-1-
580.439/639 Final Exam
Do all problems. Closed book except for a two-page cheat sheet
Problem 1 (17 points/part)
In each example at right, a current is
applied to a single node of membrane, consisting
of a capacitor C in parallel with the combination
580.439/639 Final Exam
Answer all questions. Closed book except for two pieces of paper. Point values are
indicated, for a total of 100.
Problem 1
Calcium is released from a cells internal compartment (like the ER or SR) into
the cytoplasm.
Part a) (5 poi
580.439/639 Final Exam, 2007
3 hours, closed book except for two-page cheat sheet, do all problems. Problem 4c should be done
last.
Problem 1 Short answers
Part a) (15 points) Linear cable theory predicts that EPSPs in the soma will be smaller
when produc
580.439/639 Final Exam 2008
3 hours, closed book except for 2 sheets of paper, answer all problems. 20 points for each question
plus 20 points for your name.
Problem 1
Part a) What is the difference between metabotropic and ionotropic synapses? In which
c
580.439/639 Final Exam, 2009
Three hours. Do all problems. Closed book except for two sheets of paper. 300 points total.
Problem 1
Part a) (5 points per subpart) A membrane separates two solutions containing different
concentrations of ions. There is a po
580.439/639 Final Exam 2010
3 hours, open book except for allowed cheat sheets. Do all problems
Problem 1
Consider the diagram at right which shows the
energy landscape of a receptor molecule, like
rhodopsin, in the absence of activating energy (solid
bla
580.439/639 Final Exam, 2011
3 hours, closed book except 2 sheets of paper
Do all problems.
Problem 1
Short answer questions:
Part a) Consider the cell drawn at right,
which is a component of a transport epithelium in
the inner ear. In the right-hand memb
580.439/639 Final Exam 2012
Do all problems, closed book except for 2 pages of notes. Problems marked (*) are hairy, do them
last. 20 points per problem plus 20 points for your name.
Problem 1
Consider the transport of an ion X with charge z between an
in
5 80.439 Final Exam Solutions, 1997
Problem 1
Problem 2
inside
V
Part a) T he model is drawn at
right, with the equilibrium potentials for
the three ions shown as potentials on their
batteries, computed using the Nernst
equation. The resting potential for
580.439 Final Exam Solutions, 1998
Problem 1
Part a: Equilibrium means that the thermodynamic potential of a constituent is
the same everywhere in a system. An example is the Nernst potential. If the potential
across a membrane equals an ions Nernst poten
580.439 Final Exam Solutions, 1999
Problem 1
Part a A minimal circuit model is drawn below. The synaptic terminals of cells A and B
are represented by the parallel
Cell A
combination of synaptic conductance (GS),
passive membrane conductance (gp) and
acti
580.439/639 Final Solutions
Problem 1
The answer to this problem should include a standard Hodgkin-Huxley type cable with active
conductances in the membrane compartments, a model for potassium accumulation in the
extracellular space, and a model for the
580.439/639 Final Exam, Solutions
Problem 1
Part a) For the usual circuit, a parallel capacitor and battery/resistor with a
current I injected, Kirchoffs current law gives, for t > 0.
dV
= I Gm (V Erest ) ,
dt
with initial value V (0 ) = Erest .
C
The sol
580.439/639 Final Exam solutions
Problem 1
Part a) No direct potential change because no charge is carried across the surface
membrane of the cell.
Part b) Calcium-activated channels could indirectly change the membrane
potential. The calcium-dependent po
580.439/639 Final Exam Solutions, 2007
Problem 1
Part a) Larger conductance in the postsynaptic receptor array; action potentials
(usually Ca+) propagating in the forward direction in the dendrites; amplification by
subthreshold conductances (Na+ or Ca+).
