Bio Final 06

# 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 = V mV X X mV X X x o i o i = = 25 58 ln log 2 1 2 1 1 2 ( ) ( ), at equilibrium, = 0. P P s s ∆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 V V V m i o = 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 j n n n n Joules L MPa RT C C C mole mol Ψ = − = − = − ( ) P = −Π P s Ψ = Ψ + Ψ where C n refers to the concentration of the n different solutes in the solution. 2 2 2 2 L D Dt Dt = 6 4 2 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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