Answer Key for Paper Details: S0123 / S.Y.B.Sc (Choice Base) Sem III / S2015-Life Science: Paper II Date: 26-03-2018 Time: 02:00 pm - 05:00 pm Q.P.Code: 20183 Q1 A State whether True or False:

1 – F, 2 – T, 3 – T, 4 – F, 5 – F, 6 – T, 7 – F

Q.1.B. Match the column: i)—a ii)—e iii)---d iv) ----f v)---c vi)----g vii)---b

Q. 1 C Answer The following

i) Name any two essential amino acids: Isoleucine, Histidine, Leucine, Methionine, Lysine, Phenylalanine, Tryptophan, Threonine, Valine ii) Name any two uricotelic animals: protozoans, crustaceans, platyhelminths, cnidarians,poriferans, echinoderms, fishe s, larvae / tadpoles of amphibians iii) Name any two ketone bodies: acetoacetate, beta-hydroxybutyrate, acetone

Q2 A a) Explain the different types of enzyme inhibitions with one suitable example each. 10mks Enzymes can be blocked or modulated from performing their normal role by inhibiting them using molecules. Inhibitors can be classified into three groups based on their mechanism of action: competitive, uncompetitive and mixed inhibitors. The type of inhibition can be determined through enzyme kinetic measurements.

Each of the following to be explained along with general chemical equation and Graph with Any one example

Competitive inhibition, the substrate and inhibitor cannot bind to the enzyme at the same time,

(Vmax remains constant and the apparent Km will increase

Uncompetitive inhibition, the inhibitor binds only to the substrate-enzyme complex.

Vmax to decreases andKm decreases

Non-competitive inhibition, the binding of the inhibitor to the enzyme reduces its activity but does not affect the binding of substrate.

Vmax decreases but Km remains the same

Q2 A b) What is spectrophotometry? Explain the principle, working and application

using a suitable diagram.

*Quantitation of light absorption by solutions

*Principle – based on monochromatic incident light generated by prism of complementary colour being absorbed by test solution

*Diagram indicating the following

*Working of Spectrophotometer to be explained

*Applications: Industries including semiconductors, laser and optical manufacturing, printing and forensic examination, as well in laboratories for the study of chemical substances. Spectrophotometry is often used in measurements of enzyme activities, determinations of protein concentrations, determinations of enzymatic kinetic constants, and measurements of ligand binding reactions

Q2 B a) Explain the technique of Ionexchange chromatography

Principle is to separate on basis of charge ―adsorption‖ Factors:: •Net charge of protein at the pH of elution buffer •column made of negatively charged cation exchangers •more negative the proteins faster elution •Most frequent chromatographic technique for protein purification Diagram of set up

Q2 B b) Explain the effect of i) pH and ii) Temperature on enzyme activity

Explanation of the graphs

Q2 B c) Explain the Lineweaver – Burke plot with a diagram. State its significance :

* Michaelis Mentens equation converted to reciprocal units

*Graph plotted as reciprocal of Substrate on x axis and of Velocity in the y axis

1/vo

1 / Vmax

1/[S]

- 1 / Km

Explanation of the above terms in the graph and terms indicated Significance * Straight line yields better kinetic values *Km affinity of the enzyme for the substrate, compare and predict the most suitable substrate for a particular enzyme.

*Vmax, indicates the number of substrate molecules converted into product by an enzyme molecule in a unit time when the enzyme is fully saturated with substrate.

Q2 B d) a) Write a note on i) Coenzyme and ii) Cofactor write an example each.

 Coenzymes are organic molecules that are required by certain enzymes to carry out catalysis. They bind to the active site of the enzyme and participate in catalysis but are not considered substrates of the reaction. They often function as intermediate carriers of electrons, specific atoms or functional groups that are transfered in the overall reaction. examples– NAD / FAD / CoQ/ CoA/ any other suitable  Cofactors are inorganic substances that are required by certain enzymes to carry out catalysis. examples–Fe/Zn/Cu/ any other suitable

