Homework 2 cell

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University of West Georgia *

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1101

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Chemistry

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Feb 20, 2024

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docx

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1 Homework- 2, Total points: 30, Due date: 9/18/20 by 5 p.m. Instructions: 10% of points will be deducted if homework is submitted after 5 p.m. on the due date (9-18-20) at the rate of 10% per day that you are late. Homework submitted after 5 p.m. on 9/20/20 will not be accepted or graded. Only one file submission is allowed and only one submission can be uploaded. You will not be allowed to replace file after you have submitted your final homework on Courseden. Hence carefully review your file before submission. Please submit your homework as Word file or a pdf file. No image file will be accepted. You must insert images of your calculations or image of labeled graph in the word file before submission so that your submission is one single file. Read instructions before answering questions. Failure to follow instructions will result in loss of 5 points. (Topics: Thermodynamics & Free energy, Activated carriers, enzyme catalysis and inhibitions) 1. State two major differences between a chemical catalyst and an enzyme (2 points) Enzymes are organic and chemical catalyst is inorganic. Enzymes are specific and chemical catalyst is not. 2. Why is G, but not G a criteria for predicting reaction spontaneity? Be very clear in your answer to get full credit . (2 points) G can be positive but G can be made negative by adjusting the concentrations of reactants and products. 3. A chemical reaction has a G of -20 kJ/mol. Use temperature of 37ºC and molar gas constant 8.314 X 10 -3 kJ/ degree K/mole. Calculate the Keq of this reaction under equilibrium. Show all calculation steps to get full credit including the equation you will use for calculation (2 points). ∆G 0  = -RTlog K eq ∆G =  -20 kJ/mol, R = 8.314 X 10 -3 kJ/ degree/mole T = 273 + 37 = 300K ∆G 0  = -RTlog K eq -20 = -8.314 X 10 -3  X 300 logK eq -20 = -2.494 logK eq logK eq  = 20/ 2.494 logK eq  = 8.01 K eq  = 1.02 X 10 8
2 4. The G°' for the hydrolysis of ATP to ADP + P i is approximately -31kJ/mole. Calculate the equilibrium constant for this reaction (R = 8.314 J/degree K/mole) at the cellular temperature of 37°C. If the cellular concentrations of ATP, ADP, and Pi are 8, 1, and 8 mM, respectively. Which of the following answer is correct? (2 points). A. G = 0 B. G= -48.81 kJ/mole C. G= -68.57 kJ/mole D. G= 30.89 kJ/mole 5. Study the enzyme kinetics plot given below and answer the following questions. a. Calculate the K M app and Vmax app . Show calculations to get full credit. (2+ 2 points) -6 = -1/K M app -1/-6 = K M app 0.17 µM = K M app 7= 1/ Vmax app 1/7 = Vmax app 0.14 µM / min = Vmax app b. Which type inhibition does the graph show and why ? (2 +2 points) Uncompetitive Inhibition because both K m and V max were decreased in the graph
3 6. Study the enzyme kinetics plot given below and answer the following questions. a. Calculate (approximately) the K M of the enzyme and label the answer on the plot directly with a specific arrow on the plot for both questions given below with correct units : (2 points) b. K M in the blue line plot is _____ 1 µM _________. (1 point) c. K M in the red line plot is _______ 7µM _____________. (1 point) 7. Study the picture given below and answer the following questions (Please base your answer only on the given structure and your concept. Do not base your answer on theory or “Google’ or Bing” knowledge). a. Is this compound a nucleotide? (I don’t want a chemical name!!! so don’t google!!! Base your answer on the given figure). Explain with reasons. (1+ 1 point) Yes because it has a 5-C sugar molecule, a phosphate group and a nitrogenous base. b. Does this compound have any ester bond? If yes, how many? (1 + 1 point) Yes, 3 ester bonds c. How many phosphoanhydride bonds are there in this chemical? (2 pts) One d. Label specifically (the right bond) with an arrow all the phosphoanhydride and ester bond/s present in the given structure (4 points). Red plot Blue plot
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4 Blue arrows = ester bonds Red arrows = phosphoanhydride bond