Study Guide Key for EM 109 Final Exam Spring 014 emember you will need to show your work for full credit. n the real exam always work the problems you know best first. If you get hung up on a problem, you should move on and come back to it at the end. If you have time, check over your work. To test your speed, work this study guide as if it was the exam. ow long does it take to finish? 001. Shown below is a balanced equation for the decomposition of S to form and S. S (g) (g) + S (g) [ [S a) Write an equilibrium constant expression for the reaction. K eq = [ S b) Given the equilibrium concentrations: [ S = 0.1007 M, [ = 0.019 M, and [S = 3.30 10-3 M, calculate the numerical value of Keq. -3 [0.019 [3.30 x 10 K eq = [0.1007 = 1.56 x 10-4 c) Assume the equilibrium is perturbed. When equilibrium is reestablished, the following concentrations are observed: [ = 0.0087 M and [S = 0.171 M. alculate [ S under these new conditions. earrange the equation in (a) [ S = [ [S K eq = [0.0087 [0.171-4 1.56 x 10 = 9.089 x 10-3 M [ S = 9.089 3 x 10 = 0.0950 M d) What can you say about the forward and reverse reaction rates when the system is at equilibrium? At equilibrium, the rate of the forward reaction equals the rate of the reverse reaction. k f e) ) ow is K eq defined? = K eq k r 00. a) Qualitatively, for the reaction shown in the problem above, would the G value be positive or negative? Explain your logic. G would be positive, because K eq < 1. b) Assuming I told you the reaction in 004, was relatively fast. Draw a reaction coordinate diagram that describes that system. Then draw another showing how the system would be changed in the presence of a catalyst. Be sure to clearly label all important quantities. (The Keq determined in transition state (top of hill) 004 was less than 1, so the reaction is reactant transition state (top of hill) Eact uncatalyzed favored. A relatively fast reaction means a relatively small hill. (Actually write relatively small hill next to drawing.) G o + S S S uncatalyzed reaction (small hill) eaction Progress hange in G is positive 0.1. What is the definition of a Brønsted-Lowry acid? A Brønsted-Lowry acid is a substance that can donate an +. 0.. Identify the acid and base on the reactant side of the equation shown below. Predict the productsof the reaction and indicate the conjugate of each reactant. 3 + Br 3 + Br + 4 + Br - base acid conjugate conjugate acid base G o Eact catalyzed eaction Progress + S catalyzed reaction (smaller hill) hange in G is positive
0.3.(a) Write a balanced chemical reaction for the dissociation of acetic acid ( 3 ) in water, and then (b) write the K a expression for that rxn. Acetic acid is a relatively weak acid. It is a weaker acid than formic acid. (c) Would formic acid dissociate to a greater or lesser extent in water compared to acetic acid? (d) Which acid would have the larger pk a? - [ a) 3 + 3 - + 3 + 3 [3 b) K a = 3 c) Formic acid would dissociate to a greater extent d) Acetic acid would have the larger pk a 0.4.(a) alculate the [ 3 +, [ -, and p of a 7.00 x 10-3 M solution of 3. (b) Is this solution acidic, basic or neutral? (c) Would the p be the same for a 7.00 x 10-3 M solution of acetic acid? Explain. a) 3 is a strong acid and completely dissociates, so [ 3 + = 7.00 x 10-3 M. p = -log (7.00 x 10-3 ) =.155 14 K w = [ 3 + [ - earrange to get [ - K 1x10 w = = 3 = 1.43 x 10-1 M [3 7.00x10 b) acidic (You can tell because p < 7, and/or because [ 3 + < 1 x 10-7 M.) c) The p would be higher for a solution of acetic acid, because it is a weak acid and doesn t dissociate as much as a strong acid like 3. 0.5. ow does a buffer function to keep the p of a solution relatively constant when strong acid or base is added? Make sure your answer includes equations for chemical reactions. A buffer is composed of reasonably equal amounts of a weak acid (A) and its conjugate base (A - ). The weak acid neutralizes added strong base (for example - ). A + - A - + The conjugate base, A -, neutralizes added strong acid (for example 3 + ). A - + 3 + A + 0.6. What weak acid could be used to make a buffer that was effective around p 4.5? (See table at end.) You must explain your answer thoroughly to receive full credit. An effective buffer must have reasonable quantities of both weak acid (to react with added base) and conjugate base (to react with added acid) components. Ideally [A = [A -. From the enderson-asselbalch expression, when [A = [A -, then p = pk a. Because the desired p of the buffer is 4.5, we need a weak acid with a pk a near 4.5. Looking at the accompanying list of pk a for various acids, we can see that benzoic acid, with a pk a of 4.19 would be suitable. (The second dissociation of oxalic acid, with a pk a of 4.19 would also work 0.7. Why (chemically) are you inhaling? A person inhales to take up oxygen gas,. Why (chemically) are you exhaling? A person exhales to remove carbon dioxide,. Which of these would you expect to have the most direct affect on acidosis. Explain using equations. arbon dioxide,, should have the most direct affect on acidosis, because reacts with water to form carbonic acid (a weak acid): + 3 arbonic acid dissociates in water to form 3 +, which is the direct cause of acidosis: - 3 + 3 + 3 + 0.8. Identify the functions discussed in lecture that must be achieved for living things to stay living. Briefly describe each function and indicate how it contributes to staying alive. In class we identified 6: 1. Make catalysts. Extract energy from environment 3. Extract material from the environment. 4. Store and express genetic information 5. Maintain boundries (membranes) 6. ther specialty functions See notes for details.
0.9. Draw a generalized structure for an amino acid (use may use for the side chain ). amino group represents the side chain carboxyl group + - Predominant structure at neutral p 0.10. Show how the amino acid glycine and the amino acid alanine would join to form a peptide bond. 0.11. Why would it be very appropriate for the amino acid, aspartic acid, to lie on the outer surface of a cytoplasmic protein? A cytoplasmic protein is surrounded by water which is polar. Aspartic acid s side chain is polar and negatively charged at neutral p. It is hydrophilic. 0.1. What are the levels of structure of proteins? What kinds of forces/bonds maintain this structure? Primary (1 o ) structure of proteins is the amino acid sequence. It is maintained by covalent bonds called peptide bonds. Secondary ( o ) structure is a regular repeating structure due to folding of the polypeptide chain. The main types are alpha-helix and beta sheet (either parallel or anti-parallel). Secondary structure is maintained by hydrogen bonds formed between a hydrogen (donor) attached to the nitrogen in the backbone of the chain and the non-bonding pair on the carbonyl oxygen (=) in the backbone. Tertiary (3 o ) structure refers to the location of each atom of the protein relative to every other atom in three dimensional space. This structure is maintained by hydrophobic interactions, hydrogen bonding, covalent (disulfide, -S-S-) bonds, ionic bonds and London forces. Quaternary (4 o ) structure exists only for proteins that have more than one chain or subunit. It describes the way the subunits are arranged and bind to each other. It is maintained by the same forces as tertiary structure. 0.13. Given an organic molecule, be able to circle the chiral carbons. Example with chiral carbons circled: Br l Br 0.14. Determine the amino terminal end of each protein strand shown to the right. Determine which the strand reads amino group, α-carbon, carbonyl carbon. For the strand on the left, that direction is from the bottom to the top, so the amino end is in the downward direction. For the strand on the right, the strand runs from top to bottom, so the terminal is in the upward direction.
1. Describe the two conformations of hemoglobin and describe how they aid in transport of oxygen to your tissues. emoglobin (b) has two conformations: a form with a high affinity for oxygen (the main b form in the lungs) and a form with low affinity for oxygen (the main b form present in the extremities and at lower p). b shifts back and forth between these two forms as it circulates thru the body. In the lungs, after one molecule binds to a b subunit, the affinity of the other three subunits for increases. As each subunit binds, the attraction grows stronger (a cooperative effect), so in the lungs where concentration is high, all of the subunits are bound to, meaning b can pick up the maximum load of. In the extremities, concentration is low. As the first subunit releases, the affinity for by the other subunits decreases. This allows the other subunits to more easily release. So in the tissue where concentration is lowest and where is needed most, the b molecule is able to deliver the maximum amount of oxygen.. Draw a reaction coordinate diagram for the hydrolysis of a dipeptide (like Gly-Ala). Then draw another line on the same diagram to describe the reaction when catalyzed by an efficient catalyst. transition state uncatalyzed reaction Eact uncatalyzed gly-ala + water catalyzed reaction G Eact catalyzed ΔG is negative gly + ala eaction Progress The G values on my diagram should be G o. The graph was getting crowded so I didn t include the transition state for the catalyzed reaction, although it should be indicated. 3. What happens to the rate of an enzyme catalyzed reaction at low substrate concentrations (compared to the enzyme concentration) when you double the substrate concentration? Does the same apply when you are looking at high levels of substrate concentration? Explain. At low substrate concentrations (compared to the enzyme concentration), when you double the substrate concentration, you double the rate of the reaction, because at low substrate concentrations not all of the enzyme molecules are working. At high substrate concentrations, the rate generally does not increase much, or not at all, because at high concentrations, all of the enzyme molecule s active sites are saturated (constantly filled) 4 Describe the molecular basis of sickle cell anemia, one treatment, and indicate why the trait is found at relatively high levels in some populations. ormal human b has glutamic acid (Glu) at position 6 of its beta chain. A mutation in the DA results in valine (Val) replacing Glu at the sixth position in sickle cell hemoglobin (bs). The valine side chain is much less polar than that of Glu. When bs is in its low affinity form (deoxygenated), bs starts to polymerize into long chains that protect the valine side chain from water and precipitate (come out of solution to form a solid). bs (aq) bs(s) The long rods that form lead to a change in shape and loss of flexibility of the red blood cells. This in turn leads to restricted blood flow in the capillaries and rupturing of red blood cell membranes. Ultimately this causes the broader problems associated with sickle-cell crises (pain, liver problems, etc.).
