@ProfdocDigitalLibraries <Best Online Study Materials>
SOLUTION MANUAL
Lehninger Principles of Biochemistry
8th Edition
Authors: David Nelson & Michael Cox
,@ProfdocDigitalLibraries <Best Online Study Materials>
Table of Contents
1. The Foundations of Biochemistry
2. Water
3. Amino Acids, Peptides, and Proteins
4. The Three-Dimensional Structure of Proteins
5. Protein Function
6. Enzymes
7. Carbohydrates and Glycobiology
8. Nucleotides and Nucleic Acids
9. DNA-Based Information Technologies
10. Lipids
11. Biological Membranes and Transport
12. Biosignaling
13. Bioenergetics and Biochemical Reaction Types
14. Glycolysis, Gluconeogenesis, and the Pentose Phosphate Pathway
15. Principles of Metabolic Regulation
16. The Citric Acid Cycle
17. Fatty Acid Catabolism
18. Amino Acid Oxidation and the Production of Urea
19. Oxidative Phosphorylation
20. Photosynthesis in Plants and Bacteria
21. Lipid Biosynthesis
22. Biosynthesis of Amino Acids, Nucleotides, and Related Molecules
23. Hormonal Regulation and Integration of Mammalian Metabolism
24. Genes and Chromosomes
25. DNA Metabolism
26. RNA Metabolism
27. Protein Metabolism
28. Regulation of Gene Expression
,@ProfdocDigitalLibraries <Best Online Study Materials>
The Foundations
of Biochemistry
1. The Size of Cells and Their Components
(a) If you were to magnify a cell 10,000-fold (typical of the magnification achieved using an electron
microscope), how big would it appear? Assume you are viewing a “typical” eukaryotic cell with a
cellular diameter of 50 mm.
(b) If this cell were a muscle cell (myocyte), how many molecules of actin could it hold? (Assume
the cell is spherical and no other cellular components are present; actin molecules are spherical,
with a diameter of 3.6 nm. The volume of a sphere is 4/3 pr3.)
(c) If this were a liver cell (hepatocyte) of the same dimensions, how many mitochondria could it
hold? (Assume the cell is spherical; no other cellular components are present; and the
mitochondria are spherical, with a diameter of 1.5 mm.)
(d) Glucose is the major energy-yielding nutrient for most cells. Assuming a cellular concentration of
1 mM, calculate how many molecules of glucose would be present in our hypothetical (and
spherical) eukaryotic cell. (Avogadro’s number, the number of molecules in 1 mol of a nonionized
substance, is 6.02 × 1023.)
(e) Hexokinase is an important enzyme in the metabolism of glucose. If the concentration of hexokinase
in our eukaryotic cell is 20 mM, how many glucose molecules are present per hexokinase molecule?
Answer
(a) The magnified cell would have a diameter of 50 × 104 mm = 500 × 103 mm = 500 mm,
or 20 inches—about the diameter of a large pizza.
(b) The radius of a globular actin molecule is 3.6 nm/2 = 1.8 nm; the volume of the
molecule, in cubic meters, is (4/3)(3.14)(1.8 × 10—9 m)3 = 2.4 × 10—26 m3.*
The number of actin molecules that could fit inside the cell is found by dividing the cell
volume (radius = 25 mm) by the actin molecule volume. Cell volume = (4/3)(3.14)(25 ×
10—6 m)3 = 6.5 × 10—14 m3. Thus, the number of actin molecules in the hypothetical
muscle cell is
(6.5 × 10—14 m3)/(2.4 × 10—26 m3) = 2.7 × 1012 molecules
or 2.7 trillion actin molecules.
*Significant figures: In multiplication and division, the answer can be expressed with no
more significant figures than the least precise value in the calculation. Because some of the
data in these problems are derived from measured values, we must round off the calculated
answer to reflect this. In this first example, the radius of the actin (1.8 nm) has two significant
figures, so the answer (volume of actin = 2.4 × 10—26 m3) can be expressed with no more
than two significant figures. It will be standard practice in these expanded answers to round
off answers to the proper number of significant figures.
S-1
, S-2 Chapter 1 The Foundations of Biochemistry
(c) The radius of the spherical mitochondrion is 1.5 mm/2 = 0.75 mm, therefore the volume
is (4/3)(3.14)(0.75 × 10—6 m)3 = 1.8 × 10—18 m3. The number of mitochondria in the
hypothetical liver cell is
(6.5 × 10—14 m3)/(1.8 × 10—18 m3) = 36 × 103 mitochondria
(d) The volume of the eukaryotic cell is 6.5 × 10—14 m3, which is 6.5 × 10—8 cm3 or 6.5 ×
10—8 mL. One liter of a 1 mM solution of glucose has (0.001 mol/1000 mL)(6.02 × 1023
molecules/mol) = 6.02 × 1017 molecules/mL. The number of glucose molecules in the
cell is the product of the cell volume and glucose concentration:
(6.5 × 10—8 mL)(6.02 × 1017 molecules/mL) = 3.9 × 1010 molecules
or 39 billion glucose molecules.
