MT 1 Flashcards

Question

Gibb's Free Energy Equation and implications of ∆G, ∆H, ∆S

Answer 1

∆G_sys= ∆H - T∆Ssys Where: ∆G = Gibb's free energy (kJ/mole) T = Temperature in K If ∆G \< 0, the reaction is spontaneous (exergonic) If ∆G \> 0, the reaction is non-spontaneous (endergonic) ∆H \< 0, the reaction releases heat =\> ∆G is more negative =\> more spontaneous ∆S \> 0 =\> more disorganized =\> ∆G is more negative =\> more spontaneous

Answer 2

When a non-polar molecule is added to H₂O, the water molecules are forced into a shell. This lowers entropy. However, with time, non-polar molecules come together and H₂O molecules form a shell only at the edge, and entropy increases.

Answer 3

Many biomolecules can act as weak acids and bases Behaviour of biomolecules depends on ionization state, which is determined by pH Because pH is important, it must be maintained at a certain level with buffers. pH: A measure of concentration of H⁺ in solution Buffer: A mixture of weak acid and conjugate base. It resists changes in pH. Buffering region is usually 1 pH unit on either side of pKa. pH = -log[H⁺] Scale = 0-14, where 0 is a strong acid and 14 is a strong base H₂O ⇌ H⁺ + OH^- K_w= ionization constant = 1 \* 10^-14 K_w= [H⁺][OH^-]

Answer 4

Weak acids and bases don't fully ionize in solution CH₃OOH ⇌ CH₃OO^- + H⁺ K_a = [A^-][H⁺] / [HA] = [CH₃OO^-][H⁺] / [CH₃OOH] pK_a = -log[K_a]

Answer 5

If we titrate a weak acid with a strong base (NaOH), we can calculate pH using Henderson-Haselbach equation pH = pKa + log[A^- / HA] If pH = pKa, [HA] = [A^-] If pH \< pKa, [HA] \> [A^-] If pH \> pKa, [HA] \< [A^-] There is a region where pH doesn't change much =\> buffering region

Answer 6

1. Carbonate/Bicarbonate Buffer 2. Phosphate Buffer 3. Histidine and cysteine

Answer 7

CO_2(dissolved) + H₂O ⇌ H₂CO₃(carbonic acid) ⇌ H⁺ + HCO₃^- (bicarbonate ion)

Answer 8

H₂PO₄^- (dihydrogen phosphate ion) ⇌ H⁺ + HPO₄^2-(monohydrogen phosphate ion)

Answer 9

Linear polymers built out of α-amino acids (αα) Proteins final 3D shape and function depends on its sequence of αα Each αα has different functional groups, allowing for massive diversity Proteins can be flexible or rigid Proteins can interact with each other

Answer 10

Contain central carbon (α-carbon), attached to amino group, carboxylic acid group, hydrogen atom, and unique side chain (R) Note: α-carbon is chiral

Answer 11

Chiral center: Atom with its substituents arranged so that the molecule is NOT superimpossible on its mirror image This means that there are 2 enantiomers for each amino acid (except glycine) Enantiomer: pair of molecules, each with one or more chiral centre that are mirror images of each other If the amino group is on the left, it is in the L-form (otherwise D-form) In biological systems, only L-aa's exist in proteins and all living things

Answer 12

At pH = 7, all aa's exist in zwitterion Zwitterion: ion with both (+) and (-) charge

Answer 13

The side chains (R roups)

Answer 14

Gly, G No chiral carbon Technically, not really hydrophobic

Answer 15

Aliphatic Has a second chiral center

Answer 16

Aliphatic Contains thio-ether (-S-C) group

Answer 17

Contains a ring; changes 3D structure of amino acid Still aliphatic Twist in side chain; ring structure makes it more rigid/more restrained Often introduces kinks in amino acid polypeptide chain

Answer 18

Aliphatic: Open chain structure (alkanes) 1. Glycine, Gly, G 2. Alanine, Ala, A 3. Valine, Val, V 4. Leucine, Leu, L 5. Isoleucine, Ile, I 6. Methionine, Met, M 7. Proline, Pro, P All of these are hydrophobic, often found in the center of a protein or in memebrane crossing domain

