98cMTReviewPPT

# 98cMTReviewPPT - Bio 98 Midterm Review Instructors...

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Instructors: Poulos/Cumsky Peer Tutor: Michael Chau Bio. 98 Midterm Review

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Format of this session This review session: Focuses on: A few important concepts per lecture Common questions associated with topics Does NOT focus on: Memorization-type facts Covering everything (up to Lec. 13). The outline
Format of this session Feel free to ask questions Friendly atmosphere

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I. Simple Thermodynamics Goals: Understand the variables and what they mean. Helps determine the favorability of a system. Solve simple calculations.
I. Simple Thermodynamics Equation: Δ G o rxn = H Δ o rxn - T S Δ o rxn Enthalpy ( H Δ o rxn ) (+) (-) Entropy ( Δ S o rxn ) (-) (+)

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I. Simple Thermodynamics K eq helps us predict the relative concentrations of products:reactants ] [Reactants [Products] = Keq
Question #1 Consider the following reaction: G3P 1,3 BPG K eq = 3.0x10 2 You begin with 2 mols of G3P and 3 moles of 1,3 BPG ; what are their respective concentrations at equilibrium? For chemical equilibriums AND acid-base equilibriums, it’s easier to use two system of equations: Equation 1: will be the ratio of [reactant]:[product] Equation 2: Will be the sum of the reactant and product For Chemical: Use Keq for eq. 1 For Acids/bases: Use Henderson- Hasselbach for eq. 1 Keq=[Product]/[Reactant] pH=pKa + log [A]/[HA]

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Answer #1 Use system of equations: Equation 1: Keq = [1,3BPG]/[G3P]; so 300(G3P)=(BPG) Equation 2: Moles of G3P + moles BPG = 5 moles Substitute equation 1 into equation 2 : G3P + (300G3P) = 5 moles 301 G3P = 5 moles Moles of G3P = 0.0166 Substitute the answer back into a previous equation (equation 2): 0.0166 moles of G3P + moles BPG = 5 moles Moles of BPG = 5 moles – 0.0166 moles of G3P Moles of BPG = 4.983
Question #2 With a crazy good microscope, you observe the dynamics of fatty acid aggregation in a water-filled beaker: What happens to the entropy around the fatty acids? What happens to the entropy of the system?

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Answer #2 With a crazy good microscope, you observe the dynamics of fatty acid aggregation in a water-filled beaker: What happens to the entropy around the fatty acids? Decreases What happens to the entropy of the system? Increases
II. Weak Forces Goals: Predict hydrophobic/philic interactions via thermodynamics. Hydrogen bonds. Energy as a function of radius. Non-polar interactions Charged interactions 6 R 1 = Energy R 1 = Energy

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II. Weak Forces Hydrophobic regions cluster up to prevent water molecules from reaching them This process decreases entropy of the lipids . Enthalpy changes slightly.
II. Weak Forces Hydrogen bonds form between a H (from an electronegative Donor, ie. N, O, or HF) and the lone pairs from an electronegative atom (usually N or O)

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Question #3 The energy felt between non-polar molecules is -5 kcal/ mol at 4.0 Å . What is the energy if the distance is increased to 16.0 Å.
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## This note was uploaded on 03/30/2009 for the course BIOCHEM 98 taught by Professor Mclaughlin,c during the Winter '09 term at UC Irvine.

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98cMTReviewPPT - Bio 98 Midterm Review Instructors...

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