LSM1101 Notes

LSM1101 NotesThese notes have been made with the help of the slides from NUS professors. Dr Takao, Dr Deng and Dr Ban.

Terms

1. Amphipathica. Molecule has both hydrophilic and hydrophobic groups, and can interact with,

water, other hydrophobic groups, and organic solvents2. Amphoteric

a. Molecule has the ability to accept and donate protons

Solubility

1. Hydrogen bond occurs between a hydrogen atom with a partial positive charge, and an oxygen with a partial negative charge

a. It is weak and easily brokeni. Holds ice in a hexagonal structure when cold enough

ii. Easily formed and broken when water is liquidb. Water can form 4 hydrogen bonds

i. Each oxygen can form 2, each hydrogen can form 12. Solubility of compounds depends on how they interact with water molecules

a. Hydrophilic molecules form strong bonds with water and dissolve easilyb. Hydrophobic molecules do not form strong bonds with water and resists dissolution

3. Ionic compounds consists of ions that interact with water to form hydration shellsa. Electrostatic interactions between charged ions and partial charges of waterb. Enough energy released to overcome water-water bonds and ion-ion bonds

4. Polar molecules have hydrophilic groups that have partial charges to form hydrogen bonds with water.

a. The hydrophilic groups may have stronger partial charges due to the electronegativity of the rest of the molecule.

b. Hydration shell forms allowing dissolution5. Non-polar molecules are hydrophobic due to their inability to interact with water and thus

do not dissolve easily in water.a. The lack of hydrogen bonds causes the formation of a clathrate structure around

non-polar solutes that manage to dissolvei. Clathrate is a cage-like arrangement of water molecules around the solvent

ii. Clathrate formation is not energetically favoured and do not easily form, leading to insolubility of solute.

b. Hydrophobic interaction occurs, where hydrophobic molecules clump together to reduce the surface area exposed to water, leading to a lower energy state.

i. Minimises clathrates formation, as they are smaller and required less water molecules, and is thus favoured over scattered hydrophobic solutes.

ii. Only observed in aqueous environment, as this minimises interaction with water.

iii. This is not due to attractive forces like Van der Waals, but due to the system trying to reach the lowest energy state.

6. Amphipathic molecules have both hydrophilic and hydrophobic groups

a. Their interaction with other molecules and water gives rise to complex structuresi. Hydrophobic regions clump together to minimise exposure to water

ii. Hydrophilic regions are exposed to water7. Proteins tend to be amphipathic as it gives a stable structure

a. Hydrophilic surface allows dissolution in water found in most cellsb. Hydrophobic core maintains protein shape through hydrophobic interactions

pH and Buffers

1. pH = −lg ¿¿a. Kw=¿ at 298 Kelvinb. pH+ pOH=p K w

2. Human healthy pH is around 7.4a. pH is maintained constant by homeostasisb. Needs to remain constant to provide optimum environment for enzyme function

required for life3. Enzymes at different locations in the body operate best at different pH

a. Stomach enzymes require an acidic medium4. Bronsted-Lowry definition uses protons as reference instead of electrons (Lewis)

a. Acids are proton donors, dissociating H+, producing conjugate baseb. Bases are proton acceptors, associate with H+, producing conjugate acid

5. Ka is a measure of the strength of an acida. The smaller the Ka, the weaker the acid

i. The closer the pKa to 14, the weaker the acidii. Negative pKa indicates that the acid is a strong acid

6. Henderson-Hasselbalch equation: pH=pk a+lg ¿¿a. Not applicable when A- or HA concentration is near 0

7. When mixing weak acid and its conjugate basea. Assume that further dissociation is negligibleb. Use Henderson-Hasselbalch equation to find pH

8. Buffers resists changes in pH by reacting away excess H+ or OH- added9. Buffering capacity is pH=p Ka ±1

a. When equal concentrations of acid and conjugate base are present10. Polyprotic acids have multiple buffering capacities11. The human body has several buffers

a. Phosphate system (HPO42-/H2PO4

-)i. Protects pH of intracellular and extracellular environment

b. Bicarbonate system (HCO3-/H2CO3)

i. Protects pH of bloodii. Breathing adjusts CO2 concentration, affecting H2CO3 concentration

c. Amino acids and proteinsi. Minor contribution to pH homeostasis

Amino Acids

1. In carboxylic acids, the carbon atoms are named in sequence of the Greek alphabet.a. The first, excluding the C in the functional group, is named alpha, and last is omegab. This applies to amino acids since they have a carboxyl group

i. All 20 naturally-occuring amino acids are alpha-amino acids, with the amino group attached to the alpha-carbon

2. The alpha carbon is a chiral centre, as it has 4 different groupsa. Except for glycine which has 2 identical groupsb. There are two forms of chiral molecules, dexter (right) or laevus (left)

i. Determined by CORN rule, COOH, R, NH2 ,H. With the hydrogen pointed away from the viewer

The L- form has the CO-R-N sequence going counter-clockwise The D- form has the CO-R-N sequence going clockwise

ii. Other naming systems exists, such as R-/S- or d-/l-c. L-isomers are the commonly found stereoisomer in nature

i. D-isomers can lead to disease and deathii. Gramicidin antibiotic contains D-amino acids that perforates the bacterial

membrane3. Amino acids are amphoteric, due to the amino group being able to accept a proton, and the

carboxyl group being able to donate one.a. Amino acids always have a charge when in aqueous solution, as either the amino

group, carboxyl group, or both will be chargedi. This is because the loss of charge of one of the groups will only occur at a pH

where the other group is already charged Check out the pKa to visualise the movement of protons

ii. Zwitterion occurs when the amino acid has both positive and negative charge but is electrically neutral

4. There are 20 naturally occurring amino acids, which can be organised into several groupsa. Non-polar amino acids

i. Alanine, Ala, Aii. Isoleucine, Ile, I

iii. Leucine, Leu, Liv. Methionine, Met, Mv. Phenylalanine, Phe, F

vi. Proline, Pro, P Only amino acid where the R-group covalently links with the amino

group Affects folding of amino chains

vii. Tryptophan, Trp, Wviii. Valine, Val, V

ix. These are all relatively hydrophobic due to their R-groups, but still can dissolve in water, with the R-group participating in hydrophobic interaction

x. Ala, Val, Leu, Ile, Pro are aliphatic. Phe, Trp are aromatic. Met contains sulphur

xi. They are often found in the protein core, where hydrophobic interactions causes the protein to fold

xii. The lack of sufficient non-polar amino acids leads to intrinsically disordered proteins that do not fold into a specific shape.

b. Acidic amino acidsi. Aspartic acid, Asp, D

ii. Glutamic acid, Gla, Eiii. They contain a carboxyl group that can undergo deprotonationiv. At neutral pH, they are negatively charged

c. Basic amino acidsi. Arginine, Arg, R

ii. Histidine, His, Hiii. Lysine, Lys, Kiv. They have an amino group that can undergo protonationv. Arg and Lys are positively charged at neutral pH, due to the high pKa of the

R-groupvi. Histidine has an aromatic R-group that contains Nitrogen that can accept a

proton Only the Nitrogen with the double bond can accept a proton, as it

would not disrupt the aromatic ring structured. Polar, uncharged amino acids

i. Serine, Ser, Sii. Threonine, Thr, T

iii. Tyrosine, Tyr, Yiv. Asparagine, Asn, Nv. Glutamine, Gln, Q

vi. Cysteine, Cys, Cvii. Ser, Thr, Tyr contain an alcohol group, making them polar. Asn and Gln are

amides, with the –NH2 portion unable to undergo protonationviii. Threonine has a second chiral centre in its R-group

e. Aromatic amino acidsi. Phenylalanine, Phe, F

ii. Tryptophan, Trp, Wiii. Tyrosine, Tyr, Y

f. Sulfur containing amino acidsi. Methionine, Met, M

ii. Cysteine, Cys, Ciii. Only Cys is able to form disulphide linkages, allowing two amino acids chains

to be linked together. Disulphide linkages form after the proteins have folded Helps stabilize the folded structure

g. Glycine, Gly, Gi. Smallest amino acid

Essential for the formation of collagen’s triple helix as it is the only one to have a small R-group to fit in the helix’s centre.

