Exam_2_Spring_2008_key

Exam_2_Spring_2008_key - 1 Wobble rules Base at 5 end of...

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Wobble rules Base at 5’ end of anticodon Base at 3’ end of codon G C or U U A or G A U C G I U,C,A 1
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Multiple choice (2 points each) KEY Shown below is the structure of a Drosophila gene, divided into 10 segments, designated A–J. The gene contains three exons, two introns, a promoter, and the endonuclease site. Use this for questions 1-3. 1. What segments of the gene will be represented in the initial RNA transcript (just after it is synthesized)? (If part of a segment will be represented, then include this in your answer). (1) CDEFHIJ (2) CDEFGHI (3) DEFGHI (4) DFHI (5)DFHIJ (transcription begins at the +1 and ends as the mRNA is cleaved by the endonuclease) 2. What segments of the gene will be found in the completely processed transcript? (1) CDEFHIJ (2) CDEFGHI (3) DEFGHI (4) CDFHI (5)DFHIJ (C will be the 5` UTR, D, F, and H are the exons that are spliced together and I = 3`UTR) 3. What segment of the gene will possess the translation initiation codon? (1) A (2) B (3) C (4) D (5) E The AUG will be the first codon, the beginning of exon1. 4. Which of the following is not a component of the transcription process in vivo ? (1) Shine-Dalgarno sequence (2) DNA (3) Promoter (4) . RNA polymerase (5) Hairpin loop This is part of translation initiation in prokaryotes. 5. What would be the outcome if mice injected with R cells from Avery’s experiment were also injected with heat-killed S cells incubated/treated with protease? (1) The mice die and live S cells would be recovered. (2) The mice live and live S cells would be recovered. (3) The mice die and live R cells would be recovered. (4) The mice live and live R cells would be recovered. (5) The mice die and neither R nor S cells would be recovered. Since proteins are not the genetic material, using protease will not block the transfer. 6. Suppose you isolate RNA polymerase (RNAP) from E. coli and you also purify E. coli genomic DNA. Then you mix RNAP and E. coli DNA in a test tube in the absence of NTPs. Upon treatment with the enzyme DNase, all the E. coli DNA that does not have RNAP bound to it will be degraded. The DNA that is left “protected” from the nuclease will most likely be: (1) The TATA box (2) the Shine-Dalgarno sequence (3) the -10 and -35 along with along with DNA in between the two consensus sequences (4) origin of replication (5) The AUG and nearby codons. RNAP binds to the -35 and -10 to initiate transcription. If there are no NTPs, there is no transcription. The RNAP will bind and stall. It is not “TATA” because this is eukaryotic. 2
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7. Assume you wish to radioactively label the phosphate backbone of DNA using an in vitro replication system. In a tube containing physiological buffer you mix: DNA that includes an origin of replication. The sequence of the DNA is:
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Exam_2_Spring_2008_key - 1 Wobble rules Base at 5 end of...

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