Bio Final 06 - Final Exam 06-Part 1 Biology 131 Examination...

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Final Exam 06-Part 1 Biology 131 Examination 3 April 4, 2006 You may separate the pages as needed. There is no need to turn in this part of the exam. NO CALCULATORS! ln [] o x i z RT z FZ x V x = Vm V X X mV X X x o i o i == 25 58 ln log 21 2 1 1 2 ( ) ( ), at equilibrium, = 0. PP ss ∆Ψ=Ψ −Ψ = Ψ −Ψ − Ψ −Ψ ∆Ψ ,for a monovalent cation, X. or a diavalent ion, the above equations must be divided by 2 (z=2), thus in 3, 2 and 1 dimensions, respectively. Thus, t increases with L , so t = L /6D, L /4D, L /2D. he pressure generated by the meniscus of water: P = -2T/r c , where T is the surface 1 Pa onstants, Algebraic Relationships, and Conversion Factors 100 = 2; log 1000=3, etc. log (A B) = log = 1 + 0.7 = 1.7 Avogad 6x10 As prefixes to units, , n = 10 -9 , K = 10 3 Speed c/ λ , J = -D C/ x; J = -P C F V Ca = 29 mV log (Ca o /Ca i ). For a monovalent anion such as Cl - , z = -1, so VVV mio =− V Cl = -58 mV log (Cl o /Cl i ). F = 10 5 coulomb/mol (RT/F) = 25 mV 3 2.5 10 2.5 s jn n nn Joules L MPa R TC C C mole mol Ψ=− ∑∑ () P = −Π Ps Ψ=Ψ +Ψ where C n refers to the concentration of the n different solutes in the solution. 2 2 2 2 LDD tD t = 642 t ,, T tension of water (7 x 10 -8 MPa m) and r c is the radius of curvature of the meniscus. = 1 joule/m 3 ; 1 atm = 0.1 MPa = 760 mmHg 10 m H 2 O. C log 1=0; log 10 = 1; log 2 0.3; log 3 0.5; log 5 0.7; log log (A/B) = -log(B/A) = log A – log B A + log B; thus log 50 = log 10 + log 5 log(10 x ) = x 23 ro’s number = 1 µ m = 10 -3 mm = 10 -4 cm 1 L = 10 3 mL = 10 3 cm 3 c = 10 -2 , m = 10 -3 , µ = 10 -6 For a sphere, Volume = 4 π r 3 /3, Area = 4 π r 2 , In Bio 131, π = 3. of light in vacuum, c = 3 x 10 8 m/s; energy of a quantum, E = h h = 6.6 x 10 -34 joule s 1
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Final Exam 06-Part 1 2 Nucleotide Bases. Henry’s Law: C x = K x P x . K O2 = 0.03 mL O 2 / L torr K CO2 = 0.60 mL CO 2 / L torr N H O HN N N N H N NH 2 N N H N O 2 N N H 2 O N H O O Adenine (A) Guanine (G) Uracil (U) Cytosine (C) Thymine (T) O O H OH H H H HO H OH O H H H H OH OH 1' 5' 1' 5' P O - O O - O -
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Final Exam 06-Part 1 All multiple choice questions are worth 2 points. helix e. Tertiary structure of a protein folded in a beta helix. 2. What is the physical basis of the force holding the strands together? a. Covalent bonds between the alpha carbon and amino nitrogen. b. Covalent bonds between carboxyl carbon and the amino nitrogen. c. Ionic bonds between the carboxyl oxygen and the amino nitrogen. d. Hydrogen bonds linking the amino nitrogen and the alpha carbon. e. Hydrogen bonds linking an amino nitrogen of one strand and a carboxyl oxygen of an adjacent strand. indicates a peptide bond? ? Questions 1-5 refer to the diagram to the right. 1. What does the figure to the right show? a. Primary structure of a protein. b. Secondary structure of protein folded in an antiparallel-strand beta sheet. c. Secondary structure of a protein folded in a parallel- strand beta sheet. d. Secondary structure of a protein folded in an alpha 3. Which of the numbers near the upper left of the diagram a. 1 b. 2 c. 3 d. 4 e. 5 4. What important component is left out of the diagram a. The R-groups b. The phosphate ester linkages c. Fatty acids d. Nucleotides e. G-proteins 3
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Final Exam 06-Part 1 5. Which one of the shaded boxes in the diagram
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This test prep was uploaded on 04/18/2008 for the course BIOL 131 taught by Professor Robinson during the Spring '08 term at Purdue University.

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Bio Final 06 - Final Exam 06-Part 1 Biology 131 Examination...

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