Final Examination, 2007
Biology 131
Final Examination, April 29, 2007
This examination is in two parts.
First is a multiplechoice part, and your answers will be
put on the attached Scantron sheet.
Be sure to print your name on the sheet.
The second
part of the exam consists of quantitative problems and short essays to be answered in the
space provided.
Be sure to print your name on all of those pages.
When you are
finished, turn in the Scantron sheet and pages with the written answers; you should keep
the multiplechoice questions.
Page 3 is blank and may be used for calculations.
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
ΔΨ=Ψ −Ψ = Ψ −Ψ
− Ψ −Ψ
ΔΨ
,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.
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
Spe
/
λ
,
J = D
Δ
C/
Δ
x; J = P
Δ
C
F
V
Ca
= 29 mV log (Ca
o
/Ca
i
).
For a monovalent anion such as Cl

, z = 1, so
VV
V
mio
=−
V
Cl
= 58 mV log (Cl
o
/Cl
i
).
F = 10
5
coulomb/mol
(RT/F) = 25 mV
where C
n
refers to the concentration of the n different solutes in the solution.
Thus, t increases with L
2
, so t = L
2
/6D, L
2
/4D, L
2
/2D.
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
ed of light in vacuum, c = 3 x 10
8
m/s; energy of a quantum, E = hc
h = 6.6 x 10
34
joule s
LD
D
t
D
t
=
642
t
,,
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
1