Structures in the lac-Operon:
purple: non lac-Operon region
red: Operator; binding site of the RNA polymerase
orange: Promoter; binding site of the lac repressor
yellow:
purple/blue u-shaped: Lac repressor, which the lactose binds to
green object: RNA polymerase; once the lactose binds to the lac repressor, it is free to move along the DNA so RNA can be made
yellow-red dots: lactose
yellow hay-colored string: mRNA
blue object: large subunit of ribosome
light grayish-blue object: small subunit of ribosome
orange object: B-galactosidase; converts lactose into glucose and galactose
yellow object: B-galactoside permease; transports lactose into the cell
Possible outline for the information menu
Information Menu:
Introduction
Background:
proteins:
Repressor: the lac-repressor protein represses expression of the lac-Operon until lactose attaches to it. Once the lactose attaches, it unbinds from the operator to allow RNA polymerase to bind to the promotor, and allow transcription of the structural genes.
Ribosomes: organelle consisting of two subunits, each containing RNA and protein. They translate mRNA into the amino acid sequence of B-galactoside permease, B-galactoside, and B-galactoside transacetylase proteins.
RNA Polymerase: an enzyme, when bound to the promotor region of the lac-operon, allows transcription of the structural genes.
Glucose: a carbon source; when present in the prokaryotic cell, inactivates the lac-Operon.
B-galactoside permease: an enzyme that brings lactose into the cell from the outside environment.
B-galactosidase: an enzyme with the assistance of water reduces lactose to glucose and galactose.
B-galactoside transacetylase: an enzyme whose function is unknown.
adenyl-cyclase: an enzyme that transforms ATP to cAMP. It produces higher levels of cAMP when glucose levels are low in the cell and vice versa.
Catabolite-activating protein (CAP): a protein that binds to cAMP to form the CAP-cAMP complex. The complex then binds to the CAP site and gives a higher affinity for the RNA polymerase to bind to the promotor.
Cyclic Adenosine Monophosphate (cAMP): This enzyme's production is dependent on the presence of glucose in the cell. If glucose is not present, then the molecule is created when ATP is converted to cAMP and binds to CAP to form the CAP-cAMP complex.
Components of Lac-Operon:
CAP binding site: a section of the promotor. In the absence of glucose, CAP-cAMP complex binds to the site and increases the affinity of RNA polymerase to bind to the promotor.
lac Z gene: a part of the lac-Operon structural region and specifically codes for the enzyme B-galactosidase.
lac Y gene: a part of the lac-Operon structural region and specifically codes for the enzyme B-galactoside permease.
lac A gene: a part of the lac-Operon structural region and specificaly codes for the enzyme B-galactoside transacetylase.
lac O gene: considered a regulatory gene of the lac-Operon and codes for the operator. The operator is the location where the repressor is bound until lactose attaches to the repressor. Then the repressor loses affinity for the operator and the lac-Operon is able to be expressed.
lac I gene: considered a regulatory gene of the lac-Operon and codes for the repressor protein. Expression of the lac-Operon is repressed until lactose attaches to the repressor.
lac P gene: a part of the lac-Operon regulatory region and codes for the promotor. The promotor is the binding site for the RNA polymerase and CAP-cAMP complex.
Definitions:
catabolite repression: term used when glucose is present in repressing the activity of the lac-operon.
positive control: terms used when glucose is absent and CAP binds to the CAP site, increasing the affinity of RNA polymerase to bind to the promotor.
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Questions need to be addressed when working on the storyline:
1. Overall goal and subgoals
2. Level of learning (basic to indepth)
3. Informational menu (level of information and referencing source)
4. Structure of subgoals to levels
5. How to run tests/conduct experiment and get information from experiments
main goal for students: to understand regulating gene expression in a prokaryotic cell
1. goal at Level 1: gene regulation of the lac operon / activtating-inhibiting of lac operon
2. goal at Level 2: catabolite regulation in the lac operon
3. goal at Level 3: testing knowledge given by giving mutant cells and able to distinguish which component of the lac operon or protein is mutant by the function.
beginning: will it be a diagram and a slow simulation of the lac-operon and it structures (similiar to the beginning of the ETC System. If so, this will make things consistent.)
