Laboratory Math Practice Problems

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 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 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

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 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

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

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 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

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 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

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 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 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 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 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 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

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

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

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

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