Answer 19

Contains aromatic group (phenyl ring) Participates in hydrophobic interactions 1. Phenylalanine, Phe, F 2. Tyrosine, Tyr, Y 3. Typtophan, Trp, W

Answer 20

Aromatic Is like phenylalanine but has -OH, therefore making it more reactive

Answer 21

Positively charged 1. Lysine, Lys, K 2. Arginine, Arg, R 3. Histidine, His, H Charged, so found on the surface of proteins (interacts with water)

Answer 22

Basic amino acid; has an amino group

Answer 23

Basic amino acid; guanidinium group; side chains are positively charged at pH = 7 (pKa of the side chain is greater than 10)

Answer 24

Typically considered a basic amino acid Has ionizable group with pKa ~6 That means that it can be charged or uncharged depending on its location Often found in active site of enzymes

Answer 25

Negatively charged at pH = 7 (pKa \< 4) Contains carboxylic group 1. Aspartate, Asp, D 2. Glutamate, Glu, E Charged, so found on the surface or proteins (interacts with water)

Answer 26

Not charged Can form H-bonds (hydrophilic) 1. Serine, Ser, S 2. Threonine, Thr, T 3. Cysteine, Cys, C 4. Asparagine, Asn, N 5. Glutamine, Glu, Q

Answer 27

Polar amino acid Contains hydroxy group

Answer 28

R-OH functional group

Answer 29

Polar amino acid Contains hydroxy group

Answer 30

Polar amino acid Contains sulfhydryl group (thiol group, -SH) Can form disulfide bonds with another cysteine in the same chain or another. For example, insulin has 3 disulfide bridges. Can form H-bonds, but they're weak.

Answer 31

Polar amino acid Derivative of aspartate Contains carboxyamide instead of carboxyl

Answer 32

Polar amino acid Contains carboxamide instead of carboxyl

Answer 33

pKa value of amino acids depend on the environment

Answer 34

Linear sequence of amino acids linked by peptide bonds to form a protein

Answer 35

linkage of alpha-carboxyl of one amino acid to the alpha-amino group of another amino acid, with the loss of water Not energetically favourable to form, but once formed, it is stable (requires energy to be made) Peptide bond has double bond characteristics because of resonance between the peptide bond and the carbonyl; as a result, it is planar There is no free rotation about hte peptide bond Because of steric hindrance, almost all peptide bonds are in trans-configuration (R-groups of opposite sides of the plane) except X-Proline (both cis and trans occur, but trans is preferred) The peptide bond is conformationally restrained, but the bonds between N-C_αand C_α-C=O are single and free to rotate; thus, polypeptides are flexible

Answer 36

A series of amino acids linked by peptide bonds Has polarity; amino group is on the left and the free carboxylic group is on the right Consist of repeated backbone with variable side chains All carbonyl and amino groups in the backbone can H-bond (w/ the exception of proline); has an important role in 3D structure

Answer 37

The amount of rotation; the angle between planes through two sets of three atoms Ranges from -180° to 180° N-C_αis called Φ (phi) C_α-C=O is called Ψ (psi) N-C=O is the peptide bond... no free rotation In reality, not all combinations of Φ and Ψ are allowed, because of steric hindrance that limits the number of structures a protein can adopt

Answer 38

Ramachandran plot Shows the combinations of Φ and Ψ that are allowed Areas of dark = favourable Areas of light = borderline Areas of white = not allowed Each plot is for a particular amino acid; proline and glycine would have Ramachaundron plots that look very different from the other amino acids.

Answer 39

Proteins have a series of limitations on what orientations they can adopt. They can often hold into a single structure in physiological conditions.

