EXAM TBT4135 Biopolymers 2008 Solution - NTNUfolk.ntnu.no/audunfor/7....

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EXAM TBT4135 Biopolymers December 2008

Transcript of EXAM TBT4135 Biopolymers 2008 Solution - NTNUfolk.ntnu.no/audunfor/7....

Page 1: EXAM TBT4135 Biopolymers 2008 Solution - NTNUfolk.ntnu.no/audunfor/7. semester/Biopolymerkjemi/Eksamen... · 2012-12-01 · the solvent (buffer) changed, assuming pH was kept close

EXAM TBT4135 Biopolymers December 2008

Page 2: EXAM TBT4135 Biopolymers 2008 Solution - NTNUfolk.ntnu.no/audunfor/7. semester/Biopolymerkjemi/Eksamen... · 2012-12-01 · the solvent (buffer) changed, assuming pH was kept close

1. Single- or multiple choice: Mark your answer with an X directly on the sheet (pen, not pencil) Enkel- eller flervalg: Merk svaret med X direkte på arket (Extra copy in Appendix 1) (Ekstra kopi i Appendix 1) Monosaccharide

Monosakkarid Name (incl. D/L and α,β for rings) Navn (inkl. D/L og α ,β for ringer)

Essential component (>20%) of Essentiell komponent (>20%) i

a CHO

HHO

HHO

OHH

OHH

CH2OH

D-mannose o Cellulose o Amylose o Dextran o Xanthan o Hyaluronan o Agarose o Chitosan o Alginate o Pectin o Neither/Ingen

b CHO

HO H

H OH

HO H

HO H

CH2OH

L-glucose o Cellulose o Amylose o Dextran o Xanthan o Hyaluronan o Agarose o Chitosan o Alginate o Neither/Ingen

c

β-D-glucose o Cellulose o Amylose o Dextran o Xanthan o Hyaluronan o Agarose o Chitosan o Alginate o Neither/Ingen

d O

COOH

OH

OH

OH

OH

α-D-galacturonic acid o Cellulose o Amylose o Dextran o Xanthan o Hyaluronan o Agarose o Chitosan o Alginate o Pectin o Neither/Ingen

O

H

HO

H

HO

H

H

OHH

HOH2C

OH

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e O

COOHOH

OHOH

HO

α-L-guluronic acid o Cellulose o Amylose o Dextran o Xanthan o Hyaluronan o Agarose o Chitosan o Alginate o Pectin o Neither/Ingen

f O

CH2

OH

OH

HO

O

α-3,6-anhydro-L-galactose o Carrageenan o Amylose o Dextran o Xanthan o Hyaluronan o Agarose o Chitosan o Alginate o Pectin

o Neither/Ingen

Amino acid Aminosyre

Name (incl. correct stereochemistry) Navn (inkl. rett stereokjemi)

Side chain property at pH 7 Egenskap til sidekjede ved pH 7

g H2N CH C

CH3

OH

O

L-Alanine (Ala - A) o Polar, uncharged o Nonpolar o Negatively

charged o Positively charged

h

H2N CH C

CH2

OH

O

CH2

C

OH

O

L-Glutamic acid (Glu - E) o Polar, uncharged o Nonpolar o Negatively

charged o Positively charged

i

H2N CH C

CH2

OH

O

C

NH2

O

L-Asparagine (Asn - D) o Polar, uncharged o Nonpolar o Negatively

charged o Positively charged

j

H2N CH C

CH2

OH

O

L-Phenylalanine (Phe - F) o Polar, uncharged o Nonpolar o Negatively

charged o Positively charged

Several alternatives in principle possible in column 3: Correct answer: +1 Wrong answer: -1 Ved flere mulige svaralternativer på samme spørsmål:

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Rett svar: +1 Feil svar: -1 Maximum score: 20% of total Maksimal skår: 20% av total 2.

a) Explain the structural requirements needed to form hydrogen bonds (which type of atoms, typical distances and bond angles)?

