LSM1101 Notes

LSM1101 Notes

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

Terms

1. Amphipathic

a. Molecule has both hydrophilic and hydrophobic groups, and can interact with, water, other hydrophobic groups, and organic solvents 2. 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 broken

i. Holds ice in a hexagonal structure when cold enough ii. Easily formed and broken when water is liquid

b. Water can form 4 hydrogen bonds

i. Each oxygen can form 2, each hydrogen can form 1 2. Solubility of compounds depends on how they interact with water

molecules

a. Hydrophilic molecules form strong bonds with water and dissolve easily

b. 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 shells

a. Electrostatic interactions between charged ions and partial charges of water

b. 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 dissolution

5. 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 dissolve

i. 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 structures

i. Hydrophobic regions clump together to minimise exposure to water

ii. Hydrophilic regions are exposed to water

7. Proteins tend to be amphipathic as it gives a stable structure

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

pH and Buffers

1. pH =

+

¿

H

¿

¿

−

lg

¿

a.

+

¿

H

¿

¿

−

¿

O H

¿

¿

K

_w

=

¿

at 298 Kelvin b.

pH+ pOH = p K

w

2. Human healthy pH is around 7.4

a. pH is maintained constant by homeostasis

b. Needs to remain constant to provide optimum environment for enzyme function required for life

3. Enzymes at different locations in the body operate best at different pH a. Stomach enzymes require an acidic medium

4. Bronsted-Lowry definition uses protons as reference instead of electrons (Lewis)

a. Acids are proton donors, dissociating H+_{, producing conjugate base} b. Bases are proton acceptors, associate with H+, _{producing conjugate}

acid

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

i. The closer the pKa to 14, the weaker the acid

ii. Negative pKa indicates that the acid is a strong acid

6. Henderson-Hasselbalch equation:

−

¿

A

¿

¿

¿

[

HA

]

¿

¿

¿

pH= p k

_a

+

lg

¿

a. Not applicable when A- _{or HA concentration is near 0} 7. When mixing weak acid and its conjugate base

a. Assume that further dissociation is negligible b. Use Henderson-Hasselbalch equation to find pH

8. Buffers resists changes in pH by reacting away excess H+ _{or OH}- _added 9. Buffering capacity is

pH= p K

a

± 1

a. When equal concentrations of acid and conjugate base are present 10.Polyprotic acids have multiple buffering capacities

11.The human body has several buffers a. Phosphate system (HPO42-_/H2PO4-₎

i. Protects pH of intracellular and extracellular environment b. Bicarbonate system (HCO3-_/H2CO3)

ii. Breathing adjusts CO2 concentration, affecting H2CO3 concentration

c. Amino acids and proteins

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 omega

b. 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 groups a. Except for glycine which has 2 identical groups

b. 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 death

ii. Gramicidin antibiotic contains D-amino acids that perforates the bacterial membrane

3. 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 charged

i. 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 groups

a. Non-polar amino acids i. Alanine, Ala, A ii. Isoleucine, Ile, I iii. Leucine, Leu, L iv. Methionine, Met, M

v. 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, W

viii. 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 acids

i. Aspartic acid, Asp, D ii. Glutamic acid, Gla, E

iii. They contain a carboxyl group that can undergo deprotonation

iv. At neutral pH, they are negatively charged c. Basic amino acids

i. Arginine, Arg, R ii. Histidine, His, H iii. Lysine, Lys, K

iv. They have an amino group that can undergo protonation v. Arg and Lys are positively charged at neutral pH, due to the

high pKa of the R-group

vi. 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 structure

d. Polar, uncharged amino acids i. Serine, Ser, S

ii. Threonine, Thr, T iii. Tyrosine, Tyr, Y iv. Asparagine, Asn, N

v. Glutamine, Gln, Q vi. Cysteine, Cys, C

vii. Ser, Thr, Tyr contain an alcohol group, making them polar. Asn and Gln are amides, with the –NH2 portion unable to undergo protonation

viii. Threonine has a second chiral centre in its R-group e. Aromatic amino acids

i. Phenylalanine, Phe, F ii. Tryptophan, Trp, W iii. Tyrosine, Tyr, Y

f. Sulfur containing amino acids i. Methionine, Met, M ii. Cysteine, Cys, C

iii. 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, G

i. 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 similar a. The alpha amino group

p k

a

≈ 9

b. The alpha carboxyl group

p k

a

≈ 2

c. Acidic side chain (Gla, Asp)

p k

a

≈ 4

d. Basic side chain (Lys, Arg)

p k

a

≈ 11

, (His)

p k

a

≈ 6

6. 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 produce 7. Amino acids also have an isoelectric point where it is electrically neutral

(zwitterion)

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

i.

pI=

p K

1

+

₂

p K

2 , for acidic or uncharged amino acids ii.

