BabcockHall

@OP, What does the concept of saturation mean to you?

I recommend reading Motulsky and Christopoulos, p. 256 and p. 315. It is often the case that the response Y is graphed versus [agonist], where the latter is plotted on a logarithmic scale. If one wishes to set the Hill slope to unity, then one can fit to a three-parameter equation. Y = bottom + (top - bottom)/(1 + 10^log(EC50 - X)). If one wants to include the Hill slope as the fourth parameter, then the term (EC50 - X) should be multiplied by the Hill slope.The shape of this curve is sigmoidal, and its steepness is governed by the Hill slope.

This part of your comment provides some context. In order to answer the question, we may need to make the assumption that the concentrations in vivo are proportional to the dose. Let Y be the response and X be the concentration of agonist. One equation that can be used is Y = [(a - d)/(1 + (X/c)^b)] + d. The parameter a is lower asymptote or plateau; b is the slope factor; c is the concentration of agonist that provokes a response halfway between the baseline response and the maximum response; and d is the upper asymptote or plateau. The authors of the J. Med. Chem. paper may cited previous papers; therefore, I am not certain whether or not they used this equation. How does parameter c relate to your question?

First, we are not here to do your homework or your non-homework thinking for you. You are supposed to show some sort of effort. Second, you did a poor job presenting the question. Table 2 is cut off in your diagram. Third, why did you ask for someone with a knowledge of peptide pharmacokinetics? Pharmacokinetics could be defined as "the study of the time course of drug absorption, distribution, metabolism, and excretion". The question you are asking is not entirely unrelated to that field, but we don't have any information that bears on the pharmacodynamics of either compound (and in any case your opening post assumes identical pharmacokinetics). Fourth, the answer to your opening question could be found by thinking carefully about the reply you have already been given by CharonY. Your most recent answer suggests to me that you are not approaching this problem correctly. Consider the following hypothetical. Compound A has a value of EC50 of 1 nM. Compound B has a value of EC50 has a EC50 value of 10 nM. Suppose that both are present in vivo at 100 nM. What would you conclude?

That's just a restatement of what you said in your opening post. I expected more effort than you have put in so far, such as defining EC50, or saying that EC50 was measured by examining cAMP levels. I had to go to the original paper to learn that.

Calcium ions are one member of the class of chemical species called "second messengers," which respond to hormones, the first messengers. Calcium ions activate protein kinase C, among many other effects. My answer is intended to be very general.

Is this a homework problem?

My first reaction to what you are proposing is that it is extremely ambitious for a high school project. Although docking programs can give some indication of affinity, I am not aware of programs that provide association rate constants.

Because of the chelate effect, oxalate is often better at binding metals than, say acetate, and this holds true for Mg(II). A reliable source for dissociation constants is Martell and Smith's multivolume compendium.

What are your thoughts? Once we know those, then we may be able to help you.

Do you have any acid-labile groups present?

I would put a minus sign on it, to indicate that it is hydroxide ion. And we can compare the pKa values of the conjugate acids of any two leaving groups to determine which base is stronger.

All else held equal, stronger bases make poorer leaving groups than weaker bases. Strong bases are less able to stabilize negative charge than weaker bases. Which is the stronger base?

@OP, Which is a better leaving group, hydroxide ion or water, and why?

@OP, If you have a sodium as a countering, what does that suggest about the rest of the reagent?

What I would do is look into which amino acids are catabolized into 4-carbon TCA intermediates such as fumarate, oxaloacetate, or succinyl CoA versus which ones are catabolized into pyruvate. For example, isoleucine is catabolized into acetyl CoA and propionyl CoA, and the latter compound can be converted into succinyl CoA in several steps.

