```TBT4140 Biochemical Engineering
NTNU, Fall 2012
Laboratory exercise
Group:
Ingrid Andreassen – [email protected]
Ove Øyås – [email protected]
Martin Borud – [email protected]
1
Contents
1. Plot growth curve (semi-log) OD660nm. Calculate growth rate (µ) and generation time
(g). ......................................................................................................................................................................... 3
2. Plot OD660 nm versus cell concentration (g/mL). What is the relation between OD and
dry weight (DW)? (DW/OD). ...................................................................................................................... 3
3. Plot O2 and CO2 as a function of time. Explain the graphs. Does it fit with what you
observe in the growth curve? ..................................................................................................................... 6
4. Show calculation of limiting component in medium. Which two assumptions must be
met for these calculations to be correct? ............................................................................................... 7
5. Calculate OUR [mmol/(L-hr)] using mass balance ........................................................................ 9
6. Calculate OUR [mmol/(L-hr) and kla [s-1] using dynamic method of gassing out.
Which assumptions are made when calculating kla and OUR? .................................................. 11
7. Do OUR values from dynamic method and mass balance match? If not, why?................ 14
8. Calculate yield on carbon source (one time: end of the experiment) and yield on
oxygen (x times: when you have OUR) ................................................................................................ 14
9. Calculation of the respiratory quotient (RQ) ................................................................................ 15
10. Calculate specific enzyme activity given as (µmol ONPG hydrolysed)/(min x g). ....... 16
2
1. Plot growth curve (semi-log) OD660nm. Calculate growth rate (µ) and
generation time (g).
OD (660 nm) [-]
100
10
1
0.1
0
1
2
3
4
5
6
7
8
Time from start [h]
Figure 1.1: Growth curve for Escherichia coli grown at 37 °C. OD660 nm plotted against time.
Growth rate:
m=
ln(OD2 ) - ln(OD1 ) 10,28 - 2,62
1
=
= 3,83
t 2 - t1
(5,5 - 3,5)h
h
Generation time:
g=
ln 2
m
=
ln 2
= 0,18h
1
3, 83
h
2. Plot OD660 nm versus cell concentration (g/mL). What is the relation
between OD and dry weight (DW)? (DW/OD).
Data used for plotting OD660 nm versus cell concentration and dry weight is given in Table
2.1.
3
Table 2.1: Values for OD 660 nm dry weight and cell concentration.
Dry weight [g]
OD660 nm
Cell concetration [g/ml]
0.2075
1.28
0.02075
0.2085
2.62
0.02085
0.2493
16.8
0.02493
Presuming that the relationship between OD and cell concentration is linear in the
observed range and exploiting the fact that cell concentration is known to be 0.02085
g/mL when OD660 nm is 2.62, the following formula was utilized for calculating cell
concentrations from OD measaurements:
Cell concentration =
660
∙ 0.02085 g/mL
2.62
A plot of OD660 nm versus cell concentration is shown in Figure 2.1. The data used to
create the plot are given in Table 2.2.
Table 2.2: Data for OD660 nm and cell concentration.
OD660 nm
Cell concentration [g/mL]
0.214
0.00171
0.248
0.00198
0.373
0.00297
0.516
0.00411
0.992
0.00790
1.288
0.01027
1.775
0.01415
2.616
0.02085
3.950
0.03149
5.560
0.04431
6.700
0.05340
10.28
0.08193
13.40
0.10680
15.60
0.12433
4
18.90
0.15064
16.80
0.13390
0.16
0.14
y = 0,008x + 2E-17
OD (660 nm)
0.12
0.1
0.08
0.06
0.04
0.02
0
0
5
10
Cell concentration [g/mL]
15
20
Figure 2.1: Plot of OD660 nm versus cell concentration. A fitted curve found by linear regression is
included and its equation is shown.
Data from Table 2.1 was used to produce a plot of dry weight against OD660 nm. This is
shown in Figure 2.2.
0.26
0.25
Dryweight [g]
y = 0.0028x + 0.2027
0.24
0.23
0.22
0.21
0.2
0
5
10
OD (660 nm)
15
20
Figure 2.2: Plot of OD660 nm versus cell dry weight. A fitted curve found by linear regression is
included and its equation is shown.
