FINAL EXAM
PHYS 3511 (Biological Physics)
DATE/TIME:
April 16, 2013 (9:00 a.m. - 12:00 noon)
PLACE: AT 1010
Only non-programmable calculators are allowed.
Name: ___________________________________
ID: ___________________________________
Please read the following instructions:
• This midterm has 14 pages. Make sure none are missing.
• There are three sections. Section 1 has 10 multiple choice, true/false, or short
answer questions. In section 2 you must do 4 out of 5 questions. In section 3
you must do 4 out of 5 questions.
• You must show all works, in the provided blank space below each question.
• Write your name and student ID in the provided space above.
• There is a two-page equation sheet on page 13 and 14. You may tear the
equation sheet from the midterm.
Read questions carefully before attempting the solutions.
1
SECTION I (Question 1 to 10): Do all TRUE/FALSE (circle all true statements,
can be more than one), MULTIPLE CHOICE (choose the one correct statement or
answer), OR SHORT-ANSWER (one to four sentences answer) QUESTIONS
1. (2 points) Short-Answer. In a sequence film with high-speed camera, a Northern Goldhawk
is seen catching a moth in mid-flight without any sign of change of speed of the bird. Did
the hawk find a way to violate the conservation of momentum? Explain in the space below.
2. (2 points) Multiple-Choice. A process is adiabatic if
(a) the temperature remains constant.
(b) No work is done.
(c) No heat is exchanged with the environment.
(d) The internal energy remains constant
(e) The volume remains constant.
3. (2 points) Multiple-Choice. Which of the following processes has a negative change in
entropy? Only one correct answer!
a) an increase in the number of gas particles in a box.
b) Increase in the internal energy of the particles in a box.
c) Relaxation of a stretched polymer.
d) Isothermal expansion
e) Formation of molecules from atoms, e.g., H + H ! H 2
4. (2 points) Multiple choice (circle the one correct answer) When an erythrocyte passes
through a blood capillary adjacent to an alveolus, oxygen diffusion occurs because:
A) A small temperature difference is established both sides of the membrane.
B) The moving erythrocyte causes a drag effect pulling the oxygen through the
membrane.
C) A concentration gradient (difference) of oxygen is established with lower
concentration at the erythrocyte’s side.
D) There is no concentration gradient, but oxygen moves randomly in the membrane and
ends up in the erythrocyte by chance.
Note: There’s only one correct answer.
5. (2 points) Multiple-choice Most fish can float in water? What must be true for such fish?
a. Their density is larger than that of water.
b. Their density is equal to that of water.
c. Their density is smaller than that of water.
d. No conclusions about the fish’s density can be drawn.
e. The density of most fishes is similar to that of water, but most can alter their
density to move to/from the surface of water.
2
6. (2 points) Multiple choice (circle the one correct answer) An artery is partially clogged by
a deposit on its inner wall, so that the radius around that area is smaller than in adjacent part
of the artery. Which of the following statements best describes the processes that occur
when blood rushes through this constriction? Treat blood as a Newtonian fluid.
a. Blood will rush faster through the constriction due to the equation of continuity,
causing additional wear and tear on the nearby blood vessel walls.
b. Bernoulli’s law and the equation of continuity predict a variation of the blood
pressure in the constricted zone, but the blood vessel walls prevent any adverse
effect due to the pressure variation.
c. The blood pressure in the constriction zone is lower than in the adjacent blood
vessel, causing the blood vessel to temporarily collapse at the constriction
(vascular flutter).
d. The blood pressure in the constriction zone is higher than in the adjacent blood
vessel, causing a ballooning effect of the blood vessel at the constriction
(aneurysm).
Note: There’s only one correct answer
7. (2 points) True or False Laplace’s law describes the pressure in an alveolus in the lungs in
the form Pinside ! Poutside = 2" / r , in which ! is the surface tension, and r is the radius of
curvature of the alveolus. In healthy alveoli, surfactants are used to reduce the surface
tension by coating parts of the surface. Which of the following statement(s) is (are) true.
A) It is more important for surfactants to coat alveoli with small rather than large radius.
