Laboratory Math Practice Problems

Calculate and describe how to prepare 10mL of a 1% solution of beta mercaptoethanol from pure stock.

(100%)(V)=(1%)(10ml) V= 0.1mL of beta mercaptoethanol, QS to 10mL

Calculate and describe how to prepare 500µL of 0.2µg/mL DNase from a 1µg/mL stock of DNase.

(1µg/ml)(V)=(0.2µ/mL)(500 µL) V=100 µL of 1µg/ml DNase, add 400µL of water

Calculate and describe how to prepare 500 mL of 0.5X TAE buffer from a 20X stock.

(20X)(V)=(0.5X)(500mL) V=12.5mL of 20X TAE, QS to 500mL

Calculate and describe how to prepare 75 mL of 0.5% NaCl. You have NaCl powder available (FW 58.44).

0.5% is equal to 0.5g per 100mL Formula weight is unimportant in this problem Set up an equality equation and cross-multiply: 0.5g/100mL = ? g/75mL (0.5*75)/100 = ? ? = .375g So, dissolve 0.375g of NaCl powder, QS to 75 mL

Calculate and describe how to prepare 150mL of a 5mM solution of MgCl2 from dry MgCl2 (FW 95.21).

1M=95.21g/L 5mM=0.005M (95.21g/L)(0.005)= 0.47605g/L for 0.005M (0.47605 g/L)(0.150L)= 0.0714 g...this is too small to measure, so make a stock instead: To make 1M stock of MgCl2, dissolve 95.21g, QS to 1L Dilute (1M)(V)=(0.005)(150mL) V=0.75mL or 750µL of 1M MgCl2, QS to 150mL

Calculate and describe how to make 2L of 10% bleach solution.

2L * 0.1 = 0.2 L of bleach Add 0.2 L of bleach with 1.8L of water (or you can say "add 0.2L of bleach and QS to 2L with water")

Calculate and describe how to make 10mL of a 0.5mg/mL solution of BSA from 3mg/mL BSA.

3. Another C1V1=C2V2 problem, using mg/mL as concentration units (3mg/mL)(V1) = (0.5mg/mL)(10mL) V1 = 5/3 = 1.7 mL Add 1.7 mL of 3mg/mL BSA with 8.3 mL of water

Calculate and describe how to prepare 20mL of a 10µM solution of NaCl. You have a 0.5M stock of NaCl available in the lab.

6. I know immediately to make an intermediate stock since the difference between the stock and the final concentration is so great So I'll make a 10mM stock, and I'll make 100mL of it (these are arbitrary #'s that I think are a good middle ground). The concentrated stock is 0.5M or 500 mM. (500mM)(V)=(10mM)(100mL) V=2mL of 0.5M NaCl, QS to 100mL to make 10mM stock Dilute again... (10mM)(V)=(0.010mM)(20mL) V= 0.02mL or 20µL of 10mM NaCl, QS to 20mL to get 10µM

Prepare 250 mL of hybridization solution. Hybridization Solution 50% formamide 5X SSC 100 µg/mL carrier DNA 50 µg/mL heparin 0.1% Tween You have the following available to you in the lab: Pure formamide 20X SSC 10 mg/mL carrier DNA 100 mg/mL heparin 10% Tween

9. Formamide (100%)(V)=(50%)(250mL) V = 125 mL formamide SSC (20X)(V)=(5X)(250mL) V = 62.5 mL of 20X SSC Carrier DNA (10mg/mL)(V) = (0.1mg/mL)(250mL) V = 2.5 mL of 10mg/mL carrier DNA Heparin (100mg/mL)(V)= (0.050mg/mL)(250mL) V = 0.125 mL of 100mg/mL heparin Tween (10%)(V) = (0.1%)(250mL) V = 2.5 mL of 10% Tween Combine and QS to 250 mL

