a student dissolved plasmid DNA, extracted from bacteria, in 50µl buffer.  she diluted an aliquot 100x and obtained a final absorbance of 0.275 at a wavelength of 260nm and a final absorbance of 0.156 at a wavelength of 280nm. what is the total mass of plasmid dna extracted?
210338
did you also prepare a standard solution? because the way i do it, is that i have a standard so i can compare the 2 absorbtions (same wavelength)...You know the equation..
121699
Plasmid DNA is in bacteria...they're in the shape of these circles called plasmids.\n\nChromosomal DNA is in other living creatures, and it's in the shape of chromosomes.
85809
Well, I'll start you off.\n\nAbsorbances are additive, so you subtract the blank from the measurement (.214 - .009 = .205)\n\nSo the absorbance is .205. Now you use beer's law, which is:\n\nA = ebc\n\ne is molar absorptivity (in L/mol cm)\nb is path length (cm)\nc is concentration. \n\nYou use that equation to figure out the concentration of the complex.\n\nFrom there, you can figure out the concentration of iodine based on the amount of starch used. Unfortunately, you've not provided enough information to go beyond here.
344609
You can find practical information from Qiagen website if you use their kits which are quite popular these days.  \n\nIt depends on which methods you use for isolation.  The alkali lysis (see link below) was commonly used to rupture bacteria.  Plasmid DNA (smaller in size and more soluble) is separated from baterial genomic DNA (larger and less soluble).  The plasmid DNA is then separate from proteins by phenol extraction where the proteins are removed to the phenol phase while plasmid DNA stays in the aquous phase.  The plasmid DNA is then precipated with ethanol to complete the purification.\n\nQiagen's kit and others overcome the use of phenol (which is toxic and smelly) by using a solid phase that bind DNA to seperate the contaminating proteins.  The DNA is later eluted from the solid phase to complete the purification.\n\nHere are additional links:\nhttp://biotech.biology.arizona.edu/labs/DNA_isolation_plasmid.html\nThe concept of alkali lysis method of plasmid DNA isolation is shown here:\nhttp://a32.lehman.cuny.edu/molbio_course/plasmid_protocol.htm
490557
http://www.microbesonline.org/operons/gnc205922.html
311420
Presuming that the base is monobasic.\n\na) with 25 ml added, you are at the equivalence point.  The pH is 7.\nb) after adding 25.01 ml, you will have 0.01 ml of base that did not get reacted.  The molarity of the base in solution will then be 0.1 * (0.01/50.01) = 2.0 x 10^-5.  Your pOH is 4.70, pH would be 9.30\nc) Now, it's 0.1 ml of unreacted base.  Molarity is 0.1 * (0.1/50.1) = 2.0 x 10^-4.  Your pOH is 3.70, pH is 10.30.\nd) Now 1 ml of unreacted base.  Molarity is 0.1 * (1/51) = 2.0 x 10^-3.  pOH is 2.70, pH is 11.30.\ne) Now 25 ml of unreacted base.  Molarity is 0.1 * (25/75) = 3.33 x 10^-2.  pOH is 1.48, pH is 12.52.\n\nIn each case, you have to figure out the amount of unreacted titrant and recalculate the molarity as if you were simply diluting it, using McVc = MdVd, where the volume is now the total volume of the solution (the unreacted amount plus the 25 ml of acid and the 25 ml of base that have neutralized each other).
425847
From the reaction you had heat produced\n\nQ=m*Cp*ΔT\n\nfor water Cp= 1 calorie/(gram*Celsius)\n\nYou have a total of 50+50 =100mL and since d=1.00 g/mL you have 100 g\n\nso Q =100*1*(31.43-25.32) = 611 cal\n\nyou mixed mole HCl= C*V= 1*0.05= 0.05 with\nmole NaOH =C*V =1.1*0.05 =0.055 which means that, since the stoichiometry of the reaction is 1:1, HCl is the limiting reagent\n\n0.05 mole HCl give 611 cal\n1 mole HCl gives x\nx=611/0.05= 12220 cal/mole =12.22 Kcal/mole
129428
The wavelength of absorbance for a peptide bond is 190-230nm.\n\nRetrieved from "http://en.wikipedia.org/wiki/Peptide_bond"\n\nI couldn't find any info on the other compounds - sorry!
