Bio Chem week 2 notes - dr. Berndsen PDF

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BioChem Week 2: Thermodynamics and pKa pKa, functional groups, and equilibrium  Phosphate, amine, thiol, carboxylic acid, alcohol o When deprotonated they gain a charge o Will influence week interactions because they can change charge? o When charge is changed it will increase/decrease strength and types of weak interactions  PKa- defines how well a group gives up its protons o Where there is not a lot of change when you add a base and where is can serve as a buffer  Where it goes from mostly protonated to mostly deprotonated o When pH= pKa there is equal acid and base concentrations o When pH < pKa there will be more acid than base concentrations; have more protons o When pH > pKa there will be more basic than acid concentrations; have less protons  Acid- can donate protons  Base- can accept protons (negative signs)  Equilibrium constants can be changed by more than just temperature o pKa= - logKa  Ka= base o In a polar environment (hydrophilic), makes good interactions in charged types of situation; more acid so ka will decrease and pKa will increase; o Must figure out if there is more acid or base o Does Ka increase or decrease o Does pKa increase or decrease Thermodynamics and Weak Interactions  changes in free energy determine if a reaction will occur spontaneously or nonspontaneouly  deltaG- difference in energy from where it started to where it ended o when it goes down on the curve it is spontaneous= negative  does not require a lot of energy  energy of the products is less than that of the reactants  example: two water molecules forming a hydrogen bond o hydrophobic effect- things that aren’t favorable to interact with water form clumps by themselves and fear water as much as possible  interactions of water with water are key in how the effect occurs  deltaG double dagger- difference in energy from where it started to the highest energy on the curve (transitional state; controls the rate at which something happens)  Two thermodynamic properties of the reaction o The free energy difference between the products and reactants (determines if reaction will be spontaneous)

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o The energy required to initiate the conversion of reactants to products (determines the rate of the reaction When can a spontaneous reaction occur o Only when the deltaG is negative; therefore, the reactions are exergonic o A system is at equilibrium and no net change can take place if deltaG is zero o A reaction cannot occur spontaneous if deltaG is positive. An input of free energy is required to drive such a reaction; these reactions are termed endergonic The ΔG of a reaction is independent of the path (or molecular mechanism) of the transformation. o The ΔG of a reaction depends only on the free energy of the products (the final state) minus the free energy of the reactants (the initial state) The ΔG provides no information about the rate of a reaction. o A negative ΔG indicates that a reaction can occur spontaneously, but it does not signify whether it will proceed at a perceptible rate. the ΔG of a reaction depends on the nature of the reactants (expressed in the ΔG° term of equation 1) and on their concentrations

o Units of energy o Calorie- equivalent to the amount of heat required to raise the temperature of 1 gram of water from 14.5 to 15.5 Celsius o Kilocalories- 1000 calories o Joule- the amount of energy needed to apply a 1-newton force over a distance of 1 meter o Kilojoule- equal to 1000 J o 1 kcal= 4.184 Equilibrium constant formula

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It is important to stress that whether the ΔG for a reaction is larger, smaller, or the same as ΔG°′ depends on the concentrations of the reactants and products.

Enzymes Alter only the reaction rate and not the reaction equilibrium  An enzyme cannot alter the laws of thermodynamics and consequently cannot alter the equilibrium of a chemical reaction. o enzyme accelerates the forward and reverse reactions by precisely the same factor.  Enzymes accelerate the attainment of equilibria but do not shift their positions.  The equilibrium position is a function only of the free-energy difference between reactants and products. Weak Interaction #2 1. There are many ways to draw how a single reaction could occur, however not all of them are likely. This is where thermodynamics can help by indicating how likely a reaction is to occur spontaneously. Reactions occur though a route that produces a free energy (called ΔG) which is the lowest. The net reaction will stop when the system reaches equilibrium which is when the net change in reactants and products is zero. This push towards equilibrium drives biological processes. In the table below, indicate whether a reaction with an enthalpy and entropy of the designated sign would produce a spontaneous ΔG under all conditions or only some temperatures and indicate what those temperatures would be. Use equation below to help you. a. -ΔH and +ΔS -fully spontaneous at all temperatures b. -ΔH and -ΔS -spontaneous at low temperatures c. +ΔH and +ΔS – spontaneous at high temperatures d. +ΔH and -ΔS – nonspontaneous at all temperatures 2. Biomolecule structures are formed based on the rules of chemistry. Weak interactions, thermodynamics, equilibrium, and free energy combined with sterics,

