Chapter 8: Enzymes: Basic Concepts and Kinetics

Chapter 8: Enzymes: Basic Concepts and Kinetics

• You should be able to complete the objectives once you have mastered this chapter.

• Give examples of the rate enhancements of the enzymes.

• There are examples of enzymes that convert one form of energy into another.

• The assumptions underlying the derivation are listed.

• There are examples of enzymes using each mechanism.

• The nature of an inhibitor can be determined with the help of analysis.

• Relate the functions of the coenzymes.

• Reactions arecatalyzed by binding regulatory proteins.

• There is a process that converts the energy of light into the energy of chemical bonds.

• The free energy change of a reaction allows it to happen spontaneously.

• The reaction takes place at 25.
• ]C.

• A transition state is formed along the reaction coordinate of the reaction.

• M is expressed in terms of a reaction speed.

• Myoglobin+O2 is found in muscle cells.

• 50% of the myoglobin is saturated with O2 when the partial pressure of the O2 is 50.

• The two equations have relationships.

• The turnover number for chymotrypsin is 100, while the turnover number for DNA polymerase is 15.
• This means that chymotrypsin has higher affinity than DNA polymerase.

• The inhibition of the cell wall by penicillin is a classic example of a med ically significant inhibition of an enzymatic reaction.

• Match the vitamins in the left column with their biological func tions in the right column.

• Some of the enzymes are not correct.

• Although tight binding to the substrates helps confer specificity on the reaction, it increases the activation barrier to reaction.
• The free energy of the reaction is increased when the binding to the transition state of the reaction is tight.

• Metal ion or lowmolecular weight organic molecule are known as cofactors.

• A tightly bound cofactor is what makes a prosthetic group.

• The cofactors that are bound behave like cosubstrates.

• The sun provides light energy that can be converted into chemical bond energy.
• Other examples of energy transduction include the use of an ion gradient in mitochondria to drive the synthesis of chemical bonds, and the use of the energy in ATP to cause the movement of muscles.

• Changes in the system under study and in the surroundings affect its value even if it were easy to determine.

• Chapter 8 is divided into 8 parts.
• The statements are correct.

• For a resolution of the paradoxes, see the answer to (d).

1430 cal/mol is 1 4

• 3 3 is the number of mol log10 5.

• It is incorrect because it is the hardest to form.
• The concentration of the transition state affects the speed of the reaction.

• The transition state of the reaction is tightly bound by Enzymes.

• They facilitate the formation of the transition state by binding it with high affinity.
• The transition state can be binding with hydrogen bonds and ionic and hydrophobic interactions.
• The faster the reaction, the more transition state formed.

• The rate of equilibrium is increased by the speed of the enzyme.

• It's difficult to isolating ES when it's Turnover of ES to form P. If the complex is very stable, it can be isolated in the absence of the other.
• The absence of the cosubstrate precludes turnover.
• The same consideration is given to the complexes formed by x-ray crystallography.

• A close fitting of the atoms of the side chains that make up the active site of the enzyme is required in the formation of an enzyme-substrate complex.
• Since there are different spatial arrangements of the atoms of stereoisomers, only a single stereoisomer can fit into the active site and be acted upon.

• Weak forces are involved in the binding of the two complexes.

• Question 18 can be seen.

• The equations are related because they express the Occupancy of binding sites as a function of either O2 or Subsidiary concentration.
• 50 are O2 concentrations.

• The units of [S] are mM.
• The actual concentrations are calculated using the factor 103.

• Part (c) was used to calculate 2.

• It is possible to estimate the efficiency of an enzyme.
• M,108 to109 M-1 S-1, is set by the rate of diffusion of the solution, which limits the rate at which it encounters the enzyme.
• The rate at which the substrate can reach the enzyme is the only factor that affects the M's catalytic velocity.

• It is greater for chymotrypsin than it is for DNA polymerase.

• As the concentration of the inhibitor increases, the proportion of bound substrate remains the same.

• Answer (b) is incorrect because transition state analogs are not good for corre sponding.
• They decrease rather than increase reaction rates.

• Answer (d) is incorrect because transition-state analogs mimic the shape of the transition state, which may be strained or distorted.

• The text states that a decrease of 1.36 kcal/mol in the free energy of the reaction has the effect of increasing the rate of conversion to product by a factor of 10.

• When no product has accumulated, the Michaelis- Menten model only applies to the initial velocity of the reaction.

• Concentration is what M must be expressed in.

• Two tissues, tissue A and tissue B, are tested for the activity of the enzyme X.

• Under a variety of circumstances, the activity of X, expressed as the number of moles of substrate converted to product per gram of tissue, is found to be five times greater in tissue A than in tissue B.

• The plots should be sketched on the axes.
• The pre-steady state can be assumed to be so rapid that it doesn't need to be considered.

• The data shown below can be used for an reaction.

• Suppose that the data is obtained for a reaction in the presence of and without the inhibitor X.

• Determine the type of inhibition that occurred by using double-reciprocal plots of the data.

• The data shown below can be used to derive a reaction in the presence or absence of Y.

• Use double-reciprocal plots of the data to determine the type of inhibition that occurred.

• The double-reciprocal plot is the most widely used plot for the data.
• Data points will be obtained that cluster near the vertical axis if linear increment of concentration are used.
• The intercept on the ordinate can be determined with great accuracy, but the slope of the line will be subject to considerable error, because the least reliable data points have greater weight in establishing the slope.

• Other linear plot ting forms have been created because of the limitation of double-reciprocal plots.