Q.3. A. Glycolysis- 10 steps explained steps by steps with diagram

  Glycolysis is the metabolic process that serves as the foundation for both aerobic and anaerobic cellular respiration. In glycolysis, glucose is converted into pyruvate. Glucose is a six- memebered ring molecule found in the blood and is usually a result of the breakdown of carbohydrates into sugars. It enters cells through specific transporter proteins that move it from outside the cell into the cell’s cytosol. All of the glycolytic enzymes are found in the cytosol.  The overall reaction of glycolysis which occurs in the cytoplasm is represented simply as: +  C6H12O6 + 2 NAD + 2 ADP + 2 P —–> 2 pyruvic acid, (CH3(C=O)COOH + 2 ATP + 2 NADH + 2 H+  Step 1: Hexokinase   The first step in glycolysis is the conversion of D-glucose into glucose-6-phosphate. The enzyme that catalyzes this reaction is hexokinase.  Details:  Here, the glucose ring is phosphorylated. Phosphorylation is the process of adding a phosphate group to a molecule derived from ATP. As a result, at this point in glycolysis, 1 molecule of ATP has been consumed.  The reaction occurs with the help of the enzyme hexokinase, an enzyme that catalyzes the phosphorylation of many six-membered glucose-like ring structures. Atomic magnesium (Mg) is also involved to help shield the negative charges from the phosphate groups on the ATP molecule. The result of this phosphorylation is a molecule called glucose-6-phosphate (G6P), thusly called because the 6′ carbon of the glucose acquires the phosphate group.  Step 2: Phosphoglucose Isomerase

  The second reaction of glycolysis is the rearrangement of glucose 6-phosphate (G6P) into fructose 6-phosphate (F6P) by glucose phosphate isomerase (Phosphoglucose Isomerase).  Details:  The second step of glycolysis involves the conversion of glucose-6-phosphate to fructose-6-phosphate (F6P). This reaction occurs with the help of the enzyme phosphoglucose isomerase (PI). As the name of the enzyme suggests, this reaction involves an isomerization reaction.  The reaction involves the rearrangement of the carbon-oxygen bond to transform the six-membered ring into a five-membered ring. To rearrangement takes place when the six-membered ring opens and then closes in such a way that the first carbon becomes now external to the ring.  Step 3: Phosphofructokinase   Phosphofructokinase, with magnesium as a cofactor, changes fructose 6-phosphate into fructose 1,6-bisphosphate.  Details:  In the third step of glycolysis, fructose-6-phosphate is converted to fructose- 1,6- bisphosphate (FBP). Similar to the reaction that occurs in step 1 of glycolysis, a second molecule of ATP provides the phosphate group that is added on to the F6P molecule.  The enzyme that catalyzes this reaction is phosphofructokinase (PFK). As in step 1, a magnesium atom is involved to help shield negative charges.  Step 4: Aldolase

  The enzyme Aldolase splits fructose 1, 6-bisphosphate into two sugars that are isomers of each other. These two sugars are dihydroxyacetone phosphate (DHAP) and glyceraldehyde 3-phosphate (GAP).  Details:  This step utilizes the enzyme aldolase, which catalyzes the cleavage of FBP to yield two 3-carbon molecules. One of these molecules is called glyceraldehyde-3-phosphate (GAP) and the other is called dihydroxyacetone phosphate (DHAP).  Step 5: Triphosphate isomerase   The enzyme triophosphate isomerase rapidly inter- converts the molecules dihydroxyacetone phosphate (DHAP) and glyceraldehyde 3-phosphate (GAP). Glyceraldehyde phosphate is removed / used in next step of Glycolysis.  Details:  GAP is the only molecule that continues in the glycolytic pathway. As a result, all of the DHAP molecules produced are further acted on by the enzyme triphoshpate isomerase (TIM), which reorganizes the DHAP into GAP so it can continue in glycolysis. At this point in the glycolytic pathway, we have two 3-carbon molecules, but have not yet fully converted glucose into pyruvate.  Step 6: Glyceraldehyde-3-phosphate Dehydrogenase

  Glyceraldehyde-3-phosphate dehydrogenase (GAPDH) dehydrogenates and adds an inorganic phosphate to glyceraldehyde 3-phosphate, producing 1,3- bisphosphoglycerate.  Details:  In this step, two main events take place: 1) glyceraldehyde-3-phosphate is oxidized by the coenzyme nicotinamide adenine dinucleotide (NAD); 2) the molecule is phosphorylated by the addition of a free phosphate group. The enzyme that catalyzes this reaction is glyceraldehyde-3-phosphate dehydrogenase (GAPDH).  The enzyme GAPDH contains appropriate structures and holds the molecule in a conformation such that it allows the NAD molecule to pull a hydrogen off the GAP, converting the NAD to NADH. The phosphate group then attacks the GAP molecule and releases it from the enzyme to yield 1,3 bisphoglycerate, NADH, and a hydrogen atom.  Step 7: Phosphoglycerate Kinase