1. a) Level 1: could make level one an easy level, where the student needs to guess which cell has glucose and lactose present and from the cell that only has lactose present. Students will have to run the experiment on one cell and also on the other in order to come to a conclusion.
answer dialog box: that is correct! When both glucose and lactose present, the cell will use glucose first before the lac-operon is turned on. Reason being is the glucose causes a decline of the level of cAMP in the cell. Therefore, CAP cannot form the CAP-cAMP complex needed for transcription. This is called catabolite repression resulting in efficient energy use. Return to lab guy for further instructions.
1. b) Level 1: Your goal is to identify the difference in the operation of the lac operon when glucose is present, glucose is absent, and both glucose and lactose is present. If you get stuck, refer to the information menu for help. Good luck!
Lab Guy: As you learned in the previous level, the lac-operon is not activated when glucose is present. The cell prefers glucose over the lactose. In the next level, you wil be looking at a mutant cell with only lactose present. Yet, one of the components of the lac-operon is broken and not functioning properly. You must identify the component and replace it so the lac-operon may resume work. Good luck!
Goal Box for Level 2: As you have seen from the beginning simulation and the previous level of how the lac-operon works in the presence of lactose only. In this level, there is a broken component of the lac operon. Your job is to identify the broken component and replace it so it may work again. While identifying, you will have to run many experiments and add parts (repressor, RNA polymerase, B-galactosidae, B-galactoside permease) to see which part is broken. After completing the experiments, replace the broken component and submit your conclusion.
2. Level 2: could make a moderate level. Students will need to identify the broken component. By doing this, they will run experiments by adding parts to see which one is brokent. While running the experiments, students will learn which part interacts with which component to cause what reaction. A little unsure of this part, if it can be broken down in this fashion due to the chain reaction of each sequence.
components that could be broken:
1) Repressor site: broken, doesn't allow the production of the lac repressor
2) Operator: broken, won't allow binding of the repressor
3) Promotor: broken, won't allow binding of the RNA polymerase
4) LacY: broken, doesn't produce the B-galactoside permease to transport lactose into the cell
5) LacZ: broken, doesn't produce the B-galactosidase to convert lactose into glucose and galactose
(NOTE: If this Level 2 is too similar to Level 3, it could be a broken part such as the RNA polymerase, repressor, B-galactosidase, and/or B-galactoside permease)
Answer Dialog box: Good job! You have answered correctly the promotor site was broken. The RNA polymerase binds to the promotor to produce mRNA. From this, it is able to transcribe three structural genes. Go to the lab guy to get the next task.
Lab Guy: You have just seen how the whole lac-operon system works and what is needed to keep the sequences of events flowing smoothly. Using this information and also from the first level, you will have to predict the effect of different mutants will have on the lac operon gene expression. You will want to carefully watch, because some of these will be in the presence of lactose and some will be without lactose.
3. Level 3: a harder level. Unsure of the moment on how to implement this.
1. I- gene: alter the repressor so it cannot bind to the Operator, allowing continuous transcription of the structural genes
without lactose.
2. Oc gene: repressor doesn't bind to the Operator and allows continuous transcription of the structural genes
3. Is gene: repressor remains bound to the Operator, therefore cannot transcribe mRNA even in the presence of lactose.
4. lac Z-: no glucose or galactose production from lactose
5. lac Y-: no induction because lactose will not be taken into the cell
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storyline2
main goal: to understand the regulation of gene expression in the lac-operon
1. goal at Level 1: controlling expression of lac-operon (activating/inhibiting of lac-operon)
2. goal at Level 2: to understand the importance of cAMP and CAP (?)