Answer 40

1. Determine 3D shape 2. Understand function 3. Understand dsease (ie. replacing a positiv AA with a negative AA is bad!) 4. Understand evolutionary history

Answer 41

The spatial arrangement of amino acid residues that are relatively close to each other in linear sequence of polypeptide chain (alpha helices, beta sheets)

Answer 42

Polypeptide backbond forms the inner part of a right handed helix, with the side chans sticking outwards. It is stabilized by hydrogen bonds between NH and carbonyl (C=O) of the backbone **The C=O (i) forms H-bond with N-H (i+4) --- he specifically said we need to know this** Has ideal dihedral angles of Φ = -60° and Ψ = -45° All NH and C=O in the backbone are hydrogen bonded Each aa in the a-helix increases the helix by 1.5Å (angstroms) Proline/Glycine are the worst for alpha helix R-groups in i+3 and i+4 are close to i a-helix is right-handed; left-handed in possible but not stable due to steric hindrance a-helical content vary in proteins (sometimes high, sometimes low) Keratin consists of 2 intertwined helices a-helix is shown as ribbons or rods in protein structures

Answer 43

Two or more B-strands associated as a stack of chains in an extended zigzag; form stabilized by interstrand H-bond Each aa extends B-strand by 35Å The N-H and C=O of residue "i" in one B-strand from H-bond to a single residue "j" in the other B-strand Φ= -139° and Ψ= +135°

Answer 44

N-H of residue i of one strand forms H-bond with C=O of j in another strand, but C=O of i forms H-bond with N-H of j+2 residue in another strand This makes strand shorter (3.25Å per residue) Φ = -119° and Ψ= +113°

Answer 45

Antiparallel

Answer 46

broad arrows pointing to C-terminus

Answer 47

Peptide chains reverse direction Accomplished by B-turn (common) or by larger loops (no common structure)

Answer 48

A beta barrel -- a transmembrane protein

Answer 49

The spatial arrangement of aa residues that are far apart from each other in linear sequence as well as the pattern of disulfide bonds Unique for every protein

Answer 50

Myoglobin 153 amino acids, relatively small Largely a-helices (70%) Globular Few voids (unorganized chain) In red = heme (protoporphyrin ring) = where oxygen binds Surface contour Yellow = hydrophobic amino acids; mostly in protein core Polar amino acids on the outside

Answer 51

Troponin C Ca²⁺ binding proteins 2 domains - parts of protein with defined function

Answer 52

The spatial arrangement of multiple subunits (polypeptides) and their interaction Some proteins are composed of more than one polypeptide chain (multimers) in order to function

Answer 53

Hemoglobin; Hb O2 transporter in blood of mammals 4 subunits: 2 alpha + 2 beta Hb cannot function unless it forms tetramer (4 peptides)

Answer 54

Minute virus of mice 9 VP1 + 51 VP2 = viral capside (quarternary structure) Large enough to fit DNA

Answer 55

1. Proteasome 2. Spliceosome 3. Ribosomes (+ RNA)

Answer 56

USUALLY, NOT ALWAYS

Answer 57

We cannot predict the final 3D conformation using primary structure. RIP.

Answer 58

What we do know: Folding is driven by thermodyanmics: finding the most stable complex (most negative ∆G), but the difference between folded and unfolded protein is small (~20-60 kJ/mole) Driven mostly by entropy (ie. the hydrophobic side chains are going to be exluded from water in the core, while polar amino acids are at the surface) In order to fold, hydrogen bonds must form Unpaired charged or polar groups will destabilize structures. What can we do? Though we cannot predict final 3D structures, we can predict secondary structures. Certain amino acid residues are more likely to form a-helix or B-sheets A, L, K, M form stable a-helix Proline and glycine destabilize a-helix P doesn't have H and it is rigid (steric hindrance) G is too flexible However, even peptides with the same sequence can adopt different secondary structure in different proteins Folding is usually an all-or-none process (either folded or misfolded/unfolded) Folding is cooperative (if one portion of the protein folds, it will influence how another portion of the same protein folds) We usually visualize protein folding as a free energy funnel Normal tertiary structure = native structure = functional protein In a protein's unfolded state, there are many possible structures with high free enrgy, but as a series of folding happens, free energy decreases with every structure formed. The number of possible conformations decrease until you reach the native (folded) state Note: Not all proteins have one single conformation Some might only form a final structure when bound to a substrate or regulator; some exist in equilibrium between two different structures

Answer 59

Cryoelectron microscope: uses a beam of electrons to image many, many native proteins X-ray crystallography: measure e- density (ie. myoglobin) NMR (nuclear magnetic resonance): measures the location of nuclei

