Lehninger Principles of Biochemistry 5/e

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Bioenergetics and thermodynamics
Bio-energetics is the application of
thermodynamics to biochemical systems involving a
quantitative treatment of energy exchange in
living cells, and a description of the underlying
chemical processes
A. A Review of Thermodynamics (free energy,
enthalpy, entropy) B. Phosphoryl Group Transfers
and Energy Exchanges C. Oxidation-Reduction
Reactions in Living Cells.
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• A Review of Thermodynamics
• First law of thermodynamics (Conservation of E)
• Energy is neither created nor destroyed, it
  changes from one form (chemical, ) to another
  (physical, thermal, ).
• ?E q w
• Second law of thermodynamics (Entropy, S)
• In all natural processes the entropy of the
  universe (but not necessarily the system) always
  increases.
• The total S of the universe is continuously
  increasing.
• ?Suniverse ?Ssystem ?Ssurr gt 0
• www.entropylaw.com

• Scientific definitions of Hypothesis, Theory,
  Law http//chemistry.about.com/od/chemistry101/a/
  lawtheory.htm
• Scientific laws explain things, but they do not
  describe them.

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Some energy interconversions in living organisms
Energy extraction from surroundings
Convert E to work
Living organisms preserve internal order by
taking free energy (?G, sunlight, nutrients) from
their surroundings and giving back an equal
amount of energy as heat (?H) and entropy (?S)
Release E as heat
Release less organized end-products
Ssurroundings
Produce more organized polymers
Ssystem
(Lehninger, p. 19?27)
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Second law of thermodynamics (Entropy,
S) Oxidation of glucose C6H12O6 6 O2 ? 6CO2
6 H2O energy In this process, 7 molecules are
converted to (and more randomly dispersed to) 12
molecules. ? increase in entropy (molecular
disorder).
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Gibbs Free Energy (G) The component of the total
energy of a system that can do work at constant T
and P
G H ? TS
Gibbs Free Energy change (?G)
?Gsys ?Hsys ? T?Ssys ?T?Suniverse For a
reaction occurring at constant T (S?T 0) and P,
?H is the change in enthalpy, T is the
temperature in Kelvin, and ?S is the change in
entropy.
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?G ?H ? T?S
Enthalphy (heat inside, H) the heat content of a
chemical system. ?H ?E P?V qp The
enthalpy change (?H) is the amount of heat
released or absorbed when a chemical reaction
occurs at constant pressure. ?H H(products) -
H(reactants) ?H is generally associated with
bonding, and reflects the kinds and number of
chemical bonds and noncovalent interactions
broken and formed. ?H lt 0 ? exothermic reaction,
energy released (wood burning) ?H gt 0 ?
endothermic reaction, energy absorbed (ice
melting)
Entropy (S) a measure of randomness, disorder,
freedom of motion. ?S the change in the
systems randomness.

• If products are less complex (more disordered)
  than the reactants, the reaction gains entropy
  and by convention ?S is positive.

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Spontaneity of Reaction
Systems change toward minimum free energy

• Many spontaneous reactions proceed with a
  decrease of E
• ? (?H lt 0) (Energy released)
• But some endothermic reactions (?H gt 0) occur
• spontaneously, too (ice melting, evaporation)
• ? Need new term for spontaneity ? Free Energy (G)
  

