bio test 2 practice 6

bio test 2 practice 6 - Exam 2, 2007 Biology 13 1...

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Unformatted text preview: Exam 2, 2007 Biology 13 1 Examination 2 February 27, 2007 You may wish to remove pages 1 and 2 for easy reference. If you do, be careful to leave the other pages stapled together. = -DAC/Ax; J = -PAC M"[xq Vf—‘ln Z ° FZ [x ] V = 25mV1n X0 = 58mV10g X0 ,for a monovalent cation, X. " X 4 X. l l i For a diavalent ion, the above equations must be divided by 2 (2:2), thus VCa = 29 mV log (Cao/Cai). For a monovalent anion such as Cl', 2 = -1, so V01 = -58 mV log (Clo/C1,). F = 105 coulomb/mol (RT/F) = 25 mV n=K—n ‘1’ = ‘PP + ‘PS(= P—H) A‘P = ‘1’2 4P1 = (le2 — l11,1) 4111,, — IP52), at equilibrium, A‘P = o. Joules 2C" = _25 L MPa Ir, =—RTZCJ. =—2.5-103 2C n male n ma] n where Cn refers to the concentration of the 11 different solutes in the solution. L = J6 Dtn/ 4Dt,1/2 D; in 3, 2 and 1 dimensions, respectively. Thus, t increases with L2, so t = L2/6D, L2/4D, L2/2D. The pressure generated by the meniscus of water: P = -2T/rc, where T is the surface tension of water (7 x 10'8 MPa-m) and r6 is the radius of curvature of the meniscus. 1 Pa = 1 joule/m3; 1 atm = 0.1 MPa = 760 mmHg z 10 m H20. Constants, Algebraic Relationships, and Conversion Factors log 1:0; log 10 = 1; log 2 w 0.3; log 3 z 0.5; log 5 z 0.7; log 100 = 2; log 1000=3, etc. log (A/B) = -log(B/A) = log A — log B log (AoB) = log A + log B; thus log 50 = log 10 + log 5 =1 + 0.7 = 1.7 log(10x) = x Avogadro’s number = 6x1023 ium =10'3 mm :104 cm 1 L=103 mL=103 cm3 As prefixes to units, 0 = 102, m = 103, u = 106, n = 109, K = 103 For a sphere, Volume = 4nr3/3, Area = 41112, In Bio 131, 11: = 3. Speed of light in vacuum, 0 = 3 x 108 m/s; energy of a quantum, E = hc/ls, h = 6.6 x 10'34joule s Exam 2, 2007 Nucleotide Bases. NH2 0 NH2 0 o N N <f“</I““ “Wu “H H N H N NH2 0 u 0 N o N . H H Adenlne (A) Guanine (G) CytOSine (C) Thymine (T) Uracil (U) OH Ho—CH2 _ _ I Glycerol o——P——o H HO—TH O HZC—OH FIRST SECOND POSITION THIRD POSITION W POSITION (5' END) U c A G (3' END) I Phe Ser Tyr Cys U U Phe Ser Tyr Cys C ' Leu Ser Stop Stop A Leu Ser Stop Trp G Leu Pro His Arg U I "1 C Leu Pro His Arg C Leu Pro Gln Arg A Len Pro Gln Arg G Ile Thr Asn Ser U A Ile Thr Asn Set C Ile Th: Lys Arg A Met Thr. Lys Arg 0 Val Ala Asp Gly U G Val Ala Asp Gly C Val Ala Glu Gly A Val Ala Glu Gly G Phenylalanine > (Phe) CH==CH Glutamlne _ _ yo ._ _. / \CH G‘ycma (Gm) CH1 CHI—cxNH Cysteine _CH2._5H F CH; c‘CH~CH’ (G) —H Q ' (Cys) g., c. Alanine CH Tyrosine o==c/ . — = - . \ : (NAa) : (7w) —CH1~c/CH CH\c——OH méronme "cH‘_CH2_S_CH) Praline ClHT‘CHa\CH Y ‘CHa—CH’ \ J ‘ (Pm) N-iz—CH/ ' Pl P / : Valine CH/CHI _ (<7) \CH, ~CHz—C-==CH Aspaflic acid . __CH _c¢° /CH\\ Hislidine AH}. (A59) ' ‘0‘ Tryptopnan *Cfiz'i‘fi ‘1'“ (HS) \/ D ' (Tm) cu c on c H CH . _ w \/H\ // Leucine ___CH __CH/ "3 . Glutamsc acid /0 N (Leu) ’ \ . G —CH"—CH‘-c CHI Lysme + ( I") \O L- - . ‘Lvs)—CH,—-CH,—-CHI«CH,—NH, E _ . Serine -CH,_0H K ' (Se!) . S lsoleucine cH/CH'—CH’ MSW“ awful“ —-CH «6/0 (He) ‘ \ (A19) NH 5" ' \NH rh 'ne /°"‘ I OH. -—CH,—~cH.—-CH,—NH,—c;l 2 N I > ' (1:: m ‘CH\OH- R NH; .r Exam 2, 2007 Biology 131, Examination 1, Spring, 2006 Name (PRINTED IN BLOCK LETTERS) Signature Enter your name in the spaces above and also on the following pages. Answer the questions in the space provided; do not write on the back. If your handwriting is poor, print your answers. Generally, it will be necessary to include appropriate units with numbers. If you wish to retain the possibility of a regrade, you must write in ink. 1. (20) The diagram below shows the response of an intestinal epithelial cell to vasoactive intestinal peptide (VIP). The follow questions relate to this response. Continued on p. 2. Serosa Lumen K*——é ‘ 51030mV Nev Na+ a. (3) What is the nature of VIPR? Be as detailed as possible, given the available information. m. ' 2.4 W s MW M 3f- M “LEMMA/Cw” axle (an vs? “MM/(«4 b. ( ) What is the process by which Gs becomes activated and how does it finally become in tiv t d? aCWfi/M ’(V‘Z, YIFR’ flu ¢W+v}. G5 W GDPMMQ Janka GTR Tm MWJVWW “firm?” "P M 636‘ W G" +0 GD? M Mgm 1m elm/MM) W m”; ~ Exam 2, 2007 Name (PRINTED IN BLOCK LETTERS) 1. (continued) c. (3) What is molecule 1, What does it do, and in what cellular compartment is it found? 