number Done by Corrected by Doctor Nafeth Abu Tarboush

Transcription

1 number 6 Done by أنس القيشاوي Corrected by Zaid Emad Doctor Nafeth Abu Tarboush 1 P a g e

2 In the previous lecture, we talked about redox reactions and the reduction potential briefly and how it can help us to determine the direction of electron movement, and in this lecture we will complete studying this important topic. ΔE ΔE ( E = EA ED) is the difference in reduction potential, where: EA is the reduction potential of the electron acceptor. ED is the reduction potential of the electron donor. Does it determine the direction of the reaction? We can determine the direction of the reaction depending on the potential difference by calculating ΔG which is directly related to ΔE by the equation: The difference between ΔG & ΔE is that the sign of ΔE for spontaneous reactions is +ve while ΔG is ve. You need to know that electrons will be transported across the Electron Transport chain according to the potential difference and at the end of this movement of electrons will generate energy. Before we start talking about Krebs cycle, we will discuss what we mean by the electron carrier molecules (NAD+ & FAD). There are a lot of molecules that can carry electrons, these include the Niacin (vitamin B3) based molecules and the Flavin (vitamin B2) based molecules. 2 P a g e

3 1. The Niacin based molecules ( NAD+ & NADP+ ) The only difference between NAD+ & NADP+ is the phosphate group attached to Carbon number 2 of ribose in adenosine. The electron acceptor in both molecules is Nicotinic acid which accepts two electrons in the form of hydride ion. So both of them perform the same function. 2. The Flavin based molecules ( FAD & FMN ) The only difference between FAD & FMN is that FMN doesn t have the adenosine part as FAD,so it is composed of Flavin only. Both of these molecules accept 2 electrons as 2 hydrogen atoms and electron addition occurs on the free Nitrogen atoms of the Flavin ring. 3 P a g e

4 Unlike the Niacin based molecules, Flavin based molecules accept electrons sequentially (one by one), therefore, they can pass in the state of free radicals (one unpaired electron) which is a very dangerous state that can affect any component of the cell and damage it. As a result, Flavin based molecules will always be found hidden by enzymes or other structures, that s the reason why we can determine the reduction potential of NAD+ but we can t do the same with FAD without identifying the structure to which it is attached. Additionally, the energy released by the movement of electrons from NADH to Oxygen (the final acceptor of electrons in the Electron Transport Chain) is higher than that of FADH2. 4 P a g e

5 (Kreb s, Citric Acid, TCA) Cycle *The name..? Kreb s cycle: the biochemist who discovered it. Citric acid cycle: derived from the citric acidwhich is the first molecule in the cycle. Tricarboxylic acid (TCA) cycle: because citric acid has 3 carboxylic groups. Some important concepts: Fats, proteins and carbohydrates will be digested giving a common molecule (Acytel CoA; Two-Carbon units) which will enter the cycle and finally will come out as (((single))) CO2, which is the same CO2 we exhale. This process takes place in the matrix of the mitochondria. The cycle has 8 steps, 8 reactions, 8 enzymes and 8 intermediates. When we want to study this cycle, we divide it into two halves. 5 P a g e

6 Please refer to the figure to understand the mechanism of this cycle. You also need to memorize the chemical structure of each compound. The first half of the cycle has 4 steps, and it will engage itself in trying to convert the 6-carbon molecule which was formed by joining Acetyl CoA(two-carbon unit) and oxaloacetate (4-carbon unit),into a 4-carbon molecule. The second half will try to rearrange the 4-carbon compound to form the starting 4-carbon compound (oxaloacetate). 1. Bulding up of citrate:acetyl CoA will bind to oxaloacetate to produce citrate (Citric acid). This step requires energy that comes from the detachment of the CoA. Citric acid has 6 carbons, 3 carboxylic groups. Can citric acid be oxidized? No, because citric acid has 3 carboxylic groups, and carboxylic group is the highest state of oxidation, so it won t be oxidized.it also has an alcohol group, but it won t be oxidized because it s a tertiary alcohol. The only thing we can do is change the position of OH group by an isomerisation reaction. The enzyme used to produce citrate is Citrate synthase. 2. Isomeration of citrate:in this step we change the structure by converting the tertiary alcohol into secondary alcohol (which can be oxidised easily into a ketone). The enzyme used in this step is Aconitase. This enzyme is named like this because there is an intermediate compound called Aconitate. 3. Decarboxylation of isocitrate:in this step we remove a carboxylic group (CO2) from isocitrate converting it from a 6-carbon unit molecule into a 5-carbon unit molecule (α-ketoglutarate). 6 P a g e

7 Oxidation of this secondary alcohol (isocitrate) will produce a ketone molecule, as we remove two hydrogen atoms making a double bond between carbon and oxygen. The first hydrogen atom will bind to NAD+ producing NADH. (The first NADH that comes out from the cycle). The enzyme used in this step is Isocitrate dehydrogenase. This reaction is an oxidative decarboxylation reaction. This step is the slowest step in the citric acid cycle, which means it is the rate limiting step of this cycle so it has the highest regulation among all of the steps of the cycle. 4. Decarboxylation of α-ketoglutarate:this step has the same mechanism as the previous one. We remove CO2 from α-ketoglutarate converting it from a 5-carbon unit molecule into a 4-carbon unit molecule (Succinyl CoA). When we remove the peripheral CO2, we created a peripheralcarbonyl group which is very unstable, so it will bind to CoA from the solution producing Succinyl CoA. The enzyme used in this step is called α-ketoglutarate dehydrogenase. In this step the second NADH molecule is produced. Here, the removal of hydrogen atoms from α-ketoglutarate isn t obvious as it occurs in an intermediate step. 5. Conversion of Succinyl CoA into Succinate: The first thing to do is to remove CoA. When you remove the CoA, you get a molecule which contains two carboxylic groups &two CH2 groups. The detachment of CoA will release energy that will be used in adding an inorganic phosphate to the GDP or ADP molecule producing GTP or ATP. In this step we formed ATP molecule without the need of oxygen, this process is called (substrate level phosphorylation). The enzyme that catalyses this step is Succinate thiokinase. How is this molecule different from oxaloacitate? o The difference is that we need to change the CH2 group presented in Succinate into a keto group (carbonyl group). 7 P a g e

8 6. Oxidation of Succinate: In this step we oxidise succinate into Fumarate by the removal of 2 Hydrogen atoms forming a double bond between the 2 nd and 3 rd carbons by a dehydrogenation reaction. The two hydrogen atoms will be used in reducing one FAD molecule into FADH2. The enzyme used for this step is called Succinate dehydrogenase. 7. Hydration of Fumarate: In this step we add H2O molecule in order to break the double bond producing an alcoholic compound (secondary alcohol) which is called Malate The enzyme that catalyses this step is called Fumarase. 8. Oxaloacetate regeneration: In this step, we oxidize Malate producing a double bound converting Malate into oxaloacetate, which is the starting compound for the cycle. One of the H atoms will bind to NAD+ as a hydride ion producing NADH (the 3 rd NADH produced by this cycle), and the remaining proton will go to the solution The enzyme that catalyses this step is called Malate dehydrogenase. The final products of the cycle: 1 ATP, 3 NADH, 1 FADH2, 2 CO2 This table will help you memorize the substrates, products & enzymes for each step. 8 P a g e

9 The End *Don t hesitate to contact me if u have any question regarding this lecture! The struggle you re in today is developing the strength you need for tomorrow!! 9 P a g e