5. The pKa of the amino acids are similara. The alpha amino group pka ≈9

b. The alpha carboxyl group pka ≈2

c. Acidic side chain (Gla, Asp) pka ≈4

d. Basic side chain (Lys, Arg) pka ≈11, (His) pka ≈66. Amino acids can act as precursors to important molecules

a. Essential amino acids are those the human body cannot produce and must be found from outside sources (food)

b. Non-essential amino acids are those the human body can produce7. Amino acids also have an isoelectric point where it is electrically neutral (zwitterion)

a. Affected by any charges on R-groupb. Located between two pKa

i. pI=p K1+ p K2

2, for acidic or uncharged amino acids

ii. pI=p K2+ p K3

2, for basic amino acids

8. Amino acids linked to each other via peptide bonds formed by condensation reactionsa. The peptide bond is unable to undergo protonation or deprotonationb. But can still undergo hydrogen bonding

9. Polypeptides are usually named from the N-terminal to the C-terminala. The terminals can still undergo protonation and deprotonation

10. There are also non-standard amino acidsa. They can be D-forms of amino acidsb. They can be non-alpha-amino acidsc. They have specific biological functionsd. They arise from posttranslational modification

i. Phosphorylation Adds PO4H2, that is negatively charged at neutral pH Can activate or deactivate a protein, acting as a switch

Proteins

1. Proteins are synthesized from mRNA, read in the 5’ to 3’ directiona. Produces protein in the N-terminal to C-terminal directionb. AUG is the start, methionine codonc. UAA, UGA and UAG are the stop codons that do not code for any amino acid

2. Proteins are a polymer of amino acids, with a unique sequence of thema. The primary structures ultimately determines the folding and function of the proteinb. Proteins are folded into a specific shape, that is not regular

i. There may be local sections that have regular folding, giving rise to secondary structures

3. There are several ways to determine primary structure of proteinsa. Direct sequencing through protein digestion into fragments, and then use

biochemical methods to determine amino acids present in fragmentsb. Mass spectrometry to determine mass of small protein fragments and infer amino

acids componentsc. Prediction from DNA or RNA sequence

4. Protein folding can be visualised by several methodsa. Computational models have various display modes include wireframe, ball and stick

and space-fillb. X-ray crystallography uses diffraction pattern of X-rays to determine the three

dimensional structure of the protein in a crystali. The diffraction patterns is processed to give an electron density map

ii. The electron density map is used to generate the protein structurec. The structure of an unknown protein is likely to be similar to that of a related known

one, and thus can be predicted.5. The protein needs to be correctly folded to function

a. To allow the binding of specific substratesb. To catalyse specific reactionsc. For structural proteins to have the mechanical strength, like collagen

6. Folding is driven by numerous weak interactions, assisted by molecular chaperonesa. Covalent bonds between amino acids remain unchanged

i. The only covalent bond formed is disulphide linkages after folding has occurred

Disulphide linkages can be broken by reducing agentsb. Hydrophobic interactions

i. It is the most important factorc. Hydrogen bonds

i. Give rise to secondary structuresii. Occurs between peptide linkages and R-groups

d. Ionic bondsi. Occurs between acidic and basic amino acids

ii. Affected by presence of ions from salts in solution, and pHe. Van der Waals forces

i. Weaker contributor than the others, but multiple interactions give rise to a significant effect

7. Denaturing agents like Urea or Guanidinium chloride interferes with hydrophobic interaction and hydrogen bonds, causing the protein to unfold

8. Under specific conditions, the protein can renature itself, showing that proteins know how to fold on their own

a. The unfolded protein settles into more energetically favourable conditions, before settling into the most energetically favourable configuration

b. Unfolded proteins may aggregate, thus molecular chaperones are needed inside cell, as they have high protein concentrations

i. Aggregation forms insoluble products that leads to cell deathii. The molecular chaperones prevent or reverse misfolding

They are hollow to create an environment where the misfolded protein can fold properly

Example is heat shock proteins that help proteins refold when under heat stress

9. Secondary structures are local structures in proteinsa. They are held by hydrogen bonds between peptide bonds, and disulphide linkages

i. There is strong partial charges on CO and NH, due to the double-bond characteristics caused by resonance structures

The electrons from O can travel to N, creating a full negative charge on O, and full positive charge on N

Hence, the peptide bond usually has partial charges for hydrogen bonding

This is due to the shorter than usual C-N bond length This also prevents free rotation along the C-N bond

Called the ‘amide plane’ of electron density along the bond Thus only the carbon-carbon bond adjacent to the peptide

bond can rotateRotation of Cα -C bond = Ψ (Psi)Rotation of Cα -N bond = Φ (Phi)

b. Regular structures are formed when all the amino acids have the same Psi and Phi angles

i. Alpha helix 3.6 amino acids per turn An amino acids has hydrogen bonds with the one 4 residues away All the –CO– groups point in the same direction All the –NH– groups point in the same, opposite direction to –CO– R-groups extend outwards All alpha helix are right handed helix, since left handed helix is

impossible due to steric hindrance between R-groups Glycine and Proline are rare in alpha helix

Glycine is too flexible and is unlikely to conform to the same Psi and Phi angles

Proline lacks the –NH– group to extend the alpha helixii. Beta sheet

Alternating amino acids point in the opposite direction The Psi bond is the 180 degrees different from the Phi bond Thus hydrogen bonds occurs between strands in the beta

sheet R-groups project above or below the plane of the beta sheet

Parallel

The C-terminal of each amino acid strand points in the same direction

Antiparallel The C-terminal of each amino acid strand points in different,

alternating directions Proline is rare in beta sheets, as it cannot form hydrogen bonds and

there is a constraint on the Phi angleiii. Beta turn

Usually found between antiparallel beta sheets strands There is an amide plane at right angles to the plane of the

two strands at both ends Is not a repeating structure like alpha helix or beta sheet Commonly contains proline and glycine

iv. Collagen triple helix10. Super-secondary structures are still local folds, but on a larger scale, forming motifs

a. They are a combination of alpha helices and beta sheets (with beta turns), in specific geometric arrangements

i. Beta barrel is found in Green Fluorescent Protein Made from multiple antiparallel beta sheets, with beta turns,

arranged into a cylinderii. TIM-barrel motif is found in Triose Phosphate Isomerase

Made from a sequence of beta, alpha, beta repeating secondary structures.

b. They are not unique to the protein, and can be repeated within the protein11. Another super-secondary structure is domains, which are independently folded globular

unitsa. Each domain is like a small globular protein

i. Hydrophobic core and hydrophilic surfaceii. 100 to 200 amino acids long

iii. Has a specific functioniv. Is modular, so can be combined in different combinations for a different

overall protein functionb. Trans-membrane proteins like receptor tyrosine kinases have domains that allow

them to respond to different stimulii. The extracellular domains are different to allows binding to different

signalling proteinsii. The intracellular domains are related, and have a similar function to act as

protein kinases to phosphorylate other proteins in the cell.12. Tertiary structure is the global folding of the amino acids into the specific shape of the

protein13. The evolutionary relationships of species can be determined from their proteins

a. Homologous proteins have similar amino acid sequencesi. Ortholog is when homologous proteins perform the same function in related

organisms Arises from a speciation event This reflects evolutionary relationships as each species accumulates

different mutations when branching from a common ancestor

ii. Paralog is when homologous proteins within the same organism has different functions

Arises from a duplication event Gene duplication eventually results in differing functions, as

evolutions favours the diversity of additional functions from the copy.

b. Amino acid sequence is not the only indication of relationships between proteins, structure can show relationships as well

i. Closely related species will have similar sequences and structures (homology)

ii. Distantly related species can have different sequences, but similar structures Their identical protein functions shows that the species are related

iii. Functionally distinct proteins can also be related, if they have similar structures

14. Protein types relates to their functionsa. Globular proteins

i. The proteins fold into a compact shape, with a hydrophobic core, and hydrophilic surface

They are thus soluble in waterii. They have diverse structures

Thus, have diverse functionsb. Fibrous proteins

i. The proteins largely consists of one type of secondary structure It is simple and repeated

Like alpha helices and beta sheets The hydrophobic residues are in contact with the environment

Insoluble in waterii. Thus, they usually have a structural role

iii. Collagen has a triple helix, unique to collagen 3 amino acid chains are wound around each other, with each chain

being a left-handed helix Has a characteristic amino acid sequences

-(Gly-Xxx-Pro/Hypro)n- Only Glycine has an R-group that fits in the core

c. Membrane proteinsi. Integral membrane proteins have a part of the protein within the

membrane, and a part outside 1 or more segments of the protein may span the membrane Proteins have a hydrophobic region, and a hydrophilic one

Around 20 amino acids span the membrane They are usually in an alpha helix

Polar groups in the peptide bond are involved in hydrogen bonding

Hydrophobic R-groups face the outside Receptors and transporters helps to transmit signal or molecules

across the membrane15. Quaternary structures are formed when multiple proteins are bound together by disulphide

bonds or other non-covalent interactions

a. The protein complex is made up of protein subunitsi. Subunits can be the same type or different types

16. Insulin is needed for the homeostasis of blood glucose levelsa. Is made up of two separate chains linked by disulphide linkages

i. The cell synthesises them together as one long polypeptide chain, which folds into the correct shape with disulphide linkages

ii. Protease then cleaves them to produce two linked strandsb. Since the genetic code is the same across all life, E. coli can be used to synthesise

insulini. Avoids the problem of immune reaction from using pig insulin

ii. The polypeptide chains are made separatelyiii. Denaturing and renaturing processes allow them to fold into insulin

The instructions for folding is inherent in the protein17. Prion is an infectious protein, which causes Transmissible spongiform encephalopathies

a. Prion protein is native to humans, and misfolding causes the disease phenotypei. The misfolded PrP is protease insensitive, forming insoluble fibres, leading to

deathii. Furthermore, the diseased PrP induces normal PrP to misfold

18. Other diseases like Alzheimer’s may be due to protein misfolding and aggregation, causing cell death

Enzymes

1. Enzyme is a protein catalyst that increases the velocity of a chemical reaction, without being consumed during the reaction

a. It has an active site where the substrate can bind for the reaction catalysisi. Binding of substrate is held by multiple weak forces

b. They allow for higher reaction rates, as they increase the reaction velocityc. They allow for milder reaction conditions, at lower temperaturesd. They allow for greater specificity, as there are no contaminants produced

i. As in the products are specific, giving high product yields e. They have a greater capacity for regulation

2. Enzymes fit their substrate either by lock and key model, or induced fit modela. In lock and key, the substrate has the exact shape of the active siteb. In induced fit, the binding of the substrate induces a conformational change in the

active site, giving rise to a complementary fit 3. Enzymes require cofactors to function

a. They are non-protein componentsi. Inorganic metal ions

ii. Organic molecules (coenzymes)b. They are like reactants and can get changed, but in the end, they are regenerated by

other processes in the celli. Or within the same process for metal ions

c. Prosthetic groups are cofactors that are tightly bound to proteins, and may be attached covalently.

