Unformatted text preview: Name: _________________ Student ID #:________________ MCB102 Summer 2011 – Midterm 1 Mr Prosser said: "You were quite entitled to make any suggestions or protests at the appropriate time you know." "Appropriate time?" hooted Arthur. "Appropriate time? The first I knew about it was when a workman arrived at my home yesterday. I asked him if he'd come to clean the windows and he said no he'd come to demolish the house. He didn't tell me straight away of course. Oh no. First he wiped a couple of windows and charged me a fiver. Then he told me." "But Mr Dent, the plans have been available in the local planning office for the last nine month." "Oh yes, well as soon as I heard I went straight round to see them, yesterday afternoon. You hadn't exactly gone out of your way to call attention to them had you? I mean like actually telling anybody or anything." "But the plans were on display ..." "On display? I eventually had to go down to the cellar to find them." "That's the display department." "With a torch." "Ah, well the lights had probably gone." "So had the stairs." "But look, you found the notice didn't you?" "Yes," said Arthur, "yes I did. It was on display in the bottom of a locked filing cabinet stuck in a disused lavatory with a sign on the door saying Beware of the Leopard." Douglas Adams The Hitchhiker’s Guide to the Galaxy (see midterm‐relevant instruction in picture below) Please write your name and student ID number at the top of every page – thanks! Page 2 (25 pts) ___ Page 3 (25 pts) ___ Page 4 (25 pts) ___ Page 5 (25 pts) ___ Page 6 (25 pts) ___ Page 7 (25 pts) ___ Total (150 pts) ___ Name: _________________ Student ID #:____________________ Pts ____ Short answer questions (10 questions, 5 points each) Q1. The second law of thermodynamics favors ATP hydrolysis. Why? Two molecules are made out of one – 3 pts; Resonance on phosphate – 2 pts Q2. “Glucose + phosphate = glucose‐phosphate” is a thermodynamically forbidden (for lack of a more technical word) reaction. “glucose + ATP = glucose‐phosphate + ADP” is thermodynamically favorable. Why? Gibbs energy changes are additive – the delta G of ATP hydrolysis is more negative than the delta G of glucose phosphorylation is positive, and the reactions are coupled Q3. This equation is very familiar to you: ΔG’° = ‐ RT ln K’eq It explains, why enzymes cannot change the position of reaction equlibrium. Please give that explanation in one sentence. The equation shows that the position of equilibrium is a function solely of the delta G of the reaction – which is a function of what the substrate is, and what the products are – and enzymes do not affect that. Q4. Consider the following hydrogen bond. On the drawing below, please indicate with an arrow the hydrogen bond donor and acceptor (“D” and “A”). Indicate partial charges on all the 4 atoms that are involved. Explain very briefly why the charges form the way they do. The H is the donor, the O is the acceptor. Partial positive charge on the H and the C, partial negative charge on the O and the N. The charges form the way they do because N is more electronegative than H, and O is more electronegative than C. Q5. What two hugely important biological structural motifs are stabilized by precisely the hydrogen bond you see above? The alpha helix and the beta sheet Page 2 of 7 – only this side of the page will be graded Name: _________________ Student ID #:____________________ Pts ____ Q6. What is the value of the ion product of water – in a numerical sense? (give the actual number) 1 times 10 to the ‐14 Q7. Water has a high molar heat of vaporization. Why is that useful for living organisms? Sweating is an efficient strategy for heat loss Q8. Draw the side chain of the amino acid R in its charged and uncharged form. ‐CH2‐CH2‐CH2‐NH‐CNH‐NH2 (uncharged) ‐CH2‐CH2‐CH2‐ NH‐ CNH2+ ‐ NH2 (charged) Q9. The pKa of that side group is 12.5. Calculate the fraction that is charged at cellular pH (7.2). Show your work. H‐H: pH = pKa + log (a‐/ha) | 7.2 = 12.5 + log ([uncharged]/[charged]) | Log (u/c) = 7.2‐12.5 = ‐5.3 Answer: log (c/u) = 5.3 | c/u = 199,526/1 (or Approx 99.999%). OR: c/u+c= 199,526/ (199,526+1)). Q10. Here are two binding curves. Which one shows binding associated with a lower Kd? Explain how you made that call in one sentence. Curve X. Lower