibition is lifted and production is switched back on again;; 7. Being self-regulatory in nature, negative feedback inhibition is thus important in coordinating the metabolism of cells, preventing shortage and overproduction of end products;;
HOLIDAY ASSIGNMENT: ESSAY 5. 2012 P2 Q9 (a) Outline the light dependent reactions of photosynthesis.  1. Photons of light strike photosynthetic pigment molecules in PS/PSII/PSI and energy from photons of light is passed on from pigment molecules to neighbouring pigment molecules by resonance and finally to special chlorophyll a in the reaction centres;; 2. Electrons in special chlorophyll a are excited and boosted to higher energy levels and these excited electrons are accepted by primary electron acceptors;; 3. The primary electron acceptors pass the electrons down a series of electron carriers of the electron transport chains (ETC);; 4. As electrons move down the ETC, energy is released to pump H+ from the stroma, (across the thylakoid membrane), into the thylakoid lumen; 5. Accumulation of H+ in the thylakoid lumen create an electrochemical and a proton gradient between the thylakoid lumen and stroma of chloroplast;; 6. H+ diffuse down their concentration gradient through stalked particles back to the stroma;; 7. releasing electrical potential energy which drives the phosphorylation of ADP to form ATP catalysed by ATP synthase;; 8. In non-cyclic photophosphorylation, electrons released from special chlorophyll a of PSII replaces electrons lost by special chlorophyll a of PSI and electrons from photolysis of water replaces the electrons lost from special chlorophyll a of PSII;; 9. Electrons released from special chlorophyll a of PSI combines with H+ from water to form H atom which is used to reduce NADP+ to NADPH;; 10. In cyclic photophosphorylation, electrons from special chlorophyll a of PSI passed down the ETC between PSII and PSI and return to the same special chlorophyll a of PSI;;
HOLIDAY ASSIGNMENT: ESSAY (b) Describe the effect of increasing light intensity on the rate of photosynthesis. 
[1 mark for correctly drawn and labelled graph] 1. 2. 3.
At region A, (Low light intensities) As light intensity increases, the rate of photosynthesis increases proportionally. Therefore light intensity is the main limiting factor;; Light is needed for photoactivation of special chlorophyll a in reaction centre / to excite electrons in special chlorophyll a to a higher energy level in light dependent stage of photosynthesis;; so that there is increased movement of electrons down both ETCs, resulting in greater number of ATP and NADPH synthesis for light independent reaction;;
At region B, As light intensity increases, the rate of photosynthesis increases gradually. Light intensity is becoming less of a limiting factor. Some factor other than light intensity e.g. CO2 concentration or temperature is becoming limiting;;
At region C, As light intensity increases, rate of photosynthesis levels off / becomes constant. The rate of photosynthesis (E) is at maximum. Light saturation has occurred at D. Light intensity is no longer the limiting factor, other factors e.g. temperature, carbon dioxide concentration are limiting;; This is because even though sufficient ATP and NADPH are synthesised but enzymes in the light independent reaction are temperature dependent and carbon dioxide is required for carbon fixation for synthesis of carbohydrates;;
HOLIDAY ASSIGNMENT: ESSAY 6. 2004 P2 Q10 (a) Describe the main stages of the Calvin cycle.  1. 2. 3. 4. 5. 6. 7. 8. 9.
Carbon dioxide is fixed by combining with ribulose bisphosphate (RuBP) to form a 6C unstable product;; catalysed by Rubisco/ RuBP carboxylase;; The 6C unstable product breaks down immediately into two molecules of glycerate phosphate (GP);; Each molecule of GP is phosphorylated by one molecule of ATP to one molecule of glycerate bisphosphate;; Each molecule of glycerate bisphosphate is reduced by one molecule of NADPH to form one molecule of triose phosphate (TP);; NADPH is oxidised to NADP+;; Some TP are used to regenerate the RuBP used during carboxylation;; One molecule of ATP is used to regenerate one molecule of RuBP;; Pairs of TP molecules are combined to produce an intermediate hexose sugar – glucose or fructose. (α-) Glucose can be polymerised to form starch;;
(b) Outline the role of NADP in photosynthesis.  1. 2. 3. 4. 5. 6.