B) It is more important for surfactants to coat alveoli with large rather than small radius.
C) Surfactants change the surface tension, but do not change the pressure inside the alveoli.
8. (2 points) Short Answer Explain why only very small organism ( < 0.1mm ) can rely on
passive diffusion of oxygen but human beings cannot. Then explain why some relatively
large inveterbrates such as hydra can survive by passive oxygen diffusion.
9. (2 points) Short-Answer. In the squid axon of a live squid, the resting potential is -60mV,
but when the squid dies the membrane potential (Donnan potential) becomes -38mV.
Explain the reason for the change in membrane potential in the space below.
3
10.
(2 points) True or false. Circle all true statements on a variety of biophysics topics.
a) Unlike most of its body, the whale’s fluke (tail) has no fat layer. This allows
circulated blood to dissipate heat. When a whale dies this heat dissipation
mechanism stops and the whale’s interior can heat up to 60 degree Celsius.
b) Despite the fact that the membrane potential is usually negative
!V = Vin " Vout = "60mV to !80mV , the concentration of sodium inside the
nerve cell is much smaller than outside. This is called the sodium anomaly.
c) Laplace’s law explains how water is transported upward in plants through xylem
tissues.
d) The parallel current flow of blood in artic fox maintains the cold feet, which in
turn confined low temperature blood to its extremities.
e) To a good approximation blood flowing through veins behave like an ideal fluid.
SECTION II: FULL ANSWER QUESTIONS (question 11 to 15)
Do four out of five questions on the provided space. Show all works.
11. (10 points) A Tour de France rider consumes 8000 kcal of food daily. Suppose every day
he rides his bicycle at constant speed of 25 km/ hr for six hours.
A) Assume the “drag” force on the cyclist is Fdrag = Bv 2 , where B = 1.5 kg/m. Calculate the
work done to overcome the drag. Would this account for the discrepancy 8000 kcal of energy
consumed? 1 cal = 4.184 J
B) Can the discrepancy in part A be accounted for by riding up the mountainside? To answer
this, determine the change in altitude necessary to account for the difference in part A.
4
11B continued
12.
(10 points) The diagram below is a simplified P vs V plot of lung hysteresis where
P = Palveoli ! Ppleura is the transmural pressure and V is the volume of the air in the lung. Unit
Note: 1L = 10 !3 m 3 , 1Pa = 1N / m 2
P(kPa)
inhale
2.5
inhale
exhale
1.3
3.0
3.5
V(L)
A) Calculate the work done on the gas during the inhale cycle.
B) Calculate the work done on the gas in the lung in one inhale/exhale cycle. Is the work
done positive or negative? Why?
5
12 continued
C) Calculate the change in internal energy, !U , during the cycle. Briefly justify your
answer.
D) Calculate the heat flow Q into/out of the gas in the lung. Does heat flow into or out of
the gas in the lung?
13. (10 points) Brownian Motion is the random motion of small particles in fluid (liquid or
gas), described by the Einstein diffusion equation. Polymers (long molecules such as DNA
and proteins) also undergo random motion, with size distribution rpolymer = 3N L .
A) A spherical object of radius R = 4 µ m is observed to move “randomly” by about
k T
30 µ m every minute. Use the Einstein relation D = B , the Stokes relation ! = 6"# R ,
!
and the diffusion equation to determine whether the object is living. Justify your
answer. Assume room temperature T = 300 K.
6
13 continued
B) A virus’ chromosome (DNA) is observed by a microscope to be a blob of “size” 0.8 µ m .
If the average DNA base-pair size is 0.34 nm, estimate the number of basepairs (bp) on
the DNA of the virus. How many amino acids (of proteins) would this correspond to?
14. (10 points) Diffusive transport. Consider a model of transport of nutrient vesicles
through an axon (a long slender projection of a nerve cell). Model the axon as a tube 1.5 mm
long. At one end of the axon, a process creates nutrient encapsulated in vesicles of radius
1µ m , where the vesicle is maintained at a constant number density 3.72 ! 10 24 particle / m 3 .
