Final Exam - BBMB 405 Final Examination Closed Book Part(80...

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Unformatted text preview: BBMB 405 Final Examination Closed Book Part (80 pts) May 3, 2007 Name 1. (6 pts) Explain the role of 01f in the signal transduction cascade that results in our detection of hydrogen sulfide that is release 'nto the atmosphere at sewage plants. aflmfi Mama AWHM JAM W 2. (4 pts) Explain the role of the small intestine in the synthesis and transport of retinal to your retina m from dietary beta—carotene. M fi.‘ EWW «kW W W RMM _=, M f3 M 3 ‘ ’ LL 3 OWN”. W 3. (3 pts) The GpC “boxes” in the 1 kb upstream region of the globin gene in skeletal muscle is highly methylated whereas the same region in reticulocytes (immature red blood cells) is methylated significantly less. What role does this observation suggest for the methylation of GpC boxes. MW chwavflfiemm 4. (4 pts) Coactivators regulate the action of several transcription factors by altering the binding of histones to DNA and thus the remodeling of chromatin. Explain the suggested molecular mechanism of this effect of coactivators. HAM (+3 a w J 1“ >W‘w... We MALL)” chow/21.: Cal 5. (5 pts) A chronic change from low to high iron status in your b0 will result in film—i rates of ferritin synthesis and esser or greater lesser or greater rates of transferin receptor synthesis in your liver. Justify your selections. T7":a Whit é? T / $KE~50° S W 3.2. MAS/Z; j J] wade/2:» . . Legs-g . ' . * 1W 6. Differentiate betweenTcell receptors and antibodi accordin to the follo 1ng roperties: a. (4 pts) Secretion from the synthetic cell. T034? we. MW‘ ‘ b. (4 pts) Invol ement of protein of the major histocompatibility complex (MHC) in the recognition of isolated foreign macromolecules. 24% WQMW f“ c. (4 pts) Recognition of native (intact) for ules. 11% T W * J 7. State the major biochemical role of the following in the humoral immune response. a. (3 pts) Plasma cell b. (3 Pts) Thelper cell M Z ’ 7 W ' (W W 5 W ~ 0% I a W . g 8. (3 pts) The concentration of estrogen IS usua ly grea rig—its targe issu th n in nontarge t‘ sue . IQ“ a 2 Why? True or False 9. (4 pts) The complement system facilitates removal of microorganisms from the blood. 7 Ag 1 Defend your answer. W I 1t (mg/4.24, W \AWWC “WW -WfiWWWjflfi/flgw l. 22 Antibody a. Produces monoclonal antibodies 2. Epitope b. Major antibody in external secretions 3. M2. Cytotoxic cell c. Major antibody in a primary response 4. C IgM d. Antigenic determinant 5. 4: IgA e. Malignancy of antibody cells 6. 2;; IgG f. Gamma- globulin g. Immuno globulin h. Killer cell LA) 11. (6 pts) Schematically illustrate the biochemical events by which 7—dehydrocholesterol in the skin of young children can be used to cause increased synthesis of a calcium—binding protein in the intestinal mucosa to improve calcium absorption to prevent rickets. /“\ _. E Ahaiwlp J? W ’31 3 Ca. - 12. (6 pts) Chromatin that is transcriptionally active (euchromatin) is disgerse structure, whereas chromatin that is transcriptionally inactive (heterochromatin) is compact. When the nuclei of chicken globin—producing cells were treated briefly with pancreatic DNase, the adult globin genes were selectively destroyed, but the genes for embryonic globins and ovalbumin remained intact. In contrast, when the nuclei of oviduct cells were treated with DNase, the ovalbumin genes were destroyed. Explain these results. 13. (313% A drug is %oxic?()/;1(a r>n%irfo ty of individu ls but not for the general populatioflhe W difference between these two groups is a nonsense mutation in a gene for a cytochrome P450. For ' most people, the drug is excreted conjugated to glucuronic acid. Explain the mechanism behind the Emégéhww P “('33 Oaw'da (Add, WW] Ji'i’iltm ( “OH/J hydil’fiy/ 17174.93) Lb . mm; Wm amt/W 6M M": {WWI ‘4/ f h. l“ J" _ 3‘ u it v P waif“; '7‘ f r l ‘ Pm LL WW?” y m ‘1 my 2? mummm 38mm W (3/; ‘ falgtps [Iliafiiucowmm Mr, A u a .1; I 0 Pi” 17‘ A ‘ - L' c)?" w Ni) urgiw/ W/ W, I in will ‘, my) (me/mot Mallet a WI j PM fl 1 ’ y , " ’7 r _l «f g ,1‘9' I”. 4! ' 4’ ‘w fl? ' 1V mfg; wily?) (I?! WM? 5/5 $16 f fiméc “my a all} {my gt. a can was or 4 14. F is a chemical that is essential, but in large quantities contributes to heart disease The letters A-F represent molecules in the pathway. The numbers 1-5 underneath the arrows are the enzymes that catalyze the reactions, and they are found in the liver. A is the only metabolite found in other pathways. a. (2 pts) Which enzyme would you target in the pathway to reduce the formation of F with the least amount of possible side—effects? Enzyme # Explain your reasoning. b. (2 pts) You design an inhibitor for enzyme #3 that decreases its activity by 75%, but it does not decrease the amount of F. Why is this? 0. (2 pts) You design an inhibitor for enzyme #5, but, while decreasing the amount of F present to therapeutic levels, the patients exhibit liver failure. What is the likely reason for this? 15. You are presented with the graph below from a binding experiment. The binding of a competitive inhibitor was monitored in the presence of 0, 20, and 40 uM of the natural substrate. 1 0.9 0.8 0.7 0.6 0.5 Fractional Saturation 0 | j I—I'" I f“ Ffi—T—flfi—fi—fi“ 0 1o 20 30 4o 50 60 70 80 90 100 110 120 130 140 150160 Inhibitor Concentration (nM) a. (2 pts) What is the Kd of the inhibitor? b. (2 pts) Does it bind more or less tightly to the protein than another inhibitor that has a Kd of 1 nM? More / Less (circle one) c. (2 pts) Despite the potent inhibition in in vitro binding assays, this molecule does not act as a good inhibitor when administered orally. Outline one possible scenario to explain this. Enjoy your summer! ...
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Final Exam - BBMB 405 Final Examination Closed Book Part(80...

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