Research a medication not discussed in class that utilizes enzyme inhibition for its mechanism of action. You should briefly comment on how the medicine’s regulation of the target enzyme has a therapeutic effect. Describe the type of inhibition this medication induces for its effect, not by simply listing the category, but explaining the interaction between the medication and its target enzyme. How does the presence of this medication influence the Km and Vmax of the enzyme (i. e., explain, in your own words, how the shifts observed for this kind of inhibitor make sense on the Lineweaver-Burke plot)? Based on the chemical configuration/functional groups of the chosen medicine, as well as the enzyme’s usual substrate or allosteric regulators, what can you infer about the amino acids at either the active or allosteric site of the target enzyme? This point does not have to be supported by research evidence but draw your own logical conclusions about these complementary interactions; you can be creative but be logical.

Earth's rotation cause the global convention current between the equator and the poles

Ithink is a because if say increased in electronegativity

Answerthis solution will not form a precipitateexplanationsolubility equilbrium:         pbf₂(s)  ⇄ pb²⁺(aq) + 2 f⁻(aq)the solubility product expression is    (we exclude solids from the equilibrium expression).now suppose we have the trial ion product q, which we can define as    then, we can interpret  q as the product of the concentration of the ions that we have and ksp as the product of the ion concentrations that are needed to form a saturated solution.if q > ksp, then that implies that there will be a surplus of ions that exceed the amount that is required to form a saturated solution. a precipitate will then form (the precipitate continually  forms until the surplus ions are removed and q is decreased to ksp's value).if q = ksp, the saturated solution that forms is barely saturated and the precipitate will be minimal.if q < ksp, then the ion concentration will not be enough to form a saturated solution (so no precipitate).using our information, we calculate the trial ion product:     since q = 2.7  × 10⁻⁸ < ksp =  4  ×  10⁻⁸, a precipitate will not form.