pI=

p K

2

+

₂

p K

3 , for basic amino acids

8. Amino acids linked to each other via peptide bonds formed by condensation reactions

a. The peptide bond is unable to undergo protonation or deprotonation b. But can still undergo hydrogen bonding

9. Polypeptides are usually named from the N-terminal to the C-terminal a. The terminals can still undergo protonation and deprotonation 10.There are also non-standard amino acids

a. They can be D-forms of amino acids b. They can be non-alpha-amino acids c. They have specific biological functions

d. 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’ direction a. Produces protein in the N-terminal to C-terminal direction b. AUG is the start, methionine codon

c. 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 them a. The primary structures ultimately determines the folding and

function of the protein

b. 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 proteins a. Direct sequencing through protein digestion into fragments, and

then use biochemical methods to determine amino acids present in fragments

b. Mass spectrometry to determine mass of small protein fragments and infer amino acids components

c. Prediction from DNA or RNA sequence

4. Protein folding can be visualised by several methods

a. Computational models have various display modes include wireframe, ball and stick and space-fill

b. X-ray crystallography uses diffraction pattern of X-rays to determine the three dimensional structure of the protein in a crystal

i. The diffraction patterns is processed to give an electron density map

ii. The electron density map is used to generate the protein structure

c. 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 substrates b. To catalyse specific reactions

c. For structural proteins to have the mechanical strength, like collagen

6. Folding is driven by numerous weak interactions, assisted by molecular chaperones

a. 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 agents b. Hydrophobic interactions

i. It is the most important factor c. Hydrogen bonds

i. Give rise to secondary structures

ii. Occurs between peptide linkages and R-groups d. Ionic bonds

i. Occurs between acidic and basic amino acids

ii. Affected by presence of ions from salts in solution, and pH e. 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 death ii. 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 proteins

a. 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 rotate

Rotation of C α -C bond = Ψ (Psi) Rotation of C

α

-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 helix

ii. 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 angle

iii. 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 helix

10.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 cylinder

ii. 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 protein

11.Another super-secondary structure is domains, which are independently folded globular units

a. Each domain is like a small globular protein i. Hydrophobic core and hydrophilic surface ii. 100 to 200 amino acids long

iii. Has a specific function

iv. Is modular, so can be combined in different combinations for a different overall protein function

b. Trans-membrane proteins like receptor tyrosine kinases have domains that allow them to respond to different stimuli

i. The extracellular domains are different to allows binding to different signalling proteins

ii. 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 protein

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

a. Homologous proteins have similar amino acid sequences

i. 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 functions a. Globular proteins

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

 They are thus soluble in water ii. They have diverse structures

 Thus, have diverse functions b. 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 water

ii. 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 proteins

i. 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 membrane

15.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 subunits i. Subunits can be the same type or different types 16.Insulin is needed for the homeostasis of blood glucose levels

a. 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 strands b. Since the genetic code is the same across all life, E. coli can be used

to synthesise insulin

i. Avoids the problem of immune reaction from using pig insulin ii. The polypeptide chains are made separately

iii. Denaturing and renaturing processes allow them to fold into insulin

 The instructions for folding is inherent in the protein 17.Prion is an infectious protein, which causes Transmissible spongiform

encephalopathies

a. Prion protein is native to humans, and misfolding causes the disease phenotype

i. The misfolded PrP is protease insensitive, forming insoluble fibres, leading to death

ii. Furthermore, the diseased PrP induces normal PrP to misfold 18.Other diseases like Alzheimer’s may be due to protein misfolding and

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 catalysis

i. Binding of substrate is held by multiple weak forces

b. They allow for higher reaction rates, as they increase the reaction velocity

c. They allow for milder reaction conditions, at lower temperatures d. 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 model

a. In lock and key, the substrate has the exact shape of the active site b. 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 components i. 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 cell

i. 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 pH

a. pH affects the ionization of R-groups in the protein amino acids i. This changes substrate-active site interaction, affecting

enzyme activity

5. Enzymes are affected by temperatures

a. The enzyme activity decreases at extreme temperatures due to denaturation

i. 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 energy

i. It is highly unstable and cannot be extracted

b. Activation energy is the energy required to get them to the transition state

i.

ΔG

‡ = activation energy. ‡ is double dagger

7. Unlike Chemistry, even though

K >1, ΔG

O

<0

, the reaction may not be spontaneous, due to high activation energy

a. The reaction is thermodynamically feasible, but may not be kinetically feasible

8. Enzymes lower the activation energy of the reaction, without affecting other thing like K or

ΔG

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

9. 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 ways a. 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 state

ii. 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 catalysis

i. 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 ions

iii. The covalent bond pulls electrons away from the reaction centre, allowing the other part of the substrate to be attacked c. 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 constant

3. The order of the reaction depends on the number of molecules participating in the rate-determining step

a. First-order reaction has only one molecule, second-order has two. i. The two molecules can either be the same or different b. The rate equation changes depending on the order of reaction c. 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 it b. As substrate concentration increase, the velocity approaches its

maximum, causing the rate of increase to be less than linear c. At high substrate concentrations, the velocity is independent of

substrate concentration

i. 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

k

1

i. 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 equation

b. 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 pair

ii. It is a measure of enzyme affinity for the substrate



K

M

≈

1

affinity

7. 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.

