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# 16 - LECTURE 16 ANNOUNCEMENTS 1 QUIZ 4 A 20.7 sd 5.7 B 22.8...

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LECTURE 16 ANNOUNCEMENTS 1. QUIZ 4: A 20.7 sd 5.7 B 22.8 sd 4.8 C 23.4 sd 5.2 WEDNESDAY S LECTURE TODAY 1. USING KINETICS TO COMPARE ENZYMES 2. TODAY 9/30, 2:55PM IN COMSTOCK B108: INTERVIEWING AT MEDICAL SCHOOL– WHAT ARE THEY LOOKING FOR?! 2. THE MICHAELIS-MENTEN EQUATION, V IN = (V MAX [S])/(K M + [S]) HOW TO MEASURE K M AND V MAX BY USE OF LINEWEAVER-BURK PLOT 3. ES DISSOCIATION CONSTANT K S = k -1 / k 1 1. AN ELEMENTARY RATE CONSTANT CORRESPONDS TO ONE ENERGY BARRIER TO REACH A TRANSITION STATE 2. ENZYME INHIBITION 3. PLEASE DO NOT ASK THE PROF TO ADVERTISE FOR CHARITIES, HOWEVER WORTHY THEY BE, UNLESS STUDENT EDUCATION IN HEALTH OR SCIENCE IS INVOLVED. 4. SOME ENZYMES TURNOVER THEIR SUBSTRATES IN MICROSECONDS OR FASTER (CARBONIC ANHYDRASE); OTHERS REQUIRE MUCH MORE TIME (CHYMOTRYPSIN)

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USING THE LINEWEAVER-BURK PLOT DATA [S], M V, M min 0.005 0.005 0.0025 0.0033 0.00167 0.0025 -1 FIRST STEP: CALC. RECIPROCALS 1/[S], M 1/V, M min 200 200 400 300 600 400 -1 -1 1/V 200 400 -400 -200 0 200 400 600 1/[S] X X X SECOND STEP: GRAPH I/V M = 100 M -1 x MIN; V M = 0.01 M MIN -1 -1/K M = -200 M -1 ; K M = 1/(200 M -1 ) = 0.005 M
WHAT IS AN EFFICIENT ENZYME? TURNOVER NUMBER : NUMBER OF SUBSTRATE MOLECULES CONVERTED TO PRODUCT PER UNIT TIME WHEN ENZYME IS SATURATED WITH SUBSTRATE. TURNOVER NUMBER OF SOME ENZYMES CARBONIC ANHYDRASE 600,000/sec ACETYLCHOLINESTERASE 25,000/sec PENICILLINASE 2000/sec CHYMOTRYPSIN 100/sec DNA POLYMERASE I 15/sec LYSOZYME 0.5/sec V MAX = k CAT [E] TOTAL CAN BE ELEMENTARY RATE CONSTANT k 2 , OR MORE COMPLEX. IT ANSWERS THE QUESTION: HOW FAST DOES THE ENZYME WORK? e.g. CARBONIC ANHYDRASE , k CAT = 600,000 / sec CONSIDER A SINGLE CARBONIC ANHYDRASE MOLECULE: ONCE ENZYME HAS A SUBSTRATE BOUND, PRODUCT IS FORMED IN 1/ k CAT = 1.7 x 10 -6 SEC LOOK AT TURNOVER NUMBER TABLE: CARBONIC ANHYDRASE, k CAT = 600,000 / sec WOW CHYMOTRYPSIN, k CAT = 100 / sec WIMP?

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