At the other end of the axon the vesicles are partly consumed by another process, in such a
way that the vesicle concentration is constant at 2.22 ! 10 24 particle / m 3 . Hence both ends
are at steady state.
a) Use Stokes-Einstein relations to estimate the diffusion constant, D, of the vesicles in
water at temperature 37 Celsius. Then use Fick’s law to calculate the flux density jdiffus (in
particle/m2 s). Then use the flux density equation j = cv to estimate the average “speed”,
v of the diffusion. Use the higher vesicle concentration c for the calculation of v.
7
14 continued
b) Repeat the calculation for oxygen diffusion to the same axon as in part a, and with the
same concentrations. Use data in the appendix. Compare the speed obtained in part a and
b. Briefly explain why your answer make sense or doesn’t make sense.
15.
(10 points) Thawing a spherical chicken Consider a frozen spherical chicken of radius r
)T
! Q$
that was taken out to thaw. The heat loss in time t is # &
, where ! is the
= ' 4( r 2
" t % loss
!
thickness of the heat-conducting layer, and !T = T final " Tinitial is the change in the
#T
t [1], where t is the time it takes for the
!
chicken to defrost. But the heat loss is also given by Q = Mcice !T [2], where M is the mass
temperature. This gives a heat loss of Q = ! 4" r 2
of the spherical chicken, and cice is the specific heat capacity (in JiK !1 ikg !1 ) of the ice
(frozen chicken). Assume the chicken’s radius is the same as the conducting layer.
a) By equating the heat loss equations [1] and [2] show that the time it takes the chicken to
! r 2c
defrost is t = ice ice . HINT: use M = !iceV where V is the volume of the spherical
3"
chicken. !ice is assumed to be the mass density of the chicken, which is the same as ice.
8
15 continued
b) Determine how long it would take for a 1 kg chicken to defrost.
DATA: ! " 1.6Jim #1 is #1 iK #1 , cice = 2090JiK !1 ikg !1 , !ice = 917kgim "3 . HINT: use
M = !iceV to determine the “unknown” variable.
SECTION III: FULL ANSWER QUESTIONS (question 16 to 20)
Do four out of five questions on the provided space. Show all works.
16. (10 points) Poiseuille’s flow
a) Briefly explain the difference between an ideal and a Newtonian fluid.
b) A hypodermic needle, filled with water, is 4.0 cm long and has an inner diameter of 0.25
mm. What excess pressure, !p , is required along the needle so that the flow rate of water is
3
!V
#6 m
= 1.0 " 10
? Assume Newtonian fluid.
!t
s
9
17. (10 points) Electrophoresis The electric field in an electrophoresis experiment is E =
10V/cm. The experiment seeks to identify two proteins one of radius R1 = 3nm and charge
q1 = +e , the other R2 = 2nm and charge q2 = +6e . If you want the two bands associated with
each protein to be separated by at least 10cm, estimate how long the experiment must last.
Assume that the experiment is done in water.
18. (10 points) Bernoulli Equation of Ideal Fluid In the diagram below, a water-filled
beaker has a hole of radius r = 1.75mm near its bottom, from which water is ejected. If
h = 0.5m and h1 = 1.0m . Treat the water as an ideal fluid and neglect air friction.
+y
yi =0
h1
+x
v0
h
ay = !g
yf = -h1
h2
xi = 0
xf = h2
a) Use Bernoulli’s equation to find the speed of the water stream, v0 , at the hole. Density of
water is 1000kgim !3 .
10
18 continued
b) Calculate the time it takes the water to hit the ground (i.e. at y = !h1 ).
c) Find the horizontal distance h2 where the water lands.
19. (10 points) Action Potential Consider a segment of an unmyelinated cylindrical axon of
radius 10 µ m , length 1 mm, and thickness 10nm . Useful data: ! = 5 (lipid membrane),
!membrane = 1.4 " 10 7 #im , and !axoplasm = 1.1"im .
a) If the axon has a capacitance per unit area of C a = 1.7 µ Ficm !2 , calculate the capacitance
in Farad (F). See appendix for useful geometrical relations.
b) When an action potential increases the membrane potential of the axon by !V = 100mV ,
positively charged sodium ions flows into the axon. Calculate the amount of charge, and the
amount of sodium ions Na + , that flow into the axon. Calculate the change in concentration
of Na + in unit of M. HINT: Use Q = C!V , and assume that C remains constant.
c) Calculate the time constant ! = "#membrane . Assuming that all the charges takes time ! to
flow into the axon, estimate the current (charge/time) that flows during the action potential.