?G ( T?Suniv) lt 0 ? spontaneous reaction ?G gt 0
? nonspontaneous reaction
At constant P and T, ?G represents the portion of
the total energy change that is available (free)
to do useful work.
Formation of stronger (more stable) bonds (ve
?H) and greater freedom of motion (ve ?S)
contribute to a favorable driving force for
reaction (ve ?G).
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Equilibrium Constant (Keq)
another indicator of reaction
spontaneity
Chemical reaction proceed spontaneously until it
reaches equilibrium, where ?G 0
(and Q/K 1).
Equilibrium A state of dynamic balance in which
rates of forward and reverse reactions are equal.
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A B ? C D   Keq
?G
(1)
where ?G 0 (standard
free energy change), the free energy for the
reaction under standard conditions (1 atm, all
reactants at 1 M), ?G 0 ?H 0 - T?S 0  
? ?G 0 is a given value for a particular
reaction, determined by the free
energy of formation of the reactants and the
products. ?G 0reaction ??G 0f
products ? ??G 0f reactants
(2)   At equilibrium, ?G
? ?G 0
(3)
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aA bB ? cC dD Keq CcDd/AaBb
?G is the magnitude of the free energy change
for a rxn (tendency of a reaction to move
towards equilibrium) ?G? is the standard free
energy change under standard conditions (298 K,
reactants 1 M or 1 atm pH 0) ?G?? is the
standard transformed free energy change Defined
in biochemical standard state. (H2O 55.5 M
H 10-7 M pH 7 Mg2 1 mM others
in1 M or 1atm) K?eq is the standard transformed
equilibrium constant (Biochemical standard
state used from here forward...)
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So, what do all these mean? ? Let the reaction
reach equilibrium. ? ? Keq for a reaction is
experimentally determined at equilibrium.
? Then, ?G0 for that reaction can be
calculated from Keq. ?G 0 ?
RT ln Keq Or, ?G0 for the reaction (and
subsequently Keq) can also be determined
from ?G0f of reactants and products.
?G 0reaction ??G 0f products ? ??G 0f
reactants ? Then, ?G for a particular reaction
at given concentrations of reaction
species can be determined from ?G0 and Q.
?G ?G 0 RT ln CiDi/AiBi ?
Now, one can tell which direction the reaction
will proceed spontaneously at given Ai,
Bi, Ci, and Di.
BUT, neither ?G or ?G0 wont tell us how FAST
the reaction proceeds. ? Reaction rate governed
by kinetics (?G) ? Catalysts ? Enzymes
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ATP H2O ? ADP HPO4-2 ?G'0 ?30.5 kJ/mol
R 8.314472 J K-1 mol-1 T 273 37 310
At Equilibrium, ?G0 K'eq
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Recall ?G? - RT lnKeq
Fig. 1-27
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Hydrolysis reactions tend to be strongly
favorable (Spontaneous)
Isomerization reactions have smaller free energy
changes. Isomerization between enantiomers ?G?
0.
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Consider Fructose-6-phosphate ?
Glucose-6-phosphate Starting with 9 mM F-6-P or
G-6-P we measure the final equilibrium mixture
to be 3 mM F-6-P and 6 mM G-6-P at 25? and pH
7.0. Calculate the equilibrium constant and
standard free energy change of the
reaction Remember both Keq and ?G? are
constants
Keq
?G? ? The reaction is slightly
spontaneous (F-6-P ? G-6-P), but is close to
equilibrium since this is a relatively small
negative value
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How to drive an unfavorable reaction (?G0 gt 0)?
? Reaction coupling via common intermediate

• An unfavorable reaction (ve ?G) can be driven
  spontaneously
• when coupled to a favorable reaction (ve ?G),
  
• if the sum of ?G values is negative.
• (?G Values are Additive)
• A ? B ? C (?G1? ?G2?)
• A ? C ?GT?
• Smart strategy of metabolism
• couple one favorable and one unfavorable rxn to
  drive the unfavorable reaction with the energy
  stored in other bonds
• (e.g. ATP)

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• Amino acids ? proteins (polymer) ve ?G
  (endergonic) (1)
• ATP H20 ? AMP 2Pi ?ve ?G (exergonic)
  (2)
• (2) Amino acids nATP ? proteins nAMP
• 
  Overall ?G is negative.

• Glucose Pi ? glucose 6-phosphate H20
  
• 
  ?G'0 13.8 kJ/mol (1)
• (The reverse reaction, hydrolysis of G-6-P will
  be spontaneous.)
• ATP H20 ? ADP Pi ?G'0 ?30.5
  kJ/mol (2)
• (The ?G'0 values of sequential reaction are
  additive.)
• (2)
• 
  

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Consider What happens when we couple the
formation of fructose-1,6-bisphosphate
(?G?hydrolysis ? 16.3 kJ/mol) with ATP
hydrolysis (?G?hydrolysis ? 30.5 kJ/mol)? 1.
Which reactions are involved? 2. Which direction
does each reaction proceed? 3. Arrange reactions
such that transfer groups cancel. 4.
Arithmetic. ?G? (kJ/mol)
F-6-P Pi ? F-1,6-bisP H2O 16.3
(1)
ATP H2O ? ADP Pi ? 30.5 (2)
_________________________________________