5
The slope of the trendline in Figure 2.2 gives the relationship between dry weight and
OD. The slope is determined to 0.0028 g and therefore

= 0.0028 g

3. Plot O2 and CO2 as a function of time. Explain the graphs. Does it fit with
what you observe in the growth curve?
Figure 3.1 shows the flow of oxygen and carbon dioxide as a function of time.
25
10
9
20
8
15
6
5
10
4
Flow CO2 [L/min]
Flow O2 [L/min]
7
CO2
O2
3
5
2
1
0
0
0
100
200
300
Time [min]
400
500
Figure 3.1: Flow of O2 and CO2 as a function of time.
As can be seen from the graphs in Figure 3.1, the fraction of oxygen in the outlet air
decreases during the first 5-6 hours of the experiment. The fraction of carbon dioxide
increases in this period. Moreover, the rate of change in the levels of both gases seems to
increase with time; O2 decreases and CO2 increases seemingly exponentially with time.
6
As can be seen from Figure 1.1, the time period during which this is observed
corresponds to the growth phase of the bacteria in the culture. In this period, the cells
are consuming oxygen for growth and producing carbon dioxide. As the cell
concentration increases, so does the amount of oxygen needed for doubling as well as
the amount of carbon dioxide produced, explaining the observed gas flows.
After 5-6 hours, the O2 and CO2 levels increase and decrease, respectively, fairly
abruptly. This should be an indicator that the cells have stopped growing. After a while,
the oxygen flow decreases and the carbon dioxide flow increases again, signaling that
growth has resumed. After a short while however, the oxygen level begins to rise along
with the flow of carbon dioxide, before the CO2 flow abruptly decreases again. This
particular observation is difficult to explain based on the growth of the bacteria in the
culture. A stabilization in the gas levels due to the bacteria entering the stationary phase
would be expected.
4. Show calculation of limiting component in medium. Which two
assumptions must be met for these calculations to be correct?
Data for cell composition as well as molecular weight for the components of the cell are
given in Table 4.1. Table 4.2 shows the concentration, mass, molecular mass and number
of moles of the components in the medium.
Table 4.1: The composition of the cell and the molecular weights of the elements.
Component
Composition of E. Coli
(% dry weight)
Molecular weight
[g/mol]
Glucose
53
180
N
12
14
S
1
32.1
P
3
31
Mg
0.5
24.3
Ca
0.5
40.1
K
1
39.1
Na
1
23
Cl
0.5
35.5
Fe
0.2
55.5
7
Table 4.1:The concentration, mass, molecular mass and number of moles of the components in the
medium.
Concentration
[g/L]
m
[g]
M
[g/mol]
n [mol]
NH4Cl
8
4
53.5
0.074766355
Yeast extract
2
1
-
-
KH2PO4
1
0,5
136.1
0.003673769
Na2HPO4·2H2O
6
3
178.05
0,0168492
CaCl2·2H2O
0.01
0.005
147.02
3.4009E-05
Na2SO4
1
0.5
142.06
0.00351964
FeSO4·7H2O
0.03
0.0081
278.05
2.91315E-05
Lactose
6
1.62
-
-
Glucose
20
5.4
-
-
MgSO4·7H2O
0.02
0.0054
246.51
2.19058E-05
B-medium
A-medium
Component
Table 4.3 shows the amount of the different elements in the medium, the weight
percentages of the different elements compared to the total weight of the elements and
the difference between the weight percentage and the composition of elements in E. coli.
Table 4.2: The amount of the different elements in the medium, the weight percentages of the
different elements compared to the total weight of the elements and the difference between the
weight percentage and the composition of elements in E. coli (see Table 3.1).
Element
n[mol]
m[g]
Weight %
Difference
Nitrogen
0.0748
1.0467
36.3020
24.3020
Sulfur
0.0035
0.1139
3.9507
2.9507
Phospate
0.0205
0.6362
22.0647
19.0647
Magnesium
0.0000
0.0005
0.0185
-0.4815
Calcium
0.0000
0.0014
0.0473
-0.4527
Potassium
0.0037
0.1436
4.9818
3.9818
Sodium
0.0407
0.9370
32.4953
31.4953
Chloride
0.0001
0.0024
0.0837
-0.4163
Iron
0.0000
0.0016
0.0561
-0.1439
8
As can be seen from Table 3.3, magnesium shows the largest deviation from the
expected amount needed based on the composition of the cell. Magnesium is therefore
the limiting component.
