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UNIVERSITI MALAYSIA PERLIS
UJIAN 2
Sidang Akademik 2008/2009
16 OKTOBER 2008
ERT 101 – BIOKIMIA [BIOCHEMISTRY]
Masa: 60 minit
Please make sure that this question paper has FOUR [4] printed pages including
this front page before you start the examination.
This question paper has 2 questions. Answer ALL questions.
Question 1
1.
Explain the central dogma of molecular biology and describe in details each
processes.
[40 marks]
Transfer of biological information from DNA to RNA to protein is call central dogma of
molecular biology. DNA directs its own replication to produce new DNA. The DNA of a
gene is transcribed to produce an RNA molecule that is complementary to the DNA. The
RNA sequence is then translated into the corresponding sequence of amino acids to form
a protein [10 marks]
REPLICATION
DNA replication begins at a specific sequence of nucleotides called an origin. First, a
cell removes chromosomal proteins, exposing the DNA helix. Next, an enzyme called
DNA helicase locally "unzips" the DNA molecule by breaking the hydrogen bonds
between complementary nucleotide bases, which exposes the bases in a replication fork.
Other protein molecules stabilize the single strands so that they do not rejoin while
replication proceeds. After helicase untwists and separates the strands, a molecule of an
enzyme called DNA polymerase III binds to each strand. DNA polymerases replicate
DNA in only one direction - 5' to 3' - like a jeweler stringing pearls to make a necklace,
adding them one at a time, always moving from one end of the string to the other.
Because the two original (template) strands are antiparallel cells synthesize new strands
in two different ways. One new strand, called the leading strand, is synthesized
continuously as a single long chain of nucleotides. The other new strand, called the
lagging strand, is synthesized in short segments that are later joined. DNA ligase seals
the gaps between adjacent Okazaki fragments to form a continuous DNA strand.[10
marks]
TRANSCRIPTION
Initiation - RNA polymerases - the enzymes that synthesize RNA -bind to specific
nucleotide sequences called promoters, each of which is located near the beginning of a
gene and initiates transcription.
Elongation – Like DNA polymerase, RNA polymerase links nucleotides in the 5' to 3'
direction only. RNA polymerase unwinds and opens DNA by itself; helicase is not
required. RNA polymerase does not need a primer. RNA polymerase is slower than DNA
polymerase, proceeding at a rate of about 50 nucleotides per second. RNA polymerase
incorporates ribonucleotides instead of deoxyribonucleotides. Uracil nucleotides are
incorporated instead of thymine nucleotides.
Termination – self termination and Rho dependent [ 10 marks]
TRANSLATION
Initiation
The smaller ribosomal subunit attaches to mRNA at a ribosome binding site (also known
as a Shine-Dalganno sequence after its discoverers), with a start codon at its P site.
tRNA £Met (whose anticodon is complementary to the start codon) attaches at the
ribosome's P site; GTP supplies the energy required for binding. The larger ribosomal
subunit attaches to form a complete initiation complex
Elongation
The transfer RNA whose anticodon matches the next codon delivers its amino acid to the
A site. Another protein called elongation factor escorts the tRNA along with a molecule
of GTP. Energy from GTP is used to stabilize each tRNA as it is added to the A site.
Termination - Termination does not involve tRNA; instead, proteins called release factors
halt elongation. It appears that release factors somehow recognize stop codons and
modify the larger ribosomal subunit in such a way as to activate another of its ribozymes,
which severs the polypeptide from the final tRNA (resident at the P site). The ribosome
then dissociates into its subunits. [10 marks]
2.
Considering the following mRNA sequence:
5’ UAUA AUG CUC ACU UCA GGG AGA AGC
By referring to the Genetic Code in Appendix 1, answer the following questions:
a.
What polypeptide sequence would be translated from the mRNA?
Met-Leu-Thr-Ser-Gly-Arg-Ser [10 marks]
b.
Assume that nucleotide 8 (C) is deleted from the mRNA. What
polypeptide sequence would now be translated from this modified
mRNA?
Met-Ser-Leu-Gln-Gly-Glu [10 marks]
c.
What is the DNA sequence of the modified mRNA?
