Download 7.014 Quiz I Handout

7.014 Quiz I Handout
Quiz I announcements:
• Quiz I:
Friday, February 27 12:05 - 12:55
Walker Gym, 3rd floor (room 50-340)
**This will be a closed book exam**
• Quiz Review Session:
Wednesday, February 25
7:00 - 9:00 pm
room 54-100
• Open Tutoring Session: Thursday, February 26
4:00 - 6:00 pm
room 66-156
Question 1
Shown below are the structures of three amino acids; they are shown in alphabetical order.
NH3
NH3
H C CH2
O
C
NH 3
H C CH2
C
O
O
OH
H C CH
CH3
C
O
O
O
Threonine
Phenylalanine
(Thr)
(Phe)
a) Circle the side chain group on each amino acid.
OH
Tyrosine
(Tyr)
b) Using the blanks below, rank these three in order of the hydrophobicity of their side
chains .
i)
Most Hydrophobic
ii)
Intermediate
iii)
Least Hydrophobic
c) Explain why (i) is more hydrophobic than (ii).
d) Explain why (ii) is more hydrophobic than (iii).
Question 2
From the diagram below,
O
N
O
O+
NH 3
Lys
P
O
O-
H 2N
NH
O
P
O
CH 2
N
O
N
O-
C
NH
+
NH 2
2
HO
OH
OH
O
O
-
C
Arg
NH
Tyr
C
O
O
Asp
Glu
a) Circle the strongest interaction that exists between:
i) the side chain of Lys and the phosphate group of GDP
van der Waals
covalent
hydrogen bond
ionic
2
Question 2, continued
ii) the side chain of Glu and the ribose group of GDP
van der Waals
covalent
hydrogen bond
ionic
iii) the side chain of Tyr and the guanine base of GDP
van der Waals
covalent
hydrogen bond
ionic
b) You make mutations in the GDP-binding pocket of the G protein and examine their
effects on the binding of GDP. Consider the size and the nature (e.g. charge, polarity,
hydrophilicity, hydrophobicity) of the amino acid side chains and give the most likely
reason why each mutation has the stated effect. Consider each mutation independently.
i) Arg is mutated to a Lys, resulting in a G protein that still binds GDP.
ii) Asp is mutated to a Tyr, resulting in a G protein that cannot bind GDP.
Question 3
Short answer. Some possibly useful equations can be found at the end of this exam.
a) Consider the following reaction:
(1)
A+B
C
∆G o = -1 kcal/mol
i) What is the ∆G for this reaction when [A] = [B] = [C] = 10 mM and T = 25º C? Show
your work.
ii) Suppose you add an enzyme that catalyzes reaction (1). Make each of the
following statements true by circling the appropriate underlined phrase. (4 circles
total)
becomes more
runs faster.
The forward reaction stays just as thermodynamically spontaneous and proceeds at the same rate.
becomes less
runs slower.
becomes more
runs faster.
The reverse reaction stays just as thermodynamically spontaneous and proceeds at the same rate.
becomes less
runs slower.
3
Question 3, continued
b) Consider the reaction:
(2)
X
Y
K2 =
[Y]
at equilibrium
[X]
Given the following:
(3)
ATP
(4) X + ATP
ADP + P i
o
∆G o = -7.5 kcal/mol at 25 C
Y + ADP + Pi
i) What is the Keq for the coupled reaction (4) in terms of the Keq of reaction (2), at
25° C? Show your work.
ii) Provide a plausible mechanism by which the cell could couple the energy from
reaction (3) to drive reaction (2). You need not go into chemical detail; just describe
the process clearly.
Question 4
Part 1. True or False. If false, explain why.
a) Most of the ATP produced from the aerobic metabolism of glucose results from
glycolysis.
b) In yeast cells under anaerobic conditions (no oxygen), the rate of glucose consumption is
less than that under aerobic conditions.
c) Glycolysis (glucose ----->2 pyruvate) occurs under aerobic and anaerobic conditions.
d) Under anaerobic conditions (no oxygen), the citric acid cycle degrades acetyl coenzyme A
to CO2 and H 2O, producing NADH and FADH2.
e) Electron transport produces ATP directly from the transfer of electrons from NADH
to O2.
f) In eukaryotic cells, glycolysis occurs in the cytosol, whereas the reaction of the citric acid
cycle and oxidative phosphorylation take place only in the mitochondria.
g) During fermentation, pyruvate is converted to ethanol and CO2 to use up excess oxygen.
h) During glycolysis and respiration, glucose reacts directly with O2 to form CO2 and H 2O.
4
Question 4, continued
Part 2. Short Answer
a) Pyruvate is a very versatile molecule. Give three pathways for pyruvate. For each
pathway list the end product and whether NAD+ or NADH + H+ is formed.
