CH369 exam4s02 - Nome kin",— SOC. Sec # CHEMISTRY...

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Unformatted text preview: Nome kin",— SOC. Sec # CHEMISTRY 369 Spring 2©©2 Dr. G. Barrie Kih‘o EXAM 4 NOTE : Multiple Choice Questions are 2 points each Written Answer Questions are 10 points each Multiple Choice : Wrifien Answer HHH O‘CfibOJM—A TOTAL POINTS Written Answers 1. 6.) Draw the structure of the base uracil. 0 ll Hu / CAN \ l\ b) Is uracil a purine orQrimidine? (Circle correct answer) / - c) In DNA the base equivalent to uracil is “(kiwi m (1.; d) In DNA the base equivalent to uracil forms a hydrogen bonded base pair with i\ (\Cm tf) Q, resulting in l,@§, or it hydrogen bonds for the pain (Circle correct answer) e) In DNA the covalent linkage between the monomers involves formation of two amide, (ester: )Schiff, ether, tertiary, conjugated bonds with phosphoric acid. (Circle correct afisVer) 2. Briefly define the following terms as they apply to nucleic acid metabolism. a) Replication Replication is the duplication ofDNA, giving rise4 to a new DNA ' same base sequence as the original mojecme With the ‘7 3 pts .Xfi a p b) Transcription Transcription is the production of RNA from a DNA template. The base sequence Of the DNA 15 reflected in the base sequence of the RNA product 3 pts (2) c) Translation Translation is the process by which the base sequence ofan RNA molecule (mRNA) directs the synthesis of a protein in a manner such that the amino acid sequence is reflective of the base sequence of the RNA (and thence the DNA gene coding for it). 4 pts 3. a) Using the following genetic code table translate the following sequence of an mRNA into a polypeptide sequence. Note that AUG is a start codon and UAA, UGA, and UAG are stop codons. 5’ — UGAIJEGGCAUAGGUGAUCUGUAGCG — 3’ V --"- “ — LELL Orgllaq THE ASSIGNMENT OF THE 641RIPLET CODONS INTHE 5’ —> 3’ SEQUENCE OF MRNA Base at 5’ End Middle Base of Codon Base at 3' End of Codon i e -‘ .i: t or Codon i U c A “at” .' 1 -'~"G " phe (UUU) ser ryr cys U U phe ser Iyr cys C leu ser Termination Termination A leu ser Termination trp G leu pro his arg U C leu pro his arg C leu pro gln arg A leu r0 gln arg G ile thr asn ser U A ile Lhr asn ser C ile rhr lys arg A met (and [hr lys arg G initiation) val ala asp gly U G val ala asp gly C val ala glu gly A val ala glu gly G (3) b) Using the genetic code table describe what would happen if the first codon position for lysine was changed from an A to a G in an mRNA coding for a protein . An A to G transition would change the lysine AAA and AAG codons to GAA and GAG. both ol’which encode glutamate. 2. How do you think such a change might affect the structure of the protein? This would result in an acidic residue replacing a basic residue in the protein. If the change were on the protein surface, it might have no effect on protein structure or Function, but if the change involved a buried ion pair, a catalytic residue, or some Other essential residue, its effect might be significant 4. a) Protein biosynthesis involves the formation of aminoacyl-tRNAs. Very briefly describe the two-step process by which amino acids are attached to the tRNAs. Attachment of an amino acid to a tRNA is carried out in a two-reaction process by an aminoacylitRNA synthetase (aaRS). The first reaction involves “activation” of the amino acid by its reaction with ATP to form an aminoacyl‘adenylate with the release of PP,. The mixed anhydride then reacts with tRNA to form the “charged” aminoacyl—tRNA with the release ot‘AMP. The hydrolysis of pyropliosphate (PR) ensures that the overall reaction is irreversible. 