Bubble boy disease is a deficiency in adenosine deaminase that results in the accumulation of DADP. As a resu O The specificity of ribonucleotide reductase becomes unregulated and only pyrimidines are synthesized O Allosteric binding of excess DADP cause the R1 dimer to separate into monomers O The reaction equilibrium shifts and ribonucleotide reductase starts synthesizing ribonucleotides O The overall activity of ribonucleotide reductase is inhibited The tyrosil-radical site is disrupted by binding of excess DADP
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- In the action of ribonucleotide reductase O A. a tyrosine residue removes a proton from the sugar ring O B. thioredoxin oxidizes a dithiol a tyrosine residue removes a hydrogen atom from the 2' OC position of the sugar ring a tyrosine residue donates a hydrogen atom from the 2" OD. position of the sugar ring O E. the 2' OH group is removed from the sugar ring as a water moleculeThe enzyme cytidine deaminase catalyzes the conversion of cytidine to uridine. Cytidine deaminase catalyzes the reaction through an addition of water across the cytidine 3,4-bond, forming a tetrahedral intermediate followed by the elimination of NH3 to form the product uridine. This is like the addition-elimination mechanism that we studied for adenosine deaminase. Cytidine deaminase НО. NH₂ N пOH OH cytosine R + H₂O cytidine The Km value for the substrate cytidine is 2.5 × 10-4 M, and the K; for competitive inhibition by the product uridine is 2.5 × 10-³ M. N R A reduced derivative of the product, 3,4,5,6-tetrahydrouridine was shown to be a fully reversible competitive inhibitor with a Ki of 2.4 x 10-7 M, a value approximately 10,000 times lower than that of the product uridine. NH₂ NH uridine HO. N R R = D-ribose HOH OH uridine 3NH uracil ring numbering H NH H H + NH3 H OH H 3,4,5,6-tetra- hydrouridine a.) Draw a structure of the intermediate that we predict to form during the…Many enzymes are switched "on" by attachment of a phosphate group at a specific serine somewhere on the protein (phosphorylation). The basic reaction is: E + ATP2 Ep + ADP Po SERINE PHOSPHO SERINC (Note the "squiggles" before the backone amide and carbonyl indicate the polypeptide chain continues on either side of the serine). For phosphorylation to have this effect, there has to be some equilibrium between inactive and active forms conformations of the enzyme: [Eactive] [Einactive] Einactive 2 Eactive; K* The same basic equilibrium must exist for the phosphorylated protein: [Ep,active] [Ep,inactive] EP,inactive 2 Ep,active; Kp = (a) If phosphorylation increases the measured activity of the enzyme, is K* or K larger? Why? (b) Does the phosphorylation site need to be near the site where the enzyme binds its substrate (e.g. the reactant whose chemistry it catalyzes)? Why or why not?
- Using abbreviations (not structures), write the reaction of flavin adenine dinucleotide that gives off energy (-AG). Match the items in the left column to the appropriate blanks in the equation on the right. H+ FADH → FADH3 + H+ 3/2H2 FADH3 FAD FADH 2H+ FADH2 3H+ H2The biosynthetic pathway for the two amino acides E and H is shown schematically in the Figure below: A W E 2 Z H How would this pathway change if C was mutated and non-functional?The alkaline hydrolysis of pAUGCAGC oligonucleotide produces: O A. Uridine 2'-monophosphate, uridine 3'-monophosphate, cytosine 2'-monophosphate O B. Adenosine 2'-monophosphate, adenosine 3'-monophosphate, adenosine 21,5'-bisphosphate OC. Guanosine 2'-monophosphate, guanosine 3'-monophosphate, cytosine 3'-monophosphate O D. Cytidine 3'-monophosphate, guanosine 2'-monophosphate, adenine 2'-monophosphate O E. Adenine 3,5'-bisphosphate, guanine 2,5'-bisphosphate, uridine 2'-monophosphate O F. Uridine 2'-monophosphate, uridine 3'-monophosphate, guanine 3'-monophosphate
- What is the role of his 12 in the RNase-catalyzed hydrolysis of RNA, as indicated in the Figure below? 5' His 119 RNA CH₂ H H 3 O Base 1' 12' 8-H His 12 -P-0-CH₂ a Base H H O -0-P=O H OH NH It acts as a general base, abstracting a proton from the 2' hydroxyl in order to increase its nucleophilicity. It forms an H-bond with his119 in order to stabilize the transition state. It donates a proton to improve the quality of the leaving group. It works through concerted general acid-base catalysis with his119 in order to favour the products of the reaction. Two of the above are true.You are presented with Cytidine 5' triphosphate and Thymidine 5' triphosphate. Draw and upload to Efundi these phosphorylated structures as they would be connected in a polinucleotide in the order CpT / Show the individual phosphorylated structures first then show how they combine to form the polynuleotide. Show at any one of these structures where the glycosidic bond occursIn DNA sequencing by the Sanger method, the template strand of the DNA is radioactive specific enzymes are used to cut the DNA into small pieces which are then separated by electrophoresis dideoxy ATP (ddATP) is included in each of the four reaction mixtures before enzymatic synthesis of complementary strands the role of dideoxy CTP (ddCTP) is to occasionally terminate enzymatic synthesis of DNA where Gs occur in the template strands.
- When performing his experiments on protein refolding, Christian Anfinsen obtained a quite different result when reduced ribonuclease was reoxidized while it was still in 8 M urea and the preparation was then dialyzed to remove the urea. Ribonuclease reoxidized in this way had only 1% of the enzymatic activity of the native protein. Why were the outcomes so different when reduced ribonuclease was reoxidized in the presence and absence of urea?Given the following diagram of how protein AWESOME1 binds to it's target DNA, describe the potential effects of each of the 5 mutations shown below. The wild-type sequence of a helix #1 is also shown in the blue box, and all the mutations are in helix #1 (see numbers for identifying particular residues). a helix #1 R(1)-V-I-L-Y-F-W-I-M-Y-F-S-H-Y-W-R(16) #1 Predict the consequence of the following mutations: 1) Arg(1) to Glu 2) Arg(1) to Ala 3) Phe(6) to lle 4) Trp(7) to Phe 5) Met(9) to Pro inProtease enzymes cleave proteins by hydrolyzing peptide bonds. The strategy for each type of metalloprotease begins with generating a nucleophile that attacks the peptide bond that attacks the peptide carbonyl group. O Macmillan Learning On the basis of the information provided in the figure, show the next step in the mechanism for peptide-bond cleavage by a metalloprotease. Metalloproteases H R₁ HN Zn Enz 2+ R₂ Draw curved arrows on the pre-drawn structures to show the metalloprotease mechanism. If you need to reset the structures, click More followed by Reset Drawing. Select Draw Templates Groups More B - H Enz H H с R1 | : HN O | Zn 2+ B R2 N Zn Erase