You are trying to come up with a drug to inhibit the activity of an enzyme  thought to have a role in liver disease. In the laboratory the enzyme was shown to have a Km of 1.0 x 10-6 M and Vmax of 0.1 micromoles/min.mg measured at room temperature. You developed a competitive inhibitor. In the presence of 5.0 x 10-5 M inhibitor, the apparent Km of the enzyme was found to be 1.5 x 10-5 M. What is the Ki of the inhibitor?

Given this, if you used 6g of vitamin Z powder to make 20 ml of solution, what is the % concentration of this solution? (I gave the image since I don't know if that info is needed to solve this question.)It also gives a follow-up, if you can help here too: You work in a lab as a summer student. One of your tasks is to make sure that there is enough cell culture medium containing antibiotics to grow bacteria.  One day you realize that there is only 5 ml of 10% Antibiotic stock solution in the freezer. You decide to use it all to prepare the working culture medium with 0.01% antibiotic. In the lab there is plenty of growth medium without antibiotics. (Note: dilution in medium is like dilution in water). You remember the equation to make dilutions of stock solutions. You usually use this formula to calculate the required volume of a stock solution, but you realize it can apply here as well, even though the unknown is the final volume. So, you make that dilution.  Given that each bacterial…

A purified protein sample was used in a reaction, resulting in an activity of 696.7 nmol min-1. The reaction volume was 145.0 µL and the final volume before loading the plate was 1,050 µL. The total reaction time was 4.25 min. The amount of protein used in the reaction was 4.270 µg. Calculate the specific activity of the sample (in nmol min-1 µg-1).

1 pt pt 9146 Bb 9146 Bb 1031 Class Etsy E Traps E Traps New Free Chat + ☆ 出口 keAssignment/takeCovalentActivity.do?locator-assignment-take [References] You do an enzyme kinetic experiment and calculate a Vmax of 118 μmol per minute. If each assay used 0.10 mL of an enzyme solution that had a concentration of 0.20 mg/mL, what would be the turnover number if the enzyme had a molecular weight of 128,000 g/mol? (Enter your answer to two significant figures.) turnover number = sec-1 D 1 pt Submit Answer Try Another Version 2 item attempts remaining estion stion 5 on 6 7 1pt 1 pt 1 pt 1pt 1pt 1pt 1 pt 1 pt D is the substrate concentration multiplied by the catalytic constant. KM is equivalent to the substrate concentration multiplied by the ratio of rate constants for the formation and dissociation of the enzyme-substrate complex. KM is equivalent to the substrate concentration. KM is equivalent to the substrate concentration divided by 2 A: KM is equivalent to the substrate concentration…