Chapter 6: Microbial Growth

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Chapter 6: Microbial Growth

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Chapter 6: Microbial Growth. How do bacteria grow?. Not in size Increase in population size One cell divides into 2 new cells – binary fission. Binary fission. Binary fission:. Attachment of chromosome to p.m.; replication of DNA; new p.m. and cell wall laid down between the 2 chromosomes - PowerPoint PPT Presentation

Transcript of Chapter 6: Microbial Growth

Page 1: Chapter 6: Microbial Growth

Chapter 6: Microbial Growth

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How do bacteria grow?

  Not in size

Increase in population size

One cell divides into 2 new cells – binary fission

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Binary fission

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Binary fission:

Attachment of chromosome to p.m.; replication of DNA; new p.m. and cell wall laid down between the 2 chromosomes

This is the way that each new daughter cell gets one chromosome

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Number of generations

Number of cells Log of # of cells10X = #

0 1 01 2 0.32 4 0.63 8 0.94 16 1.25 32 1.56 64 1.87 128 2.18 256 2.49 512 2.710 1024 3.011 2048 3.312 4096 3.6

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How can we describe growth?

2n = no. of cells in n generation

Generation time

Nt = N0 x 2n

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Growth problemsIf Staphylococcus aureus has a doubling time

(generation time) of 30 minutes and 5 hours have passed, how many generations have been produced?

a. How many 30 minute time chunks are in 5 hours?

= 10

ANSWER: 10 generations

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Growth problemsIf Staphylococcus aureus has a generation time

of 30 minutes and 5 hours have passed, how many bacteria will be present at the end of the time period?

a. We have already determined that 10 generations will occur.

b. 2n = # cells at n generationc. 210 = # cells at the 10th generation

ANSWER = 1024 cells

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Growth problemsIf Staphylococcus aureus has a generation time

of 30 minutes and 5 hours have passed, how many bacteria will be present at the end of the time period if we start with 3,000 cells? Nt = N0 x 2n

Nt = 3,000 x 210

= 3,072,000 cells at the end of 5 hours

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Growth curve: lag, log, stationary, and death phases

What occurs in each?

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How can population size be counted?(Advantages and disadvantages of each

method)

1. Direct methods

A. Microscopic count with hemacytometer/

Petroff Hauser counting chamber

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Petroff Hauser counting chamber

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B. Plate counts – dilution series and plates

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C. Filtration

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D. Coulter counter/flow cytometer/Fluorescence activated cell sorter (FACS)

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2. Indirect methodsA. Dry weight

B. Metabolic activity

C. Turbidity

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Turbidity

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Growth requirements of microbesA. Temperature:

ThermophilesMesophilesPsychrophiles

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B. pH: acidophilesC. Osmotic pressure (# of solutes in solution)

HalophilesD. Oxygen:

Types:Obligate aerobeFacultative anaerobeObligate anaerobeAerotolerant anaerobeMicroaerophilic

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Enzymes needed to survive in presence of oxygen:

CatalasePeroxidaseSuper oxide dismutase (SOD)

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E. Nutrients

C, N, P, S elements neededMg, Fe, etc. trace elements neededMedia:

Defined or complexSelective vs. differentialSpecial

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Mannitol salt agar – selective medium

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Blood agar – differential medium

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The End