-1-BIOS 100 - Summer, 2011Exam III, 3 August, 2011Michael Muller, Instructor This exam consists of 40 questions. Each question is worth five points. A codon chart isprovided for your convenience on the last page. Please check to see that all the pages are presentbefore you begin. Use a#2 pencil and bubble in all answers. Choose the best answer for eachquestion. Your score will be posted on the UIC Blackboard site as soon as they are in, butnot before the final, so please mark your answers on the test book and check against thekey to determine your score. Only answers on the scantron will count, though (so be careful anddon’t make mistakes when bubbling in answers!) Good luck!1. What were the results of the Hershey-Chase experiment?A. Radioactivity was detected in the pellet of the P sample only32B. Radioactivity was detected in the liquid of the P sample only32C. Radioactivity was detected in the liquid of the S sample only35D. Radioactivity was detected in the pellets of the P and S samples32 35E. Radioactivity was detected in the pellet of the P and the liquid of the S samples32 352. Which scientist(s) are incorrectly matched with their contribution to science?A. Watson & Crick - first published the correct three-dimensional structure of DNA B. Frederick Griffith - determined the molecular mechanism of bacterial transformationC. Rosalind Franklin - X-ray crystallographer who worked on the structure of DNAD. Meselson & Stahl - demonstrated that DNA replication was semi-conservativeE. All of the above scientists are correctly matched with their contribution to science3. Which of the following statements (A-D) about DNA replication is FALSE? If statements A-D are true, then choose E.A. There are multiple points of replication origin found in both prokaryote and eukaryotegenomesB. Leading and lagging strands are absent in prokaryotesC. Ligase is more active in the lagging strand than in the leading strandD. Topoisomerase breaks the phosphodiester bond in a DNA polymerE. All of the above statements about DNA replication are TRUE4. During a round of DNA replication, two nucleotides of DNA are mismatched such that thereare two purines opposite each other. Which of the below statements is true?A. There will be a bulge in the double helix because the two purines are larger than thenormal pairings of nucleotide basesB. There will be a depression in the double helix because the two purines are smaller thanthe normal pairings of nucleotide basesC. There will not be any difference in how far the sugar-phosphates are separated, butthere will be a twisting in the backbone that is not normally presentD. None of the above are true-2-5. Which of the following statements (A-D) about telomeres and telomerases is FALSE? Ifstatements A-D are true, then choose E.A. Telomeres are only found in eukaryotes with linear chromosomes - they are not foundin bacteriaB. Telomerase contains a built-in RNA primerC. Telomerase is a DNA polymerase which synthesizes DNA in a 3' to 5' directionD. Telomerase activity is limited to non-existent in most human somatic cellsE. All of the above statements about telomeres and telomerases are TRUE6. Which of the following results would you have expected after two rounds of replication hadDNA replication been conservative?14/14 ------------------------------------------------------------------------------------------------------15/14 ------------------------------------------------------------------------------------------------------15/15 ------------------------------------------------------------------------------------------------------A B C D E7. Which enzyme is incorrectly matched with one of its functions?A. DNA pol III - synthesize DNA in both the leading and lagging strandsB. DNA pol I - removes RNA primersC. Ligase - repairs broken phosphodiester bonds in a DNA polymerD. Topoisomerase - removes supercoils created by helicaseE. Telomerase - Synthesize RNA in a 3' to 5' directionUse the following non-template strand of DNA to answer questions 8-11.5' C C A G A A T G C G A C A T A A A G C G T A A A T C G C C G T 3'8. What is the third amio acid translated from the mRNA produced from this gene?A. Cysteine B. Threonine C. Valine D. Histidine E. Isoleucine9. How many amino acids are in the protein translated from the mRNA produced from this gene?A. Four B. Five C. Six D. Seven E. Eight-3-10. What will happen if a G is inserted in between the A & T which are underlined (i.e. what willbe the effect on the protein produced?)A. There will be a frame-shift mutation and the protein will have at least nine amino acidsB. There will be a frame-shift mutation which causes a premature ending of the proteinC. The insertion affects the start codon, so no protein will be transcribedD. There will be a change in a singe amino acidE. There will be no change in the translated protein11. What is the third nucleotide in the mRNA produced from this gene?A. A B. G C. C D. T E. U12. What is the amino acid attached to the tRNA with the following anticodon: 5' GGU 3'A. Glycine B. Tryptophan C. Proline D. Threonine E. None fo the above13. Which site of the ribosome contains a tRNA connected to the growing polypeptide chainafter translocation?A. The P site B. The A site C. The E site D. The X site14. In a mRNA, what is the function of the nucleotides before the start codon?A. There is no reason why there are nucleotides before the start codon, there just are. Dealwith it... B. The nucleotides before the start codon protect the start codon from being eroded orknocked off during transport from the nucleus to the cytoplasmC. The nucleotides before the start codon help the mRNA bind to the ribosome in theproper locationD. The nucleotides before the start codon contain additional start codons in case there is amutation in the typical start codon15. What is an exome?A. The roughly 1% of a person’s genome actually translates into proteinsB. A repeating unit of DNAC. The portion of a mRNA which is transcribed after the introns are removedD. A region of a promoter where transcription factors bindE. None of the above16. What would happen in an E. coli if there were a mutation in the operator so that therepressor protein would not bind to it if the allosteric site of the repressor protein wasopen but the repressor protein would be able to bind to the operator if lactose were boundto the allosteric site of the repressor