6280
The pH of DMSO should be around 7.  I would dissolve the TMB at a very high concentration (100-1000 times the concentration you would like as the final) in DMSO at pH 7 and then add just a small amount of this concentrate to your sample at pH4.5.  Otherwise, you could dissolve the TMB in DMSO at pH 7 and then titrate the pH of the TMB/DMSO solution down to 4.5 using HCl.\n\nAnother possibility would be to try adding a lower concentration of TMB to your sample.  A protocol for TMB usage in ELISA can be found on the Molecular Probes website (https://www.invitrogen.com/content/sfs/manuals/00-2023.pdf)
449778
Using the Michaelis-Menten equation.\n\nMore details on teh attached link\n\nhttp://www.le.ac.uk/by/teach/biochemweb/tutorials/michment2print.html\n\nGood luck
171163
Because they are easy to maintain, replicate quickly and can tolerate foreign DNA in the form of a plasmid. Roughly speaking, the process is:\n\na) Make plasmid from vector + insert \nb) Multiply plasmid using bacteria (you need quantities of the plasmid!)\nc) Isolate plasmid from bacteria using special protocol\nd) Use "purified" plasmid with eukaryotic cells or whatever.\n\nIf you skip (b) and (c) the quantity of the plasmid will be very small.
74172
The key problem in bacterial spores is lysis. You can use sonication (see 1st PubMed reference in Analytic Chemistry), followed by a standard DNA extraction protocol, which any molecular biology laboratory should be able to accomplish. (see 2nd website link)
406767
This is a tricky question. Perhaps reading: http://www.newton.dep.anl.gov/askasci/chem03/chem03067.htm will help.
351082
i hope i rember this from 10th grade correctly:\n\nhousehold bleach is 12% so its 0.12 M right?\n\nM( molarity) = mol/L \n\n.12 = mol/4\n\nmol = .48,  \n\nnow you get the molecular mass of NaOCl:\n\n Na: 1 x 22.99\n O: 1 x 16\n Cl: 1 x 35.45\n--------------------\n            74.44 g\n\n74.44 grams of NaOCl in a mole of that\n\nsince it was .48 mol, you times that to the molecular mass:\n\n.48 x 74.44 = 35.7312g of NaOCl in 4 liters of household bleach,\n\n------------------------------------------------------------------------------\n\npool chlorine: same procedure:\n\n30% = 0.3 M\n\n0.3 = mol/ 4 \n\nmol = 1.2 \n\n74.44 x 1.2 = 89.328 g of NaOCl in 4l of pool Chlorine,\n----------------------------------------------------------------------------\nI hope i rembered correctly, i did this last year though. This would be true if we assume that it is only an NaOCl solution.
616474
Everything is here at this link:\nhttp://genchem.chem.wisc.edu/sstutorial/Text8/Tx84/tx84.html
524944
Plasmid vectors are small circular molecules of double stranded DNA derived from natural plasmids that occur in bacterial cells. A piece of DNA can be inserted into a plasmid if both the circular plasmid and the source of DNA have recognition sites for the same restriction endonuclease.\n\nThe plasmid and the foreign DNA are cut by this restriction endonuclease (EcoRI in this example) producing intermediates with sticky and complementary ends. Those two intermediates recombine by base-pairing and are linked by the action of DNA ligase. A new plasmid containing the foreign DNA as an insert is obtained. A few mismatches occur, producing an undesirable recombinant.\n\nhttp://www.accessexcellence.org/RC/VL/GG/inserting.html
195370
Uh, the rest of the question isn't appearing. But it's an equilibrium solubility problem. Write a reaction:\n\nMgF2 --> Mg+2 + 2F-\n\nKsp = 5.52e-9\n\nFill in the values on the ICE table (I don't know them, the numbers are missing), and set an equation up.\n\nKsp = [F-]^2 * [Mg+2]\n\nSolve with the numbers from the ICE table. Remember that those concentrations in the above equation are the equilibrated concentrations, not the initial.\n\n\nHope this helped.