UB BIO 205 - Final Exam Study Guide (34 pages)

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Final Exam Study Guide



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Final Exam Study Guide

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The first part of the final will emphasize information from chapters 29 and 31-33. The rest of the exam will cover past material from chapters 2-28. Only key points from these chapters have been repeated in this study guide.


Pages:
34
Type:
Study Guide
School:
University at Buffalo, The State University of New York
Course:
Bio 205 - Fund of Bio Chemistry
Fund of Bio Chemistry Documents
Unformatted text preview:

BIO 205 Final Exam Study Guide Part I Chemical Foundations of Biochemistry Chapter 2 Stereochemistry of Biomolecules Determining R S configurations depends on the order of the four different groups attached to the central chiral carbon based on molecular weight If the order of the groups proceeds in a clockwise rotation going from 1 to 2 to 3 with the 4 th priority group going in the back then the molecule has an R configuration Remember there must be 4 different groups in order for the molecule to be chiral The only molecule that has a known d or l rotation is glyceraldehyde pictured to the right The D L naming system is based on the fact that d glyceraldehyde is also D glyceraldehyde Carbohydrates are named using the D L system All biological sugars are D which can be determined by looking at the chiral carbon that is furthest from the carbonyl in both aldoses and ketoses If the hydroxyl group of the sugar is pointing to the right then the sugar is D The D L system is applied to the chiral carbon in the backbone of an amino acid All naturally occurring amino acids are L except for glycine because it is not chiral The basic structure of an amino acid is pictured to the right Chapter 3 Weak Acids and Bases The equilibrium constant for an acid dissociation is denoted as Ka A larger Ka means that the acid is more likely to dissociate whereas a smaller Ka means that the acid will not dissociate very much The formula for pKa is pKa log Ka An acid with a large Ka will have a small pKa Things to look for on a titration curve o When the line of the graph becomes horizontal the pH is equal to the pKa At this point there is 50 of the deprotonated form and 50 of the protonated form present in the container This region on the graph is also called the buffering zone because the horizontal slope implies that the pH changes slowly as more base is added to the acid o When the line of the graph becomes vertical an equivalence point has been reached At that point the container holds 100 of the deprotonated form o Another important point that is not found on all titration curves is the isoelectric point This point exists when titrating some amino acids such as glycine At this point found at an equivalence point vertical slope the molecule present has both a positive and a negative charge at the same time creating an overall neutral molecule For glycine the isoelectric point contains the molecule H3N CH2 COO Chapter 4 Thermodynamics Gibbs Free Energy o A favorable G value is negative because a product that is more stable than the reactant will be less likely to convert back On an energy diagram this places the energy of the products below the reactants When subtracting the reactants from the products a negative and favorable Gibbs value results o A favorable reaction in terms of G is referred to as exergonic while an unfavorable reaction is called endergonic o The formula for G is G H T S products o The formula for G actual is G actual G RT ln reactants o When in equilibrium G actual 0 However the formula for G when a reaction is in equilibrium is G RT ln Keq Enthalpy o A reaction that absorbs heat is called an endothermic reaction while a reaction that releases heat is called exothermic The release of heat is favorable over absorption of heat so an exothermic reaction H is considered a favorable reaction o Bond energy is the heat energy required in order to break a molecule apart The more polar a bond the harder it is to break resulting in a high bond energy o A combustion reaction has a favorable enthalpy change due to the change in bond types A hydrocarbon and diatomic oxygen both have nonpolar bonds and carbon dioxide and water have polar bonds Since polar bond are more stable than nonpolar bonds the reaction will release energy creating a favorable negative enthalpy value Chapter 5 Water as a Solvent Polar and NonPolar Groups When determining the solubility of a molecule charged ions are always more soluble than polar molecules which are always more soluble than nonpolar molecules in an aqueous solution As stated above an easy way to charge a molecule is to change the pH of the environment For example the molecule responsible for poison ivy does not wash off with water or acid However it has a hydroxyl that can be deprotonated by adding a basic product like baking soda With a negative charge the molecule can now dissolve in solution Concanavalin A is a protein that is composed of four identical protein chains The protein chains arrange themselves so that the nonpolar portions join together towards the center When looking at the entropy ad enthalpy changes for the formation of the tetramer protein concanavalin it appears that there is a favorable enthalpy change because bonds are being formed and an unfavorable entropy change because the molecules become more ordered However the opposite is actually true because the entire solution not just the protein molecules need to be taken into account The water molecules surrounding each of the protein molecules before they join form a lot of hydrogen bonds along the protein surface When the protein molecules join together these hydrogen bonds need to be broken requiring an input of energy creating an unfavorable enthalpy change Furthermore by breaking a lot of these hydrogen bonds there is an overall increase in disorder of the water molecules creating a favorable entropy change These occurrences are known as the hydrophobic effect Chapter 6 Rate Constants and Equilibrium Constants The velocity of an irreversible unimolecular reaction can be found by calculating the change in the concentration of the product over a period of time with the units M sec or the negative change in concentration of the reactant over time The velocity can also be found by multiplying the concentration of the reactant with the rate constant also with the units M sec The formulas can be expressed as V dB dt dA dt A k It is important to note that velocity values are not constant as a reaction proceeds On the other hand the rate constant does not change hence why it is called a constant Furthermore velocity refers to the whole solution whereas the rate constant refers to one molecule The equilibrium state is independent of starting conditions and it entirely depends on the stability of the product and reactant in a unimolecular reversible reaction No matter what concentration you start with of either product or reactant there is the same amount of A and B at final equilibrium


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