1
Protein expression and purification
II. Calculation in the molecular
biosciences
Jozef Hritz, Lubomír Janda, Blanka Pekárová
Radka Dopitová
Kód předmětu: C8980
1.4. The secondary structure of proteins
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I. The molecular principles for understanding proteins – Jozef Hritz
1.4. The secondary structure of proteins
I. The molecular principles for understanding proteins – Jozef Hritz
1.4. The secondary structure of proteins
I. The molecular principles for understanding proteins – Jozef Hritz
Energetically unfavorable conformational areas
1.4. The secondary structure of proteins
I. The molecular principles for understanding proteins – Jozef Hritz
Additional energetically unfavorable conformational areas because of the presence of
carbonyl oxygen
Glycine – less sterically restricted because of lacking the side-chain
1.4. The secondary structure of proteins
I. The molecular principles for understanding proteins – Jozef Hritz
Additional energetically unfavorable conformational areas because of the presence of Cβ
Alanine
Structural preferences of the different amino acids
1.4. The secondary structure of proteins
1.4.3. Other structural features in proteins
Met, Glu, Leu,Ala + alpha - helices
Pro, Gly, Tyr - alpha helices
Val, Ile, Phe + beta sheets
Pro, Asp - beta sheets
Pro, Gly, Asp + beta turn
Met, Val, Ile - beta turn
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II. The molecular principles for understanding proteins – Jozef Hritz
1.6. The quaternary structure of proteins
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II. The molecular principles for understanding proteins – Jozef Hritz
1.3.2.1. Exact molecular mass 1.3.2.2. Isoelectric point
1.3.2. Information available from the amino acid sequence of a protein
1.3. The primary structure of proteins
http://www.expasy.ch/tools/pi_tool.html
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II. The molecular principles for understanding proteins – Jozef Hritz
1.3.2.3. Absorption coefficient 1.3.2.4. Hydrofobicity
Beer–Lambert law A = e * c * l
A is the absorbance at a particular wavelength (without
units; merely a ratio):
A = log (I0/It)
I0 is the intensity of the incident light (light striking
the cuvette).
It is the intenisty of the transmitted light (the light
leaving the cuvette).
e is the absorption coefficient (degree of absorption) (M-1
cm-1).
l is the path length of the cuvette (cm).
c is the concentration of the solution (M).
Linear form of Beer-Lambert law is valid only for lower values of A.
II. Calculation in the molecular biosciences – Jozef Hritz
Measured absorption in cuvette with 1 cm optical path-length is 0.3 at 280 nm. 10 points
Question 1: How many mg of protein you have in 5 ML of purified
protein if its molar extinction coefficient is 15000 M-1cm-1; molecular
weight 20 kDa. Advice – units!
Mass concentration equals to the molar concentration multiplied