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bonding, and hybridization from organic chemistry can be used to predict and evaluate the quality of a biomolecule structure from computational and experimental methods. The values of ΔH and ΔS are not easy to determine but equilibrium constants generally are. For biological systems, we often use concentrations of reactants and products to give a measure of spontaneity. Remember however, not all reactions involve an obvious chemical change like the production of a color. In biochemistry, often we are looking at the shift from one shape to another. In the equation below, ΔGo’ is the standard change in free energy in pure water, pH 7.0, 1 atm pressure and 25 oC. A, B are the reactant and the C, D are the products. R is the gas constant which is equal to 8.314 J/(mol K) and will be given on a test. The superscript a,b,c, and d are for reactions where the moles of that substance are different than 1. At equilibrium ΔG = 0. Note: A,B,C, and D are just placeholders for reactants and products not specific names or designations. You always have to remember the reaction direction and put the concentrations in the right spot. If ΔG = 0 at equilibrium, what might the (C*D)/(A*B) = ? * a. Keq (the equilibrium constant) b. The equilibrium constant (Keq) or any measure of the equilibrium concentrations is the ratio of products/reactions when ΔG = 0. This can also be used to calculate the ΔG0' for the reaction At equilibrium, A B, if the concentration of A (reactant) is 5 mM and the concentration of B (product) is 3 mM what is ΔGo’ in kJ/mol? (R = 0.008314 kJ/ (mol*K), T = 298 K) Is this reaction spontaneous or non-spontaneous in the standard state? a. Nonspontaneous For A + B ⇄ C at equilibrium, the [A] is 9 mM, [B] = 2 mM, and [C] = 2.5 mM. Calculate the ΔG°’ for this reaction at 298 K in kJ/mol. (Remember to convert concentrations to molar). Is this reaction spontaneous or non-spontaneous in the standard state? a. Spontaneous Would forming a weak interaction like hydrogen bonds or ionic interactions between water and DNA, have a spontaneous ΔG or non-spontaneous ΔG? a. Spontaneous b. Hydrophilic molecules spontaneously form weak interactions with water The Pauling and Corey model of DNA as a triple helix shown below, was the generally accepted model until the work of Watson, Crick, Wilkins, Gosling, and Franklin in 1952. In this model, three strands of DNA form hydrogen bonds between phosphate groups in the center of the triple helix and the nucleotide bases face outward toward the solvent. Why is this structure less likely to form than the double helical model that is currently accepted? a. Negative charges on phosphates would repel b. The interactions of the bases with bases and the water with the phosphates is stronger in double helical DNA Yasara guide a. Red= ionic interaction b. Yellow= van der waals interaction c. Green= hydrogen bond

8. What two species are formed when a water molecule becomes deprotonated? a. hydroxide (OH-) b. proton/hydronium (H+/H3O+) c. Water can be broken into hydroxide and a proton. This will be helpful when we cover hydrolysis reactions. 9. The concentration of protons in solution ([H+]) is more commonly referred to as __________ . a. -log[H+] or pH 10. What are the products and reactants in the acid dissociation reaction represented by Ka? a. conjugate base = product, acid = reactant b. The dissociation of a weak acid is: acid --> base, so base/acid in the Ka equation. Since this is an equilibrium constant it obeys the same rules of thermodynamics that we have seen previously and we can calculate free energies from it AND we can predict whether a weak acid would dissociate based on the solvent/reaction conditions. 11. Identify the charge on the group when deprotonated a. R-NH3+ = neutral b. Guanidinium in side chain of arginine = neutral c. R-OH =negative d. Imidazole in side chain of histidine = neutral e. Phenol in side chain of tyrosine = negative f. R-SH = negative g. R-COOH = negative 12. What types of conditions or surrounding environment might favor formation of the positively charged acid form of an amine? (amine = R-NH3+) a. Presence of anions b. Hydrophilic c. Water d. Presence of good hydrogen bonding molecules e. Weak interactions and forming favorable weak interactions heavily influences the ability of buffers deprotonate. 13. when pH >> pKa, it is possible to completely deprotonate a weak acid a. false b. Weak acids never completely dissociate, they exist in an equilibrium of acid and base forms. 14. If you place acetic acid into an environment with a high concentration of hexanes, the equilibrium will shift (relative to just water) to favor the _________ form of the molecule. a. Acid (protonated) b. In a hydrophobic environment, acetic acid greatly favors the acid form because the base form is charged and less stable in hexanes. 15. If you place acetic acid into an environment with a high concentration of hexanes, the Ka value (which is th equilibrium constant) will __________, which results in a _________ in the pKa value (-log(Ka)) a. Decrease, increase

b. If the Ka value decrease, then the pKa value will increase indicating that the base form is less likely to be observed in solution c. Pass Week 3  R-NH3+ ⇔ R-NH2 + H+ o How would the protonation of an amine change: o When near a negatively charged group?  near a negatively charged group, the protonated amine would be favored in order to form favorable ionic interactions. This would increase the amount of acid (NH3+) in solution, therefore, decreasing Ka (base/acid) and increasing the amine’s pKa. o When in a hydrophobic environment?  in a hydrophobic environment, the deprotonated amine would be favored to avoid unfavorable interactions b/w the positive charge and hydrophobic environment. This would increase the amount of base (NH2) in solution, therefore, increasing Ka and decreasing pKa. ...