  Phosphoglycerate kinase transfers a phosphate group from 1,3-bisphosphoglycerate to ADP to form ATP and 3-phosphoglycerate.  Details:  In this step, 1,3 bisphoglycerate is converted to 3-phosphoglycerate by the enzyme phosphoglycerate kinase (PGK). This reaction involves the loss of a phosphate group from the starting material. The phosphate is transferred to a molecule of ADP that yields our first molecule of ATP. Since we actually have two molecules of 1,3 bisphoglycerate (because there were two 3-carbon products from stage 1 of glycolysis), we actually synthesize two molecules of ATP at this step. With this synthesis of ATP, we have cancelled the first two molecules of ATP that we used, leaving us with a net of 0 ATP molecules up to this stage of glycolysis.  Again, we see that an atom of magnesium is involved to shield the negative charges on the phosphate groups of the ATP molecule.  Step 8: Phosphoglycerate Mutase

  The enzyme phosphoglycero mutase relocates the P from 3- phosphoglycerate from the 3rd carbon to the 2nd carbon to form 2-phosphoglycerate.  Details:  This step involves a simple rearrangement of the position of the phosphate group on the 3 phosphoglycerate molecule, making it 2 phosphoglycerate. The molecule responsible for catalyzing this reaction is called phosphoglycerate mutase (PGM). A mutase is an enzyme that catalyzes the transfer of a functional group from one position on a molecule to another.  The reaction mechanism proceeds by first adding an additional phosphate group to the 2′ position of the 3 phosphoglycerate. The enzyme then removes the phosphate from the 3′ position leaving just the 2′ phosphate, and thus yielding 2 phsophoglycerate. In this way, the enzyme is also restored to its original, phosphorylated state.  Step 9: Enolase

  The enzyme enolase removes a molecule of water from 2-phosphoglycerate to form phosphoenolpyruvic acid (PEP).  Details:  This step involves the conversion of 2 phosphoglycerate to phosphoenolpyruvate (PEP). The reaction is catalyzed by the enzyme enolase. Enolase works by removing a water group, or dehydratingthe 2 phosphoglycerate. The specificity of the enzyme pocket allows for the reaction to occur through a series of steps too complicated to cover here.  Step 10: Pyruvate Kinase   The enzyme pyruvate kinase transfers a P from phosphoenolpyruvate (PEP) to ADP to form pyruvic acid and ATP Result in step 10.  Details:  The final step of glycolysis converts phosphoenolpyruvate into pyruvate with the help of the enzyme pyruvate kinase. As the enzyme’s name suggests, this reaction involves the transfer of a phosphate group. The phosphate group attached to the 2′ carbon of the PEP is transferred to a molecule of ADP, yielding ATP. Again, since there are two molecules of PEP, here we actually generate 2 ATP molecules.

 Steps 1 and 3 = – 2ATP Steps 7 and 10 = + 4 ATP Net ―visible‖ ATP produced = 2.

Q.3.B. The citric acid cycle is a key component of the metabolic pathway by which all aerobic organisms generate energy. Through catabolism of sugars, fats, and proteins, a two-carbon organic product acetate in the form of acetyl-CoA is produced. Acetyl-CoA along with two equivalents of water (H2O) is consumed by the citric acid cycle producing two equivalents of carbon dioxide (CO2) and one equivalent of HS-CoA. In addition, one complete turn of the cycle converts three equivalents of nicotinamide adenine dinucleotide (NAD+) into three equivalents of reduced NAD+ (NADH), one equivalent of ubiquinone (Q) into one equivalent of reduced ubiquinone (QH2), and one equivalent each of guanosine diphosphate (GDP) and inorganic phosphate (Pi) into one equivalent of guanosine triphosphate (GTP). The NADH and QH2 generated by the citric acid cycle are in turn used by the oxidative phosphorylation pathway to generate energy-rich adenosine triphosphate (ATP). One of the primary sources of acetyl-CoA is sugars that are broken down by glycolysis to produce pyruvate that in turn is decarboxylated by the enzyme pyruvate dehydrogenase generating acetyl-CoA according to the following reaction scheme:[1] CH3C(=O)C(=O)O– (pyruvate) + HSCoA + NAD+ → CH3C(=O)SCoA (acetyl- CoA) + NADH + CO2 The product of this reaction, acetyl-CoA, is the starting point for the citric acid cycle. Below is a schematic outline of the cycle:[1] The citric acid cycle begins with the transfer of a two-carbon acetyl group from acetyl-CoA to the four-carbon acceptor compound (oxaloacetate) to form a six-carbon compound (citrate). The citrate then goes through a series of chemical transformations, losing two carboxyl groups as CO2. The carbons lost as CO2 originate from what was oxaloacetate, not directly from acetyl-CoA. The carbons donated by acetyl -CoA become part of the oxaloacetate carbon backbone after the first turn of the citric acid cycle. Loss of the acetyl-CoA-donated carbons as CO2 requires several turns of the citric acid cycle. However, because of the role of the citric acid cycle in anabolism, they may not be lost, since many TCA cycle intermediates are also used as precursors for the biosynthesis of other molecules.Most of the energy made available by the oxidative steps of the cycle is transferred as energy-rich electrons to NAD+, forming NADH. For each acetyl group that enters the citric acid cycle, three molecules of NADH are produced. Electrons are also transferred to the electron acceptor Q, forming QH2. At the end of each cycle, the four-carbon oxaloacetate has been regenerated, and the cycle continues. Q.3.B. a) Alcohol fermentation, also known as ethanol fermentation, is the anaerobic pathway carried out by yeasts in which simple sugars are converted to ethanol and carbon dioxide. Yeasts typically function under aerobic conditions, or in the presence of oxygen, but are also capable of functioning under anaerobic conditions, or in the absence of oxygen. When no oxygen is readily available, alcohol fermentation occurs in the cytosol of yeast cells. Let's explore the process of alcohol fermentation then see what it means for yeasts and for humans.

