Ch 9B Lecture

• in order to transport via RBC, you need bicarbonate
• Once in the lungs, a rapid conversion back so that CO2 can exit body

Carbonic anhydrases do this

the rate of the reaction increases. Histidine seems to be a candidate for this. But, it actually isnt. Itnstead, it is covalently bonded to a zinc atom in the active site, which interacts with H2O, which then attacks CO2

• less reactive; water does not dissociate a great deal; normal pKa of water arouund 15. 
• Interaction of water with zinc--> loss of protn --> OH (strong nucleophile)

Zn formsa strong nucleophile, lowering the pKa of water

• zn forms a strong nucleophile by removing a H--> Co2 comes in and the hydroxide attacks the carbon dioxide
• There is no disruption of tertiary structre due to the presence of negative charge. Water will displace the HCO3-

lowering the pKa of water

In physiological conditions, the reaction must go rapdily in forward and reverse direction. In tissue, it must covert to bicarbonate. In lungs, it must be converted back to CO2. In the absence of buffer, the eq. constant is way different since the diffusion of the protons is a limiting factor. There will be a disparity between substrate and product

allow the f and r reactions to happen at similar rates which should take place for us to live

because if we are producing protons at a rapid rate, those protons can't hang in the active site. They build up and something needs to get them out and hold them infor a while. So, buffers do this. Increasing buffer increases kcat

shuttle protons into and out of active site. Proton is immediately shuttled to the His residue, which is shuttled to the buffer that is present all around. It also gives up the H to g back

zn is the convergent part

• key is the substrate: dsDNA
• major enemy of bacteria are viruses. Bac evolved ability to cleave foreign DNA and have a whole suit of enzymes that can chop them up prior to them harming them

• Host DNA of bacteria must not be cut
• double stranded DNA means that both strands must be cut

• they cut somewhere within the sequence. 
• When they come apart, it results in sticky ends (small portions of single strands)

it must overcome the fact that the substrate is ds and must cleave both strands in close proximity to each other to make sure H bonds are overcome

a break in teh backbone of DNA is a nick. All cells have an enzyme called DNA ligase that goes through ligation. And, if a nick is present, it can be repaired.

the phosphodiester bond, causing a free 3'- oxygen end and a free 5'- phosphate end

two:

• nucleophile attacks and breaks a bond. Nucleophile is not activated since H is attached
• The intermediate is in the active site and covalently binds to substrate. One half of the substrate remains, but the other half does not.
• Water will attack the intermediate, hydrolyzing to produce the final products

an activated water molecule attacks directly, hydrolyzing in one step

in line displacement

• retention of stereoconfig
• inversion

it is a pentacoordinate system

they methylate the adenine nucleotide at the 5' end of hte reaction sequence. DNA methylases do this. The methyl group occupies enough space such that the substrate no longer fits prooperly in the active site.

they replaced one oxygen with sulfur and the incoming oxygen from water with oxygen-18. Water came in and resulted in inversion.

divalent cations that are not involved in the reaction mechanism. Metal chelators can inhibit the enzyme by scavenging the ions.

though enzymes bind to both cog and non, cog DNA is the only kind of DNA that gets deformed. If there is a sequence in that DNA, it is able to form a kink between the T and A bases in the sequence (if EcoRV)

• very specific H bonds with DNA to hold it therem. 
• Ben bound, DNA gets bent, foriming a kink. 

The enzyme is instantly scanning DNA and can hold on tightly. But, it will not cleave until the sequence is found, at which point attractions hol dthe DNA in place. Interactions via bininding energy cause the kink to form, lowering Ea--> DNA cut

the same elements are located in allof the enzymes. They are key to the funciton of the enzyme. REcognition sequence is not hte same, however. Difference in specificity is due to the junk and structural elements that lie around the similar elements, as well as changes in one or two amino acids in the active.

• take energy from hydrogen or ATP so t can power thingks in our cell
• we capture energy associated with it by cleaving the bond. Myosins are the answer.

fibrous region and two globular ATPase domains within which is the enzyme activity that will break bonds and liberate energy

• slime molds
• XRC
• divalent cation bound
• in line displacement

binds to ATP in a specific way so at so position the last phosphorusn near molecules that can hydrolyze it. In its absence, ATP cannot fix it

activates H2O molecule, creating a nucleophile, no basic group but serine is in the active site and donates H to one O on phosphorus and then picks up H from the H2O. ATP is acting as own base to pick u a H and make its own nucleophile

transform them to very large changes out of the periphery. Large change is equal to a step in the myosin

two different structural onformations that determine whether its ATP or ADP in the active site

the conf changes that occur in the enzyme after hydrolysis is that forces molecule out

produts don't really leave the active site, they remain there. So, the conf changes encourage them to leave

a loop that contains several glycine residues that are conserved between more closely related members ; it reacts with the phosphoryl gorups on the bound nucleotide