BIOCHEMISTRY FINAL REVIEW

1) Glycolysis is an almost universal pathway for extraction of the energy available from carbohydrates, shared among prokaryotes and eukaryotes, aerobes and anaerobes alike. By your understanding, discuss it and classify the 10 enzymes into six categories as you learnt.Glycolysis is an almost universal pathway for extraction of the energy available from carbohydrates, shared among prokaryotes and eukaryotes, aerobes and anaerobes alike. In anaerobes, glycolysis is the only significant source of energy from carbohydrates. In aerobic organisms, considerably more energy can be harvested downstream from glycolysis in the citric acid cycle. Glycolysis produces energy in the form of ATP and NADH.The intermediate products of glycolysis can be used for other pathways.-For instance, during active energy production, pyruvate, the product of glycolysis, enters the citric acid cycle. In the absence of oxygen, fermentation converts pyruvate into lactic acid. -When energy levels are high, glucose-6-phosphate, an intermediate of glycolysis, can enter glycogenesis to be stored as glycogen, or enter PPP to provide R5P for nucleic acid synthesis. PPP also produces GAP, an intermediate for glycolysis. - Glyceraldehyde-3-phosphate, which is produced by photosynthesis, is also a glycolytic intermediate, so it can be directed from this anabolic pathway into glycolysis when energy is needed-Glycogenolysis can degrade glycogen to Glucose-6-phosphate for Glycolysis to go on in situation where there is shortage of glucose.(pathway glycolysis xem cu 7)

Step 1: Hexokinase: transferaseStep 2: phosphoglucoisomerase: IsomeraseStep 3: phosphofructokinase: transferaseStep 4: aldolase: lyaseStep 5: tri-isomerase: isomeraseStep 6: triose phosphate dehydrogenase: oxidoreductaseStep 7: phosphoglycerokinase: transferaseStep 8: phosphoglyceromutase: isomeraseStep 9: enolase: lyaseStep 10: pyruvate kinase: transferase

2) You have learnt methods to determine soluble protein, total soluble sugar, calcium content and enzymatic activity (Bromelain) in a sample. Choose one and describe the principle, chemicals, experimental design and analysis for the final result.Hartree Lowry method to determine soluble protein in a sample.1. Principle: The protein content in a sample is determine based on the quantitative reaction between copper ion and phosphomolybdic/phosphotungtic acid (Folin Cioucalteu reagent) with protein. The blue color is form by 2 reactions:The bond formation between copper ion and nitrogen in peptide bonds.The reduction of phosphomolybdic/phosphotungtic acid by phenol residue of tyrosine and tryptophan amino acid in protein. The intensity of blue color is measure to determine the concentration of protein.2. Chemicals: 0.1% albumin solution is used as standard. The standard should be the purified source of the protein to be quantified or the purified source of a protein with similar structure or in the same family. 0.1% albumin solution is prepared by taking exactly 0.1g of albumin and dissolving in water to get 100 mL of solution. Solution A contains Na2CO3 and NaOH (to create alkaline condition for reduction of phosphomolybdic/phosphotungtic acid).Solution B contains CuSO4.5H2O and Sodium citrate.Solution C is the mixture between solution A and B with ratio 49:1. Solution C is easily degraded so it must be freshly prepared. When solution loses it pale blue color, it no longer can be used.Folin Cioucalteu reagent should be preserved at 4C and can be kept for 1 year. Fresh Folin reagent is yellow-orange. When the color turn yellow-green, it cannot be used. 3. Experimental design Prepare the sample: pulverize sample in blended or stone mortar Extract protein from the sample: Take the pulverized sample into stone mortar or blended with distilled water. Grind down the sample and collect extracted solutionRepeat this step several times to extract as much protein as possible. Construct standard curve:Standard curve is the graph that shows the relationship between absorbance (OD value) and the concentration of protein in the solution. Standard curve is prepared with a series of test tube with increasing concentration of standard protein. Quantify protein content:Prepare 2 types of test tube:Test tube with 100-diluted extracted protein solution. 100-diluted solution is made by adding distilled water into 100mL volumetric flask containing extracted protein solution until the marked level is reached. Test tube with 10000-diluted extracted protein solution. 10000-diluted solution is made by adding 99mL distilled water into 1mL of 100-diluted solution. Add solution C and Folin reagent respectively into all test tubes including test tubes of standard curve. Keep test tubes for 10 minutes after each addition.Dilute all test tube by adding distilled waterMeasure at A750nm.4. Analysis:Use Excel to construct the standard curve and obtain the equation (y= ax +b) showing the relationship between OD value and concentration of protein. The y-axis is OD value and x-axis is concentration. Use the obtained equation to calculate protein concentration in extracted solution. The OD value in 100-diluted test tube is out of range so it is rejected. Insert the OD value into the standard curve equation to obtain the concentration. The amount of protein in the used pulverized sample is: n= (protein concentration) 10000 (g)The amount of protein in the whole sample is: m= (n (weight of the whole sample)) (weight of the used pulverized sample) (g)

