Problem Set 1 Solutions:
1 Cell sizes: E. coli is a cylindrical bacteria with 1 M wide and 2 M long. Liver cells are approximately spherical with a diameter of 20 M. Plant palisade cells are cylinders 20 M wide and 35 M long. A) Calculate the volumes of each cells?
A) Since E.Coli is cylindrical
Volume of a cylinder = r2h, where r is half of bacterial width/diameter = 1M/2 = 0.5M and h is the length of the bacteria = *(0.5)22 = 1.57 M3 B) Since liver cells are spherical
Volume of a sphere = 4/3r3 (Radius of liver cell = 20 M/2 = 10 M) 4/3*4187 M3 C) Volume of cylindrical palisade cells = r2h
= *(10)235 = 10990 M3 B) How many E Coli will fit inside the human liver cells?
# of E. Coli that can fit in liver cells = Volume of Liver Cells/Volume of E Coli = 4187 M3/1.57 M3 = 2666 E. Coli C) How many liver cells will fit inside the palisade cells?
# of liver cells that can fit in palisade cells = Volume of Palisade Cells/Volume of Liver Cells
= 10990 M3/ 4187 M3 = 2 Liver Cells
2) . If a typical membrane is 8 nm wide, how many such membranes would have to be aligned up to see in a light microscope? How many with an electron microscope? i) Resolution/resolving power of light microscope = 0.2 nM = 200 nM # Membranes need to be aligned to be seen by light microscope = 200 nM/8 nM = 25 membranes ii) Resolution/resolving power of electron microscope = 2 nM Therefore, a single membrane will be visible
3) Human ribosomes are roughly spherical and have a diameter of 30 nm. How many ribosomes would fit in a human liver cell (from problem 1)? Volume of ribosome (spherical) = 4/3r3 , R
Radius of a ribosome = 30 nm/2 = 15 nM or 0.015 nM
= 4/3*1.413x10-5 nM3
# of ribosomes that can fit in a liver cells = volume of liver cells (from 1.2)/volume of ribosomes = 4187 nM3 / 1.413x10-5 nM3 = 2.96x108 ribosomes
4) E coli DNA is 2 nm wide by 1.36 mm long. What percentage of the cell is filled with DNA? (from problem 1.2a) Volume of E.Coli DNA (=volume of a cylinder) = r2h
Radius of E coli DNA = 2/2 nm =(2/2)/1000 M = 0.001 M
Length of E Coli DNA = 1.36 mm = 1.36*1000 M = 1360 M Therefore, volume of DNA = *(0.001)21360 = 4.27x10-3 M3 % of E.Coli filled with DNA = (Volume of E.Coli DNA/Volume of E.Coli)*100 = (4.27x10-3 nM3/1.57 M3)*100 = 0.27% of the cell
5) Facts of Life
In each of the following, indicate why the statement was considered once TRUE and it is no longer considered a FACT.
Plant and animal tissues are considered different because animal tissues do not have conspicuous boundaries that divide the cell Initially thought to be true because animal cells do not have cell walls, which made it difficult for early investigators to identify individual cells using the crude microscope available during that time. Schwann (1839) shown that this was incorrect for cartilage cells, which have well defined boundaries of collagen fibers, and later extended to all animal cells. B. Living organisms are not governed by the laws of chemistry and physics, but are subject to a “vital force” that is responsible for the formation of organic compounds. Initially thought to be true because living organisms seems to increase in complexity spontaneously. Synthesis of urea, an organic compound made by all living cells from an inorganic compound by Wohler (1828) finally disproved this concept. C. Hereditary materials are of proteins but not DNA, because DNA contains only four nucleotides as monomers Originally thought to be true...
Please join StudyMode to read the full document