Solution suggestion: Exercise no. 2
1. Size/dialysis
a. The average density of globular proteins is 1.37 g/cm3 [1]. From this we can calculate its reciprocal
value, the specific volume v2=0.73 cm3/g. The volume of a globular protein can be estimated by
calculating the volume of a sphere with the same size.
12kDa  V  14.544nm3
1/ 3
 3(14.544nm3 ) 
 3V 

r     
4
 4 


d  2r  3.0nm
1/ 3
 1.514nm
[1] Erickson, H.P. “Size and shape of protein molecules at the nanometer level determined by
sedimentation, gel filtration, and electron microscopy.” Biol Proced Online. 2009 May
b.
Contour length = 3.0 nm
Length of a monomer = 0.5 nm
DP = 3.0/0.5 = 6
Mo=6C+7O+8H=192 gmol-1
Mr=(6)(192 gmol-1)-(5)(18 gmol-1)=1062 gmol-1
2. Protein/peptide fundamentals
Estimate pI for one of the proteins A-I (choose one) given in Appendix A
1. Calculate charge fraction (α) of each amino acid and terminal ends for 1<pH<14.
Henderson-Hasselback:
(1)
(2)
(3)
1
Gives:
α =1/(1+10(pKa - pH) )
(4)
For basic groups :
α =1-1/(1+10(pKa - pH) )
(5)
Charge fraction
pKa
4
pH
Term
COOH
Term
NH3+
Asp (-)
4
Glu (-)
0
6
11
His (+)
Cys (-)
1
0
0,01
0,01
1
0
11
Tyr (-)
0
0
0
0
0
0
0
0
-0,01
-0,1
-0,5
10
12
Lys (+) 1
Arg (+) 1
1
1
1
1
1
1
1
1
1
1
1
1
0,99
1
0,9
0,99
0,5
0,9
0,1
0,5
4
10
1
2
3
4
5
6
7
8
0
0,01
-0,1
-0,5
-0,9
0,99
-1
-1
1
1
1
1
1
1
1
0
-0,1
-0,5
-0,9
-0,1
-0,5
-0,9
1
0
0,99
0
0,9
0
0,99
0,99
0,5
0
9
10
11
12
13
14
-1
-1
-1
-1
-1
-1
0,99
0,9
0,5
0,1
0,01
0
0
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
0,1 0,01
0
0
0
-0,01
0
-0,1
0
-0,5
0
-0,9
0
0
-1
0,99
-0,9 0,99
0,01
0
0,1 0,01
2
-1
0
0
Make a table. Use excel, sigmaplot or other programs.
Use eq. 4 and 5. (Check signs)
The total charge is the sum of all charge fractions multiplied with number of groups in the protein
Examples:
Charge fraction
PROTEIN A
Number
of
residue pH
Term
1 COOH
Term
1 NH3+
10 Asp (-)
0
2
0,01
0,1
1
0
1
-0,1
1
-1
1
-5
10 Glu (-)
0
-0,1
-1
-5
2 His (+)
5 Cys (-)
2
0
2
0
2
0
9 Tyr (-)
0
0
13 Lys (+) 13
6
6 Arg
(+)
13
6
Total
1
3
4
5
6
0,99
7
8
9
10
11
12
13
-1
-1
-1
-1
-1
-1
-1
1
-9,9
1
-10
0,99
-10
0,9
-10
0,5
-10
0,1
-10
0,01
-10
0
-10
-9,9
-10
-10
-10
-10
-10
-10
-10
1,98
0
1
9
9
1,8
0
1
0
0,2
0
0,02
0
0
-0,05
0
0
0
0
0
0
0
0
0
-0,09
13
6
13
6
13
6
13
6
13 12,87
6
6
11,7
5,94
0,5
0,9
6,5
5,4
2,5
4,5
1,3
3
4,5
8,1
0,13
0,6
0
4,95
8,91
0
0,06
0
10
10
0
5
9
0
0
-10
-23,6
32,86 34,8
35
-0,5 0,9
22 21,79 19,9 11,48 2,9 0,21 -0,8
-1,12
2,6
3
14
1
Here we can see that pI is between pH 6 and 7, but closer to 6.