1

_v

is plotted against

_[

1

_{S ]}

i.

_V

1

0

=

(

K

M

V

_max

)

1

[

S

]

+

1

V

_max b. The y-intercept is

_V

1

max , as the substrate concentration

approaches infinity c. The x-intercept is

−1

_K

M

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

K

_M

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

=

k

cat

[

ES]

b. Vmax=kcat

[

E

]

total

10.Catalytic efficiency:

_K

k

cat

M

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 decreases

b.

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 substrate

ii. 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

k

3

iv. The effects of the competitive inhibitor can be reduced by increasing substrate concentration

b. Uncompetitive inhibitors only bind to the enzyme-substrate complex i. This affects the enzyme conformation, reducing the catalytic

capacity

ii.

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

iii.

V

0

=

(

V

max

α

'

) [

S]

(

K

M

α

'

)+[

S ]

, where

α

'

=1+

_K

[

I

]

I '



K

I' is the dissociation constant for the formation of

ESI from ES and I

iv. The substrate binding to free enzyme is unaffected

c. Non-competitive/ mixed inhibitors bind to both the free enzyme and enzyme-substrate complex

i. 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 substrate

a. By changing the rate of transcription, translation and degradation, enzyme activity can be controlled

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

2. 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

needed

b. 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 substrate

a. 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 nucleus

c. 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 GKRP

iii. 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 factors i. Its own kinetics property

 High

K

M

ii. Binding to GKRP that inactivates it

iii. 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 activation

a. Some enzyme are synthesized in their proenzyme (zymogen) form i. These are inactivated and require cleavage to be activated

ii. Like proteolytic enzymes of digestive tract and blood clotting proteins

iii. They have either Pro- as prefix, or –ogen as suffix

b. 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 activation e. 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 clot 5. 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 activity ii. 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 substrate

e. When allosteric enzymes have multiple active site, each active site acts as a regulatory site for other active sites

i. 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 bound

f. This allows for feedback inhibition, where the high concentration of products will reduce enzyme activity, by binding to it as a negative effector

i. Need not be product directly, as it can undergo further modification before becoming the negative effector g. 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 cofactors

6. Lastly, enzyme activity can be regulated by covalent modification of the enzyme

a. Modifications include i. Phosphorylation

 Adding of phosphate group ii. Methylation

 Adding of methyl group iii. Uridylylation

 Adding of uridine base iv. Adenylylation

 Adding of adenine base

b. Phosphorylation of glycogen phosphorylase causes a conformational change into a more active state

i. This increases the rate of phosphorylation of glycogen, which extracts a glucose 1-phosphate from the glycogen chain of glucose

7. 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 AMP

ii. 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 ring

b. The iron has six liganding positions, of which the 4 liganding positions that lie on the same plane are taken up by protoporhyrin IX

i. 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 protein

3. 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 monoxide

i. 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 times

4. 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 versa i. 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 dimers

b. Each subunit has a heme group that can bind to oxygen

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

conformational change affects its interaction with the other subunits i. 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 pH

a. 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 dioxide

a. 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 dissociation

b. Carbon dioxide can also react with the N-terminal of the subunits in hemoglobin, producing carbamate that forms salt bridges which stabilise the T conformation

i. 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 cells b. 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 environments

c. Fetal hemoglobin contains a different subunit (gamma) that has a lower affinity for 2,3-BPG

i. 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

hemoglobin

a. 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

1. Chargaff’s rule states that the amount of purines equals the amount of pyrimidines in DNA

a. Purines are Adenine and Guanine

b. Pyrimidines are Thymine, Cytosine and Uracil c. Later explained by complementary base pairing 2. 3’-5’ phosphodiester bonds joins the nucleotides together

3. Beta-glycosidic bonds join the nitrogenous base to the carbon ring

4. 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 lesser 5. DNA has three conformations

a. B-DNA is commonly found i. It is right handed

ii. Its major groove is wide with medium depth iii. Its minor groove is narrow with medium depth b. A-DNA is synthesized in the lab, when DNA is dehydrated

i. It is right handed

ii. Its major groove is narrow but deep

 Still called major groove as it is the analogous location in B-DNA

iii. 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 out

iii. Its minor groove is narrow but deep

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

i. Against the direction of DNA rotation

b. 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 strand

7. In eukaryotes, DNA is compacted into chromatin

a. DNA is coiled around histone proteins to form nucleosomes i. Histone have many arginine and lysine residues that are

positive charged to bind to the negatively charged phosphates in the DNA backbone

b. Nucleosomes compact to form solenoid, which compacts to form loops, which compact to form metaphase chromosome

i. 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,XX

a. Where AA is the total number of chromosomes b. Where XX are the sex chromosomes