4. Enzymes are affected by pHa. pH affects the ionization of R-groups in the protein amino acids

i. This changes substrate-active site interaction, affecting enzyme activity5. Enzymes are affected by temperatures

a. The enzyme activity decreases at extreme temperatures due to denaturationi. Else the rate of reaction would have increased linearly, like in a non-enzyme

catalysed reaction 6. Reactants converting to products go through a transition state

a. It is their highest level of free energyi. It is highly unstable and cannot be extracted

b. Activation energy is the energy required to get them to the transition statei. ΔG‡ = activation energy. ‡ is double dagger

7. Unlike Chemistry, even though K>1 , ΔGO<0, the reaction may not be spontaneous, due to high activation energy

a. The reaction is thermodynamically feasible, but may not be kinetically feasible8. Enzymes lower the activation energy of the reaction, without affecting other thing like K or

ΔGa. They accelerate both the forward and reverse reactions, so equilibrium is not

affected9. The enzymes are complementary to the reaction’s transition state

a. If it was complementary to the reactants, the bonded state would be energetically lower than the transition state, thereby further increasing the activation energy

b. Both the enzyme and substrate interact to reach the transition state, thus lowering the energy needed

10. Enzyme catalyse reactions by 3 waysa. Acid-base catalysis

i. The enzyme adds or accepts a proton from the functional group of the substrate, making it more reactive

It helps create the partial charges formed in the transition stateii. The acidic and basic amino acids can do this, as they are have R-groups that

are either in native or conjugate acid/base state.iii. The protons accepted or donated are regenerated by water

b. Covalent catalysisi. There is a transient formation of an enzyme-substrate covalent bond

ii. Due to nucleophilic attack of enzyme nucleophile to electrophile substrate The electron rich atom is usually O, S or N, and can bond with

electron deficient atoms like C, protons or metal ionsiii. The covalent bond pulls electrons away from the reaction centre, allowing

the other part of the substrate to be attackedc. Metal ion catalysis, with the ion from the environment

i. The metal ions bind to both the substrate and enzyme, orientating the substrate

ii. The metal ion stabilize negative charges during transition state The positive charges help reduce repulsion between a negative

group and electron pairs on attaching nucleophiles.iii. The metal ion can promote nucleophilic attack, by allowing the electrons to

be passed along a chain of water The water gets ionized while passing the charge to the metal ion The hydrogen bond network, a series of formation and breaking of

hydrogen bonds as the electrons are passed.iv. Metal ions also allow for carbonic anhydrase

The metal ion accepts a OH- ion produced from water The OH- ion attacks nearby bound substrate CO2 to form HCO3

- The catalytic site is then regenerated by another water molecule

Kinetics

1. The step with the higher activation energy, is the slowest step. Thus it is the rate-determining step

a. Unlike the transition state, the intermediate product between reaction steps can be extracted

2. The velocity of the reaction is the change in concentration of reactant or product per unit time

a. Velocity unit is M per second.i. Short essay question answers needs to contain units

b. v=d [P]

dt=

−d [R]dt

, where R is reactant, and P is product

c. v=k [R], where k is the rate constant3. The order of the reaction depends on the number of molecules participating in the rate-

determining stepa. First-order reaction has only one molecule, second-order has two.

i. The two molecules can either be the same or differentb. The rate equation changes depending on the order of reactionc. Third-order is rare, as simultaneous collision of three molecules is rare

4. While chemical reaction rates can continue to increase indefinitely with increasing concentration, enzyme-catalysed reactions have a plateau

a. At low substrate concentration, velocity is linearly proportional to itb. As substrate concentration increase, the velocity approaches its maximum, causing

the rate of increase to be less than linearc. At high substrate concentrations, the velocity is independent of substrate

concentrationi. Other factors still affect velocity, like temperature, pH and enzyme

concentration

5. Michaelis-Menten Equation: V 0=V max [S]K M+[S ]

a. At the initial reaction velocity, there is no reverse reaction of product becoming substrates, and the concentration of enzyme-substrate complex is constant

i. The rate of formation of the ES complex is equal to its dissociation rate

b. K M is the Michaelis constant: K M=k−1+k 2

k1i. It is equal to the rate constant of the breakdown of ES (forward to form

products, and backwards to substrate), divided by the rate constant of the formation of ES

ii. Unit is s−1

M−1 s−1=M

c. V max is the maximum velocity, and occurs at high substrate concentration where all the enzymes are in enzyme-substrate complexes.

6. K M is also the initial substrate concentration when the initial reaction velocity is half its maximum

a. This is derived from the Michaelis-Menten equationb. The equation also shows how the reaction is zero-order at high substrate

concentration, and first-order at low substrate concentration

c. The smaller the K M value of the enzyme, the lower the substrate concentration needed to achieve maximal velocity

i. It is unique for each enzyme-substrate pairii. It is a measure of enzyme affinity for the substrate

K M ≈1

affinity7. The disadvantage to the Michaelis-Menten plot is the difficulty in finding V max and K M, as

large substrate concentrations are needed for the curve to plateau.a. Even then, V max might not be obvious

8. The alternative is to draw the Lineweaver-Burk plot, that is a double-reciprocal plot

a.1v

is plotted against 1

[S ]

i.1V 0

=( K M

V max) 1[S ]

+ 1V max

b. The y-intercept is 1

V max, as the substrate concentration approaches infinity

c. The x-intercept is −1KM

i. It is negative, as it is found behind the y-axis, since there is no natural way for the velocity to increase above V max

ii. It is derived from the equation

d. The gradient of the slope is KM

V max

9. k cat is the turnover number, the number of substrate molecules converted to product per enzyme molecule, per unit time, when the enzyme is saturated with substrate

a. V 0=kcat [ES ]b. V max=k cat [E ] total

c. k cat=k2

10. Catalytic efficiency: kcat

KM

a. It is the efficiency of the enzyme at low substrate concentrations

b. V 0≈k cat

K M

[ E ]total [S ]

11. Irreversible inhibition is when the inhibitor forms covalent bonds with the functional groups at the active site of the enzyme

a. V max decreases, as effective enzyme concentration decreasesb. K M decreases, with the K M decreasing proportionally to V max

12. Reversible inhibition is when the inhibitor associates reversibly with the enzyme at various sites

a. Competitive inhibitors binds to free enzyme at the active site, competing with substrate binding

i. They are structural analogues of the substrateii. They increase the K M of the enzyme without affecting V max

iii. V 0=V max[S ]

α K M+[S], where α=1+

[ I ]K I

K I is the dissociation constant for the formation of EI from E and I

K I=k−3

k3iv. The effects of the competitive inhibitor can be reduced by increasing

substrate concentrationb. Uncompetitive inhibitors only bind to the enzyme-substrate complex

i. This affects the enzyme conformation, reducing the catalytic capacityii. V max and K M decreases, with the K M decreasing proportionally to V max

The, enzyme is unable to catalyse the reaction as effectively, decreasing V max proportionally

K M decrease as well, as k 2 is reduced k 2 is equal to V max at high substrate concentration, thus the

decrease is proportional

iii. V 0=(V max

α ' )[S ]

(K M

α ' )+[S], where α

'=1+[ I ]K I

'

K I' is the dissociation constant for the formation of ESI from ES and I

iv. The substrate binding to free enzyme is unaffectedc. Non-competitive/ mixed inhibitors bind to both the free enzyme and enzyme-

substrate complexi. The V max, will definitely decrease, as ESI affects catalytic activity

ii. Change in K M depends on affinity of inhibitor with E and ES If K I is greater, then K M will increase

If K I' is greater, then K M will decrease

If they are equal, there will be no change in K M

Pure non-competitor

iii. V 0=(V max

α ' )[S ]

(αK M

α ' )+[ S]

Enzyme regulation

1. Enzyme activity is dependent on their concentration and affinity for the substratea. By changing the rate of transcription, translation and degradation, enzyme activity

can be controlledb. There are also modifications that can be done to the enzyme to regulate their

activity2. Glucose levels in the cell are controlled by hexokinases I to IV, which phosphorylate the

glucose to prevent their transport out of the cell.a. Hexokinase I to III are found in most tissues except liver

i. They have a low K M , that allows for utilisation of glucose when blood glucose is low, by trapping the glucose in the cell

ii. Their V max is low as well, excessive levels of glucose is not neededb. Hexokinase IV (glucokinase) is found in the liver cells

i. They have a high V max to allow the uptake of large amounts of glucose from the blood

ii. They also have a high K M to allow for the phosphorylation to occur mainly at high blood glucose concentrations

iii. This buffers the level of blood glucose, by removing excess glucose and storing them in the liver.

