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

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

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

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.

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).