by molecular weight of the protein. 8
points
Unit M correspond to the mol.L-1 6
points
g.L-1 = mg.ml-1 4 points
Please solve a problem.
II. Calculation in the molecular biosciences – Jozef Hritz
A = εcl
c [M] = A/εl = 0.3 / (15000 M-1cm-1 *1 cm) = 0.00002 M = 0.02 mM
C [g.L-1] = 0.00002 mol.L-1 * 20000 g.mol-1 = 0.4 g.L-1 = 0.4 mg.ml-1
m [mg] = C*V = 0.4 mg.ml-1 * 5 ml = 2 mg
Solution:
12
II. Calculation in the molecular biosciences – Jozef Hritz
1.7. Forces contributing to the structures and interactions of proteins
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Bond strenght
Covalent bond energy
kJ/mol
1101001000
Ionic bond
Hydrophobic interaction
Hydrogen bond
Van der Waals interaction
C=C
610
C=O
740
C-C
415
C-0
360
O-H
465 S-S
250
10-20
5-15
5-10
0.5-5
II. Calculation in the molecular biosciences – Jozef Hritz
1.7. Forces contributing to the structures and interactions of proteins
1.7.1. Ionic (electrostatic) interactions
1.7.2. Hydrogen bonds
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Covalent bonds
II. Calculation in the molecular biosciences – Jozef Hritz
Combination rules for Lennard-Jones potential
sij =
sii +s jj
2
eij = eiiejj
Eij
LJ
= 4eij
sij
rij
æ
è
çç
ö
ø
÷÷
12
-
sij
rij
æ
è
çç
ö
ø
÷÷
6é
ë
ê
ê
ù
û
ú
ú
II. Calculation in the molecular biosciences – Jozef Hritz
er = 80 e = ?
intermed.
“vacuum”, ε~1?
but absence
of intermed.
dipoles can
only increase
interaction…
Electrostatic interactions at different media
ε0= 8.85*10-12 C*N-1*m-2, q=1.6*10-19 C
Eij
el
=
1
4pe0er
qiqj
rij
II. Calculation in the molecular biosciences – Jozef Hritz
Relative permittivity:
- vacuum: εr = 1
- water: εr = 80
- protein εr ~ 4
Eij
el
=
1
4pe0er
qiqj
rij
;ke =
1
4pe0
Eij
el
=
ke
er
qiqj
rij
Constant Value
Avogadro's Number NA = 6.022 × 1023 mol-1
Faraday Constant F = 96 485.33 C mol-1
Molar Gas Constant R = 8.314 J mol-1 K-1
Coulomb's Constant ke = 8.987 × 109 N m2 C-2
Speed of Light (Vacuum) c = 299 792 458 m s-1
Boltzmann Constant kb = 1.38 × 10-23 J K-1
Charge on a Proton/Electron e = 1.602 × 10-19 C
Standard acceleration of gravity g = 9.8 m s-2
Planck's Constant h = 6.6 × 10-34 m2 kg / s
Useful Physical/Chemical Constants
II. Calculation in the molecular biosciences – Jozef Hritz
𝐸 𝑝
𝑡𝑜𝑡𝑎𝑙 = ෍
𝑖,𝑗
𝐸𝑖𝑗
𝑒𝑙
+ 𝐸𝑖𝑗
𝐿𝐽
Ions are separated by 0.28 nm distance. 20
points
Question 2: What is the total potential energy of system consisting
of salt ions Na+ and Cl- in kJ/mol?For the electrostatic energy
contribution consider the permittivity of water environment
Coulombic energy term correspond to the energy per particle, not per mol 16 points
N.m = J 12 points
Be careful, which energy term is positive and which negative
8 points
Please solve a problem.
Mw[g/mol] σ[Å] ε[kJ/mol]
Na+ 23 2.3 0.5
Cl- 35 4.3 0.5
II. Calculation in the molecular biosciences – Jozef Hritz
sij =
sii +sjj
2
;eij = eiiejj 18 points
Solution:
19
  11
612
1
612
.18.97.229.7.2
8.2
3.3
8.2
3.3
.5.0*44
−−
−
=−=
=














