Harpers

8 / CHAPTER 2of hydrolytic reactions when needed. Proteases catalyzethe hydrolysis of proteins into their component aminoacids, while nucleases catalyze the hydrolysis of thephosphoester bonds in DNA and RNA. Careful controlof the activities of these enzymes is required to ensurethat they act only on appropriate target molecules.Many Metabolic Reactions InvolveGroup TransferIn group transfer reactions, a group G is transferredfrom a donor D to an acceptor A, forming an acceptorgroup complex A–G:D− G + A = A− G + DThe hydrolysis and phosphorolysis of glycogen representgroup transfer reactions in which glucosyl groupsare transferred to water or to orthophosphate. Theequilibrium constant for the hydrolysis of covalentbonds strongly favors the formation of split products.The biosynthesis of macromolecules also involves grouptransfer reactions in which the thermodynamically unfavoredsynthesis of covalent bonds is coupled to favoredreactions so that the overall change in free energyfavors biopolymer synthesis. Given the nucleophiliccharacter of water and its high concentration in cells,why are biopolymers such as proteins and DNA relativelystable? And how can synthesis of biopolymersoccur in an apparently aqueous environment? Centralto both questions are the properties of enzymes. In theabsence of enzymic catalysis, even thermodynamicallyhighly favored reactions do not necessarily take placerapidly. Precise and differential control of enzyme activityand the sequestration of enzymes in specific organellesdetermine under what physiologic conditions agiven biopolymer will be synthesized or degraded.Newly synthesized polymers are not immediately hydrolyzed,in part because the active sites of biosyntheticenzymes sequester substrates in an environment fromwhich water can be excluded.Water Molecules Exhibit a Slight butImportant Tendency to DissociateThe ability of water to ionize, while slight, is of centralimportance for life. Since water can act both as an acidand as a base, its ionization may be represented as anintermolecular proton transfer that forms a hydroniumion (H 3 O + ) and a hydroxide ion (OH − ):+ −HO 2 + HO 2 = HO 3 + OHThe transferred proton is actually associated with acluster of water molecules. Protons exist in solution notonly as H 3 O + , but also as multimers such as H 5 O 2 + andH 7 O 3 + . The proton is nevertheless routinely representedas H + , even though it is in fact highly hydrated.Since hydronium and hydroxide ions continuouslyrecombine to form water molecules, an individual hydrogenor oxygen cannot be stated to be present as anion or as part of a water molecule. At one instant it isan ion. An instant later it is part of a molecule. Individualions or molecules are therefore not considered. Werefer instead to the probability that at any instant intime a hydrogen will be present as an ion or as part of awater molecule. Since 1 g of water contains 3.46 × 10 22molecules, the ionization of water can be described statistically.To state that the probability that a hydrogenexists as an ion is 0.01 means that a hydrogen atom hasone chance in 100 of being an ion and 99 chances outof 100 of being part of a water molecule. The actualprobability of a hydrogen atom in pure water existing asa hydrogen ion is approximately 1.8 × 10 −9 . The probabilityof its being part of a molecule thus is almostunity. Stated another way, for every hydrogen ion andhydroxyl ion in pure water there are 1.8 billion or 1.8 ×10 9 water molecules. Hydrogen ions and hydroxyl ionsnevertheless contribute significantly to the properties ofwater.For dissociation of water,+[ H ][ OH− ]K =[ HO 2 ]where brackets represent molar concentrations (strictlyspeaking, molar activities) and K is the dissociationconstant. Since one mole (mol) of water weighs 18 g,one liter (L) (1000 g) of water contains 1000 × 18 =55.56 mol. Pure water thus is 55.56 molar. Since theprobability that a hydrogen in pure water will exist as ahydrogen ion is 1.8 × 10 −9 , the molar concentration ofH + ions (or of OH − ions) in pure water is the productof the probability, 1.8 × 10 −9 , times the molar concentrationof water, 55.56 mol/L. The result is 1.0 × 10 −7mol/L.We can now calculate K for water:+ − −7 −7[ H ][ OH ] [ 10 ][ 10 ]K = =[ H2O][ 55. 56]−14 −16= 0. 018 × 10 = 1. 8 × 10 mol / LThe molar concentration of water, 55.56 mol/L, istoo great to be significantly affected by dissociation. Ittherefore is considered to be essentially constant. Thisconstant may then be incorporated into the dissociationconstant K to provide a useful new constant K w termedthe ion product for water. The relationship betweenK w and K is shown below: