EXAM 1 ANSWER KEY Fall 2011 ANSWER KEY EXAM 1, Fall 2011 The exam was worth 100 points. All questions (except 51 for version B) were worth 2 points. The columns Va & Vb corresponds to answers for Version A & B. Mean = 76.4, Stdev = 12.3, Median score = 78, Range 34-98. A+ = 98-100, A =92-97, A- = 90- 92, B+ = 88-89, B = 84 - 87, B- = 80-83, C+ = 76-79, C = 72 - 75, C- = 62 - 71, D+ = 58-61, D= 54- 57, D- = 51 - 53. F = 50 or less. Va Vb a b a b a b a b a b a b a b a b a b 1 D C 6 C B 11 A D 16 B A 21 A D 26 B B 31 E E 36 C D 41 A E 46 C C 2 B C 7 A A 12 E E 17 C E 22 D B 27 C C 32 D B 37 E E 42 B E 47 C C 3 A D 8 C A 13 E E 18 B C 23 E B 28 A A 33 D C 38 A C 43 A E 48 D D 4 E A 9 D B 14 A E 19 A A 24 D A 29 D D 34 E D 39 C C 44 E E 49 C B 5 B B 10 C C 15 D C 20 D D 25 E D 30 B D 35 B B 40 D A 45 A B 50 C C Most commonly missed questions. (3, 13, 14, 20, 21, 24, 25, 27, 28, 29, 33, 37, 38, 40, 41, 42, 43, 45, 48, 50) 1) Enzymes lower Ea, they do not change free energy, equilibrium, etc. 2) Phospholipids are free to move laterally with their hydrophobic tails in the interior of the membrane (the greasy portion). 3) Since interior is hydrophobic the R groups are expected to be hydrophobic. 4) NADH + H+ donates electrons to an organic molecule to regenerate the NAD+ to allow glycolysis to continue. 5) Anabolic is synthetic/building up reactions. 6) Facilitated diffusion is powered by a electrochemical gradient and requires the use of proteins to transport the molecule. 7) Chlorophyll is certainly a photosynthetic pigment that is green and is found in both two forms (Chl a and Chl b forms). 8) Light reactions generate NADPH + H+ and ATP to power the Calvin cycle. 9) Allosteric inhibition means the inhibitor is binding at a site other than the active site (i.e. at the allosteric site). 10) ATP is produced in both mitochondria and chloroplasts via chemiosmosis. 11) The double bond in unsaturated fatty acids introduce a kink which prevents tight packing at lower temperatures. 12) The primary messenger is outside the cell and this information is relayed to the interior via the G protein coupled receptor (GPCR) and a G protein alters the activity of an effector – ultimately affecting the levels of second messengers which convey the message throughout the cytoplasm. E is the best answer. 13) During glycolysis glucose is phosphorylated and eventually 2 pyruvates are produced along with 2 (NADH + H+) and 2 Net ATP molecules. 14) Estradiol is the only steroid hormone listed. 15) See the comment for 1. Both the forward and reverse reactions go faster. Thus at equilibrium you don’t change the ratios of products to reactants. Lets say it is 20,000 –with an enzyme at equilibrium you might convert 1000 products to reactants and convert 1000 reactants to products but without enzyme it would be 2 products to reactants and 2 reactants to products. Thus you aren’t changing equibrium concentrations. 16) See 12 as well. B is False. 17) The Keq states products exceed reactants so the ΔG is negative and the reaction is spontaneous. Since ΔG is = ΔH –TΔS, a change in temperature would change Δ G. 18) There are many similarities of respiration in bacteria and eukaryotes; a notable exception being the presence of Mitochondria in eukaryotes. 19) ATP synthase activity could be present in Mitochondria or chloroplasts but liver tissue should only have mitochondria. 20) Since this involves the interaction of 2 sub-units it must be quaternary structure. 21) Intermediate filaments are strands of proteins, not globular protein units assembled together. 22) Animal cells would use gap junctions for transferring molecules, plants would use plasmodesmata. 23) O2 comes from water (H2O). 24) Since it an intermediate late in the pathway inhibiting an early step this would be feedback inhibtion. 25) Lysine has the + charge and an R group with a + charge could potentially substitute with minimal consequences but a – charge would have maximal consequences. 26) A kinase should add a phosphate group and it requires an OH group such as that found only in R group II. 27) C and E are similar but not identical. Note that at equilibria there should be both forward and reverse reactions. 28) This experiment wasn’t designed very well. Note that question 29 shows a much better experiment where at some concentration pretty much all of the receptors have bound to hormone. For 28 changing the concentration 1000 fold didn’t decrease binding by 1,000 fold so most of the hormone is binding non-specifically and thus as you add more hormone you keep increase binding and that binding will include both labeled and unlabeled hormone. Choice B (thyroid hormone) is out because there is no reason to suspect thyroid hormone. Choice C is out because the unlabeled hormone should compete with the labeled hormone. Choice E is out because there is no reason to assume this involves IP3. The answer must be either B or D. D makes no real sense because if there is unlabeled hormone how can there be any signal (we are measuring the label). 29) Graph D illustrates the saturation point of the receptors. 30) Gradients represent potential energy which can be coupled to do work. 31) All three listed processes require the presence of functional tubulin. EXAM 1 ANSWER KEY Fall 2011 32) Plant cells transfer via plasmodesmata. 33) Water is split but also reformed. Thus H is transferred to O2 to form water and also to reduce CO2. 34) All of these properties are dependent upon hydrogen bonding. 35) Since the bacteria cluster near yellow and green these wavelengths must correspond to the presence of O2 which is a is a product of photosynthesis. Thus yellow and green promote photosynthesis. 36) Electrons move from water to PSII to the ETS to PSI to NADP+ to sugars eventually. 37) Lysosomes contain degradative enzymes. 38) 3 X C16H22O2 = C48H66O6. Glycerol = C3H8O3. Thus together it should be C51H74O9 but we lose 3 H2O molecules during the condensation reaction. End value = C51H68O6. 39) Protein denaturation refers to the loss of conformation/structure. 40) The RNA is antiparallel and follows complementary base pairing rules. Thus it is 3” CUAAUGU – 5’. For the exam the 5’ end was written on the left hand side. 41) DNA synthesis occurs within the nucleus. You should be familiar with protein production and movement through the endomembrane system. 42) Since oxygen is the terminal electron acceptor it must be very electronegative. 43) Receptors (Receptors plus signal) are turned over by endocytosis. 44) Both Ct and Mt are believed to have evolved due to endosymbiosis and both produce ATP via chemiosmosis and contain their own DNA and ribosomes. 45) As O2 decreases photorespiration would be expected to decrease. 46) C-4 plants use PEP carboxylase to fix the CO2. They add CO2 to PEP to form a 4C compound. 47) In Mt FADH2 enters at complex II, not complex I. All of the complexes are multi-subunit complexes that are integral in the membrane. NADH2 enters at complex I so more H+ can be pumped and thus more ATP can be made. Throughout the chain there are coupled reduction/oxidation reactions. 48) The molecule is a ribonucleotide (rATP) which should be found in RNA (but NOT DNA) and is the primary currency of the cell. There is a N glycosidic bond linking the base to the ribose sugar. 49) In order for the yeast cells to allow glycolysis to continue they must be able to regenerate NAD+. They do this through a couple of reactions and eventually produce ethanol. 50) NADPH is used to reduce 3PGA (acid) to eventually form sugars (aledhyde, note the reduction).