(3)
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Exergonic rxn system releases free energy ?G is
Endergonic rxn system gains free energy ?G
is
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Q5. Experimental Determination of ?G? for ATP
Hydrolysis A direct measurement of the standard
free-energy change associated with the hydrolysis
of ATP is technically demanding because the
minute amount of ATP remaining at equilibrium is
difficult to measure accurately. The value of
?G? can be calculated indirectly, however, from
the equilibrium constants of two other enzymatic
reactions having less favorable equilibrium
constants Glucose 6-phosphate H2O ? glucose
Pi Keq 270 ATP glucose ? ADP
glucose 6-phosphate Keq 890 Using this
information for equilibrium constants determined
at 25 C, calculate the standard free energy of
hydrolysis of ATP. Answer The reactions, if
coupled together, constitute a futile cycle
that results in the net hydrolysis of ATP
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6. Difference between G and G Consider the
following interconversion, which occurs in
glycolysis (Chapter 14) Fructose 6-phosphate ?
glucose 6-phosphate Keq 1.97 (a) What is
?G? for the reaction (Keq measured at 25
C)? (b) If the concentration of fructose
6-phosphate is adjusted to 1.5 M and that of
glucose 6-phosphate is adjusted to 0.50 M, what
is ?G? (c) Why are ?G? and ?G different? (a) At
equilibrium, ?G? (b) ?G Q ?G
(c) ?G? for any reaction is a fixed
parameter because it is defined for standard
conditions of temperature (25oC 298 K) and
concentration (both F6P and G6P 1 M). In
contrast, ?G is a variable and can be calculated
for any set of product and reactant
concentrations. ?G is defined as ?G? (standard
conditions) plus whatever difference occurs in ?G
on moving to nonstandard conditions.
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B. Phosphoryl Group Transfers and Energy Exchanges
Factors Contributing to Free Energy Release 1.
Reactant bond strain from electrostatic repulsion
is relieved. 2. Products stabilized by i)
ii) iii) iv) Lets look at some
examples (a-e) (a ATP b PEP c 1,3-bisPG
d PC e thioester)
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Chemical basis for the large free-energy change
associated with ATP hydrolysis.
? Greater hydration of ADP Pi relative to ATP
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• a) ATP the donation of energy by ATP generally
  involves a covalent participation of ATP in the
  reaction driven
• Why is ATP a Suitable Energy Storage Unit and
  Donor?
• ATP is kinetically stable _at_ pH 7 (it has a high
  Ea for hydrolysis and so requires an enzyme,
  therefore the reaction can be regulated)
• Cellular ATP, ADP, Pi are much lower than 1
  M, and ATP is complexed to Mg2 (Figure 13-12)
• - In fact, we can calculate the phosphorylation
  
• potential ?GP (-50 to - 65 kJ/mol see Ex.
  13-2
• Table 13-5) - ?GP, which is affected by
  ATP, ADP,
• Pi, is the relevant value in living cells

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Also see Table 13-6 A ?G? lt -25 kJ/mol is
considered high energy, but compared to other
phosphate donor and acceptors, ATP is intermediate
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b) Phosphoenolpyruvate (PEP) hydrolysis of a
phosphate ester - tautomerization of pyruvate to
keto form is mostly responsible for the high ?G?
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PEP can be used to regenerate ATP (more
negative ?G?)
?G? (kJ/mol) _______________________________
_____________ The ability of ATP to give and
receive, makes it the UNIVERSAL ENERGY
CURRENCY in all living cells
phosphoenolpyruvate H2O ? pyruvate Pi -61.9
ADP Pi ? ATP H2O 30.5
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c) 1,3-bisphosphoglycerate hydrolysis of acyl
phosphate - ionization resonance of product
responsible for the large negative ?G?
2-iii.
2-iii.
2-i.
Ionization (loss of H)
2-iv. solvation
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d) Phosphocreatine (PC) hydrolysis of P-N bond -
resonance stabilization of creatine contributes
to the large, negative ?G?
2-iii.
2-iii.
2-iv. solvation
. For all reactions with phosphate transfer
(a-d), resonance stabilization of Pi contributes
to -ve ?G? value
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• e) Thioesters
• Acetyl coenzyme A
• (acetyl-CoA or -CoASH)
• ? Hydrolysis of acetyl-CoA
• Acyl group used for
• transacylation
• condensation
• oxidation-reduction

