Holiday Assignment - Essay

7/18/2019 Holiday Assignment - Essay

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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. [6]

Structure of amino acids

1. Each amino acid consists of a central α-carbon atom bonded covalently to fourgroups;;

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 acid4. 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;;

;;




(b) Explain what is meant by primary, secondary, tertiary and quaternary structureof a named protein. [8]

1. The primary structure of a protein molecule refers to its number, type and sequenceof amino acids held together by peptide bonds in the linear strand of polypeptidechain;;

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 146amino acids;;

3. Secondary structures are geometrically regular repeating structures, including α-helixand the β-pleated sheet, stabilised by hydrogen bonds between groups in the mainchain of the polypeptide;;

4. Tertiary structure is the three-dimensional conformation of a polypeptide maintainedby ionic bonds, hydrogen bonds, disulphide bonds and hydrophobic interactions;;

5. Each polypeptide chain of haemoglobin coils into an α helix which is then foldedupon itself into a rough spherical globular protein ;;

6. The shape of the haemoglobin is held by interactions between the R-groups of itsamino acids - hydrogen bonds, ionic bonds, disulphide bonds and hydrophobicinteractions;;

7. Quaternary structure consists of an aggregation of 2 or more polypeptide chains heldtogether by hydrophobic interactions, disulphide linkages, hydrogen bonds and ionicbonds;;

8. Haemoglobin’s four polypeptide chains pack closely together, held together byhydrophobic interactions, disulphide linkages, hydrogen bonds and ionic bondsresulting in a nearly spherical haemoglobin molecule;;

(c) Outline how the structure of a named globular protein is related to its specific

function. [6]

1. Haemoglobin is a globular protein found in red blood cells that functions to transportoxygen;;

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 sphericalshape;;

3. Each polypeptide chain contains a (non-proteinaceous) prosthetic haem group andeach haem group contains an iron, Fe2+, which binds a molecule of oxygen;;

4. The four polypeptide chains pack closely together, held together by hydrophobicinteractions, 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 oxygenmolecules at a time;;

6. Since haemoglobin is in contact with solution, their hydrophobic R-groups point intothe centre of the molecule and their hydrophilic R-groups point outwards to formhydrogen bonds with water, hence maintaining its solubility in water;;

7. Due to its allosteric nature, the haemoglobin molecule can undergo drastic changesin shape as it binds O2 to make it easier for further O2 to bind;;




2. 2007 P2 Q8

(a) Describe the main struc tural features of cellulose and collagen. [8]

Cellulose1. 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;;

Collagen5. 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 arestaggered;;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. [8]

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 chemicalbonds 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 thesubstrate 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-substratecomplexes are formed;;

10. Temporary bonds such as ionic bonds, hydrogen bonds and hydrophobic interactionsbetween 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;;




(b) Explain how pH affects the rate of an enzyme catalysed reaction. [8]

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 reactiondecreases;;

6. as the change in pH alters the ionic charge of the acidic and basic R groups on theamino acids at the active site of the enzyme;;

7. the ionic bonds and hydrogen bonds that help to maintain the specific shape of the activesite of the enzyme are disrupted;;

8. causing a loss of 3D conformation of enzymes’ active site, loss of 3D conformation ofenzymes 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 offormation of products;;

(c) Explain the effect of non-competitive inhib itors on enzyme activity. [4]

1. well-annotated graph;;

2. non-competitive inhibitor has no structural similarity to the substrate and binds to theenzyme at site other than the active site (allosteric site);;

3. upon binding of the non-competitive inhibitor to the enzyme, the enzyme 3Dconformation is changed such that its active site's conformation is altered and substratecan no longer bind to the active site;;

4. formation of such enzyme-inhibitor complexes prevents the formation of E-S complexesand formation of products and decreasing the rate of reaction;;

5. an increase in substrate concentration will not reverse the inhibition/reduce the effect ofinhibition, even at very high substrate concentration, the maximum rate of reaction in thepresence of non-competitive inhibitor is lower than that of reaction in the absence of non-competitive inhibitor;;

6. allosteric inhibition can be reversible or irreversible and some non-competitive inhibitorsbind to active site irreversibly;




4. 2005 P2 Q8(a) Describe the effect of inhibi tors on the rate of enzyme activ ity. [8]

1. Enzyme inhibitors are molecules which prevent enzymes from catalysing the reactions,hence reducing the rate of reaction and inhibition can be competitive or non-competitive;;

Competitive inhibitor (max 4m)2. competitive inhibitor is structurally similar to the actual substrate and can fit into the

active site of the enzyme;;3. 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 ofproducts;;

4. enzyme-inhibitor complexes are formed instead as competitive inhibitor competes withthe actual substrate for binding at the active site;;