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20. (10 points) Signal transmission in myelinated nerve
a) Briefly explain how signals can be transmitted a great distance in myelinated (insulated)
nerve cell, where most of the nerve is encapsulated by thick myelin sheath, with periodic
sections of unmyelinated nerve called the Ranvier nodes. Why is the speed of signal
transmission in myelinated nerve, generally, faster than in unmyelinated nerve?
b) Consider a myelinated nerve cell with resistivity !membrane = 1.4 " 10 7 #im and
!axoplasm = 1.1"im , radius 100 µ m and membrane thickness 50 µ m . Calculate the space
(electrotonus decay) constant ! .
c) Suppose it takes a potential change of !V = 8mV to stimulate an action potential with
maximum potential change !Vmax = 80mV . Let the distance between two adjacent Ranvier
nodes be 200a, where a is the radius of the nerve cell. If a Ranvier node is stimulated by an
action potential how many Ranvier nodes are triggered? Briefly explain why this
calculation shows that signal transmissions are more stable in myelinated nerve.
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Useful Equations
Kinematics of 2d trajectories: x-direction x f = xi + vx t , vx ! constant; y-direction
m
1 2
gt , g = 9.8 2 , v fy = viy ! gt ; i ! initial, f ! final.
2
s
! net
!
!
!
Newton’s Laws F = ! F = 0 (equilibrium); F net = ma (Nonzero net force); 3rd Law if object
A exerts a force on object B, B exerts a force on A of equal magnitude and opposite in direction.
!
!
Newton’s law for rotation ! = I" . Friction f s ! µ s n , f k = µ k n . Fluid friction: 1) turbulent
!
D
!
Fdrag = ! Av 2 ; 2) laminar flow Fdrag = ! v (v speed), ! is drag coefficient. Momentum: P = mv ;
2
!
in a collision of isolated particles total momentum is Ptotal constant; Newton 2nd law in terms of
!
! !
!
dp
1
momentum Fnet =
. KE = mv 2 ; work: Force*distance, W = F • S = (F cos ! )s = F|| s ;
dt
2
1
1
W net = "K = mv 22 ! mv12 ( W net is net work); gravitational PE, U = mgh ; conservation of
2
2
energy Etotal = U PE + K ! constant. Unit of energy Joules ! J Power P = work / time = Fv , in
J / s ! watt (W).
ATP hydrolysis (produces ADP) releases 29kJ/mol of heat. Reverse ADP ! ATP requires
29kJ/mol. POLYMER SIZE (DNA, RNA,..): rpolymer = 3N L .
y f = yi + viyt !
THERMODYNAMICS:1st law !U = Q + W ; 2nd law isolated system evolves to state of
Vf
maximum entropy. W = !P"V or W = ! " P dV ; PV = NkBT = nRT , kB = 1.381 ! 10 "23 J / K ,
Vi
R = 8.314 J/K*mol, Avogadro number N A = 6.023 ! 10 23 particle/mole . Isothermal
(
)
process !U = 0,W = NkBT ln Vi / V f ;Isobaric W = !P"V ; adiabatic PV 5 / 3 ! constant,
VT ! constant, Q = 0. U = (3 / 2)nRT . Entropy S = Q / T ; Enthalpy H = U + PV; Gibbs free
energy G = H ! TS , !G = !H " T !S , stable state has lowest Gibbs free energy; spontaneous
N
if !G < 0 . Osmotic pressure ! = posm = ck BT , c = is the number concentration of solutes.
V
kT
kT
Stokes-Einstein Relation: drag coefficient ! = 6"# R ; diffusion coefficient D = B = B ;
!