• The formation of a bond at a break in the phosphodiester of a duplex DNA molecule is accomplished by the DNA ligase.
• The free energy of hydrolysis ATP is the energy source for the formation of the phosphodiester bond.
• An intermediate in the reaction is a modified form of the enzyme that is bound to a side chain.
• The intermediate is formed by the reaction of E+ATP.
• The next step is the transfer of theAMP from the enzyme to thephosphate on the DNA.
• The last step of the reaction is the formation of a bond between the free enzyme and the phosphodiester bond.

• Section 8.2.2 has an equation used to find answers.

• Chapter 8 has a value of 4.

• 4 0 is 10 3.

• + 7 is the 1n [FBP] number.

• 10 3 is 7.
• Under standard conditions, the concentration term in the equation is much smaller than 1.0.
• The concentration is kept low by the removal of G3P.

• The rate of the reverse reaction must increase by a factor of 10.
• The rate at which equilibrium is attained is affected by the rate at which enzymes affect the equilibria of processes.
• The rate of the reverse reaction must increase by the same factor if the forward reaction rate is increased.

• M is 0 8.
• M is 0 2.

• A concentration four times greater than the Michaelis constant yields 80% of maximal velocity.

• It is two steps as being irreversible.
• At the initial stage of the reaction, P and E can't recombine to give ES at a higher rate if P is present.

• The magnitude of its rate constant as well as the concentrations of P and E are variables that affect the -2 step.

• The amount of enzyme X must be five times greater in tissue A than in tissue B.
• The activity of the enzymes is related to the concentration.

Slope is 0

• The inhibi tion is competitive.

• There is a plot of data showing the effects of an drug.

• The free enzyme is combined with the inhibitor.
• The competition for the same binding site on the enzyme makes it impossible for a competitive inhibitor to combine with ES.

• The slope without X is zero.

• Substituting for these values yields zero.

• A double-reciprocal plot of data shows the effects of an Y.

• It may combine with both of them.

• The slope without Y is zero.

• Substituting for these values yields zero.

• See figure 9.

• We can conclude that the measurements are reliable.

• Hanes-Woolf plots show the effects of competitive and non-competitive drugs.

• No inhibitor 15.
• Q increases the rate of reaction, so it's an activator.
• Both E and ES are combined.

• The first part of the double-displacement reaction does not involve DNA.

• In the absence of additional information, you would think that the enzyme had losteric properties because it was binding to a different site than the active one.

• The turnover number is 31.1Y=106 mol S-1 s 5Y10-8 mol E.

• The acyl-enzyme intermediate formed by penicillinase is transferred to water rather than to the terminal glycine of the bridge.

• The M is 3.1Y.

• The Michaelis- Menten equation tells us the fraction of en zyme molecule bound to the substrate.

• The sum of [EI] +[E] is only 0.757 of the total.

• The ratio is equal, 0.73/0.49 and 33.8/22.6.

• M is the same as without a drug.

• This is not competitive inhibition.

• See 4(e) for the solution.

• The x axis is plotted on.
• One is extrapolating to S.

• The slope of 2 is greater than the slope of 1.

• The side chains of the hydrogen-bond donors are arginine, asparagine, glutamine, histidine, and serine.

• More free energy of activation is needed to convert S to S++, if a Mutant Enzyme binding a Substrate, 100-fold as tightly as the native one.
• The speed of the reaction will be slowed down by a factor of 100 due to the increase in theDG++.

• The ratio of KM/Vmax is not affected.

• M + [jS] where j is the concentration.

• Increasing [I] 1/[S] b.

T [ES](KM/[S]+[I]/KI)

• The rate equation is v.

• The slope of the double reciprocal plot is the same in the presence of an uncompetitive inhibitor.

• By substituting [S]+KM into the Michaelis- Menten equation, we can show that.

• The apparent Vmax will change with dif ferent amounts of enzyme, and the center graph will remain the same.
• The correct answer is the center graph.

• The double-reciprocal plot will form a second line near the 1/v axis.

• The decrease in reaction speed could be due to an allosteric inhibition at a second binding site.
• The binding affinity of the second site could be lower than that of the catalytic site.

• The mechanism suggests that H+ is acting as a competitor.
• The vo will equal Vmax, independent of pH, if the concentration is sufficiently high.
• The observed vo will follow a curve with a pK of 6.0.

• Half of the enzyme will be in the E: form and the reaction speed will be 1/2 of Vmax.

• Theidase is not stable at 37.
• ]C.

• The thermal unfolding is partially protected by the PLP coenzyme.

• The rate of denaturation is slower when PLP is bound.

• The page was left blank.

• In the previous chapter, you learned that the activity of the enzymes is based on their ability to stabilizing the transition states of chemical reactions.

• These specific enzymes are used as models to show the fundamental principles of enzyme catalysis.
• The principles used by the enzymes show how basic chemistry can be used to perform reactions.
• Before reading this chapter, a review of Chapter 3 would be helpful because of the chemical properties of the substrates and the interactions of the enzymes with them.
• In addition, refresh your understanding of the concepts presented in Chapter 8.

• You should be able to complete the objectives once you have mastered this chapter.

• There are four strategies commonly used by enzymes.

• Explain why pep tide bonds are resistant to hydrolysis.

• Examples of each of the other catalytic mechanisms can be found on the list.

• Explain the relationship of CO2 to aerobic metabolism and how most of the CO2 generated by peripheral tissues is transported to the lungs.

• The physiologic requirement for the reaction that causes CO2 is indicated.

• A water molecule can be activated to attack CO2.