NADP act as a coenzyme (for dehydrogenase);; NADP acts as hydrogen atom/hydrogen ion and electron carrier and carries the H removed from the substrates by dehydrogenases;; and is reduced to NADPH in non-cyclic photophosphorylation of the light dependent reactions;; NADPH provides the reducing power for Calvin cycle;; NADPH carries hydrogen atom / electrons and hydrogen ion from light dependent reaction to Calvin cycle;; For the reduction of GP to TP / reduction of glycerate bisphosphate to TP in the stroma of chloroplast;; leading to the regeneration of NADP (for subsequent light dependent reactions);;
(c) Explain how ATP is synthesized using light energy in photosynthesis.  1. Photosynthetic pigments such as chlorophyll a, chlorophyll b and carotenoids are embedded on the thylakoid membranes and arranged in photosystems i.e. PSI and PSII;; 2. Photons of light strike photosynthetic pigment molecules in PS/PSII/PSI and energy from photons of light is passed on from pigment molecules to neighbouring pigment molecules by resonance and finally to special chlorophyll a in the reaction centres;; 3. Electrons in special chlorophyll a is excited and boosted to higher energy levels and these excited electrons are accepted by primary electron acceptors;; 4. The primary electron acceptors pass the electrons down a series of electron carriers of the electron transport chains (ETC);; 5. As electrons move down the ETC, energy is released to pump H+ from the stroma, (across the thylakoid membrane), into the thylakoid lumen; 6. Accumulation of H+ in the thylakoid lumen create an electrochemical and a proton gradient between the thylakoid lumen and stroma of chloroplast;; 7. H+ diffuse down their concentration gradient through stalked particles back to the stroma;; 8. releasing electrical potential energy which drives the phosphorylation of ADP to form ATP catalysed by ATP synthase;;
HOLIDAY ASSIGNMENT: ESSAY 7. 2006 P2 Q7 (a) Outline the role of NAD and NADP in cells.  Role of NAD 1. Act as a coenzyme (for dehydrogenase) and hydrogen atom / hydrogen ion and electron carrier and is reduced to NADH during glycolysis, link reaction and Krebs cycle in aerobic respiration;; 2. NADH carries electrons and protons / hydrogen atoms to electron transport chain at inner mitochondrial membrane to be used in oxidative phosphorylation;; 3. Leading to the regeneration of NAD for subsequent glycolysis, link reaction and Krebs cycle to proceed;; Role of NADP 8. NADP act as a coenzyme (for dehydrogenase) and hydrogen atom / hydrogen ion and electron carrier and carries the H atoms removed from the substrates by dehydrogenases;; 9. NADP is reduced to NADPH in non-cyclic photophosphorylation of the light dependent reactions to provide the reducing power for Calvin cycle;; 10. NADPH carries hydrogen atom / electrons and hydrogen ion from light dependent reaction to Calvin cycle for the reduction of GBP to TP in the stroma of chloroplast;; 11. leading to the regeneration of NADP for subsequent light dependent reactions;; (b) Describe how photophosphorylation differs from oxidative phosphorylation. 
1. 2. 3. 4. 5. 6.
Features Location;; Accumulation of protons in;; Involvement of light energy;; Energy conversion Involvement of oxygen;; Involvement of water;;
Sources of energy for ATP synthesis;; Establishment of proton gradient;;
Photophosphorylation Thylakoid membrane of chloroplast Thylakoid space
Oxidative Phosphorylation Inner mitochondrial membrane Intermembrane space
Required to energise the electrons in special chl a Light energy Æ chemical energy Oxygen released as a by-product
Water molecules is split to produce H+, oxygen and provides replacement electrons for PSII For non-cyclic reaction: water For cyclic reaction: PS I For non-cyclic reaction: NADP+ For cyclic reaction: PS I Energy for making ATP comes from light H+ pumped inwards, from stroma to thylakoid space / lumen
Chemical energy Æ chemical energy Oxygen is used as the final electron acceptor Water released as a by-product
NADH, FADH2 Oxygen Energy for making ATP comes from glucose oxidation processes H+ pumped outwards, from mitochondrial matrix to intermembrane space
HOLIDAY ASSIGNMENT: ESSAY 8. 2008 P2 Q8 (a) Describe how Calvin cycle differs from the Krebs cycle. 