6"# R
3/2
Einstein diffusion: x 2 = 2Dt (1D), and r 2 = 6Dt (3D). Diffusion data D = 10 !9 m 2 / s for
small molecules like oxygen; ! = 10 "3 Pais for water, and for blood ! = 2.5 " 10 #3 Pais .
"c
Concentration gradient flux: Fick’s law j = !D
in particle / sim 2 , c is in particle / m 3 ;
"x
continuity flux density j = cv , v is the average speed of the fluid. Total flux is I = jA in unit of
particle / s , where A is the cross-section area of flow medium. Volume flux jv = L p !p (in m/s),
(
)
where L p is filtration coefficient in mis !1Pa !1 , with total volume flow jv A (in m 3 is !1 ).
Fourier law of heat flow: Q / t = ! A"T / L , ! is the thermal conductivity coefficient.
13
Heat Capacity: C =
Heat
Q
=
in J/K or Q = Mc!T , c is the specific heat
change in temperature !T
capacity in JiK !1 ikg !1 .
Surface equations: Atmospheric pressure patm = 1.01 ! 10 5 Pa ; Gauge pressure
pgauge = pabsolute ! patm ; Buoyant force Fbuoyant = ! fluidVobject g ; surface energy (surface tension)
between two phases ! = "E / "A , unit J / m 2 ! N / m ; Pascal Law p = patm + ! gd , d is the
depth measured from surface downward; Laplace Formula !P = 4" / R (hollow bubble), and
!P = 2" / R (air bubble in liquid).
Fluid Flow equation: Equation of continuity !V / !t = Av = constant , v is the average fluid
speed; Bernoulli’s law for an ideal fluid p + (1 / 2 ) !v 2 = constant ; Poiseuille’s law for
!V
" 4 !p
m3
=
rtube
, (in
)
!t 8#
l
s
!
!
Electricity: electric force due to electric field F = qE , E in units of Volts/m = V/m; electric
Newtonian fluid in a cylindrical tube
(
)
field of plate in vacuum E = ! / " 0 , ! 0 = 8.85 " 10 #12 C 2 / Nim 2 , and in material
E = ! / " , " = #" 0 , where ! is the dielectric constant; electric potential difference between two
+/- plates !V = Ed , d is the plate separation. Capacitance; C = ! 0 A / b in vacuum, unit is Farad
or F; C = ! A / b in material; Q = C!V ; capacitance per unit area C a = C / A = ! / b , and charge
stored per unit area in a capacitor q a = q / A = C!V / A = C a !V . Also !Eenergy = q!V .
Electrophoresis: terminal velocity vt = qE / ( 6!" R ) of a macromolecule of charge q.
Electric Current: resistance R = ! L / A (unit ! ); resistivity ! (unit !im ); conductance
G = 1 / R (unit !"1 ); conductivity ! = 1 / " (unit !"1m "1 ); conductance per area g = G / A (unit
(
)
!"1m "2 ); axoplasm resistivity ! = ( kBT ) / Dq 2 c .
Membrane rest potential: Nernst potential Vi Nernst = (Vin ! Vout ) =
(
)
(
)
kBT ci,out
; Ohmic leakage
ln
ze
ci,in
(
)
Ohmic
= VKNernst
! "V / RK + , I Na
= VNaNernst
! "V / RNa+ , I ClOhmic
= ! VClNernst
! "V RCl !
current I KOhmic
+
+
+
+
!
!
coulomb C
! ! ampere ! A
s
s
Action Potential: Electrotonus decay constant ! = ab"membrane / 2 "axoplasm ; time constant
!V is the membrane rest potential. Unit of current (I) is
! = "#membrane ; Signal potential decay !V / !VHH = exp" ( x / # ) , !VHH is initial electric potential
stimulation, !V potential at distance x from initial stimulation; Signal propagation speed
v = ! /" .
DATA e = 1.6 ! 10 "19 C ; 1M = 1mole / L ; 1L = 10 !3 m 3 ; Avogardo Number
N A = 6.023 ! 10 23 particle / mole .
4
Geometrical Relations: 1) sphere, surface area A = 4! r 2 and volume V = ! r 3 ; 2) cylinder
3
2
surface area A = 2! rL and volume V = ! r L .
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