 The Process of Alcohol Fermentation  The basic equation for alcohol fermentation shows that yeast starts with glucose, a type of sugar, and finishes with carbon dioxide and ethanol. However, to better understand the process, we need to take a look at some of the steps that take us from glucose to the final products.  The process of alcohol fermentation can be divided into two parts. In the first part, the yeast breaks down glucose to form 2 pyruvate molecules. This part is known as glycolysis. In the second part, the 2 pyruvate molecules are converted into 2 carbon dioxide molecules and 2 molecules of ethanol, otherwise known as alcohol. This second part is called fermentation.  The main purpose of alcohol fermentation is to produce ATP, the energy currency for cells, under anaerobic conditions. So from the yeast's perspective, the carbon dioxide and ethanol are waste products. That's the basic overview of alcohol fermentation. Now, let's examine each part of this process in greater detail.

 In the first part of this process, each glucose molecule is broken down into 2 pyruvate molecules. Pyruvate, or pyruvic acid, is an amino acid and will help form ethanol. In the process of breaking glucose down to form pyruvate, several molecules known as electron acceptors are involved.  Electron acceptors are molecules whose job is to give and take the electrons released when a chemical reaction takes place. During this first part, an electron acceptor molecule called NAD+ is reduced to form NADH, gathering up the electrons released by breaking one glucose down to 2 pyruvate molecules. This exchange of electrons that occurs while glucose is being broken down is essentially what helps build ATP.  The conversion of glucose to pyruvate creates a net total of 2 ATP. While this isn't as much ATP as aerobic respiration can produce, it's enough to keep the yeast alive until oxygen is available. This first part may look familiar because it's essentially glycolysis, or the first stage of aerobic respiration.  If oxygen were present, then the pyruvate molecules would enter a mitochondrion to undergo the remainder of aerobic respiration. However, in alcohol fermentation, the pyruvate instead stays in the cytosol, the gooey interior space of the cell. This is where the second part of our reaction, the conversion of pyruvate to ethanol, will take place. Q.3.B.b.)

Chemiosmosis is the movement of ions across a semipermeable membrane, down their electrochemical gradient. An example of this would be the generation of adenosine triphosphate (ATP) by the movement of hydrogen ions across a membrane during cellular respiration or photosynthesis.

 The Chemiosmosis Theory and the Generation of ATP by ATP Synthase. Thechemiosmotic theory explains the functioning of electron transport chains. According to this theory, the tranfer of electrons down an electron transport system through a series of oxidation-reduction reactions releases energy .  Chemiosmosis is one of the processes by which ATP is synthesized. In eukaryotes, it takes place in the mitochondria during cellular respiration and in the chloroplasts during photosynthesis. In prokaryotes, it occurs in the cell membrane.

Q.3.B.c)

 The enzyme cytochrome c oxidase or Complex IV, EC 1.9.3.1 is a large transmembrane protein complex found in bacteria and the mitochondrion of eukaryotes. It is the last enzyme in the respiratory electron transport chain of mitochondria (or bacteria) located in the mitochondrial (or bacterial) membrane.  Complex IV (Cytochrome c oxidase). This final complex in the electron transport chain accomplishes the final transfer of the electrons to oxygen and pumps two protons across the membrane. This makes a total of 10 protons across the membrane for one NADH into the electron transfer chain. ATP Synthase.  Cytochrome c oxidase: The enzyme cytochrome c oxidase or Complex IV, is a large transmembrane protein complex found in bacteria and the mitochondrion of eukaryotes. It is the last enzyme in the respiratory electron transport chain of mitochondria (or bacteria) located in the mitochondrial (or bacterial) membrane. It receives an electron from each of four cytochrome c molecules, and transfers them to one oxygen molecule, converting molecular oxygen to two molecules of water. In the process, it binds four protons from the inner aqueous phase to make water, and in addition translocates four protons across the membrane, helping to establish a transmembrane difference of proton electrochemical potential that the ATP synthase then uses to synthesize ATP.

