How can precisely obtain a dillution with 1 bacteria per ml? There is standard inocullum with 1 bacteria /ml?
928716
1.What bacteria are we talking about (E.coli? Which strain?)\n2.There is no such standard solution for any type of bacteria.\n3.Why on earth would you need such a dilution? \n4.What type of facilites do you have access to (You can isolate single cells by FACs provided you have a proper marker and this expensive equipment)
185951
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)
827835
Do it in a laminar flow hood. Follow aseptic techniques.\nDepending on your use, you may suspend your bacterial culture in an appropriate medium.\n\nIf you dont have a laminar flow hood, do it the traditional way...\nwash your hands, then put alcohol.\nwork on a clean table - clean it with alcohol.\nlight an alcohol lamp in the area to clean air.\nopen your culture tube near the light.\ntransfer aseptically the bacterial culture to cuvette\n  - do it near light to ensure no contaminants from air\ncover the cuvetter near the light.
265173
You are correct using the formula:\n\nC1*V1 = C2 * V2\n\nwhere C is the concentration and V is the volume.\n\n\nNow, you know that you want a concentration of 0.39M solution of NaCl in a 100 ml sample.\n\nSo, the equation should read:\n\n0.82 * V1 = 0.39 * .1 L\n\n0.82*V1 = 0.039\n\nV1 = .047L\n\nso, take 47 ml of the .82M NaCl and dilute it up to 100 mL.  That will give you the 0.39 M NaCl.
1186593
Mix with a 0% solution in the relation 33:2\n\nSo use 2 units of your solution (like 2 liters) and 33 units of the 0% solution (like 33 liters).\n\nThat's all.
1081509
A cubic centimeter (cc) is always equal to a millileter, nomater what material you are measuring. But a cc or mL or pure water at STP (standard temperature and pressure) is equal to one gram, which is what I think Mac C was hinting at. It was all devised this way when they made the metric system.
740769
USE DIFFERENT STAINING FOR BACTERIA EX. GRAM STAINING. LOOK UP DIFFERENT BACTERIA STAINING
746766
You could do serial dilutions and plate from the dilutions and then extract one of the plugs from the plate and let it sit in some media with a bacterial culture and incubate till it clears. Then put the culture in tube with chloroform and place on ice for 15 mins. Centrifuge it for 20 min at 25000 rpm and remove the supernatant and that will give you a bacteriophage stock.
349946
Intracellular K+ should be about 140 mM (140 mEq/L)\nExtracellular K+ should be about 4 mM (4 mEq/L)\n\nTherefore, the ratio of intracellular to extracellular is about 35:1
722185
From this patent\n\nwww.freepatentsonline.com/6962719.html\n\n\n\nThe total extract described in EXAMPLE 1 was loaded in a dialysis bag with molecular weight cut off 500 (Spectra/Por®, Cat. No. 131 057). Both ends of the bag were sealed with clamps. The bag was placed in a bucket contained deionized water, in which the ratio of the supernatant and deionized water was 1:10. The total extract was dialyzed at 4° C. with stirring thrice each for 20 hours. The solution remaining in dialysis bag was collected and concentrated under reduced pressure to give the extract.
1006885
I think you are leaving out the first step.\n\nRemember that 12.0 M means 12 moles of HCL per liter of aqueous solution.\n\nTo understand the equations so that you can use them correctly, first think that you will have to add some amount of water to some amount of the concentrated solution to make it the dilute solution.\n\nso, 12 mol/liter , assuming you had a liter of the solution, you keep the same number of moles and add water\n\nhow much water do you add to get to the weak, .6 mol/liter concentration\n\n12 mol/ (1+x) liters = .6 mol/liter\n\n12/(1+x)=.6\n12=.6(1+x)=.6+.6x\n11.4=.6x\n19 = x\n\nso if you had a liter of the solution at 12 M, you would have to add 19 liters of water and then you would have 20 liters of solution at the diluted .6 M\n\nbut you don't need that much of the final solution, so you use the same ratio of water addition, but only start with enough concentrated to end with 300 ml\n\nthe ratio is 19 to 1\n\nso 1/20 * 300 = 15   you need 15 ml of the conc. solution\n   19/20 * 300 = 285  you add 285 ml of water \n\nthe total is 300 ml of the .6 M solution\n\nnow look back at the equations and you will see where they came from and how to apply them properly
773889
Suspected trick question.  The volumes included are irrelevant.  You know the molecular weight (66000 Da) and you know how many micromoles you need total in each solution.  What does it matter if the one micromole is in 1 ml or 10K gallons?  You know it only contains one micromole.  Divide the molecular weight by 10^6 and you're done.  (or you have not stated the problem correctly)
816577
I generally follow the recommendations provided by our safety officer at my academic lab.\n\nIf your liquid cultures can afford the extra volume, you can just add a lot of deionized h2o, that should stop the growing (by lysing the cells).  Then toss in the bio-hazard bag, for the autoclave.  \n\nOr, if you have access to an autoclave, then autoclave the culture, and you can probably dump down the drain.\n\nCheck with your safety officer!!\n\nIt was fun thinking about your question.  Pass on the good Karma!
258981
The convertion is with reference to a certain medium...\n\nIf the medium is ordinary fresh WATER ...\n\n1 gram (gm.) = 1 millilitre (ml.)\n\n1,000 milligram (mg.) = 1 gm.\n\nTherefore:\n\nIt will take 1,000 mg. to produce 1 ml. of water.\nIt will take 1,000 gm. to produce 1 liter of water.
95494
well this aint an answer but wat did u do it on which bacteria? cuz thats where im lost idonno which one to do it on
921065
you know you have got 152g total. So 1.8g/ml means you have got 152/1.8 for volume.Now just use MW KOH. Molarity will be moles KOH per litre, molality will be moles KOH/kg of solvent and Normality in this case will be the same as molarity but written  with symbol N.\n\nAs an aside the density of this solution would not be anywhere near 1.8g/ml, are you sure you have this right?
741771
0.5M = 500mM\nv x c=v'xc'\nv x 500 = 50 x 1\n0.1mL of the 0.5M solution diluted to 50mL will be 1mM