-1-BIOS 100, Cells and Organisms Name:Exam III, 6 November, 2010Michael Muller, InstructorThis exam consists of 42 very simple questions lovingly dispersed over the following eightpages. Each question is worth 4 points apiece. The exam is out of 150 points. Bubble in yourlast name first. Choose the best answer for each question. Good luck!Use the following information on GFP and IPTG to answer questions 1 - 3Green florescent protein (GFP) is a protein which will glow green when it is activated. Abioengineer has linked the activation of this protein to activation of the lac operon so that whente lac operon is active at any level, the bacteria will glow green. IPTG (Isopropyl â-D-1-thiogalactopyranoside) is a chemical which can bind to the allosteric siteof the repressor protein of the lac operon and cause the same conformation change as lactose. IPTG is not broken down by â-galactosidase (or any other enzyme relative to this problem). 1. You have four E. coli cultures indicated below, how many of the cultures will have activeGFP and be glowing green?Culture 1: No lactose or glucose Culture 3: Lactose present, no glucoseCulture 2: No lactose, glucose present Culture 4: Lactose and glucose presentA. One cultures B. Two cultures C. Three cultures D. Four Cultures2. You now create a mutant E. coli which lacks an operator. You create four cultures of thismutant E. coli as seen in question one. How many of the cultures will have an activeGFP and be glowing green?A. One cultures B. Two cultures C. Three cultures D. Four Cultures3. You have two cultures of E. coli, one in which you add IPTG and one in which you addlactose. Neither culture has glucose added. What will be the experimental observationsof these two cultures?A. Neither culture will have active GFP.B. The IPTG culture will have constant GFP activation, the lactose culture will not haveactive GFPC. The lactose culture will have constant GFP activation, the IPTG culture will not haveactive GFPD. Both cultures will have initial GFP activation. The GFP in the lactose-added culturewill express constant activation, the GFP in the IPTG-added culture willeventually become inactive.E. Both cultures will have initial GFP activation. The GFP in the IPTG-added culturewill express constant activation, the GFP in the lactose-added culture willeventually become inactive-2-4. Which of the following statements are true about an E. coli which possessed a mutation sothat CAP protein could never bind to cAMP?A. This E. coli would never express the lac operonB. This E. coli would only express the lac operon if lactose and glucose were bothpresentC. This E. coli would only express the lac operon if lactose were present and glucosewere absentD. Expression of the lac operon in this E. coli would only be regulated by lactose;glucose concentration would not increase or decrease expression of the lac operonE. All of the above statements about the lac operon are FALSE5. What would happen to the trp operon if the repressor protein were mutated so that it wouldpermanently bind to tryptophan?A. The trp operon would be permanently turned onB. The trp operon would be permanently turned off.C. The trp operon would be turned on, but could be turned off if external tryptophanwere addedD. The trp operon would be turned of, but could be turned on if external tryptophan wereaddedE. None of the above.6. How many of the below steps of eukaryotic gene expression take place in the nucleus?Chromatin remodeling Translational controlTranscriptional control Post-translational modification of Post-transcriptional control and RNA processing proteinsA. One B. Two C. Three D. Four E. Five7. What would happen to the length of a protein if the introns were not removed?A. The protein would be larger B. The protein would remain the same sizeC. The protein would be smaller D. None of the above8. How would removing the silencers which act upon a gene producing the B-subunit ofhemoglobin in a worm affect gene transcription?A. It would decrease or possibly stop transcription of this geneB. The gene would be transcribed, but at a lesser rate than that of an unmutated geneC. The gene would be transcribed. The maximum rate of transcription in the mutatedorganism would be the same as the maximum rate of transcription of anunmutated organismD. The gene would be transcribed. The maximum rate of transcription in the mutatedorganism would be the greater than the maximum rate of transcription of anunmutated organismE. None of the above-3-9. The ë repressor protein of a bacteriophage has a helix-turn-helix motif. What can you sayabout this protein?A. It is a transcription factorB. It is a repressor protein for an operonC. It accelerates the rate of binding of RNA polymeraseD. It unwinds DNAE. It binds to DNA. That is all you can say about it.10. Which of the following is not a component of post-transcriptional control in eukaryotes?A. Removal of introns B. HAT activityC. Addition of a 5' G cap D. Addition of a 3' poly-A tailE. All of the above are examples of post-transcriptional control in eukaryotes11. What is the ploidy of the cell shown below:A. 2n = 4B. 2n = 6C. 2n = 8D. 2n = 12E. 2n = 1612. Which of the cells below properly illustrates a 2n=4 cell in Metaphase of Mitosis?A. B. C. D. E. 13. Which of the cells below properly illustrates a 2n=8 cell in Metaphase II of Meiosis?A. B. C. D. E.-4-14. Which of the cells below properly illustrates a 2n=4 cell in Anaphase I of Meiosis?A. B. C. D. E. 15. When do the homologous chromosomes separate during meiosis?A. Metaphase I B. Anaphase I C. Metaphase II D. Anaphase II16. Which of the following statements (A-D) about non-disjunctions is FALSE? If statementsA-D are true, then choose E.A. Non-disjunctions result when chromosomes do not separate properly during meiosisB. Organisms formed from fertilization involving a gamete with a non-disjunction arerare since these organisms usually die in development.C. A non-disjunction which occurs on chromosome 21 during anaphase I will have adifferent outcome as a non-disjunction which occurs on chromosome 21 duringanaphase IID. The rate of meiotic events with a non-disjunction increases exponentially with age inwomen but not in menE. All of the above statements about non-disjunctions are TRUE17. What is the function of crossing over?A. To increase genetic diversity in the gametesB. To ensure