Equation 7a of the text, (max/KM1) [S]-(v'max/KM2) [P] 1+ [S]/KM1 + [P]/KM2 gives the expression for the rate of formation of product by a modified version of the Michaelis-Menten mechanism in which the second step is also reversible. Derive the expression and find its limiting behaviour for high and low concentrations of substrate.
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![Equation 7a of the text,
(Umax/KM1) [S] (v'max/KM2) [P]
1+ [S]/KM1 + [P]/KM2
-
gives the expression for the rate of formation of product by a modified version of the Michaelis-Menten
mechanism in which the second step is also reversible. Derive the expression and find its limiting behaviour for
high and low concentrations of substrate.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fd97245cf-b189-4bba-96f3-67d0dd7020a2%2Fadf8bdc1-42c8-4625-888f-b5ce221ba5a1%2Facdi908_processed.png&w=3840&q=75)
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- The specific activity of a pure preparation of pyruvate kinase (PK) assayed in the direction of pyruvate formation (forward direction) is 220 U/mg at 25 C and at pH 7.5. This form of PK is a homotetramer (MW 240,000) with one active site per subunit. What is the kcat for PK in the forward direction?Residue Asn 204 in the glucose binding site of hexokinase IV was mutated, in two separate experiments, to either Ala or Asp. The Asn → Ala mutant had a KM nearly 50-fold greater than the wild-type enzyme, and the Asn → Asp mutant had a 140-fold greater KM value than the wild-type enzyme. These mutations impact the intermolecular interactions between the enzyme and the glucose substrate.The amide functional group of the Asn side chain can form with the hydroxyl groups of the glucose substrate and can potentially function as either a . The methyl group of Ala cannot participate in hydrogen bond formation, which explains the in glucose affinity as indicated by the higher KM for the mutant enzyme. The side chain of Asp could potentially serve as a , but…Residue Asn 204 in the glucose binding site of hexokinase IV was mutated, in two separate experiments, to either Ala or Asp. The Asn → Ala mutant had a KM nearly 50-fold greater than the wild-type enzyme, and the Asn → Asp mutant had a 140-fold greater KM value than the wild-type enzyme. These mutations impact the intermolecular interactions between the enzyme and the glucose substrate.The amide functional group of the Asn side chain can form (dipole-dipole interactions, hydrodgen bonds, London Dispersion Interactions, or Ion-Dipole Interactions) with the hydroxyl groups of the glucose substrate and can potentially function as either a (hydrogen bond donor and/or acceptor, hydrogen bond donor, or hydrogen bond acceptor) . The methyl group of Ala cannot participate in hydrogen bond formation, which explains the (increase or decrease)…
- Calculate the Km of the maltase reaction with inhibitor X (If inhibitor X changes maltase activity to a Vo of 0.10 mM per minute when [S] = 0.125 mM, and a Vo of 0.25 mM per minute when [S] = 0.50 mM). 0.10 mM 0.20 mM 0.50 mM 1.0 mM 2.0 mMA. Lineweaver-Burk plot of the enzyme with increasing amounts of substrate in the absence or the presence of the inhibitor is shown below. Graph A : x-intercept Graph B : x-intercept = - 0.012, y-intercept = 0.8 Graph C : x-intercept = - 0.027, y-intercept = 0.8 Graph D : x-intercept = - 0.039, y-intercept = 0.8 - 0.007, y-intercept = 0.8 Graph A 4 Graph B Graph C Graph D 1 -0,04 -0,02 0,00 0,02 0,04 1/[Substrate] (uM) (i) Which graph indicates an enzymatic reaction without inhibitor? (ii) Which type of inhibitor is it? Briefly explain. (iii) Which graph indicates the highest concentration of inhibitor? (iv) Calculate the Vmax and Km of the graph showing an enzymatic reaction with the lowest concentration of inhibitor. Show the steps of calculation and unit in your answers. Keep 2 decimal places in your answers. 1/Rate (umol/min)*The enzyme glucose oxidase isolated from the mold Penicillium notatum catalyzes the oxidation of 3-D-glucose to D-glucono-6- lactose. This enzyme is highly specific for the ß anomer of In spite of this glucose and does not affect the a anomer. specificity, the reaction catalyzed by glucose oxidase is commonly used in a clinical assay for total blood glucose that is, for solutions consisting of a mixture of 3- and a-D- glucose. What are the circumstances required to make this possible? Aside from allowing the detection of smaller quan- tities of glucose, what advantage does glucose oxidase offer over non-enzymatic oxidizing agents like Tollens reagent? *Is B-D-glucosamine a reducing sugar?
- Substrate A has a Km of 45uM and a kcat of 100/s with trypsin and a Km of 540mM and a kcat of 2/s with chymotrypsin. Substrate A is a better substrate of [trypsin/chymotrypsin] and likely contains a/an/some [aromatic/acidic/basic] residues(s) in its sequence. Chymotrypsin contains a [glycine/serine/aspartate] in its specificity pocket which likely [repel/attract] the residues in substrate A.The Y150F (Tyr 150 to Phe) mutant of ß-lactamase catalyzes the reaction with a keat of 3.1 * 10³ s¹ at pH = 7.0 and 37 °C. Assuming this decrease in keat is due to the loss of transition state stabilization by the Y150F mutant, calculate the AAG* between the wild-type and Y150F enzymes. Show your work and state any assumptions you make in solving this problem.The researchers working on happyase discover that the compound STRESS is a potent competitive inhibitor ofhappyase. Addition of 1 nM STRESS increases the measured Km for SAD by a factor of 2. What are the values for α and α′ under these conditions?
- Consider the role of Histidine in the Serine protease mechanism and sketch a plot showing the predicted pH profile of chymotrypsin which has a pH optimum of approximately ~8. The pk, for the His in the catalytic triad is 7.3 in free chymotrypsin which increases to greater than 8 with a bound peptide. Be sure to label the plot axes and indicate the pka of His on the plot,In 1966, Ferdinand showed that a random-order ternary-complex mechanism for a two-substrate enzyme-catalysed reaction can lead to sigmoidal kinetics being observation in the absence of cooperative binding. Discuss this scenario so that it is clear why a plot of Vo versus [AXo] at constant [Bo] will be sigmoidal.Synthesis of the purine nucleotides AMP andGMP proceeds by a branched pathway starting with ribose5-phosphate (R5P), as shown schematically in FigureQ3–4. Using the principles of feedback inhibition, proposea regulatory strategy for this pathway that ensures an ade-quate supply of both AMP and GMP and minimizes thebuildup of the intermediates (A–I) when supplies of AMPand GMP are adequate.