3. goal at Level 3: testing knowledge by giving mutant cells and identifying the mutant gene in the lac operon by running tests and performing assays
Introduction: In Prokaryotic (bacteria) cells, gene transcription is regulated by controlling expression of the operon. An operon is a part of the chromosome where genes are clustered performing coordinated functions. An example is the lac-Operon. Go ahead and observe the simulation of the activated lac-Operon. Mouse over proteins involved in the lac-Operon to learn their functions and names. You will want to do the same for the genes in the lac-Operon. When you feel comfortable with the lac-Operon, report to the Lab Guy for your first assignment.
Lab Guy: I am glad you are back. After studying the lac-Operon system, Dr Monroe needs your expertise. He has two mediums containing E.coli cells. One of the mediums has lactose only, while the other has lactose and glucose. The problem is he forgot what medium has what. Go to medium A and B and run tests to find out. When you have concluded, submit your report. If you have trouble, you can go back to the simulation of the lac-Operon. Otherwise, there is your Information menu for guidance. Good luck!
<when clicking on Lab Guy, will give user options to do the following:>
go to Lab
go to medium A
go to medium B
go to simulation
submit report
-Student will run tests on levels of cAMP, B-galactosidease, lactose permease, thiogalactoside, lactose inside the cell, lactose outside the cell, and RNA permease.
EXAMPLE
medium A <lactose only>
perform tests
cAMP <99.7%>
B-galactosidease <92.0%>
lactose permease <92.0%>
thiogalactoside <92.0%>
lactose inside cell <75.0%>
lactose outside cell <82.0%>
medium B <lactose and glucose>
perform tests
cAMP <9.2%>
B-galactosidease <21.0%>
lactose permease <15.0%>
thiogalactoside <13.0%>
lactose inside cell <0.0%>
lactose outside cell <82.0%>
submit report
medium B: glucose & lactose and medium A: lactose
medium B: glucose & lactose and medium A: glucose & lactose
medium B: lactose and medium A: lactose
medium B: lactose and medium A: glucose & lactose
After answering correctly: You are correct! When lactose and glucose are present, the cell will prefer glucose over lactose. Since lactose is not used, the lactose is unable to bind to the repressor, releasing it from the operator site. Thus, the lac-Operon is never activated and transcription is never initiated.
As you discovered, a good indicator of glucose present is the level of cAMP. When glucose is present, the level of cAMP is low. When glucose levels drop, cAMP levels increase. cAMP plays an important role in the lac-Operon. When cAMP is bound to CAP, it forms a complex that is able to bind to the CAP site, increasing the affinity of RNA polymerase to the promotor site. Thus, enabling a high level of transcription. Go back to the Lab Guy for your next assignment.
Lab Guy: Thank goodness you are back! Dr Sylvia need your help. She has two bacteria cells with mutant components of the lac-Operon. These mutant components pertain to cAMP and CAP. Using the previous knowledge acquired, help her identify the mutant components. First, you must run tests and perform assays to conclude the mutant component. When you feel confident, go ahead and submit your report. If you have trouble, you can go back to the simulation in the beginning and also utilize the Information menu. Good luck!
<when clicking on Lab Guy, will give user options to do the following:>
go to Lab
go to cell A
go to cell B
go to simulation
submit report on cell A
submit report on cell B
-student must pick running tests with glucose, lactose, and/or glucose & lactose which will run a simulation with the added carbon source. Students can also test the level of CAP and cAMP to come to the correct conclusion.
cell A <defected CAP>
carbon source run tests
add glucose
add lactose & glucose
add lactose
perform assays
CAP <2.0%>
cAMP <89.0%>
cell B <defected CAP site>
carbon source run tests
add glucose
add lactose & glucose
add lactose
perform assays
CAP <91.0%>
cAMP <89.0%>
answering cell A correctly: Good job! You have answered CAP as the mutant component! The cell lacks CAP, therefore the cell can only metabolize glucose. CAP needs to be present so it can bind to cAMP to form cAMP-CAP complex. This complex binds to the CAP site and increases the affinity of RNA polymerase to the promotor region.
answering cell B correctly: Good job! You have correctly identified the mutant component as the CAP binding site. This prevents the cAMP-CAP complex to bind to the site. Thus, preventing affinity for the RNA polymerase to bind to the promotor. As a result, this cell is only able to metabolize glucose. Go back to the Lab Guy for your next assignment.