(e) The concentration ratio of glucose/hexokinase is 0.001 M/0.00002 M, or 50/1, meaning that
each enzyme molecule would have about 50 molecules of glucose available as substrate.
2. Components of E. coli E. coli cells are rod-shaped, about 2 mm long and 0.8 mm in diameter. The
volume of a cylinder is pr2h, where h is the height of the cylinder.
(a) If the average density of E. coli (mostly water) is 1.1 × 103 g/L, what is the mass of a single cell?
(b) E. coli has a protective cell envelope 10 nm thick. What percentage of the total volume of the
bacterium does the cell envelope occupy?
(c) E. coli is capable of growing and multiplying rapidly because it contains some 15,000 spherical
ribosomes (diameter 18 nm), which carry out protein synthesis. What percentage of the cell
volume do the ribosomes occupy?
Answer
(a) The volume of a single E. coli cell can be calculated from pr2h (radius = 0.4 mm):
3.14(4 × 10—5 cm)2(2 × 10—4 cm) = 1.0 × 10—12 cm3 = 1 × 10—15 m3 = 1 × 10—15 L
Density (g/L) multiplied by volume (L) gives the mass of a single cell:
(1.1 × 103 g/L)(1 × 10—15 L) = 1 × 10—12 g
or a mass of 1 pg.
(b) First, calculate the proportion of cell volume that does not include the cell envelope,
that is, the cell volume without the envelope—with r = 0.4 mm — 0.01 mm; and h = 2 mm
— 2(0.01 mm)—divided by the total volume.
Volume without envelope = p(0.39 mm)2(1.98 mm)
Volume with envelope = p(0.4 mm)2(2 mm)
So the percentage of cell that does not include the envelope is
p(0.39 mm)2(1.98 mm) × 100
———— = 90%
p(0.4 mm)2(2 mm)
(Note that we had to calculate to one significant figure, rounding down the 94% to 90%,
which here makes a large difference to the answer.) The cell envelope must account for
10% of the total volume of this bacterium.
(c) The volume of all the ribosomes (each ribosome of radius 9 nm) = 15,000 × (4/3)p(9 ×
10—3 mm)3
The volume of the cell = p(0.4 mm)2(2 mm)
So the percentage of cell volume occupied by the ribosomes is
15,000 × (4/3)p(9 × 10—3 mm)3 × 100
————— = 5%
p(0.4 mm)2(2 mm)
SOLUTION MANUAL
Lehninger Principles of Biochemistry
8th Edition
Authors: David Nelson & Michael Cox
,@ProfdocDigitalLibraries <Best Online Study Materials>
Table of Contents
1. The Foundations of Biochemistry
2. Water
3. Amino Acids, Peptides, and Proteins
4. The Three-Dimensional Structure of Proteins
5. Protein Function
6. Enzymes
7. Carbohydrates and Glycobiology
8. Nucleotides and Nucleic Acids
9. DNA-Based Information Technologies
10. Lipids
11. Biological Membranes and Transport
12. Biosignaling
13. Bioenergetics and Biochemical Reaction Types
14. Glycolysis, Gluconeogenesis, and the Pentose Phosphate Pathway
15. Principles of Metabolic Regulation
16. The Citric Acid Cycle
17. Fatty Acid Catabolism
18. Amino Acid Oxidation and the Production of Urea
19. Oxidative Phosphorylation
20. Photosynthesis in Plants and Bacteria
21. Lipid Biosynthesis
22. Biosynthesis of Amino Acids, Nucleotides, and Related Molecules
23. Hormonal Regulation and Integration of Mammalian Metabolism
24. Genes and Chromosomes
25. DNA Metabolism
26. RNA Metabolism
27. Protein Metabolism
28. Regulation of Gene Expression
,@ProfdocDigitalLibraries <Best Online Study Materials>
The Foundations
of Biochemistry
1. The Size of Cells and Their Components
(a) If you were to magnify a cell 10,000-fold (typical of the magnification achieved using an electron
microscope), how big would it appear? Assume you are viewing a “typical” eukaryotic cell with a
cellular diameter of 50 mm.
(b) If this cell were a muscle cell (myocyte), how many molecules of actin could it hold? (Assume
the cell is spherical and no other cellular components are present; actin molecules are spherical,
with a diameter of 3.6 nm. The volume of a sphere is 4/3 pr3.)
(c) If this were a liver cell (hepatocyte) of the same dimensions, how many mitochondria could it
hold? (Assume the cell is spherical; no other cellular components are present; and the
mitochondria are spherical, with a diameter of 1.5 mm.)
(d) Glucose is the major energy-yielding nutrient for most cells. Assuming a cellular concentration of
1 mM, calculate how many molecules of glucose would be present in our hypothetical (and
spherical) eukaryotic cell. (Avogadro’s number, the number of molecules in 1 mol of a nonionized
substance, is 6.02 × 1023.)
