(1) A 1/10 serial dilution of a bacterial culture was performed in tubes \"A\" t
ID: 142629 • Letter: #
Question
(1) A 1/10 serial dilution of a bacterial culture was performed in tubes "A" through "E" (dilutions were made in that order beginning with tube "A" like in the handout slide titled "Spread Plating"). 100 microliters of the diluted culture from the final tube "E" was spread onto a plate. Following incubation of the plate, growth of the bacteria produced 65 colonies on that plate. What is the concentration of viable cells in tube "B" (in cells per milliliter)? (2 pts)
(2) Penicillin is an antibiotic that inhibits an enzyme located in the cytoplasmic membrane (the enzyme is required to synthesize peptidoglycan). In general, explain why Gram-positive bacteria are sensitive to penicillin and Gram-negative bacteria are not? (2 pts)
(3) The 40x objective has a numerical aperture of 0.65. What would be the smallest distance (in m) between two objects that can be resolved when using that objective with light at 470 nm? (2 pts)
Explanation / Answer
(1)
Answer:
Original CFU/ml= ?
Dilution factor plate E= 1/10 x 1/10 x 1/10 x 1/10 x 1/10=10^-5
Number of colonies= 65
Volume of sample =0.1 mL to the plate
cfu/ml = (no. of colonies)/( dilution factor x volume of culture plate)
cfu/ml = (65)/ (10^-5 x 0.1)
cfu/ml = 6.5 x 10^7 cfu/mL
the concentration of viable cells in tube "B"
=Dilution factor of tube B x original number of cells
=10^-2 x 6.5 x 10^7
=6.5 x 10^5 cells/mL
(2)
Gram-negative bacteria have a lipopolysaccharide and protein layer which surrounds the peptidoglycan layer of the cell wall whereas in Gram-positive bacteria the peptidoglycan is not protected by an outer membrane. Therefore, Gram-positive bacteria are more sensitive to penicillin.
)3)
Resolution (d)= (0.61 x lambda)/NA
Given
NA=0.65
Lambda=470nm
d=?
d=(0.61 x 470)/0.65 =441nm =0.44uM
the smallest distance (in m) between two objects that can be resolved when using that objective with light at 470 nm is 0.44uM
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