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The Exam: Friday Sept. 2 Will start at what time 8 AM Will end at 10:55 AM, sharp Format

The Exam: Friday Sept. 2 Will start at what time 8 AM Will end at 10:55 AM, sharp Format No calculator needed. Know scientific notation: 10 3 = 1000 10 -3 = 0.001, 1/10 6 = 10 -6 8 M - = 8 per molar 8 M -. sec - = 8 per molar per second Closed notes/Closed book

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The Exam: Friday Sept. 2 Will start at what time 8 AM Will end at 10:55 AM, sharp Format

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  1. The Exam: • Friday Sept. 2 • Will start at what time 8 AM • Will end at 10:55 AM, sharp • Format • No calculator needed. • Know scientific notation: 103 = 1000 10-3 = 0.001, 1/106 = 10-6 • 8 M- = 8 per molar 8 M-.sec- = 8 per molar per second • Closed notes/Closed book • Strongly weighted towards lectures. • Goal: Class average of 87% This document at:

  2. Corrections to Homework + Key: • #9 E is also false • #14 Correct answer is E • #84 D is also a correct answer (chaperones DO affect folding rate) • #87: dissociation constant… not “equilibrium constant” • Ignore the “why doesn’t myoglobin form • sickle cell-like polymers?” question. • Cool Cool Logarithms: • Log(x/y) = Log x – Log y • Log (x . y) = Log x + Log y • Log xy = y . Log x • Same rules apply to ln

  3. Since the equilibrium constant for peptide hydrolysis • is extremely favorable (in favor of bond cleavage), how can • it be that some proteins (like some of the proteins in the lens • of our eyes) are chemically stable for many years? Concept of Kinetic Entrapment

  4. 57. Which of the following is NOT a possible function of the proteolytic processing of a protein? • a. plays roles in the process of metastasis • b. plays roles in protecting protein from hydrolysis • c. plays roles in peptide hormone production • d. plays a role in zymogen activation • e. plays roles in blood clotting cascade

  5. Homologs: Proteins that share significant sequence homology. Orthologs: Different organisms, homologous, same function. Paralogs: Same organism, homologous, but different functions.

  6. 91 Protein sequence motifs: a. May be used as intracellular organellar zip codes. b. May be intimately linked to protein function. c. One example is the helix-loop-helix motif. d. May be signals to direct posttranslational modification. e. May serve as “signatures” which allow membership in a particular protein family to be predicted.

  7. Which class of 1 substrate/1 inhibitor • kinetic scheme is most similar to the random • sequential scheme for a 2 substrate enzyme? Random sequential reaction

  8. Noncompetitive Inhibition (Your text refers to this as “mixed” inhibition).

  9. Lipid Modification of Proteins

  10. Other Forms of “Protein Stability”

  11. Prion Diseases A Special Class of Amyloid-Associated Disorders

  12. The Human Prion Protein

  13. Speculative 3-D Structure of the PrPSc (scrapies form) Experimental 3-D Structure of Core Domain of PrPC (healthy form).

  14. Competing models for the molecular basis of prion infectivity.

  15. Medical History Of Prion Disorders No need to memorize.

  16. Prion Diseases: Transmissable Spongiform Encephalopathies (TSEs) • Scrapies (sheep) • Bovine Spongiform Encephalopathy (BSE, Mad Cow Disease) • Kuru • Creutzfeld-Jakob Disease (CJD) • Gerstmann-Straussler-Scheinker Syndrome • Fatal Familial Insomnia (which is not always familial) • These disease are characterized by the formation of PrPSc-containing • “spongiform” deposits in the brain (amyloid-like), ataxia, and a variety • of other devastating symptoms that lead to death.

  17. Prion diseases can be triggered by: • Inherited and (most likely) sporadic mutations. • Transfer of infectious seed from one organism to another. • surgical procedures involving organs, tissues, fluids, or • molecules from infected organism • ingestion of tissue from infected organism. • Disease and Culture

  18. Making Binding Data Linear intercept = 1/Kd slope = -1/Kd fR/[L] 0 0 1.0 fR [L] fR = Kd + [L] y = m.x + b m = slope (constant) b = y intercept (constant) y= exptl. variable x= exptl. Variable i.e.: fR/[L] = y fR = x m = -1/Kd b = 1/Kd fR + fR [L] 1/Kd = 1 = [L] Kd fR Kd + fR [L] fR 1/Kd – fR/Kd = fR Kd + fR [L] [L] 1 = [L] fR Kd + fR 1 = [L] fR/[L] = 1/Kd – fR/Kd Scatchard Equation

  19. Hemoglobin has 4 binding sites, which are positively homocooperative. The Hill equation: [L]a fR = Kd,apparent + [L]a fR log = a. log[L] - logKd,apparent (1 – fR)

  20. (this has been corrected since the 8/31 review session– the mgs were right all along, but I had left out the 5 when I showed how it was calculated) 150 lbs X 454 grams/lb = 68,100 grams = 68 kg Assume density = 1 g/ml So… 68.1 kg = 68 liters = volume 1 micromolar = 10-6 molar 68 liters X 300 gram/mol X 10-6 moles/liter X 5 = 0.122 grams = 122 mgs

  21. It’s My Function!

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