Solution-Explain the kinetics of bacterial growth during | Homework Helpers
Problem – E. coli grows faster on the monosaccharide glucose than it does on the disaccharide lactose for two reasons:
(1) lactose is taken up more slowly than glucose and
(2) lactose must be hydrolyzed to glucose and galactose (by _-galactosidase) before it can be further metabolized. When E. coil is grown on a medium containing a mixture of glucose and lactose, it shows complex growth kinetics (Fig I squares). The bacteria grow faster at the beginning than at the end, and there is a lag between these two growth phases when they virtually stop growing. Assays of the concentrations of the two sugars in the medium show that glucose falls to very low levels after a few cell doublings (Fig. I circles), but lactose remains high until near the end of the experimental time course. Although the concentration of lactose is high throughout the experiment, _-galactosidase, which is regulated as part of the lac operon, is not induced until more than 100 minutes have passed (Fig. 1 triangles).
a. Explain the kinetics of bacterial growth during the experiment. Account for the rapid rate of initial growth, the slower rate of final growth, and the delay in growth in the miIDle of the experiment.
b. Explain why the lac operon is not induced by lactose during the rapid initial phase of bacterial growth.
