Suggested solution Ex. 4
1. Polysaccharide chemistry Galactomannans are polysaccharides having a main chain of 1,4-linked β-D-mannose (Man) and side chains consisting of single α-D-galactose (Gal) linked to C6 of the mannose (Fig. 4.38 in textbook). a) Write the Fisher formulae of free D-mannose and D-galactose
D-mannose D-galactose
1 4 b) Ring conformations ( C4 or C1)? 1 4
C4 or C1 is decided by how many of the OH-group and the CH2OH on C5 that are in equatorial or axial position. The molecule is most stable if it has these big groups in equatorial position.
4 β-D-mannose: in C1 we have only one OH that is axial – this is the most stable form.
4 α-D-galactose: in C1 we have two axial OH – this is the most stable form. c) Identify axial and equatorial linkages in galactomannans
M to M: eq-eq (diequatorial) G to M: ax-eq (axial-equatorial) 2. Alginates a) Most industrial alginates are water soluble at pH 6-7 but precipitate when pH is lowered below ca. 3. Explain briefly why.
- pKa to the carboxylate group is about 3,5. Below pH 3,5 we start to obtain COOH instead of COO . (See “1.2 Alginates” page 19 for more details) b) Alginates are most stable against degradation at pH 5-8, but become degraded at lower or higher pH values, especially if heated. Explain briefly why?
At low pH the alginate can be acid hydrolyzed and at high pH it is victim for β-elimination (or alkaline hydrolysis; β-elimination gives an unsaturated compound, but hydrolysis doesn’t). See compendium “5.1. Degradation of polysaccharides” for more details. c) Ca-alginate gels are never used for protein electrophoresis. Why not? (Explain briefly).
In electrophoresis you don’t want your gel to have any charge. In an alginate gel, cationic (positively charged) proteins can’t travel true the gel.
(Ca-alginate gels are also inhomogeneous if not made in the right way, and hence not fitted to separate proteins by size. )
3. Cellulose a) What is the difference between Cellulose I and Cellulose II?
Cellulose I Cellulose II Parallel chains Antiparallel chains More stable (stabilizing H-bonds between the layers) Slight tilting of the chains compared to cellulose I (see figure in the Compendium “1.4 Cellulose and its derivatives”) b) Cellulose is insoluble in water, but most cellulose derivatives such as CMC and HEC are soluble. Why?
The high degree of crystallinity is responsible for the poor solubility. Cellulose derivatives have substituents attached to the hydroxyls that interfere with the H-bonding and results in solubility in water.
4. Chitosan a) Define FA for chitosans
GlcNAc means N-acetylated β-D-glucosoamine. nGlcNAc is the number of N-acetylated residues, and nGlcNAc + nGlcN = total number of residues.
b) Why is a chitosan with FA = 0.1 soluble at pH 4 but insoluble at pH 8?
+ + -NH3 -NH2 + H pKa = 6,2 (for FA = 0,01, would be higher for FA = 0,1) At pH 4: all GlcN residues are positively charged.
At pH 8: almost none of the GlcN residues are positively charged.
5.Polysaccharide degradation A linear (unbranced) polysaccharide (A-A-A-A-A-A-A-A-A-A-A…………A-A-A) is degraded by two different enzymes: Enzyme 1 (E1) degrades randomly, whereas Enzyme 2 (E2) cleaves off a single monosaccharide (A) at a time from the non-reducing end. a) What is the non-reducing end? b) Why is it called non-reducing?
http://wiki.answers.com/Q/What_is_a_non-reducing_sugar: Reducing and Non-reducing Sugars: Sugars which are capable of reducing the oxidizing agents Tollen's reagent (AgO), Fehling reagents (CuO) or Ferricynide in alkaline solution are called the Reducing Sugars. Examples include glucose, fructose, maltose and lactose. Those sugars which are unable to reduce oxidizing agents such as those listed above are called non-reducing sugars. Sucrose is a non-reducing sugar. Carbohydrates that can reduce oxidizing agents are reducing sugars. In the instance of disaccharides, structures that possess one free unsubstituted anomeric carbon atom are reducing sugars. The end of the molecule containing the free anomeric carbon is called the reducing end, and the other end is called the nonreducing end. So non-reducing sugars that cannot reduce oxidizing agents. http://en.wikipedia.org/wiki/Biochemistry: Sugar polymers are characterised by having reducing or non-reducing ends. A reducing end of a carbohydrate is a carbon atom that can be in equilibrium with the open-chain aldehyde or keto form. If the joining of monomers takes place at such a carbon atom, the free hydroxy group of the pyranose or furanose form is exchanged with an OH-side-chain of another sugar, yielding a full acetal. This prevents opening of the chain to the aldehyde or keto form and renders the modified residue non-reducing.
The end that can open is called the reducing end and the other end is called the non-reducing end.
c) Assume DPw = DPn = 1000 t=0. Calculate DPn and DPw after the cleavage of 3 linkages. i) Randomly
After 3 random chain breaks we have 4 new chains that can have any length, but the sum of their lengths will always be equal to 1000. Also we can assume that the weight will be approximately equally distributed among the 4 new chains because of the random nature of the cleavages.
Therefore DPw = DPn= 1000/4 = 250. ii) End-wise
After 3 chain breaks we have one very long chain and 3 very short chains. The sum of their lengths is still equal to 1000. However the weight will be almost only carried by the one very long chain and we can thus assume that DPw is equal to the DP of the very long chain and that the 3 very short chains do not afftect DPw.
Thus DPn = 1000/4 = 250 and DPw = 997.
d) How can we by analysing the changes in Mw distinguish between the two modes of degradation (random or end-wise)?
ni – moles of i-mer = Ni/Navo (Ni – molceules of i-mer) wi – Mass (weight) of i-mer Mi – Molar mass of i-mer (g/mol)
So when we monitor the change in Mw over time, you can plot Mw against time and from the plot you will see if it is randomly degraded or degraded from the end of the molecule.
100000 80000 60000 40000 E1 20000 E2 0 0 5 10 15 Time (relatively)
E1: End-wise degradation E2: Random degradation