Complete the following reaction: P680 + light →_______→ O P680*; P680 + H+ O P680*; P680* + e O P680; P680* O P680*; P680*
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- At what time dark reaction takes place?In the experimental setup described in the attached figure (Figure 2.5 of the textbook), the middle panel shows two compartments with 10 mM KCI on the left (inside) and 1 mM KCL on the right (outside), separated by a membrane that is permeable to K+. What would initially happen if you replaced the KCI solutions with NaCl solutions (1 mM on the left or inside and 10 mM on the right or outside)? (A) Inside 1 mM KC Voltmeter Outside 1 mM KCI Permeable to K Nonet Bux of K (B) Initial conditions Initially Inside Outside 10 mM KC 1 mM KC Net flux of K from inside in outside → At equilibrium --58 mY Inside Outside 10 mM KCI 1 mM KC Flux of K from inside to outside balanced by opposing membrare potential Membrane potential (A) -116 [KL] 4 tended change O Na+ would move up its concentration gradient from the left (inside) to the right (outside) compartment. O Na+ would move down its concentration gradient from the right (outside) to the left (inside) compartment. O Na+ would not move because…A research group discovers a new version of happyase, which they call happyase*, that catalyzes the chemical reaction: HAPPY-SAD The researchers begin to characterize the enzyme. The researchers determined the kcat of the enzyme to be 452 s-1. In a separate experiment with [E;) at 1.5 nM and [HAPPY) at 38 µM, the researchers find that Vo is equal to 320 nM s-1. What is the measured Km of happyase* for its substrate HAPPY? Make sure to include units in the answer.
- An enzyme is found that catalyzes the reaction X ⇌ Y. Researchers find that the Km for the substrate X is 4 μM, and the kcat is 20 min−1.(a) In an experiment, [X] = 6 mM, and V0 = 480 nM min−1. What was the [Et] used in the experiment?(b) In another experiment, [Et] = 0.5 μM, and the measured V0 = 5 μM min−1. What was the [X] used in the experiment?(c) The compound Z is found to be a very strong competitive inhibitor of the enzyme, with an α of 10. In an experiment with the same [Et] as in (a), but a different [X], an amount of Z is added that reduces V0 to 240 nM min−1. What is the [X] in this experiment?(d) Based on the kinetic parameters given above, has this enzyme evolved to achieve catalytic perfection? Explain your answer briefly, using the kinetic parameter(s) that define catalytic perfection.Light independent means that light is required to complete the reaction. True False ---------------------------------------------------------- Light dependent means that light is not required to complete the reaction. True FalseFrom data in Table 13-6, calculate the ΔG′° value for the following reactions:(a) Phosphocreatine + ADP → creatine + ATP(b) ATP + fructose → ADP + fructose 6-phosphate
- An enzyme is found that catalyzes the reaction X = Y. Researchers find that the Km for the substrate X is 4 µM, and the kcat is 20 min-1. (a) In an experiment, [X] = 6 mM, and Vo = 480 nM min¬1. What was the [E] used in the experiment? (b) In another experiment, [E] = 0.5 µM, and the measured Vo = 5 µM min1. What was the [X] used in the experiment? (c) The compound Z is found to be a very strong competitive inhibitor of the enzyme, with an a of 10. In an experiment with the same [E] as in (a), but a different [X], an amount of Z is added that reduces Vo to 240 nM min. What is the [X] in this experiment?In an uncatalyzed reaction, KF(uncatalyzed) = 2.5 X 10-8/s and Keq is 5 X 10². For the same reaction in the presence of an enzyme, KF(catalyzed) = 2.5 X 10²/s. Remember that Keq = KF/KR = k₁/k-1. (a) What is the rate enhancement of the catalyzed reaction in the forward direction? (b) What is the value for KR(uncatalyzed)?Which of the following is TRUE under the following conditions: the enzyme concentration is 2.5 nM, substrate concentration is 75 nM, the KM = 150 nM, and the Vmax = 20 nmol/min a) The rate of the reaction is 20 nmol/min! b) The rate of the reaction is between 10 nmol/min and 20 nmol/min. c) The rate of the reaction is 10 nmol/min. d) The rate of the reaction is below 10 nmol/min. e) The rate cannot be determined from the above information.
- Show how to bring about the following chemical transformation. This transformation might require one or several steps. CH3 O O CH3 OH CH3 OH CH3 OH K CH3 OH CH3 OH H3PO4 heat H3PO4 heat H₂O/H₂SO4 H3PO4 heat CH3 CH3 H₂O/H₂SO4 H₂ Pt CH3 H3PO4 heat CH₂ H₂O/H₂SO4 CH3 CH3 CH3 CH3MAP= CO x TPR / 80 I don't understand the value of TPR how do I get this number?A research group discovers a new version of happyase, that catalyzes the chemical reaction HAPPY = SAD. The reseachers begin to characterize the enzyme. In the first experiment, with [E] at 4 nM, they find that the Vmax is 1.6 µM/s. Based on this experiment, what is the kcat for happyase? 1/s