(e) Hexokinase is an important enzyme in the metabolism of glucose. If the concentration of hexokinase
in our eukaryotic cell is 20 mM, how many glucose molecules are present per hexokinase molecule?
Answer
(a) The magnified cell would have a diameter of 50 × 104 mm = 500 × 103 mm = 500 mm,
or 20 inches—about the diameter of a large pizza.
(b) The radius of a globular actin molecule is 3.6 nm/2 = 1.8 nm; the volume of the
molecule, in cubic meters, is (4/3)(3.14)(1.8 × 10—9 m)3 = 2.4 × 10—26 m3.*
The number of actin molecules that could fit inside the cell is found by dividing the cell
volume (radius = 25 mm) by the actin molecule volume. Cell volume = (4/3)(3.14)(25 ×
10—6 m)3 = 6.5 × 10—14 m3. Thus, the number of actin molecules in the hypothetical
muscle cell is
(6.5 × 10—14 m3)/(2.4 × 10—26 m3) = 2.7 × 1012 molecules
or 2.7 trillion actin molecules.
*Significant figures: In multiplication and division, the answer can be expressed with no
more significant figures than the least precise value in the calculation. Because some of the
data in these problems are derived from measured values, we must round off the calculated
answer to reflect this. In this first example, the radius of the actin (1.8 nm) has two significant
figures, so the answer (volume of actin = 2.4 × 10—26 m3) can be expressed with no more
than two significant figures. It will be standard practice in these expanded answers to round
off answers to the proper number of significant figures.
S-1
, S-2 Chapter 1 The Foundations of Biochemistry
(c) The radius of the spherical mitochondrion is 1.5 mm/2 = 0.75 mm, therefore the volume
is (4/3)(3.14)(0.75 × 10—6 m)3 = 1.8 × 10—18 m3. The number of mitochondria in the
hypothetical liver cell is
(6.5 × 10—14 m3)/(1.8 × 10—18 m3) = 36 × 103 mitochondria
(d) The volume of the eukaryotic cell is 6.5 × 10—14 m3, which is 6.5 × 10—8 cm3 or 6.5 ×
10—8 mL. One liter of a 1 mM solution of glucose has (0.001 mol/1000 mL)(6.02 × 1023
molecules/mol) = 6.02 × 1017 molecules/mL. The number of glucose molecules in the
cell is the product of the cell volume and glucose concentration:
(6.5 × 10—8 mL)(6.02 × 1017 molecules/mL) = 3.9 × 1010 molecules
or 39 billion glucose molecules.
(e) The concentration ratio of glucose/hexokinase is 0.001 M/0.00002 M, or 50/1, meaning that
each enzyme molecule would have about 50 molecules of glucose available as substrate.
2. Components of E. coli E. coli cells are rod-shaped, about 2 mm long and 0.8 mm in diameter. The
volume of a cylinder is pr2h, where h is the height of the cylinder.
(a) If the average density of E. coli (mostly water) is 1.1 × 103 g/L, what is the mass of a single cell?
(b) E. coli has a protective cell envelope 10 nm thick. What percentage of the total volume of the
bacterium does the cell envelope occupy?
(c) E. coli is capable of growing and multiplying rapidly because it contains some 15,000 spherical
ribosomes (diameter 18 nm), which carry out protein synthesis. What percentage of the cell
volume do the ribosomes occupy?
Answer
(a) The volume of a single E. coli cell can be calculated from pr2h (radius = 0.4 mm):
3.14(4 × 10—5 cm)2(2 × 10—4 cm) = 1.0 × 10—12 cm3 = 1 × 10—15 m3 = 1 × 10—15 L
Density (g/L) multiplied by volume (L) gives the mass of a single cell:
(1.1 × 103 g/L)(1 × 10—15 L) = 1 × 10—12 g
or a mass of 1 pg.
(b) First, calculate the proportion of cell volume that does not include the cell envelope,
that is, the cell volume without the envelope—with r = 0.4 mm — 0.01 mm; and h = 2 mm
— 2(0.01 mm)—divided by the total volume.
Volume without envelope = p(0.39 mm)2(1.98 mm)
Volume with envelope = p(0.4 mm)2(2 mm)
So the percentage of cell that does not include the envelope is
p(0.39 mm)2(1.98 mm) × 100
———— = 90%
p(0.4 mm)2(2 mm)
(Note that we had to calculate to one significant figure, rounding down the 94% to 90%,
which here makes a large difference to the answer.) The cell envelope must account for
10% of the total volume of this bacterium.
(c) The volume of all the ribosomes (each ribosome of radius 9 nm) = 15,000 × (4/3)p(9 ×
10—3 mm)3
The volume of the cell = p(0.4 mm)2(2 mm)
So the percentage of cell volume occupied by the ribosomes is
15,000 × (4/3)p(9 × 10—3 mm)3 × 100
————— = 5%
p(0.4 mm)2(2 mm)