3) The citric acid cycle is a central metabolic pathway that completes the oxidative degradation of fatty acids, amino acids, and monosaccharides. Based on your understanding, clarify the above statement.The citric acid cycle is a central metabolic pathway that completes the oxidative degradation of fatty acids, amino acids, and monosaccharides.During aerobic catabolism, these biomolecules are broken down to smaller molecules that ultimately contribute to a cells energetic or molecular needs. In early metabolic steps, From monosaccharides (a six-carbon sugar) in glycolytic pathway with the activity of the pyruvate dehydrogenase yield a two-carbon fragment -an acetyl groupAcetyl group is linked to a large cofactor known as coenzyme A (or CoA). Next, during the citric acid cycle that acetyl-CoA is oxidized to carbon dioxide with the reduction of the cofactors NAD+ and ubiquinone. Fatty acids are source of Acetyl CoA When a cells metabolic needs increase, free fatty acids enter the mitochondrion where the degradative reactions called Beta oxidation takes place. In each round of Beta oxidation, a fatty acid is shortened by two carbon atoms and a release of a free acetyl-CoA molecule Acetyl-CoA initiates the citric acid cycle Amino Acids Typically, the amino group of an amino acid is removed in a deamination reaction. The remaining carbon skeleton is broken down to various a product depending on which of the twenty amino acids is undergoing catabolism. In some cases, the remaining carbon skeleton is broken down to acetyl-CoA or to pyruvate, which is then converted to acetyl-CoA. Alternatively, a citric acid cycle intermediate such as a-ketoglutarate may result. In all cases, the citric acid cycle plays a large role in breaking down the amino acid skeleton to carbon dioxide. For example, catabolism of lysine yields carbon dioxide and acetyl-CoA, while glutamate breaks down to a-ketoglutarate, carbon dioxide, and acetyl-CoA. Acetyl-CoA initiates the citric acid cycle.

4) You have learnt metabolic pathways of carbohydrate, protein, lipid and nucleic acid, show your overview by drawing the relation among their metabolism.

ku draw thi nhng vit ra cho c hiu hnh cho d thuc d v-Glucose is oxidized by glycolysis, an energy-generating pathway that converts it to pyruvate. In the absence of oxygen, pyruvate is converted to lactate. In the presence of oxygen, pyruvate is further degraded to form acetyl-CoA. Excess glucose is converted to its storage form, glycogen, by glycogenesis. When glucose is needed as a source of energy or as a precursor molecule in biosynthetic processes, glycogen is degraded by glycogenolysis.Glucose can be converted to ribose-5-phosphate (a component of nucleotides) and NADPH (a powerful reducing agent) by means of the pentose phosphate pathway. Significant amounts of energy in the form of ATP can be extracted from acetyl-CoA by the citric acid cycle and the electron transport system.Acetyl-CoA is generated from the breakdown of fatty acids and certain amino acids. When acetyl-CoA is present in excess, a different pathway converts it into fatty acids. That pathway is called lipogenesis.Glycerol also enters the Gluconeogensis in bypass step 2.The process Gluconeogenesis is a process that builds up glucose from pyruvate and noncarbohydrate precursor such as lactate, amino acid, glycerol,

