PROBLEM SET 5.1: BLOOD
ANSW ER KEY
1. The value of Lp
for the red blood cell is about 1.8 x 10-11 cm3
dyne-1 s-1
. Its surface area is about
... [Show More] 1.35
x 10-6 cm2
(Solomon, Methods in Enzymology, 192-222 , 1989).
A. W hat is the initial osmotic flow if the osmolarity inside is initially 300 mOsM and the
osmolarity outside is 275 mOsM? (Assume F = 1.0 for all solutes)
The appropriate equation here is
QV
= A Lp
()P - F)B)
In the particular example no information is given about the hydrostatic pressure difference across the
mem brane, so we will assume there is none. We are given values for A and for Lp
and F and for the
osm olarity. So we calculate
Qv
= 1.35 x 10-6 cm2
x 1.8 x 10
-11 cm3 dyne-1 s-1 (RT (0.300 - 0.275))
W e can calculate )B = RT)C as .082 L atm m ol-1 °K-1 x 310°K x 0.025 = .6355 atm
W e need to convert this to units of dyne cm-2 in order to use the units of Lp
that are given. Fortunately,
we already did this last semester: 1 mm Hg = 1334 dyne cm-2. Since 1 atm = 760 m Hg, we have 1
atm = 1.01 x 106
dyne cm-2 so we can multiply this in our equation for Qv
:
Qv
= 1.35 x 10-6 cm2
x 1.8 x 10
-11 cm3 dyne-1 s-1 x .6355 atm x 1.01 x 106
dyne cm-2 atm-1
= 1.56 x 10-11 cm3
s-1 = 15.6 x 10-12 cm3
s-1
B. If the volume of the cell is 100 x 10-12 cm3
, how long would it take to double its volume
provided that the osmotic pressure and area of the membrane and Lp
did not change?
The cell would double its volume if another 100 x 10-12 cm3
of water cross the mem brane. If the initial
flow is maintained, this would occur in
100 x 10-12 cm3
/ 15.6 x 10-12 cm3
s
-1 = 6.41 s
The purpose of this calculation is to convince you that these fluid shifts are very fast.
C. In the case described, how much water would be required to enter the cell to equilibrate the
osmotic pressure between inside and outside? Assume that the outside bath is essential
infinite so that its osmotic pressure is kept constant.
In this case there has to be enough water entering the cell to dilute its contents from 300 mOsM to
275 mOsM, at which point the flow of water would cease. To calculate this volume of water, we
assume that the intracellular contents cannot leave the cell. The total amount of osmolytes is just V0
x C, where C is the concentration of osmolytes and V0
is the initial volume of the cell. After some
volume has entered the cell, the new concentration is V0
x C/V, where V = V0
+ )V, and )V is the
volume of water that enters the cell. This new concentration would be equal to the outside osm olarity.
Here we write
V0
x C/ V = Cout [Show Less]