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October 1, 2012 Glycolysis steps 1-5

Today we start looking at the metabolic pathway of glycolysis. In the textbook, this information is catagorized as how cells harvest energy, but cells harvest carbon as well with these reactions. A question that every cell asks moment by moment is: What do I need now, carbon or energy? Intermediates from our catabolic pathways provide precursors for other biochemicals, but they also provide energy in the form of reducing potential.

With digital technology, you can go on line and pull down images of the glycolytic pathway. Nearly every biology textbook has these images. Your goal this weak is NOT to memorize these metabolic reactions. Your goal is to understand what is happening. I would like you to also learn something of the enzymes being used.

Glycolysis describe the splitting of glucose into two three-carbon Pyrvate molecules. This pathway consists of 10 reactions that carry out this splitting by inducing specific changes into the molecule. The first five steps are classified as preparatory, or energy consumptive. In these steps, we are preparing the molecule of glucose for the first split.

Glucose is chemical stable. Glucose does not spontaneously explode or degrade. Chemical stability also implies that it does not react easily. So, we need to make it more reactive. We also need to get it into the correct configuration for splitting. That is the goal of the first four steps.

In step 1, we use the enzyme hexokinase (specifically Glucokinase). The root word here is kinase. A kinase is an enzyme that adds a phosphate group to a molecule. In this case, hexokinase is an enzyme that adds a phosphate group to a six carbon sugar, in this case glucose. You will notice in the above picture that the enzyme is referred to as glucokinase. Glucokinase is a specific hexokinase (Hexokinase IV), and is found in specific mammalian cells such as the intestines, liver, pancreas, and brain.

Hexokinase is generally attached to the glucose carrier found in the cell membrane. When glucose is brought into the cell, hexokinase adds a phosphate group to the sixth carbon. Glucose 6-P refers to a glucose molecule with a phosphate on the sixth carbon. In order to carry out this phosphorylation, we use ATP. ATP transfers a phosphate to the sixth carbon of glucose. (enzymatically, how would this happen? Would you need both ATP and Glucose in the active site?)

Why do we need to phosphorylate glucose? This is an important question, and something you should ask for every metabolic reaction. Why do we need to do it? What is the end product? You should also ask yourself questions about the enzyme.

Why do you phosphorylate glucose?

Glucose is stable, so the addition of phosphate with its -2 charge causes an electrical instability in the molecule.
Glucose 6P is more reactive than Glucose.
Glucose 6P can not leave the cell through the Glucose Carrier (they are different molecules now).
Glucose 6P does not interfere with the concentration gradient of Glucose (they are different molecules, each with a concentration gradient).
Glucose then remains high on the outside of the cell, but almost zero inside of the cell (incredibly strong concentration gradient).

What about the enzyme?

Is it regulated?

Unidentified allosteric regulation.

What is the structure?

465 amino acids

What does the active site look like?

Active site fits Glucose and ATP.

What about the activity?

This first reaction has a ΔG^o' of approximately -4 kcal/mol. This indicates a spontaneous reaction, but as we have discussed above, the reaction is not spontaneous, so why such a large negative ΔG^o'? Coupled Reactions. You are seeing the overall reaction in the above image, but there are two reactions taking place: removal of Phosphate from ATP, and the addition of Phosphate to Glucose. Remember that in a lab, ATP yeilds a large release of energy. In biology, this energy is coupled to the building of the phosphate bond to glucose. ATP helps to drive the reaction of the stable glucose molecule.

In the second reaction of glycolysis, the enzyme phosphoglucose isomerase (Glucose-6-phosphate isomerase) causes the glucose molecule to rearrange slightly into the the hexose isomer fructose.

Glucose and Fructose have the same chemical formula, but different chemical structures. Notice that glucose forms a six sided ring, with the sixth carbon outside of the ring. Fructose forms a five sided ring, with the 1^st and 6^th carbons outside of the ring structure. Why is that important?

The ΔG^o'for this reaction is +0.4 kcal/mol; so in a lab it is not expected to proceed at significant levels. Enzymes though allow for this reaction to proceed at significant rates. Of course, a build up of fructose-6-phosphate would disrupt the dynamic equilibrium we want, so we move this product to become the substrate of the next reaction.

Reaction 3 is catalyzed by phosphofructokinase (remember this from last week?). This enzyme will add a phosphate group onto the 1^st carbon, and the reaction has a ΔG^o'of -3.4 kcal/mol.

Notice that the free energy is less than in the first reaction. Why? Think about what is happening; we are adding a second phosphate to a molecule that already has a phosphate. We are adding a large -2 charge to a molecule that already has a large -2 charge. This is not energetically favorable, so more of the energy from ATP is needed to get this reaction to go.

Now look at the resulting molecule: Fructose 1,6-bis phosphate. There are two carbons sticking out of the ring structure. In the image, you are seeing them on the same plane (above the ring). The phosphates have a -2 charge. What are they doing to the molecule? Fructose 1,6-bisphophaste

(Hint: Molecular Tension). To the right is another image of fructose-1,6,-bisphophate to help you visualize the resulting tension in this molecule. NOTE: bis is an archaic prefix to indicate two or second instance of something. In modern English, we usually us bi-, but organic chemistry (and more importantly IUPAC) still uses the bis- prefix.

In the fourth reaction, we see the splitting of the glucose molecule. The molecular tension is focused on the covalent bond between the 3^rd and 4^th carbon. Aldolase is the enzyme that catalyzes this reaction. This reaction deals with some specific organic chemistry, so we will hold off on the full discussion at this time.

The goal of the reaction is the splitting of glucose into two 3-carbon isomers: Glyceraldehyde 3-P and Dihydroxyacetone Phosphate (DHAP). The problem here is that DHAP is a metabolic dead in in our current pathways. We do no want to loose the energy and carbon held in DHAP, so we need to convert it into something useful. That is the goal of reaction 5.

NOTE: the ΔG^o'of reaction 4 is generally reported at +5.7 kcal/mol. Remember that these numbers are from isolated reactions. Covalent bonds are strong and stable; even with the molecular tension of 2 phosphate groups, the bond between carbon 3 and 4 does not want to break. That is where the enzyme aldolase comes in. If you remember from lecture, we talked about what can happen at an active site. One of the options was inducing physical stress and charges. What would happen if you had complementary positive charges for the phosphate pulling away from the fructose? What would happen if you physically pulled these regions apart? Would it be easier to break the covalent bond?
In reaction 5, we convert DHAP into a more usable form, namely Glyceraldehyde 3-Phosphate (G3P is a common acronym). The enzyme that carries out this reaction is shown in the above diagraph: Triosphosphate Isomerase.

At the end of these first five steps, two phosphates have been added to a hexose in order to destabilize the sugar and cause it to break in a very specific manner (between carbon 3 and 4). The result is that we have two 3-carbon sugars: G3P. Notice the oxidation state of the molecule. There are a number of hydrogens attached directly to carbon. There is energy that we can harvest in the form of reducing potential. Tomorrow we will continue with steps 6-10.

Daily Challenge

In your own words, describe steps 1-5 of glycolysis. Discuss how the enzymes make the reactions possible, including how they can couple reactions (this is important for any reaction with ATP). Discuss how the stepwise change in the molecule allows for the localization of the bond that needs to be broken. Why don't we just break that bond immediately? Remember, this will become part of your next milestone paper.
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