Download CHM 112

CHM 112
GAGE
SUMMER I 1999
1.
NAME _____Key____________________
DATE___________________________________
EXAM 4
Match the name of the compound in Column A with its classification in Column B.
__E__1. keratin
(16)
A. monosaccharide
B. disaccharide
C. polysaccharide
D. amino acid
E. polypeptide
F. enzyme
G. triglyceride
H. phospholipid
I. sphingolipid
J. glycolipid
K. steroid
L. prostaglandin
__C__2. amylopectin
__F__3. serine transaminase
__L__4. PGB2
__D__5. threonine
__H__6. lecithin
__K__7. estradiol
__B__8. maltose
2.
Supply the most appropriate word or phrase for each of the following statements.
a.____non-competetive_____
a.
b.____select any three from list_____
type of enzyme inhibition where a molecule alters
the active site of the enzyme from another point on
the enzyme
b.
three functions that proteins perform
___________________________
___________________________
c._____amino acid___________
c.
small biochemical that can act as a buffer
d. _____electrophoresis________
d.
technique for separation of proteins
e. _____fatty acid
e.
long, straight chain carboxylic acid
f.
two classes of substances that denature proteins
f. ___heat, UV light
________
_______
___heavy metals, pH change____
g. _substrate concentration, pH, tempg.
(20)
factor that affects enzyme rate
3.
Classify each of the following molecules as specifically as possible by type (i.e.
monosaccharide, phospholipid...) from the structure.
(18)
O
OH OH
H2N
CH
C
OH
H
H OH
O
H
CH 2
H
C
H
O
HO
O
OH
H
O
OH
HO
H
H
OH
H
H
NH 2
______amino acid______________
____________disaccharide____________
O
O
OH
CH 3
HO
H2
C
C
CH 3
H
N
CH
C
CH
N
H
CH 2OH
CH 3
CH 3
CH
CH
C
NH 2
O
HO
CH 3
C
HO
O
O
________steroid______________
______________dipeptide___________
O
C
O
O
H3C
C
(
CH 2)16
(CH 2)17CH 3
O
O
C
H2
O
C
H2
C
(CH 2)18CH 3
_________wax___________________
4.
CH
O
C
O
C17H31
(CH 2)16CH 3
__________triglyceride_____________
Identify the type of glycosidic linkage in each of the following carbohydrates:
H OH
H
H
O
HO
OH O
H
OH
O
H
HO
H
H
H
_______alpha-1, 4_______________
O
HO
HO
H
(6)
H OH
H OH
O
HO
H
H
OH
H
OH
H
H
H2C
H
O
HO
H
HO
H
H
OH
OH
________beta-1,6__________________
5.
Complete the following biochemical reactions with appropriate products.
O
O
H2N
CH
C
+
OH
H2N
CH
C
O
O
OH
H2N
CH
C
(15)
H
N CH
C
OH
CH 2
CH 3
CH 2
CH 3
CH
CH 3
CH
CH 3
CH 3
CH 3
O
H2C
O
(CH 2)16CH 3
O
C
CH2OH
CH
O
O
C
H
C (CH 2)14CH 3
O
O
H2O/H+
----------->
CH3
CH2CH 2
P
N
(CH 2)16CH 3
HO
C
HO
C (CH 2)14CH 3
CHOH
O
O
C
CH3
O
H
O
CH2CH 2
P
CH3
O
CH3
N
CH3
O
O
CH3
HOH 2C
O
C
C
O
HO
CH 3
NaOH
NR - non-saponifiable lipid
-------------------------------->
O
6.
a.
A polysaccharide has a molar mass of about 400 g/mol and is soluble in water
while a triglyceride with the same molar mass is highly insoluble. Explain why.
Lipids have large, non-polar hydrocarbon sections which are not attracted to water.
Carbohydrates have multiple hydroxyl groups that form hydrogen bonds easily with water
so the interactions, and thus the solubility, are greater.
(5)
b.
Explain how insoluble lipids can be transported through the bloodstream.
(5)
Lipids attach to transport proteins such as serum albumin. Since proteins have many side groups
that are polar, proteins are more easily dispersed/moved in a water environment. Protein acts as a
carrier.
c.
What is the tertiary structure of a protein and how is it maintained?
(5)
The tertiary structure of a protein is the “folding” or compacting of the coiled or pleated
structure. The tertiary structure is maintained by disulfide bridges (between amino acids with
thiol groups), hydrogen bonds (between polar groups), salt bridges (between amines and
carboxylic acid groups), and hydrophobic or dispersion forces (between non-polar groups).
d.
Explain from a chemical perspective why cellulose is indigestible while starch is
even though they are of comparable masses and are made of the same monomers.
(5)
Cellulose is held together with beta-1,4 glycosidic linkages. Human enzymes are designed to
attack mainly alpha linkages. Humans have no enzyme that will hydrolyze this linkage.
e.
From an energy perspective, explain how an enzyme is able to increase the rate of
a chemical reaction.
(5)
An enzyme allows a different route or mechanism for the conversion from reactant to product.
The enzymatic mechanism has a lower activation energy (energy barrier). Since less energy is
necessary to overcome the barrier, the reaction will occur faster or the rate increases.