BMS 3008: Integrated Biomedical Sciences - Lecture Notes

Introductory Session

Module Overview

This module covers molecular and cellular techniques, practical skills, experimental design, and data assessment.

Learning Objectives:

• Understanding basic molecular and cellular techniques and their applications.

• Basic calculations and principles of experimental design.

• Design of experiments involving patients (clinical trials).

• Data assessment.

Module Content

The module consists of 5 lectures: 4 from Luke Gaughan and 1 from Prof Julie Irving.

• The module is self-contained, requiring minimal extra reading.

• All lecture content can be examined.

• Lectures use case studies to introduce biological concepts and techniques.

Key Expectations:

• Understand how and why experimental approaches are used.

• Interpret resultant outputs.

• You will not be required to memorise names of proteins or drugs.

Assessment

The exam will consist of 7 questions with sub-questions, completed in 3 hours.

• No essay questions, but some questions may require longer answers.

• Elements of calculations, data interpretation, and reasoning.

• Practice questions will be available on Canvas.

• Unfortunately, there are no past papers.

Lecture Breakdown (Luke Gaughan)

Lecture 1

• Introduction to common techniques in biomedical research (Western blotting, ChIP).

• Revision of calculations: molarity, making solutions, and dilutions.

Lecture 2

• Case study: anti-cancer drugs in a lab setting.

• Additional techniques and lab-based calculations.

Lecture 3

• Detailed walkthrough of a lab project: investigating a protein in cell differentiation.

• Introduction to clinical trials: types, ethical considerations.

Lecture 4

• CRISPR technology: explanation and applications in health and disease.

Contact

• Module Leader: Dr Luke Gaughan (luke.gaughan@ncl.ac.uk)

• For module content queries, email directly rather than using Canvas.

Lecture 1: Introduction to Molecular and Cellular Biology Techniques

Techniques Covered

• Western blotting

• Chromatin immunoprecipitation (ChIP)

• Mammalian cell culture

Western Blotting

A protein detection technique using specific antibodies to detect a protein of interest from a cell lysate.

Main Steps:

1. SDS-polyacrylamide gel electrophoresis (SDS-PAGE): separates proteins by size.

2. Protein transfer onto a high-affinity protein-binding matrix (e.g., nitrocellulose).

3. Sequential incubations with primary and secondary antibodies to detect the target protein.

SDS = sodium dodecyl sulphate

SDS-PAGE

• Protein mixtures (cell lysate) are denatured in SDS solution by boiling.

• SDS gives each protein an overall negative charge.

• A polyacrylamide gel resolves the proteins according to size.

• Proteins migrate towards a positive electrode.

• Larger proteins are retarded more in the gel than smaller ones.

Protein Transfer

• Proteins are electro-transferred from the gel onto a membrane, such as nitrocellulose.

Western Blotting Procedure

1. Membrane blocking:

   • Nitrocellulose membrane has a high affinity for protein.

   • Milk proteins non-specifically bind and block aberrant antibody binding in subsequent steps.

2. Primary antibody incubation:

   • Antibody incubated with membrane.

   • The primary antibody binds to its target 'epitope' on the protein of interest.

3. Secondary antibody incubation:

   • Secondary antibodies bind to specific species of primary antibodies.

   • The secondary antibody is conjugated to horse-radish peroxidase enzyme (HRP).

4. Detection of bound Ab complex:

   • The horse-radish peroxidase conjugate emits light upon substrate addition.

Control for Western Analysis

• Proteins expressed at equal levels in cells (e.g., α-Tubulin and β-Actin) monitor loading between samples.

• Protein concentrations of each sample determined using a spectrophotometer (Bradford assay) prior to loading on a gel. Equal quantities of protein can be loaded onto a gel.

Example

• Investigating protein levels between sample arms of an experiment requires a control to confirm equal loading amounts.

Chromatin Immunoprecipitation (ChIP)

Basic Principles

• ChIP analyzes protein function and epigenetic changes during transcription.

• It can be applied to study cellular events that require chromatin processing, such as DNA replication and repair.

• Relies on enrichment of a target protein using an antibody in immunoprecipitation.

Transcription and Epigenetics

• Transcription: single-stranded RNA generated from a complementary DNA sequence.

• RNA polymerase II transcribes protein-coding genes into RNA.

• Transcription Factors facilitate the loading of RNA polymerase onto target gene promoter elements.

Chromatin Structure

• DNA is largely inaccessible due to packaging.

• Proteins facilitate access of RNA polymerase to DNA are recruited to promoter elements.

• The basic unit of chromatin is the nucleosome (147bp of DNA wrapped around a histone octamer).

• 2 copies of histones H2A, H2B, H3, and H4 make up the core protein complex.

• For transcription to take place, remodelling of nucleosomes must occur to permit access for RNA polymerase binding.

Histone Modification

• Modification of histone proteins regulates compaction of nucleosomal DNA.

• Enzymes modify histone proteins (e.g., methylation and acetylation).

• These post-translational modifications regulate transcriptional rate by permitting more or less RNA polymerase loading, respectively.

Transcriptional Effects:

• Histone acetylation activates.

• Histone methylation on histone H3-Lysine 4 activates.

• Histone methylation on histone H3-Lysine 9 represses.

• Histone acetyltransferases catalyse histone acetylation.

• Histone methyltransferases catalyse histone methylation.

ChIP Experiment

• To determine if the Estrogen Receptor (ER) is bound and activates transcription to the BMS-1 gene.

