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Lehninger Principles of Biochemistry Sixth Edition Solution Manual

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The Foundationsof Biochemistrychapter1S-11.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 electronmicroscope), how big would it appear? Assume you are viewing a “typical” eukaryotic cell with acellular diameter of 50mm.(b)If this cell were a muscle cell (myocyte), how many molecules of actin could it hold? Assumethe 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 4/3pr3.)(c)If this were a liver cell (hepatocyte) of the same dimensions, how many mitochondria could ithold? Assume the cell is spherical; no other cellular components are present; and themitochondria are spherical, with a diameter of 1.5mm.(d)Glucose is the major energy-yielding nutrient for most cells. Assuming a cellular concentration of1 mM(i.e., 1 millimole/L), calculate how many molecules of glucose would be present in ourhypothetical (and spherical) eukaryotic cell. (Avogadro’s number, the number of molecules in 1mol of a nonionized substance, is 6.021023.)(e)Hexokinase is an important enzyme in the metabolism of glucose. If the concentration of hexokinasein our eukaryotic cell is 20mM, how many glucose molecules are present per hexokinase molecule?Answer(a)The magnified cell would have a diameter of 50104mm500103mm500 mm,or 20 inches—about the diameter of a large pizza.(b)The radius of a globular actin molecule is 3.6 nm/21.8 nm; the volume of themolecule, in cubic meters, is (4/3)(3.14)(1.8109m)32.41026m3.*The number of actin molecules that could fit inside the cell is found by dividing the cellvolume (radius25mm) by the actin molecule volume. Cell volume(4/3)(3.14)(25106m)36.51014m3. Thus, the number of actin molecules in the hypotheticalmuscle cell is(6.51014m3)/(2.41026m3)2.71012moleculesor 2.7 trillion actin molecules.*Significant figures:In multiplication and division, the answer can be expressed with nomore significant figures than the least precise value in the calculation. Because some of thedata in these problems are derived from measured values, we must round off the calculatedanswer to reflect this. In this first example, the radius of the actin (1.8 nm) has two significantfigures, so the answer (volume of actin2.41026m3) can be expressed with no morethan two significant figures. It will be standard practice in these expanded answers to roundoff answers to the proper number of significant figures.c01TheFoundationsofBiochemistry.qxd12/6/124:09PMPageS-1

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