Q.3.B.d)

Pyruvate degradation occurs in the mitochondria. Pyruvate is produced by glycolysis in the cytosol, while PDH and all subsequent degradative steps are located in the mitochondria. Therefore, pyruvate needs to be transported from the cytosol to the mitochondrial matrix.

 Pyruvate dehydrogenase complex (PDC) is a complex of three enzymes that converts pyruvate into acetyl-CoA by a process called pyruvate decarboxylation. Acetyl-CoA may then be used in the citric acid cycle to carry out cellular respiration, and this complex links the glycolysis metabolic pathway to the citric acid cycle. Pyruvate decarboxylation is also known as the "pyruvate dehydrogenase reaction" because it also involves the oxidation of pyruvate.  This multi-enzyme complex is related structurally and functionally to the oxoglutarate dehydrogenase and branched-chain oxo-acid dehydrogenase multi- enzyme complexes.  Pyruvate dehydrogenase (E1)

 Initially, pyruvate and thiamine pyrophosphate (TPP or vitamin B1) are bound by pyruvate dehydrogenasesubunits. The thiazolium ring of TPP is in a zwitterionic form, and the anionic C2 carbon performs a nucleophilic attack on the C2 (ketone) carbonyl of pyruvate. The resulting hemithioacetal undergoes decarboxylation to produce an acyl anion equivalent (see cyanohydrin or aldehyde-dithiane umpolung chemistry, as well as benzoin condensation). This anion attacks S1 of an oxidized lipoate species that is attached to a lysineresidue. In a ring- opening SN2-like mechanism, S2 is displaced as a sulfide or sulfhydryl moiety. Subsequent collapse of the tetrahedral hemithioacetal ejects thiazole, releasing the TPP cofactor and generating a thioacetate on S1 of lipoate. The E1-catalyzed process is the rate-limiting step of the whole pyruvate dehydrogenase complex.  Dihydrolipoyl transacetylase (E2)

 At this point, the lipoate-thioester functionality is translocated into the dihydrolipoyl transacetylase (E2) active site, where a transacylation reaction transfers the acetyl from the "swinging arm" of lipoyl to the thiol of coenzyme A. This produces acetyl- CoA, which is released from the enzyme complex and subsequently enters the citric acid cycle. E2 can also be known as lipoamide reductase-transacetylase.  Dihydrolipoyl dehydrogenase (E3)

 The dihydrolipoate, still bound to a lysine residue of the complex, then migrates to the dihydrolipoyl dehydrogenase (E3) active site where it undergoes a flavin-mediated oxidation, identical in chemistry to disulfide isomerase. First, FAD oxidizes dihydrolipoate back to its lipoate resting state, producing FADH2. Then, + a NAD cofactor oxidizes FADH2 back to its FAD resting state, producing NADH.

Q. 4 A) Explain in detail any one of the following: 10mks i) Beta – oxidation of fatty acids The individual reactions involved in the degradation of fatty acids by -oxidation are as follows: 1. Oxidation: Oxidation of the fatty acyl CoA to enoyl CoA forming a trans 2 – double bond on the fatty acyl chain and producing FADH2 (catalyzed by acyl CoA dehydrogenase).

2. Hydration: Hydration of the trans 2 – enoyl CoA to form 3 –hydroxyacyl CoA (catalyzed by enoyl CoA hydratase).

3. Oxidation: Oxidation of 3-hydroxyacyl CoA to 3 –ketoacyl CoA producing NADH (catalyzed by hydroxyacyl CoA dehydrogenase).

4. Thiolytic Cleavage or Thiolysis: Thiolysis of 3 – ketoacyl CoA by a secind CoA molecule,giving acetyl CoA and an acyl CoA shortened by two carbon atoms (catalyzed by  - ketothiolase ).

OR

ii) Enzymatic steps involved in the production of urea.