1L 1. Amino acid + ATP aminoacyl—AMP + PP, AminoacyliAMl’ + IRNA aminoaeyl—tRNA + AMP Amino acid + tRNA + ATP : aminoacyl—IRNA + AMP + PP, b) In terms of the lac operon describe by means of a diagram and a paragraph of discussion what you think would happen to the activity of the operon when lactose levels are low and glucose levels are high. See below for diagram (see pp 298-301) 5 pts 2.5 pts When lactose levels are low the lac operon will not be induced. The repressor will be bound to the operator and a block RNA polymerase binding and therefore transcript High glucose will also block expression of the lac operon by catabolie repression. When [glu] are high [CAMP] will be low therefore CAP will not bind to promoter and will thereford decrease lac protein synthesis [A‘th r ‘Amx v\ J\ vwfl’jufioft‘i“ >8 hiatus ‘\X B fiend \\ Q, —\ ) T V if ts ion. 5 pts (Q) Li k-\\:*YI! \‘\l\Y1 Control sites structural genes Regulatory gene m f when [lactose] low Repressor binds to #0 control Site / Re pressor protein \, Catabolite Repression v ‘ CAMP CAD + cAMP M? CAP cAMP When glucose low CAP CAP site RNA polymerase structural genes 1‘ Site l l l 5“ ‘" ' i Only { CAFlc‘AMP combination binds to CAP site to enhance RNA polymerase bindin. *“* ~ ' g RMFA \C’Q‘A \D‘NL J‘\“" ‘ Hml'rfi; \‘ \"' (J~\\\"AHLL«\V\I~Al‘ 5. Describe how restriction enzymes that produce “sticky ends” can be used to insert a piece of foreign DNA into a bacterial plasmid so that large amounts of the foreign DNA can be rapidly synthesized in a bacterial host. Include in your answer a description of the techniques used to detect whether bacterial colonies contain the appropriate recombinant plasmid :J-PC. (a W MOL l0 \\0\ K .2 In ‘5 a 6599 \ mam \ a; . P‘m‘d vas: \s Nan UV ForeignDNA " ’ Cleave with same \ afflii’lfil'ffiflflfl; I ‘04:) BE; WM Q M Mix plasmid and foreign DNA; sticky ends bond forming recombinant DNA Seal gaps wuh DNA lignsc Recomhin an I. DNA FIGURE I0.5 The muhndolugy lor producing reA \ [mnlnlmr‘l DNA )( QB ‘ CQC; vxfi Pvlm‘k' (-x'lll'im (unvlm [Lgnlm InmLm- pludluuun Amp): .11”. n \l\I_|HU/ Fun-w] DNA wilh ~1irLV mm inwm-rl :u [fin gvm- inn B*§LII1|HH\‘II‘:|\A‘ puulmlmn Mix barn-rm] (‘rllx .mIl phxnmh \Lun Imrm ml gunmm' FIGURE I0.I9 “WHHL’ 1m I)“ IlllHlleJHlm ml Imr x-ign l)\ \ mm A |u.u m ml phan l’mnl l‘. n :]n mw ul rh me‘ u! M:- plxlmml In llu' [mun 11ml mmm: ll\l'(l [u uhmm 1hr I)\ \ In hr' My Fuwign DNA I!I\("Il(‘(l l‘mcign D\';\ xml illSk‘lH‘d No pluumd in u-ll M'I'lt'd‘ Immlmmlwnnt ('1?th mm Hum |>.u lx'lm nu l(‘ll2\( u llm'~ m .unpuIlhlnunminmg mwlm DNX mln flu |}1A\HllllHMHII\.4[!\V‘K Wm... bywnlmnultw m mm lln- u<~m~ lm AlllpH IHm HWHHHI L' umln-uwl, («1]me KIHHUI mqlnn Nahum ‘1 plmnml .1-r wmmu m [hx‘ 41min _ (*(mlniniug rcugvnl llLu run ,umlm- _4 bllu~ ('HIUI' ullt‘H [mrl (In HUI pnuluu- Mm (nlnlnrx (Mk l|1.I| .uqum- A NA» / K§: _ / , é ' Alnlmilhn- at”: I containing V _ mulluln mlrl \mhnnl impugn I)\.\ 'Mr H’NIV mm 1m IIH' AHHIHIHH x JIM] pumiuu a hlm- ( nlul. \xhvu ‘IN « vHx Ih‘vl m (1mm :I menul mm Iln- 1m mun [IV \ ‘uw l('\i\l}ll1l In Amp“ \Elm .mri (In um ( L'”~(1()ll()l puuhm- (.(‘I]\‘1llk‘ Imlsmm (wllx .ll‘l' wnmm‘ m lelw.