Henderson-Hasselbalch BMB 460 - Spring 2016- group 1

What is the Hendersen-Hasselbalch equation? pH=pka + log ([A-]/[HA])

What are its components? pH= -log of the relative amount of free H and OH ions in a solution pka= -log of acid dissociation constant [A-]= concentration of conjugate base in a solution [HA]= concentration of weak acid in a solution

Acetic Acid-Acetate Pair as a Buffering System

“Buffering action is the consequence of two reversible

reactions taking place simultaneously and reaching their points of equilibrium as governed by their equilibrium

constants, Kw and Ka.”

Nelson, D. L., Lehninger, A. L., & Cox, M. M. (2008). Lehninger principles of biochemistry (6th ed.). New York: W.H. Freeman.

any weak acid “HA”

corresponding “A-”

Various Uses of the Henderson-Hasselbalch Equation

1) Calculating pH with known weak Acid/ conjugate base ratio 2) Calculating A- / HA ratio with known pH and pKa

a)  Ratio of HA to Conjugate Base b)  Ratio of Salt to Acid c)  Ratio of Ionized and Un- ionized forms of the molecule of

interest 3) Determining level of ionization of amino acid side chains based on pKa and conditions

Relevance & Application

• Buffers needed in pH sensitive situations, such as biological systems. pH- buffers affect: •  Charge of proteins (Amino Acid side chain ionization) •  Charge of DNA/RNA (Phosphate backbone) •  Enzymes (have an optimal pH)

Applications: •  Calculating pH of a solution •  Determining buffer concentrations •  Calculating ionized/un- ionized concentrations •  Calculating pKa using pH

Problem 1

Calculation of pH from Molar Concentrations What is the pH of a solution containing 0.12M of NH4Cl and 0.03M of NaOH (pKa of NH4

+/NH3 is 9.25)?

pH = pKa + log([A-]/[HA])

Problem 2

Problem 3 Amino acid ionization: Calculate the fraction of histidine that has its imidazole side chain protonated at pH 7.3. The pKa values for histidine are pK1=1.8, pK2 (imidazole)=6.0, and pK3 = 9.2.

pH = pKa + log([A-]/[HA])

7.3 = 6.0 + log([His]/[HisH+])

1.3 = log([His]/[HisH+])

antilog(1.3) = ([His]/[HisH+]) = 20 ( 20:1 His/HisH+ )

His = 20/21 = 0.952 = 95.2% HisH+ = 1/21 = 0.0476 = 4.8% unprotonated protonated

If you are having trouble with these problems and/or still need more of a review, see section 2.3 of the book (pg. 63).

Recommended practice problems: Chapter 2, #9, #12, #18, #22, #31, and #32. --- try: 7 and 27 6th edition

Amino Acids Review

What is an Amino Acid?

alpha Carbon

Amine Group

Carboxylic Acid Group

Variable R Group

Amino Acids (think at neutral pH)

Nonpolar R Groups : Gly, Ala, Pro, Val, Leu, Ile, Met

Aromatic R Groups: Phe, Tyr, Trp (F,Y,W) -absorb at 280nm

Polar, Uncharged R Groups: Ser, Thr, Cys, Asn (N), Gln (Q)

Positively Charged R Groups : Lys (K), Arg (R)- sometimes His,

Negatively Charged R Groups: Asp (D), Glu (E)

www.neb.com

Why does R group chemistry matter ‘more’ for enzymes?

Can you draw glycyl-glycine?

Most common Amino Acids involved in enzyme chemistry and R group:

 Hydroxyl (-OH) –ST and tyr (Y)

  Thiol (-SH) - C

 Carboxylic acid – COO- glu (E) and asp (D)

 Amino -NH3+

lys (K)   Imidazole pKa =6.0

H ONLY residue ‘buffer’ at 7

Nucleophilic Functional Groups (e- donors)

Negatively charged oxygen

Negatively charged sulfhydryl

Carbanion

Uncharged amine group

Imidazole

Hydroxide ion

Functional groups act as proton

donors or acceptors in the active sites of

enzymes to help catalyze chemical

reactions in biological systems.

