Chapter 12 Answers - Study Guide Chapter 12 Membrane...

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Study Guide Chapter 12: Membrane Transport --- ANSWERS 12-2 The two basic properties governing the likelihood of whether a molecule will diffuse through a lipid bilayer are the size of the molecule and the charge of the molecule. A smaller molecule will be more likely to diffuse through the lipid bilayer than a larger molecule. A nonpolar (hydrophobic) molecule will be more likely to diffuse through the lipid bilayer than a polar molecule, which is more likely to diffuse through the lipid bilayer than a charged molecule. A. benzene (small nonpolar vs. larger uncharged) B. ethanol (polar vs. charged) C. glucose (large polar vs. very large highly charged) D. O 2 (nonpolar vs. polar) E. adenosine (polar vs. highly charged) 12-3 Choice (d) is the correct answer. Because the lipid bilayer is permeable to carbon dioxide and ethanol, destroying membrane proteins is unlikely to affect their exit (choices (a) and (b)). The lipid bilayer is also permeable to water (choice (e)). On the other hand glucose requires a membrane transport protein to be imported into the cell. ATP, which is a highly charged molecule, also requires a transport protein to cross a membrane and thus could not be lost from the cell by simple diffusion (choice (c)). 12-5 A. Molecule B is more likely to utilize a carrier protein. From the graph, the yeast cell’s uptake of molecule B levels off when the concentration of carbon source is high; this indicates a saturation point for the uptake of molecule B. Carrier proteins work by specifically binding their solutes and transferring the solute molecules across the membrane one at a time, down the concentration gradient of the solute. Thus, the rate at which a solute can be transported by a carrier protein is limited by the number of carrier proteins in the membrane and will reach a maximum when the solute concentration is high enough for solute molecules to saturate the carrier proteins. Since
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