Box 55 Diauxic growth

When E. coli grows in a medium containing glucose and lactose, it preferentially metabolises the former since it is more energy efficient to do so. The cell has a regulatory mechanism that suppresses the synthesis of lactose-metabolising enzymes until all the glucose has been used up (see Chapter 11). At this point a second lag phase is entered, while the lactose metabolising enzymes are synthesised. Such growth is termed diauxic, and the resulting growth curve is characteristically biphasic.



where N0 is the number of cells at the start of exponential growth, Nt is the number of cells after time T, L is the length of the lag phase and n is the number of doubling times that have elapsed. n is therefore equal to T/Td, where Td is the doubling time. Substituting into the first equation:

This can be expressed more conveniently by using logarithms to the base 2 (don't worry too much about how this is done!):

Thus, if we know the number of cells at the start and end of a period of exponential growth, we can calculate the doubling time. See Box 5.6 for a worked example. We can also determine the mean growth rate constant (K); this is a measure of the number of doublings of the population per unit time, and is equal to 1/ Td *.

Many antibiotics such as penicillin (Chapter 14) are only effective when cells are actively dividing, since they depend on disrupting new cell wall synthesis.

You may also come across the instantaneus growth rate (p), which equals

0 0

Post a comment