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.
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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.
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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

Daily Newsletter: October 31, 2012 - Codons, Anticodons & Amino Acids

Daily Newsletter

October 31, 2012

Codons, Anticodons & Amino Acids

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Translation is the process of "reading" mRNA, and using the code to construct a protein. But what is the code? The nucleotide language of mRNA can be divided into codons. Three sequential nucleotides that represent a genetic (nucleotide) word. So, how do you read this code or nucleotide language?
In the image to the right, you hav
e have sequential nucleotides divided up into codons. Notice that AUG is listed as Codon 1. This is important! AUG is the Universal Start Codon. Nearly every organism (and every gene) that has been studied uses the three ribonucleotide sequence AUG to indicate the "START" of protein synthesis (Start Point of Translation).

As we will see tomorrow, it takes more than a start codon to initiate transcription, but for now just remember that this is the codon that indicates the START point of the instructions on how to make a protein.
The start codon established the Reading Frame for translation. From the start codon, every three sequential nucleotides will be viewed as a codon. This is critical! Mutations can affect reading frames. For example, if a nucleotide is inserted between codon 2 and 3 (G G), would you have the same reading frame down stream? What if you deleted the first nucleotide of codon 4? What is the effect of changing the reading frame? What would happen to the resulting protein?

Insertions and deletions can change reading frames, but point mutations can also occur. In this case, one nucleotide is change to a different nucleotide. What would happen if the final nucleotide of condon 3 were changed to a C? To an A? How about the second nucleotide in codon 4? Change the U to an A, what happens?

Each codon is a "genetic word," and refers to a specific amino acid (thus changes to these words can result in changes to final proteins). The tRNA is the agent of translation. On one end of the tRNA, you will find an anti-codon. Anti-codons are complimentary to codons. Example: Codon 1 reads AUG. The corresponding tRNA would have an anticodon reading UAC. (Question: Would these be antiparallel?). Codon 2 reads ACG, so the anticodon would read UGC. Oppisite the anticodon, you will find a binding site for a specific amino acid.

An amino acid can be attached to the free 3' end of the tRNA. There is a class of enzymes capable of attaching an amino acid to a tRNA: Aminoacyl tRNA Synthetase. Below is a very basic cartoon of how an amino acid is added to a tRNA.
Note that an ATP is needed to complete the binding. There is an Aminoacyl tRNA Synthetase for each tRNA-Amino Acid combination.
Below is a diagram showing the pairing of codon to anticodon. The diagram also contains a version of the Genetic Code table, showing the relationship between codon and amino acid.

Note that three codons are referred to as STOP codons: UAA, UAG, and UGA. These are used to terminate translation; they indicate the end of the gene's coding region. What would happen if you lost a Stop codon?

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Daily Challenge

In your own words, describe the genetic code and how codons/anticodons work to relate the genetic code to amino acids. Discuss what can happen if the coding sequence is change by either changing, adding or deleting a nucleotide.
Link to Forum

