bioe10-fall06-mt1-Kumar_Conboy-soln

bioe10-fall06-mt1-Kumar_Conboy-soln - BioE 10 Introduction...

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BioE 10 Introduction to Biomedicine for Engineers Midterm Exam Fall 2006 Name Solutions SID Write your name and SID on the top of each page! If you need extra space, use the back of the sheet.  No computers or electronic communications devices allowed. SCORE (for instructors only) Question 1:  30 Question 2:  30 Question 3:  20 Question 4: 15
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NAME: SID:   Question 5:  20 Question 6:   15 TOTAL 130
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NAME: SID: 1.  The following peptide sequence is derived in part from the extracellular matrix protein  laminin, and may be used to coat biomaterial surfaces to make them suitable for cell  adhesion: Glycine-Glutamate-Glycine-Tyrosine-Isoleucine-Glycine-Serine-Arginine A.  Write out this sequence using three-letter  amino acid code. Gly-Glu-Gly-Tyr-Ile-Gly-Ser-Arg (5 points, subtract one for each incorrect AA, or if sequence is in wrong order) B.  Write out this sequence using one-letter  amino acid code. GEGYIGSR (5 points, subtract one for each incorrect AA, or if sequence is in wrong order. C. Suppose that the pKa of Glutamate is 4.0, the pKa of Arginine and Serine are both 9.0,  the pKa of tyrosine is 10.0.  Use the Henderson-Hasselbalch equation to calculate the  fraction of charged tyrosine residues and the fraction of charged serine residues at pH. 9.5.   Henderson-Hasselbalch: pH = pKa + ln  log  ([base]/[acid]) Rearrange to get [base]/[acid] = exp  10^  (pH-pkA) For both ser and tyr, the  base  is the charged species, so we want [base]/([base]+[acid])  For Tyr, [base]/[acid] = 10^ exp  (9.5-10) = 0.606 .316 , and for Ser, [base]/[acid] = 10^ exp  (10- 9.5) = 1.65 3.16 Converting these to fractions, we get  f Tyr-  = 0.316 606 /(1+0.60 31 6) = 0.38 24  and f Ser-  =  1.65 3.16 /(1+1.65 3.16 ) = 0.62 .76 .   (15 points, 5 for using the HH equation and 5 for each of the calculations) D. What will be the net  charge on the majority  of peptide molecules at pH 7?  It is OK to  answer this question without doing any calculations, as long as you explain your reasoning. Because all of the pKa's of the titratable groups are far (2+ points) from the pH (7), these  groups will all be essentially all charged or uncharged.  Thus, we simply add up the charges  using the predominant form at pH 7:
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NAME: SID: NH3 + G E G Y G S R COO- +1 0 –1 0 0 0 0 +1 –1 Net charge is  zero (5 pts: 3 points for understanding the pH-pKa concept, 2 points for the correct final charge,   -1 for forgetting the termini charges )
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NAME: SID: 2.  The enzyme lactase catalyzes the degradation of the dietary sugar lactose into the  monosaccharides glucose and galactose.  People that cannot produce this enzyme at  sufficient levels are unable to digest lactose, a condition known as lactose intolerance that 
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This note was uploaded on 09/16/2009 for the course BIO ENG 10 taught by Professor Lar during the Spring '08 term at University of California, Berkeley.

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bioe10-fall06-mt1-Kumar_Conboy-soln - BioE 10 Introduction...

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