Biomedical Engineering Reference
In-Depth Information
10
0
X
Lactose
Glucose
10
-1
S, g/L
or
X, g/L
10
-2
10
-3
10
-4
10
-5
0
5
10
15
Time, h
FIGURE 10.14
Diauxic growth curve for E. coli on glucose and lactose.
2 h after inoculation, cells are growing rapidly, glucose is being consumed, and lactose is not
being utilized. At 7 h, cell mass accumulation is zero. All the glucose has been consumed. At
10 h, the culture is growing and lactose is being consumed, but the rate of growth (cell mass
accumulation) is less than at 2 h. Explain what is happening with intracellular control to
account for the observations at 2, 7, and 10 h. What rate of
-galactosidase formation would
you expect to find in the culture at these times in comparison to the basal rate (which is
b
<
1%
of the maximum rate)?
Solution:
At 2 h, the lac-operon is fully induced, since lactose converted to allolactose combines with
the lac i repressor protein, inactivating it. With the repressor protein deactivated, RNA poly-
merase is free to bind to the promoter but does so inefficiently. Glucose levels are still high,
which results in higher levels of ATP and low levels of AMP and cAMP. Consequently, little
cAMP-CAP complex is formed, and the interaction of the lac promoter with RNA poly-
merase is weak in the absence of cAMP-CAP. The rate of
b
-galactosidase formation would
be slightly increased from the basal level
d
perhaps 5% of the maximal rate.
At 7 h, the glucose has been fully consumed. The cell cannot generate energy, and the level
of ATP decreases and cAMP increases. The cAMP-CAP complex level is high, which
increases the efficiency of binding RNA polymerase to the lac promoter. This increased
binding leads to increased transcription and translation. The rate of
b
-galactosidase forma-
tion is maximal and much higher than the basal rate or the 2 h rate. However, the cells
have not yet accumulated sufficient intracellular concentrations of
b
-galactosidase and lac
permease to allow efficient use of lactose and rapid growth.
At 10 h, the intracellular content of proteins made from the lac-operon is sufficiently high
to allow maximal growth on lactose. However, this growth rate on lactose is slower than on
glucose, since lactose utilization generates energy less efficiently. Consequently, the cAMP
level remains higher than when the cell was growing on glucose. The level of production
of
-galactosidase is thus higher than the basal or 2-h level.
Irrespective of whether an enzyme is made from a regulated or constitutive gene, its
activity in the cell is regulated. Let us now consider control at the enzymatic level.
b
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