Daily Newsletter: October 30, 2012 - Transcription

Daily Newsletter

October 30, 2012 Transcription

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Transcription is the genetic process where a single strand of DNA acts as a template for the construction of a complementary RNA strand. Generally when talking about transcription, we will be talking about the formation of messenger RNA (mRNA), which carries the code for one gene to a ribosome where it is translated into a protein.
DNA holds the "permanent" copy of the genes needed to make a functional organism (nothing is really permenant). Think of DNA as a locked safe where you hold all your company's blueprints, patents and documented procedures. You don't want to loose these, or risk that they might be changed. You only bring them out to make copies of them, then they go back to the safe. This is what happens with your DNA. You keep it tightly locked up (in a double-helix that is coiled around histones, and then possibly supercoiled), and open it up only when you NEED to make a copy. Notice how NEED is highlighted? Do you think it might be an important concept?
In eukaryotic DNA every gene starts with a promoter. This is a sight of ~8 nucleotides visible in the major groove of DNA. The transcription complex recognizes this sequence as a "START" indicator. The main core of the transcription complex will be RNA polymerase. This enzyme works to build a strand of RNA complementary to DNA. The name polymerase indicates that it is involved with dehydration synthesis polymerazation reactions (taking one nucleotide, and adding it to a growing chain of nucleotides). Like DNA polymerase, RNA polymerase builds in the 5' to 3', and builds phosphodiester linkages between nucleotides.
But RNA polymerase can not act alone. In eukaryotic systems, initiation factors are needed to recognize the promoter region, and then to correctly align the RNA polymerase. To the left is a great picture showing the initiation complex and the RNA polymerase II holozyme (RNA polymerase II with all associated protein structures). You are not responsible for knowing all of the factors needed to initiate eukaryotic transcription, but you do need to start understanding the concept that it takes multiple factors to identify a promoter and start RNA polymerase. What do you think you need in order to recognize a specific sequence of nuclotides?
As you can see, TATA Binding Protein (TBP) is the first structure to attach to DNA. It recognizes the TATA sequence in the major groove of the DNA double helix. It then forces the the DNA to bend, and acts as a signal to other enzymes directing interactions with DNA. A cascade of reactions occur to then produce the Preinitiation Complex, which ensures that the transcription complex is positioned correctly over the Transcription Start Site, and begins the unwinding (sometimes referred to as denaturation) of the double helix. The Transcription Complex then begins to read the template strand of DNA, and makes an RNA copy (Elongation). [NOTE: Bacteria use proteins known as sigma factors to help find promoter regions and initiate transcription. There are different sigma factors linked to different environmental and physiological states, such as the Heat Shock Sigma factor which alter's the bacteria's ability to deal with higher temperatures)]
Elongation works due to base complementarity. Ribonucleotide triphophates are brought into the transcription complex, and are added to the free 3' end of the growing RNA strand. During the elongation phase, the RNA polymerase continues to add nucleotides to the growing RNA strand.
At some point, the RNA polymerase comes to a termination sequence. We are not going to spend a lot of time on termination (you are not held responsible for the various models). There are a couple different models of eukaryotic transcription termination. The main feature is that there is a signal sequence of deoxyribonucleotides in DNA that signals the end of transcription. Once this signal sequence is found, RNA polymerase is removed and the new transcript (new RNA molecule) is released.
mRNA processing: Once transcription is complete, in eukaryotes, the RNA needs to be processed. The following is a quick reference for mRNA processing:

• 5' capping: To protect the mRNA from ribonucleases (RNA degrading enzymes) that attack the 5' end, 7-methylguanosine is added to the 5' end. Usually, the 5' ribonucleotide is replace by this compound. Additionally, methyl groups can be added to the sugar-phosphate backbone to further protect the mRNA.
• Polyadenylation: In maturing RNA to mRNA, a poly-A tail is added (usually after cleaving off a small section of the 3' end). This process adds ~250 adenyls to the 3'end of the molecule. This is needed to stabilize the molecule and facilitate export through the nuclear pores. As mRNA is translated, the poly-A tail gets shorter. When short enough, the mRNA is degraded. Thus, the polyadenylation (poly-A tail) is responsible for setting a time limit to the mRNA. 
• Splicing: The RNA is composed of both coding (exon) and non-coding (intron) regions. To mature into mRNA, the introns have to be removed, and the remaining exon spliced together. This job is the responsibility of the splicosomes.

• The above image is a quick reference to the effects of splicing.

 

• The above image is a quick reference to the effects of the splicosome.

Once RNA has been processed (matured), it is ready to be used in translation (protein synthesis). NOTE: Bacterial RNA does not undergo processing. The bacterial RNA transcript is immediately translated.

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Daily Challenge

Transcription In your own words, discuss the process of transcription, and the formation (maturation) of mRNA. Remember that we have focused on eukaryotic transcription. Briefly, how does prokaryotic (specifically bacterial) transcription differ from eukaryotic transcription?
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Daily Newsletter: October 29, 2012 - Ground Rules for Gene Expression

Daily Newsletter

October 29, 2012

Ground Rules for Gene Expression

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Last week, we introduced the concept of the Central Dogma, which in the broadest sense encompasses the genetic mechanisms of Replication, Translation and Translation. In the strictest sense it the Central Dogma describes gene expression: Information encoded in the nucleotides of DNA being use to construct proteins. The two core genetics processes involved in gene expression are Transcription (synthesis of RNA) and Translation (synthesis of proteins).

Before digging into each process, let's talk a little about what is at stake here. DNA holds our genetic history. It holds codes on how to build an organism, but what does that really mean?

As we saw at the start of the semester, phospholipids can naturally form bilayers. They can even form spherical structures that create two fluid compartments, outside vs. inside.


