Question

1. A chemical reaction "A" has a DG of -10 kcal/mole. Calculate the key for the reaction A. Show the equation that you have used and all the calculations to get full credit. (2 pts). 2. State two common features among most of the activate carriers that you have studied in class. (2 points) 3. Study the sequence of the peptide given below and answer the following questions based on the peptide sequence given below. AGLRHTICHMKDCFY (10 points) A. Name the acidic amino acids in this peptide B. Name the basic amino acid/s in this peptide C. Name the polar aromatic amino acids in this peptide D. Name the hydrophobic aromatic amino acids in this peptide E. Name the hydrophobic non-aromatic amino acid/s in this peptide 4. State three differences between a motif and a domain that were taught in class. (3 points) 5. Which of the following parameter dictates whether a reaction can occur spontaneously or not-- DG'or DG? Explain your answer with very specific reasons. (3 points) 6. A chemical reaction has a DGº of -20 kJ/mol''. Calculate the Keq of this reaction under equilibrium. You will use the equation given below to start your calculation and will show stepwise derivation of the final equation that you will use to solve your problem. DO NOT copy the final equation from a book or google. I need to see every step in the calculation. Show all calculations to get credit (8 points). Temperature=37°C and R 8.314 J mol deg Equation given: DG-DGº +RT Ln (product] /[Reactant] 7. Study the enzyme kinetics plot given below and answer the following questions. Please show on the graph by using lines how you figured out the Ky values for the two plots, in addition to answering the questions given below. Points will be deducted if you do not show your work on the graph. R16 8 10 14 16 18 20 34 34 DSUM Calculate approximately: (2 +2 points) (I want the numerical values as answers here with correct unit) 1) the Ky of the enzyme shown in the blue plot 2) the Ky of the enzyme shown in the red plot

Answer 1

1) ∆G° = -10Kcal/mol = -10000 cal/mol (1kcal = 1000cal)

= -10000×4.184 J/mol = -41840 J/mol

∆G° = -RTln(K) where R = 8.314J/K.mol

-41840 = -8.314×298×ln{K)

ln(K) = 16.8875

K = e^(16.8875)

= 2.158×10⁷