metabolism-3

lecture – three

absorption into cells:

          within cells, these molecules are further degraded into still simpler molecules containing two to four carbon atoms. these fragments (acetyl-coa for example) face one of two alternatives:

          they may proceed up various metabolic pathways and serve as the building blocks of, for example, sugars and fatty acids. from these will be assembled the macromolecules of the cell:

–              polysaccharides

–              fats

–              proteins

–              nucleic acids

          or the molecules in this pool of two- to four-carbon fragments may be still further degraded - ultimately to simple inorganic molecules such as carbon dioxide (co2), h2o, and ammonia (nh3).

          this phase of catabolism releases large amounts of energy (in the form of atp). one use to which this energy is put is to run the anabolic activities of the cell.

the energetics of transport across membranes:

          the movement of any molecule or ion down - or up - a concentration gradient involves a change in free energy, ?g ("delta g")

          down releases energy so ?g is negative

          up consumes energy so ?g is positive

          the amount of free energy released or consumed can be calculated from the equation

          active transport of glucose:

v  filtration of the blood in the glomeruli of the kidneys produces a nephric filtrate with a concentration of glucose the same as that of the blood (~ 5 mm). all of this glucose is normally reclaimed by transport

v  from the fluid within the proximal tubule

v  across the plasma membrane at the apicalsurface of the epithelial cells lining the tubule and

v  across the plasma membrane at the basolateral surface of the cell into the interstitial fluid, and on back into the blood. . as the process continues and more and more glucose is removed from the fluid, the concentration gradient up which the glucose must be pumped - by active transport - increases.

problem:

•                what is the free energy needed to move glucose back from the tubular fluid to the blood when the concentration in the tubular fluid has dropingped to 0.005 mm?

•                the problem is to pump glucose into the cell (where it is about 0.5 mm) and then across the plasma membrane at the basolateral surfaceof the cell into the interstitial fluid, where the glucose concentration is 5 mm (the same as in the blood). so the total gradient through which the glucose must be pumped is 0.005 mm , 5 mm.

•                ?g = (2) (310) x ln (5/.005)
= 620 ln (1000) = (620) (6.91) = + 4284 cal/mole
= + 4.3 kcal/mole

•                where is the needed energy to come from?