Question

In: Biology

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

  1. 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

  1. 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)

Solutions

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

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