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 >