Solutions to Selected End of Chapter 7 Problems
15. Reducing sugars (= aldol sugars). N-acetyl-β-D-glucosamine, has the N-acetyl acetylating the
amino group attached to carbon 2 of glucosamine. Carbon 1 in the hemiacetal and when in the linear
form is an aldehyde and therefore reducing. What is the state of carbon 1 in gluconate? It is the
oxidized form, a carboxylic acid and therefore non-reducing. Lastly, GlcN(α1 1α)Glc. Both number
one carbons are no longer hemiacetals and are in the glycosidic link making these glucoses nonreducing.
17. The physical properties differences between alpha- or beta-(14) glycosidic linked
polysaccharides of glucose: glycogen and cellulose: one very soluble and the other not so. Cellulose
can be cotton, paper, etc. are not water soluble…parallel series of intermolecular H-bonds aggregate
to long insoluble fibers. Whereas glycogen (and starch) can be dissolved in warm water because it
has a helical structure by intramolecular H-bonds and it is branched. Check out the structure of
cellulose (fig 7-20) compared to starch (fig 7-18).
18. Dimensions of cellulose and amylose, each MW 200,000 daltons. Cellulose is going to be very
long compared to the amylose helix and winding.
19. Growth rate of bamboo is faster than hair (Chapter 3 and ZZTop). Bamboo can grow at 0.3
m/day. The glucose units in cellulose are arranged linearly along the length of the plant. Assume a
glucose unit is 0.5 nm of the cellulose chain. How many glucoses are added per second to each
chain? First convert 0.3 m/day to nm: 0.3 x109 nm
[ (0.3 x 109 nm) / day] / 0.5 nm/sugar = 6 x 108 sugars/day
(6 x 108 sugars/day) x (1 day / 24 hrs ) x (1 hr / 60 min) X (1 min / 60 sec) = 6944 sugars/sec.
Pretty spectacular enzymes! Considering they are not just producing one chain, but millions
across the bamboo stem. Biology has a real biochemical beauty.
20. I don’t know about eating bustards, but apparently some Greeks do. Panic flight of some bird’s
glycogen breakdown produces 120 μmole/min of glucose-1-phosphate per gram of muscle. These
birds (quail, grouse, pheasants) contain 0.35% glycogen by weight. How long can they carry out a
panic flight? The MW of glucose is 180, a nice round number. But, we will use 162. Why?
So, lets just look at 1 gram of flight muscle. That gram (1000 mg) has only 3.5 mg of glycogen.
This 3.5 mg of glycogen contains:
3.5 mg
162 mg/mmole
= 0.022 mmoles of glucose
That is = 22 μmoles of glucose. Time of flight then is 22 μmoles glucose / 120 μmoles/min = 0.18 min
Or 11seconds.
23. Heparin, a highly negatively charged glycosaminoglycan (un-numbered fig pg 261) is used as an
anticoagulant (it coats plastic and tubing that are used in medical devices that come in contact with
blood). The anticoagulant properties are due to its binding to antithrombin-III and inhibitor of blood
clotting which causes a conformational change to make antithrombin-III an even more effective
inhibitor of clot formation. What amino acids of antithrombin-III are likely involved in this tight
binding?
The positively charged amino acids at pH 7.4…lysine and arginine are the candidates to be attracted
to the carboxyl and sulfonyl groups. It turns out, it is mostly lys.
27. Determination of branching in a polysaccharide. The basis of the procedure is to take the pure
natural polysaccharide and exhaustively methylate all R-OHs. This is done by reacting the
polysaccharide with methyl iodide at an alkaline pH, thereby converting R-OH to R-O-CH3. This is
then followed by acid hydrolysis of all the glycosidic bonds. The carbons involved in the glycosidic
link are not methylated, only open alcohols are methylated. So, the major product is 2,3,6-trimethylD-glucose. The not methylated on carbons are 1 and 4 because they are part of the major α 1,4
glycosidic link
a. When this is done with amylopectin, most of the product is 2,3,6-trimethyl-D-glucose. But
there is a small amount of 2,3 methyl-D-glucose; this is the sugar that did not get methylated at
the 6 position because it was in a glycoside branch link: this was the sugar that was α 1,4 in its
chain, but also it’s 6 position was connected to carbon 1 of the first branch chain glucose.
b. The 258 mg sample of amylopectin after methylation and hydrolysis yielded 12.4 mg of 2,3
dimethyl-D-glucose. Glucoses in amylopectin, MW = 162. So let figure how many moles of
glucose are present in the sample:
258 mg / (162 mg/mmole) = 1.59 mmoles
The yield of the dimethyl product (MW = 208) was 12.4 mg, lets figure the amount of it:
12.4 mg / (190 mg/mmole) = 0.065 mmoles
% of branch points in this amylopectin sample is:
0.065 mmoles /1.59 mmoles = 4.1%
So that means about 4 out of 100 sugars are in a branch point. Or one branch about every 25
sugars.
Lets take this problem a little further: a small amount of 2,3,4,6 tetramethyl-D-glucose was
found. Where in the polysaccharide is that? These are the non-reducing ends!
Ahhhh, this has been a sweet chapter.