For these calculations to be correct we must assume that there are no side reactions, the
that the given cell composition is correct and that all the elements are continually
consumed in equal amounts. Also, we should assume that 1 mol of lactose yields 1 mol of
glucose.
5. Calculate OUR [mmol/(L-hr)] using mass balance
Our is calculated from the following equation
OUR =
F
( xin  xout )
V
where F is the flow, V is the volume, xin is the oxygen ratio into the fermenter and xout is
the oxygen ratio out. The preferred unit for flow is mmol/h. To convert the unit of the
measured value, L/h, into mmol/h, ideal gas law is assumed and the following formula
employed:
V
p
t
F
1000 mmol/h
R T
The resulting flow is
F
0.4 1.049
1000  1.65 mmol/h
0.082  310
Data used in the calculation of OUR is shown in Table 5.1. A plot of OUR against time is
shown in Figure 5.1.
9
Table 5.1: Measured and calculated data needed to calculate OUR.
Time from start [h]
Weight [g]
O2, out (%)
Volume [L]
OUR [mmol/L·h]
0
10356.6
20.6
0.7900
0.00731
0.5
10343.6
20.38
0.7887
0.01192
1
10343.6
20.19
0.7887
0.01590
1.5
10326.3
20.08
0.7871
0.01824
2
10323.2
19.53
0.7868
0.02978
2.5
10304.4
18.29
0.7850
0.05591
3
10305.2
18.69
0.7850
0.04750
3.5
10277.2
18.29
0.7823
0.05610
4
10290.4
17.18
0.7836
0.07938
0.09
0.08
OUR [mmol/L·h]
0.07
0.06
0.05
0.04
0.03
0.02
0.01
0.00
0
1
2
3
Time [h]
Figure 5.1: Plot of OUR against time.
10
4
5
6. Calculate OUR [mmol/(L-hr) and kla [s-1] using dynamic method of gassing
out. Which assumptions are made when calculating kla and OUR?
To calculate kla, the dynamic method of gassing out was used. DO was recorded over
time after turning off the air flow into the fermenter. At a DO of 5%, the air flow was
turned back on and DO recorded until a stable value was reached.
The values obtained from the first gassing out are shown in Figure 6.1.
60
DO (% of maximum) [-]
50
40
30
y = -0.4306x + 39.227
R² = 0.9694
20
10
0
0
50
100
150
200
250
Time [s]
300
350
400
450
Figure 6.1: DO plotted against time for the first gassing out. A trendline found by regression is
added to the initial, linear part of the curve. The equation of the trendline and its coefficient of
determination, R2, are also shown.
OUR was calculated from the slope of a trendline found by performing linear regression
on the initial, linear part of the curve shown in Figure 6.1. The negative value of this
slope gives OUR with units s-1. To determine OUR with the desired units mmol/L·h, the
following formula was used:
[
mmol
[s−1 ]
1
]=
∙ 3600 s/h ∙ , ·
L∙h
100

CL,max is given by the following empirical equation:
CL,max  14.161  (0.3943  T )  (0.007714  (T 2 ))  (0.0000646  (T 3 ))  6,86 mg/L
using T = 37 °C. From this, OUR was determined to be 6.65 mmol/L·h.
The kla value can be found using the following equation
11
kla (t2  t1 )  ln(
CL ' CL1
)
CL ' CL 2
where CL1 is the CL value at t1 and CL2 at t2. CL’ was found to be 52%. From this equation
it can be seen that a plot of ln(′ − 1 )/(′ − 2 ) against time should result in a linear
plot with kla as slope. A plot of ln(′ − 1 )/(′ − 2 ) against time for the first gassing
out is shown in Figure 6.2.
0.8
ln(CL'-CL1)/(CL'-CL2)
0.7
0.6
0.5
y = 0.0038x
R² = 0.9661
0.4
0.3
0.2
0.1
0
0
20
40
60
80
Time [s]
100
120
140
160
Figure 6.2: Plot of (′ −  )/(′ −  ) against time for the first gassing out. A trendline found
by linear regression is included and the equation of the trendline as well as its coefficient of
determination, R2, are shown.
