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Exam1_key_F2009 - BioC 4332 Fall 2009 Exam l — October 6...

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Unformatted text preview: BioC 4332 Fall 2009 Exam l — October 6, 2009 100 points total NAME KE Z ID# PLEASE NOTE THAT THIS EXAMINATION HAS 10 PAGES CONTAINING 32 QUESTIONS. Multiple Choice Questions Circle the best answer. (2 points per question) 1) DNA melts at a lower temperature if: a. AT base pairs are replaced by GC base pairs. b. the DNA is made longer. c. the DNA is composed of simple repeats. mismatches are added to the DNA. e. the concentration of sodium ions is increased. 2) Which of the following statements describing B DNA is FALSE? a. right handed b. base stacking contributes to its stability c. 3.4 nm per helical turn .20 bp per helical turn e. 146 bp wrapped around histone octamer core 3) Nearly half of our DNA is composed of: a. BXOHS. b. introns. c. simple repeats like those found at centromeres and telomeres. retrotransposons. e. retroviruses. 4) An ORF: a. contains no stop codons in any reading frame. b. contains introns. @shows the codons in DNA form. C]. is the entire sequence of a cDNA. e. are always on the same strand of DNA. 5) Which statement describing LINE transposition is FALSE? 3. requires a string ofT residues in the genome b. requires reverse transcriptase ©requires the involvement of SINE genes d. requires transcription of an existing element e. increases the number of genomic copies 6) Which statement describing gene families is TRUE? a. coded on the same DNA strand b. coded on the same DNA molecule 0. are orthologs .can expand in number by unequal crossing over e. all copies are transcribed and translated into functional products 7) DNA twists around a histone octamer core: a. 2 1/2 times. clockwise. c. with electrostatic interactions between the DNA and glutamic and aspartic acid residues of the proteins. d. in the A form. a. contacting only H3 and H4 and not H2a and H2b. 8) Which of the DNA sequences is not a palindrome? a‘ ATTAAT ATTTTA c. ATATAT d. AAATTT e. AATATT 9) Which enzyme does not utilize ATP hydrolysis for energy? @ Topoisomerase l b. Topoisomerase ll c. DNA Helicase d. DNA Ligase e. SWIISNF chromatin remodeling complex 10) Which of the following proteins does not localize to double strand breaks in the DNA? d. MRN complex e. Rad51 11) ATM phosphorylates all of the following EXCEPT: c. Chk2 (Checkpoint kinase 2) d. H2AX e. p53 my 12. — 16. (2 points for each answer) Match the polymerase with a descriptor. You may use only one letter per answer although you can use a letter for more than one answer. 12. ‘F DNA polymerase 6 a. does not need a primer b. needs neither a primer nor a 13. reverse transcriptase template 0. carries its own template 14. (12’ primase d. used to make cDNA 15. a E. coli DNA polymerase I e. removes primers from Okazaki fragments 16. C/ telomerase f. synthesizes Okazaki fragments g. polymerizes over damaged bases 17. - 20. (1 point for each answer) Fill in the appropriate description. 17. The hydrolysis of the N-glycosidic bond leaves a(n) mind HAL) ‘5in - f (11533.; :gagm mazfls that is repaired error-free by the bJ-se 41:;di 9m repair system. oltwwI/S 18. UV light createsT—TCX'Mu/s {firimdme in DNA that are repaired error- free by the Vladwhu wagon repair system. 19. X- radiation causesdg able g-(L‘a lg JDZLKLESn DNA that are repaired (amass every 3 error-free by that! meII/Iodn Ol/ULL repair sy 20. Gaps in DNA are repaired by the two enzymes-D39: {20% WLLU‘M and ML lL%E>Q§b . 21. — 24. (2 points for each answer) Fill in with either 5’-3' or 3’-5' or both. 21. 5f- ‘5 I direction of hydrolysis by proofreading exonuclease 22. SI "' 5" direction of polymerization by DNA polymerase 23. S/ ”'5’ direction of hydrolysis by primer removal exonuclease ! . . . . . 24. SI *‘ E dIrectIon of travel by a helIcase enCIrclIng the lagging strand template 25. (2 points for each answer) Starting from a replication origin that has been licensed to fire in a eukaryotic nucleus, list six enzyme catalyzed steps that constitute making the second Okazaki fragment and joining it to the first. D0152 (me/hm LMLM Erbium: ) :lzl/‘lmm fillr 99%me g Igflk Pg'gimasc g ELAOF meb A. I KEY 26. -— 29. (12 points total) In the following open reading frame give a biochemical (change in DNA) (1 point) and a genetic (change in amino acid sequence) (2 points) descriptor to each change. (The descriptors single base change or SNP cannot be used for an answer.) ATG TAT CGC AAA CAG 123 456 789101112131415 26. Change of T6 to a G Mme/ma n hon$éflggd (biochemical) (genetic) 27. Change of C7 to a T-Jnr‘am'sa. fig fl Ml “De—14g g (biochemical) g(enetic) 28. Change ofCQ toaT :IMQLJVLuQ Eagle”? Zg'no Manger: (biochemical) (genetic) eel/I Cr)- 29. Removal ofA12 deg; *3 gm. ( mdel ) (biochemical) (genetic) 131 2nd 3rd position position posilion (5’ end) (3’ end) i U C G i Phe Ser Cys Phe Ser Cys Leu Leu Ser Trp Leu Pro Arg Leu Pro Arg Leu Pro Arg Leu Arg Ile Ser Ser Ile Arg Met Arg Val Gly Val Gly Gly Val Gly KEV 30. (10 points total) (8 points) Draw the chemical structure of 5’-3’ GC. (2 points) Why is this sequence complementary to itself? \ O Afr-o A 3 ol // \H/\NH- O l xP’fiO ”Hi «*0 l l\\ O W Th 1 4H2, 0 \M/\\Q l/ \/ (I) Beams; W \ CQWW'F 0% st- G—C—B’ rP‘L—‘O ' I / ,0 ( l5 5 «CG-pg 0" O g’«QC——5’- M Swm am O’N—‘JE’EQM' 8 WMOM_ KEY 31. (12 points total) The diagram shows a portion of chromosomal DNA encoding a gene with two exons (dark boxes). Distances between portions of the DNA are shown In kb. The heavy lines with arrows are Pstl sites. (6 points) Using the box to represent the cDNA, indicate the total length of the cDNA and the map positions of Pstl sites. (6 points) In the gel electrophoresis lane show the sizes in kb of the fragments detected in a Southern blot of genomic DNA hydrolyzed with Pstl and probed with the cDNA. l1.51/53..01101 l15t éé::-:§é direction of electrophoresis Key 32. (6 points) The nucleoside shown below is AZT (azidothymidine), one of the first HIV drugs. Once transported into a cell, AZT will be made into the triphosphate and DNA polymerase will incorporate it into DNA. Its incorporation inhibits further polymerization. Why? 0 CH /”\ Agl l5; lmwaov‘aM 6&3 wmrm 3‘ NH I li/H'O b015- OWLL 0 vaM 12.4, LNAO P “i s 10 ...
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