Final Exam 06Part 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