Charge fraction
PROTEIN D
Number
of
residue pH
Term
1 COOH
Term
1 NH3+
3 Asp (-)
1
2
4
5
6
7
8
9
10
11
12
13
14
-0,1 -0,5
-0,9
-0,99
-1
-1
-1
-1
-1
-1
-1
-1
1
1
1
1
0 -0,03 -0,3 -1,5
1
-2,7
1
-2,97
1
-3
0,99
-3
0,9
-3
0,5
-3
0,1
-3
0,01
-3
0
-3
0
-3
0
0
0
0
0
0
0
0
0
0
0,9
0,5
0,1
0,01
0
0
0
0
0
0
0
0
0
0
0
0 -0,01
3
0 Glu (-)
0
0
0
1 His (+)
1
1
1
0 Cys (-)
0
0
0
0
0
0
0
0
0
2 Tyr (-)
0
0
0
0
0
0
0
0
-0,2
-1
-1,8
-1,98
-2
1 Lys (+)
1
1
1
1
1
1
1
0,99
0,02
0,9
0,5
0,1
0,01
0
0
2 Arg (+)
2
2
2
2
2
2
2
2
1,98
1,8
1
0,2
0,02
0
5 4,96
4,6
1,3
0,54
0,1
-0,01
0,24
-1,4
-3,8
-5,58
-5,96
-6
Total
0
0,99
2,99
Here we can see that pI is approximately 8.
2. Plot total charge as a function of pH
Example:
30
isoelctric point
20
10
0
-10
0
2
4
6
8
10
12
14
16
Protein A
Protein D
-20
-30
-40
pH
4
Isoelectric points for all the proteins:
Protein
pI
A
6,1
B
9,9
C
3,7
D
8,0
E
8,0
F
9,3
G
6,9
H
10,35
I
4,42
3. Radius of gyration
a) Calculate RG for the following molecules. The monomers all have a mass of 200 g/mol, and each
linkage is 0.5 nm.
Since m1 = m2 = m3 = and so on, we have:
n = 7 (count the monomers)
a 2 + b2 = c 2
Pythagoras:
Sorce:
Wikipedia
a.
Lengths from center of mass, ri :
2x
2x
2x
0,5 nm
1,0 nm (2*0,5 nm)
1,5 nm (3*0,5 nm)
b.
5
Lengths from center of mass, ri :
2x
0,5 nm
2x
nm
2x
pythagoras
nm
c.
Lengths from center of mass, ri :
2x
4x
0,5 nm
nm
pythagoras
d.
Lengths from center of mass, ri :
6x
0,5 nm
b) Assume a molecule flips between a) and b) above, so that at any time 50% is in state a) and 50% is
in state b) (each state equally probable).
Calculate the root-mean-square radius of gyration <RG2>0.5 for the molecule (time average =
ensemble average)
Wikipedia:
In the case of a set of
values
, the RMS value is given by this formula:
For us, this give:
6
4. Shape of biopolymers in solution
a) Determine the shape (sphere, rod or random coil) of biopolymers A, B and C given the following
data:
Biopolymer 1
Biopolymer 2
Biopolymer 3
M (g/mol)
RG (nm)
RG (nm)
RG (nm)
50 000
5.0
18.0
8.0
200 000
7.9
36.0
32.0
800 000
12.6
72.0
128.0
1. Log transformation
M
Log M
Biopolymer 1
Biopolymer 2
Biopolymer 3
RG
RG
RG
Log RG
Log RG
Log RG
50000
4,69897
5
0,69897
18
1,255273
8
0,90309
200000
5,30103
7,9
0,897627
36
1,556303
32
1,50515
800000
5,90309
12,6
1,100371
72
1,857332
128
2,10721
2. Plot log RG against log M
log RG = b log M + const.
2,5
y = x - 3,7959
2
1
1,5
y = 0,5x - 1,0942
2
3
1
Linear (1)
y = 0,3334x - 0,8681
0,5
Linear (2)
Linear (3)
0
4
4,5
5
5,5
6
log M
3. When you don’t have excel (like on the exam) to find the slope, use two sets of the values from
your log transformation to calculate it. Example for polymer 1:
Or you can check if the constant, in log RG = b log M + const., is the same when you try with the
different RG’s and M’s.
7
4. You need to know the relation between the slope and shape of biopolymers:
Shape
Sphere
Rod
Random coil
Slope
1/3
1
1/2
b) Estimate C∞ for the random coil biopolymer. Assume M0 = 162 (as for glucose) and l (monomer
length) = 0.5 nm
Equations for this problem:
M
n
50000
RG <r^2> <r^2>/l^2 <r^2>/nl^2
308,642
18
1944
7776
25,19424
200000
1234,568
36
7776
31104
25,19424
800000
4938,272
72
31104
124416
25,19424
C infinite
140000
120000
100000
80000
60000
C infinite
40000
20000
0
0
1000
2000
3000
4000
5000
6000
n
8
C infinite
30
25
20
15
C infinite
10
5
0
0
200000
400000
600000
800000 1000000
M
5. Titration/Polyelectrolytes
a. You need to use equation (4) from problem 1.
α =1/(1+10(pKa - pH) ) = 1/(1+10(4.3-4.7)) = 0.715
b. Assume the following reaction is in equilibrium
When NaCl is added, there is a new equilibrium since R-COO- groups associate with
Na+. The result is a lowering of the pKa of the carboxyl groups and thereby the pH is
lowered.
9