580.439/639 Final Exam 2008, Solutions
Problem 1
Part a) Ionotropic receptors contain a channel that is directly gated open by activating the
receptor. Metabotropic receptors are coupled to a second-messenger system, like a G-protein, that
indirectly affe
580.439/639 Final Solutions, 2009
Problem 1
Part a) The operating principle is that the Nernst equation holds simultaneously for all
permeant ions:
V = EA =
RT Aout
RT Bout
RT N out
ln
= EB =
ln
= ! EN =
ln
zA F Ain
zB F Bin
zN F N in
meaning that concent
580.439/639, Solutions to Final Exam 2010
Problem 1
The figure at right shows definitions of the rate
constants for use in this problem.
Part a) Steady state means the fraction of
channel in each state is fixed in time. As argued in
class, that means the
580.439/639 Final Exam Solutions, 2011
Problem 1
Part a) In steady state, the net flux of sodium will be zero so that 3a = p. Here a and p have
units of moles/(s cell), so that a is the rate of consumption of ATP, but any reasonable flux measure
would be
580.439/639 Final Exam Solutions, 2012
Problem 1
Part a) At equilibrium there must be no change in free energy of the consituents involved in
the active transport. For this system
RT ln( Xi ) + zFVi + H = RT ln( Xe ) ,
so
Xi
= e ( zFV + H )/ RT .
Xe
Part
580.439/639 Midterm Exam, 2000
Answer all questions. Closed book except for 2 sheets of paper.
Problem 1
A
6
5
G (kcal mol-1)
Part A of the figure at right
shows the energy barriers computed
for permeation through a part of the
KcsA channel by qvist and
L
580.439/639 Midterm Exam
October 20, 2004
1.5 hours, do all problems, closed book except for a single sheet of paper.
Problem 1
Part a) Sketch the structure of the KcsA channel, showing its relationship to the cell
membrane. In particular, show the parts
580.439/639 Midterm Exam, 2005
1.5 hours, answer all questions, closed book except for one sheet of paper. 9 points for
each question part, 1 point for your name.
Problem 1:
Part a) The ion concentrations in two extracellular fluid spaces of the mammalian
580.439/639 Solutions to Homework #1
Problem 1
Part a) At equilibrium, the electrochemical potentials of molecules in the A and B states
will be equal, so
A = B
al
al
or
0A + RT lnCA + FVA = 0B + RT lnCB + FVB
al
al
al
al
Note that zal = +1 and that it ha
580.439/639 Homework #2 Solutions
Problem 1
Part a) Using the principle of conservation of mass, the time rate of change of
concentration in the ith well is given by the difference between the flux into the well and the flux
out of the well. Let Ci be the
580.439/639 Homework #3 Solutions
Problem 1
Part a) If the membrane is depolarized, the number of open gates increases in the HH
K+ channel; in the model, this means that the gating charges move to the outer membrane
surface. With this definition of gatin
580.439/639 Solutions to Homework #4
Problem 1
Part a) During the steady portions of a voltage clamp beginning at time 0, the HH
parameter h satisfies the following equations:
dh
H (V2 )
= h (V2 ) h
a nd
h(0) = h (V1 )
dt
where V1 is the potential to whi
580.439/639 Solutions to Homework #5
Problem 1
Part a) The isoclines for this system are the equations:
dx
=0
dt
y = x 3 + 3x
dy
=0
dt
y = 3x + 4.5
The isoclines are plotted below (lines marked d()/dt=0).
The arrows show the directions of flow in the phas
580.439/639 Homework #6 Solutions
Problem 1
Part a) The derivation of the cable equation is the same up to the specification of the
membrane impedance. That is, we can start from
2
1 V = i (x,t,V) = membrane current/unit length of cylinder
m
ri +re x2
the
580.439/639 Homework 7 Solutions
Problem 1
Part a) For the cylinder model:
q = 1+ s m
x
L=
=
x
a
2Gm Ri
G =
1
2Gm
= a3 2
ri
Ri
where s is the Laplace transform variable in units of Hz. In class, s was dimensionless, because the
transform was done in term