Proteins in our body have nitrogen pool or amino acid pool which is a component of tissue protein and also produce NH3. NH3 enter Urea cycle.The urea cycle describes the conversion reactions of ammonia into urea. The urea cycle and CAC are linked (common connections are Fumerate, Oxaloacetate, alpha-ketogluterate)5) Oxidative phosphorylation utilizes the chemical energy of these reduced molecules from glycolysis and C.A.C to produce ATP. How is your opinion and discuss it.Oxidative phosphorylation happen in mitochondrion, is the process by which electrons from the reduced cofactors NADH and ubiquinol are funneled in a stepwise manner to oxygen. Electrons flow much like electricity through the concomitant formation of a proton gradient. In the end the investment of reduced cofactors results in the production of ATP.Reduced electron carries NADH and ubiquinol are produced during glycolysis and citric acid cycle, as well as fatty oxidation pathway. During the cellular process of respiration oxidative phosphorylation utilize the chemical energy of these reduced molecules to produce ATP.6) Pentose phosphate pathway is to provide reduced NADPH for synthetic reactions and ribose-5 phosphate for nucleic acid synthesis. By your understanding, discuss about oxidative and non-oxidative pathways. The pentose phosphate pathway is an alternative metabolic pathway for glucose oxidation in which no ATP is generated. It occurs in cytosol Two phases: oxidative phase and non-oxidative phase:a. Oxidative : NADPH is generated when glucose 6-phosphate is oxidized to ribose 5-phosphateb. Non-oxidative involves the isomerization and condensation of a number of different sugar molecules: three-, four-, five-, six-, and seven-carbon sugarsDuring the remaining reactions of the pathway transketolase and transaldolase catalyze the interconversions of trioses, pentoses, and hexoses.The intermediates in this process that are useful in other pathways i. fructose-6-phosphate (glycolysis)ii. glyceraldehyde-3-phosphate (glycolysis)7) Gluconeogensis is the pathway to form new glucose from simpler molecules, called noncarbohydrate precursors, happening mainly in liver (~90%) and kidneys from pyruvate, lactate, glycerol, amino acids, and TCA cycle intermediates. The purpose is to make glucose when blood glucose level is low. By what you have learnt, choose one precursor and discuss how to result in glucose.

In humans the main gluconeogenic precursors lactate, glycerol, alanine and glutamine. Lactate produced by active skeletal muscle and erythrocytes(or red blood cells). Erythrocytes lack mitochondria and can never oxidize glucose completely. In contracting skeletal muscle, the rate at which glycolysis produces pyruvate exceeds the rate at which the citric acid cycle oxidizes it. Lactate is a dead end in metabolism. It must be converted back into pyruvate before it can be metabolized. The lactate that enters the liver is oxidized to pyruvate. Pyruvate in the liver is converted into glucose by the gluconeogenic pathway. Glucose then enters the blood and is taken up by skeletal muscle. Gluconeogensis is the pathway to form new glucose from simpler molecules, called noncarbohydrate precursors, happening mainly in liver (~90%) and kidneys from pyruvate, lactate, glycerol, amino acids, and TCA cycle intermediates.Gluconeogenesis pathway consists of 11 steps which are reverse of glycolysis and there are 3 bypass steps- which use different enzymes and mechanisms compare with glycolysis. (ci bng l so snh bypass steps) (hnh cho d hnh dung Gluconeogenesis)

8) When referring to the glycogen metabolism, three contents must be remembered as glycogenesis, glycogenolysis or regulation related to insulin, glucagon & epinephrine (adrenalin). Among them, choose one and discuss it.General Intro about glycogen: A carbohydrate Large number of glucose by -1,4 and -1,6-glycosidic bondsStorage of glycogen: In muscle forms as cytosolic granules with a diameter of 10 to 40 nm - the same as the size of ribosome. Concentration (liver cells) > concentration (muscle cells) Total amount of glycogen in muscle cells > in the liver. In the uterus, glycogen is stored during pregnancy to nourish the embryo. In vagina, glycogen is secreted and then converted into lactic acid to maintain the acidic environment in order to protect vagina from outside bacteria infections. In kidneys, brain and white blood cells. Glycogenesis Excess of glucose Glycogen is synthesized from glucose-6-phosphate. Glycogen can either enter into pentose-phosphate pathway or contribute to the process of glycolysis.Involves 3 steps:1. Synthesis of glucose-1-phosphate2. Synthesis of UDP-glucose3. Synthesis glycogen from UDP-glucose