Membranes though are passive. As a selectively permeable barrier, only certain materials can cross. Proteins add functionality to the membrane. By embedding proteins, you can chance the permeability of the membrane. This is how cells balance what is on the inside, and what is on the outside. Membrane proteins can also have enzymatic or signal functions. Proteins add functionality to the membrane.

A common expression is that DNA holds the code to make an organism. The meaning of this phrase lies in the concept that by making proteins, we make phospholipid membranes and cells functional. From DNA, cells can build metabolic pathways, produce various chemical compounds, anchor with other cells, and in multicellular complex life, we even have the development of special cellular roles that work together to form a composit whole.

The concept of how we go from DNA to RNA and then Proteins is one of the most critical concepts in biology! Today we are going to focus on some of the basics, the Ground Rules, of genetics.

All genetic processes work due to base complementarity. If you know the base complementarity rules, then the foundations of genetics will make sense. At times, this may seem repetitious, but I really want you to get these terms and concepts.
Genes are sometimes referred to as the unit of heredity, and with good reason. A gene is a segment of DNA that holds the code to make a protein (NOTE: or functional RNA, such as trasfer RNA). In modern biology, we refer to gene products, which are just the expressed macromolecules coded by a gene.
Remember, a gene product can be either a proteins or functional RNA (e.g., tRNA). Functional RNA does not code for proteins, instead, these RNA strands have some function in cellular metabolism, most notably in the genetic process of Translation. Examples include tranfer RNA (tRNA), ribosomal RNA (rRNA), and small nuclear RNA (snRNA).

All genes have non-coding portions that are critical for the correct transcription (synthesis of RNA). These non-coding areas are critical for regulation and aligning the transcription enzymes (e.g., RNA polymerase). Below is a graphic that shows structure of a gene. The promoter of a gene is a sequence of DNA upstream of the actual code (coding region) that indicates the "Start" point for transcription. This is how your cell knows where to begin transcription. The loss of the promoter means that the gene will no longer be expressed.

In Eukaryotic cells, a common promoter is a DNA sequence that reads TATAAA, and is better known as the TATA-Box. In bacteria, the common promoter is TATAAT, bettern known as the Pribnow Box (Pribnow-Schaller box). In both cases, the promoter is found in the Major Groove of the DNA molecule. As can be seen in the image to the right, the major groove is wide enough to "see" the base pairs. The base pairs have an electrochemical profile, and thus can respond to other chemicals (via van Der Waals forces). Thus, the major groove is a place where proteins (and other compounds) can bind to DNA! The promoter sequences are found in the major groove. The image to the right shows a bacterial promoter event. One of the factors needed to start transcription (by recognizing the promoter) has bound into the major groove. This recognition event is needed to identify the start point of a gene. The Transcription Initation Complex will then begin to form at this site, and begin the transcription of the gene. 

Many genes are regulated, meaning they can be turned on and off. Beyond a promoter, a regulated gene will typically have a non-coding region known as the Operator. The operator is located down stream of the promoter (meaning it will be between the promoter and the coding region). Regulatory proteins can bind to the operator, preventing transcription. Remember, cells are masters at energy conservation. They will not begin producing proteins that are unnecessary. Gene regulation is a common activity of Signal & Receptor systems. The image below is a good visual of the promoter & operator systems. House keeping genes are those that are needed for the general function of the cell, and can include genes for glycolysis, citric acid cycle, and ribosomes. These genes are always ON, and are referred to as constituative genes. 

Messanger RNA (mRNA) is a molecule of RNA that cares the gene code for the construction of a protein. mRNA is sent to the Ribosome in order to produce a protein. The code for constructing a protein is in Nucleotide Language, meaning the code is a code of nucleotides. Specifically, the code in mRNA is in the ribonucleotide language (A, U, G, C). In order to make a protein, it is necessary to Translate the ribonucleotide language into the language of proteins, i.e., amino acid sequcences. 

In order to translate, you need an agent of translation. This agent of translation must be a molecule that contains both ribonucleotides and amino acids (think of it as the nucleotide-amino acid dictionary). A specific ribonucleotide sequence must directly correspond to an amino acid, just as in translating human languages requires word for word relationships. This concept of a direct nucleotide to amino acid relationship is the basis of the Genetic Code.

