A fictional cubed-shaped bacterium, Bacterius cubis, occupies a volume of 2.3 fe

Question

A fictional cubed-shaped bacterium, Bacterius cubis, occupies a volume of 2.3 femtoliters. This particular type of bacteria is known to communicate with its own species by secreting a small molecule called bactoX (MW = 116.0 g/mol). Answer the following questions regarding this bacterium.

A) Each bacterium contains 4.960 × 103 bactoX molecules that can be secreted. How many moles of bactoX are present in a 2.8-L sample volume that contains 3682000 bacterial cells?

B) Calculate the molarity (mol/L) of bactoX in the 2.8-L sample volume if all of the bacteria were to simultaneously secrete all of the bactoX molecules contained within their cell bodies into the vial.

C) Scientists discovered that bactoX will induce a luminescence response in B. cubis when the concentration in a sample reaches 5.00 × 10-12 M. How many molecules of bactoX must be present in solution to initiate luminescence in a 1.8-L sample? How many individual bacteria cells does this represent (the number of bactoX molecules per cell is given in part A)?

D) What volume (in nanoliters) would the bacteria cells in part C occupy if tightly packed (assume there is no space between the cells)?

Explanation / Answer

Ans. A. Concentration of bactoX = 4.960 x 103 molecules / bacteria  

Now,

Number of bactoX in 3682000 bacteria = [BactoX] x number of bacteria

                                                = (4.960 x 103 molecules / bacteria) x 3682000 bacteria

                                                = 1.826 x 1010 molecules

Moles of bactoX = No. of molecules / Avogadro number

                                    = (1.826 x 1010 molecules) / (6.023 x 1023 molecules/ mol)

                                    = 3.032 x 10-14 mol

#B. Total volume of solution = 2.8 uL = 2.8 x 10-6 L

Now,

Molarity of bactoX = moles / Volume of solution in liters

                                    = (3.032 x 10-14 mol) / (2.8 x 10-6 L)

                                    = 1.083 x 10-8 moles/ L

                                    = 1.083 x 10-8 M

#C. Volume = 1.8 uL = 1.8 x 10-6 L

[BactoX] in cell = 5.00 x 10-12 M = 5.00 x 10-12 moles/ L

Molecules of bactoX = [bactoX] x Vol. of bacterial cell x Avogadro number

                        = (5.00 x 10-12 moles/ L) x (1.8 x 10-6 L) x (6.023 x 1023 molecules/ mol)

                        = 5.421 x 106 molecules

Now,

No. of bacterial cells = No. of bactoX / (No. of bactoX/ cell)

                                    = 5.421 x 106 / (4.960 x 103/ cell)

                                    = 1.093 x 103 cells

#D. Volume of bacterial cell = 2.3 fL = 2.3 x 10-15 L

Required volume of packing bacterial cells =

Volume of a bacterial cell x No. of bacteria

= (2.3 x 10-15 L/ cell) x 1.093 x 103 cells

= 2.514 x 1012 L                                           ; [1 nL = 10-9 L]

= 2.514 x 10-3 nL

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