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recap6 (RECAP 6.2 (1.Describe an experiment showing a role for…

- - - - The hydrophilic “heads” of fatty acids are the polar ends, and the hydrophobic “tails” are the nonpolar ends. So the heads tend to associate with water molecules, and the tails away from water molecules. Placed in an aqueous environment, the fatty acids will arrange themselves so that their tails interact with one another while their heads face the water of the environment and cytoplasm, forming a bilayer.
    - - An integral membrane protein is embedded in the phospholipid bilayer by hydrophobic interactions with the lipid interior. It must have amino acids with hydrophobic R groups to insert into the nonpolar fatty acid tail region of the membrane bilayer. A peripheral membrane protein lacks hydrophobic regions and does not interact with the hydrophobic core of the phospholipid bilayer. Instead it is usually bound to the membrane indirectly by interactions with integral membrane proteins or directly by interactions with lipid polar head groups.
    - - Both freeze-fracturing and cell fusion experiments indicate that nonpolar membrane proteins are inserted into the hydrophobic interior of the lipid bilayer. Cell fusion experiments also show that the proteins can move in the plane of the membrane.
    - - To measure membrane fluidity, label a small amount of a lipid or protein with a dye and allow it to incorporate into the membrane of a cancer cell and a noncancer cell. This may make a localized labeled spot on the cells. The localized region will be seen to diffuse over the cells over time. In the cancer cell, this rate of diffusion may be faster.
- - - - The properties that affect diffusion across a membrane are size and mass (smaller is faster), electric charge (less polar is faster), and concentration gradient (the higher the gradient, the faster the rate of diffusion).
    - - If blood is hypotonic, water will enter red blood cells, causing them to swell and perhaps burst. (This is called hemolytic anemia.)
    - - The hydrophobic cell membrane lipids are relatively impermeable to charged ions such as K+. A channel will allow diffusion of K+ out of the cell until equilibrium is reached, with equal concentrations inside and outside the cell.
- - - - Substances tend to diffuse and reach equilibrium, where they are evenly distributed in the environment. Energy is needed to overcome this natural tendency.
    - - An antiporter transports two substances in opposite directions. In the case of the Na+–K+ pump, Na+ is transported across the cell membrane outward, and K+ is transported across the cell membrane inward.
    - - In primary active transport, ATP hydrolysis supplies the energy needed for transport against a concentration gradient. In secondary active transport, the energy comes from a gradient set up by a different (secondary) active transport.
    - - Both active transport and facilitated diffusion speed up transport across a membrane by using a protein that binds to the substance transported. However, active transport is an energy-requiring process that transports a substance against its concentration gradient and diffusion tendency. By contrast, facilitated diffusion does not require energy and transports a substance along its concentration gradient and in conjunction with its tendency to diffuse.
- - - - Phagocytosis involves a large cell membrane–derived vesicle forming around mac-romolecules. Pinocytosis involves smaller vesicles forming around water and its dissolved solutes.
    - - In receptor-mediated endocytosis, a molecule binds to a cell membrane protein receptor and causes the membrane to form a vesicle around the molecule and receptor. The vesicle with its contents enters the cell and usually fuses with a lysosome. An example is the endocytosis of lipoprotein particles from blood into liver cells.
    - - Diatom wall components move from the Golgi apparatus to the cell wall by exocytosis.