Glycogenesisistheformationofglycogenfromglucose.Glycogenissynthesized dependingonthedemandforglucoseandATP(energy).Ifbotharepresentinrelatively highamounts,thentheexcessofinsulinpromotestheglucoseconversionintoglycogen forstorageinliverandmusclecells.(Itmeansthatinsulininhibitsglycogenolysisby inhibitingGlycogenphosphorylaseenzymeactivity.) Inmusclecells,glycogendegradationservestoprovideanimmediatesourceof glucose-6-phosphateforglycolysis,toprovideenergyformusclecontraction. Inlivercells,themainpurposeofthebreakdownofglycogenisforthereleaseof glucoseintothebloodstreamforuptakebyothercells.Thephosphategroupofglucose-6-phosphateisremovedbytheenzymeglucose-6-phosphatase,whichisnotpresentin myocytes,andthefreeglucoseexitsthecellviaGLUT2facilitateddiffusionchannelsin the hepatocytecellmembrane.

9) Fatty acids are an important energy source as their yield over twice as energy as an equal mass of carbohydrate or protein. The two main pathways of fatty acid metabolism are oxidation and fatty acid synthesis. oxidation result in the formation of reduced cofactors and acetyl-CoA molecules, which can be further catabolized to release free energy. By what you have learnt, discuss a round of oxidation and focus on the changing of each reaction, the energy yield, and name of enzymes.The activated fatty acid is called a fatty acyl-coenzyme A, or fatty acyl-CoA. -In the first step of oxidation, an acyl-CoA dehydrogenase catalyzes the oxidation of the acyl group, resulting the formation of a double bond between carbons two and three. The two electrons removed from the acyl group are transferred to an FAD prosthetic group. These electrons are transferred to ubiquinone through a series of electron transfer reactions.-In the second step of oxidation, a hydratase adds a molecule of water across the double bond produced in the first step. -In the third step of oxidation, another dehydrogenase catalyzes the oxidation of the hydroxyacyl group. In this case, NAD+ is the cofactor. The fourth and final step of oxidation is called thiolysis. In this step, a thiolase catalyzes the release of acetyl-CoA from the ketoacyl-CoA.Energy yield: One round of oxidation yields three productsone ubiquinol cofactor, one NADH cofactor, and one molecule of acetyl-CoA.

-During oxidative phosphorylation, 1 ubiqunol -> 2 ATP; 1 NADH -> 3ATP, 1 actyl-CoA -> 12ATP => total 17 ATP, and the net ATP is 15 (2 was used for fatty acid activation)10) Call the names of below fatty acids and calculate the energy yield followed by equation of oxidation

Name : acid panmitic Calculate the energy yield of this fatty acid: There are 16 carbon in this fatty acid (n=16), so the round number of it is: (n/2)-1 = (16/2) 1 = 7 (rounds) One round of oxidation produces the equivalent of 17 molecules of ATP. In the 1st round, 2 ATP molecules were used for activation step. The final product of complete oxidation is an additional molecule of acetyl-CoA, which are equivalent to12 molecules of ATP. The total energy yield of this fatty acid is: (16/2 1 ) x 17 + 12 2= 129 ATP molecules.

Name: The name of this fatty acid is Linoleic acid (Linoleoyl-CoA)Calculate the energy yield of this fatty acid: There are 18 carbon in this fatty acid (n=18), so the round number of it is: n/2 1 = 18/2 1 = 8 One round of oxidation produces the equivalent of 17 molecules of ATP. Double bonds at odd-numbered position cost 2 ATP molecules. Double bonds at even-numbered position cost 3 ATP molecules. In the 1st round, two ATP molecules were used for activation step. The final product of complete oxidation is an additional molecule of acetyl-CoA, which are equivalent to12 molecules of ATP. The total energy yield of this fatty acid is: 8 x 17 + 12 2 2 3 = 141 ATP molecules.

11) . About nitrogen metabolism, choose one and discussa. - Essential & non-essential amino acids (my cu di kh w, fi hc urea cycle ht nn nh chn cu ny >