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Get essential exam info, corrections to homework, key concepts, and study materials in this comprehensive guide for biochemistry students. Covering topics like protein stability, prion diseases, and enzyme kinetics.
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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:
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
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
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
Homologs: Proteins that share significant sequence homology. Orthologs: Different organisms, homologous, same function. Paralogs: Same organism, homologous, but different functions.
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.
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
Noncompetitive Inhibition (Your text refers to this as “mixed” inhibition).
Prion Diseases A Special Class of Amyloid-Associated Disorders
Speculative 3-D Structure of the PrPSc (scrapies form) Experimental 3-D Structure of Core Domain of PrPC (healthy form).
Competing models for the molecular basis of prion infectivity.
Medical History Of Prion Disorders No need to memorize.
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.
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
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
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)
(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