2-i.
2-iv. solvation
2-iii.
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How does ATP react? SN2 nucleophilic displacement
by oxygen (OH, or COOH), nitrogen (creatine, Arg
or His) transfer of phosphoryl
Inorganic pyrophosphatase catalyzes PPi ? 2 Pi
?G? -19 kJ/mol Therefore, adenylation
reactions are even more favorable!
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• Transphophorylation Reactions
• All (deoxy)nucleoside triphosphates are
  energetically
• equivalent to ATP
• Nucleoside diphosphate kinase catalyzes the
  synthesis of all (d)NTPs (given ATP and (d)NDPs)
  
• (deoxy)nucleoside triphosphates (d)NTPs
• ATP NDP (dNDP) ? Mg2? ADP NTP (dNTP) ? G?
  0
• If ? G? is close to zero, what might drive the
  reaction?

ATP/ADP is normally high in
the cell, driving the
reaction by mass action
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Transphophorylation Reactions b) Restoration of
ATP (ATP) ?G? (kJ/mol) Adenylate kinase
2 ADP ? Mg2? ATP AMP 0 Guanylate kinase
GTP ADP ? Mg2? ATP GDP 0 Creatine
kinase ADP PCr ? Mg2? ATP Cr
-12.5 (CK can respond to a sudden demand of the
cell for ATP CK 30 mM 10 x
ATP) 1,3-Bisphosphoglycerate calculate
?G? Phosphoenolpyruvate previous example
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C. Oxidation-reduction reaction
Many biochemical oxidation-reduction reactions
involve transfer of two electrons In order to
keep charges in balance, proton transfer often
accompanies electron transfer In many
dehydrogenases, the reaction proceeds by a
stepwise transfers of proton ( H ) and hydride (
H- )
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2C. Oxidation-Reduction Reactions in Living Cells

• Importance of Oxidation-Reductions in Living
  Cells
• Electron transfer is equally as important to
  metabolism as phosphoryl group transfers ...
• LEO says GER
• Lose Electrons Oxidation
• Gain Electrons
  Reduction
• Electrons flow from metabolic intermediates to
• ? electron carriers
• acceptors with higher electron affinities with a
  release
• of energy
• The Energy transducers (NADH, NADPH, FAD, FMN)
  convert the energy of electron flow into
  metabolic work.

Lehninger, pp. 507-516
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How can electron flow accomplish biological work?
Consider living cells to be a biological circuit,
and glucose as a source of electrons (glucose
O2) to ultimately provide energy for ATP
synthesis that in turn drives biological
processes (flagellar motion, muscle contraction
) Can the Energy Harnessed be
Measured? Electromotive force (emf) force
proportional to the difference in electron
affinity of the chemical compounds i.e. emf can
accomplish work
Lehninger, pp. 507-516
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Recall The Oxidation-reduction
half-reactions Consider the following reaction
for reducing sugars by a cupric ion R-C(O)-H
4 OH- 2 Cu2 ? R-C(O)-OH Cu2O 2
H2O This can be split into two 1/2
reactions 1) and 2) General Rules A
compound more rich in H is more reduced A
compound more rich in O is more oxidized
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FIG. 13-22 Oxidation states of carbon in the
biosphere.
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How are Oxidations and Dehydrogenations
Accomplished in Biological systems? Dehydrogenase
s often catalyze biological oxidation reactions,
accompanied by a dehydrogenation reaction
Their Modes of Electron Transfer 1. Direct
transfer as electrons (Fe/Cu example - text) 2.
As hydrogen atoms (proton (H) an electron
(e-)) AH2 ? A 2e- 2 H i.e.
FADH2 3. As a hydride ion (proton 2 e- is H-)
with two e-s i.e. NAD 4. Direct combination
with oxygen R-CH3 ½ O2 ? R-CH2-OH
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Reduction Potentials and Electron
Affinity Standard Reduction Potential (E ?) a
measure (volts) of the relative affinity of the
electron acceptor of each redox pair for
electrons. The Standard half cell (Figure
13-23) H e- ? 1/2 H2 is assigned E ? 0
V Under standard conditions (1 M solute, 1 atm
(101.3 kPa) gas), e-s flow through the external
circuit from the half-cell of lower standard
reduction potential to that of higher standard
reduction potential Just as for ?G?, standard
conditions are considered pH 7.0, and the
standard reduction potential at pH 7.0 is ?E ?
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Convention The half-cell with the stronger
tendency to acquire e-s has a more ve E ?
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EXAMPLE H2 ? 2 H 2 e- (-)
0.000 O2 2 H 2 e- ? H2O2
0.295 H2 O2 ? H2O2
0.295 OR ?E