Krebs cycle Matrix of the mitochondrion (Occurs in all aerobically respiring cells) Released by oxidative decarboxylation - 4 CO2 lost per glucose molecule Involves oxidation e.g. citrate undergoes oxidative decarboxylation to form αketoglutarate NADH is formed, NAD+ is reduced to NADH ATP is synthesized by substrate level phosphorylation
Type of reaction;;
Fate of hydrogen atom carrier;; ATP;;
Oxaloacetate was regenerated
Nature of reaction;;
Types of hydrogen atom carrier;;
Catabolic reaction (breakdown of pyruvate) NAD+ & FAD
Calvin cycle Stroma of the chloroplast (Occurs in plant cells/ algae / bluegreen bacteria) Used for carboxylation - 6 CO2 fixed by combining with 6 RuBP Involves reduction e.g. GBP is reduced to triose phosphate
NADPH is used, NADPH is oxidised to NADP+ ATP is used in the phosphorylation of GP to GBP and in regeneration of RuBP Ribulose biphosphate was regenerated Anabolic reaction (formation of triose phosphate or starch) NADP+
(b) Explain the small yield of ATP under anaerobic conditions in both yeast and mammals.  1. 2. 3. 4. 5. 6. 7. 8.
Oxygen acts as the final electron acceptor at the end of the electron transport chain (ETC) and combines with electrons and H+ to form water catalysed by cytochrome oxidase;; In the absence of oxygen during anaerobic conditions, there is no acceptance of electrons at the end of the ETC and NADH, FADH2 and electron carriers along the ETC will all remain reduced and link reaction and Krebs cycle will stop;; No electron flow along ETC will occur and the build up of proton / electrochemical gradient between the intermembrane space and mitochondrial matrix for synthesis of ATP will be disrupted;; During anaerobic respiration, glycolysis followed by fermentation occurs in the cytosol;; Alcoholic fermentation occurs in yeast cells whereby pyruvate was converted to ethanol and carbon dioxide and lactate fermentation occurs in mammalian cells whereby pyruvate was reduced to lactate;; During fermentation, NAD+ is regenerated in order for glycolysis to continue;; There is a net gain of 2 ATP per glucose molecule only in glycolysis via substrate level phosphorylation;; during the conversion of glycerate bisphosphate (GBP) to glycerate phosphate (GP) and GP to pyruvate;;
HOLIDAY ASSIGNMENT: ESSAY (c) State the similarities between ATP production in mitochondria and chloroplasts and suggest why these similarities exist.  Similarities 1. Both involved an electron transport chain where electrons are transferred down the electron transport chain comprising of a series of electron carriers of progressively lower energy levels via redox reactions;; 2. Both involved the release of energy during electron flow down the ETC which is used to pump protons across membranes - inner membrane of mitochondria / thylakoid membrane of chloroplast;; 3. Both involves the generation of a proton gradient across the membrane - inner membrane of mitochondria / thylakoid membrane of chloroplast;; 4. During chemiosmotic synthesis of ATP in both mitochondria and chloroplasts, protons diffuse down their concentration gradient through stalked particles and releases electrical potential energy which drives the phosphorylation of ADP to form ATP catalysed by ATP synthase;; Reason for the similarities 5. Both mitochondria and chloroplasts were descendants of prokaryotic organisms that took up residence inside the host-cell precursors of eukaryotes and end up having a symbiotic relationship with the host cell;; 9. Outline the structure of DNA.  1. DNA molecule is a double helix of 2 complementary polynucleotide strands/chains;; 2. The two strands coil around each other in a right-handed double helix;; 3. The strands are antiparallel i.e. run in opposite directions (one strand runs in the 5’ to 3’ direction while the complementary strand runs in the 3’ to 5’ direction);; 4. Each strand consists of very long chain of nucleotides, a nucleotide is made up of a deoxyribose sugar, phosphate group and a nitrogenous base - Adenine, Thymine, Cytosine or Guanine;; 5. Each strand consists of a sugar-phosphate backbone with the nucleotides arranged in sequence, held together by phosphodiester bond between C3 of the sugar of one nucleotide and C5 of the sugar of the adjacent nucleotide;; 6. The nitrogenous bases are arranged as side groups of the chains (oriented toward the central axis);; (Extra pt) 7. The width between the 2 backbones is constant (2nm) , equals to the width of 1 base pair i.e. 1 purine + 1 pyrimidine;; 8. The bases pair with bases of the opposite strand via hydrogen bonds between complementary bases. There are 2 hydrogen bonds between adenine and thymine, and 3 hydrogen bonds between cytosine and guanine;; 9. The base pairs are 0.34nm apart in the DNA helix, therefore, there are 10 base pairs per turn, hence a complete turn of the helix is 3.4nm;; 10. Base pairing is complementary, i.e. A-T and C-G. The base-pairing is specific and the 2 strands are complementary (no of G = no of C, no. of A = no. of T);;
HOLIDAY ASSIGNMENT: ESSAY 10. 2011 P2 Q9 (c) Outline the main features of DNA replication  1. Semi-conservative replication begins at origins of replication;; 2. The two parental DNA strands separate due to the breaking of hydrogen bonds between complementary bases;; 3. Both strands act as templates for the synthesis of new complementary DNA strands;; 4. each new DNA molecule contains one original DNA strand and one newly synthesised DNA strand;; 5. Helicase causes the DNA molecule to unwind and unzip and the hydrogen bonds between complementary bases to break, causing the DNA strands to separate;; 6. single-strand DNA binding proteins bind to the 2 separated parental DNA strands to stabilise the single-stranded DNA formed;; 7. so that the unwound region can serve as template for synthesis of a new complementary strand;; (same point as pt 5. award once) 8. Primase catalyses the formation of a short RNA primer – the start of a new strand in the 5’ to 3’ direction (to initiate the replication process);; 9. DNA polymerase then binds to the RNA primer and adds nucleotides to the free 3’ end of the RNA primer;; 10. DNA polymerase can only work in one direction from 5’ to 3’, and can only add nucleotides to 3’ end of an existing strand;; 11. Free DNA nucleotides attach by complementary base pairing via hydrogen bonds with the complementary base in the DNA template;; 12. There are 2 hydrogen bonds between adenine and thymine, and 3 hydrogen bonds between cytosine and guanine;; 13. DNA polymerase also carries out proof-reading;; (extra point) 14. DNA polymerase catalyses the formation of phosphodiester bonds between adjacent nucleotides;; 15. Synthesis of newly synthesised strand occurs in 5’ to 3’ direction, thus the DNA template is read in the 3’ to 5’ direction as the DNA is antiparallel;; 16. Leading strand is synthesised continuously because the DNA polymerase is moving in the same direction as the unwinding of DNA;; 17. Lagging strand is synthesised discontinuously because the DNA polymerase is moving in opposite direction as the unwinding of DNA, resulting in Okazaki fragments;; 18. RNA nucleotides of RNA primers are replaced with DNA nucleotides by another DNA polymerase;; 19. DNA ligase seals the gaps between the DNA fragments by catalysing the formation of phosphodiester bonds between adjacent nucleotides to form a continuous strand;; 20. At the end of replication, each new DNA molecule contain 1 original DNA strand and 1 newly synthesised DNA strand;; (same point as pt 6. award once)