5. At low substrate concentration, the frequency of enzyme-substrate collisions is similar tothe 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;;6. 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;;

7. competitive inhibition is reversible and can be overcome by high substratesconcentration, as substrates can out-compete the inhibitors for binding to the active siteand allow the maximum rate of reaction to be reached;;

Non-competitive inhibitor (max 4m)8. 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 substratecan no longer bind to the active site;;

10. formation of such enzyme-inhibitor complexes prevents the formation of E-S complexesand formation of products, decreasing the rate of reaction;;

11. an increase in substrate concentration will not reverse the inhibition/reduce the effect ofinhibitor, even at very high substrate concentration, the maximum rate of reaction in thepresence of non-competitive inhibitor is lower than that of reaction in the absence ofnon-competitive inhibitor;;

12. allosteric inhibition can be reversible or irreversible and some non-competitive inhibitorsbind to active site irreversibly;;




(b) Cells have metabolic pathways commonly made up of sequences of enzymecatalysed reactions.Explain, with examples, the advantages of such sequences of enzyme catalysedreactions. [6]

1. Cells have metabolic pathways commonly made up of sequences of enzyme catalysedreactions in which the product from one reaction acts as the substrate for the nextreaction;;

2. Each reaction is catalysed by different enzymes and this allows intermediates to godown different pathways;;

3. The enzymes which catalyse such chain reactions often form a linear series bound tomembranes within the cell, constituting a multi-enzyme complex, and such closeproximity between enzymes is efficient since collisions between enzymes and theirsubstrates 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, toform a 6C unstable intermediate which splits immediately to 2 molecules of GP.

OR

b) Glycolysis in cytosol of cellsIn glycolysis, glucose is phosphorylated by ATP to glucose 6-phosphate, catalysedby hexokinase. Glucose-6-phosphate is isomerised by a different enzyme to becomefructose-6-phosphate.

OR

c) Krebs cycle in matrix of mitochondrionIn Krebs cycle, acetyl-coA combines with oxaloacetate to form citrate, catalysed bycitrate synthase. Citrate undergoes oxidative decarboxylation to become α-ketoglutarate, catalysed by citrate dehydrogenase

5. For many metabolic pathways, the final product of the pathway is usually an allostericinhibitor of one of the earlier enzymes in the pathway, and such inhibition of an earlierstage 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 furtherproduction, preventing wastage of resources; when the product is used up, the inhibitionis 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 endproducts;;




5. 2012 P2 Q9

(a) Outline the light dependent reactions of photosynthesis. [7]

1. Photons of light strike photosynthetic pigment molecules in PS/PSII/PSI and energy fromphotons of light is passed on from pigment molecules to neighbouring pigment moleculesby 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 andthese excited electrons are accepted by primary electron acceptors;;

3. The primary electron acceptors pass the electrons down a series of electron carriers ofthe 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 protongradient between the thylakoid lumen and stroma of chloroplast;;

6. H+ diffuse down their concentration gradient through stalked particles back to thestroma;;

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 PSIIreplaces electrons lost by special chlorophyll a of PSI and electrons from photolysis ofwater 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 formH atom which is used to reduce NADP+ to NADPH;;

10. In cyclic photophosphorylation, electrons from special chlorophyll a of PSI passed downthe ETC between PSII and PSI and return to the same special chlorophyll a of PSI;;




(b) Describe the effect of increasing light intensity on the rate of photosynthesis. [6]

[1 mark for correctly drawn and labelled graph]

1. At region A, (Low light intensities) As light intensity increases, the rate of photosynthesis increases proportionally.Therefore light intensity is the main limiting factor;;

2. Light is needed for photoactivation of special chlorophyll a in reaction centre / to exciteelectrons in special chlorophyll a to a higher energy level in light dependent stage ofphotosynthesis;;

3. so that there is increased movement of electrons down both ETCs, resulting in greater

number of ATP and NADPH synthesis for light independent reaction;;

4. At region B, As light intensity increases, the rate of photosynthesis increases gradually. Lightintensity is becoming less of a limiting factor. Some factor other than light intensity e.g.CO2 concentration or temperature is becoming limiting;;

5. At region C, As light intensity increases, rate of photosynthesis levels off / becomes constant. Therate of photosynthesis (E) is at maximum. Light saturation has occurred at D. Lightintensity is no longer the limiting factor, other factors e.g. temperature, carbon dioxide

concentration are limiting;;6. This is because even though sufficient ATP and NADPH are synthesised but enzymesin the light independent reaction are temperature dependent and carbon dioxide isrequired for carbon fixation for synthesis of carbohydrates;;




6. 2004 P2 Q10(a) Describe the main stages of the Calvin cycle. [8]

