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Part li Cytotoxicity Tertest the effect of Drug A on breast cancer cells, pertformed cytotoxicity assay (WST1) was Drug _ is prepared in DMSO as concentrations of s...

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Part li Cytotoxicity Tertest the effect of Drug A on breast cancer cells, pertformed cytotoxicity assay (WST1) was Drug _ is prepared in DMSO as concentrations of solvent, and serial dilution is done to obtain different Dryne A (500, 300 , 200, 100 , 75_ added to each well_ and 25 pM) with stock bottle 5OuL of drug to be you started with has The absorbance of the different concentration of 1OOOpM results obtained_ wells was then read on the ELIZA machine and the followingConcentration SOOpM 300

Part li Cytotoxicity Tertest the effect of Drug A on breast cancer cells, pertformed cytotoxicity assay (WST1) was Drug _ is prepared in DMSO as concentrations of solvent, and serial dilution is done to obtain different Dryne A (500, 300 , 200, 100 , 75_ added to each well_ and 25 pM) with stock bottle 5OuL of drug to be you started with has The absorbance of the different concentration of 1OOOpM results obtained_ wells was then read on the ELIZA machine and the following Concentration SOOpM 300 pM 200 HM 100 pM 75 HM of Drug 50 pM (pM) 25 pM Absorbance 3.065847 2.908687 2.85327 2.686384 .894488 (OD) 1.310001 0.003333 Describe in details the protocol used to perform this experiment starting from plating of cells (without numbers) then through serial dilution (with calculations) up until the reading: What could be said about Drug _ A? Why? (Illustrate if possible) control, another plate with only DMSO (without Drug was prepared: What would be the shape of the curve (DMSO concentration vs: Absorbance) if DMSO has no effect on the breast cancer cells? What if it had killing effect? What if it had proliferative effect?



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A variation of the indicator-dilution method (see preceding problem) is used to measure total blood volume. A known amount of a tracer is injected into the bloodstream and disperses uniformly throughout the circulatory system. A blood sample is then withdrawn, the tracer concentration in the sample is measured, and the measured concentration [which equals (tracer injected)/(total blood volume) if no tracer is lost through blood vessel walls] is used to determine the total blood volume. In one such experiment, $0.60 \mathrm{cm}^{3}$ of a solution containing $5.00 \mathrm{mg} / \mathrm{L}$ of a dye is injected into an artery of a grown man. About 10 minutes later, after the tracer has had time to distribute itself uniformly throughout the bloodstream, a blood sample is withdrawn and placed in the sample chamber of a spectrophotometer. A beam of light passes through the chamber, and the spectrophotometer measures the intensity of the transmitted beam and displays the value of the solution absorbance (a quantity that increases with the amount of light absorbed by the sample). The value displayed is 0.18. A calibration curve of absorbance $A$ versus tracer concentration $C$ (micrograms dye/liter blood) is a straight line through the origin and the point $(A=0.9, C=3 \mu \mathrm{g} / \mathrm{L}) .$ Estimate the patient's total blood volume from these data.

Here. The solution for the answer is that we know that a beer slaw formula is as that equal toe PCL here, from the giving daughter concentration that s C physical to 0.0 05 a.m. Heartland that is L one m absorption. That is a equal to 0.372 Moeller absorption coefficient, That is e equals is we have to find discussion, Mark. I'm university em universe on. We have to find it. So the from the creation beers the equation that is equal toe e c l Here we put it, the values in the equation and sold for proficient at every on we get the value is 7440 AM Universe cm university. Now, therefore, the moral absorption coefficient that is equal is about 7440 a. M university, um, universe. Now calculation off the number off tyrosine residues is as follows that mhm 280 and m equal to number off throughout crept often. Residue s multiplied by 5500 plus number off tyrosine residue is multiplied by 14 90 Here by putting the values in the equation and solving we get the value is five point eight. Therefore, the number off the tyrosine residues the protein is five pointed or approx Earth six in the round off. So this is the solution in detail, step by step. Please go through this. Thank you.

This question. We're looking at cytochrome C release from mitochondria in response to UV light and in a tissue. This takes quite a long time. It appears to be slow, and we're asking, Is it because each cell releases said crime, see slowly? Or is it because individual cells release all of it at once, but that different cells are being triggered at different times? And to look at that, we were looking at two specific cells. The picture is in the textbook, and we're watching them, um, uncle time lapse, what to be pictures indicate as far as which model, um, we're likely to be seeing. So the most important thing about the data given in the text book is the timeframe we're looking at. So the time frame is only of a few minutes. And why is that important? It's important because each cell has released all of this at supremacy from the mitochondria in six minutes for a and eight minutes would be so that really favors the second hypothesis that individual cells acts very quickly but only triggered after a certain amount of time. Each so favours. Second hypothesis, this should trigger after different amounts of time, but then release site of room, see rapidly

The concentration off. So any time that is the effect of the mutation on enzyme activity. The maximum rate achieve decreases when there is increasing of struck concentration. So V max is decreased and chemise increased Here, the wild type of Imax's 13 Onda more minute universe and universe the wild type came 2 45. both here in this both TV Max on came on the crest Wild type man. Why did type be Max is what they want and when they're turned worse on getting voice on the my father while the type Kim is 104. No helm. You turned to be Max. Where is Yeah, B match is 8.3 and yeah, minute inverse. And she in verse mutant k m a 69. Um, I think he was really affect off by the new CEO. Eight of the wild type mutant. And time is shown in a given figure expression. While that is much more sensitive to melon etc. Okay, The part the most prominent effect off the mutation on the properties of enzyme are as follows the A mutant. This list about 65% off the activity on the Nativity off the world with respect profoundly dine. It is waste 50% with respect of parameter c away also Then you off million Elsa in inhibitory already percent on the world's type but does not have on effect on them You turn in time a part of the functional Rohloff glutamate three in Canada tiene aside. Transfer is I seem to play a more significant role in relation My Ma Elena CEO a then in cattle Isis.


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