Daily Newsletter: November 6, 2012 - Lac Operon

Daily Newsletter

November 6, 2012 Lac Operon

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One of the most well studied gene regulation systems is the Lac Operon found in Escherichia coli. To understand this system, it is important to first understand that bacteria do not experience Transcription and Translation in exactly the same way as eukarotes (the mechanics are very similar, but there are some distinct differences).
The core difference between prokaryotes and eukaryotes is the nucleus. Since bacteria lack a nucleus, transcription takes place in the cytosol. There is no need for a tag to get out of the nucleus, so there is no need for capping. Also, there are no introns, so no need for splicing. Basically, there is no processing of RNA to make mRNA. There is no pre-mRNA. When a gene is transcribed, the transcript is mRNA.















Since there is no membrane separating transcription from translation, you can couple these to processes. As mRNA is made, it can be translated (polyribosome).

Bacteria have a single circular molecule of DNA (a genophore, not a chromosome). They have to conserve their genetic space, so bacteria combine genes for a metabolic process in a single mRNA. IMPORTANT: bacteria can combine genes for a metabolic pathway into a sequential sequence with a single promoter. Thus, when you transcribe, you get all the genes for a given metabolic pathway.
The word OPERON describes this unique arrangement of prokaryotic genes: One promoter and one operator for a given series of metabolically linked genes.
The lac Operon holds three genes that give the cell the ability to take in and use the sugar LACTOSE. [Image Note: Bacteria have two promoter regions, -10 & -35, for a given gene or operon]


For E. coli, glucose is the preferred sugar. When glucose is present, there is no need to use lactose: these genes are not transcribed. When there is no glucose, E. coli has to use other sugars. IF lactose is present, the genes for lactose utilization will be made. Conversely, if there is no lactose, they genes remain locked down.

• Glucose Present: No transcription
• Glucose Absent: Minimal transcription
• Glucose Absent, Lactose Present: Transcription of the lac operon.

NOTE: In this example, there are two ways to control the expression of a gene or operon:

1. You can block the operator of the gene. This prevents RNA polymerase from making RNA.
2. You can alter the promoter (or the interaction between transcription factors and DNA) to prevent binding of the Transcription Complex (RNA polymerase).

Negative Transcription Regulation (Repression) in the lac operon: There is a repressor for the lac operon. This is a protein that can bind to the operator of the lac operon (that region immediately downstream from the promoter). This creates a physical block that prevents RNA polymerase from transcribing.

As you can see in the image to the right, the lac repressor is a protein that is constitutively (always) expressed. This indicates that the lac operon is normally turned OFF. Notice that the gene for the regulatory protein is upstream from the lac operon.
The lac repressor (LacI) binds to the major groove of the DNA at the operator. As you can see in the annotated image of the repressor, you have a DNA-binding region (active site), and a regulatory domain. The regulatory domain has the ability to bind to Lactose (the inducer).

You must have a way to unlock the operon, or to put it another way, to inactivate the repressor. An inducer is a ligand that can bind to a regulatory domain, changing the shape of the regulatory domain, and thus inactivating the DNA-binding domain. In the case of the the lac repressor, the ligand is allolactose (a derivative of lactose). When allolactose binds to the lac repressor, the repressor is inactivated, and the operon cleared. This is seen in the image below.

Therefore, the lac operon is partially regulated by the presence of lactose in the environment. If there is no lactose in the environment, then there is no need to transcribe the three genes needed to use lactose.
Remember, we don't want to expend energy for things we don't need. Below is the size of the three gene products:

• β-Galactosidase 1,024 Amino Acids
• β-Galactoside Permease 418 Amino Acids
• β-galactoside transacetylase 203 Amino Acids

Total Amino Acids, 1645. Using the assumption of 4 ATP per amino acid added to a protein, that makes 6580 ATPs needed just for protein synthesis. With this many amino acids, you are also looking at 4935 nucleotides (remember 3 nucleotides = codon = 1 amino acid), plus at least 3 stop codons. Each nucleotide added to a transcript takes the equivalent of 1 ATP, so you are looking at 4944 ATP minimum to make the transcript. Just to make one example of each protein (gene product), you are looking at 11,524 ATP. Do you just make one example of each protein? NO! Do you have though the idea that this is energy consumptive? Would you make it if there was no lactose around?

Tomorrow we will look at the positive regulation of the lac operon. Remember, you still will not transcribe this operon if glucose is present. You only transcribe when glucose is absent. So how does the operon know when glucose is absent? That will be our discussion tomorrow.