1. Carbon dioxide is fixed by combining with ribulose bisphosphate (RuBP) to form a 6Cunstable product;;

2. catalysed by Rubisco/ RuBP carboxylase;;3. The 6C unstable product breaks down immediately into two molecules of glycerate

phosphate (GP);;4. Each molecule of GP is phosphorylated by one molecule of ATP to one molecule of

glycerate bisphosphate;;5. Each molecule of glycerate bisphosphate is reduced by one molecule of NADPH to form

one molecule of triose phosphate (TP);;6. NADPH is oxidised to NADP+;;7. Some TP are used to regenerate the RuBP used during carboxylation;;8. One molecule of ATP is used to regenerate one molecule of RuBP;;9. 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. [6]

1. NADP act as a coenzyme (for dehydrogenase);;2. NADP acts as hydrogen atom/hydrogen ion and electron carrier and carries the H

removed from the substrates by dehydrogenases;;3. and is reduced to NADPH in non-cyclic photophosphorylation of the light dependent

reactions;;4. NADPH provides the reducing power for Calvin cycle;;5. NADPH carries hydrogen atom / electrons and hydrogen ion from light dependent

reaction to Calvin cycle;;6. For the reduction of GP to TP / reduction of glycerate bisphosphate to TP in the stroma

of chloroplast;;7. leading to the regeneration of NADP (for subsequent light dependent reactions);;

(c) Explain how ATP is synthesized using light energy in photosynthesis. [6]

1. Photosynthetic pigments such as chlorophyll a, chlorophyll b and carotenoids areembedded on the thylakoid membranes and arranged in photosystems i.e. PSI andPSII;;

2. Photons of light strike photosynthetic pigment molecules in PS/PSII/PSI and energyfrom photons of light is passed on from pigment molecules to neighbouring pigmentmolecules 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;;




7. 2006 P2 Q7(a) Outline the role of NAD and NADP in cells. [7]

Role of NAD1. 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 inaerobic respiration;;

2. NADH carries electrons and protons / hydrogen atoms to electron transport chain atinner mitochondrial membrane to be used in oxidative phosphorylation;;

3. Leading to the regeneration of NAD for subsequent glycolysis, link reaction and Krebscycle to proceed;;

Role of NADP8. NADP act as a coenzyme (for dehydrogenase) and hydrogen atom / hydrogen ion and

electron carrier and carries the H atoms removed from the substrates bydehydrogenases;;

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. [7]

Features Photophosphorylation Oxidative Phosphorylation

1. Location;; Thylakoid membrane of chloroplast Inner mitochondrial membrane

2. Accumulation ofprotons in;;

Thylakoid space Intermembrane space

3. Involvement of lightenergy;;

Required to energise the electrons inspecial chl a

Not required

4. Energy conversion Light energy chemical energy Chemical energy chemical energy

5. Involvement ofoxygen;;

Oxygen released as a by-product Oxygen is used as the final electronacceptor

6. Involvement ofwater;;

Water molecules is split to produce H+,oxygen and provides replacementelectrons for PSII

Water released as a by-product

7. Electron donors;; For non-cyclic reaction: water

For cyclic reaction: PS I

NADH, FADH2

8. Electron acceptors;; For non-cyclic reaction: NADP+

For cyclic reaction: PS I

Oxygen

9. Sources of energy for ATP synthesis;;

Energy for making ATP comes fromlight

Energy for making ATP comes fromglucose oxidation processes

10. Establishment ofproton gradient;;

H+ pumped inwards, from stroma tothylakoid space / lumen

H+ pumped outwards, frommitochondrial matrix to intermembranespace




8. 2008 P2 Q8(a) Describe how Calvin cycle differs from the Krebs cycle. [7]

Features Krebs cycle Calvin cycle

1. Site;; Matrix of the mitochondrion

(Occurs in all aerobically respiringcells)

Stroma of the chloroplast

(Occurs in plant cells/ algae / blue-green bacteria)

2. Carbon dioxide;; Released by oxidativedecarboxylation - 4 CO2 lost perglucose molecule

Used for carboxylation - 6 CO2 fixedby combining with 6 RuBP

3. Type of reaction;; Involves oxidation e.g. citrateundergoes oxidativedecarboxylation to form α-ketoglutarate

Involves reduction e.g. GBP isreduced to triose phosphate

4. Fate of hydrogen atomcarrier;;

NADH is formed, NAD+ is reducedto NADH

NADPH is used, NADPH is oxidisedto NADP+

5. ATP;; ATP is synthesized by substratelevel phosphorylation

ATP is used in the phosphorylationof GP to GBP and in regeneration ofRuBP

6. Starting point;; Oxaloacetate was regenerated Ribulose biphosphate wasregenerated

7. Nature of reaction;; Catabolic reaction (breakdown ofpyruvate)

Anabolic reaction (formation oftriose phosphate or starch)

