MEDSTUDY.COM
Solutions Manual for Lehninger Principles of Biochemistry
5ed. (Freeman, 2008 ) ( PDFDrive )
M
Biophysic (Harvard University)
ED
ST
U
D
Y
Studocu is not sponsored or endorsed by any college or university
Downloaded by Edward John ()
,2608T_ch01sm_S1-S12 1/31/08 9:40PM Page S-1 ntt 102:WHQY028:Solutions Manual:Ch-01:
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
M
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
ED
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
ST
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?
U
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.
The radius of a globular actin molecule is 3.6 nm/2 = 1.8 nm; the volume of the
D
(b)
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 ×
Y
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
Downloaded by Edward John ()
,2608T_ch01sm_S1-S12 2/2/08 7:21AM Page S-2 ntt 102:WHQY028:Solutions Manual:Ch-01:
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
M
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?
ED
(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):
ST
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.
U
(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.
D
Volume without envelope = p(0.39 mm)2(1.98 mm)
Volume with envelope = p(0.4 mm)2(2 mm)
Y
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)
Downloaded by Edward John ()
, 2608T_ch01sm_S1-S12 1/31/08 9:40PM Page S-3 ntt 102:WHQY028:Solutions Manual:Ch-01:
Chapter 1 The Foundations of Biochemistry S-3
3. Genetic Information in E. Coli DNA The genetic information contained in DNA consists of a
linear sequence of coding units, known as codons. Each codon is a specific sequence of three deoxyri-
bonucleotides (three deoxyribonucleotide pairs in double-stranded DNA), and each codon codes for a
single amino acid unit in a protein. The molecular weight of an E. coli DNA molecule is about
3.1 × 109 g/mol. The average molecular weight of a nucleotide pair is 660 g/mol, and each nucleotide
pair contributes 0.34 nm to the length of DNA.
(a) Calculate the length of an E. coli DNA molecule. Compare the length of the DNA molecule with
the cell dimensions (see Problem 2). How does the DNA molecule fit into the cell?
(b) Assume that the average protein in E. coli consists of a chain of 400 amino acids. What is the
maximum number of proteins that can be coded by an E. coli DNA molecule?
Answer
(a) The number of nucleotide pairs in the DNA molecule is calculated by dividing the molec-
ular weight of DNA by that of a single pair:
(3.1 × 109 g/mol)/(0.66 × 103 g/mol) = 4.7 × 106 pairs
M
Multiplying the number of pairs by the length per pair gives
(4.7 × 106 pairs)(0.34 nm/pair) = 1.6 × 106 nm = 1.6 mm
The length of the cell is 2 mm (from Problem 2), or 0.002 mm, which means the DNA is
ED
(1.6 mm)/(0.002 mm) = 800 times longer than the cell. The DNA must be tightly coiled
to fit into the cell.
(b) Because the DNA molecule has 4.7 × 106 nucleotide pairs, as calculated in (a), it must
have one-third this number of triplet codons:
(4.7 × 106)/3 = 1.6 × 106 codons
ST
If each protein has an average of 400 amino acids, each requiring one codon, the number
of proteins that can be coded by E. coli DNA is
(1.6 × 106 codons)(1 amino acid/codon)/(400 amino acids/protein) = 4,000 proteins
4. The High Rate of Bacterial Metabolism Bacterial cells have a much higher rate of metabolism
than animal cells. Under ideal conditions some bacteria double in size and divide every 20 min,
U
whereas most animal cells under rapid growth conditions require 24 hours. The high rate of bacterial
metabolism requires a high ratio of surface area to cell volume.
(a) Why does surface-to-volume ratio affect the maximum rate of metabolism?
D
(b) Calculate the surface-to-volume ratio for the spherical bacterium Neisseria gonorrhoeae (diameter
0.5 mm), responsible for the disease gonorrhea. Compare it with the surface-to-volume ratio for a
globular amoeba, a large eukaryotic cell (diameter 150 mm). The surface area of a sphere is 4pr2.
Y
Answer
(a) Metabolic rate is limited by diffusion of fuels into the cell and waste products out of the
cell. This diffusion in turn is limited by the surface area of the cell. As the ratio of
surface area to volume decreases, the rate of diffusion cannot keep up with the rate of
metabolism within the cell.
(b) For a sphere, surface area = 4pr2 and volume = 4/3 pr3. The ratio of the two is the
surface-to-volume ratio, S/V, which is 3/r or 6/D, where D = diameter. Thus, rather than
calculating S and V separately for each cell, we can rapidly calculate and compare S/V
ratios for cells of different diameters.
S/V for N. gonorrhoeae = 6/(0.5 mm) = 12 mm—1
S/V for amoeba = 6/(150 mm) = 0.04 mm—1
S/V for bacterium 12mm—1
—— = —— —1 = 300
S/V for amoeba 0.04 mm
Thus, the surface-to-volume ratio is 300 times greater for the bacterium.
