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Nucleotides

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Nucleic Acids

Nucleotide Structure: The following image from Wikipedia’s image gallery shows the basic structure of the nucleotide and the five nitrogenous bases.

The central component of all nucleotides will be a pentose sugar (5-carbon sugar). We will either see ribose or 2’deoxyribose as the sugar (the second carbon has one less oxygen than ribose). Off of the 5′ carbon of the sugar, you will find a phosphate group attached, while on the 1′ carbon, you will find a nitrogenous base. [NOTE: remember the numbering of carbon atoms in carbohydrates from yesterday? Do you see why the numbering is important?]
There are five nitrogenous bases, divided into two categories: Purines and Pyrimidines. Notice that the purines are a composite of two ring structures, while the pyrimidines are a single ring structure. When you take organic chemistry and biochemistry, the importance and complexity of these ring structures will be further discussed. At present, just become aware of their respective shapes and sizes (and the inclusion of nitrogen).

As with amino acids, the nucleotide contains a functional group: the nitrogenous base. Just like the side chain in an amino acid, the nitrogenous base will play an important part in the function of this biomolecule. The Sugar-Phosphate then becomes the backbone of the molecule (line the Amino-Chiral Carbon-Carboxyl of an amino acid). We will in later weeks that the sugar-phosphates of nucleotides will create the strands of DNA and RNA. The nitrogenous bases then playing an information role.

Base Complementarity:

The nucleic acids are referred to as informational biomolecules (biopolymers). This is because the sequence of nucleotides carries information on how to build RNA and Proteins. One of the central foundations of genetics (i.e., how it all works), is base complementarity. Here we are looking at the interactions between purines and pyrimidines:

A links with T through 2 hydrogen bonds.

G links with C through 3 hydrogen bonds.

A to T G to C

 U has the binding properties of T, but is only found in RNA.
T is never found in RNA, only DNA.
NOTE: base complementarity is a critical concept to remember. All genetic processes rely on base complementarity!

Directionality

When we get to genetics, we will be talking about the directionality of the nucleic acids. For example, we will talk about DNA being built from the 5′ to 3′. This is in reference to the carbon atoms in the ribose or deoxyribose. The 5′ holds a phosphate, while the 3′ holds an open -OH (hydroxyl) group. This concept of directionality is critical, and you are warned to learn how it works, and what the terms represent.
As with all biopolymers, monomers are added together through dehydration synthesis, and separation is through hydrolysis. When synthesis occurs, the 5′ phosphate links to the 3′ -OH, forming a phosphodiester bond.

Daily Challenge

The challenge today is to understand the history of the discovery of DNA.  Look up the following researchers and read about their discovery, how it was done, and the importance of the discovery.  In addition, watch the TED Talk from James Watson “How we discovered DNA.”  What are your impressions?

Carbohydrates and Molecular Interactions

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I would like you to bookmark a website that will be helpful for you as you move through biology and biochemistry: Molecular Interactions.   This website is a product of the Loren Dean Williams’s lab at Georgia Tech, and is an excellent resource to help you understand the molecular interactions that allow for the structure and function of biomolecules.  To start, look at the following sections of the page:

Carbohydrates

While carbohydrates are mainly used as chemical energy storage, carbohydrates are also used as modifiers of proteins and in forming cellular receptors and anchors. One of your goals is to gain a good understanding of the structure of carbohydrates, and a little about their naming.
A topic that will come up throughout the semester is how carbons are numbered in carbohydrates. This is important as we will find carbohydrates being components of monomers and when we move through the carbohydrate catabolism. The following image shows the linear form of glucose, and the two possible cyclic (pyranose ring) isomers.

http://images.tutorcircle.com/cms/images/44/glucose.png

 

The formation is based on aldehyde chemistry so we will leave some of this discussion to organic chemistry and biochemistry. For our purpose this semester, what is important is that we number carbons from the aldehyde. Notice in the above diagram that carbon 1 is to the left of the oxygen, we go around to carbon 5, and then carbon 6 is outside of the ring. If you see the expression 3′, it is referring to the third carbon. 5′ the fifth carbon. 6′ the sixth carbon, and so forth.
Notice also, that when the ring was formed, there were differences in the groups coming off of carbon 1. These differences are important and can influence how the sugar is metabolized. We say that these different forms are isomers (if you don’t know what an isomer is, look it up and add the definition to your notebook).
One critical difference comes when linking two monosaccharides together to form disaccharides and polysaccharides. For instance, here is maltose:

Maltose
https://chemstory.files.wordpress.com/2013/06/dokeo14.png

This is an α 1-4 glycosidic linkage. We have an α Maltose (look at carbon 1) bound from carbon 1 to carbon 4. Since the maltose on the left-hand side is α at the 1 carbon, we form an α linkage. In comparison, look at cellobiose:

https://biochemphilic.files.wordpress.com/2013/03/cellobios.gif
 
Cellobiose has a β 1-4 glycosidic linkage. The designation of β comes from the sugar unit that donates carbon 1 to the bond.
 

So, what is the big deal? Maltose is digestible by humans, cellobiose is not. Just this slight isomeric difference changes the metabolism.

All carbohydrate monomers are connected through glycosidic linkages, whether it is a disaccharide, oligosaccharide or a polysaccharide. Make sure that you learn the different types of carbohydrates.

Molecular Visualization

Here are four different ways of visualizing α-D-Glucose.  The first is a ball-n-stick model, with the black spheres representing carbon, the red oxygen, and the grey hydrogen.  This is a useful way to begin understanding the 3-D orientation and structure of the molecule.  If you notice, the other models do not always explicitly state where the carbon is located.  Instead, the where lines intersect are where you would find carbon.  This is done to create a simplified diagram that allows you to see the geometry of the molecule.
https://upload.wikimedia.org/wikipedia/commons/8/8a/Alpha-D-Glucose.png
Glucose model – rotatable in 3 dimensions is a good place to go to gain a good visual impression of glucose.
 
Please start gaining a good visual of the biomolecules.

Challenge

Draw the structure for Ribose and Deoxyribose.   Number the carbon atoms.
Structurally, what is different between ribose and deoxyribose?
What are the differences in the electrochemistry of the molecule (where do you find charged and partially charged regions for example).