1M W WWW KWW-%.LhJ-XL FLY“) ¥PM I43”). MM MMMWW M4 WWW“ d. (2) Draw the chemical structure of molecule 3, and indicate what cellular compartment it is foundin. A’ LAM?!“ @- 4M WWW'L e. (3) What is molecule 4 and what does it do? Be as specific about the process as you can. IMQ QL’EWWA.WWMMWWW M $31.43? chP{ U? f. (3) What is the consequence of the action of molecule 4 on CFTR? Include here the general physiological consequences, as well as the specific effects on CFTR’s function. 1-9;] a Q‘pQ—md Mun; PWI$W~ g. (3) What is the mode of action of cholera toxin (CTX) and what are the consequences of the action of CTX on the general physiology of the individual? CTX MW m‘rh W 0,53% kawfi m thflv 4.,> WWQT . Thu/W Exam 2, 2007 Name (PRINTED IN BLOCK LETTERS) 2. (10) a. The strength of contraction of a muscle cell is regulated by Ca2+, and higher Ca2+ produces stronger contractions. Explain how a drug such as ouabain that blocks the Na—K ATPase would affect the strength of contraction of a heart muscle cell. Mention both the effect on contraction and the mechanism. 2+. 4° 0% tar-7M C“ N M 769/»: W emu 7U» Na+W . | Nq‘r _ WM 7th aNawK A-‘rpma. CW QweQ‘ ' 26% MM r064 M'Hwe 3; (W mew a} WWW- . Signal melecme Pl 4,5-bisphosphate w G-protein—linked . [p|(4,5)92; receptor activated / phosphotipase C-B activated Gq inosito! ‘ a subunit 1,4,5-trisphosphate §% ‘ g b. What is calmodulin and What would be the consequence for calmodulin’s activity in a cell as a result of the signaling process indicated in the diagram above? mpray ABQM,WWW.MMWW M of :93 M afiwamwwwc WW) Maw} w wa MW W“' Exam 2, 2007 Name (PRINTED IN BLOCK LETTERS) 3. (10) Plant and animal cells differ fundamentally in the mechanism they use for establishing the membrane potential and critical electrochemical ionic gradients. Animal Cell * Plant Cell K*Nw a. In the diagrams above, show the main primary active transport process for each cell type. Your answer should Show the location of the enzymes, the identity of the substances that are transported and the direction they are moved, and energy source. b. Also on the diagrams above, show one specific example for each cell type of secondary active transport (we have discussed several). Again, show the location of the transport systems, the substances transported and the direction the substances are moved. Exam 2, 2007 Name (PRINTED IN BLOCK LETTERS) 4. (15) Make a diagram of one of the plant cells that is involved in the absorption of nutrients and water from the soil and identify the cell. Be sure to show the main anatomical feature that relates to its function, and name the organ and tissue type that the cell is part of. Describe the mechanism by which these cells take up K+, Cl‘, and H20 in the space provided. You will need to mention ion channels, pumps (including primary active transport and secondary active transport), protons, and water potential in your answer. [We “‘7‘- w" a. (5) Put your diagram J and identification of the cell here. Grade“: (00"— R w TiSJu—e T3 9e ‘. Dermal 017k l—lw‘w b. (2) Approximate value of Vm and the mechanism of its generation (only one or two sentences are needed to answer this and the following questions): \/ .7 450+» - IsoMV. 5%)» W {mm}? “ 1,67% H*-ATPW c. (2) Mechanism of K+ uptake: \Im a me «An» AMVK;+\W, \-<* WW (1. (2) Mechanism of Cl' uptake: C \— W M «pm/Z M m WW qul-ive Cl‘ wait—44m M10 1»? MW WM MH" ‘4‘“ e. (2) Mechanism of H20 upt e: . W :va WWWJQM) So gum m W ’W} 41')“ ' f. (2) Plants must accumulate phosphorus from soil in the form of phosphate (HPO42'). How might they accomplish this? ®6M M” /J‘7”""“*P“"jr “AHA H+‘ Exam 2, 2007 Name (PRINTED IN BLOCK LETTERS) 5. (15) a. The radius of curvature of the air-water interface in a xylem element is about 20 pm. To what height (in meters) can water be raised by xylem capillarity? QT z —— .2- =1--Io‘7r*\?a~m : _7x 10—37191 "_._——-p rt. 10 x I040?