Respuesta :
Answer:
i) 769 E. coli cells should be laid end to end would fit across a pinhead of diameter 0.5 mm.
ii) [tex]4.1466\times 10^{-18} m^3[/tex] is the volume of an E. coli cell.
iii) The surface area of E coli cell is [tex]1.8578\times 10^{-11} m^2[/tex] and surface area to volume ratio of an E. coli cell is [tex]4,480,297.110 m^{-1}[/tex].
Explanation:
Diameter of the Escherichia coli cell ,d= 2.85 μm = 0.00285 mm
1 μm = 0.001 mm
Length of the Escherichia coli cell ,h= 0.650 μm = 0.000650 mm
Let the number of Escherichia coli cell laid end to end to total length of 0.5 mm be x
[tex]x\times h =0.5 mm[/tex]
[tex]x=\frac{0.5 mm}{0.000650 mm}=769.23\approx 769 [/tex]
769 E. coli cells should be laid end to end would fit across a pinhead of diameter 0.5 mm.
Radius of the Escherichia coli cell r = d/2 = [tex]\frac{0.00285 mm}{2}=0.001425 mm[/tex]
Length of the Escherichia coli cell ,h = 0.000650 mm
Volume of the cylinder = [tex]\pi r^2h[/tex]
Volume of the E coli cell: V
[tex]V=3.14\times (0.001425 mm)^2\times 0.000650 mm[/tex]
[tex]V=4.1466\times 10^{-9} mm^3=4.1466\times 10^{-18} m^3[/tex]
[tex]1 mm^3=10^{-9} m^3[/tex]
Total surface area of cylinder :[tex]2\pi r (r+h)[/tex]
Area of the E coli cell: A
[tex]A=2\times 3.14\times 0.001425 mm\times (0.001425 mm+0.000650 mm)[/tex]
[tex]A=1.8578\times 10^{-5} mm^2=1.8578\times 10^{-11} m^2[/tex]
[tex]1 mm^2=10^{-6} m^2[/tex]
Surface area to volume ratio of an E. coli cell:
[tex]\frac{A}{V}=\frac{1.8578\times 10^{-11} m^2}{4.1466\times 10^{-18} m^3}[/tex]
[tex]=4,480,297.110 m^{-1}[/tex]
