Serial dilution is a simple yet efficient technique to determine the number of cells or organisms in a concentrated sample. First, take a portion of the sample and does serial dilution on it. Repeat the steps until the cells can be observed under the microscope when the diluted sample was observed.
Proteins such as antigens and other biological materials can, under proper conditions, physically bind to the plastic material composing the wells of the ELISA plate. The coating procedure must be done carefully. If too little antigen is used, bare spots will permit antibody or other protein to stick, leading to a false-positive reaction. If too much antigen is used, the excess will be able to bind SLE antibody from patient sera but then will be washed away, creating a false-negative reaction.
The addition of antigen is the crucial first step in the chain of recognition events between antigen and antibody that will end with the formation of color from the enzyme bound to the second antibody. An ELISA may be subject to many errors. One is that the biological and chemical reagents used in ELISA can change with time.
Another is that the ELISA is not always conducted under appropriate conditions. To rule out such problems, two controls are used. One control should always produce a positive response if the reagents and conditions are correct. The second control should never produce a positive response.
If either control sample fails to react as expected, then the results for the patients' samples cannot be trusted and the assay must be repeated. Any serum from a patient that contains the antibody for SLE will recognize the antigen in the well and bind to it.
Each serum sample contains many different types of antibodies, but because they are so specific in how they react, usually no antibody will recognize the SLE antigen except the SLE antibody. Washing helps remove any antibody that did not react with the SLE antigen in the well.
When the fluid is removed from the well, antibody that has reacted with antigen remains attached to the well surface. Unreacted (unbound) antibody may also remain in the well in the small amount of fluid that is left behind. This unbound antibody must be removed, because the anti-human antibody added in the next step will recognize and react with any antibody remaining in the well, regardless of whether that antibody is specific for the SLE antigen. A reaction with non-SLE antibody will produce a false-positive result. The second antibody, unlike the first, does not recognize the SLE antigen. Instead, rabbit anti-human antibody reacts with human antibody.
SLE antibody is a human antibody that may be present in a well because it is being held by antigen. The second antibody (from rabbit) will therefore recognize this antibody and bind to it. If the well has not been washed thoroughly, other human antibody may still be there and will also react with the second antibody.
Reaction of a non-SLE human antibody with the second antibody will produce a false-positive result.
This book is licensed under a license. See the license for more details, but that basically means you can share this book as long as you credit the author (but see below), don't make money from it, and do make it available to everyone else under the same terms. This content was accessible as of December 29, 2012, and it was downloaded then by in an effort to preserve the availability of this book. Normally, the author and publisher would be credited here. However, the publisher has asked for the customary Creative Commons attribution to the original publisher, authors, title, and book URI to be removed. Additionally, per the publisher's request, their name has been removed in some passages.
More information is available on this project's. For more information on the source of this book, or why it is available for free, please see. You can browse or download additional books there. To download a.zip file containing this book to use offline, simply. Learning Objective.
Learn how to dilute and concentrate solutions. Often, a worker will need to change the concentration of a solution by changing the amount of solvent. The addition of solvent, which decreases the concentration of the solute in the solution.
Is the addition of solvent, which decreases the concentration of the solute in the solution. The removal of solvent, which increases the concentration of the solute in the solution.
Is the removal of solvent, which increases the concentration of the solute in the solution. (Do not confuse the two uses of the word concentration here!) In both dilution and concentration, the amount of solute stays the same. This gives us a way to calculate what the new solution volume must be for the desired concentration of solute. From the definition of molarity, molarity = moles of solute liters of solution we can solve for the number of moles of solute: moles of solute = (molarity)(liters of solution) A simpler way of writing this is to use M to represent molarity and V to represent volume. So the equation becomes moles of solute = MV Because this quantity does not change before and after the change in concentration, the product MV must be the same before and after the concentration change.
Using numbers to represent the initial and final conditions, we have M 1 V 1 = M 2 V 2 as the The mathematical formula for calculating new concentrations or volumes when a solution is diluted or concentrated. The volumes must be expressed in the same units. Note that this equation gives only the initial and final conditions, not the amount of the change. The amount of change is determined by subtraction. Example 9 If 25.0 mL of a 2.19 M solution are diluted to 72.8 mL, what is the final concentration? Solution It does not matter which set of conditions is labeled 1 or 2, as long as the conditions are paired together properly.
Using the dilution equation, we have (2.19 M)(25.0 mL) = M 2(72.8 mL) Solving for the second concentration (noting that the milliliter units cancel), M 2 = 0.752 M The concentration of the solution has decreased. In going from 25.0 mL to 72.8 mL, 72.8 − 25.0 = 47.8 mL of solvent must be added. Test Yourself A 0.885 M solution of KBr whose initial volume is 76.5 mL has more water added until its concentration is 0.500 M. What is the new volume of the solution?
Answer 135.4 mL Concentrating solutions involves removing solvent. Usually this is done by evaporating or boiling, assuming that the heat of boiling does not affect the solute. The dilution equation is used in these circumstances as well. Chemistry Is Everywhere: Preparing IV Solutions In a hospital emergency room, a physician orders an intravenous (IV) delivery of 100 mL of 0.5% KCl for a patient suffering from hypokalemia (low potassium levels).
Does an aide run to a supply cabinet and take out an IV bag containing this concentration of KCl? It is more probable that the aide must make the proper solution from an IV bag of sterile solution and a more concentrated, sterile solution, called a stock solution, of KCl. The aide is expected to use a syringe to draw up some stock solution and inject it into the waiting IV bag and dilute it to the proper concentration. Thus the aide must perform a dilution calculation. © Thinkstock If the stock solution is 10.0% KCl and the final volume and concentration need to be 100 mL and 0.50%, respectively, then it is an easy calculation to determine how much stock solution to use: (10%) V 1 = (0.50%)(100 mL) V 1 = 5 mL Of course, the addition of the stock solution affects the total volume of the diluted solution, but the final concentration is likely close enough even for medical purposes.
Medical and pharmaceutical personnel are constantly dealing with dosages that require concentration measurements and dilutions. It is an important responsibility: calculating the wrong dose can be useless, harmful, or even fatal!