ExamICollectionII - CHEM 153L: Exam I Collection Part II...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
CHEM 153L: Exam I Collection Part II Summer 2009 Question 1 Concepts of the Protein Assay 20 pts. a) Although ionic bonds are energetically stronger than nonpolar bonds, the order of noncovalent bond interactions between the protein and the coomassie dye are opposite (This is also true for most inter-molecular interactions between macromolecules in physiological conditions). Explain why ionic interactions between the dye and protein are the last to form while the nonpolar interactions are considered the driving force of the first interaction. Limit your answer to 2-3 sentences. Water’s concentration is sufficiently abundant to prevent ionic interaction to form between (+4) the net negative charge of the blue form dye to bind to the net positive charge of the protein (+2). Nonpolar regions of the both molecules unfavorably interact with water, and this weak solubility easily allows the blue form dye and the hydrophobic pockets of the protein to interact (+4 for nonpolar regions and weak interaction with water giving rising to more easily formed interactions). Once these van der waals (i.e. nonpolar bonds) interactions are formed, the negative charge of the blue form dye and the positive charges of the proteins are now in close enough proximity to allow for the ionic bond to stably form (+2). b) How is the interaction between the polypeptide chain and the dye able to stabilize the blue form while the interaction between an equivalent mass of nonpolar amino acids and dye unable to stabilize the blue form? Explain in 1-2 sentences. Denatured protein provides a 3-D nonpolar pocket that allows for the blue form dye to be shield from the acidic environment (+4) while nonpolar amino acids lack such a pocket and allow for the dye to protonated even if it where to temporally interact (+4). Question 2: Suppose you set out to perform the enzyme assay by monitoring the reduction of NAD + as opposed to how we normally measure enzymatic activity. Answer the following questions relative to this change in our enzyme assay. a) If you used the same concentrations of NAD + and lactate relative to our normally designed enzyme assay of NADH and pyruvate, then how would this negatively affect our ability to measure the initial velocity and why? Explain in less than 2-3 sentences. The initial velocity would be difficult to obtain because the absorbance change in both the rate and to equilibrium would be so minute (+4) due to the favorability of forming NAD+ as opposed to NADH (+4). Even if [E] concentration was low enough and rate was slow enough,
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
CHEM 153L: Exam I Collection Part II the low change in absorbance relative to the equilibrium absorbance would contain significant spectrophotometer error resulting in an unreliable initial velocity (+4 for spectrophotometer error). b) Explain exactly how the phosphate buffer would work under this direction of the reaction to
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 02/13/2010 for the course BIOCHEM 153L taught by Professor Kim during the Fall '09 term at UCLA.

Page1 / 9

ExamICollectionII - CHEM 153L: Exam I Collection Part II...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online