Yeast gene expression lab using-galactosidase vectors that can be completed in

one 2 hour laboratory session.

Stephanie C Schroeder, PhDAssistant Professor, Dept. of Biological SciencesWebster University, St. Louis, MO

BackgroundBIOL 3050/3051 is a 16-week sophomore level Genetics Lecture and Laboratory course.

Both lecture and lab are taught by same instructor (me)

The lab meets 3 hours, once per week.

The lab classroom is shared between other classes, so is not accessible to students outside of the lab period.

Lab ScheduleTitle/Topic Model OrganismMeiotic Recombination SordariaMonohybrid/Dihybrid Corn/Drosophila/YeastSex Linked Traits DrosophilaGenetic Linkage DrosophilaTransformation Yeast(importance of CEN/ARS)Population Genetics DrosophilaDNA repair (UV mutagenesis) YeastMitochondrial DNA HumanGene expression Bacteria and Yeast

Background

This lab was designed to demonstrate two principles of gene expression in eukaryotes

1. Differences between positive and negative acting elements

2. Synergy with multiple promoter elements.

• Title: Eukaryotic Gene Expression

• Purpose: To perform b Galactosidase assays on yeast strains containing plasmids with different promoters to determine the activity of these promoter fragments.

Strains were generated in lab of Tony Weil, at Vanderbilt University.

Constructs for Part One

lacZA B C

T Wild Type

lacZA B C

T

lacZA B C

T

x Deletion ElementA

x Deletion Element B

Results from Original Paper Identifying Three Cis-Acting Elements

AA

B

C

Part Two: Synergy

Element C was subcloned into the XhoI site at -178

1 copy2 copies3 copies 4 copies

Synergy

Pre Lab Prep

Week before : (I have students make the Z buffer and ONPG)1. Streak out strains from glycerol stock onto Sc-ura plates2. Make Z buffer, ONPG (wrap in foil), Na2CO3

Day before lab: (I do)Inoculate 5 ml cultures of Sc-ura with strains

Morning of lab:1. Read Absorbances at 600 nm for overnight cultures2. Aliquot yeast into 1.5 ml tubes for students3. Aliquot SDS, Chloroform, Na2CO3

Protocol For Students (They work in pairs)

You have 8 tubes (labeled 1-8) which contain 0.2 mL each of the yeast strains listed in Table I.

At the beginning of your lab session, I will tell you the A600 for each strain, to enter into your Table I.

1. Vortex the cultures to mix the cells.. 2. Add 0.8 ml Z buffer to each tube. (P-1000 set to 080)3. Add 50 ul chloroform to each tube (P-200 set to 050)4. Add 20 ul 0.1% SDS to each tube (P-20 set to 200)5. Vortex each tube for 10" 6. Add 200 ul ONPG (1mg/ml in Z buffer +BME). Vortex the tubes to mix. 7. Time reaction at Room temperature (30 minutes from time added to tube 1) 8. Stop with 0.5ml 1M Na2CO3.

9. Pellet 5 minutes in eppendorf centrifuge10. Carefully transfer upper layer to a cuvette (chloroform will “etch” or dissolve the

cuvettes) 11. Read the Absorbance at 420 nm (blank with tube that says Blank))

Taken from Cold Spring Harbor Yeast Genetics Lab Handbook (1989)

Cell culture Absorbance at 600

Volume (mls) Time Absorbance at 420 Activity Units

1. No Promoter 0.2 30

2. Wild Type Promoter

0.2 30

3. Deletion DNA Element A

0.2 30

4. Deletion DNA Element B

0.2 30

5. CYC1 vector alone 0.2 30

6. CYC1 + 2 copies element C

0.2 30

7. CYC1 +3 copies element C

0.2 30

8. CYC1 + 4 copies element C

0.2 30

Table I:

Calculate activity using the following formula:

Activity Units= Absorbance at 420 nm x 1000 = Velocity/mL A600 x 30 minutes x 0.2 mL (200uL)

Write-up/ ConclusionsConstruct ONE graph (bar graph) with the Activity Units for each construct on the Y axis and the names of the promoter constructs on the X axis. Use this graph to answer the following questions.

Act

ivit

y

Typical results from student experiments

Questions:

1.In the equation to determine the -galactosidase activity for our samples, why do we divide by the Absorbance at 600 nm? 2.What fold difference is there in -galactosidase activity between sample 1 (empty vector) and sample 2 (full length promoter)?

3.Do you think the DNA sequence Element A is a positive acting element or a negative acting element based on the difference between sample 2 (full length) and sample 3 (deletion of Element A)? Explain your rationale.

4.Do you think the DNA sequence Element B is a positive acting element or a negative acting element based on the difference between sample 2 (full length) and sample 4 (deletion of Element B)? Explain your rationale.

5.Samples 5-8 are constructs in which I inserted sequences from the TBP promoter into a DIFFERENT promoter (CYC1) to see their effect in a heterologous system. Based on the differences between samples 6,7, and 8 (2, 3 and 4 copies of Element C) compared to sample 5 (CYC1 alone), is the sequence Element C a positive or negative acting element? Explain your rationale6.Based on your answers above, what do you think would happen if we put the sequences Element A from the TBP promoter into the CYC1 promoter? Based on the activity of the CYC1 promoter alone, do you think we could really measure a difference?

Assessment

• Classroom discussion (lecture and lab)• Students answers to questions• Ability to apply to lecture exam/cumulative

final