3’ ATAT TAC AGT GAA GTC CCT CTT CG 5’ [10 marks]
[30 marks]
3.
Protein may have up to four levels of structure. For each type, describe the kind of
bonding interactions that maintain that type of each structure.
Primary: Covalent; amide bond (peptide bond)
Secondary: Noncovalent; primarily hydrogen bonding
Tertiary: Noncovalent; hydrogen bonding, ionic, hydrophobic
Covalent; disulfide bonds
Quaternary: Noncovalent; hydrogen bonding, hydrophobic interactions, ionic
[30 marks]
Question 2
1.
Explain the principle of gel electrophoresis. How this technique can be used in
protein purification and DNA separation.
Gel electrophoresis is the technique that uses electric current to separate different
size molecules in a porous gel. Smaller molecules move more easily through the gel
pores than do larger molecules. In practical, A porous gel is often made from agarose (a
polysaccharide isolated from seaweed), which is melted in a buffer solution and poured
into a plastic mold. As it cools, the agarose solidifies, making a gel that looks something
like stiff gelatin. Small indentions called wells are made at one end of the gel to hold
sample solutions. Gel is submerged in a chamber filled with a salt solution that conducts
electricity. Samples and standard are loaded into slots made in the agarose (from
seaweed) gel.
Gel electrophoresis can be used to purify protein because protein-molecules are
electrically charged, hence they move in an electric field. Protein molecules separate
from each other because of differences in their net charge. Molecules with a positive net
charge migrate toward the negatively charged electrode (cathode). Molecules with a net
negative charge will move toward the positively charged electrode (anode). Standard
solution is normally used to determine the molecular weight of the samples.
Gel electrophoresis can also be used to separate DNA because the phosphate groups in
the DNA backbone carry negatively-charged oxygen- giving DNA molecule an overall
negative charge. In the electrophoresis chamber, the negatively-charged DNA moves
toward the positive pole. Small DNA fragments migrate more rapidly than do large ones
and, with time, the fragments separate on the basis of their size. The distance that each
fragment migrates depends on its size. Typically, DNA fragments of known length (a
marker sample) are placed in another well. By comparing the migration distance of the
unknown fragments with the distance traveled by the marker fragments, one can
determine the approximate size of the unknown fragments.
[30 marks]
2.
Compare between competitive inhibitors and uncompetitive inhibitors.
[20 marks]
A competitive inhibitor typically has structural similarities to the normal substrate for
the enzyme. Thus it competes with substrate molecules to bind to the active site. The
competitive inhibitor binds reversibly to the active site.
At high substrate concentrations the action of a competitive inhibitor is overcome
because a sufficiently high substrate concentration will successfully compete out the
inhibitor molecule in binding to the active site. Thus there is no change in the Vmax of
the enzyme but the apparent affinity of the enzyme for its substrate decreases in the
presence of the competitive inhibitor, and hence Km increases.
In uncompetitive inhibition, the inhibitor binds only to the enzyme-substrate complex,
and not the free enzyme.
Adding more substrate to the reaction results in an increase in reaction velocity,
but not to the extent observed in the uninhibited reaction. Uncompetitive inhibition is
most commonly observed in reactions in which enzymes bind more than one substrate. In
uncompetitive inhibition both Km and Vlmx are changed, although their ratio remains the
same.
3.
Consider the following reaction:
Glucose + O2
gluconate + H2O2
What is the enzyme that can be used to speed up the reaction? To which class is
this enzyme belongs to?
[20 marks]
The enzyme that can be used to speed up the reaction is “glucose oxidase”. This enzyme
belongs to oxidoreductase class. . In the reaction, glucose is oxidized to gluconate, and
oxygen is reduced to hydrogen peroxide
4.
Consider the following equation:
What is the name of this equation and explain each notation (Vmax, Km, S) in
related to enzyme kinetics?
[30 marks]
The name of the equation is : Michaelis Menten equation.
S= substrate
V = velocity
Vmax= Maximum velocity
Km= Michaelis menten constant, Km is considered a constant that is characteristic of the
enzyme and the substrate under specified conditions. It may reflect the affinity of the
enzyme for its substrate. The lower the value of Km, the greater the affinity of the
enzyme for ES complex formation.
Appendix 1