BMB401 F06 T4andT5 KEY

BMB401 F06 T4andT5 KEY - I, (Name) _ have neither received...

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I, (Name) _________________________________________ have neither received nor given help BMB 401/506-Fall 2006 Fourth Examination November 30, 2006 KEY During the exam you should not be in possession of a beeper or a cellular phone. Please place your beeper and/or cellular phone in your book bag (after you turn it off!!!) and place the bag against the wall of the room. THESE ARE STANDARD SINGLE ANSWER MULTIPLE CHOICE QUESTIONS. THE POSSIBLE ANSWERS ARE LABELED A,B,C,D AND E. SELECT THE SINGLE BEST ANSWER AND MARK THE APPROPRIATE CIRCLE ON THE SCANTRON ANSWER SHEET. BUBBLE IN YOUR NAME AND ID# (C# ONLY PLEASE) . There are 35 questions. You will receive full credit for 34 correct answers. 1. RNA molecules that exhibit catalytic activity are called A. mRNAs B. Ribonucleases C. Ribosomes D. Ribozymes E. Ribonucleotides 2. Which of the following is true about a circular double- stranded DNA genome that is determined by chemical means to be 21 percent adenosine? A. The genome is 10.5% guanosine. B. The genome is 21% guanosine. C. The genome is 29% guanosine. D. The genome is 51% guanosine. E. The base composition of guanosine in the genome cannot be determined from the information given. 3. Common lesions found in DNA after exposure to ultraviolet light are A. Pyrimidine dimers B. Single strand DNA breaks C. Base deletions D. Purine dimers E. Base insertions 4. Activation of transcriptionally silent genes by 5-azacytidine to treat prostate cancer is an example of. .. A. ...metabolic manipulation. B. ...manipulation of protein. C. ...modification of the genome. D. ...germline engineering. E. ...gene therapy. 5. Germline modification… A. …creates heritable genetic changes. B. …requires homologous recombination. C. …is the most common form of gene therapy. D. ...always repairs genetic diseases. E. …reactivates silenced genes. 6. Which of the following is most likely to lead to a loss of gene function? A. A missense mutation in the open reading frame B. A change from a TAA codon to a TAG codon in the coding region C. A change from T to C in the promoter region D. A frameshift mutation in the coding region E. A sequence change in the 3' untranslated region 7. An E. coli strain lacking DNA polymerase I would be deficient in DNA. .. A. B. ...methylation C. ...splicing D. ...degradation E. ...transcription 8. Which of the following DNA sequences is a palindrome? A. 5’– ATATTATA –3’ B. 5’ GTATATAC –3’ C. 5’– GAGTGCAC –3’ D. 5’– CTAGGATC –3’ E. 5’– CTATATAC –3’ 9. Drawn below is part of the sense strand of a gene. The DNA sequence shown encodes the last amino acids of a protein that is 380 amino acids long. The bracketed codon indicates the correct reading frame of this gene. Which of the following codons could be mutated by a single nucleotide substitution to extend the length of the gene to encode 381 amino acids? A.
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This document was uploaded on 11/04/2011 for the course BMB 401 at University of Miami.

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BMB401 F06 T4andT5 KEY - I, (Name) _ have neither received...

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