PATHWAY
END PRODUCTS
NAD + or
NADH + H+
ADDITIONAL ATPs
GENERATED
1.
2.
3.
b) Explain briefly how is ATP formed during photosynthesis.
c) Explain briefly, what carbon fixation is and what process in plants leads to carbon
fixation.
Part 2. Short Answer, continued
d) When exposed to light, plant cells show net absorption of CO2 and net production of O2.
In the dark, they show net production of CO2 and net absorption of O2.
i) What biochemical process is responsible for the plant's absorption of O2 and
production of CO2 in the dark? Explain briefly.
ii) Does this process continue when the plant is exposed to light? If so, why aren't
net production of CO2 and absorption of O2 seen under these conditions? Explain
briefly.
5
Shown below is the portion of the diagram from page 8 which shows the electron
transport pathway. In this pathway, electrons are transferred from NADH to O2; the energy
of this reaction is coupled to pumping of H+.
drug X
+
2e
NADH + H
H2 O
(FADH2 )
2 eelectron transport
2 eNAD +
(FAD)
proton
pumping
H+ (in space A)
H+ (in
space B)
O2
2 e-
drug Y
Two drugs, drug X and drug Y, can pick up electrons from particular intermediates in this
pathway. This is shown above.
e) You treat cells carrying out respiration with a saturating dose of drug X, so that all the
electrons which would normally continue along the pathway are captured by drug X.
Under these conditions:
• Will the cells continue to consume O2?
• Will the cells continue to produce CO2?
• Will the rate of ATP synthesis increase or decrease or stay the same?
Explain your reasoning.
f) You treat cells carrying out respiration with a saturating dose of drug Y, so that all the
electrons which would normally continue along the pathway are captured by drug Y.
Under these conditions:
• Will the cells continue to consume O2?
• Will the cells continue to produce CO2?
• Will the rate of ATP synthesis increase or decrease or stay the same?
Explain your reasoning.
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STRUCTURES OF AMINO ACIDS
O
O
C
H
C
CH3
H
NH3
+
C
H
C
CH2CH2CH2 N
CH2 SH
H
NH3
+
O
H
N
C
O
+
C
H
H
CH2
NH3
+
N
C
O
S
CH3
H
NH3
+
METHIONINE
(met)
C
C
O
CH2CH3
H
C
CH2
H
CH3
NH3 OH
+
THREONINE
(thr)
C
CH2
H
H
H
C
NH3
+
H
H
H
O
O
H
C
H
H
C
C
H
O
O
C
CH2
OH
NH3
+
H
TYROSINE
(tyr)
H
OH
SERINE
(ser)
C
H
CH2
NH3
+
PROLINE
(pro)
H
N
NH3+
O
O
H C CH2
CH2
H
N
CH2
H +
H
CH2CH2CH2CH2
LYSINE
(lys)
O
O
C
NH3
+
CH3
C
H
TRYPTOPHAN
(trp)
H
CH3
LEUCINE
(leu)
H
O
C
C
NH3
+
CH2
C
GLYCINE
(gly)
H
PHENYLALANINE
(phe)
O
O
O
C H
H
C
NH3
+
C H
NH3
+
O
C
CH2CH2
H
NH2
C
O
O
C
C
C
C
O
O
ISOLEUCINE
(ile)
HISTIDINE
(his)
H
C
CH2CH2
GLUTAMINE
(glN)
O
C H
O
C
O
C
O
ASPARTIC ACID
(asp)
O
NH3
+
NH3 CH3
+
H
H
O
H
O
CH2 C
NH3
+
O
C
GLUTAMIC ACID
(glu)
C
H
C
NH3
+
CYSTEINE
(cys)
O
CH2CH2
O
C
NH2
ASPARAGINE
(asN)
O
C
H
CH2 C
O
C
C
C
NH2
+
C
O
NH3
+
O
O
C
H
C
H
NH2
O
O
C
ARGININE
(arg)
O
O
O
NH3
+
ALANINE
(ala)
O
O
O
H
CH3
C
C
NH3
+
H
CH3
VALINE
(val)
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Thermodynamics:
For the reaction:
K eq
A + B ←
→ C + D with ∆Go as its standard free energy
at equilibrium:
 [C][D]
∆Go = _ RT ln
 [A][B]
if T = 25 º C then
where:
if T = 37 º C then
under any conditions:


kcal
mol
kcal
RT = 0.61
mol
RT = 0.59
∆G = ∆Go
 [C][D]
+ RT ln
 [A][B]


Solutions to Practice Quiz I
Question 1
a)
NH 3
H C CH2
C
O
O
(Phe)
b)
NH 3
OH
CH
CH3
C
O
O
(Thr)
H C
i) most hydrophobic:
ii) intermediate
iii) least hydrophobic
NH 3
H C CH2
O
OH
C
O
(Tyr)
phenylalanine
tyrosine
threonine
c) Phe is more hydrophobic than tyr because tyr has a hydrophilic -OH group that phe
lacks (-OH can form H-bonds).
d) Tyr is more hydrophobic than thr because, although both have -OH’s, tyr has more
non-polar CH’s than thr.