Downloaded by Edward John ()
Solutions Manual for Lehninger Principles of Biochemistry
5ed. (Freeman, 2008 ) ( PDFDrive )
M
Biophysic (Harvard University)
ED
ST
U
D
Y
Studocu is not sponsored or endorsed by any college or university
Downloaded by Edward John ()
,2608T_ch01sm_S1-S12 1/31/08 9:40PM Page S-1 ntt 102:WHQY028:Solutions Manual:Ch-01:
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
M
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
ED
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
ST
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?
U
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.
The radius of a globular actin molecule is 3.6 nm/2 = 1.8 nm; the volume of the
D
(b)
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 ×
Y
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
Downloaded by Edward John ()
,2608T_ch01sm_S1-S12 2/2/08 7:21AM Page S-2 ntt 102:WHQY028:Solutions Manual:Ch-01:
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
M
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?
ED
(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):
ST
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.
U
(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.
D
Volume without envelope = p(0.39 mm)2(1.98 mm)
Volume with envelope = p(0.4 mm)2(2 mm)
Y
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)
Downloaded by Edward John ()
, 2608T_ch01sm_S1-S12 1/31/08 9:40PM Page S-3 ntt 102:WHQY028:Solutions Manual:Ch-01:
Chapter 1 The Foundations of Biochemistry S-3
3. Genetic Information in E. Coli DNA The genetic information contained in DNA consists of a
linear sequence of coding units, known as codons. Each codon is a specific sequence of three deoxyri-
bonucleotides (three deoxyribonucleotide pairs in double-stranded DNA), and each codon codes for a
single amino acid unit in a protein. The molecular weight of an E. coli DNA molecule is about
3.1 × 109 g/mol. The average molecular weight of a nucleotide pair is 660 g/mol, and each nucleotide
pair contributes 0.34 nm to the length of DNA.
(a) Calculate the length of an E. coli DNA molecule. Compare the length of the DNA molecule with
the cell dimensions (see Problem 2). How does the DNA molecule fit into the cell?
(b) Assume that the average protein in E. coli consists of a chain of 400 amino acids. What is the
maximum number of proteins that can be coded by an E. coli DNA molecule?
Answer
(a) The number of nucleotide pairs in the DNA molecule is calculated by dividing the molec-
ular weight of DNA by that of a single pair:
(3.1 × 109 g/mol)/(0.66 × 103 g/mol) = 4.7 × 106 pairs
M
Multiplying the number of pairs by the length per pair gives
(4.7 × 106 pairs)(0.34 nm/pair) = 1.6 × 106 nm = 1.6 mm
The length of the cell is 2 mm (from Problem 2), or 0.002 mm, which means the DNA is
ED
(1.6 mm)/(0.002 mm) = 800 times longer than the cell. The DNA must be tightly coiled
to fit into the cell.
(b) Because the DNA molecule has 4.7 × 106 nucleotide pairs, as calculated in (a), it must
have one-third this number of triplet codons:
(4.7 × 106)/3 = 1.6 × 106 codons
ST
If each protein has an average of 400 amino acids, each requiring one codon, the number
of proteins that can be coded by E. coli DNA is
(1.6 × 106 codons)(1 amino acid/codon)/(400 amino acids/protein) = 4,000 proteins
4. The High Rate of Bacterial Metabolism Bacterial cells have a much higher rate of metabolism
than animal cells. Under ideal conditions some bacteria double in size and divide every 20 min,
U
whereas most animal cells under rapid growth conditions require 24 hours. The high rate of bacterial
metabolism requires a high ratio of surface area to cell volume.
(a) Why does surface-to-volume ratio affect the maximum rate of metabolism?
D
(b) Calculate the surface-to-volume ratio for the spherical bacterium Neisseria gonorrhoeae (diameter
0.5 mm), responsible for the disease gonorrhea. Compare it with the surface-to-volume ratio for a
globular amoeba, a large eukaryotic cell (diameter 150 mm). The surface area of a sphere is 4pr2.
Y
Answer
(a) Metabolic rate is limited by diffusion of fuels into the cell and waste products out of the
cell. This diffusion in turn is limited by the surface area of the cell. As the ratio of
surface area to volume decreases, the rate of diffusion cannot keep up with the rate of
metabolism within the cell.
(b) For a sphere, surface area = 4pr2 and volume = 4/3 pr3. The ratio of the two is the
surface-to-volume ratio, S/V, which is 3/r or 6/D, where D = diameter. Thus, rather than
calculating S and V separately for each cell, we can rapidly calculate and compare S/V
ratios for cells of different diameters.
S/V for N. gonorrhoeae = 6/(0.5 mm) = 12 mm—1
S/V for amoeba = 6/(150 mm) = 0.04 mm—1
S/V for bacterium 12mm—1
—— = —— —1 = 300
S/V for amoeba 0.04 mm
Thus, the surface-to-volume ratio is 300 times greater for the bacterium.
Downloaded by Edward John ()