“ ‘l NIW (HIM bhw mlm lmx arc m .mnlnnnr and .mulmun and (In HUI 7 wwmm lu .Impiulhn pmrluu' hhu- mlur pllldllK'l' bluv mlor u Wm‘ 6. a) What is meant by the statement that DNA is replicated in a “semiconservative manner”? Contrast this mode with conservative replication. See p 275—276 In semiconservative replication a parent DNA molecule (double helical) gives rise to two daughter molecules, each of which contains one of the parent strands and a newly synthesized daughter strand. % t + In a conservative mechanism the daughter molecule would consist of 2 newly synthesized strands. (é ‘9 § * ,wa b) List- briefly describe 2 molecular changes that take place in the processing of IIIIQIA‘ (PAKJ‘Y finch Emu/1&8 QJ‘ (fr/Lkmsixdst-QKB semiconservative Any two ofthe following (see p 295-296) 1) capping : addition of a modified guanylate residue to the 5’ end of eukaryotic mRN A 2) polyadenylation : addition of a poly A sequence to the 3’ end of the mRNA 3) splicing of coding sequences (eukaryotes only) = removal of intervening sequences (introns) and linking together of the exons ( = coding regions) 4) alternative splicing mechanisms (not covered in class but OK for an answer 5) trimming : removal of sequences from either the 5’ or 3’ end Multiple Choice 1. In DNA biosynthesis the monomers involved in the polymerization reaction are a. purine and pyrimidine bases b. nucleoside monophosphates c. deoxynucleoside monophosphates d. nucleoside diphosphates @deoxynucleotide triphosphates 2. The filndamental differences between RNA and DNA are a. they differ in only the bases b. bases, ribose units, and the phosphodiester linkage c. bases, ribose units, and the glycosidic bond type bases and the ribose units only 3. What will be the Trrl or melting of DNA that has a cytosine plus guanine content of 30% compared to that of DNA that has a cytosine plus guanine content of 50% if both are heated under the same experimental conditions? a. higher @ lower c. the same d. impossible to determine from the information given 4. One of the most important ways in which eukaryotic DNA differs from that of prokaryotes is a. prokaryotic DNA is complexed to proteins whereas eukaryotic DNA is not @eukaryotic DNA is complexed to proteins whereas prokaryotic DNA is not 0. DNA synthesis in eukaryotes takes place in the opposite direction from that in prokaryotes d. there is no requirement for a primer in the synthesis of eukaryotic DNA 5. The genetic code is said to be degenerate. This means that a. each codon codes for more than one amino acid b each anticodon can interact with many different triplet sequences in the mRN A. which may differ in any or all of the three nucleotides @nany of the amino acids are coded for by several different codons d. the code is universally used by virtually all species 6. Protein synthesis in prokaryotes always starts with a. a methionine residue 3 formylmethionine residue c. a cysteine residue d. no specific residue 7. There is no codon for the amino acid hydroxyproline, but this amino acid is a prominent feature of collagen structure. Which of the following is a likely explanation? a. Hydroxyproline is substituted for praline afier translation by a cut- and patch mechanism. @Proline is covalently modified to give hydroxyproline afier translation. 0. There is an alternative mechanism for synthesis of proteins that contain hydroxyproline. d. It is not possible to form a hypothesis from the information given. 