−





=
















−








=
molkJmolkJ
A
A
A
A
molkJ
rr
E
ij
ij
ij
ij
ij
LJ
ij

e
II. Calculation in the molecular biosciences – Jozef Hritz
𝐸𝑖𝑗
𝑒𝑙
=
𝑘 𝑒
𝜀 𝑟
𝑞𝑖 𝑞 𝑗
𝑟𝑖𝑗
𝑁𝐴 =
9 ∗ 109
𝑁𝑚2
𝐶−2
80
− 1.6 ∗ 10−19
𝐶 1.6 ∗ 10−19
𝐶
0.28 ∗ 10−9 𝑚
6.022 ∗ 1023
𝑚𝑜𝑙−1
=
= −6.2 ∗ 103
𝐽𝑚𝑜𝑙−1
= −6.2𝑘𝐽𝑚𝑜𝑙−1
𝐸 𝑝
𝑡𝑜𝑡𝑎𝑙
= 𝐸𝑖𝑗
𝑒𝑙
+ 𝐸𝑖𝑗
𝐿𝐽
= −6.2𝑘𝐽𝑚𝑜𝑙−1
+ 9.18𝑘𝐽𝑚𝑜𝑙−1
= 3𝑘𝐽𝑚𝑜𝑙−1
According what principle there should be the melting of an ice?
II. Calculation in the molecular biosciences – Jozef Hritz
The crude meaning of entropy and free energy
Where is the higher probability to find one particular molcule of air here in our
classroom (pc) or in astronomy observatory at Lomnicky Stit (pL)?
pc/pL = [Vcexp(-Ec/kT)] / [VLexp(-EL/kT)]
= [exp(-(Ec-T*k*ln(Vc))/kT)] / [exp(-(EL-T*k*ln(VL))/kT)]
where S = k*ln(V) is entropy;
more general: S = k*[logarithm of the number of accessible states of the particle ]
and F = E – T*S is (Helmholtz) free energy (for system at constant volume)
pc/pL = exp(-Fc/kT) / exp(-FL/kT) = exp(-(Fc-FL)/kT)
II. Calculation in the molecular biosciences – Jozef Hritz
Difference between Helmholtz and Gibbs free energy
Helmholtz free energy Gibbs free energy
V=const P=const
F=E-TS G=H-TS=F+PV
E enthalpy H=E+PV
chemical potential: μ=G/N
probability  exp(-F/kT) ➔max
F ➔min
probability  exp(-G/kT) ➔max
G ➔min
Typical value of PV for liquids at atmospheric pressure is very small in comparison
with the thermal energy of the body. Therefore very often the term “
Let’s estimate the work against the external atmospheric pressure related to the insertion
of 1M of water.
T=const:
II. Calculation in the molecular biosciences – Jozef Hritz
Let’s estimate the work against the external atmospheric pressure related to the
insertion of 1M of water.
1M of water: 18 g*mol-1, 18cm3*mol-1
at atmospheric pressure pV = 105 Pa * 18 * 10-6 m3*mol-1 = 1.8 J*mol-1
It is negligible value in comparison to the thermal energy.
II. Calculation in the molecular biosciences – Jozef Hritz
G=H-TS; dG=dH-TdS-SdT
At T=const (dT=0)
at equilibrium, G=min, dG=0, dH-TdS=0
It implies:
T = dH/dS
Definition of heat capacity: Cp=(dH/dT)P=const
S=-(dG/dT)P=const
II. Calculation in the molecular biosciences – Jozef Hritz
Free energy – driving force
Typical Examples:
(1) Conformational changes
(2) Affinity of the ligand binding
(3) Solubility
(4) Lipophilicity
(5) Protein in folded vs unfolded state
In the computational simulations it is easy to calculate the potential or kinetical energy
of the system. However, it is much more challenging to calculate the free energy differences.
II. Calculation in the molecular biosciences – Jozef Hritz
26
Cl Cl
OH OH
DGbind(1)
DGbind(2)
DG21(bound)
DG21(free)
1
2
( ) ( )
( ) ( )
bind bind bindG G G
G free G bound
DD = D - D
= D - D21 21
2 1
Thermodynamic cycle for binding affinities
II. Calculation in the molecular biosciences – Jozef Hritz
Hydrophobic bond between non-polar particles
Hydrophobic free energy increases almost proportionally to the accessible surface
area of the non-polar molecules.
Its value is ~85-105 J*mol-1*Å-2
II. Calculation in the molecular biosciences – Jozef Hritz
~ -50 Å2 for each polar atom
The accessible surface area of amino-acid side-chains and their hydrophobicity
Adopted from Schulz G.E., Schirmer R.H. Principles of protein structure. 1979
Hydrophobic effect is the principle component of the protein 3D structure formation/stability.
II. Calculation in the molecular biosciences – Jozef Hritz
Role of water
H-bond energy is ~5kcal*mol-1=21kJ*mol-1
in water H-bond energy is only ~5-6 kJ*mol-1
Large difference in increase of protein stability by forming of intraprotein H-bond
in vacuum vs. water!!!
Second look on the hydrogen bonding
II. Calculation in the molecular biosciences – Jozef Hritz
Difference in entropy of water molecule.
At the melting temperature 273 K
increase in entropy is compensated by decrease in energy
II. Calculation in the molecular biosciences – Jozef Hritz
Melting heat of ice: 334 J*g-1 ( 80 cal*g-1)
334 J*g-1 * 18 g*mol-1 = 6.012 kJ*mol-1 (1.5 kcal * mol-1)
II. Calculation in the molecular biosciences – Jozef Hritz
Conclusions
II. Calculation in the molecular biosciences – Jozef Hritz
• Free energy difference is the most important determinant
of behavior in solutions
• Free energy has enthalpic and entropic part
• The increasing importance of entropic part with the increasing
temperature
• Probability of certain state is given by Boltzmann formula
• Binding affinity depends on both – bound and unbound state
in water
Email: jozef.hritz@ceitec.muni.cz