Spring08_MidtermI_Exam_KEY

# Spring08_MidtermI_Exam_KEY - MCB102 MIDTERM EXAM February...

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MCB102 MIDTERM EXAM February 27, 2008 Name ________________________________ Student ID_____________________________ Mean: 72.19 Standard deviation: 14.44 1 Question Points 1 (10 pts) 2 (14 pts) 3 ( 8 pts) 4 (10 pts) 5 (15 pts) 6 (10 pts) 7 (08 pts) 8 (15 pts) 9 (10 pts) Total _____/100

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Some information you may need: Henderson-Hasselbalch equation: pH = pK a + log([A - ]/[HA]) Michaelis-Menten equation: v o = V max [S]/K m + [S] 1. (10 points) While working in a research lab over Spring break, your supervisor asks you to make a buffer solution. She knows that you learned how to make buffers in MCB102 and asks you to come up with your own recipe. You are asked to make 500 mL of phosphate buffer at 0.5 M concentration and final pH of 6. You have the following solutions on the chemical shelf in the laboratory: 1M H 2 PO 4 - (pK a of dihydrogen phosphate is 6.86) 0.5M KOH How much of each solution would you mix to make the buffer? (You will have to add the appropriate amount of water to fill up to 500 mL at the end) Ans: 3 points total for 250mL 1M H 2 PO 4 2- : For 500 mL 0.5M phosphate: (0.5M H 2 PO 4 - )(500 mL) / 1M H 2 PO 4 - = 250 mL Same points if no calculator (can do this by hand) 6 points total for 61mL 0.5M KOH: 2 points: 6 = 6.86 + log A-/HA A-/HA = 0.14 Could leave as 10^-0.86 if no calculator 1 point: HA + A- = (0.5L)(0.5M) = 0.25 mol Same points if no calculator 2 mL of 1M H 2 PO 4 2- mL of 0.5M KOH mL of H 2 O
2 points (solving for moles): A- = 0.0307 mol have to be converted 1 point (converting to mLs): (0.0307 mol KOH) / (0.5 M) = 0.061 L KOH = 61 mL No calculator: describe in words that you would use the 2 equations with 2 variables to solve for moles of A- then add equivalent moles of KOH. Then you calculate volume of KOH by dividing moles by concentration. 1 point for 189mL water (if the other volumes were wrong, you could still get this point) 500 mL – 250 mL – 61 mL = 189 mL H 2 O 2. (14 points) (1 point each, no partial credit) Fill in the following blanks with the single-letter code for the amino acid that best fits the description: __S___ Plays the role of the initiating nucleophile during peptide bond cleavage catalyzed by the catalytic triad of chymotrypsin __M___ Cyanogen bromide cleaves polypeptides on the C-terminal side of this amino acid __K___ In addition to R and H, this amino acid is positively charged at neutral pH __W___

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## This note was uploaded on 10/26/2008 for the course MCB 102 taught by Professor Staff during the Spring '08 term at University of California, Berkeley.

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Spring08_MidtermI_Exam_KEY - MCB102 MIDTERM EXAM February...

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