PracticeExam1_key - Name_KEY_ EID_ CH n339K Exam 1 10 June...

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Name__________ KEY ______________________________ EID__________________________________________ 1 CH n339K Exam 1 10 June 2008 125 points total 15 pts. 1 . a . A spheroidal bacterium with a diameter of 1 µ m contains two molecules of a particular protein. What is the molar concentration of the protein? V = 4 3 " r 3 ; r (in cm) = 0.5 μ m " 1 m 10 6 μ m " 100 cm 1 m = 0.5 " 10 # 4 cm V = 4 3 (0.5 # 10 $ 4 cm ) 3 = 4 3 # 1.25 # 10 $ 13 cm 3 = 5.2 # 10 $ 13 cm 3 = 5.2 # 10 $ 13 ml = 5.2 # 10 $ 16 L 2 molecules " 1 mol 6.02 " 10 23 molecules " 1 5.2 " 10 # 16 L = 6.4 " 10 # 9 M = 6.4 nM 10 pts. b . This same cell contains 200,000 molecules of the free amino acid tyrosine. What is the internal tyrosine concentration? This could be calculated the same way as in part a: 200000 molecules " 1 mol 6.02 " 10 23 molecules " 1 5.2 " 10 # 16 L = 6.4 " 10 # 4 M = 0.64 mM Or, an easier approach is to recognize that the tyrosine concentration must be 10 5 times greater than the protein concentration (200,000/2). Thus, the internal tyrosine concentration is (6.4 x 10 -9 )10 5 = 6.4 x 10 -4 M = 0.64 mM 10 pts. c . Each cell contains 6 x 10 -2 fg of tyrosine. What is the molecular weight of tyrosine?
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This note was uploaded on 11/25/2009 for the course CH 53890 taught by Professor Raymonds during the Spring '09 term at University of Texas at Austin.

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PracticeExam1_key - Name_KEY_ EID_ CH n339K Exam 1 10 June...

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