8. Types of hydrogenatom carrier;;

NAD+ & FAD NADP+

(b) Explain the small yield of ATP under anaerobic conditions in both yeast and

mammals. [8]

1. 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 cytochromeoxidase;;

2. In the absence of oxygen during anaerobic conditions, there is no acceptance ofelectrons at the end of the ETC and NADH, FADH2 and electron carriers along the ETCwill all remain reduced and link reaction and Krebs cycle will stop;;

3. No electron flow along ETC will occur and the build up of proton / electrochemicalgradient between the intermembrane space and mitochondrial matrix for synthesis of ATP will be disrupted;;

4. During anaerobic respiration, glycolysis followed by fermentation occurs in the cytosol;;

5. Alcoholic fermentation occurs in yeast cells whereby pyruvate was converted to ethanoland carbon dioxide and lactate fermentation occurs in mammalian cells wherebypyruvate was reduced to lactate;;

6. During fermentation, NAD+ is regenerated in order for glycolysis to continue;;7. There is a net gain of 2 ATP per glucose molecule only in glycolysis via substrate level

phosphorylation;;8. during the conversion of glycerate bisphosphate (GBP) to glycerate phosphate (GP) and

GP to pyruvate;;




(c) State the simi larities between ATP production in mitochondria and chloroplastsand suggest why these similarities exist. [5]

Similarities1. Both involved an electron transport chain where electrons are transferred down the

electron transport chain comprising of a series of electron carriers of progressivelylower energy levels via redox reactions;;

2. Both involved the release of energy during electron flow down the ETC which is usedto pump protons across membranes - inner membrane of mitochondria / thylakoidmembrane of chloroplast;;

3. Both involves the generation of a proton gradient across the membrane - innermembrane 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 andreleases electrical potential energy which drives the phosphorylation of ADP to form ATP catalysed by ATP synthase;;

Reason for the similarities5. Both mitochondria and chloroplasts were descendants of prokaryotic organisms thattook up residence inside the host-cell precursors of eukaryotes and end up having asymbiotic relationship with the host cell;;

9. Outline the structure of DNA. [8]

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 insequence, held together by phosphodiester bond between C3 of the sugar of onenucleotide and C5 of the sugar of the adjacent nucleotide;;

6. The nitrogenous bases are arranged as side groups of the chains (oriented toward thecentral axis);; (Extra pt)

7. The width between the 2 backbones is constant (2nm) , equals to the width of 1 basepair i.e. 1 purine + 1 pyrimidine;;

8. The bases pair with bases of the opposite strand via hydrogen bonds betweencomplementary bases. There are 2 hydrogen bonds between adenine and thymine, and3 hydrogen bonds between cytosine and guanine;;

9. The base pairs are 0.34nm apart in the DNA helix, therefore, there are 10 base pairs perturn, 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 2strands are complementary (no of G = no of C, no. of A = no. of T);;




10. 2011 P2 Q9(c) Outline the main features of DNA replication [8]

1. Semi-conservative replication begins at origins of replication;;

2. The two parental DNA strands separate due to the breaking of hydrogen bonds betweencomplementary 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 synthesisedDNA strand;;

5. Helicase causes the DNA molecule to unwind and unzip and the hydrogen bondsbetween complementary bases to break, causing the DNA strands to separate;;

6. single-strand DNA binding proteins bind to the 2 separated parental DNA strands tostabilise 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 ofthe RNA primer;;

10. DNA polymerase can only work in one direction from 5’ to 3’, and can only addnucleotides to 3’ end of an existing strand;;

11. Free DNA nucleotides attach by complementary base pairing via hydrogen bonds withthe complementary base in the DNA template;;

12. There are 2 hydrogen bonds between adenine and thymine, and 3 hydrogen bondsbetween cytosine and guanine;;

13. DNA polymerase also carries out proof-reading;; (extra point)

14. DNA polymerase catalyses the formation of phosphodiester bonds between adjacentnucleotides;;

15. Synthesis of newly synthesised strand occurs in 5’ to 3’ direction, thus the DNA templateis read in the 3’ to 5’ direction as the DNA is antiparallel;;

16. Leading strand is synthesised continuously because the DNA polymerase is moving inthe same direction as the unwinding of DNA;;

17. Lagging strand is synthesised discontinuously because the DNA polymerase is movingin opposite direction as the unwinding of DNA, resulting in Okazaki fragments;;

18. RNA nucleotides of RNA primers are replaced with DNA nucleotides by another DNApolymerase;;

19. DNA ligase seals the gaps between the DNA fragments by catalysing the formation ofphosphodiester 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 1newly synthesised DNA strand;; (same point as pt 6. award once)

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