Chem 142C_Midterm1_2009_Key

Chem 142C_Midterm1_2009_Key - Name:_Perm #_ Chem 142C/242C...

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Name:_________________________Perm #__________ Chem 142C/242C April 21, 2009 Midterm #1 Problem Points/Possible 1 /10 2 /10 3 /5 4 /5 5 /7 6 /5 7 /12 8 /6 9 /8 10 /5 10 /5 10 /5 10 /14 10 /8 _______________ TOTAL /100
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1) Briefly (1-2 sentences) describe the biochemical role of the following players in methyl-directed mismatch repair in E. coli . (10 pts) Rec J nuclease degrade strands of DNA in the 3’-5’ direction Mut H (when it is complexed with MutL-MutS) cleaves the unmethylated strand GATC Sequence The sequence at which MutH cleaves the unmethylated strand DNA Helicase Unwinds the DNA so that the exonucleases can degrade the unmethylated strand. Exonuclease VII degrades strands of DNA in the 5’-3’ direction 2) Make a diagram of the reaction catalyzed by DNA polymerase that occurs between deoxyribose at the end of a DNA chain and the 5 ' phosphates of a deoxyribonucleoside triphosphate. Include the chemical structure of the phosphate group, indicate the locations of the sugar and base, and show the rearrangements of electrons that occur. (10 Points)
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3) Briefly explain how DNA polymerase is able to discriminate against incorrect base pairing. (5 points) Isostericity: G and A are isosteric (superimposable) and T and C are isosteric. The geometry of each of the correct base pairs can be detected by the active site of DNA polymerase. 4) What are transposons and why are they so tightly regulated? (5 points) A method of recombination, transposons are segments of DNA that move, or “jump”, from one place on a chromosome to another. Unlike homologous recombination, transposition does not inlvolve DNA sequence homology and can insert into a new location at random. If a transposon inserts into the middle of an important gene encoding for an essential protein, this can kill the organism. This is why it is so tightly regulated. 5) Briefly explain the difference between base-excision repair and nucleotide-excision repair. (7 points) Base excision involves removing only the defective base from the DNA by cleavage of the N -glycosidic linkage of the base to deoxyribose. This leaves an apurinic or apyrimidinic site, which must then undergo additional repair processes. Nucleotide excision involves removing the defective base together with its deoxyribose and phosphate (as well as some neighboring nucleotides) by cleavage of phosphodiester bonds in the DNA chain. 6) Rank the terms below on a scale from 1-5. One being the least compact and five being the most compact (5 points) Nucleosome __1_____ 30nm Fiber___3____ Chromatid ___5____ “beads on a string” __2____ Rosette____4___
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7) Calculate values for the following topological properties of a closed-circular DNA molecule containing 2,500 base pairs (for simplicity, assume there are 10 base pairs per turn in the relaxed DNA). (12 points) (a) The linking number when the DNA is relaxed Ans: 250 (a) The linking number when the DNA has been underwound by 10 enzymatic turnovers of DNA
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Chem 142C_Midterm1_2009_Key - Name:_Perm #_ Chem 142C/242C...

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