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Daily Challenge

In your own words, describe how lactose (allolactose) is used to regulate the transcription of the lac operon. In your discussion, make sure that you explain the concept of an operon, and discuss the differences between eukaryote and prokaryote gene transcription.
Link to Forum

Special Edition: November 6, 2012 - Evolution and Gene Expression

Special Edition

November 6, 2012 Evolution and Gene Expression

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Suggested Reading

Prior to accepting the daily challenge, read the following articles.

• Mammalian gene expression. 
• Neutral Theory.

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Evolution has its base in genetics. All variation starts with mutation. The addition, deletion or change in the nucleotide sequence of a coding region can result in protein changes. Any mutation, if it has an expressed effect, then becomes subject to the mechanism of Natural Selection.

A great example of this nucleotide alteration is seen in Sickle Cell Anemia. A change in one nucleotide causes a conformational change in the β-hemoglobin chain, which causes the entire expression of the disease (the altered hemoglobin molecule is known as HbS). Review the β-hemoglobin gene sequence and see how just one point-mutation can have such a strong effect. Also note, this is a change in the third nucleotide of a codon.

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Optional Challenge

Your task today is to discuss the relationship DNA changes and Gene Expression has with evolution. Sit with the concept for a moment, and come up with a coherent discussion. Feel free to use models such as Sickle Cell Anemia. Remember, this alteration in gene expression sets up the diversity of the population. Natural Selection then acts on the new phenotype.
Link to Forum

Daily Newsletter: November 5, 2012 - Introduction to Gene Regulation

Daily Newsletter

November 5, 2012

Introduction to Gene Regulation

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Suggested Reading

These brief articles are a supplement to the readings from your textbook on Gene Regulation. You do not have to finish these articles today, but they will help you understand gene regulation at a deeper level. They also make great references for your next milestone paper.

• Regulation of Transcription and Gene Expression in Eukaryotes
• Operons and Prokaryotic Gene Regulation
• Positive Transcription Control: The Glucose Effect
• Negative Transcription Regulation in Prokaryotes
• Obesity, Epigenetics, and Gene Regulation
• Environmental Influences on Gene Expression

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• Do we express all of our genes at the same time? Why?
• Do we need all of our genes expressed all the time? Why?
• Why do we have so many genes?

These are just a few of the questions you need to start asking yourself. Humans have hundreds of thousands of genes. Many are needed all the time (constitutive), but others are only needed when the cell get's certain signals. So how do we control the expression of all this genetic knowledge?

During mitosis, for example, did you see the production of DNA polymerase and the replication complex during the start of G1, or did you only see it after you passed the first restriction point? Do we keep DNA polymerase around just in case we are going to do some nuclear division? or do we unlock its expression only when needed?

Consider: The first restriction point determines if you are going to prep for division. When you have enough cyclin-dependent kinase available, you pass the restriction point. CDK signals the cell to get ready for division. How does this signal work? It changes gene expression (i.e., we activate regulated genes).

Think about the human body and homeostasis. Think about hormones. Are you always producing everything, or do you need to trigger some events? Could that trigger then be a regulated gene?

Remember that you need at minimum the equivalent of 4ATP per amino acid incorporated into a protein. Add to this 1 ATP equivalent for each nucleotide during transcription. You should quickly realize that gene expression is energy expensive.
Your goal today is to start reading about gene regulation, and more specifically, come to an understanding of the necessity of gene regulation.

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Daily Challenge

Why do we need gene regulation? Today, reflect on the need and use of gene regulation. Why would an organism need to have some genes that it could turn on or off? Why would you need to control gene expression? Can the environment affect gene regulation? Can gene regulation affect evolution?

Daily Newsletter: November 1, 2012 - Translation

Daily Newsletter

November 1, 2012 Translation

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The core concept of translation is the connecting of a codon to an amino acid. As we saw yesterday, this is accomplished with the Transfer RNAs. What that leaves us with then is the actual mechanism of amino acid polymerization.

Initiation of Translation

Protein function is determined by the sequence of amino acids. This sequence allows the protein to fold into the correct configuration to produce activity. Any variation in the sequence can produce alterations to function, or even result in non-functionality. In order to generate the correct sequence, we must first establish the correct reading frame of codons. We must first find the start codon on mRNA.

