a Escherichia coli cells are about 2.85 µm long and 0.705 µm in diameter. i) How many E. coli cells laid end to end would fit across a pinhead of diameter 0.5 mm? Number of cells = 175 cells ii) Calculate the volume of an E. coli cell, assuming the cell to be cylindrical in shape. The volume of a cylinder is: Volume Tp? h Cell volume = 6.67x10^-19 |m3 iii) Calculate the surface area and surface area to volume ratio of an E. coli cell. The surface area of a cylinder is: Surface area = 2rrh + 2rr? Surface area = 5.36x10^-12 m2 Surface area:Volume = 8.04x10^6 m
a Escherichia coli cells are about 2.85 µm long and 0.705 µm in diameter. i) How many E. coli cells laid end to end would fit across a pinhead of diameter 0.5 mm? Number of cells = 175 cells ii) Calculate the volume of an E. coli cell, assuming the cell to be cylindrical in shape. The volume of a cylinder is: Volume Tp? h Cell volume = 6.67x10^-19 |m3 iii) Calculate the surface area and surface area to volume ratio of an E. coli cell. The surface area of a cylinder is: Surface area = 2rrh + 2rr? Surface area = 5.36x10^-12 m2 Surface area:Volume = 8.04x10^6 m
Elementary Geometry For College Students, 7e
7th Edition
ISBN:9781337614085
Author:Alexander, Daniel C.; Koeberlein, Geralyn M.
Publisher:Alexander, Daniel C.; Koeberlein, Geralyn M.
ChapterP: Preliminary Concepts
SectionP.CT: Test
Problem 1CT
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Transcribed Image Text:a Escherichia coli cells are about 2.85 um long and 0.705 um in diameter.
i) How many E. coli cells laid end to end would fit across a pinhead of diameter 0.5 mm?
Number of cells
175
cells
ii) Calculate the volume of an E. coli cell, assuming the cell to be cylindrical in shape.
The volume of a cylinder is:
Volume
Tp² h
Cell volume
= 6.67x10^-19
3
m
iii) Calculate the surface area and surface area to volume ratio of an E. coli cell.
The surface area of a cylinder is:
Surface area =
2rrh + 2Tr?
Surface area = 5.36x10^-12
m2
1
Surface area:Volume
= 8.04x10^6
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