Electrophilic Functional Groups (e- acceptors)

Carbon atom of carbonyl group

Protonated imine group

Phosphorous of a phosphate group

Proton

Common Reactions Involving Amino Acids

Oxidation-Reduction Reactions: require COENZYMES – amino acids not good to oxidize/reduce

Many of these reactions involve the loss of two electrons and one or two hydrogen ions..

Group Transfer Reactions:

Transfer of acyl, glycosyl and phosphoryl groups are common. This reaction occurs through the addition of a nucleophile, ex. hydroxyl group, to an electrophile such as a carbonyl carbon. An example of an enzyme that catalyzes this type of reaction is a kinase.

Common Reactions Involving Amino Acids Aldol and Claisen Condensations and Decarboxylations:

These reactions would be impossible without the environment created from the functional groups. Electronegative atoms such as oxygen and nitrogen can stabilize required intermediates.

Aldol condensations occur by the nucleophilic alpha carbanion, stabilized by an adjacent carbonyl group, attacking a electrophilic carbonyl carbon.

Claisen condensations are similar except the alpha carbanion is stabilized by an adjacent thioester.

Decarboxylations occur through the formation of an alpha carbanion also stabilized by an adjacent carbonyl group.

Relation between pka and pH

For each functional group in an amino acid, when:

pH>pKa the functional group becomes deprotonated. EXAMPLES?

pH<pKa the protons bind to the functional group.

EXAMPLES?

Titration Curves for Amino Acids with non-ionizing R

Amino acids have a zwitterionic form (dipolar ion) which allow them to act as either an acid or base at different pH values

pI (Isoelectric point): pH where the net charge of the amino acid is 0.

pKa: The pH at which the rate of protonation of the R group is in equilibrium.

Titration Curves for Amino Acids with ionizing R

Histidine is of interest, because its ionizable side chain has a

pKa near neutral.

At biological pH, Histidine can be positively charged or uncharged.

Facilitates enzyme-catalyzed reactions by acting as a

proton donor or acceptor.

Essentials of Enzymes By: Kevin, Remy, Fufei, Stephen, Amy, Ben

Enzyme (E): a biological catalyst that increase reaction rates without being consumed in the reaction (While usually proteins, a small group of RNA based enzymes called ribozymes also exist)

Substrate (S): a molecule that binds to an enzyme and undergoes a reaction

Active site: the location (usually a groove/cleft or pocket) on an enzyme

that binds the substrate and contains the key Catalytic residues that carry out the chemistry - catalyze chemical transformation of S to P

E + S ⇌ ES ⇌ EP ⇌ E + P

Enzyme-catalyzed reactions  Enzymes increase the RATE of the reaction.

 ΔG unchanged -

 Activation energy decreases.

through noncovalent interactions between E and S

Binding energy: the free energy that is released by the formation of weak interactions between a complementary substrate and enzyme.

Figure 6.3

Enzyme-Substrate interactions

Figure 6.4 Figure 6.5

ENZYMES require native protein structure- Proper primary, secondary, tertiary, and sometimes quaternary structure formation are critical to enzymatic activity.

Cofactors: inorganic ions (Fe2+) or CoENZYMES - a complex organic/metal-organic molecule known as a coenzyme (Biocytin) may be required

Regulatory enzymes: increased/decreased activity in response to certain signals and display a sigmoid model, reflecting cooperative interactions. (do not follow Michaelis Menten equation but can be related by Km0.5)

Allosteric enzymes: function by noncovalent binding to regulatory compounds such as metabolites/cofactors & differ by containing active and regulatory binding sites Example : ATCase, Amino acid residue modification - Phosphorylation

Enzyme Kinetics- what assumptions were required to derive M-M equation?