The agent of translation is Transfer RNA (tRNA). In tRNA, there is a direct physical correspondence between a 3 nucleotide sequence (anti-codon) and an amino acid. To the right are common ways of illustrating tRNA, with the 3^rd image being the most common way of drawing the molecule. In the image, each molecule has a region known as the anticodon; this region will interact with mRNA. At the 3' end of the molecule, a specific amino acid will be bound. 

On the mRNA, the code is broken down into codons (think of these as genetic words). Codons consist of 3 adjacent nucleotides. Codons are complimentary to anticodons found on tRNA. Each tRNA has a specific anticodon-amino acid relationship, so each codon then specifies an amino acid. The genetic code is NOT ambiguous. There is a direct correspondence between codon and amino acid; the tRNAs make sure of this.

The ribonucleic language is divided into 64 3-nucleotide words known as codons. Condons specify though tRNA an amino acid. The Genetic Code is thus the translation scheme between codons and amino acids. [NOTE: another way to describe the genetic code is in terms of a computer algorithm]. Below is a rather unique way of viewing the genetic code. It is an excellent way of visualizing the number of redundancies in the code.

The genetic code is redundant, which means that there are multiple codons (3 nucleotides) that specify the same amino acid. For example, around the 12 o'clock position of the above chart, you see the amino acid glycine. The codons GGU, GGC, GGA and GGG all specify Glycine. Phenylalanine is specified by UUU and UUC. There are only a few amino acids, such as methionine, that are specified by a single codon (in the case of methionine it is AUG).

The presence of redundancies means that some alterations in the gene sequence are silenced (silent mutation). For example, changing GGU to GGA does not change the specified amino acid (Glycine). This is a silent mutation. Changing UUC to UUA may cause a problem (point mutation), but both Leucine and Phenylalanine are hydrophobic, so the variation may be minor. Chaing CAC to CAG though has more impact as you are changing the positive histidine to a polar glutamine (you loose the full positive charge of histidine). Remember, chaning amino acids can easily change the way a protein folds. REMEMBER: The genetic code has redundancies, and this will limit some problems with mutation.

Below is a more classic way to represent the genetic code, in the form of a table. The way the table is arranged, you can easily see the various redundancies in the system. In both representations, notice that there are three codons that specify STOP. These stop codons, UAA, UAG and UGA are essential for the termination of protein synthesis. In the image below, you will notice AUG has been tagged as the initiation (start) codon. All protein synthesis begins with the code AUG. We will talk more about this later in the the week.

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Daily Challenge

Your challenge today is just to discuss the concept of the Central Dogma. Don't worry about mechanisms yet. Consider the implication of the central dogma. In your own words, express why it is important. What are the important features? You can use questions above in "Today's Topic" as a way of focusing your thoughts.
Link to Forum

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One final image

Below is a variation of the genetic code diagram shown above. In this image, the amino acids are drawn out and color coded by their type (basic, acidic, polar and nonpolar). Look over the chart. How easy is it to change amino acid type? If you change the first nucleotide in a codon, what is your chance of changing the amino acid type? What about the second codon? What about the third codon. An optional forum is open for you to explore these concepts.
Link to Forum

Daily Newsltter: October 26, 2012 - Mendel, Meiosis and Evolution

Daily Newsltter

October 26, 2012

Mendel, Meiosis and Evolution

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With meiosis we see the mechanism of inheritance first hinted at by Mendel. Gregor Mendel discussed how each person has two "copies" of the information that gives rise to a physical trait (genes), and that there were variations in that information which gave rise to different expressions of these traits (alleles).

Mendel's work gave biologists an understanding of inheritence, and an understanding of chromosomes and nuclear division helps to understand Mendel's inheritence model and the variations of that model. With cross-over in Prophase I, and the random seperation of homologs in Anaphase II, we can see the recombination of parental genes that result in genetically unique gametes (sperm and egg). This results in genetically unique individuals each generation.

Diversity is one of the greatest advantages of sexual reproduction. While >99% of our genes are held in common within a species, there is sufficent genetic variation to allow provide a species with a broad adaptaion range. Remember, the goal of evolution and survival resides not in the individual, but in the population. Those with genetic, metabolic, or physiological advantage will have a greater chance to leave offspring; thus the species survives and thrives.

In class, I mentioned that when you get to the core of biology, it is ultimately all about sex. Not the physical act, but the concept. The recombination of genes allows for diversity. Some organisms spend most of their life in haploid, or non-sexual reproductive states. They only expend the energy for sexual reproduction when they need to insure the survival of their offspring.

