Concept explainers
Answers to all problems are at the end of this book. Detailed solutions are available in the Student Solutions Manual, Study Guide, and Problems Book.
The Dimensions of Prokaryotic Cells and Their Constituents Escherichia coli cells are about 2 μm (microns) long and 0.8 μm in diameter. (Section 1.5)
a. How many E. coli cells laid end to end would fit across the diameter of a pinhead? (Assume a pinhead diameter of 0.5 mm.)
b. What is the volume of an E. coli cell? (Assume it is a cylinder, with the volume of a cylinder given by
c. What is the surface area of ail E coli cell? What is the surface-to- volume ratio of an E coli cell?
d. Glucose, a major energy-yielding nutrient, is present in bacterial cells at a concentration of about 1 mM. What is the concentra¬tion of glucose, expressed as mg/mL? How many glucose molecules are contained in a typical E. coli cell? (Recall that Avogadro’s number
e. A number of regulatory proteins are present in E. coli at only one or two molecules per cell. If we assume that an E. coli cell contains just one molecule of a particular protein, what is the molar concentration of this protein in the cell? If the molecular weight of this protein is
f. An E coli cell contains about 15,000 ribosomes, which carry out protein synthesis. Assuming ribosomes are spherical and have a diameter of 20 nm (nanometers), what fraction of the E. coli cell volume is occupied by ribosomes?
g. The E coli chromosome is a single DNA molecule whose mass is about
(The answer to this question is a reasonable approximation of the maximum number of different kinds of proteins that can be expected in bacteria.)
To determine:
The number of E. coli cells per a pinhead.
Introduction:
Escherichia coli is a bacterium which lives in the intestine of healthy animals. E. coli genome consists of just one circular DNA.
Explanation of Solution
Given information:
Length of a cell
Diameter of pinhead
Formula used:
Number of E. coli per pinhead = Diameter of the pinhead/ Length of a cell
Calculation:
Number of E. coli cells that can be fitted to a pinhead
Number of E. coli cells that can be fitted to a pinhead
To determine:
The volume of an E. coli cell.
Introduction:
Escherichia coli is a bacterium which lives in the intestine of healthy animals. E. coli genome consists of just one circular DNA.
Explanation of Solution
Given information:
Length of a cell
Diameter of a cell
Formula used:
Calculation:
The volume of an E. coli cell
To determine:
The surface area and surface to volume ratio of an E. coli cell.
Introduction:
Escherichia coli is a bacterium which lives in the intestine of healthy animals. E. coli genome consists of just one circular DNA.
Explanation of Solution
Given information:
Length of a cell
Diameter of a cell
Formula used:
Surface area = (Surface area of a circle
Surface area =
Surface to volume ratio = Surface area/ Volume
Calculation:
Surface area of an E. coli cell
Surface to volume ratio of an E. coli cell
Thus, surface area of an E. coli cell
And, surface to volume ratio of an E. coli cell
To determine:
The concentration of glucose in mg/mL and the number of glucose molecules in an E. coli cell
Introduction:
Escherichia coli is a bacterium which lives in the intestine of healthy animals. E. coli genome consists of just one circular DNA.
Explanation of Solution
Given information:
Glucose concentration
Molecular weight of glucose
Formula used:
Number of moles = Concentration
Calculation:
Concentration of glucose
Moles of glucose in an E. coli cell
Number of glucose molecules in an E. coli cell
Thus, concentration of glucose
And, number of glucose molecules in an E. coli cell
To determine:
The molar concentration of the protein in the cell
Introduction:
Escherichia coli is a bacterium which lives in the intestine of healthy animals. E. coli genome consists of just one circular DNA.
Explanation of Solution
Given information:
Molecular weight of protein
Formula used:
Molar concentration = Moles/ Volume
Calculation:
Number of moles
Molar concentration
Protein concentration
Thus, molar concentration
Protein concentration
To determine:
The fraction of the E. coli cell volume which is occupied by ribosomes.
Introduction:
Escherichia coli is a bacterium which lives in the intestine of healthy animals. E. coli genome consists of just one circular DNA.
Explanation of Solution
Given information:
Number of ribosomes
Diameter of ribosome
Formula used:
Volume of a ribosome (cylinder)
Fractional volume = Volume of ribosome/ Volume of cell
Calculation:
Volume of one ribosome
Volume of 15,000 ribosome
Fractional volume
Thus, fractional volume
To determine:
The total length of E. coli chromosome and compare this length with the overall dimensions of E. coli cells, number of nucleotide pairs contains in the DNA, and number of different proteins can be encoded by E. coli chromosome.
Introduction:
Escherichia coli is a bacterium which lives in the intestine of healthy animals. E. coli genome consists of just one circular DNA.
Explanation of Solution
Given information:
Mass of E. coli chromosome
Average molecular weight of a nucleotide pair
Distance of a nucleotide pair
Number of amino acids is an average protein
Calculation:
Number of moles of base pairs in DNA
Length
Length DNA/ Length E. coli
Number of base pairs per protein
Number of different proteins
Thus,
Total length of E. coli chromosome
Length DNA/ Length E. coli
Number of nucleotide pairs
Number of different proteins
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Chapter 1 Solutions
Biochemistry
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- BiochemistryBiochemistryISBN:9781305577206Author:Reginald H. Garrett, Charles M. GrishamPublisher:Cengage Learning