Calculate and describe how you will prepare 200mL of 10X Buffer A. The recipe and available reagents are below: 10X Buffer A 1.37 M NaCl 1 M Na2HPO4 20 mM KH2PO4 0.01% Sodium Azide You have available in the lab: 5M NaCl Na2HPO4 powder (FW 140.95) 1M KH2PO4 10% Sodium Azide

Buffer problems are just a series of individual calculations, go step by step. The final volume for the whole thing is 200mL, so use that volume as V2 in all calculations NaCl: C1V1=C2V2 problem (5M)(V1)=(1.37M)(200mL) V1 = 274/5 = 54.8 mL of 5M NaCl Na2HPO4 : You are starting from powder here, so you'll be adding grams of powder rather than mL of a stock solution 1M = 140.95 g/L, you need 200mL or 0.2L (140.95 g/L)(0.2L) = 28.2 g of Na2HPO4 powder KH2PO4 : C1V1=C2V2, convert 20mM to 0.02M (1M)(V1) = (0. 02M)(200mL) V1 = 4 mL Sodium azide: % means we have solutions, can use C1V1=C2V2 (10%)(V1)=(0.01%)(200mL) V1 = 0.2mL of 10% sodium azide To prepare, put it all together: 54.8 mL of 5M NaCl 28.2 g of Na2HPO4 powder 4 mL of 1M KH2PO4 0.2 mL of 10% sodium azide QS to 200 mL with water NOTE: You must specify how many g or mL or whatever unit OF which stock. You can't just say "add 54.8 mL of NaCl" because that doesn't make sense. We have to know, 54.8 mL of what concentration of NaCl solution.

Calculate and describe how to make 300mL of 1XTE buffer from 50X TE stock.

C1V1=C2V2 (50X)(V1) = (1X)(300mL) V1 = 300/50 V1= 6 mL Add 6 mL of 50X TE buffer, QS (quantity sufficient) to 300mL with water

Calculate and describe how to prepare 30 mL of a 25 mM solution of KCl. You have a 1M stock of KCl.

C1V1=C2V2 problem, just watch your units Convert 25mM to 0.025M (conversely you can convert 1M to 1000mM, just keep units the same on both sides) (1M)(V1)=(0.025M)(30mL) V1= 0.75 mL To prepare, add 0.75 mL of 1M KCl, QS to 30 mL with water

From a 250µL solution of purified DNA, you dilute 5 µL of DNA into a total volume of 500µL of water. You make the following readings on the spectrophotometer: A260 = 0.300 A280 = 0.150 A. What is the concentration of the stock DNA? B. How many µL of your stock DNA solution would give you 10µg?

Figure out the dilution factor first: 500 total/5ul DNA = 100 (1 to 100 dilution), then use equation (A260)(50 ug/mL)(dilution factor) = DNA concentration in ug/mL (0.300)(50ug/mL)(100) = 1500 ug/mL is the concentration of the DNA stock To figure how many uL of stock gives 10ug, first convert concentration to ug/uL: (1500 ug/mL)/1000 = 1.5 ug/uL Then figure: 1.5 ug/uL times how many uL is equal to 10 ug? 10 ug/ (1.5ug/uL) = 6.67 uL So 6.67 uL of the DNA stock contains 10 ug total

Calculate and prepare 150µL of Buffer B. Buffer B 100mM Tris-HCl 500 mM KCl 40 mM MgCl2 500µg/mL proteinase K You have available in the lab: 1M Tris-HCl 5M KCl 100mM MgCl2 10 mg/mL proteinase K

Step by step, smaller volumes here, but same principles Tris-HCl: convert 100mM to 0.1M (1M)(V1) = (0.1M)(150uL) V1= 15uL KCl: convert 500 mM to 0.5M (5M)(V1) = (0.5M)(150uL) V1 = 15 uL MgCl2: can keep both units in mM (100mM)(V1)=(40mM)(150uL) V1= 60 uL Proteinase K: use C1V1=C2V2 with mg/mL as concentration units, convert 500ug/mL to 0.5mg/mL AND convert the total volume of 150 uL to 0.150 mL. Need to keep units the same for everything to cancel out properly (10mg/mL)(V1) = (0.5mg/mL)(0.150mL) V1 = 0.0075 mL OR 7.5 uL To prepare: Add 15 uL of 1M Tris-HCl Add 15 uL of 5M KCl Add 60 uL of 100mM MgCl2 Add 7.5 uL of 10mg/mL proteinase K Add 52.5 uL of water

Make 250mL of 10% solution of NaCl from NaCl powder.