B) Answer any two of the following: (10) i) Explain: Glucose – Alanine cycle

ii) Explain : Carnitine Shuttle Small and medium chain acyl CoA molecules (upto 10 carbon atoms) are readily able to cross the inner mitochondrial membrane by diffusion. However, larger chain acyl CoAs do not readily cross the inner mitochodrial membrane, and require a specific transport mechanism. To achieve this, the longer chain acyl CoAs are conjugated to the polar carnitine molecule, that is found in both plants and animals. This reaction, catalyzed by an enzyme on the outer face of the inner mitochodrial membrane (carnitine acetyltransferase I ), removes the CoA and substitutes it with a Carnitine molecule. The acylcarnitine is then transported across the inner mitochondrial membrane by a carnitine / acylcarnitine translocase. This integral membrane protein transports acylcarnitine molecules into the mitochondrial matrix and free carnitine molecules out. Once inside the mitochondrial matrix the acyl group is transferred back onto CoA, releasing free carnitine, by the enzyme carnitine acyltransferase II which is located on the matrix side of the inner mitochondrial membrane.

iii) What is Deamination reaction? Explain using suitable example. Glutamate is transported from the cytosol to the mitochondria, where it undergoes oxidative deamination catalyzed by L-glutamate dehydrogenase (GD). GD can employ NAD+ or NADP+ as cofactor and is allosterically regulated by GTP and ADP. The combined action of the aminotransferases and GD is referred to as transdeamination. A few amino acids bypass the transdeamination pathway and undergo direct oxidative deamination. Glutamate dehydrogenase is a complex allosteric enzyme and is present only in the mitochondrial matrix.

iv) How is a fatty acid converted to is fatty acyl – CoA form? Give the reactions involved. Fatty acid breakdown occurs in the cytosol of prokaryotes and in the mitochondrial matrix of eukaryotes. Before entering the mitochondrial matrix, the fatty acid is Activated by forming a thioester link with CoA. This reaction is catalyzed by acyl CoA synthetase (also called fatty acid thiokinase) which is present on the outer mitochondrial membrane, and uses a molecule of ATP. The overall reaction is irreversible due to the subsequent hydrolysis of PPi to two molecules of Pi.

Q.5. Write short Notes on – Any Four 20 mks

Q. 5 a) Gel Filtration Significance:

Gel filtration chromatography seprarates proteins, peptides, and oligonucleotides on the basis of size. Molecules move through a bed of porous beads, diffusing into the beads to greater or lesser degrees. Smaller molecules diffuse further into the pores of the beads and therefore move through the bed more slowly, while larger molecules enter less or not at all and thus move through the bed more quickly. Both molecular weight and three- dimensional shape contribute to the degree of retention. Gel Filtration Chromatography may be used for analysis of molecular size, for separations of components in a mixture, or for salt removal or buffer exchange from a preparation of macromolecules. Significance: Molecules separate by size, Smllest elute out first, Polymer beads nonreactive Column, eluted using suitable buffer – small molecules eluted first

Q. 5 b) Six classes of Enzymes: 1. Oxidoreductase (Class 1) 2. Transferase (Class 2) 3. Hydrolase (Class 3) 4. Lyase (Class 4) 5. Isomerase (Class 5) 6. Ligase (Class 6)

Oxidoreductase: Oxidoreductases catalyze oxidation reduction reactions. At least one substrate becomes oxidized and at least one substrate becomes reduced. Transferase: Transferases catalyze group transfer reactions- the transfer of a functional group from one molecule to another. Hydrolase: In hydrolysis reactions, C-O, C-N, and C-S bonds are cleaved by addition of - + H2O in the form of OH and H to the atoms forming the bond. Lyase: Lyases cleave C-C, C-O, C-N, and C-S bonds by means other than hydrolysis or oxidation. Isomerase: Isomerases just rearrange the existing atoms of a molecule, that is, create isomers of the starting material. Ligase: Ligases synthesize C-C, C-S, C-O, and C-N bonds in reactions coupled to the cleavage of high energy phosphate bonds in ATP or some other nucleotide.

Q. 5 c) Galactosemia is a disorder that affects how the body processes a simple sugar called galactose. A small amount of galactose is present in many foods. It is primarily part of a larger sugar called lactose, which is found in all dairy products and many baby formulas. The signs and symptoms of galactosemia result from an inability to use galactose to produce energy.

 Researchers have identified several types of galactosemia. These conditions are each caused by mutations in a particular gene and affect different enzymes involved in breaking down galactose.