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Walter Nernst Derived the Following
Equation Nernst equation (E also depends on
activities of species) E E ? (RT)lne-
acceptor nF e- donor where
n is the number of electrons transferred and F is
the Faraday constant 96 480 J/V?mol R is the gas
constant and T is temperature in kelvin (K) Just
as for ?G?, standard conditions are considered
pH 7.0, and the standard reduction potential at
pH 7.0 is ?E ?
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Can We Convert the EMF to a Free Energy Value?
The free energy change (?G) is ? to ?E ?G
-nF?E or ?G? -nF?E ? where n of
electrons transferred in the reaction Using
Table 13-7, and the concentrations of the
reacting species, we can calculate the free
energy change.
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Consider Crotonyl-CoA H NADH ? butyrylCoA
NAD
Half reactions E? (V) Crotonyl-CoA 2H
2e- ? butyrylCoA -0.015
NADH ? NAD H 2e- (reverse sign)
0.320

OR use ?E? and ?G?

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What if Crotonyl-CoA and NADH are 1 M, and
butyrylCoA and NAD are 0.2 M? Can we still
calculate a free energy change?
First we must calculate the E values corrected
for the non-standard concentrations
EcrotonylCoA
RT/F (8.315 J/Kmol)(298.15 K)/(96480 J/V mol)
0.0257 V 0.026 V
ENAD

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What next?
?E
AND ?G
The complete oxidation of glucose (?G? -2840
kJ/mol) is conducted in a series of controlled
reactions and the energy transferred to ATP
synthesis via e- carriers
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Universal Electron Carriers
1. Fe-S Proteins (Chapter 19) 2. Quinones
(Chapter 19) 3. Coenzymes and Prosthetic
Groups a) Nicotinamides -NADH, NADPH
(coenzymes) Source Niacin b) Flavin
Nucleotides FAD (prosthetic group) Source
Riboflavin
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NAD 2e- 2 H ? NADH H NADP 2e- 2 H
? NADPH H

• Substrate gives up 2 H atoms
• Each oxidized nucleotide accepts a hydride ion
• (proton 2 electrons)
• H is released to the medium

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• More Versatile than NAD(P)H
• Accepts 1 or 2 electrons with
• one or two H (each is a hydrogen transfer)
• b) Reduction potential (-0.4 V to 0.06 V)
  changes based on binding strength

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Chapter 13 Summary
In this chapter, we learned that the rules of
thermodynamics, and organic chemistry still apply
to living systems. For example

• Group transfer reactions are favorable when the
  free energy of products is much lower than the
  free energy of reactants. In biochemical
  phosphoryl transfer reactions, the good phosphate
  donors are destabilized by electrostatic
  repulsion, and the reaction products are often
  stabilized by resonance.
• Unfavorable reactions can be made possible by
  chemically coupling a highly favorable reaction
  to the unfavorable reaction. For example, ATP
  can be synthesized in the cell using energy in
  phosphoenolpyruvate.
• Oxidation-reduction reaction commonly involve
  transfer of electrons from reduced organic
  compounds to specialized redox cofactors. The
  reduced cofactors can be used in the
  biosynthesis, or may serve as a source of energy
  for ATP synthesis.