The slope of the trendline in Figure 6.2 gives kla = 0.0038 s-1 for the first gassing out.
The values obtained from the second gassing out are shown in Figure 6.3.
12
35
Do (% of maximum) [-9
30
25
y = -0.6425x + 39.133
R² = 0.9837
20
15
10
5
0
0
50
100
150
200
250
Time [s]
Figure 6.3: DO plotted against time for the second gassing out. A trendline found by regression is
added to the initial, linear part of the curve. The equation of the trendline and its coefficient of
determination, R2, are also shown.
For the second gassing out, OUR was determined to 9.92 mmol/L·h.
A plot of ln(′ − 1 )/(′ − 2 ) against time for the second gassing out is shown in
Figure 6.4.
0.9
ln(CL'-CL1)/(CL'-CL2)
0.8
0.7
0.6
0.5
0.4
0.3
y = 0.0085x
R² = 0.9837
0.2
0.1
0
0
20
40
60
80
Time [s]
100
120
140
160
Figure 6.4: Plot of (′ −  )/(′ −  ) against time for the second gassing out. A trendline
found by linear regression is included and the equation of the trendline as well as its coefficient of
determination, R2, are shown.
The slope of the trendline in Figure 6.4 gives kla = 0.0085 s-1 for the second gassing out.
13
The following assumptions were made in the calculations:
1. The liquid phase is well mixed
2. The response time of the dissolved oxygen electrode is much smaller than (1/kla)
3. The measurement is performed at sufficiently high stirrer speed to eliminate
liquid boundary layers at the surface of the oxygen probe
4. Gas-phase dynamics can be ignored
7. Do OUR values from dynamic method and mass balance match? If not,
why?
No, the OUR values determined by the dynamic method of gassing out do not match
those obtained from the mass balance (see Table 5.1 and Figure 5.1). The difference
between the values is very big, about a factor of 100. It is not known precisely what
causes this deviation. One or more of the assumptions made in the calculations in
section may not be valid or the empirical formula used to calculate CL,max may be very
inaccurate at 37 °C (it is determined for 36 °C). the assumption of ideal gas in section 5
could also be invalid. It does however seem unlikely that any of these errors should
result in a difference of the magnitude that is actually observed.
It seems probable that an invalidity in the mass balances is responsible for the deviation,
as significant gas and liquid leakages did occur several times during the experiment. It is
plausible that the loss of mass from these leakages has made the mass balance
calculations invalid.
8. Calculate yield on carbon source (one time: end of the experiment) and
yield on oxygen (x times: when you have OUR)
To calculate the yield on carbon source, we must know the total mass of the cells in the
reactor. Therefore, the relation between OD and dry weight used in section 2 is used. At
the end of the experiment, the OD is … which gives a cell concentration of … . Assuming
that the reactor volume is constant – a necessary but probably highly incorrect
assumption – the bacterial biomass at the end of the experiment would be. Given the
initial concentrations of glucose and lactose in the medium, it is known that 20.8 g of
carbon sources were used. The yield on carbon source can thus be calculated as follows:
=
g ,
=
g  + g
=
g cells
g carbon
The yield on oxygen can be calculated from
14
2 =
µ

9. Calculation of the respiratory quotient (RQ)
15
16
10. Calculate specific enzyme activity given as (µmol ONPG hydrolysed)/(min
x g).
Change in absorbance at 420 nm as a result of ONPG hydrolysis by β-galactosidase was
measured over time for each of three samples taken during the experiment. Data
obtained from these measurements are given in Table 10.1, 10.2 and 10.3 and plots of
these data are shown in Figure 10.1, 10.2 and 10.3.
Table 10.1: Measured absorbance at 420 nm with time for samples 1, 2 and 3.
Absorbance (420 nm)
Time [s]
Sample 1
Sample 2
Sample 3
10
20
-9
192
15
13
-1
284
20
11
8
376
25
11
23
460
30
11
36
538
35
10
50
610
40
9
62
675
45
11
71
733
50
12
80
781
55
10
89
819
60
9
99
848
65
9
105
871
70
9
110
890
75
9
119
905
80
9
125
917
85
9
133
928
90
9
152
937
95
9
166
945
100
9
177
951
17
105
9
190
957
110
9
205
960
115
8
223
963
120
9
234
965
125
9
243
968
130
9
255
967
135
9
267
959
140
11
276
959
145
10
284
958
150
9
293
960
155
9
303
965
160
9
313
968
165
9
318
967
170
9
318
961
175
9
320
957
180
9
324
956
Absorbance (420 nm)
25
20
15
10
5
0
10 20 30 40 50 60 70 80 90 100 110 120 130 140 150 160 170 180
Time [s]
Figure 10.1: Plot of absorbance at 420 nm against time for sample 1.