3. Enzyme activity can be regulated by localization, by isolating the enzyme from the substratea. Glucokinase activity in liver cells is regulated by Glucokinase regulatory protein

(GKRP)b. GKRP will competitive inhibit the binding of glucose to hexokinase IV

i. Once bound, GKRP inactivates glucokinase by bringing it into the nucleusc. Thus formation of more GKRP-GK complexes reduces enzyme activity

i. GKRP’s affinity for GK is increased in the presence of Fructose 6-phosphate, another sugar source

ii. Whereas, higher levels of glucose reduces the overall activity of GKRPiii. Thus, at low blood glucose levels, Hexokinase IV activity is reduced as it is

bound to GKRP Prevents the liver from competing with other organs for glucose

d. Overall, glucokinase activity is regulated by three factorsi. Its own kinetics property

High K M

ii. Binding to GKRP that inactivates itiii. The level of expression of glucokinase gene

In response to insulin levels, as insulin induces more glucokinase production to reduce blood glucose levels

4. Enzyme activity is also regulated by cleavage activationa. Some enzyme are synthesized in their proenzyme (zymogen) form

i. These are inactivated and require cleavage to be activatedii. Like proteolytic enzymes of digestive tract and blood clotting proteins

iii. They have either Pro- as prefix, or –ogen as suffixb. Autocatalysis may occur to speed up activation at desired site

i. Where the product acts as a cofactor to the cleaving enzyme Enteropeptidase requires trypsin to cleave trypsinogen into trypsin

c. Self-digestion can also occur to further cleave excess amino acids

i. Chymotrypsinogen is partially activated through trypsin cleavage to give pi-chymotrypsin

ii. Pi-chymotrypsin will self-digest to produce alpha-chyymotrypsin that is fully activated

d. For proteases, trypsin, once activated, will cleave several other proenzymes to activate them

i. Thus this is like a cascade of enzyme activatione. Enzyme activation can occur through different stimulus

i. Blood clotting is triggered by either external damage to tissue surface, or internal trauma

ii. Both start different cascade reactions that eventually produce Factor Xa

iii. Factor Xa cleaves prothrombin into thrombin, allowing it to cleave fibrinogen into fibrin

iv. Fibrin then aggregates into the clot5. Enzyme activity is regulated by allosteric regulation

a. Through the non-covalent binding of effectors at a regulatory site other than the active site

b. Protein usually consists of multiple subunits, and the binding of effectors changes their conformation

i. Negative effectors inhibit enzyme activityii. Positive effectors promote enzyme activity

iii. When conformation changes, the shape of the active site changes, affecting its affinity to the substrate.

c. Homotropic effectors are the substrates that bind to the regulatory site to affect the binding of substrate to the active site

d. Heterotropic effectors are effectors other than the substratee. When allosteric enzymes have multiple active site, each active site acts as a

regulatory site for other active sitesi. Thus the binding of one substrate will influence the binding of substrate to

the other active site (Cooperative substrate binding) Cooperativity can be positive or negative A sigmoid curve (S shape) results when active site affinity keeps

switching depending on number of substrates already boundf. This allows for feedback inhibition, where the high concentration of products will

reduce enzyme activity, by binding to it as a negative effectori. Need not be product directly, as it can undergo further modification before

becoming the negative effectorg. Other enzymes can be in the inactive state until the required cofactor (positive

effector) binds to the regulatory sites to expose the active sites.i. The catalytic subunits can also detach or rejoin the regulatory subunits

depending on presence of cofactors6. Lastly, enzyme activity can be regulated by covalent modification of the enzyme

a. Modifications includei. Phosphorylation

Adding of phosphate groupii. Methylation

Adding of methyl groupiii. Uridylylation

Adding of uridine baseiv. Adenylylation

Adding of adenine baseb. Phosphorylation of glycogen phosphorylase causes a conformational change into a

more active statei. This increases the rate of phosphorylation of glycogen, which extracts a

glucose 1-phosphate from the glycogen chain of glucose7. Enzyme can be regulated by a combination of the above methods

a. Glycogen phosphorylase can be activated by phosphorylation (covalent modification) or binding of effectors (allosteric regulation)

i. Positive effector is AMPii. Negative effector is ATP and glucose-6-phosphate

Haemoglobin and Myoglobin

1. Heme is a complex of an organic molecule (protoporphyrin IX) and ferrous iron (Fe2+)a. The iron is held in the centre ‘space’ of the molecule by four Nitrogen atoms in the

porphyrin ringb. The iron has six liganding positions, of which the 4 liganding positions that lie on the

same plane are taken up by protoporhyrin IXi. Thus 2 more liganding positions, above and below the plane, are available

for bonding to other ligands In myoglobin, one of the protein domains is the fifth ligand, allowing

oxygen to bind to the 6th and last slot.2. Myoglobin is an oxygen binding protein that comes in different forms

a. Deoxymyoglobin contains ferrous iron (Fe2+) in the heme group, but the 6th ligand position is vacant and available for binding to oxygen

b. Oxymyoglobin contains ferrous iron (Fe2+) in the heme group, with the 6th ligand position occupied by oxygen

c. Metmyoglobin contains ferric iron (Fe3+) in the hematin group, with the 6th ligand occupied by water

i. Due to the oxidation of the iron, the heme turns into hematin, preventing binding to oxygen

ii. The metmyoglobin thus cannot function as an oxygen transport protein3. The globin component helps to protect and control the heme group

a. Heme can bind to oxygen by itself, but it can also freely bind to carbon monoxide with a higher affinity

b. The globin group cradles the heme group, protecting the iron ion from further oxidation

c. The globin group also decreases the affinity of heme for carbon monoxidei. This is done by steric hindrance, where the amino acid residues near the 6th

ligand position forces carbon monoxide from its preferred perpendicular alignment

The carbon monoxide has to bend at a 120 degrees angle instead, thereby making it less likely to bind

It’s affinity decreases from 25000 times that of oxygen, to 250 times4. The binding of oxygen to the iron ion in heme drags it towards the oxygen molecule.

a. This then pulls on the amino acid residue in the globin chain, leading to a conformational change

5. Hemoglobin consist of 4 myoglobin subunits (tetramer), two alpha chains, and two beta chains

a. Each alpha chain is in contact with two beta chains, and vice versai. Little alpha-alpha chain interaction and vice versa

ii. A alpha chain interacts with one of its beta chain neighbours, forming 2 alpha-beta dimers altogether

Strong hydrophobic interactions between the two chains in the dimer

Weak hydrogen and ionic bonds between the two dimersb. Each subunit has a heme group that can bind to oxygen

i. Thus each hemoglobin molecule can bind 4 oxygen moleculesc. When one myoglobin subunit binds to an oxygen molecule, the conformational

change affects its interaction with the other subunitsi. The change in quaternary structure weakens the bonds between the two

alpha-beta dimers The hemoglobin changes from the deoxy form (T state) to its oxy

form (R state)ii. In the T conformation, oxygen can only bind to the accessible heme groups

in the alpha chains Steric hindrance prevents binding to the beta chains’s heme

iii. Thus there is cooperative binding of oxygen, as transiting to the R conformation allows binding to the heme groups in the beta chains

6. Hemoglobin and myoglobin have different properties (affinity for oxygen)a. Hemoglobin has a steep oxygen dissociation curve at the oxygen concentrations

occurring in tissues.i. Thus oxygen can be released when hemoglobin reaches the tissues

ii. This steepness also allows it to respond to small changes in tissue oxygen pressure, releasing more or less oxygen when needed

7. Hemoglobin’s affinity for oxygen is affected by pHa. The Bohr effect, where greater acidity (lower pH) leads to lower affinity for oxygen

i. Greater oxygen pressure is needed for saturation, and less oxygen is transported to the tissues

ii. However, more of the oxygen in hameglobin is released due to the lower affinity, as less oxyhemoglobin leaves the tissues

b. The binding of protons (H+) to the hemoglobin promotes the dissociation of oxygen, thus reducing affinity for oxygen binding

i. The protons will induce the hemoglobin into the T form, hindering binding of oxygen

By promoting ionic bonding between amine and carboxylic acid Carboxylic acid is usually negatively charged as the pH does not fall

to low levels enough for it to protonate Thus lower pH will cause the amino group to protonate

8. Hemoglobin’s affinity for oxygen is also affected by carbon dioxidea. Carbon dioxide dissolved in the blood reacts with the water to give carbonic acid

i. This reduces the blood pH, as protons are produced from the dissociationb. Carbon dioxide can also react with the N-terminal of the subunits in hemoglobin,

producing carbamate that forms salt bridges which stabilise the T conformationi. This also produces protons that further decrease the pH

c. Thus the higher carbon dioxide concentration near the tissues promote the release of oxygen from the hemoglobin due to the two effects that favour the T conformation

9. Lastly, Hemoglobin’s affinity for oxygen is affected by 2,3-Bisphosphoglycerate (BPG)a. 2,3-BPG is an abundant organic phosphate in red blood cellsb. 2,3-BPG binds to the positively charged cavity only present in deoxyhemoglobin

i. The cavity is formed by the two beta chains, and is positively charged due to the amino acid residues facing the cavity

ii. This reduces hemoglobin’s affinity for oxygen, as the T conformation is stabilised

The oxygen dissociation curve is shifted to the right, meaning more oxygen is released at the tissues

Needed for high altitude environmentsc. Fetal hemoglobin contains a different subunit (gamma) that has a lower affinity for

2,3-BPGi. This increases its affinity for oxygen

ii. It binds to oxygen released by the maternal hemoglobin, allowing for the fetal tissues to be supplied with oxygen