 Classic galactosemia, also known as type I, is the most common and most severe form of the condition. If infants with classic galactosemia are not treated promptly with a low-galactose diet, life-threatening complications appear within a few days after birth. Affected infants typically develop feeding difficulties, a lack of energy (lethargy), a failure to gain weight and grow as expected (failure to thrive), yellowing of the skin and whites of the eyes (jaundice), liver damage, and abnormal bleeding. Other serious complications of this condition can include overwhelming bacterial infections (sepsis) and shock. Affected children are also at increased risk of delayed development, clouding of the lens of the eye (cataract), speech difficulties, and intellectual disability. Females with classic galactosemia may develop reproductive problems caused by an early loss of function of the ovaries (premature ovarian insufficiency).

 Galactosemia type II (also called galactokinase deficiency) and type III (also called galactose epimerase deficiency) cause different patterns of signs and symptoms. Galactosemia type II causes fewer medical problems than the classic type. Affected infants develop cataracts but otherwise experience few long-term complications. The signs and symptoms of galactosemia type III vary from mild to severe and can include cataracts, delayed growth and development, intellectual disability, liver disease, and kidney problems.

Q 5 d. Electron transport inhibitors

 ETS inhibitors act by binding somewhere on the electron transport chain, literally preventing electrons from being passed from one carrier to the next. They all act specifically, that is, each inhibitor binds a particular carrier or complex in the ETS. Irreversible inhibition results in a complete stoppage of respiration via the blocked pathway. Competitive inhibition allows some oxygen consumption since a "trickle" of electrons can still pass through the blocked site. Although it allows some oxygen consumption, competitive inhibition prevents maintenance of a chemiosmotic gradient, thus the addition of ADP can have no effect on respiration.  Whatever the mechanism of inhibition, an electron transport inhibitor can block respiration specifically along the NADH pathway, along the succinate pathway, or along the pathway that is common to both routes of electron entry. Careful addition of inhibitors to mitochondria on specific substrates can reveal the sites of inhibition. Some combinations of inhibitors enable demonstration of alternative entry points to the electron transport system.  Rotenone-Rotenone is still used as an insecticide, but is not available for general use. It is toxic to wildlife and to humans as well as to insects. The location of inhibition by this competitive inhibitor of electron transport can be worked out by testing its ability to block respiration via the NADH versus succinate pathway.  Antimycin-The antimycin that we use in research was formerly known as antimycin A. The latter term has been dropped since only one antimycin is used in the literature. The binding site for antimycin can be narrowed considerably using combinations of substrates inlcuding succinate, NADH or glutamate, and the dye TMPD (N,N,N',N'- tetramethyl-p-phenylenediamine) along with ascorbic acid.  Cyanide-Cyanide is an extremely effective reversible inhibitor of cytochrome oxidase. A concentration of 1 mM KCN is sufficient to inhibit oxygen consumption by mitochondria from a vertebrate source by >98%. For a nominally 2 ml chamber, a convenient concentration for the stock solution would be 0.5M (20 µl produces a 2.5 mM final concentration).  Mitochondria from some sources have cyanide resistant pathways. KCN solutions are volatile, so that a dilute solution left open to the atmosphere will quickly lose its potency. Concentrations greater than 1 mM have been known to cause uncoupling. In the presence of TMPD we have seen a dramatic increase in oxygen consumption upon the addition of excess cyanide, using a Clark electrode. Indications were that a non- biological mechanism was responsible.  Cyanide is one of the most deadly compounds in a laboratory. Stocks of the dry chemical should be stored under lock and key. As we know from the Tylenol incidents of a number of years ago, a 500 mg capsule can hold enough cyanide to kill a person. Because of its volatility, exposure to fumes from large quantities is hazardous.  Malonate-Malonate (malonic acid) has long been known to inhibit cellular respiration. Among the key observations made in the 1930s investigations into the nature of cellular respiration was that the addition of fumarate, malate, or oxaloacetate to cell preparations resulted in the accumulation of succinate in the presence of malonate. Malonate is in fact a competitive inhibitor, and although we treat it as an inhibitor of electron transport it really is an enzyme inhibitor.  Uncoupling agents-Uncoupling is defined as a condition in which the rate of electron transport can no longer be regulated by an intact chemiosmotic gradient. The condition is differentiated from electron transport inhibition by the fact that in the latter case, bypassing the block can restore the gradient. In uncoupling, the electron transport system is uninhibited due to complete and irreversible dissipation of the chemiosmotic gradient.  2,4-Dinitrophenol-The compound 2,4-dinitrophenol (DNP) acts as a proton ionophore, that is, it binds protons on one side of a membrane, and being fat-soluble it drifts to the opposite side where it loses the protons. Actually, the associations/dissociations are random, but the probability of binding is greatest on the side of the membrane with greatest proton concentration, and least on the side with the lesser concentration. Thus, it is impossible to maintain a proton gradient with sufficient DNP in the system.  DNP is known to have mixed actions, that is, it produces other effects in addition to uncoupling. DNP gradually inhibits electron transport itself as it is incorporated into mitochondrial membranes. The effects appear to depend on concentration of DNP and of mitochondria, and vary from one preparation to the next.  Back in the 1930s DNP was touted as an effective diet pill. Indeed, the uncoupling of electron transport from ATP synthesis allows rapid oxidation of Krebs substrates, promoting the mobilization of carbohydrates and fats, since regulatory pathways are programmed to maintain concentrations of those substrates at set levels. Since the energy is lost as heat, biosynthesis is not promoted, and weight loss is dramatic. However, to quote Efraim Racker (A New Look at Mechanisms in Bioenergetics, Academic Press, 1976, p. 155), ..."the treatment eliminated not only the fat but also the patients,...This discouraged physicians for awhile..."  It is not a good idea to mess with cellular metabolism.  Carbonyl cyanide p-[rifluoromethoxyl]-phenyl-hydrozone (FCCP)  This agent is, in fact, a pure uncoupler. It acts as an ionophore, completely dissipating the chemiosmotic gradient, leaving the electron transport system uninhibited. It is also expensive.  Oligomycin-Oligomycin, an antibiotic, acts by binding ATP synthase in such a way as to block the proton channel. That is the mechanism by which oligomycin inhibits oxidative phosphorylation. Experimentally, oligomycin has no effect on state IV respiration, that is, it has no direct effect on electron transport or the chemiosmotic gradient. On the other hand oligomycin prevents state III respiration completely. To draw the conclusion that an agent is an inhibitor of ATP synthase (inhibitor of oxidative phosphorylation), the above conditions must be demonstrated experimentally and unequivocally.  It takes awhile for the effects of oligmycin to show up. Attempts to interrupt state III respiration by adding oligomycin may fail because of the delay.

