Download CHM 20 EXAM 3 – REVIEW Name Ms Dang Indicate whether each

CHM 20
Ms Dang
EXAM 3 – REVIEW
Name
1. Indicate whether each of the following is D-glyceraldehyde or L-Glyceraldehyde, assuming that the
horizontal bonds point toward you and the vertical bonds point way from you
2. What sugar is the C-3 epimer of D-xylose? C-5 of Dallose?
D-Ribose and L - Talose
3. What other monosaccharide is reduced only to the alditol obtained from the reduction of D-Galactose?
4. Identify the following sugars:
An aldopentose that is not D-arabinose forms D-arabinitol when it is reduced with NaBH4
D-lyxose
A ketose, when reduced with NaBH4, forms D-altritol and D-allitol
D-psicose
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5. The reaction of D-ribose with one equivalent of methanol plus HCl forms four products. Give the
structures of the products and their names
6. What is the main structural difference between
Amylose and cellulose?
Amylose has -1,4’-glycosidic linkage, whereas cellulose has β-1,4’-glycosidic linkage
Amylopectin and glycogen
Glycogen and amylopectin have the same kind of linages, but glycogen has a higher frequency of -1,6’glycosidic linkages
7. People with type O blood can donate blood to anyone, but they cannot receive blood from everyone. From
whom can they not receive blood?
People with type O blood can receive blood only from other people with Type O Blood, because Type A,
B, and AB blood have sugar components that Type O blood does not have
8. Draw the form in which each of the following amino acids predominately exists at physiological pH (7.3)
a. Aspartic acid
b.
Glutamine
c. arginine
9. Draw the structural formula of the tripeptide met-ser-cys
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10. Draw the different ionic structure of this tripeptide at pH 2.0, 10.0.
11. What kinds of changes are necessary to transform a protein having a predominantly -helical structure into
one having a β-pleated sheet structure?
Both the α-helix and the β-pleated sheet are examples of secondary structure. These structures are held
together by hydrogen bonding between the amide carbonyl and the amide hydrogens. To change from one
to the other, the protein must become denatured so that the hydrogen bonds are disrupted. The protein will
then exist in a random coil state and may then form a new secondary structure as new hydrogen bonds are
formed within a chain or between chains.
12. How would a protein that resides in the interior of a membrane fold, compared with the water-soluble
protein?
It would fold so that its nonpolar residues are on the outside of the protein in tact with the nonpolar
membrane and its polar residues are on the inside of the protein
13. Explain why the following are not found in -helix: two adjacent glutamates, two adjacent aspartates, or a
glutamate adjacent to an aspartate
In each case, the two adjacent amino acids each have a negative charge. Two adjacent side chains will be close
to one another if they are in the helix, so they will repel each other if they have the same charge
14. Why does the body need so many different enzymes?
There are thousands of different kinds of biochemicals in an organism and thousands of
reactions are required to synthesize and degrade the compounds. The mild conditions
inside the cell are not conducive to fast reaction rates so each reaction requires an enzyme
for catalysis under physiological conditions.
15. Monoamine oxidases are important enzyme in brain chemistry. Judging from the name, which of the
following would be suitable substrate for this class of enzyme
Substrates for the monoamine oxidases usually have only one amino group. Therefore,
based only on the name of the enzymes, both (a) and (b) should be active substrates.
Compound (c) contains a nitro group, not an amino group. In actuality, the monoamine
oxidases also prefer primary and secondary amines and aromatic compounds as
substrates, so it is expected that (a) is the best substrate.
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16. What term do we use to refer to (a) the protein portion and (b) the organic molecule portion?
a) The protein portion of the enzyme is called the apoenzyme.
(b) The organic molecule portion is called a coenzyme.
17. At a very concentration of a certain substrate, we find that, when the substrate concentration doubles, the
rate of the enzyme-catalyzed reaction also doubles. Would you expect the same finding at a very high
substrate concentration? Explain
At very low concentrations of substrate, the rate of an enzyme-catalyzed reaction increases in a linear
fashion with increasing substrate concentrations. At higher concentrations of substrate, the active sites of the
enzyme molecules become saturated
18. The hydrolysis of glycogen to yield glucose is catalyzed by the enzyme phospharylase. Caffeine, which is not
a carbohydrate and not a substrate for the enzyme, inhibits phosphorylase. What kind of regulatory
mechanism is at work?
At first, one might expect the inhibition of phosphorylase action by caffeine to be a case of traditional
noncompetitive inhibition because the inhibitor apparently does not bind at the active site. However,
glycogen phosphorylase is known to be an allosteric enzyme and the classical terms competitive and
noncompetitive are not used to define inhibition. The term negative modulator is used to define the
action of caffeine on the allosteric enzyme, phosphorylase.
19. Which LDH isozyme is monitored in the case of a heart attack?
Lactate dehydrogenase, which is tetrameric, has two different kinds of subunits called H and M. Thus,
LDH exists as isoenzymes (isozymes). The subunit that dominates in heart cells is the H type. Therefore,
the isoform, H4, is monitored in the diagnosis of a heart attack.
20. Explain why transition-state analogs are potent inhibitors?
A transition state analog is a molecule that has a shape that fits the active site of an enzyme almost
perfectly. However, being an analog and not the real transition state, no reaction is catalyzed. Therefore,
the transition state analog acts as an inhibitor, and its perfect fit makes it a potent one.
21. Can the product of a reaction that is part of a sequence act as an inhibitor for another reaction in the
sequence? Explain
The regulation process described in the problem is called feedback inhibition and is a common
mechanism for controlling metabolic reactions. Consider the reaction sequence:
A
B
C
D
E
F
The letters refer to metabolites or intermediates that are produced in the pathway to synthesize
compound F. In feedback inhibition, compound F acts as an inhibitor for the enzyme that converts A to
B. This type of inhibition becomes important when enough F has been produced for the cell.
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22. Draw the following nucleotides
a. Deoxycytidine-3’-monophosphoate (dCMP)
(dTTP)
b.
Deoxythymine-5’-triphosphate
c. Uracil-5’-diphosphate (UDP)
23. RNA and DNA refer to nucleic acids. Which part of the molecule is acidic?
The name nucleic acid comes from the fact that the nucleosides are linked by phosphate
groups, which are the dissociated form of phosphoric acid.
24. What type of bond connects the base to the ribose in GTP?
The bond that connects the base to the ribose in GTP is a β-N-glycosidic bond
25. What are the higher order structures of DNA that eventually make up a chromosome?
DNA is wound around histones, collectively forming nucleosomes that are further wound
into solenoids, loops, and bands.
26. Draw the structure of adenine and thymine, and show with a diagram the two hydrogen bonds that stabilize
C-G pairing in nucleic acids
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27. When cytosine is deaminated, uracil if formed. Uracil is a naturally occurring base. Why would the cell use
base excision repair to remove it?
Deamination of the base cytosine transforms it to uracil which creates a mismatch. What
was a C-G pair now becomes a U-G mismatched pair that must be removed by BER.
28. What 12-nucleotide primer would you use in the PCR technique when you want to amplify a gene whose
end is as follows: 3’TACCGTCATCCGGTG5’
The 12-nucleotide primer to use is: 5’ATGGCAGTAGGC3’.
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