10. Sickle Cell Anemia is caused by a mutation in the gene coding for hemoglobina. This results in a protrusion in the beta-chain subunit

i. The protrusion can fit in hydrophobic pocket present only in deoxyhemoglobin

ii. This causes the polymerisation of HbS, forming insoluble fibers when oxygen is released

b. The formation of fibers deforms the red blood cell, hindering its movement through tiny blood capillaries, reducing blood flow, preventing efficient oxygen transport

Nucleic Acids

1. Chargaff’s rule states that the amount of purines equals the amount of pyrimidines in DNAa. Purines are Adenine and Guanineb. Pyrimidines are Thymine, Cytosine and Uracilc. Later explained by complementary base pairing

2. 3’-5’ phosphodiester bonds joins the nucleotides together3. Beta-glycosidic bonds join the nitrogenous base to the carbon ring4. DNA double helix structure results in a major groove and a minor groove

a. Major groove where the vertical distance between a backbone and the next one is bigger

i. Due to the angle of the glycosidic bonds between the carbon ring and nitrogenous base

b. Minor groove where the vertical distance is lesser5. DNA has three conformations

a. B-DNA is commonly foundi. It is right handed

ii. Its major groove is wide with medium depthiii. Its minor groove is narrow with medium depth

b. A-DNA is synthesized in the lab, when DNA is dehydratedi. It is right handed

ii. Its major groove is narrow but deep

Still called major groove as it is the analogous location in B-DNAiii. Its minor groove is broad but shallow

c. Z-DNA occurs when there are CpG repeats (p stand for phosphate)i. It is left handed

Due to glycosidic bonds for pyrimidines being in the syn-position All glycosidic bonds are in the anti-position in A and B-DNA

Can be transitioned to from B-DNA, by the rotation of those bonds May be linked to regulation of gene expression Methylation of cytosine promotes this transition

ii. Its major groove is flattened outiii. Its minor groove is narrow but deep

6. In prokaryotes, their closed loop DNA can undergo supercoilinga. It is usually negatively supercoiled

i. Against the direction of DNA rotationb. The topology of supercoiling is determined by the Link number

i. Link number = number of twists + number of writhing Twists is the turning about the helical axis

Can be positive or negative Related to tension

Writhing is the number of turns around the superhelical axis Can be positive or negative Related to supercoiling

Any combination of Twists and Writhing is possible so long as they equal the link number

ii. Link number can only be changed by topoisomerase that cut the strand7. In eukaryotes, DNA is compacted into chromatin

a. DNA is coiled around histone proteins to form nucleosomesi. Histone have many arginine and lysine residues that are positive charged to

bind to the negatively charged phosphates in the DNA backboneb. Nucleosomes compact to form solenoid, which compacts to form loops, which

compact to form metaphase chromosomei. Chromosomes have dark and light banding when stained

GC rich regions are light bands that usually contain more genes AT rich regions are dark bands that usually contain less genes

ii. The location on a chromosome can be written in the format AA q/p BB Where AA is the chromosome number q or p refers to the shorter or longer arm Where BB is the region and band

Number is counted from the centromere and may not start from 1

8. Karyotypes are written in the format AA,XXa. Where AA is the total number of chromosomesb. Where XX are the sex chromosomes

9. RNA is usually single stranded, but can form secondary and tertiary structuresa. Not all bases needs to be complementary paired, thus giving a budge that is the

secondary structurei. The junction secondary structure of tRNA allows for base pairs to stick out of

the molecule, to interact with the codons on mRNA

The arms of the junction structure interact with each other, giving rise to a tertiary structure

ii. The base pairs can either stick out externally or internally, thereby giving a bulge

b. mRNA can interact with metabolites, causing a change in conformation, which can restrict access to the translation initiation site

c. Double-stranded RNA can also control gene expression through RNA interferencei. Dicer (a protein) cleaves dsRNA to produce siRNA (small interfering)

siRNA, also known as silencing RNA, is still double strandedii. RNA-induced silencing complex (RISC) binds to siRNA and separates it

RISC-siRNA complex uses the siRNA as a template to bind to mRNA and cleave it to disrupt gene expression

10. DNA replication occurs in the 5’ to 3’ direction, with DNA polymerase adding nucleotides to the 3’ end of an existing strand

a. The existing 3’ OH group carries out a nucleophilic attack on the phosphate group of the incoming nucleotide

b. DNA polymerase also has exonuclease activity that can remove basesi. 3’ to 5’ exonuclease is used for proofreading, for the excision of mismatched

bases The proofreading ability of DNA polymerase III in prokaryotes

ii. 5’ to 3’ exonuclease is used for replacement of bases, from the nick onwards Such as the replacement of the RNA primer by DNA polymerase I in

prokaryotes11. DNA replication in prokaryotes

a. Initiation occurs at the origin of replication (oriC)i. DnaA binds to the oriC, which is AT rich

ii. DnaA recruits DnaB which acts as helicase to unwind the DNAiii. DNA gryase helps to prevent supercoiling as the DNA unwindsiv. Single strand binding proteins help to prevent single strands from reanneling

b. DNA polymerase III holoenzymes consists of several subunits i. Alpha, epsilon and theta subunits synthesizes DNA in the 5’ to 3’ direction

ii. Beta subunit functions as a clamp to keep the complex associated with DNA This increases the polymerisation rate by preventing dissociation

iii. Other subunits are the clamp loader complex that loads the clamp onto the DNA periodically every 1000 nucleotides onto the lagging strand

c. Primase adds the RNA primer on the lagging strandd. DNA polymerase I replaces the RNA with DNA, using a 5’ to 3’ exonuclease

i. It also has proofreading abilities.e. DNA ligase seals the nick between nucleotides, joining the Okazaki fragments

together to from a continuous daughter strandi. The replacement of the last nucleotide of the RNA primer does not

automatically join it to the next fragment12. DNA replication in eukaryotes

a. Eukaryotes have several oriCb. Origin recognition complex (ORC) binds to the origin of replication to prepare the

strands for replicationi. During the G1 phase, ORC recruits replication activator protein (RAP) to bind

to itself

ii. RAP recruits replication licensing factors that bind along the DNA moleculeiii. This pre-replication complex (ORC/RAP/RLF) primes the DNA for replication

c. During S phase, cyclin-dependent kinases (CDKs) phosphorylate the complex to activate DNA replication by recruiting DNA polymerase and primase

i. This phosphorylation also degrades the RAP and RLF to prevent formation of another pre-replication complex, preventing further DNA replication

ii. Polymerase delta synthesises the daughter strand on both the leading and lagging strands

iii. Polymerase alpha is associated with primase for DNA replication on the lagging strand

This DNA polymerase cannot proofread Replicates DNA only on the lagging strand for a length of base pairs,

before polymerase delta replaces it and continues synthesis Before polymerase delta can reach the RNA primer of the next

fragment, FEN-I nicks the RNA primer, allowing RNase-H1 to remove it

Polymerase delta then replicates to the DNA portion of the next Okazaki fragment, with DNA ligase sealing the nick to join the fragments

iv. Both DNA polymerases are help in place by proliferating cell nuclear antigen (PCNA)

PCNA is loaded by replication factor Cd. Replication proteins (RPA) help stabilize the single strands during replicatione. Telomerase can overcome the 3’ end replication problem, by lengthening the

telomeresi. This prevents cellular senescence

ii. Telomerase is a ribonucleoprotein Its RNA component is complementary to the telomeric repeats

Telomeres have tandem repeats of G-rich regions The RNA binds with a portion exceeding the telomere

The catalytic subunit telomerase reverse transcriptase (TERT) extends the 3’ end of the leading strand using the RNA template that is not bound to the DNA

The lagging strand is then extended by the same DNA replication machinery

13. DNA in an organism can be shuffled around by recombination processesa. Homologous recombination occurs during meiosis or during repair of double

stranded breaksi. Crossing over and exchange of DNA occurs between homologous sequences

ii. There is a formation and resolution of a Holliday junction RecBCD contains helicases that unwinds both strands of DNA,

producing a 3’ invading end RecBCD consists of three subunits, Rec B, Rec C and RecD

RecA binds to the 3’ invading strand and helps direct strand invasion It binds to both the single stranded 3’ invading strand, and

the homologous double stranded DNA It travels along the homologous chromosome until it finds

the complementary region

The invading strand displaces the complementary strand in the homologous chromosome

DNA ligase then joins the invading strand with the invaded strand

Branch migration then occurs where the holiday junction moves further away from the point of invasion, causing more DNA to be swapped between the two homologous chromosomes

RuvABC complex (has three subunits) has helicases in RuvA and RuvB to unwind each homologous chromosome, allowing the branch to migrate

Resolution of the holiday junction occurs when the DNA twists to form a 4-way junction

RuvC resolvase then cleaves the holiday junction either vertically or horizontally

The alleles downstream of the junction may or may not be exchanged depending on the cleavage

b. Transposons are mobile genetic elements that can change their position within the genome

i. Class I are retrotransposons that proliferate by a copy and paste method The RNA of the transposon is synthesised Reverse transcription occurs to produce a DNA molecule

complementary to the RNA The reverse transcriptase may be encoded for by the

transposon The DNA molecules then invades the DNA chromosome at another

location away from the original transposon This increases the number of transposons in the chromosome

ii. Class II are DNA transposons that move about by cut and paste method The transposon is flanked by repeating sequences that contain a

recognition sequence It is cut and extracted from the DNA chromosome by transposase

that produce blunt ends Transposase is encoded for by the transposon

Another location along the DNA that contains the recognition sequence is cut by enzymes that produce sticky ends