Answer 60

Modifications to proteins after it has been synthesized 1. Phosphorylation - adding phosphate to amino acids with -OH (ie. Ser, Tyr, Thr); ie. signal transduction to activate or deactivate proteins 2. Glycosylation - addition of sugars to a residue (usually Asn and Ser); ie. cell recognition (immune system); ie. glycoprotein 3. Hydroxylation - usually protein (hydroxyproline); ie. collagen - triple helix 4. Carboxylation - addition of a carboxyl group; glutamate (ie. blood and clotting) 5. Acetylation - addition of acetyl group; lysine - regulation of gene expression (epigenetics) 6. Proteins can be trimmed (ie. trypsinogen --\> trypsin)

Answer 61

1. Hydroxylation 2. Carboxylation 3. Glycosylation 4. Phosphorylation

Answer 62

Biological macromolecule that acts as a catalyst for biochemical rxns. Usually proteins.

Answer 63

Chemical that increases the rate of rxn without being consumed

Answer 64

of molecules of substrate converted to product per molecule of enzyme per second

Answer 65

Enzymes are very specific; they will catalyze only one specific reaction or a set of rxns Specificity is based on a series of weak interactions between substrate and enzyme, especially in the active site The shape of the enzyme determines specificity and function

Answer 66

Digestive enzyme Cleaves peptide bond on the carboxyl side of Lys or Arg

Answer 67

Cleaves any peptide bond

Answer 68

Involved in blood clotting Cleaves Arg (or Gly) peptide bonds

Answer 69

Region of an enzyme that binds the substrate. It contains the residues that directly participate in the rxn. The nonaxtive site residues are important for structure of the protein or regulation of enzyme (ie. phosphorylation/inhibitors)

Answer 70

1. They are clefts in the enzyme made up by residues from all over the primary structure 2. Takes up a small volume of an enzyme 3. Water is usually manipulated or excluded from an active site -- This changes the behaviour of the resiudes 4. Substrate is bound by weak interactions in active site 5. There is a partial complimentarity between substrate and active site

Answer 71

1. Lock and Key - Active site is complimentary to match to a substrate 2. Induced Fit - Binding of a substrate causes the active site to assume a matching shape

Answer 72

Enzymes may require cofactors in order to function Cofactor: inorganic ion or small organic compound required for enzyme activity

Answer 73

Cofactor - only inorganic ions (ie. Mg2+) Coenzyme - organic compounds (ie. FAD)

Answer 74

Cofactor that is tightly bound to an enzyme (heme in myoglobin)

Answer 75

Enzyme that requires a coenzyme but doesn't have the cofactor (no function)

Answer 76

Enzyme that requires a cofactor and has the cofactor (functional)

Answer 77

This is a cofactor that is found in tap water and needed for DNA synthesis

Answer 78

1. Enzymes do not alter ∆G; they obey laws of thermodynamics; do not change how spontaneous rxn is 2. Enzymes do not alter the final equilibrium of products to reactants 3. Enzymes do speed up rxns; enzymes accelerate rxns by decreasing the activation energy (∆G^‡) by facilitating the formation of the transition state (X^‡) The energy needed to get to X^‡is known as activation energy (∆G^‡) ∆G^‡_s→p= G_x^‡ - G_s → forward rxn ∆G^‡_p→s= G_x^‡ - G_p→ reverse rxn Only a fraction of substrate (s) will have enough energy to form X^‡ The activation energy cotnrols the rate (rate limiting step) Activation energy is not a part of the ∆G of the rxn ∆G^‡_cat \< ∆G^‡_uncat This is how enzymes work: S+E ⇌ ES ⇌ EP ⇌ E + P Enzymes will interact with transition state such that the activation energy is lowered: the rxn will speed up as a greater fraction of S has energy to react (to convert to X^‡)

Answer 79

It comes from the enzyme binding and stabilizing the transition state in the active site. Active site is most complimentary to X^‡

Answer 80

∆G_B = binding energy Defined as the energy derived from the non-covalent interactions between the enzyme and substrate

Answer 81

We can measure the rate (or velocity) of this rxn as eithr disappearance of substrate A over time or the appearance of product (P) over time. V = -∆A/∆t = ∆P/∆t (usually expressed as moles/unit of time) If we measure the disappearance of A, then we can express the rate as directly related to [A] V = k[A] k - rate constant (not an equilibrium constant!) k = proportionality constant that relates to [S] at a specific temperature. This is known as first order rxn and k has units of sec^-1 or min^-1

Answer 82

2A → P where v = k[A]² A + B → P where v = k[A][B] These 2nd order rxns usually have k as units of M^-1sec^-2 or M^-1min^-1 or moles^-1sec^-1

Answer 83

Series of tubes with fixed amount of enzyme, mixed with different amount of substrate. Then they measured the amount of product formed over time.