Daily Challenge

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Today, you are to reflect upon what you have learned so far regarding metabolism, Mendelian inheritence, and meiosis. Build a picture of evolution based upon the need for a diverse population. Use an example to help illustrate the need for diversity. As a starting point, look to endangered species (why is it a problem when you have few individuals left?) or look at what happens when you have genetically similar trees struck by disease (Pine Bark Beetle infestations).
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Video Challenge

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Today, I'm going to give you a video challenge. Below is a link to a video about animal genitalia (penises to be percise). The video is informative, but at times very vulgar (so be warned). The video's narrator makes some interesting, and humorous, comments about evolution.

What you are to do is reflect upon the views of evolution discussed. What insights can you glean from the narrator and the topic, and what errors are made (even jokingly) about evolution.

The World's Most Terrifying Penises: The Echidna

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Special Edition: October 25, 2012 - DNA Replication

Special Edition

October 25, 2012 DNA Replication

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Textbooks have a tendency to make replication one of the most complext topics covered. With a tendency to throw all the current research and understanding at students, they rarely take a step back and try to explain it. This newsletters has two goals: 1) to help biology students understand DNA replication, and 2) to show you want is expected from a Biology Freshman/Sophmore.



Central Dogma of Biology

 Before we get into replication, let's take a step back and look at the three core genetic processes, a.k.a., the Central Dogma. The central dogma describes the flow of genetic information in a cell. The core idea is INFORMATION. You may recall some of our early discussion on DNA, and about base complimentarity and the directionality of the molecule. This will become important rather quickly, but for just this moment, I want you to concentrate on the fact that DNA carries information. Information on how to build RNA and Proteins, both of which will produce the phenotype (expression) of the cell. For this reason, DNA, RNA and Proteins are considered Informational Macromolecules. This means that the sequence of monomers contains infomation, e.g., instructions on how to build RNA and Proteins. Since it is critical, the sequence of nucleotides carries information.
As you can see from the diagram of central metabolism shown above, there are three processes: Replication, Transcription and Translation. Think about those words. They are words used in reference to languages and documents. When you replicate a docuement, you want to ensure that you are getting a faithful (or even exact) copy of the original.

When you transcribe, you are moving from one medium, e.g., spoken word, to another medium, e.g., text. If you watch news shows, they will tell you that transcripts of the show are available. Court reports make transcripts of the trial. You are taking the language from one medium (in our case DNA) to another medium (RNA). The language is still the same (i.e., nucleic acid lanugae), just in a different form. Does the transcript have to be 100% correct? You want it to be, but it is not as exacting as a replication.
Translation is where you change languages. Unless you're fluent in another language, you will need somoene to help translate, or at least a good translation dictionary. Now you are moving the context from one language (nucleic acid) to another language (amino acids).

All of these processes rely on one key feature of nucleic acids: BASE COMPLIMENTARITY. In DNA: A complements T, and G complements C. In RNA: U complements A, and G still complements C.

Replication

In replication our goal is to take one molecule of DNA and make two daughter molecules of DNA that are identical to the first. Even the best replication processes can produce errors, but our goal is to be error free. This takes precision! As DNA is long, we also need this to be a fast process.

The enzyme that is ultimately responsible for replication is DNA Polymerase. [NOTE: there are multiple types of DNA Polymerase, but for now you just need to understand the core concept common to the DNA Polymerase family.] DNA polymerase is only one component of the Replication Complex, which is a complex association of proteins needed to successfully complete the replication event. Your goal at this time is to concentrate on DNA polymerase; we will talk about some of the other components later.
DNA is a double stranded molecule, in which the strands are anti-parallel. This means that one strand starts at the 5' end → 3' end, while the the other strand is revered. This can be seen in the image to the right. A common way of saying this is that we read DNA in the 5' → 3' direction. Why is this important? DNA Polymerase can only read DNA in the 5' → 3' direction, and can only build a new strand in the 3' → 5' direction. Before we go on, let's look another time at how deoxyribonucleotides are polymerized.


To the left you will see a generic image showing DNA Polymerase adding nucleotides to a growing DNA strand. Look carefully: A deoxyribonucleotide triphosphate (dTTP) is being added. The 5' phosphate of the new nucleotide is what will be used to form the phosphodiester bond. DNA Polymerase requires a free 3' end on which to grow the new DNA strand. To the right is another image that will help you with this concept. At this point, the critical thing to remember is that DNA Polymerase will need a free 3' end on which to add a new nucleotide.
This requirement to build in only one direction (3' → 5') creates a problem for the DNA process: the two strands read in opposite directions, and each must be replicated. It was noticed that one strand appears to replicate continuously, while the opposite strand appears to replicate discontinuously.

