Given the data below (spread plate counts in CFUs/mL and OD reads for S. aureus) create...

Given the data below (spread plate counts in CFUs/mL and OD reads for S. aureus) create a graph of time v. CFUs/mL, time v. OD = 600nm and OD = 600nm v. CFUs/mL (2 pts). Then calculate the growth rate and the generation time for this organism. (1 pt) Turn in images of your graphs and show your calculation work.

Time (h)	CFUs/mL	OD = 600nm
0	1x10⁷	0.021
1	1.8x10⁷	0.041
2	7x10⁷	0.175
3	2.4x10⁸	0.463
4	1x10⁹	1.530
5	1.2x10⁹	1.790
6	1.3x10⁹	1.920

Say we create a graph of OD = 600nm v. CFUs/mL for S. aureus… What is the OD = 600 nm reading measuring? And are there any factors that could affect the results from assuming that a particular OD corresponds with a specific bacterial concentration? (1 pt)

Expert Solution


Slope of the linear equation of the semi-logarithmic graph	m =	0.3768	per hour
m corresponds to the growth rate (mu)
Equation for Growth Rate (mu)	=	ln2/td
td (Doubling Time)	=	ln2/mu
		ln2/0.3768

	=	1.839562581	hours

gladiator answered 3 years ago

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