Answer 84

Leoner Michaelis Maud Menten

Answer 85

1. We only examine early times in the rxn when [P] is low. We can ignore reverse rxn. E + S ⇌ ES → P 2. [S] \>\> [E], we can assume that [ES] doesn't affect [S] 3. A steady state exists such that the rate of [ES] formation = rate of [ES] consumption

Answer 86

We can calculate the initial velocty, V₀ (initial rate, μM/time) Based on an earlier assumption, we can describe how initial velocity depends on [S] V₀ = V_max ([S]/([S}+Km)) [S] = concentration of substrate V₀= initial velocity V_max = maximum velocity of rxn, when all the active sites are saturated with subtrate K_M = Michaelis constant; [substrate] at which the enzyme catalyzed rxn proceeds at 1/2 Vmax rate. It is a rate constant.

Answer 87

K_M = K_-1 + K₂ / K₁ When [S] \<\< K_M =\> V₀ = V_max ([S]/KM) [S] = K_M =\> V₀ = Vmax (K_M/2K_M) = 1/2 V_max [S] \>\> K_M=\> V₀ = Vmax ([S]/[S]) = Vmax K_M is an important characteristic of enzymes; it's a rough estimate of affinity of the enzyme to its subtrate K_Mhigh = affinity is low K_Mlow = affinity is high It depends on the type of enzyme, pH, temperature, ionic strength

Answer 88

Maximum rate of enzyme It will change with [enzyme]

Answer 89

Lineweaver-Burk plot

Answer 90

1. Irreversible inhibition 2. Competitive inhibition 3. Non-competitive inhibition 4. Uncompetitive inhibition

Answer 91

No, because they change depending on the concentration of enzyme used.

Answer 92

We can use K_cat!

Answer 93

The minimum amount of subtrate an enzyme can convert into product in a given time (min^-1 or sec^-1) K_cat = V_max/[E]_T [E]_T = enzyme total concentration = [E] \* # of active sites Higher number = better / more efficient enzyme

Answer 94

K_cat / K_M sec^-1M^-1

Answer 95

AKA suicide inhibitor Inhibitor stays boundt o enzyme for long periods (often forever). Usually covalently attached in active site =\> inhibitor blocks active site

Answer 96

- inhibitor binds to the active site and competes with the substrate for it - only one can bind at a time - V_max doesn't change but amount of subtrate needed to reach V_max -- as well as half of V_max -- is increased (K_M increases) - Adding inhibitor increases K_M but V_max remains unchanged - Can be overcome by adding more subtrate - In a double-reciprocal plot, Y-intercept stays the same; X-intercept increases (K_M increases, because -1/K_Mi \> -1/K_M =\> K_Mi \> K_M

Answer 97

- Inhibitor binds at a site other than the active site - Inhibitor doesn't block active site (ie. substrate can bind to active site, but products are not formed) - can't compete by adding more substrate - K_M doesn't change, but V_max decreases - In effect, you are lowering the number of functional enzymes - In double-reciprocal plot: x-intecept is the same, but y-intercept is different (increases)

Answer 98

- Inhibitor only binds to ES, forming ESI complex - Like non-competitive inhibition, this results in lower V_max (lowering # of functional enzymes) - But it also decreases K_M because as ES → ESI =\> [ES] ↓ ! This promotes increased substrate binding to enzyme, lowering K_M - In the double reciprocal plot: both X and Y-intercepts change =\> parallel lines -- x-intercept decreases (moves left) and y-intercept increases (moves up)

Answer 99

Uncompetitive

Answer 100

Noncompetitive

Answer 101

Competitive

MT 1 Flashcards

(146 cards)