To make a percent solution, remember % = g/100mL So, 10% means 10g/ 100mL, which is 0.1g/mL You need 250 ml of this, so (250mL)(0.1g/mL) = 25 g To prepare, dissolve 25 g of NaCl and QS to 250 mL with water

Calculate and describe how to prepare 300mL of a 5mM solution of Na2HPO4 from dry Na2HPO4 (MW 140.95)

You know the molecular weight is 140.95, so that means: 1M Na2HPO4 = 140.95 g/L You want 5mM, which is 0.005M, so multiply 140.95 by 0.005 That equals 7.0475 g/L for a 0.005M solution So how many grams do you need for 300 mL? (7.05g/L)(0.3L) = 0.211 g of Na2HPO4 powder To prepare, dissolve 0.211 g of Na2HPO4 powder and QS to 300mL with water

You have 200µL of DNA. You make a 1:100 dilution and read the absorbance of the dilution on the spectrophotometer and get the following absorbances: A260 = 0.175 A280 = 0.100 a. What is the concentration of the DNA solution? b. How much DNA is in 5µL of the DNA? c. What volume of DNA will contain exactly 6 µg?

a. (0.175)(50ug/mL)(100)= 875ug/mL b. 875ug/mL=?ug/0.005mL ?= 4.38 ug c. 875ug/mL=6ug/?mL ?=0.00686 mL or 6.86 µL

A solution of DNA is known to be 500 pmol/mL. What is the DNA concentration expressed as µM?

µM = µmol/L 500 pmol/mL x 1 nmole/ 1000pmol x 1 umole/1000nmole x 1000mL/1L =0.5 uM

Calculate and describe how to prepare 250mL of 50% isopropanol from a 95% isopropanol solution.

C1V1=C2V2 using percentages as concentration units (95%)(V1)=(50%)(250mL) V1 = 12500/95 = 131.6 mL To prepare, add 131.6 mL of 95% isopropanol, QS to 250mL with water

Prepare 1mL of 10X EcoR1 enzyme buffer. The recipe for 10X EcoR1 enzyme buffer is below: 10X EcoR1 Enzyme Buffer 5 mM NaCl 5 mM Tris-HCl 1 µM KCl 0.025 % Triton X-100 You have the following available to you in the lab: NaCl powder (FW 58.44) 5M Tris 200 mM KCl 10% Triton X-100

NaCL This is in powder form, so for such a small volume, a concentrated stock is needed Make 1M NaCL Dissolve 58.44g , QS to 1L Dilute (1M)(V)= (0.005M)(1mL) V = 0.005 mL or 5 µL of 1M NaCl Tris (5M)(V) = (0.005M)(1mL) V = 0.001 mL or 1 µL of 5M Tris KCl: This will be more than a 1000-fold dilution, so I'll make an extra intermediate stock Dilute 200mM KCL to 100mL of 200uM (0.2mM). The volume and stock chosen here are arbitrary (200mM)(V)=(0.2mM)(100mL) V= 0.1mL or 100uL of 200mM KCL, QS to 100mL for a 200uM stock Then dilute again to get to 1uM (200uM)(V)=(1uM)(1mL) V= 0.005 mL or 5 µL of 200uM KCL Triton X-100 (10%)(V) = (0.025%)(1mL) V = 0.0025mL or 2.5 µL of 10% Triton X-100