Chm - 2 a K M = V max = K M = V max = b Lineweaver-Burke...

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Chm. 327 - W 2011 - Homework #5 1. Enzyme classification a) i. 6.3.2.12 – Dihydrofolate synthase ii. 2.3.1.36 – D-amino-acid N-acethyltransferase iii. 4.2.3.1 – Threonine synthase iv. Alchohol dehydrogenase – 1.1.1.1 v. Chymotrypsin – 3.4.21.1 vi. Retinol isomer – 5.2.1.7 b. i. What is the other name? Aldehyde reductase ii. How many PDB entries are there for ADH? 91 How many from yeast? 1 What is/are the YADH (YeastAlcoholDehydrogenase) PDB code(s)? 2hcy iii. How many PDB entries are there for human ADH? 26 How many subunits does 1agn have? One NAD+ per subunit. NAD is the coenzyme for 1agn, what is the cofactor? Zinc or iron Estimate how many sheets and helices are in 1agn from the diagram in the Uniprot section. 4 sheets and 17 helices iv. What is its E.C. number if ADH uses NADP as an acceptor? (You’ll have to go back in the EBI site.) 1.1.1._
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Unformatted text preview: 2. a) K M = V max = K M = V max = b. Lineweaver-Burke graphical method gives most accurate Km, Vmax. c. 337 s-1 4. a) b) In the presence of a competitive inhibitor, it takes a higher substrate concentration to achieve the same velocities that were reached in its absence. So while V max can still be reached if sufficient substrate is available, one-half V max requires a higher [S] than before and thus K m is larger. With noncompetitive inhibition, enzyme molecules that have been bound by the inhibitor are taken out of the game so • enzyme rate (velocity) is reduced for all values of [S] , including • V max and one-half V max but • K m remains unchanged because the active site of those enzyme molecules that have not been inhibited is unchanged....
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