R1-XH :::Y-R2 , where X is a electronegative atom (O, N) carrying a proton, and X is another electronegative atom having at least one free electron pair. XH: typically 1 Å (0.1 nm), H::Y typically 1.8 Å, Angle R1XH: ca 180°, H::YR2: ca 140° b) What is the thermodynamic basis for the formation of hydrogen bonds

(such as in α-helices) in aqueous systems? Explain briefly (max 100 words)

When a H-bond is formed between two segments (e.g. amino acids), two H-bonded water molecules which were bound to the segments are usually released: -S-H2O + H2O-S → -S :: S- + 2H2O. The water molecules form two novel H-bonds with other water molecules (bulk water). Thus, the number of H-bonds is generally unchanged. Therefore, the total ΔH ≈ 0. The driving force is the increase in entropy (ΔS > 0) when the water molecules are released into bulk water.

c) High concentrations (6-8 M) of urea are known to denaturate many

proteins in aqueous solutions. Explain briefly the thermodynamic basis for such denaturation (max 100 words)

The urea/water system has high free energy (G = H-TS). In 6 M urea there is about 6-7 H2O molecules per urea molecules. These water molecules are highly ordered, i.e. they have low entropy (S), which is thermodynamically unfavorable. By breaking H-bonds in dissolved proteins and thus form new segment-water H-bonds, the system reaches lower free energy. Breaking intramolecular H-bonds denaturates the protein.

3.

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A linear (unbranched) polysaccharide is subjected to random degradation, for instance by acid hydrolysis. The processed is monitored using a capillary viscometer by regularly recording the flow-through-time of the solution,. Key parameters: c = concentration of polysaccharide (g/ml) x = degradation time (sec) ts = flow-through-time of pure solvent (sec) tx = flow-through-time of polymer solution at a degradation time x

a) Explain, step-by-step, how the rate constant of the reaction (k) can be determined from a series of experimental data (tx values obtained for different x values)

Step 1: The specific viscosity is calculated for each measurement: ηsp = (tx-ts)/ts, and by dividing by c, ηsp/c Step 2: The intrinsic viscosity is calculated according to Hugginsʼ equation: ηsp/c = [η] + kʼ[η]2c. kʼ must be known. NOTE: kʼ is the Hugginsʼ constant and not the rate constant (k) Step 3. The molecular weight is calculated from [η] by the MHS equation: [η] = KMw

a. K and a must be known Step 4. 1/M is calculated for each measurement and plotted against the degradation time, according to the equation: 1

Mw

=1

Mw,x=0

+kt

2M0

(M0 = the monomer equivalent weight, must be

known).

Step 5: The data are fitted to a straight line, from which the slope is calculated. Slope = k/(2M0), from which k is found. b) Which fundamental properties of a polymer/solvent system are

expressed by the intrinsic viscosity?

The intrinsic viscosity expresses 1) vh: The specific hydrodynamic volume (volume occupied per gram of macromolecule) and 2) A form factor (ν), through the equation [η] = νvh c) Why should viscosity measurements be performed at relatively low

shear rate when they are used to find the shape and extension of polymers in solution?

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At high shear the polymers may deform and align with the flow, leading to reduced viscosity. This is also called shear thinning. The molecules only keep the native shape when investigated at sufficiently low shear rate: the Newtonian rang.

d) If the polysaccharide has a weight average molecular weight (Mw) of

200.000 Da at x = 0 min, and 100.000 Da after 60 min of degradation, what was Mw after 30 min of degradation?

According to the equation above (a, step 4) we find k/M0 : 1

Mw

=1

Mw,x=0

+kt

2M0

1

100.000=

1

200.000+k60

2M0

k

M0

=1

100.000!

1

200.000

"#$

%&'2

60= 1.67 (10!7

Inserting for 30 min:

1

Mw,30

=1

200.000+k30

2M0

=1

200.000

!"#

$%&+ 1.67 '10(7( )

30

2

= 7.5 '10(6

)M

w,30= 133.000

4. The figure shows results (intrinsic viscosity versus molecular weight) obtained by size-exclusion chromatography (SEC) with on-line concentration detector, multi-angle light scattering detector, and viscosity detector. The two lines correspond to alginate (upper) and hyaluronan (lower).

a) Explain briefly [max 100 words and (preferentially) a simple sketch] the principle of polymer separation in SEC