The original strand that reads 5' → 3' can be used by DNA polymerase to continuously produce the new 3' → 5' strand (the antiparallel complement to the original strand). We refer to the continuous synthesis strand as the Leading Strand. The other original strand, which read 3' → 5' cannot be copied continuously. A section will have to be exposed, replicated, and then another section exposed. This strand is constructed discontinuously, and is reffered to as the Lagging Strand.The image above shows the leading and lagging strand. Notice that the leading strand is replicating toward the Replication Fork(where the original strands seperate). As more DNA unwinds and opens, DNA synthesis continues down the leading strand.
The lagging strand though has to build in a start-stop action, producing Okazaki fragments. These fragments have to be sealed (phosphodiester bonds) together before DNA can rewind into the α-helix.

In the image above, you will notice a number of enzymes on the lagging strand. These enzymes are required for the initiation of DNA polymerazation, and then sealing the fragments. Looking at the image, you will see an enzyme called Primase, and a structure in red known as a Primer. Another restriction on DNA polymerase is that it must have a free 3' end from which to start building. DNA Polymerase is prevented from building a DNA strand from nothing. Something (a primer) must be in place upon which DNA polymerase can build. The Primer is constructed from RNA, and is a temporary scaffold upon which DNA polymerase can start working.

Eventually the primer will need to be moved. This is where you need to learn a little more about DNA polymerase. There are multiple DNA polymerases in eukaryotic systems. The general work horse of replication is DNA Polymerase III (DNA pol III), which is used to make long strands of DNA. DNA Polymerase I (DNA pol I) is used to replace primers (it is also used in repair functions). Even after the RNA primer is replaced with DNA, there is still a gap between fragments. Ligase is the enzyme used to create a phosphodiester bond between fragments, thus sealing the new sugar-phosphate backbone of the synthesized strand.

The result of DNA replication is that one molecule one DNA was used to create two new molecules of DNA. The two strands of the original DNA molecule became the template from which to build new complimentary strands of DNA. This is referred to as semi-conservative replication, as each new molecule has one strand from the original molecule, and one freshly synthesized complimentary strand.

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Optional Challenge

In your own words describe the purpose and process of replication. While you are encouraged to deepen your knowledge of replication, you are only required at this time to go into as much detail as was presented in this newsletter.
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Link to Lecture Reflection Forum

Daily Newsletter: October 25, 2012 - Meiosis

Daily Newsletter

October 25, 2012 Meiosis

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On Tuesday, the topic discussed was mitosis. We saw that mitosis was standard nuclear division. When coupled with cytokinesis, we saw that the goal was taking one parental cell and creating two identical daughter cells. With meiosis, we will again experience nuclear division, but with a different goal than mitosis. The goal of meiosis is to produce gametic (reproductive) cells. As such, we will 'always' couple meiosis with cytokinesis.

Goal of Meiosis: From one diploid parental nuclei, generate four genetically unique haploid nuclei.

Goal of Meiosis-Cytokinesis: From one diploid parental cell, generate four genetically unique haploid cells.

The production of gametic cells requires two nuclear division events: a reduction division, and an equatorial division.
Important Note: All forms of nuclear division have the same basic four stages. Prophase is the set-up, Metaphase is when chromosomes align, Anaphase is when chromosome/chromatid seperate, and Telophase is when the nucleus returns to "norm".
The first meiotic divison is a reduction division, and is reffered to as Meiosis I. As the term reduction implies, we are "reducing" something. Specifically, during the reduction division the cell is reducing chromosomal number. REMEMBER: a diploid organism (2N) carries two copies of each chromosome. During a reduction division, the copies are separated and moved to opposite sides (poles) of the cell.

Homologous Chromosomes: This term refers to the two copies of each chromosome. For example: Chromosome 1 in humans is the largest of the chromosomes. Individuals have two copies of chromosome 1: one from their mother and one from their father. These two individual examples of chromosome 1 are considered homologous. Homologous chromosomes carry the same genes, but each homolog (one copy) will have a unique set of alleles (one allele for each gene).
The image to the right is an example of of two different homologous chromosomes. Notice that at the same gene is found at same location on each homolog (a member of a homologous pair). As shown in the first homologous pair, one homolog carries allele A for the first labelled gene, while the second homolog carries allele a.