Figure 10.1 shows the change in absorbance with time for sample 1, which was taken
approximately 2.5 hours into the experiment, before the depletion of glucose. The
18
absorbance seems to stabilize and no growing trend can be observed. This indicates no
or very low enzyme activity. In other words,

≈ 0 s −1

This result is as expected, as β-galactosidase should not be present at the beginning of
the experiment.
350
Absorbance (420 nm)
300
250
y = 2.171x - 33.532
R² = 0.995
200
150
100
50
0
0
10 20 30 40 50 60 70 80 90 100 110 120 130 140 150 160 170 180
-50
Time [s]
Figure 10.2: Plot of absorbance at 420 nm against time for sample 2 with trendline added to the
linear part of the curve. The equation of the trendline and its coefficient of determination, R2, are
also shown.
Figure 10.2 shows the change in absorbance with time for sample 2 with a trendline
added to the linear part of the curve. The absorbance grew throughout the experiment,
although a plateau seems to be reached around the last four data points. The
instantaneous change in absorbance with time is given by the slope of the trendline,
meaning that

= 2.171 s−1

19
1200
Absorbance (420 nm)
1000
800
600
y = 74.183x + 145.64
R² = 0.9902
400
200
0
10 20 30 40 50 60 70 80 90 100 110 120 130 140 150 160 170 180
Time [s]
Figure 10.3: Plot of absorbance at 420 nm against time for sample 3. Maximum absorbance is
estimated from the mean of the values in the plateau of the graph and indicated as a horizontal
line. A fitted curve found by regression is included in the linear part of the graph. The equation of
the trendline and its coefficient of determination, R2, are also shown.
Figure 10.3 shows the change in absorbance with time for sample 3 with a trendline
added to the linear part of the curve. The maximum absorbance at 420 nm was
determined from an average of the values in the plateau of the graph:
= 962.3
Also, the slope of the trendline in Figure 10.3 gave

= 74,18 s−1

The initial concentration of ONPG was calculated as follows:
µmol
1 1 2.5 mL ∙ 1mL
2 =
=
= 0.278 µmol/mL
2
9 mL
mol
µmol ONPG0 = 2 ∙ 2.5 mL = 0.278 µ mL ∙ 2.5 mL = 0.695 µmol
From this, V0 was calculated for the second and the third sample. For sample 2:
20
0 =
µmol ONPG0 Abs 0.695
∙
=
∙ 2.171 ∙ 60 = 0.09408 µmol/min
Absmax

962.3
For sample 3:
0 =
µmol ONPG0 Abs 0.695
∙
=
∙ 74.18 ∙ 60 = 3.214 µmol/min
Absmax

962.3
For sample 1, V0 = 0, as dAbs/dt = 0.
Once again it is assumed that the relationship between OD and cell concentration is
linear in the observed range and the fact that cell concentration is known to be 0.02085
g/mL when OD660 nm is 2.62 is exploited. It is also known that 0.3 mL of the sample is
mixed with buffer solution to prepare it for OD measurement:
Weight of biomass = 0.02085
g
∙
∙ 0.3 mL
mL 2.62
For sample 2, OD is 2.62 and the weight of biomass is simply
0.02085
g
∙ 0.3 mL = 0.00625 g
mL
For sample 3, OD is 13.4 and the weight of biomass is
0.02085
g 13.4
∙
∙ 0.3 mL = 0.0320 g
mL 2.62
The specific enzyme activity is given by
Specific enzyme activity =
0
Weight of biomass
For sample 2, this gives
Specific enzyme activity =
0.09408
= 15.06 µmol/min ∙ g
0.00625
For sample 3:
Specific enzyme activity =
3.214
= 100.4 µmol/min ∙ g
0.0320
21
```