May be specific or non-specific depending on the transposase, so long as they contain the recognition sequence

The transposon is then inserted into this position The gaps from the sticky ends are then filled in by DNA polymerase

and ligated by DNA ligase The number of transposons stay constant and only move about

14. DNA mutation occurs due to errors or mutagensa. Spontaneous errors can occur during DNA replication through substitution,

insertion, deletion, depurination or deaminationi. Substitution can either be transitions or transversions

Transitions causes a purine to be substituted by the other purine, and a pyrimidine by another pyrimidine

Transversion causes a purine to be substituted by either pyrimidine and vice-versa

These wobble base pairing is non-Watson-Crick paring If not corrected, it is incorporated into the daughter strand

after DNA replication The wild type strand produces a wild type daughter strand,

while the mutant strand produces a mutant daughter strand that does not experience wobble base pairing

ii. Depurination results in a loss of a G nitrogenous base If not repaired, the damaged strand will always produce a mutant

daughter strand with a random nucleotide at the damaged position This leads to several types of mutant chromosomes that

continue to proliferate since the damaged strand produces a different mutant daughter strand with different nucleotide used as the damaged location

iii. Deamination removes an amine group from the nitrogenous base A cytosine base is converted to a uracil A methylated cytosine base is converted to a thymine

b. Mutagens are UV radiation, HNO2, ionisation radiation and alkylating compoundsi. UV radiation causes the formation of pyrimidine dimers

ii. HNO2 causes deaminationiii. Ionisation radiation and alkylating compounds causes chemical modification

of bases15. DNA repair mechanisms correct the mutations by removing the damage

a. Base excision repair (BER) places a single damaged basei. The damaged nitrogenous base is removed by DNA glycosylase that breaks

the glycosidic bond, forming an Apurinic/apyrimidinic (AP) siteii. AP endonuclease then creates a nick to remove the sugar ring and

phosphate group of the damaged nucleotideiii. The gap is then repaired by DNA polymerase that adds the correct

nucleotideiv. DNA ligase seals the nick between the new nucleotide and the sugar

phosphate backbone downstream.b. Nucleotide excision repair (NER) replaces a section of DNA, usually for bulky lesions

like pyrimidine dimersi. NER complex binds to the damaged section and the area around it

It separates the two strands which are stabilized by proteins Single strand binding proteins for prokaryotes RPA for eukaryotes

It then cleaves the ends at either side of the damaged DNA, removing a section of the DNA strand

Then recruits DNA polymerase and DNA ligase to use the remaining strand as a template for repair and ligation

c. Mismatch repair (MMR) corrects incorrectly paired bases that were missed during DNA replication

i. The template parent strand is identifiable since it is methylatedii. MutS binds to the mismatched base and recruits MutL and MutH

MutH binds to the DNA molecule at a hemimethylated site away from the damage and is connected to MutS through MutL

Hemimethylated since only the parent strand is methylatediii. MutH then nicks the unmethylated daughter strandiv. An exonuclease then degrades the DNA from the MutH nick until the

mismatched base on the daughter strand Different exonuclease is recruited depending on whether the nick is

a 3’ or 5’v. DNA polymerase then fills in the gap and DNA ligase seals the nick

d. Non-homologous recombination repairs a double stranded break without a use of a template, resulting in error-prone repair.

i. The double-stranded break recruits Ku70/Ku80 proteins (a heterodimer)ii. The Ku proteins recruits accessory proteins

Artemis:DNA-PK nucelease complex resects the damaged DNA ends It a complex consisting of the artemis protein with DNA-

dependent protein kinases DNA polymerase fills in the gaps DNA ligase ligates the ends

e. Homologous recombination repair uses the undamaged homologous copy as a template for repair

i. It can only occur when there is a homologous chromosome present after S phase

ii. ATM kinase detects the double stranded break and recruits the MRN complex

MRN complex consists of Mre11, Rad50 and Nbs1iii. Rad52 is then recruited to process the ends to form 3’ single stranded endsiv. BRCA2 recruits Rad51 which binds to the single strands to promote strand

invasion Rad51 is homologous to RecA in prokaryotes

v. Rad54 helps Rad51 to find the homologous region during strand invasionvi. DNA polymerase fills in the gaps and DNA ligase fills the nicks

vii. Resolvase resolves the Holliday junction

Lipids

1. Lipids are organic molecules that have certain characteristics and functionsa. Characteristics

i. Low solubility in waterii. High solubility in nonpolar solvents

iii. Mostly hydrocarbon in natureb. Functions

i. Energy storageii. Cellular membranes

iii. Biological signals2. Lipids are made from fatty acids with hydrocarbon chains that terminate with a carboxyl

groupa. They have an even number of carbons between 14 and 24b. The hydrocarbon chains can be either saturated or unsaturated

i. Cis double bonds introduces kinks in the hydrocarbon tail, preventing close packing that lowers the melting point

c. Named in the manner x:y n-zi. Where x is the number of carbon, and y is the number of double bonds

ii. If the fatty acid is unsaturated, z is carbon number that contains the first double bond when counted from the methyl end

Note that when naming, the first carbon is the one in the carboxyl group

3. Triacylglycerols are formed from the esterification between three fatty acid molecules and one glycerol molecule

a. A simple triacylglycerol consists of three of the same fatty acids, while in mixed triacylglycerol they are different

b. It functions as an energy storage compound, in adipose tissuei. It is broken down by lipases through hydrolysis to produce energy

ii. It can react with NaOH to form soap through saponification The strong base causes base catalysed hydrolysis

c. It is not found in lipid membranes4. Glycerolphospholipids consists of a glycerol-3-phosphate with two fatty acid chains

a. The hydrophobic fatty acid tails and polar phosphoryl group makes it a good component for lipid membranes as it is amphipathic

b. The phosphoryl group can also be bonded to another group, giving rise to different forms of glycerolphospholipids

i. Phosphatidic acid (PA): -Hii. Phosphatidylethanolamine (PE): -CH2CH2NH3

+

iii. Phosphatidylcholine (PC): -CH2CH2N(CH3)3+

iv. Phosphatidylserine (PS): -CH2CH(NH3+)COO-

v. Phosphatidylinositol (PI): ionositol (a 6 carbon ring) Can be phosphorylated to form signalling molecules

vi. Phosphatidylglycerol (PG): -CH2CH(OH)CH2OH Diphosphatidylglycerol can be formed by another glycophospholipid

bonding with the substituent glycerolc. When hydrolysed by phospholipases, signalling molecules are formed

i. Phospholipase A2 can produce arachidonic acid that is a precursor for prostaglandins, leukotrienes and thromboxanes

Prostaglandins are cyclic fatty acids that have hormone-like effects like promotion of uterine contractions

Leukotrienes are produced by leukocytes and regulate immune responses by acting as inflammatory mediators

Thromboxanes is involved in clot formation as a vasoconstrictor5. Sphingolipids are derivatives of sphingosine

a. Sphingosine is a 18-carbon amino alcohol, and the amino group forms an amide bond with a fatty acid molecule

i. This gives rise to ceramide, a parent compound for sphingolipidsb. Different groups can react with the alcohol group on ceramides

i. Sphingomyelin are formed with phosphocholine or phosphoethanolamine Differs from phosphatidylcholine and phosphatidylethanolamine It is the choline group with a phosphoryl group, without the glycerol

or fatty acids

ii. Cerebrosides are formed with simple sugarsiii. Gangliosides are formed with complex sugars with at least one sialic acid

bonded to any residue of the sugarc. Sphingomyelin is the most common sphingolipid, forming 10% to 20% of the lipid

membranesd. Gangliosides are found on the cell surface of lipid membranes and serve as receptors

i. The oligosaccharide bind to ligands Ligands like toxins or viruses

ii. Fatty acyl chains anchors the ganglioside to the lipid membrane6. Terpenes are a diverse group of lipids derived from 2 or more units of isoprene molecules

a. Isoprene is a five carbon molecules that can polymerase repeatedly due to it having two carbon bonds

b. Monoterpenes are made from two isoprene moleculesi. Found in plants

ii. A component for flavour or odorc. Diterpenes are made from four isoprene moleculesd. Triterpenes are made from six isoprene molecules

i. Squalene is a triterpene that is a precursor to steroids7. Steroids are derivatives of cyclopentanoperhydrophenanthrene that is built from isoprene

a. The most common steroid is cholesterol which is also a precursor for other steroidsi. Like hormones or bile acids

b. Cholesterol is a constituent of plasma membranesi. It buffers the fluidity of the cell membrane

At cellular temperatures, it decrease fluidity by stabilising the extended conformation of the hydrocarbon tails of fatty acids with its fused ring structure

At lower temperatures, it increases fluidity by preventing the close packing of the lipids as it separates the groups

