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- АCTIVITY 5.2.4 Create a graph illustrating the activity of an enzyme or rate of reaction with the different factors affecting enzyme activity: 1) concentration of enzyme, 2) concentration of substrate, 3) temperature, 4) pHP3D.2 In biological cells, the energy released by the oxidation of foods is stored in adenosine triphosphate (ATP or ATP“).The essence of ATP's action is its ability to lose its terminal phosphate group by hydrolysis and to form adenosine diphosphate (ADP or ADP): ATP* (aq) + H,O() → ADP* (aq) + HPO (aq) + H,O*(aq) At pH = 7.0 and 37°C (310K, blood temperature) the enthalpy and Gibbs energy of hydrolysis are A,H =-20kJ mol and A,G=-31 kJ mol", respectively. Under these conditions, the hydrolysis of 1 mol ATP“(aq) results in the extraction of up to 31kJ of energy that can be used to do non- expansion work, such as the synthesis of proteins from amino acids, muscular contraction, and the activation of neuronal circuits in our brains. (a) Calculate and account for the sign of the entropy of hydrolysis of ATP at pH = 7.0 and 310K. (b) Suppose that the radius of a typical biological cell is 10µm and that inside it 1x 10ʻ ATP molecules are hydrolysed each second. What is the power density of…The turnover number for an enzyme that approximates Michaelis-Menten kinetics is known to be 500 min^-1. From the results shown in the table, enumerate Km and total amount of enzyme present. What is the Km for this enzyme? What is the Vmax for this enzyme? And what is the [E]T for this enzyme?
- Glucosidase I catalyzes hydrolysis of specific glucosidase I is a synthetic trisaccharide, glucose-al-2- glucose-al-3-glucose-a-O(CH₂) #COOCH3. Kinetic measurements oligosaccharides containing glucose. obtained using this trisaccharide as substrate in the deoxynorjirimycin at concentrations of 50 μM (), 100 μM absence (x-x) and presence of the inhibitor 1- A) were used to prepare the (-), and 200 μM (4 Lineweaver-Burk plot below: b) Page 3 12) 7. a) V/V (nmol/hr)-1 1.S 1.0- 0.5 1/Trisaccharide (mM)-! Estimate the values for Vmax and KM for the trisaccharide substrate in the absence of the inhibitor. 0.0 -1.0 0.0 One substrate for 1.0 2.0 Determine whether inhibition by 1-deoxynorjirimycin is competitive, non-competitive or neither.Que:- An Michaels - Menten Kenetirs eneyme obeying the following parameters K2= 1•4 x10%M's" , K,= 2:5x1o°5' gave - 1 K3 = l-5 x10s ca) So this enzyme ctalytically pufet ? Explain your heasoning . (b) What is Vmaz , ik 1-4 nmol/me and saturabing eveyrme are usecd ? substrateSubstrate KM (M) N-Acetylvaline ethyl ether 8.8 X 10 -2 N-Acetyltyrosine ethyl ether 6.6 X 10-4 Which substrate has the higher apparent affinity for the enzyme? Explain. Which substrate is likely to give a higher value for Vmax?
- 1. pH Effects a. In the enzyme mechanism of lysozyme, two acidic amino acid residues, Asp52 and Glu35, are critical for catalytic activity. If we assume normal side chain pKa values for Asp (pKar = 3.90) and for Glu (pKar = 4.07), what proportion of enzyme molecules will have both Asp52 and Glu35 in the correct ionization state at pH 5.0 (the pH optimum for lysozyme)? b. Are the traditional pKa values likely to be correct within the protein? What pKa changes might be present within lysozyme?Initial rate data for an enzyme that obeys Michaelis–Menten kinetics areshown in the following table. When the enzyme concentration is 3 nmolml-1, a Lineweaver–Burk plot of this data gives a line with a y-intercept of0.00426 (μmol-1 ml s). (a) Calculate kcat for the reaction.(b) Calculate KM for the enzyme.(c) When the reactions in part (b) are repeated in the presence of 12 μM ofan uncompetitive inhibitor, the y-intercept of the Lineweaver–Burk plotis 0.352 (μmol-1 ml s). Calculate K′I for this inhibitor.A synthetic substrate, the para-nitrophenylacetate (PNPA), is used to monitor enzyme activity of protein P. The product of the hydrolysis reaction absorbs at 410 nm with a molar extinction coefficient of 4 000 M-¹.cm-¹. 1- Write the reaction catalyzed by the protease P using the pNPA substrate. 2- The enzyme extract is too concentrated and a 1/300 dilution is needed for enzyme tests. Considering that you would need at least 600 µL of diluted enzyme extract for activity tests, indicate which volume of buffer and enzyme extract you must use for the dilution.
- An enzyme has a rate enhancement of 1.3x106. Calculate the value of ΔΔG‡ at 25.0 °C in kJ mol-1. (Hint: be sure to pay attention to units and signs.) (R = 8.3145 J mol-1 K-1)11 X For the hydrolysis of glucose-6-phosphate, glucose-6-P + H₂O = glucose + Pi AG'°= -13.8 kJ/mol and K'eq = 3.81x10-³ at 25°C. Therefore the [glucose-6- phosphate]/[glucose] ratio is about 1/260 at equilibrium. What is the [glucose-6- phosphate]/[glucose] ratio at equilibrium when this reaction is coupled to ATP hydrolysis in the coupled reaction glucose + ATP = glucose-6-phosphate + ADP and [ATP] = ADP]? For ATP hydrolysis AG'°= -30.5 kJ/ mol at 25°C. 0/1 Point Earned 9/10 Attempts Remaining1 pt pt 9146 Bb 9146 Bb 1031 Class Etsy E Traps E Traps New Free Chat + ☆ 出口 keAssignment/takeCovalentActivity.do?locator-assignment-take [References] You do an enzyme kinetic experiment and calculate a Vmax of 118 μmol per minute. If each assay used 0.10 mL of an enzyme solution that had a concentration of 0.20 mg/mL, what would be the turnover number if the enzyme had a molecular weight of 128,000 g/mol? (Enter your answer to two significant figures.) turnover number = sec-1 D 1 pt Submit Answer Try Another Version 2 item attempts remaining estion stion 5 on 6 7 1pt 1 pt 1 pt 1pt 1pt 1pt 1 pt 1 pt D is the substrate concentration multiplied by the catalytic constant. KM is equivalent to the substrate concentration multiplied by the ratio of rate constants for the formation and dissociation of the enzyme-substrate complex. KM is equivalent to the substrate concentration. KM is equivalent to the substrate concentration divided by 2 A: KM is equivalent to the substrate concentration…