IMPORTANT NOTE: Remember that what we are seeing are chomosomes. The two chromatids of a given chromosome are the products of DNA replication. The image to the left should help you remember that you start with a Maternal DNA molecule and a Paternal DNA molecule. After DNA replication, these will consense into maternal and paternal chrmosomes. It is critical that you remember the difference between Chromosome and Chromatid.

Prophase 1 (prophase of Meiosis I) is the most critical stage of meiosis. Prior to the reduction division (which occurs in Anaphase 1), a recombination event will occur between the maternal and paternal chromosomes. Recombination is a genetic event in which homologous genes on two different DNA molecules are swapped between the molecules. A section of each DNA molecule is cleaved and then bound to a new molecule of DNA. The resulting molecules will still have the same genes, but the alleles they carried have been swapped.

This recombination event begins when homologous chromosomes are brought together during prophase/prometaphase I. The homologs possess DNA sequence similarities, and are able to bind to each other (the homologs chemically recognize each other). Regions where Cross-Over (recombination) can occur begin to over lap, forming a visible structure known as a chiasma (pl. chiasmata). During cross-over(recombination), alleles are swapped between the two chromosomes, with the end result being genetically unique allele patterns on each chromosome. CRITICAL POINT: the result is two unique chromosomes! This step is critical for maintaining diversity in diploid (eukaryotic) cell systems. Ever generation inheriets a unique genetic composite of maternal and paternal alleles.

This event is one of the most critical ways that eukaryotic organisms ensure the diversity of their populations. Why?

After this recombination event, Chromosomes will line up the equitorial line (metaphase plate) so that there is one homolog on either side of the equitorial line. RECALL: in mitosis, chromosomes lined up so that there was a chromatid on either side of the metaphase plate; now in meiosis I, there is a homolog chromosome on either side. Why is this important?

Metaphase in Mitosis

Meiosis I

As you can see in the image of meiosis I, the alignment of chromosomes during Metaphase is critical. During anaphase I, the freshly recombined homologs are seperated. The cell moves from a diploid (having 2 sets of chromosomes, 2N), to a haploid state (having only 1 set of chromosomes, 1N). This is the reduction division!
Mitosis II occurs like mitosis, which is classified as an equitorial division. In an equitorial division, the chromatids are seperated. (see the difference?) NOTICE: each cell produced in meiosis I now undergo meiosis II.

During the reduction division (meiosis I), you went from a diploid to a haploid state. This is done by aligning the chromosomes in metaphase so that there is a homolog on either side of the metaphase plate. During anaphase, the homologs are seperated.

During the equitorial division (meiosis II), each chromosome is seperated into the individual chromatids. In metaphase II, chromosomes are aligned so that the chromatids are on either side of the metaphase plate (or to put it another way, they are alinged down the centromere). During anaphase II, the sister chromatids are then seperated.

The result: 4 genetically unique haploid cells!

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Daily Challenge

We have discussed the main stages of nuclear division, and you have readings from your textbook and supplemental reading on meiosis. In your own words, describe the process of meiosis, complete with a discussion of synapsis, chisamata, and cross-over. Why is the recombination event so critical to population diversity? How could recombination affect evolution?
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Daily Newsletter: October 24, 2012 - The Cell Cycle

Daily Newsletter

October 24, 2012 The Cell Cycle

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Reading:

The Eukaryotic Cell Cycle
This is an excellent overview of the cell cycle. This is considered supplemental to what is in your textbook.

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The cell cycle describes the life stages of a cell. It starts just after cytokinesis, and continues until the next cytokinetic (or mitotic) event. The period when a cell is growing between division events is known as interphase.

Interphase is divided into three distinct steps: G₁ Phase, S Phase, and G₂ Phase.

The G₁ Phase stands for Gap 1 Phase, and begins just after cytokinesis. Gap refers to no visible change in the cell. Another good term for G is growth, and that is usually what happens during this phase. The cell is building primary products, compounds such as amino acids, phospholipids, carbohydrates, triglycerides. These are describe as primary products as they are the cellular biochemicals needed to increase the cells size (biomass). Thus we generally consider the G₁ phase as the timethe cell increases in size. Another way to say this is that a major goal in this phase is the development of biomass.
[NOTE: In multicellular organisms, some cells will stop mitosis. They are locked in what is known as the G₀ phase. The G₀ phase is similar to the G₁ Phase, but the cell will never level leave this physiological stage. After reading this newletter, I want you to consider a question: Why would you stop in the G₀ Phase if you were never going to replicate again?]

