Dithiothreitol (DTT) is a compound with molecular weight of 154.3 Daltons, which means that one mole of DTT has a mass of 154.3 grams, and a 1.0 molar solution (abbreviated , with indicating moles per Liter) has of DTT dissolved in water to a total volume of . Let’s say I give you an aqueous solution that is (millimolar) DTT. Calculate the concentration of DTT in units of mass per volume.
Problem 2:
Let’s say you have on your lab bench the following four stock solutions:
Tris ()
glycerol
EDTA
DTT
Suppose you wish to prepare a buffer solution that has the following composition: Tris glycerol, EDTA, DTT (carefully read these units, and carefully read the stock solutions’ units also!) Prepare a table showing precisely what volume of each stock solution (and of water) you would dispense into a single beaker to
The Correct Answer and Explanation is:
Problem 1: Convert mM DTT to mg/mL
Given:
- Molecular weight of DTT = 154.3 g/mol
- Solution concentration = X mM (millimolar), where 1 mM = 0.001 mol/L
We want to convert this into mg/mL.
Let’s say the solution is 1.0 mM DTT for illustration (since you left the value blank). Then: 1.0 mM=0.001 mol/L\text{1.0 mM} = 0.001 \text{ mol/L} Mass of DTT per L=0.001 mol×154.3 g/mol=0.1543 g=154.3 mg\text{Mass of DTT per L} = 0.001 \text{ mol} \times 154.3 \text{ g/mol} = 0.1543 \text{ g} = 154.3 \text{ mg} Since this is per liter (1000 mL),concentration in mg/mL=154.3 mg1000 mL=0.1543 mg/mL\text{Since this is per liter (1000 mL)}, \text{concentration in mg/mL} = \frac{154.3 \text{ mg}}{1000 \text{ mL}} = 0.1543 \text{ mg/mL}
🔹 Final formula:
For C mM DTT solution: Concentration in mg/mL=C×154.31000=C×0.1543\text{Concentration in mg/mL} = C \times \frac{154.3}{1000} = C \times 0.1543
Problem 2: Buffer Preparation Table
Stock solutions:
| Compound | Stock Concentration |
|---|---|
| Tris | 1.0 M |
| Glycerol | 100% (v/v) |
| EDTA | 0.5 M |
| DTT | 1.0 M |
Desired final buffer composition:
- 100 mM Tris
- 10% glycerol
- 1 mM EDTA
- 1 mM DTT
- Final Volume: 100 mL
Step-by-step calculations:
1. Tris (100 mM from 1.0 M stock):
Use dilution formula C1V1=C2V2C_1V_1 = C_2V_2 (1.0 M)⋅V1=(0.1 M)⋅(0.1 L)⇒V1=0.01 L=10 mL(1.0 \text{ M}) \cdot V_1 = (0.1 \text{ M}) \cdot (0.1 \text{ L}) \Rightarrow V_1 = 0.01 \text{ L} = 10 \text{ mL}
2. Glycerol (10% from 100% stock): 10% of 100 mL=10 mL10\% \text{ of } 100 \text{ mL} = 10 \text{ mL}
3. EDTA (1 mM from 0.5 M stock): (0.5 M)⋅V1=(0.001 M)⋅(0.1 L)⇒V1=0.0002 L=0.2 mL(0.5 \text{ M}) \cdot V_1 = (0.001 \text{ M}) \cdot (0.1 \text{ L}) \Rightarrow V_1 = 0.0002 \text{ L} = 0.2 \text{ mL}
4. DTT (1 mM from 1.0 M stock): (1.0 M)⋅V1=(0.001 M)⋅(0.1 L)⇒V1=0.0001 L=0.1 mL(1.0 \text{ M}) \cdot V_1 = (0.001 \text{ M}) \cdot (0.1 \text{ L}) \Rightarrow V_1 = 0.0001 \text{ L} = 0.1 \text{ mL}
5. Water to reach 100 mL: Total so far: 10+10+0.2+0.1=20.3 mL⇒Water needed=100−20.3=79.7 mL\text{Total so far: } 10 + 10 + 0.2 + 0.1 = 20.3 \text{ mL} \Rightarrow \text{Water needed} = 100 – 20.3 = 79.7 \text{ mL}
📋 Final Table: Buffer Composition
| Reagent | Stock Conc. | Final Conc. | Volume to Add |
|---|---|---|---|
| Tris | 1.0 M | 100 mM | 10.0 mL |
| Glycerol | 100% | 10% | 10.0 mL |
| EDTA | 0.5 M | 1 mM | 0.2 mL |
| DTT | 1.0 M | 1 mM | 0.1 mL |
| Water | — | — | 79.7 mL |
| Total | — | — | 100.0 mL |
✍️ Explanation
Buffer preparation in the lab involves mixing known volumes of concentrated stock solutions to achieve desired final concentrations. Each component is diluted based on the equation C1V1=C2V2C_1V_1 = C_2V_2, which maintains molarity during dilution.
For example, to make 100 mL of buffer containing 100 mM Tris from a 1.0 M stock, we solve:
(1.0 M)⋅V=(0.1 M)⋅0.1 L(1.0 \text{ M}) \cdot V = (0.1 \text{ M}) \cdot 0.1 \text{ L} → V=10 mLV = 10 \text{ mL}.
This logic applies to all reagents. Glycerol, often expressed in percentage (v/v), simply requires adding the corresponding percentage volume (10% of 100 mL = 10 mL). For EDTA and DTT, both used at low concentrations (1 mM), very small volumes (0.2 mL and 0.1 mL respectively) are needed from concentrated stocks.
After calculating all component volumes, subtract their total from 100 mL to determine the amount of water required to reach the final desired volume. This ensures the buffer’s integrity and accuracy. This approach not only saves time but also ensures reproducibility across experiments, critical in biochemical and molecular biology workflows.
