Most plasmid isolation kits (see LAB5) are purchased so each buffer that you use
ID: 1075526 • Letter: M
Question
Most plasmid isolation kits (see LAB5) are purchased so each buffer that you use (i.e. Resuspension Buffer, Lysis Buffer and Neutralization Buffer) has been prepared for you to save time. However, most of us spent the first few days of our undergraduate lab careers making our own buffers because it was cheaper than using a kit and most microbiology professors are "old school." Fortunately, you won't have to make these solutions from scratch but you should know what goes into each buffer. Here you have some recipes for the three main buffers that you will be using during the plasmid isolation lab later in the quarter. Let us know what we need in order to make a 250 mL working stock for each buffer. And just so you know, while the Neutralization Buffers and Resuspension Buffers have to be kept at 4°C (in order to preserve acetate and RNaseA activity, respectively), the Lysis Buffers need to be kept at RT or else the Sodium Dodecyl Sulfate (SDS) will come out of solution and make your molecular life quite difficult. Resuspension Buffer 1 M Tris-HCI, pH 8 (MW 157.6) 0.5 M EDTA (MW 372.2) Glucose (MW 180.2) RNase A Final Concentration 25 mM 10 mM 50 mM 100 g/mL Amount Needed (250 mLs Lysis Buffer 1 M NaOH (MW 40) SDS (MW 288.4) Final Concentration 200 mM 196 (w/v) Amount Needed (250 mLs Neutralization Buffer Potassium Acetate (MW 98.1) 17.4 M Glacial Acetic Acid (MW 60.1) Final Concentration 3 M 11% (v/v) Amount Needed (250 mLs)Explanation / Answer
Ans. 1. Resuspension Buffer:
# Tris-HCl
Using C1V1 (Stock solution) = C2V2 (Finally diluted solution)
Or, 1M x V1 = 25 mM x 250.0 mL
Or, 1M x V1 = 0.025 M x 250.0 mL ; [1 M = 1000 mM]
Or, V1 = (0.025 M x 250.0 mL) / 1M
Hence, V1 = 6.25 mL
Therefore, required volume of 1M Tris-HCl = 6.25 mL
# 0.5 M EDTA
Using C1V1 (Stock solution) = C2V2 (Finally diluted solution)
Or, 0.5M x V1 = 0.010 M x 250.0 mL
Or, V1 = (0.010 M x 250.0 mL) / 0.5 M
Hence, V1 = 5.0 mL
# Glucose
Given, Molarity of glucose = 50.0 mM = 0.050 M
Volume of solution = 250.0 mL = 0.250 L
Required moles of glucose = Final molarity x Final volume of solution in liters
= 0.050 M x 0.250 L
= 0.0125 mol
Required mass of glucose = Required moles x Molar mass
= 0.0125 mol x (180.2 g/ mol)
= 2.2525 g
# RNase A
Required amount = Required concertation x Volume of solution
= (100 ug/ mL) x 250.0 mL
= 25000 ug
= 25.0 mg
= 0.025 g
#2. Lysis Buffer:
# NaOH
Using C1V1 (Stock solution) = C2V2 (Finally diluted solution)
Or, 1.0 M x V1 = 0.200 M x 250.0 mL
Or, V1 = (0.200 M x 250.0 mL) / 1.0 M
Hence, V1 = 50.0 mL
# SDS
Required amount = Required concertation x Volume of solution
= (1 g / 100 mL) x 250.0 mL
= 2.5 g
# Neutralization Buffer:
# Potassium Acetate:
Required moles of K-acetate = Final molarity x Final volume of solution in liters
= 3.0 M x 0.250 L
= 0.750 mol
Required mass of K-acetate = Required moles x Molar mass
= 0.750 mol x (98.1 g/ mol)
= 73.575 g
# Glacial acetic acid
Required amount = Required concertation x Volume of solution
= (11.0 mL / 100 mL) x 250.0 mL
= 27.5 mL
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