9. RNA is usually single stranded, but can form secondary and tertiary structures

a. Not all bases needs to be complementary paired, thus giving a budge that is the secondary structure

i. 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 interference

i. Dicer (a protein) cleaves dsRNA to produce siRNA (small interfering)

 siRNA, also known as silencing RNA, is still double stranded

ii. 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 bases

i. 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 prokaryotes

11.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 DNA iii. DNA gryase helps to prevent supercoiling as the DNA unwinds iv. 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 strand

d. 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 strand

i. The replacement of the last nucleotide of the RNA primer does not automatically join it to the next fragment

12.DNA replication in eukaryotes a. Eukaryotes have several oriC

b. Origin recognition complex (ORC) binds to the origin of replication to prepare the strands for replication

i. 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 molecule

iii. 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 C

d. Replication proteins (RPA) help stabilize the single strands during replication

e. Telomerase can overcome the 3’ end replication problem, by lengthening the telomeres

i. 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 processes a. Homologous recombination occurs during meiosis or during repair of

double stranded breaks

i. 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 mutagens

a. Spontaneous errors can occur during DNA replication through substitution, insertion, deletion, depurination or deamination

i. 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

compounds

i. UV radiation causes the formation of pyrimidine dimers ii. HNO2 causes deamination

iii. Ionisation radiation and alkylating compounds causes chemical modification of bases

15.DNA repair mechanisms correct the mutations by removing the damage a. Base excision repair (BER) places a single damaged base

i. The damaged nitrogenous base is removed by DNA glycosylase that breaks the glycosidic bond, forming an Apurinic/apyrimidinic (AP) site

ii. AP endonuclease then creates a nick to remove the sugar ring and phosphate group of the damaged nucleotide

iii. The gap is then repaired by DNA polymerase that adds the correct nucleotide

iv. 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 dimers

i. 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 methylated

ii. 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 methylated

iii. MutH then nicks the unmethylated daughter strand

iv. 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 Nbs1 iii. Rad52 is then recruited to process the ends to form 3’ single

stranded ends

iv. BRCA2 recruits Rad51 which binds to the single strands to promote strand invasion

v. Rad54 helps Rad51 to find the homologous region during strand invasion

vi. 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 functions

a. Characteristics

i. Low solubility in water

ii. High solubility in nonpolar solvents iii. Mostly hydrocarbon in nature b. Functions

i. Energy storage ii. Cellular membranes iii. Biological signals

2. Lipids are made from fatty acids with hydrocarbon chains that terminate with a carboxyl group

a. They have an even number of carbons between 14 and 24 b. 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-z

i. 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 tissue i. 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 membranes

4. 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): -H

ii. Phosphatidylethanolamine (PE): -CH2CH2NH3+ iii. Phosphatidylcholine (PC): -CH2CH2N(CH3)3+ iv. Phosphatidylserine (PS): -CH2CH(NH3+_)COO

-v. Phosphatidylinositol (PI): ionositol (a 6 carbon ring)

vi. Phosphatidylglycerol (PG): -CH2CH(OH)CH2OH

 Diphosphatidylglycerol can be formed by another glycophospholipid bonding with the substituent glycerol

c. 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 vasoconstrictor

5. 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 sphingolipids

b. 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 sugars

iii. Gangliosides are formed with complex sugars with at least one sialic acid bonded to any residue of the sugar

c. Sphingomyelin is the most common sphingolipid, forming 10% to 20% of the lipid membranes

d. 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 membrane

6. 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 molecules i. Found in plants

ii. A component for flavour or odor

c. Diterpenes are made from four isoprene molecules d. Triterpenes are made from six isoprene molecules

i. Squalene is a triterpene that is a precursor to steroids

7. Steroids are derivatives of cyclopentanoperhydrophenanthrene that is built from isoprene

a. The most common steroid is cholesterol which is also a precursor for other steroids

i. Like hormones or bile acids

b. Cholesterol is a constituent of plasma membranes i. 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 bilayers

a. The amphipathic lipids will form spontaneous structures in water i. Either a monolayer micelle

ii. Or bilayer vesicle

 A cell membrane is a very large vesicle b. 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 enzymes

i. They diffuse laterally rapidly, but transversely very slowly ii. Flippase uses ATP to move lipids from the outer layer to the

inner layer

iii. Floppase uses ATP to move lipids from the inner layer to outer layer

iv. 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 acids

9. The lipid bilayer follows a fluid mosaic model with integral and peripheral proteins and other components