BabcockHall
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Everything posted by BabcockHall
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Something to bear in mind is that formal charges and oxidation numbers are both tools to help us keep track of electrons and charges. They make different assumptions about which atom owns the electrons.
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Please give us your thoughts first, and perhaps we can help. I would start by ruling out some choices.
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This topic really belongs in homework help. In that section, you may expect help but not answers. My hint is that you first make sure you have a solid grasp of ionic bonds and intermolecular forces.
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A Question About Endorphines
BabcockHall replied to Carl Fredrik Ahl's topic in Anatomy, Physiology and Neuroscience
It might take me or someone else some time to find it, but I seem to recall reading somewhere that the connection between runner's high and endorphins has never been established. -
We are zooming, too. I am trying to learn how to use a Wacom Intuos, but it's the old dog/new tricks problem.
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HCl is soluble in many solvents. I used to buy a solution of at least 1 M in dioxane IIRC.
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It would not be a true solution; it would be heterogeneous. It could be that the problem is imperfect, as opposed to being a trick question. As someone who has written my fair share of homework problems, I can say that it easy to overlook difficulties like this. Nevertheless, the amount of HCl needed to neutralize it is still something that could be calculated.
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The calcium hydroxide should come into solution as one adds more HCl. I doubt that it is a trick question.
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What's wrong with Bradford?
BabcockHall replied to Fanipal's topic in Biochemistry and Molecular Biology
You wrote, "An optical density of samples with different protein content constituted from 0.0237 to 0.0933." Is this correct? One problem that I see is that if your unknown has a larger absorbance than 0.0933, you will be doing an extrapolation, not an interpolation. I generally filter the Bradford solution the same day as I use it. When I want the highest accuracy, I prepare a standard graph on the same day as the unknown.- 1 reply
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Thank you; that is very helpful. I think that I had been assuming implicitly that the two antibodies (soluble versus immobilized) recognized the same epitope. I now wonder whether the antibody at the Test position versus the gold-particle labeled antibody recognize the same or different epitopes. Perhaps the answer is different, depending on which test is used.
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link Good Morning Everyone, I have been studying various forensic test kits for body fluids, and a number of them from Abacus Diagnostics or Independent Forensics use lateral diffusion immunochromatography. In a sandwich format the antigen forms a complex with both the soluble, labeled antibody and the antibody immobilized onto a membrane at the test (T) position. I can see how this could easily work with a protein such as hemoglobin, which is a tetramer of two alpha and two beta subunits. It is less clear how this assay works with human salivary alpha-amylase, which is a monomer from what I can gather. In other words I can see how the sandwich forms when there are multiple epitopes, but it is less clear to me what is happening when there might be only one epitope.
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Enzyme Kinetics: Derive Michaelis-Menten Equation
BabcockHall replied to Biochemstudneedshelp's topic in Homework Help
Try expressing [ES] in terms of [E] and then proceeding with the differential equation. -
Enzyme Kinetics: Derive Michaelis-Menten Equation
BabcockHall replied to Biochemstudneedshelp's topic in Homework Help
There is an alternate derivation that assumes that E, S, and ES are in quasi-equilibrium. It is actually an easier derivation than steady-state. -
You might look at the book Criminalistics by Richard Saferstein.
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SDS PAGE denatures proteins, and not every protein that is denatured can be renatured. If you need active enzymes and you want to separate by molecular weight, then gel filtration might be an appropriate technique. However, it is often the case that obtaining a pure protein requires more than one method of separation.
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My own experience is that the shape of the standard curve depends upon the identity of the protein. However I generally put little trust in absorbance values of greater than 1.0 in the Bradford assay, even using a research-grade instrument, let alone one that is not. The problem is that there is not as much excess Bradford reagent as one moves higher in protein concentration. A standard curve should be made, preferably with the same protein that is being assayed.
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Oguri K et al. (1995) Japanese J. Toxic. Environ. Health 41(4):274-279 were not able to isolate a cocaine-containing complex by silica gel chromatography. They suggested that the complex is too labile. However, they determined the stoichiometry between cocaine and cobalt to be 2:1 in one complex. They proposed a structure involving two bidentate cocaine ligands and two isothiocyanate (Figure 2 in their paper). My only reservation about their proposed structure is that the isothiocyanate may coordinate through nitrogen, not sulfur. I am still working on the acid-base aspects of this question.
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"An elucidation of the reaction mechanisms involved has been published [56] . lt assigns the formula (alkaloid H)2 [Co(SCN)4] to the relatively water-insoluble blue complexes formed in neutral-to-basic solutions, and the formula [Co(alkaloid)2] (SCN)2 to the more water-soluble, brownish-red to pink complexes formed in acid-to-neutral solutions. Solubilities in water and chloroform are given for complexes formed with 29 alkaloids and nine metal thiocyanates." Schlesinger Presumably (alkaloid H) is the protonated form of cocaine or other compounds. Experimentally, there are blue crystals initially formed in the reaction, then the solution turns pink upon addition of HCl and that addition of chloroform leads to a blue organic layer in the Scott test. What I don't understand is why the cationic species (the conjugate acid) is in the organic layer and the neutral species (conjugate base) is in the water layer. Yet all of the reference material (one textbook and a couple of articles) have stated some variation of the passage above. I am in the process of acquiring additional references, including (56) above. Does anyone have any thoughts?
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Thank you for some food for thought. It seems to me that a volatile solvent (assuming it is essentially gone before the disk is put into place) solves many of the problems that you mentioned, possibly all of them. Or am I mistaken? We are in the process of making additional compounds and testing the ones that we already made.
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Inasmuch as we are working in collaboration, we are probably limited to disk diffusion for now. Because the compounds are soluble in DMSO, I was going to try acetone next for one of them. That way the acetone will simply evaporate, leaving the compound behind. The other compound came into a solution of K2HPO4. If we did our calculations correctly, we created a phosphate buffer.
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I am collaborating with a microbiologist to test some compounds to see whether or not they possess antibacterial properties. These compounds are either neutral or bear a single negative charge near pH 7, and they have aromatic rings. The compounds we have tested so far are soluble in DMSO, but modestly soluble in ethanol (less than 10 mg/mL), and probably not soluble in unbuffered water. The assay begins with adding the compound to a disk that is placed in contact with bacteria. My question concerns solvents that would be appropriate to use to dissolve the compounds. I assume that one control is to try the solvent without the compound of interest. Are there any solvents that are known to be so toxic in this sort of assay that they are best avoided altogether? If two solvents had about the same polarity, is it better to choice the more volatile or less volatile solvent? Is there any other property that we should consider? So far I looked in a microbiology laboratory manual, but I did not find anything. Perhaps there is a good reference that someone could suggest. Thank you.