The enzyme PFK was studied in mosquitos. The authors found that: PFK Vmax was 4.5 μmol/min with F-1,6-BP (along with ATP) as the substrates PFK KM using F-1,6-BP and ATP was 1.48 mM F-2,6-BP was a powerful activator for PFK PFK was inhibited by ATP PFK inhibition by ATP could not be overridden by AMP On the graph below, draw: a. A Michaelis-Menten curve for the normal mosquito PFK enzyme b. A Michaelis-Menten curve for the PFK enzyme activated by F-2,6-GP C. A Michaelis-Menten curve for the PFK enzyme inhibited by ATP d. A Michaelis-Menten curve for the PFK enzyme when ATP and AMP are both present
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- Using the ActiveModel for phosphofructokinase (Trypanosoma), describe the difference between the APO1, AP02, and holoenzyme conformations.Figure 27.3 illustrates the response of R (ATP-regenerating) and U (ATP-utilizing) enzymes to energy charge. a. Would hexokinase be an R enzyme or a U enzyme? Would glutamine: PRPP amidotransferase, the second enzyme in purine biosynthesis, be an R enzyme or a U enzyme? b. If energy charge = 0.5: Is the activity of hexokinase high or low? Is ribose-5-P pyrophosphokinase activity high or low? c. If energy charge = 0.95: Is the activity of hexokinase high or low? Is ribose-5-P pyrophosphokinase activity high or low?V-class proton pumps run backward relative to the F-class ATP synthase. Consider the cartoon, which shows the conformations of the beta-subunits and ATPIADP + Pj of the F-class synthase. Which of the following associations between the conformation of the beta subunit and ATP/ADP + P¡ is correct for V- Binding Change Mechanism loose binding ADP+P ATP ATP class pumps? C repeat ADP + P, ADP АТР tight binding АТР +P оpen АТР O The open conformation releases ATP. Hydrolysis of ATP to ADP + P¡ drives the change from tight to loose. O Binding of ADP + P¡ drives change from open to loose. Hydrolysis of ATP to ADP + Pj drives the change from open to loose.
- Pls send solution fast within 10 minutes and i will give like for sure Solution must be in typed form Calcium ions bind to the SERCA Ca2+ -ATPase, which has two identical calcium sites, in two stages with apparent equilibrium constants K1=7×105M−1 and K2=2×106M−1 . a.) Calculate K. Note: K is not 0.1*1013 or 14*1011 or 1*1012. b.) Calculate τ.How is cigarette smoking related to elastase function? Methionine sulfoxide is formed by cigarette smoke in the antitrypsin protein, which leads to the full manifestation of the disease in individuals heterozygous for emphysema. Oxidation of methionine 358 by cigarette smoke of the elastase inhibitor active site leads to an increased inhibition of elastase and subsequent overaccumulation of mucus in the lungs. The inhibitor of elastase is axidized by cigarette smoke so elastase doesn't function properly in healthy individuals. Cigarette smoke modifies the elastase methionine residues such that the enzyme activity is blocked.The enzyme mutase which is important for the synthetic of tyrosine and phenylalanine in saccharomyces cerevisiae has been studied as an example of an allosteric enzyme. Tyrosine acts as a negative effector for this enzyme. What effects would you see on the action of the enzyme were you to increase the concentration of tyrosine? The chorismate mutase would shift to its R conformation The curve showing the kinetics or chorismate mutase would shift to the right The curve showing the kinetics of chorismate mutase would shift to the left The chorismate mutase would become saturated more rapidly
- Drag the terms on the left to the appropriate blanks on the right to complete the sentences. Terms can be used once, more than once, or not at all. Submit dCTP GMP AMP GTP dTTP activation inhibition dGTP My Answers Give Up UTP + Gln + ATP → CTP + Glu+ ADP + P₁. Cofactor: dCMP + H₂O →→→→DUMP + NH3. Cofactor: CDP+ [2H] → dCDP+ H₂O. Cofactor: Effect: Effect: Effect: Reset Help1. Which expression below shows hemoglobin bound to a proton a. HbH+ b. HbO₂H+ c. HbBPG d. None of the above 2. Does hemoglobin, from you answer choice in question number 1, exist in the high affinity state or the low affinity state? What is hemoglobin's heterotropic ligand, based upon your answer choice in question number 1? 3. What type of the four catalysis reactions referenced in class is depicted by Asp52 in the diagram to the right? Mur2Ae H AcN H AcÑ Glu Foo Asp82 (1) HH CH₂OH Glu 00 ⁰0 Asp2 H НО OH Н GlcNAc CH₂OH H OH H H NAC First product2. The two diagrams to the right il- lustrate plots of steady-state ki- netic studies to characterize the in- teraction of heart muscle phos- phofructokinase-1 with a non-phy- siological, synthetic substrate fruc- tose-6-sulfate. Because the kcat is smaller than that for the natural 5 0.8- NH = 3.5 0.6- 0.4- 0.2- 10 μΜ 20 μΜ 48 µM substrate, higher enzyme concen- trations could be used. The results show the influence of increasing 2 0.2- 0.4- concentrations of ATP on the initial -0.6- > velocity of the enzyme catalyzed reaction in the presence of no -0.8F 4 12 20 28 36 44 52 60 68 76 84 92 .2 2.0 2.4 ΑMP () , 10 μΜ AMP (+ ) 20 μΜ AMP (-), and 48 µM AMP (-). [AΤP] (μΜ) log[ATP] (µM) (a) . Write the reaction in words catalyzed by the enzyme for the alternative substrate, describe how ATP interacts with the enzyme in the case of no AMP (•). (b) ! with respect to the binding of AMP and ATP to the allosteric effector sites on the enzyme. Explain the physical significance of the displacement…
- What advantage is there to phosphoglycerate kinase having an open and closed protein configuration? It allows water to be trapped in the active site along with the substrate. O Changing of the configuration of the enzyme makes the reaction exergonic. The induced-fit mechanism maximizes accessibility of active site without sacrificing hydrophobic environment. It forces covalent binding of the substrate to the enzyme active site.In an enzymatic reaction, the equations that correspond with and without inhibition according to linewear-burk are calculated, these are with inh: 3.5x + 6 = y without inh: 3.5 + 10 = y. calculate alpha factorThe steps of the chymotrypsin mechanisms are listed below (1-7). Put the steps of chymotrypsin mechanism in the correct order. Figure representing chymotrypsin mechanism is given for reference. a.The portion (N-terminal end) of original substrate with the new C terminus diffuses away b. Substrate binding c. His 57 catalyzes removal of H from Ser 195 hydroxyl; Ser 195’s nucleophilic O attacks carbonyl C of substrate; tetrahedral intermediate is formed d. Water binding; water is deprotonated by His 57; resulting OH nucleophilically attacks carbonyl of remaining substrate; tetrahedral intermediate is formed e. His 57 donates H to N of…