Compare the primary role of Mg21 in the mechanism of restriction endonucleases with that of Zn21 in the carbonic anhydrases.
Discuss the ways in which restriction enzymes use binding energy.
The chemical functions of the Mg21-nucleotide complex can be rationalized.
The free energy is created by many weak intermolecular interactions.
The three enzymes trypsin, elastase, and chymotrypsin are related to one another.
The proteolytic-enzyme class is long in the left column.
Provide an example.
Match the molecule in the first column with the item in the second column.
Several different carbonic anhydrases coordinate Zn21 in their active sites using the side chains of His and Cys.
For instance, a DNA injection is used to destroy invading, exogenous DNA.
Most of the P-loops are made of P.
chymotrypsin is resistant to digest itself.
In the left column, indicate the appropriate transition state or chemical entity in the right column that has a postulated involvement in the catalytic mechanism.
The composition of the genome is 50% G+C.
The reaction of nucleoside monophosphate kinase is shown in the primary text.
Estimate the equilibrium constant for the reaction.
Trypsin and chymotrypsin fail to cleave proline bonds.
Trypsin won't cleave a peptide at a Lys-Pro junction.
Many enzymes can be protected against thermal denaturation.
Give an explanation for this phenomenon.
S-adenosylmethionine (AdoMet) is used as the methyl donor in a reaction that releases the remnant S-adenosylhomocysteine (AdoH).
The full potential of carbonic anhydrase can only be supported by a sufficient concentration of protons.
As a result, the enzyme has evolved to use buffers as acid-base catalysts.
Some of the buffers have large dimensions that make them hard to penetrate into the active site.
The buffers support efficient catalysis despite being excluded from the active site.
In addition, ionizable groups are involved in the delivery and removal of protons.
The site-specific mutagenesis that substituted some of these residues with other ionizable amino acids failed to inactivate the enzyme.
When the transition state of the reaction is fully formed, there are interactions between the two components.
Favorable interactions between the enzyme and the substrate in its ground state before the transition state can hinder the reaction by lowering the valley preceding the activation barrier in the reaction coordinate diagram if they do not contribute to binding the transition state.
A good competitive inhibitor forms strong interactions with the enzyme, but can't develop a transition state.
The carbonyl-carbon-to-amide nitrogen link partial double-bond character makes it more stable to hydrolysis.
There is a partially negatively charged carbonyl oxygen that decreases the susceptibility of the carbon atom to nucleophilic attack.
The aspartic acid carboxylate is ionized and has a neg ative charge.
The mechanism of lysozyme doesn't involve a nucleophilic attack.
The other three enzymes have an activated serine hydroxyl and react to form a complex with DIPF.
The structure of the enzymes at the sites where they interact with the side chains determine their specificity.
Normal and pathological physiology in humans are affected by specific saponins.
ACE is involved in blood-pressure regulation.
Overactivity of ACE causes hypertension.
A specific protease is needed for human immunodeficiency virus maturation.
It is possible to limit HIV infections with the inhibition of this protease.
One of the compo nents of the CO2 buffering system is dissolved CO2, which is a volatile gas.
The total concentration of dissolved CO2 H2CO3 is represented by 2CO3].
Carbonic acid that forms H1+bicarbonate is replaced by CO2 with water.
The observed pKa value of carbonic acid in a solution in equilibrium with CO2 in the lungs is approximately 6.1, not 3.5, because of its equilibrium with dissolved CO2, which exceeds it in concentration by 1000-fold.
The true pKa value of carbonic acid is 3.5, but it behaves in gas transport in mammals as if the value were 6.1.
The dissociation of bicarbonate has a pKa value of 10.3.
The bicarbonate ion is formed when the carbonyl carbon of CO2 is attacked.
The dissociation of the Zn21 from the water facilitates the creation of the reactive substrate.
The zinc ion has a positive charge that pulls electrons from the oxygen of the bound water, weakens the bonds to its hydrogen atoms, and promotes the dissociation of a protons to form an enzyme bound hydroxyl.
5C/-GAATTC-3C/ is the same strand.
The strands of duplex DNA are not the same.
The palindromic sequence has two-fold symmetry.
If you pierce the two strands between the AT sequence in each strand with a 180 degree rotation of the duplex molecule, you will generate a starting configuration of atoms.
A restriction enzyme hydrolyzes a sequence of genes.
The same sequence is recognized by the same partner.
The restriction site is immune to the restriction enzyme.
The host's DNA is protected by the methylase.
If unmethylated, the invading DNA will be cleaved and destroyed by less specific nucleases.
Only target DNA, Mg21, and water are required.
Unmethylated target DNA and S-adenosylmethionine are required.
There is an identical strand in 17 b, d, and e.
The carboxyl groups can bind Mg21 effectively.
The P-loops are very mobile.
The imidazole ring of histidine can be used as a catalyst.
chymotrypsin and carbonic anhydrase illustrated the first and third functions.