ne of the treatments for sickle-cell disease is hydroxyurea. This drug works by inducing the synthesis of the fetally produced gamma ( ) chain, that can partially replace the beta ( ) chain. This reduces the incidence of a sickle-cell crisis. Sickle-cell heterozygotes (individuals who have one wild type (wt) and one sickle cell b gene) are resistant to malaria. This resistance may account for the relatively high frequency of the sickle cell trait in individuals from areas where malaria is prevalent. 7. What is the entral Dogma of Molecular Biology? It shows the direction and flow of information. transcription translation DA A Protein replication reverse transcription 8. Why is it important that DA be able to replicate itself millions of times without error? What feature of DA structure is particularly important with regard to avoiding errors? DA needs to be able to replicate itself without errors, because an error if not corrected results in a mutation. Mutations are rarely positive for the cell or organism, sometimes neutral and often have negative consequences. Mutations can result in the formation of ineffective proteins, can cause disease states, and cancer. The structural feature of DA that helps avoid mutations is that DA is double stranded with complementary bases. This means that (1) each strand contains the information in each sequence, and () copying errors can be detected early on as a mismatch that can be recognized and repaired before replication. 11. Describe where DA is located and where proteins are synthesized. What molecule and processes exist that allow the transfer of this information from one location to the other? DA is located in the nucleus. Proteins are synthesized on ribosomes in the cytoplasm. The information encoded in the DA is carried from the nucleus to the cytoplasm via an A molecule (called ma). This A is transcribed from DA and processed in the nucleus and then passes thru the nuclear membrane into the cytoplasm and moves to the ribosomal machinery where it is translated to protein. leading strand 1. Draw a DA molecule being replicated. Include the direction the DA is unwinding, the polarity of the parent and daughter strands and the direction the daughter strands are being synthesized. daughter strands parental strands lagging strand direction of unwinding and of replication 13. You are attending a family reunion, and the subject of mitochondrial DA comes up (as it often will at such gatherings). ow likely is it that you have the same mtda sequence as your cousin Luke? Luke is your mother s sister s son. Briefly explain your logic. 100% likely. Mitochondrial DA is inherited maternally in humans. Therefore both your mother and her sister inherited their mtda from their mother, and (barring mutation) both you and Luke will inherit that same mtda.
14. Draw a cartoon model of DA indicating how the three major pieces are connected. Indicate the location of hydrogen bonds. See drawing a right. TE: ydrogen bonds need to be dotted or dashed. I couldn t figure out how to do that on my picture A hydrogen bonding T P4 P 4 15. Given the following DA sequence, what would the complementary A sequence be? Indicate the polarity of the A. A T G G A G T G G T A A 3 3 U A U G A G A U U 5 P 4 T G A P4 P4 P 4 16. o question. 17. If a researcher determined that DA from a new organism was 8.3% G, what would be the percentages of A, T and in the DA of this creature? G and are complementary, so the % of must equal the % of G. G So, G is 8.3% and is 8.3%. Total is 56.6%, so that leaves 100-56.6 or 43.4% for A and T. The bases A and T are complementary, so the % of A must equal the % of T. So each percentage is half of 43.4%, or 1.7% So, A is 1.7% and T is 1.7%. 18. Is the nucleic acid strand shown on the right A or DA? ow do you know? Draw an arrow that represents the 5 to 3 direction. The structure shown on the study guide is a strand of DA. You can tell because the carbon of the has an bonded to it, rather than an. ote: I have labeled some of the carbons in the s. 1' ' P 4' - P -
Supplemental information Periodic Table, electronegativity chart, the structures of the side chains of the 0 common amino acids, and a copy of the genetic code. K = + 73.15 Kw = 1.0 x 10-14 p = pk a + log ( ) Go = -T ln Keq = 0.080578 (L atm/(k mol) = 8.3145 J/(K mol) Acid Dissociation onstants as pkas From hemistry by McMurray & Fay, 4th ed. Acid Formula pk a1 pk a pk a3 acetic 3 4.74 acetylsalicyclic 8 7 3.5 ascorbic 6 8 6 4.10 benzoic 6 5 4.19 boric 3 B 3 9.4 carbonic 3 6.37 10.5 chloroacetic l.85 citric 6 8 7 3.15 formic 3.74 hydrocyanic 9.31 hydrogen peroxide 11.6 hydrosulfuric S 7.00 ~19 hypobromous Br 8.70 hypochlorous l 7.46 hypoiodous I 10.64 iodic I 3 0.77 oxalix 4 1.3 4.19 phenol 6 5 9.89 phosphoric 3 P 4.1 6.1 1.3 phosphorous 3 P 3.00 6.58 saccharin 7 5 3 S 11.68