8. Cellular membranes are made from lipid bilayersa. The amphipathic lipids will form spontaneous structures in water

i. Either a monolayer micelleii. Or bilayer vesicle

A cell membrane is a very large vesicleb. Lipid bilayers have asymmetric distribution of lipids

i. The outer layer has more bulky lipids like gangliosides and sphingomyelins Glycolipids and glycoproteins also face outside for cell recognition

ii. The inner layer has more compact lipids Because it bends more in a spherical cell membrane

iii. The loss of asymmetry will have consequences The flipping of PS from the inside to outside is observed in apoptosis

c. The lipids move about either by diffusion or catalysed movement by enzymesi. They diffuse laterally rapidly, but transversely very slowly

ii. Flippase uses ATP to move lipids from the outer layer to the inner layeriii. Floppase uses ATP to move lipids from the inner layer to outer layeriv. Scramblase uses Ca2+ ions to randomly flip lipids transversely

d. The lipids either behave like a gel or liquid crystal depending on composition and temperature

i. When it is gel-like it is thicker with a smaller surface area, and it less fluid

ii. When it is a liquid crystal, it is thinner with a larger surface area and is more fluid

iii. Proportions of unsaturated fatty acids and cholesterol will affect fluidity More unsaturated fatty acids cause disorder that increases fluidity More cholesterol increase order and rigidity by stabilizing the

straight chain arrangement of saturated fatty acids9. The lipid bilayer follows a fluid mosaic model with integral and peripheral proteins and other

componentsa. The evidence of fluidity comes from diffusion of differently labelled fused cell

membranes, and recovery of photobleached region of labelled cell membraneb. While the lipids bilayer forms a boundary between the cell and the environment, the

other components perform other functionsi. Controlled metabolite transport

ii. Signal reception and transmissioniii. Enzymatic reactionsiv. Contact with other cellsv. As anchor for cytoskeleton

c. Integral proteins span the entire membrane through hydrophobic segmentsi. Hydropathy index measures the hydrophobicity of a region of the protein

through the amino acid residuesii. It is used to find transmembrane domains as they have more hydrophobic

amino acid residues Integral proteins can have either a single pass transmembrane

region, or multi-pass transmembrane region The multi-pass can be due to alpha-helices entering and

leaving the cell membrane repeatedly Transmembrane domains are usually alpha helices or beta sheets

d. Peripheral proteins are non-covalently associated with polar groups of lipids and integral proteins

i. They can be associated directly or indirectly They associate directly if they have a region that forms ionic and

hydrogen bonds, or have a hydrophobic loops that enters the cell membrane

They associate indirectly through association with integral proteinse. Some membrane proteins can be covalently-linked to lipids to anchor them to the

membrane, forming lipoproteini. They can be amide linked or thioester linked

10. As the lipid bilayer forms an partially permeable boundary, membrane transport can either be unmediated or mediated

a. Unmediated transport is by passive diffusion of small, uncharged molecules down a concentration gradient without carriers or proteins

i. The rate is proportional to the differences in concentrationb. Mediated transport can be facilitated diffusion or active transport by membrane

protein transportersi. Transporters can be uniport, symport or antiport

Uniport transporters only allow one molecule to move in one direction

Symport transporters allow two molecules to move in the same direction

Antiport transporters allow two molecules to move in opposite directions

ii. Facilitated diffusion uses transporters to mediate passive diffusion down a concentration gradient

This increases the rate of diffusion The transporter has a passive channel or can undergo

conformational change No ATP hydrolysis despite conformational change Binding of target molecules triggers the conformational

changeIt gets released through diffusion at the less

concentrated side The transporters can become saturated limiting the rate of diffusion

The rate will still be greater than that of unmediated transport

The transporters can be highly selective and gated Allows a specific molecule to move in only on direction Selectivity can be due to charges attracting a specific target

molecules, and narrow diameter to exclude larger molecules Gates can respond to pressure, ligands, signals and voltage

iii. Primary active transport moves molecules against a concentration gradient with ATP hydrolysis

ATP is used to phosphorylate the transporter that causes a conformational change

The added phosphate is then hydrolysed to restore the original conformation

iv. Secondary active transport uses existing ion gradient to drive the unfavourable transport of the target molecule

ATP is used to maintain the ion gradient instead of transporting the target molecule

The target molecules moves when the ion moves, either through a symport or antiport

The Na+/glucose symport allows glucose to be transported against its concentration gradient through the Na+ gradient that is maintain by Na+/K+ ATP pump

Carbohydrates

1. Carbohydrates are the most abundant biological molecule containing 3 carbons or morea. There is one carbonyl group (ketone or aldehyde) and the each other carbon

contains a hydroxyl groupb. General formula is (CH2O)n, where n is greater than or equal to 3c. They perform several functions

i. As an energy sourceii. As structural materials

iii. For signal recognition

2. Monosaccharaides are classified depending on their carbonyl group, whether it is an aldose or ketose, and the number of carbons

a. They exist as stereoisomers and bend polarized lighti. Due to the non-carbonyl or terminal carbons being chiral centers

ii. The isomer is drawn using Fischer projection The C=O bond is drawn facing the right The stereoisomer form is determined by the highest numbered

chiral carbon The carbon furthest from the carbonyl group that is not

terminal (since terminal has two hydrogens) It is D(+) if the hydroxyl group is on the right, the same side

as the carbonyl group It is L(-) if the hydroxyl group is on the left, the opposite side

of the carbonyl groupiii. The number of stereoisomers of aldoses are determined by 2n-2, where n is

the number of carbons Since the carbonyl carbon and terminal carbon are not chiral centres

iv. The number of stereoisomers of ketoses are determined by 2n-3, where n is the number of carbons

Since there are two terminal carbons that are not chiral centresb. Monosaccharides usually exist in cyclic form, since the carbonyl group can react with

the hydroxyl group to form a hemiacetal or hemiketali. Majority of monosaccharides exist in cyclic form

Either in boat or chair conformation Chair conformation is the one taught in CM1401 and is more

stable than boat Only aldohexoses can have the bulky terminal carbon in the

equatorial positionii. The carbon originally with the carbonyl group becomes the anomeric carbon

as the new hydroxyl group can project above (beta) or below (alpha) the ring Formed in equal proportions For aldoses it is carbon 1, for ketoses it is carbon 2 Cyclic ketoses have a terminal carbon bonded to the anomeric

carbon rather than a hydrogeniii. The carbonyl group will attack the second last carbon instead of the terminal

carbon because it is more stable to do so Monosaccharides usually form 5-sided (furan) or 6-sided rings

(pyran) Aldopentose will form furanose Aldohexose will form pyranose

c. Monosaccharides have several derivatives i. It can be oxidised to form sugar acid

ii. It can be reduced to form sugar alcoholiii. It can undergo esterification to form sugar esteriv. One hydroxyl group can be removed to form deoxy-sugarv. It can undergo amination to form amino sugar

3. Disaccharides are form when the anomeric carbon in a cyclic monosaccharide forms a glycosidic bond with another monosaccharide

a. An alpha glycosidic bond causes the other monosaccharide to have the hydroxyl group on its anomeric carbon to be on the same side as the attacking monosaccharide

b. A beta glycosidic bond causes the other monosaccharide to be flipped such that the hydroxyl group on its anomeric carbon is on the opposite side as the attacking monosaccharide

c. Alpha or beta glycosidic bond can only be formed when attacking any carbon other than the terminal one, since it is achiral

i. Alpha or beta-1,4-glycosidic bond but only alpha-1,6-glycosidic bond for glucose

4. Polysaccharides are formed from the linkage of multiple monosaccharides and their derivatives

a. Branching can be present through glycosidic bonds forming on the terminal carbonb. The polysaccharide can also be linked to other compounds to form glycoproteins

and glycolipids5. Starch and glycogen are formed from glucose polymerising into linear amylose and branched

amylopectina. Amylose is formed from only glucose alpha 1,4 linkages while amylopectin also

contain glucose alpha 1,6 linkagesb. Starch branches about one every 12-30 residues, while glycogen branches once

every 8-12 residuesc. The linear amylose can form helical structures, like alpha helix in proteins, due to the

bond angle in alpha glycosidic bondsi. Iodine can be inserted into the amylose helix causing starch to be stained

d. They function as storage compounds as the alpha-glycosidic bonds can be hydrolysed at the reducing end to release glucose units

6. Cellulose is formed from glucose polymerising into sheets through beta-1,4 glycosidic bondsa. These bonds are straight thus allowing the formation of extended linear chains

i. Interchain hydrogen bonds allows the formation of sheetsii. Intersheet hydrogen bond gives cellulose structural strength

iii. Intrachain hydrogen bonds can also form7. Polysaccharides can be linked to peptides, forming peptidoglycans, in bacterial cell walls,

leading to gram positive or gram negative bacteriaa. If the peptidoglycan are exposed, the bacteria is gram positive.b. Else if the peptidoglycan is sandwiched between two lipid bilayers, the bacteria is

gram negativec. The crosslinkage between peptidoglycans, either directly or indirectly, confers

structural rigidity8. Glycoproteins are proteins that are covalently linked to oligosaccharides or polysaccharides

groupsa. Linked by O- or N-linkages

i. O-linked glycoproteins are found in mucin and cell surface glycoproteins The polysaccharides help to keep the protein extended, or keep a

globular head extendedii. N-linked glycoproteins are used for sorting signals, for protein folding and

stabilization or cell surface recognitionb. Glycosylation of proteins occur in the endoplasmic reticulum and golgi network

i. O-glycosylation occurs only in the Golgi apparatus

ii. N-glycosylation occurs in the ER and Golgic. Proteoglycans are a class of glycoproteins that consists primarily of

glycosaminoglycansi. Glycosaminoglycans are polymerised amino sugars with carboxyl or sulphate

groups replacing the hydroxyl groupsii. They are attached hyaluronic acid in cartilage, and interact with water to

provide flexibility and shock resistanceiii. Proteoglycans can also bind to growth factors to increase local

concentrationiv. They can also bind to the extracellular matrix for cell adhesion