Vₒ = Vmax[S]/Km + [S]

Vₒ: initial velocity Vmax: maximum velocity [S]: initial substrate concentration Km: Michaelis- Menten rate constant, or the rate of breakdown to reactants & products/formation of enzyme substrate complex

From this equation: Km = [S] when Vₒ = 1/2Vmax At low [S], Vₒ is linearly dependent on [S], at high [S] Vₒ=Vmax, there is a plateau and the enzyme is saturated Kcat = Vmax[Et] - Kcat: turnover number, or the amount of substrate converted to product when an enzyme is saturated (per unit of time)

Lineweaver-Burk Michaelis-Menten

Vo = Vmax[S]/Km + [S]

Vmax = kcat [Et]

Km = kcat+k-1

k1

When Kcat is truly rate limiting (very small

relative to K-1)

Km ≈ k-1/k1

Substitute for Vmax = Kcat[Et]

Vo = kcat [Et] [S]/Km + [S]

When [S] is <<< than Km?

Vo = kcat [Et] [S]

Km

Enzyme Efficiency?

+ modulator

- modulator

Bioenergetics Group-4

Bioenergetics

The study of energy transformations in living systems.

The means by which energy from fuel metabolism (or light capture) is coupled to a cell’s energy requiring reactions.

Enthalpy and Entropy

H is enthalpy, the heat contained in a system -ΔH is exothermic, heat exits the system. +ΔH is endothermic, heat enters the system

S is entropy, a quantitative expression of randomness in a system.

Gibbs  Free  Energy  

• ΔG  =  Δ  H  –  TΔS                  =  G  (Final  State)  –  G(Ini9al  State)  It’s  a  measure  of  the  distance  of  a  system  from  its  equilibrium  

Gibbs  Free  Energy  • ΔG  =  Δ  H  –  TΔS                  =  G  (Final  State)  –  G(Ini9al  State)  

• What  ΔG  tells:  The  direc9on  of  the  reac9on(s):  which  is  more  favorable  

• What  ΔG  doesn’t  tell:  1.  The  pathway  of  reac9on(s)                                                                                        2.  The  speed  of  reac9on(s)  

∆G (k

J/m

ol)

Enzymatic reactions of glycolysis – actual free energy change

Free energy change predicts Direction but NOTHING about RATE or pathway

Gibb’s Free Energy The energy available to do work in a reaction during a constant temp and pressure.

Keq is ([C]C[D]D/[A]A[B]B) when the reaction is at equilibrium.

Keq ΔG Then the Reaction

>1 <0 Proceeds Forward

1 0 Is at Equilibrium

<1 >0 Proceeds in Reverse

ΔG’ vs ΔG°’ Chemical reactions proceed spontaneously until equilibrium is achieved.

ΔG’=ΔG°’+RTln([C]C[D]D/[A]A[B]B) ΔG°’: the biochemical standard free energy change, is a constant given reaction and set of conditions ΔG’ is the free energy available from the reaction at the substrate and product concentration actually occurring in the system.

glucose 1 phosphate glucose 6 phosphate

fructose 6 phosphate glucose 6 phosphate

Glucose 1 phosphate fructose 6 phosphate

Δ G’° = -7.3 kJ/mol

Δ G’° = -1.7 kJ/mol

Δ G’° = ? kJ/mol

Coupling reactions

Using [product]/[reactant] concentration to drive a reaction

∆G = ∆G’°+ RT ln [C][D]/[A][B]

glucose 6 phosphate fructose 6 phosphate Δ G’° = +1.7 kJ/mol

[fructose 6 phosphate] [glucose 6 phosphate]

To achieve ΔG = - 3.4 kJ/mol

= ???

NTP-Nucleoside Triphosphate Phosphate is a good leaving group and the removal of the

third “gamma” phosphate has a high -ΔG°’ value.

NTPs are used to transport energy.

Phosphoryl transfer occurs- though reactions may look like NTP ‘hydrolysis’

WHY are NTPs a preferred option

for energy storage/use?

Reducing Equivalents Reducing Equivalent- Single electron equivalent participation in an oxidation - reduction reaction.

Biological fuel molecules are generally enzymatically dehydrogenated to lose two equivalent at a time- hydride ions or two hydrogens

Each oxygen atom can accept two reducing equivalents passing from substrate to oxygen.