When the cell gets a signal to divide, the physiology of the cell changes. Metabolic pathways for creating deoxyribonucleotides are unlocked (dATP, dTTP, dGTP, dCTP). The cell will use these deoxyribonucleotides to synthesize DNA (Replication). The production of enzymes needed to replicate DNA will also be begin. (Think back to cell signaling).

During the S Phase, cellular metabolism is focused on DNA replication (S stands for DNA Synthesis). This is a complex event, takes time, and has consequences if there is an error; so most of the cell's work is dealing with DNA replication.

Once S Phase has begun, the cell has committed its self to the process of Nuclear Division and Cytokinesis. You never replicate DNA without moving toward Nuclear division, DNA replication without nuclear division usually results in cell termination.

After DNA replication, the cell needs to build all of the proteins and compounds used during Nuclear Division and Cytokinesis. This is the goal of the G₂ Phase. The G₂ Phase will continue until the required components are constructed, and the cell receives the signal to continue. At this point the cell moves into the M phase (which stands for either mitosis or meiosis).

Regulation

The processes of the cell cycle is tightly regulated.
What happens in unregulated or uncontrolled cell growth?

The regulation is based on Signals (yes, we're back to signals). To maintain homeostasis, an organism must replace certain cells during its life time. For instance, humans replace skin and mucous membranes constantly. Hormones, such as insulin-like growth factors, are signals used to make sure the body maintains its self by replacing cells.

Inside of cells, there is an internal signal system based on the protein family Cyclin Depdendent Kinases (CDK). Note, this is a family of proteins involved in cell cycle regulation (they also have a few other functions). As the name implies, the protein needs a Cyclin to function. For example, CDK2 requires Cyclin E during G₁. (NOTE: Cyclinn is a family of protein signal molecules assoiated with the cell cycle).

CDKs are always produced in mitotically active cells. (What term do we use for a protein that is always produced?)
Cells don't always have Cyclin, but instead produce them in response to a signal. [Note: The signals differ depending on if you are dealing with multicellular, colonial or singal celled eukaryotes, so we are not going to get into specifics. As these control vital processes that could cause damage to cells, e.g., think cancer, they have a very complex signal transduction. Think of it this way, the cell has to have get numerous "permissions" before it starts producing Cyclin, so it is highly regulated. For the purpose of this class, we will just say that it is controlled by "Growth Factors".]

Throughout the cell cycle, there are time points referred to as Check Points. These check points are where regulation occurs. For example, to move from the G₁ Phase to the S phase, you need to produce a set of cyclins to induce the activity of CDKs. The Cyclin-CDK complexes can then phosphorylate proteins. Why would you need to phosphorylate proteins?

During the G₁ Phase, the check point determines if you turn on the production of deoxyribonucleotides and the production of replication complex enzymes. There are check points in the G₁ Phase, S Phase, G₂ Phase, and M Phase. Many of these check points are determinations of the health of the cell, or the correctness of DNA replication and chromosome condensation. The image to the right shows the major check points, as well as the cyclin needed to activate (Cyc D to make deoxyribonucleic acids and the replication complex, Cyce to start the S phase, Cyc A to make sure replication is occuring properly, Cyc B to make sure the cell is ready for mitosis). Note: the check point in mitosis (M phase) is not shown. The M phase check point is to ensure that chromosomes have properly condensed and migrated.
As you can tell, the check points are there to make sure that the process is occuring properly. Consider the Cyclin/CDK system as the Quality Assurance & Quality Control (QA/QC) officer of the cell.

G_o Phase (in case you missed it the first time around)
Some cells become non-mitotic at a given point in development. For example, nerve cells stop dividing, as do cardiac muscle cells. When a cell becomes non-mitotic, it shifts from the G₁ Phase to the G_o Phase. The cell can be locked from mitosis by blocking the genes for either CDK or Cyclin.

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Daily Challenge

In your own words, describe the cell cycle. Then answer the following question: How are CDKs related to cancer?
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Optional Reading

If your really into the regulation of the cell cycle, try out this article:
Cyclins and Cell Cycle Regulation