"Most likely" is an odd choice of terms. Amino acids are classified as glucogenic (glycogenic), ketogenic, or both. Once anything is tranformed into pyruvate, I don't see what would prevent its further conversion. Therefore, I don't know what you mean by "...cannot be transformed from oxaloacetate to pyruvate." Can you explain? Can alanine be transformed into pyruvate? Once you have the answer to that, you will be on your way. BTW your title is a bit confusing. If there is a deficiency in this enzyme or in its allosteric activator, gluconeogenesis is likely to be impaired. I am not sure that I see the connection between that and the question of how the amino acids are glycogenic. Are you attempting to distinguish between those that are catabolized to pyruvate versus those that are catabolized to oxaloacetate?

@OP, Why not write out the light-dependent and light-independent reactions for us? Then we can discuss them.

The various isotopes produce can produce a kinetic isotope effect that is useful in understanding the mechanisms of reactions, including enzyme-catalyzed reactions. The various isotopes have different spectroscopic properties that are especially evident in nuclear magnetic resonance spectroscopy.

That might be the product. The starting material is a triacylglycerol.

Do you want to measure the amount of each via protein quantitation, or do you want to measure the rate at which an enzyme catalyzes a particular reaction (often called its activity)? You may need to fractionate your sample with standard protein purification techniques.

Do you mean benzene and toluene? What is the reason for your question?

“Anion gap = [Na+] – ([Cl-] + [HCO3-]) “The normal value is about 12 meq/liter (range 8 to 16). The anion gap estimates the unmeasured anions in the plasma and is normally composed of polyanionic plasma proteins such as albumin (1 g/dL of serum protein possesses negative charge equivalent to 1.7 to 2.4 meq/liter), phosphate, sulfate, lactate, and other organic anions. “The anion gap is particularly useful in evaluating metabolic acidosis. Elevated values indicate that the acidosis is due to ingestion or generation of a fixed acid, stronger than H2CO3, at rates that exceed the rate at which the anion can be excreted from the body. Examples are diabetic ketoacidosis, in which acetoacetic and b-hydroxybutyric acids are generated; lactic acidosis; and renal failure, in which the rate of generation of strong acids is normal but the anions the acids, e.g., phosphate and sulfate, cannot be normally excreted. Ingestion of methanol, which generates formic acid, ingestion of ethylene glycol, which yields oxalic acid, and salicylate intoxication all produce high anion gap acidosis.” p. 188 in Principles of Biochemistry: Mammalian Biochemistry, 7th ed., Smith E, Hill, RL, Lehman IR, Lefkowitz, RJ, Handler P, White A, McGraw-Hill, 1983 Thank you. I found the passage above to be helpful. I have not yet looked into the delta-delta gap. I have been examining the reactions I wrote out in the pdf (I have not yet figured out how to format them for this thread). If we take acetyl CoA as the starting material and imagine that only beta-hydroxybutyrate is formed, then no net protons are consumed or produced. If we instead imagine that only acetoacetate is formed, then one proton is produced. This is equivalent to four protons per palmitoyl CoA, the precursor to acetyl CoA. On the other hand, seven protons are produced in the beta-oxidation of one palmitoyl CoA to eight acetyl CoA molecules.

Do you recognize the starting material or the alcohol?

When the Krebs' cycle is too slow to consume all of the acetyl CoA from beta-oxidation, some of the acetyl CoA is used to make ketone bodies. When the three ketone bodies, acetone, acetoacetate, and beta-hydroxybutyrate are in abnormally high concentrations, this is ketosis. Type I (and occasionally Type II) diabetics suffer from diabetic ketoacidosis, in which the blood pH lowers (there are three grades of severity). What is unclear to me is the cause of the lowered pH. My working hypothesis is that the acid may be generated when the fatty acyl esters are oxidized to acetyl CoA (beta-oxidation), because each round should produce one proton. When I looked at the pathway to produce ketone bodies, I became convinced that the textbook presentations do not always write balanced equations for this pathway. The attached file is modified from one textbook: I added protons to reactions 2 and 4. Ketone_body_pathway_v1.pdf