Week 1 Lab: Pipetting

Performing Serial Dilution

Dilution Practice 1

Dilution Practice 2

1. The stock concentration of a mystery substance (MS) is 5 M. If the molar mass of MS is 532 g/mol, what’s the concentration of the stock concentration in g/mL? To make your life easier, you can use one of many online calculators.

Answer 1:
The stock concentration of MS is 2.66 g/mL.
Steps:
1. Convert moles to grams. A 5 M solution contains 5 moles per liter: 5 mol/L * 532 g/mol = 2660 g/L
2. Convert grams per liter to grams per milliliter: 2660 g / 1000 mL = 2.66 g/mL

2. You will perform a serial dilution to get 100 uM of MS. Devise a plan to dilute a 5 M MS solution to 100 uM. How many dilution steps will we need? Which tubes should we use? Which pipettes?

Answer 2:
Three dilution steps are needed using P10 & P1000 pipettes with three 1.5 mL microcentrifuge tubes.
Steps:
1. Calculate the total dilution factor.
5 M = 5 mol/L. The target is 100 µM = 1 × 10^-4 mol/L. Total dilution factor = (5 mol/L) / (1 × 10^-4 mol/L) = 50,000.
2. Break this into three steps.
50,000 = 100 × 100 × 5, so we will do a 1:100 dilution, then another 1:100 dilution, then a 1:5 dilution.
3. Use three 1.5 mL microcentrifuge tubes (Tube 1, Tube 2, Tube 3).
• Tube 1: 1:100 dilution from the 5 M stock. Add 990 µL diluent + 10 µL of 5 M MS (P1000 + P10).
   Final concentration in Tube 1 = (5 M) / 100 = 0.05 M = 50 mM.
• Tube 2: 1:100 dilution from Tube 1. Add 990 µL diluent + 10 µL from Tube 1 (P1000 + P10).
   Final concentration in Tube 2 = (50 mM) / 100 = 0.5 mM = 500 µM.
• Tube 3: 1:5 dilution from Tube 2. Add 800 µL diluent + 200 µL from Tube 2 (both with P1000).
   Final concentration in Tube 3 = (500 µM) / 5 = 100 µM.<br

3. Fill out the following chart to prepare a final reaction with 60 uL reaction volume. Why did we make 100 uM MS if we actually need 40 uM MS? Why not prepare 40 uM in serial dilutions?

Answer 3 Part 1 -
Check with C₁V₁ = C₂V₂:

• Loading dye: (6X) · V₁ = (1X) · (60 µL) → V₁ = 10 µL
• MS: (100 µM) · V₁ = (40 µM) · (60 µL) → V₁ = (40/100) · 60 µL = 24 µL
• dH₂O: 60 µL total − 10 µL dye − 24 µL MS = 26 µL

Answer 3 Part 2 -
We made 100 µM MS as a working stock because:

1. Pipetting accuracy: It’s much easier and more accurate to pipette volumes like 10–25 µL from a 100 µM stock than to pipette tiny volumes from a much more dilute 40 µM stock.

2. Flexibility: A 100 µM stock can be used to make many different final concentrations (e.g., 10, 20, 40, 50 µM) simply by adjusting the volume added. A 40 µM stock would only be useful for reactions requiring exactly 40 µM.

3. Separation of tasks:
Preparing a 100 µM intermediate stock allows you to complete the serial dilution step once. After that, setting up reactions becomes a simple mixing step rather than repeating dilution calculations every time.

4. Cleaner math for the final mix: Using a 100 µM stock makes the final reaction calculation simple. Going from 100 µM to 40 µM in a 60 µL reaction is a single C₁V₁ = C₂V₂ step, and it uses pipetting volumes that are accurate and easy to measure.

Week 2 Lab: DNA Gel Art

Gel Art - Restriction Digests and Gel Electrophoresis

Protocol | Part 0: Designing your Gel Art
Protocol | Part 1a: Preparing a 1% agarose electrophoresis gel
gel protocals

Protocol | Part 1b: Restriction Digest
Protocol | Part 2: Gel Run
Protocol | Part 3: Imaging Your Results with a Transilluminator

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Week 3 Lab: Opentrons Art

Lab 3 - Opentron Art

design

Opentrons

Week 6 Lab: Restriction Enzyme Digest I - MiniPrep

Week 7 Lab: Restriction Enzyme Digest II