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Question 2
a)
i)
ii)
iii)
van der Waals
van der Waals
van der Waals
covalent
covalent
covalent
hydrogen bond
hydrogen bond
hydrogen bond
ionic
ionic
ionic
b)
i) Arg and Lys are both positively charged, thus the ionic interaction with the
phosphate group is preserved. The side chains of both amino acids are also of
similar size.
ii) Tyr is much larger than Asp. Although Tyr can form a hydrogen bond,
GDP will no longer fit into the binding pocket. The Tyr side chain is also
much more hydrophobic than the Asp side chain.
Question 3
a)
i) (remember that 10mM = 0.01M)
 [C] 

0.01 
∆G = ∆G 0 +RT ln
becomes : ∆G = −1 +( 0.59) ln
 [A][ B]
 (0.01)(0.01)
kcal
mol
ii) Both the forward & reverse reactions stay just as thermodynamically
spontaneous and both run faster.
= −1 + (0.59) ln(100) = +1.71
b)
i)
K4 =
[Y][ADP][Pi ]
[ADP][Pi ]
= K2
= K 2e
[X][ATP]
[ATP]
 −∆G0 
 RT 
= K 2e
 7.5 
 0.59 
= 3.3x10 5 (K 2 )
ii) Several possibilities, here are 2:
(1) common intermediate: ATP + Q ---> ADP + Pi + Q* then X + Q* --> Y + Q
(2) both reactions catalyzed by the same enzyme so that the energy is coupled
Question 4
Part 1
a) False. Most of the ATP produced from the aerobic metabolism of glucose results from
respiration.
b) False. In yeast cells under anaerobic conditions (no oxygen), the rate of glucose
consumption is greater than that under aerobic conditions.
c) True.
d) False. The citric acid cycle does not occur under anaerobic conditions. The statement is
true as written of aerobic conditions.
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Question 4
Part 1, continued
e) False. Electron transport from NADH to O 2 proceeds through many intermediates
leading to the production of a proton gradient which drives ATP synthesis.
f) True.
g) False. Pyruvate is converted to ethanol and CO 2 to regenerate NAD + .
h) False. Glucose and its derivatives are oxidized by NAD+ and FAD. The resulting
NADH and FADH 2 carry electron to O 2 via the electron transport chain, producing H 2O.
Part 2
a) Pyruvate is a very versatile molecule. Give three pathways for pyruvate. For each
pathway list the end product and whether NAD+ or NADH + H+ is formed.
1. fermentation
Lactic acid
NAD +
none
2. fermentation
ethanol
NAD +
none
CO2
NAD +
34
3. respiration
b) ATP is generated by photophosphorylation using the energy from the electrochemical
proton gradient formed by the passing of electrons through the photosystems.
c) Carbon fixation is changing CO 2 to an organic form by covalently binding it to an organic
molecule, usually a sugar. One example is seen in mesophyll cells where a CO 2 is added to
the sugar ribulose-1,5-bisphosphate (RuBP) by the enzyme rubisco to form 3phosphoglycerate. This is the first part of the Calvin-Benson Cycle, also referred to as the
dark reactions, because they are light-independent.
d) When exposed to light, plant cells show net absorption of CO2 and net production of O2.
In the dark, they show net production of CO2 and net absorption of O2
i) What biochemical process is responsible for the plant's absorption of O2 and
production of CO2 in the dark? Explain briefly.
Respiration. The plant cells are using O 2 to oxidize stored carbohydrates to CO 2, in
order to produce energy.
ii) Does this process continue when the plant is exposed to light? If so, why is no net
production of CO2 and absorption of O2 seen under these conditions? Explain briefly.
Yes, the plant always needs energy from respiration to perform cellular reactions. In
the light, the rate of photosynthesis is greater than the rate of respiration, so result is
net O 2 production and net CO 2 absorption. In the dark, there is no photosynthesis, so
the basal respiration level predominates (O 2 absorption and CO 2 production).
e) won’t consume O 2; will produce CO 2; ATP will decrease because no H + pumped.
f) won’t consume O 2; will produce CO 2; ATP will stay the same because H + still pumped.
10