Question

A fictional cubed-shaped bacterium, Bacterius cubis, occupies a volume of 2.9 femtoliters. This particular type of bacteria is known to communicate with its own species by secreting a small molecule called bactoX (MW 1214 g/mol). A. Each bacterium contains 5530 bactoX molecules that can be secreted. How many moles of bactoX are present in a 1.0 L sample volume that contains 2.620 x 10" bacterial cells? moles: mol B. Calculate the molarity (mol/L.) of bacto X in the 1.0 ML sample volume if all of the bacteria were to simultaneously secrete all of the bactoX molecules contained within their cell bodies into the vial. concentration: C. Scientists discovered that bactoX will induce a luminescence response in B. cubis when the concentration in a sample reaches 5.60 x 10 M. How many molecules of bactoX must be present in solution to initiate luminescence in a 2.8 pl. sample? How many individual bacteria cells does this represent? The number of bactoX molecules per cell is given in part A. molecules: molecules cells: cells D. What volume, in nanoliters, would the bacteria cells in part Coccupy if tightly packed? Assume that there is no space between the cells. volume:

Answer 1

is to calculate the following The objective questions. Aus Given that, concentration of bacto X =5530 molecules/ bacteria. Number of bactox in 2.620 x 106 bacteria = - (Bacto X J x number of bacteria =(5530 molecules/bacteria)x(2.6207166) 1.45x10° molecules. Moles of bacto X = No.of molecules / Avogadro number. = (1.45 xidmolecules)/(6.023xio "molecules/ mol2 =[2.41Xio mok]

Total volume of solution aloo ul. Fluox16L. Molarity of bacto X = moles/volime of Solution in liters. = (2-418101 mol) / (160x16&L) 2:41x108 molesIL -12:41x188 m] -12. Given that, volume =2.8 ML. a 2,8x10° L. [Bacto x3 in cell = 5.50x16" M. = 5.50x1072 molesi Moles of bacto x =[bacto XJ x vol. of bactarial cell x Avogadro number.

= (5.50 x1612 moles/L)*(218x10€ LIX (6.023x183 molecules /mol) = 19.27xic“ melecules. No.of bacterial cells < NO.Cf bacto X/CNO. Of . bacto x/cell) = 9,27x1ő molecules/5530 . = (1.68x10² cells volume of bacterial cell = 2.95L =2098103 L. Required volume of packing bacterial cells Evolume of a bacterial cell x No. of bacteria = (2987645 L)x(1•68X10% cells) =4.8781012 L (since inl=1092) (4.87 x 103 mL