Joseph Fourier University - Grenoble 1
2013-2014
Final Exam BIO 121 (1st Session)
Exam conditions:
- 2H,
- no document allowed,
- no calculator allowed,
- no cell phone allowed
Answer to the questions on this stack of papers
100 points for the whole exam
SUBJECT A (25 points)
(You should be answering to these questions in 30 minutes or less)
Titration and analysis of a bacterial oligosaccharide
The properties of oligosaccharides are widely used in different industrial areas, in the
food-processing industry (as texturizing reagents) and in the pharmacological field
(biocompatible and therapeutic substances). Indeed, the interactions between proteins
and polysaccharides have an essential role in many biological and pathological
processes (such as thrombosis, inflammation, metastases, sterility …)
Your goal is to do the titration after purification of an oligosaccharide from
bacteria (named A) and to analyze precisely its monosaccharide composition.
Experiment 1: Titration of the purified oligosaccharide A (12 points total)
Its total hydrolysis, in acidic conditions, leads to the following sugars (the stoechiometry
obtained is indicated in brackets). The reaction is illustrated below:
Oligosaccharide A
(1 equivalent)
H2SO4, 2h at ebullition
D-glucose (1 equivalent)
+
C2 epimer of D-glucose (1 equivalent)
+
N-acetyl-D-galactosamine (1 equivalent)
1
Knowing the stoechiometry of each of the single monosaccharides, it is possible
indirectly to do the titration of oligosaccharide A. You decide to use a colorimetric test
with an enzyme: the glucose oxidase (G.O.D.)
a- Complete in the frame provided the reactions of this test (5 points).
Flavine adenine dinucleotide
Step 1
FAD
FADH2
Haworth representation
D-Glucuronic acid
Step 2
Step 3
Reduced chromogen colourless
b- Glucose titration after total hydrolysis of the oligosaccharide (7 points).
The monosaccharide solution is in a first step diluted 100 times before the titration. Four
absorbance measurements are done in parallel. The results obtained are 0.425; 0.430;
0.435 and 0.950.
After the selection of the absorbance values, calculate the absorbance average
obtained experimentaly: ____________ (1 point).
Justify your choice with a few words (1 point):
In this titration, what is the molecule that absorb at 525 nm?
Answer: ________________________________ (1point).
2
During the test, you also carried out a standard curve with known glucose concentration
represented below.
courbe étalon du dosage du glucose par
la glucose oxydase
Title:
1,2
absorbance à 525 nm
1,0
0,8
0,6
0,4
0,2
0,0
0
20
40
60
80
100
120
[glucose] en mg/L
Give a title to this figure (fill the frame provided on the top of the curve)
(1 point).
With the informations indicated on the standard curve, calculate the initial
concentration of glucose in your hydrolyzed oligosaccharide sample. (3 points).
Justify your calculation.
En parallèle de ce test, vous avez réalisé une courbe étalon de concentrations connues
3
Experiment 2: Structural characterization of oligosaccharide A (13 points total)
You decide to determine the type of osidic bond that connect the monosaccharides
characterized previously with a permethylation technique. The analysis of the mixture
obtained, allows the identification of the modified s listed below (in equal quantities):
 2,3,4-tri-O-methyl-D-glucose
 2,3,4,6-tetra-O-methyl-D-ose to be deduced
 1,3,6-tri-O-methyl-N-acetyl-D-galactosamine
Represent according to Haworth and in the pyranose form these 3 monosaccharides
(7 points).
α-2,3,4-tri-O-methyl-D-glucose (2 points)
[Tapez une citation prise dans le document ou la synthèse d'un passage intéressant. Vous pouvez
placer la zone de texte n'importe où dans le document. Utilisez l'onglet Outils de zone de texte
pour modifier la mise en forme de la zone de texte de la citation.]
α-2,3,4,6-tetra-O-methyl-D-ose to be deduced (2 points)
(give its name) : _____________ (1 point)
α-1,3,6-tri-O-methyl-N-acetyl-D-galactosamine (2 points)
4
Using the permethylation results, deduce the structure of the oligosaccharide studied.
Indicate the type of osidic bond implicated in the connection of the different
monosaccharides knowing that they are all in the α anomeric form (10 points).
5