“ :5,“ m"?— 7;. ~'L MW z—‘Om .qx30 02km Z 0.1M - dim b. What radius of curvature at an air-water interface is required to raise water just to the top of a 100 In tree? ._ , 7,! x lo”? M Fa, ' W '9 z _ was», C. “ € —- I “Va. 03' : (9. H/M-m 2 ~l [‘4ch c. Is the pressure generated by the radius of curvature that you calculated in b. sufficient to meet the water needs of the 100 m tree? Why or why not? fivAiWLMWWCA‘h—E (1. Where are the radii of curvatures that are needed to supply the water needs of trees found? Mention the organ and the subcellular structures that are involved. . Mam zen. memtmttm cm). e. How are the radii of curvatures discussed in d. maintained and what cells are involved in regulating this process? What changes are caused by the operation of these cells in daylight and at night? . . ’ . ' £4 MWwWWMfWOth)|MD~m\ owe. mum Wd\u”“‘°°' ".4, M c W Exam 2, 2007 Name (PRINTED IN BLOCK LETTERS) 6. (15) The photographs above show lettuce seeds that were exposed to different regimes of alternating treatments of 670 nm light and 730 nm light before being placed in the dark and allowed to germinate. a. (4) What statement can you make about the light regimes of A and B? A: W Why’d Wm W M 117M((‘:1'0"“‘) B: m Lad/033M mytqu lmwbym (430%) b. (3) What molecule is responsible for these effects and what is the general chemical nature of the molecule? HW-W hW‘M"& M“ c. (4) What is the effect of 670 nm and 730 nm light on the molecule, in terms of the molecule’s biological activity? Use the terms “Pr” and Pfr” here to identify the two forms of the molecule. (9an M; Pp +9 Egr- 1—39nm W5 Er l° P“ d. (4) If a seedling is grown in red light, which form of the molecule is produced? What can you infer about the effect of that form of the molecule on auxin (IAA) production, based on what you have learned in Bio. 131? Explain. Exam 2, 2007 Name (PRINTED IN BLOCK LETTERS) First foliage leaf Coleoptile Shoot apical Scutellum 7. (10) The figure to the right shows the overall effect of gibberellin (GA) on the seed. a. (4) When GA encounters the cells of the cellular layer, what must it FIRST do to trigger responses? What substance is secreted by the cells of the cellular layer in response to GA, what does that substance act on in the endosperm, and what is the product of that action? i. First action ofGA? ivwk E M 0424-1 wk In; W“ A?” WW ii. Substance secreted Ama Egg 9 iii. Target of the secreted substance? iv. Product? g‘mpge mafia/u b. (6) After stimulation by GA, the cells of the cellular layer have a rapid response and a slower response, brought about by two separate pathways. In the spaces below, mention the key events of the two pathways. Well labeled diagrams may be useful. Rapid Response Slower Response Ca2+? AWWM 961“?va WW “waer X} T'me ~ WWW? °6 . .w wwmw? ; W‘WWWN Exam 2, 2007 Name (PRINTED IN BLOCK LETTERS) 8. (5) Plants require nitrogen from the soil. In legumes, the roots become infected with Rhizobium bacteria, which convert atmospheric nitrogen to a useable form. What is the chemical form of atmospheric nitrogen? N; i NBA) > +- To what chemical form do the bacteria convert nitrogen? H; (n‘ H ‘5) Other bacteria in the soil convert the nitrogen into a form that is non-toxic to plants and can be taken up by the plants. What is that chemical form? fl 0; . To what chemical form do plants convert nitrogen in the chloroplasts? N H 3 . What is the major direct (not final) product that plants synthesize from the form of nitrogen mentioned above? Mm MW. 11 ...
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bio test 2 practice 6 - Exam 2, 2007 Biology 13 1...

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