Lehninger Principles of Biochemistry 6th Edition

Published by W.H. Freeman
ISBN 10: 1-42923-414-8
ISBN 13: 978-1-42923-414-6

Chapter 1 - The Foundations of Biochemistry - Problems: 2


A) 1.11 x $10^{-12}$ g B) $4.94$% C) $36.5$%

Work Step by Step

A) We are given the density of a single E. coli cell (g/L); so if we can determine the volume of the cell, then we can find the cell's mass. Given values: $h=2$ x $10^{-6}$m $d=0.8$ x $10^{-6}$m , therefore $r=0.4$ x $10^{-6}$m $$Volume (V)=πr^{2}h$$ So, $V=π(0.4$ x $10^{-6})^{2}(2$ x $10^{-6})$ $m^{3}$ $=1.01$ x $10^{-18} m^{3}$ Then, convert $m^{3}$ to $L$: $1.01$ x $10^{-18} m^{3} * (L/1$ x $10^{-3} m^{3})=1.01$ x $10^{-15}L$ Last, use the density to convert volume to mass: $1.01$ x $10^{-15}L*(1.1$ x $10^{3}g/L)=1.11$ x $10^{-12}g$ B) To solve this problem we need to know the volume of the cell with the envelope ($V_{o}$) and the volume of the cell without the envelope ($V_{i}$) $$V_{env}=V_{o}-V_{i}$$ To determine $V_{o}$: -We already calculated this volume in part A: $V_{o}=1.01$ x $10^{-18}m^{3}$ To determine $V_{i}$: -If the envelope is 10 nm thick, then $r_{i}=r_{o}-10$ x $10^{-9}m$ $r_{o}=0.4$ x $10^{-6}$ so, $r_{i}=3.94$ x $10^{-7}m$ -Using the equation for volume then we get: $V_{i}=π(3.9$ x $10^{-7})^{2}(2$ x $10^{-6})m^{3}=9.56$ x $10^{-19} m^{3}$ Armed with $V_{o}$ and $V_{i}$ we can find $V_{env}$: $V_{env}=1.01$ x $10^{-18}-9.56$ x $10^{-19}=4.96$ x $10^{-20}m^{3}$ Now, to find the total volume of the bacterium that the envelope occupies: $V_{env}/V_{cell}=4.96$ x $10^{-20}/1.01$ x $10^{-18}=0.0494=4.94$% C) To solve this, figure out the volume of a single ribosome ($r=18$ x $10^{-9}m$): $$V_{sphere}=\frac{4}{3}πr^{3}$$ $V_{rib}=\frac{4}{3}π(18$ x $10^{-9})^{3}m^{3}=2.44$ x $10^{-23} m^{3}$ Therefore the volume of 15,000 ribosomes equals: $15,000*2.44$ x $10^{-23}=3.66$ x $10^{-19}m^{3}$ So, the percentage of the cell volume occupied by ribosomes equals: $V_{ribs}/V_{cell}=3.66$ x $10^{-19}/1.01$ x $10^{-18}=0.365=36.5$%
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