The chymotrypsin cleaves the bonds whose C-terminal amino acid is adja cent to non-polar aromatic and methionine.
The self-hydrolysis of native, folded chymotrypsin is very inefficient because of the buried residues.
chymotrypsin acts most effectively on partially degraded and denatured proteins during digestion.
A cell with a restriction/modification system has both strands of a parent DNA molecule.
After a short time, the newly synthesized daughter strand would be unmethylated.
The endonuclease can only be stopped by one methyl group on a restriction site.
Each base has a 0.25 chance of appearing at any position in the sequence if it has equal proportions of A, C, G, and T. The RV site is six bases long.
The Keq value is close to one because the b and l phosphorous atoms are broken in the same way as the a and b phosphorous atoms of the product.
The l phosphoryl group would likely be transferred to water.
The inability of the Mutant to close the lid would allow water into the active site where it would react with the ATP to form Pi and ADP.
The imino acid proline can't be accommodated in the binding sites of trypsin, chymotrypsin, or carboxypeptidase A because of its ring structure.
The proteases fail to cleave proline bonds.
Weak bonds formed with groups on the enzyme help to protect it against thermal denaturation when the active site is occupied.
Functional groups in the active site are needed to distinguish the enzyme from other similar molecules and position it for a productive reaction.
During the development of the transition state, the catalyst must react with a specific chemical bond of the substrate.
The ability to convert a substrate to a product is determined by ground-state interactions and the chemistry of the reaction.
The initial reaction measured a single turnover and there was no excess free enzyme in the solution.
When AdoMet was pre-programmed with labeled AdoMet before being mixed with excess DNA and unlabeled AdoMet, a burst was detected.
If the radiolabeled AdoMet had been removed from the enzyme before reacting, its specific activity would have been decreased 50-fold by the unlabeled AdoMet in the solution, and the maximum inclusion would have been 2% of that seen in the burst experiment.
The result doesn't prove that the reaction is ordered with the order of binding being AdoMet first and DNA second.
It shows that AdoMet can be used in the reaction after DNA binding.
The order of addition of the substrates might be random with either AdoMet or DNA binding first.
There were two questions that were derived from Gumport, R. I. and Reich, N.O.
The buffer is too large to be able to access the active center of the enzyme.
The protons supplied by these buffers reach the reaction center by being transported or shuttled through a network of ionizable groups.
The formation of different networks of hydrogen-bonded water molecule probably resulted in the malleability of the positioning of the active site.
The networks form in various shapes to accommodate the different positions of the side chains.
The answer deals with the different behavior of chymotrypsin.
The second step (release of N-acetyl-phenylalanine and free chymotrypsin) is slower than the first step.
The relative rates of the two steps are almost the same, so there is no burst.
Normal subtilisin cleaves more slowly than the Ala-64 subtilisin because it lacks critical histidine in the active site.
The missing histidine can be compensated for by the histidine in substrate B.
The statement is not true.
There is a serious impairment for the enzyme that only one or the other can cause.
A reasonable prediction is that the trypsin would be re-sembled by the mutant protease.
The predicted hydrolyze peptide bonds would be those whose carbonyl groups are from either lysine or arginine.
imidazole can diffuse into the active site of carbonic an hydrase and substitute for the protons shuttle function of His 64 near the zinc ion.
The loss of the side chain of His 64 can't be compensated for by large molecular buffers because of their steric bulk.
The chance of finding a unique restric tion site of length 10 is about once per million base pairs of DNA.
Most viral genomes that contain only 50,000 base pairs have a low chance of having a recognized site.
The increased rate wouldn't be beneficial.
The host cell would not benefit from a faster rate of hydrolysis because only a small number of cuts of an invading foreign DNA molecule will be enough to inactivate it.
Turnover number for restriction endonucleases is more important than specificity.
There would be no benefit in the absence of the gene.
The host cell's own DNA would be digest by the restriction endonuclease.
There are both forward and reverse reactions.
Knowledge of an equilibrium constant for the reaction will be required for the answer.
The products are approximately isoenergetic with the two substrates.
An equilibrium constant of one is assumed.
At equilibrium, let [ATP] be x.
There are two answers.
It is impossible that x-1 is unreasonable.
The concentrations ofATP,ADP, andAMP are all 0.300 mM.
The zinc will be removed from the active site.
The carbonic anhydrase is inactive without zinc.
N-acetyl-lysine is an analogue of Molecule A.
The posi tively charged e-ammonium group will bind with trypsin.
The B-O group is likely to bind in the oxyanion hole.
Because there is no bond to be cleaved, the inhibitor will remain bound to the enzyme and will interfere with the binding of natural substrates.
The hemiacetal can be formed with one molecule of alcohol.
It is reasonable that an aldehyde derivative of a peptide would react with the serine -OH group to form a hemiacetal.