Respiration

1. Glucose is an important sugar molecule as it is used for respirationa. The catabolism of glucose produces intermediates for synthesis of substrates for

energy production2. Glycolysis is the first part of cellular respiration

a. Glycolysis modifies the glucose for several usesi. It can produce pyruvate for anaerobic respiration to produce ATP without

oxygenii. It can produce acetyl-CoA for use in the citric acid cycle for aerobic

respiration to generate ATPiii. It can also provide intermediates for other biosynthesis and regulation

processesb. Glycolysis occurs in all cells, in their cytoplasm

i. Some cell have no mitochondria and solely rely on glycolysis for ATP production

c. Glycolysis occurs in two main stepsi. Glucose is first trapped in the cell, and then cleaved into two three carbon

units Glucose is phosphorylated using ATP to form glucose-6-phosphate

that cannot freely leave the cell Hexokinase catalyses the irreversible reaction Hexokinase is product inhibited

A specialized form, glucokinase in liver allows high activity without product inhibition

Glucose-6-phosphate is then converted to its isomer, fructose-6-phospahte

By phosphor-glucose isomerase Fructose-6-phosphate is then further phosphorylated to form

fructose-1,6-diphosphate By phosphor-fructokinase I This is an irreversible step and a major point of regulation

Fructose-1,6-diphosphate is then cleaved to form glyceraldehyde 3-phosphate (G3P) and dihydroxyacetone phosphate (DHAP)

By aldolase DHAP can be converted to G3P by triose phosphate isomerase G3P is used in the next part of glycolysis

ii. ATP is then generated through substrate level phosphorylation

G3P is oxidised by NAD+ and Pi to give 1,3-bisphosphoglycerate By glyceraldehyde 3-phosphate dehydrogenase Also produces 1 unit of NADH for every unit of G3P

1,3-BPG then undergoes substrate level phosphorylation to produce 3-phosphoglycerate

By phosphoglycerate kinase Also converts 1 unit of ADP to ATP for every unit of G3P

3-phosphoglycerate is then converted to a higher energy intermediate, phosphoenol-pyruvate

Isomerism by phosphoglyceromutase Dehydration by enolase

Phosphoenol-pyruvate then undergoes another substrate level phosphorylation to produce pyruvate

By pyruvate kinase Also converts 1 unit of ADP to ATP for every unit of G3P This is another major point of regulation

d. Overall, glucose+2NA D+¿+2Pi+2 ADP →2 pyruvate+2 NADH+2 H+¿+2 ATP+2H 2O ¿ ¿

i. Glycolysis consumes 2 ATP but generates 4 ATP for a net gain of 2 ATP.ii. 2 units of NADH is also generated per unit of glucose that will be used to

regenerate NAD+ in either anaerobic or aerobic respiration3. Under anaerobic conditions, pyruvate is reacted to form different products to regenerate

NAD+ for glycolysis to continuea. In animals, lactate is produced from pyruvate with the use of NADH

i. The reaction is catalysed by lactate dehydrogenaseb. In yeast, ethanol is produced from pyruvate with the use of NADH

i. The reaction is catalysed by alcohol dehydrogenase4. The tricarboxylic acid (TCA) cycle, aka citric acid cycle, generates NADH and FADH2 for

oxidative phosphorylation (which uses oxygen and regenerates NAD+) in the mitochondriaa. The TCA cycle performs several functions

i. It generates energy directly with GTP, and indirectly with NADH and FADH2

ii. It provides intermediates for biosynthesisiii. It provides feedback regulation to other pathways through regulating citrate

levelsb. The enzymes for the TCA cycle are located in the mitochondria

i. Succinic dehydrogenase (SDH) is found as part of a complex in the mitochondrial inner membrane, but does not span the membrane

ii. All other enzymes are found in the matrixc. The TCA cycle links with glycolysis through the dehydration of pyruvate to acetyl-

CoAi. The pyruvate dehydrogenase (PDH) complex consists of three enzymes

E1 requires thiamine pyrophosphate as coenzyme E2 requires lipoate and coenzyme-A as coenzymes E2 requires FAD and NAD+ as coenzymes

ii. This is an irreversible reaction that is a major point of regulationiii. This reaction produces 1 unit of NADH with every unit of pyruvateiv. Overall: pyruvate+NA D+¿+Coenzyme A→ AcetylCoA+CO 2+NADH+H+¿¿ ¿

d. Acetyl-CoA will then enter the TCA cycle

i. It first undergoes an irreversible condensation reaction with oxaloacetate to form citrate

With citrate synthase and coenzyme A as by-product There is product inhibition by citrate

ii. Citrate is then rearranged into its isomer, isocitrate By aconitase Isocitrate has two out of three carboxyl groups available for

decarboxylationiii. Isocitrate undergoes oxidative decarboxylation to form alpha-ketoglutarate

By isocitrate dehydrogenase Uses NAD+ and produces NADH, H+ and CO2

Is rate limiting and a major point of regulationiv. Alpha-ketoglutarate undergoes further oxidative decarboxylation to form

succinyl-CoA By alpha-ketoglutarate dehydrogenase

A complex of three enzymes that require the same prosthetic groups as PDH

Uses coenzyme-A and NAD+, and produces NADH, H+ and CO2

v. As succinyl-CoA has a high energy thioester bond, substrate level phosphorylation occurs to generate GTP and succinate

By succinate thiokinase Uses GDP and Pi, and produces GTP and coenzyme A

vi. Succinate gets oxidized to fumarate By succinate dehydrogenase, found in the inner mitochondrial

membrane Uses FAD and produces FADH2

vii. Fumarate gets isomerised to malate By fumarase

viii. Malate gets oxidised to oxaloacetate, regenerating it as a catalyst By malate dehydrogenase Uses NAD+ and produces NADH, H+

ix. Overall: AcetylCoA+3NA D+¿+FAD+GDP →2C O 2+Coenzyme A+3 NADH+3 H +¿+FAD H2+GTP¿ ¿

5. Oxidative phosphorylation generates ATP from NADH and FADH2 from glycolysis and the TCA cycle

a. The necessary protein complexes are found in the inner membrane of the mitochondria

i. The lipid bilayer is impermeable to protons, NADH and FADH2

ii. NADH produced outside of the mitochondria during glycolysis enters through 2 shuttles

Glycerol-3-phosphate converts NADH + H+ from the cytoplasm into FADH2 in the mitochondria

Maltate-asparate transports NADH + H+ from the cytoplasm into the mitochondria matrix as it is

b. Oxidative phosphorylation works by generating a proton motive force through the pumping out of protons from the mitochondria

i. The electron transport chain uses two carriers, Coenzyme Q and cytochrome C

ii. NADH + H+ reacts with complex I, NADH-Q oxidoreductase, and CoQ to produce NAD+ and CoQH2

In the process, 4 protons are pumped out by the complexiii. FADH2 reacts with complex II, succinate-Q reductase, and CoQ to produce

FAD and CoQH2 In the process, no protons are pumped out by the complex

iv. CoQH2 from both processes then react with complex III, Q-cytochrome c oxidoreductase, and cytochrome c, to produce CoQ and reduced cytochrome c

In the process, 4 protons are pumped out by the complexv. 2 units of reduced cytochrome c then reacts with complex IV, cytochrome c

oxidase, and 1 atom of oxygen to produce oxidised cytochrome c and water In the process, 2 protons are pumped out

vi. The proton gradient across the inner mitochondrial membrane causes protons to flow through ATP synthase that catalyses the formation of ATP from ADP and Pi

ATP synthase has two major areas The F0 pore forms a channel for the movement of protons

cross the inner membrane The stalk in the F1 headpiece then uses the energy from the

movement of protons to rotateThis induces conformational changes in the subunits

in the F1 headpiece, driving the release of ATPEach rotation of the stalk requires 10 to 14 protons,

and produces 3 ATP moleculesvii. Each molecule of NADH results in 2.5 molecules of ATP produced

viii. Each molecule of FADH2 results in 1.5 molecules of ATP produced6. Overall, one molecule of glucose produces 2 ATP, 2 GTP, 10 units of NADH + H+, 2 units of

FADH2

a. Using the glycerol-3-phosphate shuttle, the 2 units of NADH outside the mitochondria is converted to FADH2

b. Overall, 28 units of ATP and 2 units of GTP is producedi. 30 units of ATP if the outside NADH is brought in by the maltate-asparate

shuttle instead7. There are several major points of regulation for respiration

a. Phosphorylation of fructose-6-phosphate to fructose-1,6-diphosphateb. Substrate level phosphorylation of phosphoenol-pyruvate to pyruvatec. Dehydration of pyruvate to acetyl-CoA in the matrix for the TCA cycled. Oxidative decarboxylation of isocitrate to form alpha-ketoglutarate

Questions

1. Difference between beta hairpin and beta turn?2. Are domains considered to be a type of super-secondary structure, or a tertiary structure?

Can proteins have more than one tertiary structure?3. Why does gene duplication in paralog eventually gives rise to different functions?4. Why does negative effector binding to an allosteric protein increase Vmax?

Miscellaneous

1. Concentration of water is 55 M, since there are 55 moles of water in 1 litre of watera. 1000g/18 = 55.55556