In SEC the separation medium consists of porous particles. The pore size distribution corresponds to the range of molecular size of the sample. When passing through the column the largest molecules are completely or partly excluded because they cannot enter the pores, or have at least short residence times. Smaller molecules penetrate more easily into the pores, have longer residence times inside the pores, and have therefore

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longer retention times. Therefore, large molecules elute first, and the elution time (retention time) increases with decreasing molecular size. Since it is the physical size and not the molecular weight which determines the elution time, macromolecules with different shapes have different calibration curves (log M vs V).

b) On basis of the figure in Appendix 3, determine the exponent in the

MHS constants in the two cases

By reading the figure few find (using any two values): Alginate: M = 100.000 [η] ≈ 450 ⇒ log M = 5.00, log [η] = 2.65 M = 1.000.000 [η] ≈ 3.300 ⇒ log M = 6.00, log [η] = 3.52 Exponent a = slope in log-log plot = (3.52-2.65)/(6-5) = 0.87 Hyaluronan: M = 100.000 [η] ≈ 420 ⇒ log M = 5.00, log [η] = 2.62 M = 1.000.000 [η] ≈1.800 (approx) ⇒ log M = 4.00, log [η] = 3.26 Exponent a = slope in log-log plot = (3.26-2.62)/(5-4) = 0.64

c) Comment the results

Exponents are 0 for solid spheres, 1.8 for rigid rods, and 0.5-0.8 for random coils in good solvents and θ-solvents, respectively. Values for HA are within the RC range, whereas for alginate the exponent is just outside the RC limit. The larger value for alginate shows it is stiffer and more expanded than HA.

5.

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The figure/table shows results obtained by light scattering of a polysaccharide:

a) Estimate the molecular weight and the second virial coefficient of the polysaccharide

We first extrapolate the data in the figure to zero angle, and find 6.5E-6 for c1 and 8.5E-6 for c2. According to the theory, the following equation applies for the data at zero angle: Kc

R!=0

=1

M+ 2A

2c

Thus, a new plot is made, with extrapolated values plotted against c:

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The intercept equals 1/M, and the slope equals 2A2 Intercept: 4.55E-06 Slope: 1.00E-02 M 220,000 (g/mol) A2: 5.00E-03 (ml mol-1 g-2)

b) Which average of the molecular weight is obtained if the

polysaccharide is polydisperse?

Weight average (Mw) c) Define the average in b) in terms of Mi and ci.

Mw=

ciM

i

i

!

ci

i

!

d) Provide proof for the molecular weight average

At the limit c→0 and θ→0, the light scattering equation is: Kc/R = 1/M ⇔ R = KcM The Rayleigh factor R is additive ⇒ R = ΣRi = ΣKciMi = KΣciMi Dividing and multiplying by c = Σci we obtain directly (according to the definition in c) above):

R = K ciM

i=

i

! Kc

K ciM

i

i

!

ci

i

!= Kc

1

Mw

Thus, LS provides Mw for polydiperse samples, regardless of the molecular weight distribution

e) Given the biopolymer contains 50% uronic acids, would the molecular

weight and the second virial coefficient change if the ionic strength of

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the solvent (buffer) changed, assuming pH was kept close to 7? Explain briefly (max 100 words)

The molecular weight would normally not change unless the change in ionic strength led to association or dissociation of chains. The high content of uronic acids (with pKa in the range 3-4) shows the biopolymer is a polyelectrolyte at pH 7. The second virial coefficient of polyelectrolytes has a significant contribution from the Donnan term. A2 (Donnan) is roughly inversely proportional to the ionic strength. Thus, A2 would indeed change if the ionic strength changed.

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APPENDIX 1 Copy of table in question 1 APPENDIX 2 Numerical data used in question 5

Raw data: (Kc/R!) as a function of c and !:

! sin2(!/2) c1 (g/ml) c2 (g/ml)

0.0002 0.0004

!1 30 0.07 7.08E-06 9.24E-06

!2 60 0.25 8.53E-06 1.11E-05

!3 90 0.50 1.05E-05 1.37E-05

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APPENDIX 3 [Note: log 100 = 2, log 1000 = 3 etc]