Lab Guy: You are not done yet! Dr Monroe needs your assistance again. He has five mutant E.coli cells and is unable to identify the mutant gene in each. You are to run tests and then identify the mutant gene of the lac-Operon. If you have trouble, you can go back to the simulation. Otherwise, you can refer to the Information menu above. Good luck!
<when clicking on Lab Guy, will give user options to do the following:>
go to Lab
go to cell A
go to cell B
go to cell C
go to cell D
go to cell E
go to simulation
submit report on cell A
submit report on cell B
submit report on cell C
submit report on cell D
submit report on cell E
-student will have to add a carbon source when running tests. They also have to test levels of B-galactosidease, lactose permease, and thiogalactoside
cell A <mutant I gene-repressor does not block RNA polymerase, therefore mRNA is continually produced and translated>
(maybe lactose is unable to bind to the repressor to show it is the repressor)
carbon source run tests
add glucose
add lactose & glucose
add lactose
perform tests
B-galactosidease <99.0%>
lactose permease <99.0%>
thiogalactoside <99.0%>
cell B <lac Oc mutant-lactose is present/absent, repressor unable to bind to operator, continually allowing expression>
(should be able to notice this in the simulation)
carbon source run tests
add glucose
add lactose & glucose
add lactose
perform tests
B-galactosidease <99.0%>
lactose permease <99.0%>
thiogalactoside <99.0%>
cell C <lac P- gene-no expression of operon because RNA polymerase cannot bind)
(Should be able to tell in simulation)
carbon source run tests
add glucose
add lactose & glucose
add lactose
perform tests
B-galactosidease <2.0%>
lactose permease <2.0%>
thiogalactoside <2.0%>
cell D <lac Z- gene- no glucose and galactose production from lactose>
carbon source run tests
add glucose
add lactose & glucose
add lactose
perform tests
B-galactosidease <2.0%>
lactose permease <67.0%>
thiogalactoside <71.0%>
cell E <lac Y- gene- no induction, because lactose unable to be taken into cell>
(can show one simulation without lactose permease produced and stop)
carbon source run tests
add glucose
add lactose & glucose
add lactose
perform tests
B-galactosidease <74.0%>
lactose permease <3.0%>
thiogalactoside <71.0%>
After answering cell A correctly: Good job! You have correctly answered lac I gene as the mutant gene. It produced a mutant repressor that could not bind to the operator site. Thus, it was unable to block RNA polymerase from the promotor site. Therefore, mRNA is continually produced and translating proteins. I gene is a Regulator gene of the lac-Operon. When there is a mutation in a regulator gene, it affects expression of all structural genes in an operon.
After answering cell B correctly: Good job! You have correctly answered lac O gene as the mutant gene. This caused the operator site to be defected, and the repressor was unable to bind to the operator. This occurs in the presence and absence of lactose. So, it will continually allow expression of operon. O gene is a Regulator gene of the lac-Operon. When there is a mutation in a regulator gene, it affects expression of all structural genes in an operon.
After answering cell C correctly: Good job! You have correctly answered lac P gene as the mutant gene. This caused the promotor site to be defected, and the RNA polymerase was unable to bind. There was no experession of the operon. P gene is a Regulator gene of the lac-Operon. When there is a mutation in a regulator gene, it affects expression of all structural genes in an operon.
After answering cell D correctly: Good job! You have correctly answered lac Z gene as the mutant gene. Z gene was unable to transcribe for the protein, B-galactosidease that breaks lactose into glucose and galactose.
After answering cell E correctly: Good job! You have correclty answered lac Y gene as the mutant gene. Y gene was unable to transcribe for the protein, lactose permease that takes lactose into the cell from the outside environment.