1) The kinetic parameter of an enzyme is:

K_m = 0.15 mM for substrate A
K_m = 1.2 mM for substrate B
V_max is same for both the substrate

a) Calculate the initial velocity (Vo) for each substrate in terms of V_max when:

i) [S] = 0.15 mM
ii) [S] = 1.5 mM
iii) [S] = 15 mM

b) Which substrate has greater affinity for the given enzyme?

2) The following data was obtained for an enzyme-catalyzed reaction. Estimate the K_m and V_max for this enzyme and find what will be the V_o at [S] = 2.5 x 10^-5 M?

[S], M Vo (nmoles/min/mg/protein)

6.25 x 10^-6 15

7.50 x 10^-5 56.3

1.50 x 10^-4 60

1.00 x 10^-3 74.5

1.00 x 10^-2 75

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Nelly StrackeLv2

12 Apr 2019

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An allosteric interaction between a ligand and a protein is one in which: 74. changes in Km are negligible, so Km can be treated as a constant. C. binding of a molecule to a binding site affects binding of additional molecules to the same site. binding of a molecule to a binding site affects binding properties of another site on the protein. binding of the ligand to the protein is covalent. multiple molecules of the same ligand can bind to the same binding site. A. D. VO Vmax. E. varying [S] has no effect on Vo. 78. The steady state assumption, as applied to B. enzyme kinetics, implies: Km = Ks. the enzyme is regulated the ES complex is formed and broken down at equivalent rates. the Km is equivalent to the cellular substrate concentration A. B. C. C. D. D. E. the mximum velocity occurs when two different ligands can bind to the same binding site. E. the enzyme is saturated. The following data were obtained in a study of an enzyme known to follow Michaelis-Menten kinetics: 79. 75. One of the enzymes involved in glycolysis aldolase, requires Zn2+ for catalysis. Under conditions of zinc deficiency, when the enzyme may lack zinc, it would be referred to as the: Substrate added vo COmol/min) (mmol/L) A. apoenzyme. B. coenzyme. C. holoenzyme. D. prosthetic group. E. substrate 0.8 217 325 488 647 76. 1,000 The role of an enzyme in an enzyme- catalyzed reaction is to: bind a transition state intermediate, such that it cannot be converted back to substrate. ensure that all of the substrate is converted to product ensure that the product is more stable than the substrate. increase the rate at which substrate is converted into product. make the free-energy change for the reaction more favorable. A. The Km for this enzyme is approximately: B. C. D. E. A. 1 mM B. 1,000 mM. C. 2 mM D. 4 mM E. 6 mM 80. In competitive inhibition, an inhibitor: binds at several different sites on an enzyme. binds covalently to the enzyme. binds only to the ES complex. binds reversibly at the active site. lowers the characteristic Vmax of the enzyme A. B. C. D. E. 77. The benefit of measuring the initial rate of a reaction VO is that at the beginning of a reaction: A. [ES] can be measured accurately. B. changes in [S] are negligible, so [S can be treated as a constant

1. If Vmax for an E catalyzed reaction is 3.0 Î¼mole/min in thepresence of 1.5 micrograms of enzyme (mol. Wt 30,000) What is theturnover number? Express answer in sec-1

2. An enzyme catalyzed reaction whereby substrate S isisomerized to product P, using 2.5 mg of enzyme, mol wt 125 000,was found to have a Km of 3 x 10^-3 M and a maximumvelocity of 275 mmole/min. What is the turnover number?

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[S], M	Vo (nmoles/min/mg/protein)
6.25 x 10^-6	15
7.50 x 10^-5	56.3
1.50 x 10^-4	60
1.00 x 10^-3	74.5
1.00 x 10^-2	75