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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