A fictional cubed-shaped bacterium, Bacterius cubis, occupies a volume of 2.1 fe

Question

A fictional cubed-shaped bacterium, Bacterius cubis, occupies a volume of 2.1 femtoliters. This particular type of bacteria is known to communicate with its own species by secreting a small molecule called bactoX (MW = 130.9 g/mol). Answer the following questions regarding this bacterium A) Each bacterium contains 9.780 x103 bactoX molecules that can be secreted. How many moles of bactoX are present in a 1.6-L sample volume that contains 6.790 × 106 bacterial cells? Number mol B) Calculate the molarity (mol/L) of bactoX in the 1.6- 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. Number C) Scientists discovered that bactoX will induce a luminescence response in B. cubis when the concentration in a sample reaches 5.10 × 10-12 M. How many molecules of bactoX must be present in solution to initiate luminescence in a 1.7-i sample? How many individual bacteria cells does this represent (the number of bactoX molecules per cell is given in part A)? Number Number Scroll down to view the rest of the question. molecules cells

Explanation / Answer

A. Concentration of bactoX = 9.780 x  103 molecules / bacteria  

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

= ( 9.780 x  103 molecules / bacteria) x 6790000 bacteria

= 6.640 x 1010 molecules

Moles of bactoX = No. of molecules / Avogadro number

  = (6.640 x 1010 molecules) / (6.023 x 1023 molecules/ mol)

= 1.102 x 10-13 mol

B. Total volume of solution = 1.6 micro L = 1.6 x 10-6 L

Molarity of bactoX = moles / Volume of solution in liters

   = (1.102 x 10-13 mol) / (1.6x 10-6 L)

= 6.88 x 10-8 moles / L = 6.88 x 10-8 M

C. Volume = 1.8 uL = 1.7 x 10-6 L

[BactoX] in cell = 5.10x 10-12 M = 5.10 x 10-12 moles/ L

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

= (5.10 x 10-12 moles/ L) x (1.7 x 10-6 L) x (6.023 x 1023 molecules/ mol)

= 5.221 x 106 molecules

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

= (5.221 x 106 molecules) / (9.780 x  103 per cell)

= 0.533 x 103 cells

D. Volume of bacterial cell = 2.3 fL = 2.1 x 10-15 L

Required volume of packing bacterial cells = Volume of a bacterial cell x No. of bacteria

= (2.1 x 10-15 L/ cell) x (0.533 x 103 cells)

= 1.119 x 1012 L = 1.119x 103 nL

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