8. RNA is hydrolyzed in basic solution but DNA is not. This occurs because a. thymine is found in DNA, but uracil is not b. DNA is double stranded, while RNA is single stranded ©DNA contains a 2’-deoxyribose, while RNA does not d. RNA has modified bases, while DNA does not 9. The sequence of bases in the part of a complementary strand of DNA that pairs with 5'- adenine - cytosine - adenine - guanine -3' is 3l_ _ _ _ _5| (i) (ii) (iii) (iv) a. cytosine — adenine - cytosine - thymine b. guanine - thymine — thymine - cytosine c. thymine - adenine - thymine - cytosine (d) thymine - guanine — thymine — cytosine 10. In double-stranded DNA, hydrogen bonding between the bases on the two strands typically occurs between: a. adenine and guanine @ adenine and thymine c. adenine and cytosine d. adenine and uracil e. adenine and inosine 11. Transfer RNAs are involved in: a. carrying nucleotides to the ribosome for messenger RNA synthesis Cb) carrying amino acids to the ribosome for protein synthesis c. carrying amino acids to peroxisomes for degradation d. carrying glucose residues to the Golgi apparatus for polysaccharide synthesis e. carrying nucleotides to the nucleus for coenzyme biosynthesis 12. This molecule is a substrate analog for: a. dATP b. dGTP cdeTP @dTTP 13. The significance of 64 possible codons is: a. The majority of codons are nonsense and do not code for any amino acid. b. The code can overlap yet always encode an amino acid. Co) The majority of codons code for some amino acid, since each amino acid can be coded by several triplets. d. Commas are present to indicate the appropriate grouping of triplets. 14. Restriction endonucleases, such as EcoRl, cleave: a. by removing purine bases fiorn the polynucleotide chain b. by removing pyrimidines from the polynucleotide chain c. only when it is single-stranded @} by breaking phosphodiester linkages in double-stranded DNA 6. only when a cell dies 15. In order to be able to isolate plasmids containing a foreign gene of interest. cloning vectors usually contain a "selectable marker. " These selectable markers are ofien: inverted repeats restriction sites origins of replication . telomeres @ antibiotic resistance genes more 16. DNA ligases are used to: a. cleave double-stranded DNA at specific restriction sites @join the ends of foreign DNA to the free ends of cleaved plasmid DNA c. cleave single-stranded segments of DNA fragments to yield blunt ends d. convert DNA sequences to RNA sequences e. transcribe RNA sequences to DNA sequences 17. How is human insulin produced by genetically engineered bacteria? a. this procedure cannot be done bi DNA that codes for the insulin precursor is introduced into bacterial plasmids c. DNA that codes for both the polypeptide chains is introduced into a single population of bacteria @DNA that codes for each of the two polypeptide chains is introduced into two difierent populations of bacteria 18. The biological role of restriction enzymes is to: a. aid recombinant DNA research b. restrict the size of DNA in certain bacteria degrade foreign DNA that enters a bacterium d. restrict the damage to DNA by ultraviolet light e. make bacteria resistant to antibiotics 19. In the polymerase chain reaction Q) it is possible to amplify small amounts of DNA without cloning. b. conditions must be carefully controlled to prevent explosions. c. reaction mixtures must be kept chilled at all times. d. all of the above 20. A feature of the Sanger-Coulson or dideoxy method for DNA sequencing is a. chemical modification of all bases b. no need to use gel electrophoresis c. use of triple-helical DNA @ the use of DNA synthesis ...
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This note was uploaded on 03/30/2010 for the course CH 369 taught by Professor Kbrowning during the Spring '07 term at University of Texas.

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CH369 exam4s02 - Nome kin",— SOC. Sec # CHEMISTRY...

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