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SOLUTIONS MANUAL FOR LEHNINGER PRINCIPLES OF BIOCHEMISTRY 5 TH EDITION. (FREEMAN, 2025 ) BY NELSON D.L., COX M.M| QUESTIONS AND CORRECT ANSWER S 2025| A+ GRADE100% PASS €15,26
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SOLUTIONS MANUAL FOR LEHNINGER PRINCIPLES OF BIOCHEMISTRY 5 TH EDITION. (FREEMAN, 2025 ) BY NELSON D.L., COX M.M| QUESTIONS AND CORRECT ANSWER S 2025| A+ GRADE100% PASS
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SOLUTIONS MANUAL FOR LEHNINGER PRINCIPLES OF BIOCHEMISTRY 5 TH EDITION. (FREEMAN, 2025 ) BY NELSON D.L., COX M.M| QUESTIONS AND CORRECT ANSWER S 2025| A+ GRADE100% PASS

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  • solutions manual for lehninger principles of bioch
  • the foundations of biochemistry 1 the size of cel
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lOMoARcPSD|37668344

SOLUTIONS MANUAL FOR LEHNINGER PRINCIPLES OF
BIOCHEMISTRY 5 TH EDITION. (FREEMAN, 2025 ) BY
NELSON D.L., COX M.M| QUESTIONS AND CORRECT
ANSWER S 2025| A+ GRADE100% PASS

, chapter



1
The Foundationsof 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 Answer s to r
off Answer s to the proper number of significant figures.

, lOMoAR cPSD| 37668344




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 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

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