Q 5 e) Give the ATP yield during complete oxidation of a 10C fatty acid. C10 saturated acyl CoA, would be completely degraded into 5 molecules of acetyl CoA by 4 rounds of degradation leading to the overall equation: + Palmitoyl CoA + 4 FAD + 4 NAD + 4 CoA + 4 H2O  5 Acetyl CoA + 4 FADH 2 + 4 NADH + 4 H+

+ 5 Acetyl CoA + 4 FADH2 + 4 NADH + 4 H

• 1 Acetyl CoA in TCA gives: + 1 GTP (1 ATP), 3 NADH + H , 1 FADH2 • 5 Acetyl CoA in TCA give: 5 GTP (05 ATP)= 05 ATP 15 NADH + H+ = 15 x 2.5 = 37.5 ATP 5 FADH2 = 5 x 1.5 = 7.5 ATP • 5 Acetyl CoA in TCA: TOTAL = 50 ATP

+ 5 Acetyl CoA + 4 FADH2 + 4 NADH + 4 H

5 Acetyl CoA in TCA = 50 ATP

4 NADH + 4 H+ = 4 x 2.5 = 10 ATP

4 FADH2 = 4 x 1.5 = 6 ATP

TOTAL = 66 ATP – 1 ATP (For Activation)

Gross Energy generated = 65 ATP

Q.5 f) Explain the process of transport of ammonia in the form of glutamine to liver. Nitrogen is the fourth most important contributor (after carbon, hydrogen and oxygen) to the mass of living cells. Atmospheric nitrogen is abundant but is too inert for use in most biochemical processes. Only a few microorganisms have the capacity to convert into biologically useful forms (such as NH3) and as such amino groups are used with great economy in biological systems.

Amino acids derived from dietary proteins are the source of most amino groups. Most of the amino groups are metabolized in the liver. Some of the ammonia that is generated is recycled and used in a variety of biosynthetic processes; the excess is either excreted directly or converted to uric acid or urea for excretion. Excess ammonia generated in extrahepatic (i.e., other than liver) tissues is transported to the liver in the form of amino groups, as described below, for conversion to the appropriate excreted form. The coenzyme pyridoxal phosphate (PLP or PALP) participates in these reactions. Two amino acids, glutamate and its amide form glutamine, play crucial roles in these pathways. Amino groups from amino acids are generally first transferred to a a-ketoglutarate in the cytosol of liver cells (= hepatocytes) to form glutamate. Glutamate is then transported into the mitochondria. In muscle, excess amino groups are generally transferred to pyruvate to form alanine. Alanine is another important molecule in the transport of amino groups, transporting them from muscle to the liver

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