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26. You have purified fabulous enzyme that scers to decrcase anxicty during exams: Size exclusion chromatography shows that it has mass of 100 kD. You take some of ...

Question

26. You have purified fabulous enzyme that scers to decrcase anxicty during exams: Size exclusion chromatography shows that it has mass of 100 kD. You take some of the 100 kD mass protein from the size exclusion column, boil it in sodium dodecyl sulfate and perform SDS-PAGE You get two bands having molccular weights of 40 kD ard 30 kD with the 30 kD) band heing twice aS dark aS the 40 kD band.Which is the best explanation of these data?A prolease has cut your prolein into (wO fragments, restrict

26. You have purified fabulous enzyme that scers to decrcase anxicty during exams: Size exclusion chromatography shows that it has mass of 100 kD. You take some of the 100 kD mass protein from the size exclusion column, boil it in sodium dodecyl sulfate and perform SDS-PAGE You get two bands having molccular weights of 40 kD ard 30 kD with the 30 kD) band heing twice aS dark aS the 40 kD band. Which is the best explanation of these data? A prolease has cut your prolein into (wO fragments, restriction nuclease has cul our protein into [WO fragments our enzyme 1S heterodimer. It consists of two different subunits: Your enzyme is heterotrimer. It consists of three different subunits, two of which have the same mass. The data conflict and make no sense



Answers

You have isolated the proteins from two adjacent spots after two-dimensional polyacrylamide-gel electrophoresis and digested them with trypsin. When the masses of the peptides were measured by MALDI-TOF mass spectrometry, the peptides from the two proteins were found to be identical except for one (Figure $Q 8-2$ ). For this peptide, the mass-to-charge $(m / z)$ values differed by $80,$ a value that does not correspond to a difference in amino acid sequence. (For example, glutamic acid instead of valine at one position would give an $m / z$ difference of around $30 .$ ) Can you suggest a possible difference between the two peptides that might account for the observed $m / z$ difference?

So we have a solution off Volume 10 mill that has one milligram per mil off constitution of our enzyme. Um and so the formula for calculating the minimum molecular weight is as follows Minimum molecular weight people to total volume of solution multiplied by so talk. Conch of enzyme find that by total amount of heavy metal, I am required. So we substitute in the values. So the random, um, molecular weight does it gets one milligram milk. It was quite by 10 male, but by 9.342 times 10th, minus three Malamala, we got a volume off 29 1000 so the minimum molecular weight of the enzyme obtained 29,000. So we assume that the enzyme contains only tight writable S H groups per molecule. Are we issued only as one of those? So the minimum molecular way of the enzyme has obtained us. The heavy metal ions provided are sufficient to react with limited number off. So for hydroxyl groups that are needed to inactivate the whole enzyme

All right. So for this problem were given three different proteins and were asked to separate um, from being see. All right, so a is found in the matter Condra matrix and has a molecular weight of 60,000 Daltons. And it's ice Electric Point is at a pH of 6.5 protein B is embedded in the mitochondrial membrane, has a molecular weight 60,000 Daltons and has a nice electric point of 7.5 protein C is found in paroxysms and has a molecular weight $100,000. And I saw that your point at a Ph of seven point thought so my first step for isolating this would be to get the proteins on their own, not not attached or embedded into anything, so that we can make use of their different molecular weights and ice electric points. So first I'm gonna homogenize these with by putting me in a blender with some detergent. The blender will open up these organelles, allowing us to give the proteins, and the detergent will separate any embedded proteins from membranes. So a second I think I'd use size exclusion chromatography in size. Exclusion chromatography, of course, eludes bigger, larger sized molecules before smaller size molecules because the smaller size molecules get in the small pores in the stationary phased. So this will loot, see first so we can and elude A and B at the same time. Second, so we get rid of C separate. See this way. So now we're loved with a mixture of A and B. I think I'll do ion exchange chromatography at a pH of 6.5. Now what happens at a pH of 6.5 at a pH of 6.5 protein, A will have a have no net charge. It will be neutral, so it will get through the column first, at a pH of 6.5 protein be. We'll have extra protons because it's in a more acidic solution, so it will have an overall positive charge. So in this case, when I say ion exchange, I'd want to do a cat ion exchange chromatography so that so that our protein be we'll get caught up in the stationary phase that way will dilute protein. A first and protein be will still be attached to the second stationary phase or dilute later, you can do this process various ways. For instance, you could do it at a pH of 7.5, and which case be would come out first in a would be stuck in the stationary phase. Either way, you've separated it.

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.

We have a protein synthesis system that is synthesizing a protein designated. We also know that protein A has four trips and sensitive areas that are equally spaced in the protein and on digestion with trips in the following peptide is yielded a one a two, a three, a four and a five peptide. A one is the amino terminal peptide. Yeah, and a five is the car box all terminal peptide. Mm. Last time We know that our system requires four minutes to synthesize a complete protein At T equals zero, which is our time. We add all 20 amino acids that each are carrying a carbon 14 label. When t equals one minute, we isolate the intact protein a cleave it with trips in and then isolate the five peptides. The most heavily labeled peptide would be a five. Since it is car box. All terminal peptide. Yeah, when Teague was three minutes, the order of the labelling of peptides from greatest to least, would be a five, a four, a three and a two. And by conducting the experiment, we see that the direction of protein synthesis would be synthesis from the amino terminus to the car boxing terminus, which would be the ordering of the peptides from least to greatest. The


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