Bacteriology 102 – Exp. 5.4:


By definition, bacterial growth is cell replication – i.e., growth of the culture. Most species of bacteria replicate by binary fission, where one cell divides into 2 cells, the 2 cells into 4, the 4 into 8, etc. If this cell division occurs at a steady rate – such as when the cells have adequate nutrients and compatible growing conditions – we can plot numbers of cells vs. time such as on the graph at right. Before too long, we will need to extend the paper vertically as the population continues to double. For a culture where cells divide every 20 minutes, one cell can result in 16,777,216 (i.e., 2^{24}) cells after just 8 hours – barring nutrient depletion or other growthaltering conditions.  
If we were to convert our vertical axis to a logarithmic scale – as on the graph at right – we will not need as many sheets of graph paper, and we will find that a steady rate of growth is reflected as a straight line. (On the vertical axis, the same distance on the paper is covered with each doubling.) This type of graph paper is called semilogarithmic graph paper on which we will be plotting our class results. The numbers we plot will fall on the graph at the same place the logarithms of these numbers would fall when plotted on conventional graph paper.  
The example at right shows the type of graph we may obtain from our class data. NOTE: As absorbance is already a logarithmic value, it makes no sense to plot it on the same sheet of semilog paper as CFU/ml – unless one wants to merely demonstrate the general upward trend of both graphs. But there is no way that growth rate and generation time can be determined by an absorbance graph so posted. On linear paper one will get a real correlation between the log of CFUs/ml vs. time and absorbance vs. time.) Rather than "connecting the dots," we draw the best straight line among our CFU/ml points to represent the phases of growth – lag, exponential, and the start of the maximum stationary phase. For the growth rate formula we are about to use, we need to choose two points on the straight line drawn through the exponential phase, also making note of the time interval between them. As we will be converting our numbers to logarithms for the formula, why not choose two points for which the logs are easy to obtain? (For example, the log of 1X10^{10} is simply 10.)
Using the first formula, we find the growth rate which is the number of generations (doublings) per hour:
With the second formula, we find the generation time which is the time it takes for the population to double: GENERATION TIME (t_{gen}) = 1/k = 1/2.21 = 0.45 hour/gen = 27 min/gen When we graph the CFUs/ml and absorbance on the same graph, we would hope to see an upward trend for both. Sometimes the absorbance continues to rise after the CFUs/ml level off into the maximum stationary phase. What would be the cause of that? With a clear graph, one should be able to determine the generation time without the use of formulas. Just look for a doubling of the population and the time it takes for that to happen. For example – in the above graph – the time it takes to go from 3 X 10^{9} to 6 X 10^{9} appears to be approximately 30 minutes, which is close to the generation time determined above. The laboratory manual referred to herein is referenced here. Check out the growth curve questions on pages 157158. 
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