ChIP Procedure

1. Cross-link protein and DNA with formaldehyde.

2. Lyse cells and sonicate to fragment chromatin.

3. Add antibody-bound beads to bind protein-DNA complexes (immunoprecipitation).

4. Reverse cross-links between protein-DNA and Proteinase K treat samples to remove protein from mixture.

5. Purify DNA and set up quantitative PCR or DNA sequencing.

Key Steps Explained:

• Cross-linking: Formaldehyde introduces covalent cross-links between DNA and protein.

• Sonication: High-frequency sonic waves break phosphodiester bonds within DNA.

  • Ideal size between 200-500 base-pairs of DNA.

• Immunoprecipitation: Antibodies target a specific protein or control IgG are mixed with magnetic beads.

• Cross-link reversal and proteinase K treatment: Salt buffer and heat reverse covalent cross-links, Proteinase K degrades protein.

Control IgG Antibodies

• Are critical to confirm specific protein binding to target genomic loci.

• Distinguish specific antibody-protein interactions from non-specific protein immunoprecipitation.

PCR Analysis

• Primers specific to candidate gene promoter elements (e.g. BMS-1) are utilised in PCR.

• Enrichment of transcription factor binding is compared to control IgG.

Potential Outcomes

• More ER-DNA complexes in the +E2 arm due to ER-DNA interactions only occurring in the presence of estrogen. - Estradiol

• More products in QPCR compared to the untreated (-E2) arm.

Data Interpretation

• To determine recruitment of a protein of interest by comparing the experimental arms.

Further Analysis

• Assess activation status of the BMS-1 gene.

• Analyse histone acetylation and methylation levels.

ChIP Summary

• Key approach to understanding transcription factor function and influence of epigenetics on chromatin metabolism.

• Use of specific antibodies allows discriminate immunoprecipitation of cross-linked protein-DNA complexes that yields DNA which is selectively bound by protein of interest: - To identify if a protein of interest interacts with a particular genomic locus (e.g. ER binding to BMS1 gene). - To assess histone modifications at genomic loci to gauge their transcriptional competency.

• ChIP linked with total DNA sequencing provides a snapshot of global transcription factor/histone modification enrichment across the genome.

Part 2: Basic Laboratory Calculations

Importance

Calculations are required for:

• Preparation of solutions for Western blotting

• Calculating dilutions of stock solutions for drug treatments

Units

Units used in chemistry/biology will aid your calculations.

Molarity (M)

A mole of a substance = molecular weight in grams

Molarity is a measure of concentration of a solution:

A 1 Molar (M) solution is the molecular weight of the substance in grams dissolved in a litre of water

• e.g. Glycine Mol. Wt = 74; 1M Glycine solution = 74g in a litre of H2O

Molarity Units

• 1 M Glycine = 74 g/litre = 74 mg/ml = 74 μg/μl = 74 ng/nl

• $1 M = 1000 mM = 1000000 μM = 1000000000 nM$

Examples of Making and Diluting Solutions

Dilution Factor

$[Initial] / [Final] = Dilution Factor$

Where [] = Concentration

Example Calculation

We have a 1 M Tris stock solution. Calculate the volume of 1 M Tris to make a 5 mM Tris solution in total volume of 500 ml

1 M / 5 mM = 1,000 mM / 5 mM = 200

Final Volume / Initial Volume = Dilution Factor

500 ml / Initial Volume = 200

500 ml / 200 = 2.5 ml 1M Tris

Finally, add 497.5 ml $H_2O$

Second Example Calculation

Make a 2 M sucrose stock solution in 1 litre and use to make a 50 mM sub-stock in a total of 300 ml Mol Wt. Sucrose = 342

1 M = 342 g/l

2 M = 684 g/l

[Initial] / [Final] = Dilution Factor

2 M / 50 mM = 2,000 mM / 50 mM = 40

Final Volume / Initial Volume = Dilution Factor

300 ml / Initial Volume = 40

300 ml / 40 = 7.5 ml

Finally, add 292.5 ml $H_2O$

Dilution

Final

Percentages

Definition

• Percentage is calculated as weight/volume (W/V) for a substance.

• Percentage is calculated as volume/volume (V/V) for a solution.

• A 100% solution is denoted as 100 g (or 100 ml) in 100 ml solvent

Importance

• Ignore the molecular weight- it is not required for these calculations!

Examples

• 100 % sucrose solution = 100 g/ 100 ml

• 50 % sucrose solution = 50 g/ 100 ml

• 0.25 % sucrose solution = 0.25 g/ 100 ml

Stock Solution Calculation

Make up a stock 15 % solution of glycerol in 3 litres and then dilute to 0.03 % in 500 ml: 15% = 15 g/ 100 ml in 3 litres

[450 g Glycerol in 3 litres]

3000 ml / 100 ml = 30 X 15g = 450g

Dilution of stock:

[Initial] / [Final] = Dilution Factor

15 / 0.03 = 500

Initial Volume = 500 ml = 1 ml glycerol in 499 ml $H_2O$

Second Example Calculations

Make up a stock 25 % solution of sucrose in 4.5 litres and then dilute to 0.5 % in 400 ml:

25% = 25 g/ 100 ml in 4.5 litres

[1125 g sucrose in 4.5 litres ]

(4500 ml / 100 ml) = 45 X 25g = 1125g

Dilution of Stock

[Initial] / [Final] = Dilution Factor

25 / 0.5 = 50

Initial Volume = 400 ml / 50 = 8 ml sucrose in 392 ml $H_2O$

Contact information

Still unsure? : luke.gaughan@ncl.ac.uk