Kd means tighter binding, and curve X shows theta of 0.5 being reached at a lower amount of ligand Page 3 of 7 – only this side of the page will be graded Name: _________________ Student ID #:____________________ Pts ____ Questions that require a slightly longer answer (25 points each) Please draw the chemical structure of the following short polypeptide – feel free to rotate the page sideways if you wish. HAMLET 1. Please show all atoms and all the bonds they form. 2. For all ionizable groups, please show the form that is the most prevalent at normal cellular pH (7.2). The pKa of the side chain of histidine is 6.00. Page 4 of 7 – only this side of the page will be graded Name: _________________ Student ID #:____________________ Pts ____ Consider the schematic below, and give one‐sentence answers to each of the questions that follow. What do the dotted lines represent? Ion pairs (2 pts) What do the things that these dotted lines represent – actually do from a functional perspective? Stabilize the T state of hemoglobin (3 pts) Some of these lines are very long, and some are short. Some are straight, and some look like an arc. Since they all represent the same thing – why are they have different shapes and lengths in the schematic? It’s a two‐dimensional representation of a 3‐dimensional structure (5 pts) Below is a set of empty Cartesian coordinates. Please label the axes and draw binding curves to oxygen of the following (number the curves accordingly): 1. Myoglobin 2. Hemoglobin in the T state 3. Hemoglobin in the R state Now, answer briefly two questions. Curve 1 is different from curves 2 and 3 in two specific ways. Why is that? Myoglobin’s curve is not sigmoidal b/c it does not exhibit cooperativity; the curve is further to the left b/c it’s a tighter binder than either T or R hemoglobin (5 pts) Curve 2 is different from curve 3. Why? Oxygen affinity for hemoglobin differs in the T and R state (5 pts) Page 5 of 7 – only this side of the page will be graded Name: _________________ Student ID #:____________________ Pts ____ Consider the reaction intermediate below. Note that it has been annotated with three numbered dotted‐line boxes. Please answer questions below, numbered according to the cognate box. 1: this bond is typical of a reaction intermediate that often occurs in enzymatic catalysis. In a general sense, this is a bond links what and what? (“a carbon and an oxygen” is technically correct, but not the answer being looked for here) It’s an inte mediate between the enzyme and the substrate or positioning the nucleophilic oxygen in er
close proximity to the electrophilic carbonyl carbon for attack forming a tetrahedral intermediate (5 pts) 2: this enzyme proteolyses peptide bonds next to aromatic amino acids. Here we can see the reason, why. What is that reason? There is a hydrophobic pocket on the enzyme into which the aromatic ring fits (5 pts) 3: this enzyme loses activity substantially at pH below 6. Focusing on the area around box 3 – why?
at low pH, the imidazole of the histidine will be protonated on precisely the nitrogen that is shown forming the hydrogen bond to water (5 pts), and as a consequence water deprotonation will not happen and the OH‐ will not form (10 points) Page 6 of 7 – only this side of the page will be graded Name: _________________ Student ID #:____________________ Pts ____ Here is another empty set of Cartesian coordinates. Label the axes and draw two “initial enzyme velocity vs substrate concentration” curves for: 1. An enzyme with a Km of 3 mM and Vmax of 100 μmoles/minute 2. The same enzyme, but in the presence of a competitive inhibitor that increases the Km 2‐fold. If you answered the question correctly, you will see that the two curves overlap in a specific section of the graph. Write out the Michaelis‐Menten equation, please, and use it to briefly explain, why that overlap occurs. M‐M V0= Vmax[S]/(Km+[S]) (5 points) The curves overlap at high substrate – when [S]>>Km, the substrate competes out the inhibitor (5 points), and the equation above simplifies to V ≈ Vmax (5 points). Page 7 of 7 – only this side of the page will be graded ...
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This note was uploaded on 07/21/2011 for the course MCB 102 taught by Professor Staff during the Summer '08 term at Berkeley.
- Summer '08