The small subunit of the Ribosome (40s in eukaryotes) is built to find the Start Codon (AUG) and will align the full ribosome with the correct reading frame. A number of proteins will help the alignment, and in the formation of the full (holozyme) Ribosome.
The diagram below shows the overall formation of the initiation complex, complete with a tRNA (the yellow structure with a pink circle attached). Again, the function of this replication complex is to find the start codon and set the reading frame for the Ribosome. Notice that the large ribosomal subunit (60s in eukaryotes) only attaches after AUG has been found. As before, it is not neccesary at this academic level to memorize all of the factors involved. What is more critical is that it is a multifactor system designed to find the correct start point, and thus the same reading frame.

Key Feature: Notice that the first tRNA is already linked to the small subunit. Why? It's anticodon is complimentary to the a codon on mRNA. Specifically it has the anticodon for the start codon (AUG). So we are using base complementarity to find AUG.

Elongation

The polymerization of amino acids occurs during elongation. This is where the P and A sites become important (NOTE: P and A sites are the active sites of the enzyme). P stands for Peptidyl, while A stands for Aminoacyl. These are chemical terms,which shows the orientation of the amino acid. The exit site, represented by E, is not an active site. Consider it a disposal point for spent tRNAs. [NOTE: you may also find references to a fourth site where the tRNA first comes into the comples. Don't worry about this optional site.]

The P and A sites reveal a single codon on mRNA, and can hold a single complimentary tRNA. During elongation, when you have a filled P and A site, the amino acid from the P site will be linked to the amino acid in the A site. This is a process that you will have to visualize, so use the diagram below as reference:

The amino end of the amino acid is free. The carboxyl end is attached to the tRNA. Starting at the top of the above diagram, the growing amino acid chain is attached to a tRNA in the P site. A new tRNA with an aminom acid (charged tRNA) is brought into the A site. Using GTP, the Ribosome (large subunit) takes the growing peptide chain and links (carboxyl to amino) it to the individual amino acid in the A site. When this is done, the entire ribosome shifts downstream to the next codon (the new codon appears in the A site).

The spent tRNA that started in the P site is now moved to the E site, where it is removed from the ribosome. NOTE: It takes 2 GTP to create the peptide bond, then another GTP to move the ribosome. So a total of 3 GTP are used in one 'round' of Ribosomal action. REMEMBER THIS! In addition, it took a triphosphate to charge tRNA (so a total of 4 for each amino acid added to the polymer).
When both the P and A sites have charged tRNA, the growing chain from the P site is added to the single amino acid in the A site. The ribosome shifts, and the process continues. This elongation process of adding amino acids (amino acid polymerization) will continue until a STOP codon is reached (UAA, UAG and UGA).

Question: How much ATP will you need to expend to make a protein with 100 amino acids? How about a 150 amino acid protein? GOAL: Recognize and be able to articulate why protein synthesis is an energy consumptive process, and be able to discuss why it is critical for cells to regulate energy consumptive processes.

Termination

To create a functional protein, translation must end with the appropriate amino acid. If translation stops to soon, the protein will be to short and many not bend (configure) correctly. If it is too long, then it may not bend (configure) correctly. Termination is a critical process. Termination begins when a STOP CODON (UAA, UAG and UGA) is reached. In eukaryotes, a releasing factor is used to seperate the ribosomal subunits. KEY CONCEPT: The stop codon signals the end of the coded message.

Once completed, proteins can be further modifed as fits their function (such as adding sugars). This is known as post-translational modification. The image below shows the posttranslational modifications needed in the production of insulin. Production starts with a ribosome bound on the Rough Endoplasmic Reticulum (RER). Processing will occur in the RER and in the Golig body. This is only one example of posttranscriptional modification, and a majority of proteins require such modifications before they are functional. [NOTE: as a general rule, there is less extensive posttranscriptional modification in prokaryotes, but they have numerous proteins that do require modification].

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Daily Challenge

In your own words, describe the process of translation. Discuss initiation, elongation and termination. Make sure that you discuss the P and A site, as well as the importance of the start and stop codon. Afterwards, give a BRIEF discussion on how this is an energy consumptive process that needs to be regulated.
Link to Forum

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Optional Challenge: Genomics and Proteomics

Read the following articles:
A brief guide to genomics - NIH fact sheet
Transcriptome - NIH fact sheet
Proteomics
The genome can be seen as the genetic potential of an individual (think Genotype), while the proteome shows what is actually produced at a given time, under a given condition (consider this the phenotype). Provide a discussion of the importance of genomic and proteomic studies in modern biological resarch, and make sure that you provide a description of both the geneome and the proteome of an organism.
Link to Forum