Unlock document.
This document is partially blurred.
Unlock all pages and 1 million more documents.
Get Access
Chapter 14
Problem. 14.1 What is P for a single cell or homogeneous population with the equivalent of
50 monomer plasmids at division if multimerization results in 40% of plasmid DNA as
dimers and 16% as tetramers?
Plasmid – free cells / division
In this part of the question we calculate how the distribution of plasmids in a population alters
P. Let
P
be the population – averaged value. Then
ii
P f P=
; Pi< 1 & fi< 1 where fi is the
fraction of the population in category i with a plasmid # j. Pi is the probability that cells with
plasmid # j will form a plasmid – free cell upon division. For example, category 1 is 4% of
the population with an average plasmid # of 6 (since the data give a range of plasmid number
Note that only the first two terms are significant – the others are very small
Note also that the average copy number is about 20. If the probability of plasmid loss was
calculated based on the population averaged copy number then:
6
P 1.9 10−
=
–a difference of
3 orders of magnitude!
Problem 14.2. As a basis for the calculation we assume @ t = 0 that n+ = 1000 and n– = 0.
Further μ+ = μ– for
i
n−
where i = 1, 2, or 3;
4
n−
= dead cell
Problem 14.3. Total cells required:
( )
39 1.2 0.0005 1
1 1.2 0.0005
f1 1.2 0.0005 2
0.64 f 1 f 0.36
++−
−+
−−
=− −
= = − =
Problem 14.4. Several choices could be well-defended. Since post-translational processing is
not required, efficiency and productivity become critical. If the protein can be resolabilized
Problem 14.5. See D. A. Lauffenburger, Biotech.Prog.1: p53 (1985) for more details
Let: n+ = # plasmid – containing cells
d
n C n
−−
−−
+→
For a chemostat
( )
1
m
S
S
dn dt Dn 1 P n
KS
+
+ + +
+
+
= − + −
+
(1)
11
mm
SS
SS
dn dt Dn n P n
K S K S
k Cn
−+
−+
− − − +
+−
+ +
= − + +
++
−
(2)
dC dt DC n+
= − +
(3)
( )
11
1
mm
0
SS
SS
dS dt D S S n n
K S K S
+−
+−
+−
+ +
= − − −
++
(4)
11
SS
YY
+−
Problem 14.6. Plot f– vs time. It gives a sigmoidal curve case 1
Plot ln f– vs # generations – yields straight lines (correlation coeff 0.99) with following
eqns:
D = 0.3 h–1 ln f– = 0.451 t – 5.637
D = 0.67 h–1 ln f– = 0.183 t – 5.759
11
m
1
@D 0.67h R 0.18gen 0.0032
0.00576gen
−−
−
= =
=
Comparing to example 14.2
The addition of amino acids decreases stability!
Problem 14.7. The key point is that in a large volume the formation of a single plasmid –
free cell “contaminates” the whole volume. In the gel – or any device which seqregates a
large volume into many distinct sub units – cross centamination from one sub unit to another
is not allowed. Even if a plasmid – free cell has a very large growth advantage, it can only
effect change in its own vessel or cavity.
( )
g
n
f0
Z Z 1 P
+=−
(1)
where ng = # generations and
f
Z+
is the number of cavities with only plasmid – containing
cells. In the cavities with plasmid – free cells the ratio of n+ to n– will be determined by P and
by ∆μ or = μ– /μ+ exactly as it would be in a large well-mixed reactor. To estimate f– (or 1 –
f+) in a well-mixed vessel:
Solving eq. 2 with this initial condition yields
( )
( )
( )
1 P t 1 P t t
e e e
1P
++
−
−+ − −
+
+−
+−
− −
Define
g
t
nln 2
+
=
(with μ+ in units of h–1)
gg
# cavities with # cavities with
Increase Increase in
only plasmid- one or more
in cell
ncontaining cells plasmid-free
cell # in
at n 7.64 cells at n 7.64
cavities
+
=+
==
# of
plasmid-plus
cells
(8)
Using eq. 1 to determine # cavities with only plasmid-containing cells,
g
n
0g
ln n n n ln 2 or 2 ,
+==
and eq 4 for increase in plasmid-containing cells in the
prescence of plasmid-free cells gives:
( ) ( )
( )
gg
g
nn
n1 P t
o0
n N 1 P 2 N 1 1 P e +
−
+ + +
= − + − −
(9a)
or noting
( ) ( )
g
n 1 P
1 P t
e2
+−
− =
( ) ( )
( )
gg
gg
nn
n n 1 P
o
n N 1 P 2 1 1 P 2 −
++
= − + − −
(9b)
We can show that n+ for case 2 (eq 9b) is always n+ for case 1 (eq 4) or
Also we can show that n– (case 2) is n– (case 1); i.e. using eq. 5
g
Since n+ (case 2) n+ (case 1)
( ) ( )
( )
g
n1 P t t
o
P
n N 1 1 P e e
1P
+−
−
+
−+
+−
= − − −
− −
reduces when we note that μ– t = ln 2 ng
o
1P
−+
− −
Problem 14. 8. This operating strategy can make use of the stoichiometeric approachs
described in chapter 7. A constant cell composition and RQ will be assumed. The rate of CO2
Problem 14.9. The most general case allows cells to segragate plasmids imperfectly (see
K.D. Wittrup & J. E. Bailey, Biotechnol.Bioeng.31: 304 (1988)). Here we assume perfect
segregation or partitioning of plasmids and an exact derabhing of plasmids from birth to
division.
Let n(i) = # cells with i plasmids at division & P (o, i) = probability of forming a plasmid-free
cell from a cell with i plasmids
o i i
i2
=
Problem 14.10. Since n–0 0 we can not use eq. 14.51
Starting with eqs. 14.44 & 14.45 and initial condctions of n+o = 0.95 No and n–o = 0.05 No we
This is a trial & error solution as we do not know t that gives Nfinal = 4 × 1016 cells (at least
trial & error is easier).
final
0.9998 0.69 1.0
−
You can make first estimates of t using just plasmid-containing & plasmid-free cells in
inoculum. t+ ≈ 10 h t– ≈ 9.9 h → so the two terms about equal. After two trials t = 9.15 h
gives
( )
13 16
final
N 2094 15 1883 10 3.992 10 cells= + + =
16
16
final
n2.094 10
N 3.992 10
+
+
Problem 14.11. a) no = 1 × 103 ml/L ∙ 1 × 1010 cells/mL ∙ 1L = 1013 cells
nfinal = 1 × 103 mL/L ∙ 5 × 109 cells/mL ∙ 2 × 104 L = 1017 cells
g
n 13.3
=
1.00.95 1.053
−
+
= = =
b) Similar to prob. 14.10 as n–0 0 then
( )
1 P t
0
n n e +
−
++
=
;
( )
( )
1 P t tt
o0
Pn
n e e n e
1P
+
− − −
++
−−
+−
= − +
− −
17 0.995 0.15 6 17
n 0.994 1 10 e 2.44 10 cells
+= =
that contain plasmid 6 hours after induction
( )( )
( )( )
( )
17 0.9995 0.15 6 1.0 6
17 1.0 6
17 17
t
0.0005 0.15 0.994 10
n e e
0.9995 0.15 1.0
0.00593 10 e
0.035 2.39 10 2.42 10 cells
n2.44
f 0.50
n n 2.44 2.42
−
+
+−
=−
−
+
= + =
= = =
++
