Answer to each other. Answer 2: Myo-inositol monophosphates

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Last updated: May 16, 2019

Answer 1Fourspiral chains of keratin twist to form keratin. Spiral chains cross linked byhydrogen and disulfide bonds, changes of these linkage determine whether hairis curly or straight.Tochange the arrangement disulfide bonds of Keratin chains is to be disrupted,reducing agents would oxidize keratin by disrupting the covalent bond betweencysteine molecules and favors formations of bonds between cysteine and hydrogen,which is a weak covalent bond easily broken at this step hair can be rearranged.

In presence of oxidizing agent, the hydrogen ions will be removed so thatcysteine residues forms disulfide bonds. Notall the broken disulfide bonds do not reform again and form weak spots inkeratin. So, the bonds formed were weak and likely to break and allows proteinto come back to original state.Some type of hairs is curly as high number of disulfide bonds areformed along different sites of nearby chains when compared to straight hairand cause keratin to fold, and follicle shape alsoa determining factor which helps in bringing hair to each other.

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Answer 2:Myo-inositol monophosphates catalyzes the hydrolysisof inositol monophosphate is inhibited by Lithium is an Uncompetitive inhibition. Uncompetitive inhibition in which enzymesubstrate complex is binded with inhibitor (Lithium) to prevent formation offinal product as it takes longer for the substrate orproduct to leave the activesite. At higher concentration of substrate, the inhibitorworks well.Penicillin binds to DD-transpeptidase and prevents it frombinding and cross-linkingpeptidoglycans in the bacterial cell wall, which will cause the bacterial cellto burst. This is an example of Irreversibleinhibitor .An irreversibleinhibitor will bind to an enzyme so that no other enzyme-substrate complexes can form.

It will bind tothe enzyme using a covalent bond at the active site whichtherefore makes the enzymedenatured.COX-2 catalyzesformation of prostaglandins which is competitively inhibited by NSAIDs. Competitive inhibition inwhich binding of an inhibitor prevents binding of substrate to enzyme.

This isdone by blocking binding to active site of substrate. Addition of substrate displacesinhibitor from the active site of enzyme and increase binding of substrate toenzyme, this alters only the Km, leaving the Vmax unchanged. Answer 3: ?G values which are negative values indicate forwardreaction and positive values indicate reverse reaction strongly favored. Inabsence of inorganic phosphate which indicates that hexokinase (step 1) andphosphofructokinase (Step 3) favor forward reactions. But Aldolase (step 4) andphosphoglycerate mutase (step 8) never favor reaction even in presence ofinorganic phosphate as it has positive ?G but as a result of indirect couplingthese reactions are become favorable. Step 10 (PEP to pyruvate) is favorable thatlow concentration of all intermediates prior to this pushes reaction in forwarddirection to produce new substrate. Steps6, 7, and 10 releases sufficient energy to drive the formation of NADH (step 6)or the formation of ATP (steps 7 and 10).Step 1 is formation of Glucose 6 phosphate (G6P) fromglucose with use of ATP molecule which is irreversible reaction.

It iscatalyzed by hexokinase which is feedback inhibited by G6P, so phosphorylationof glucose is controlled depending upon concentration of formed G6P. ?G is negative as the ATP used as phosphatedonor. The glucose is locked up in cell and It also enables the glucose to go on to step 2 of glycolysis.Step3 isconversion of fructose 6 phosphate to fructose 1,6 bisphosphate catalyzed byphospho fructokinase by utilizing ATP. Phosphofructokinase is also activated byfructose?2,6? bisphosphate, which is formed from fructose?1?phosphate by asecond, separate phosphofructokinase enzyme phosphofructokinase II. This step is activated under low energy whencell has high AMP and ADP and fructose1,6 bisphosphate.

Step1 and 3 consumes ATP energy to take but prepares substrates for the productionof energy in later steps. Step7 Involves 1,3-bisphosphoglycerateto 3- phosphoglycerate catalyzes by phosphoglycerate Kinase. The first ATPforming step in glycolysis. ATP is formed by transfer of a phosphoryl groupfrom 1,3-bisphosphoglycerate to ATP is called as substrate levelphosphorylation. Coupling of step 6 with step 7 produces one NADH per glucoseare to be reoxidized to regenerate NAD that is needed for the oxidation ofglyceraldehyde 3 phosphate.Step 4 is formation of trioses G-3 P (glyceraldehyde3 phosphate) and DHAP (Dihydroxy acetone phosphate) from fructose 1,6bisphosphate catalyzed by Aldolase which has high positive gibbs free energy. Aldolasereaction is a near equilibrium reaction in cells, which shows concentrationof fructose 1,6 bisphosphate is highly relative to two triose groups G-3 P andDHAP.

The concentration of fructose 1,6 bisphosphate is very high in thecells which is formed due to high -ve (negative) gibbs free energy. Theconcentration of these trioses is low in cells when compared to fructose1,6 bisphosphate. Flux due to high concentration of fructose 1,6 bisphosphatewhich goes to next steps for pyruvate synthesis.Answer 4:In TCA cycle total of 20 ATPproduced from 2 Acetyl COA molecules. For one it produces 10 ATPs.Conversion of Isocitrate to alfa Ketoglutarate produces one 1NADHEnzyme: isocitrate dehydrogenase.

Conversion of Alphaketoglutarate to succinyl CoA produce 1 NADH Enzyme: alpha ketoglutarate dehydrogenase.Oxidation of Malate forms oxaloacetate, 1 NAD+ is reduced to NADH. Enzyme: malatedehydrogenase.

 1 NADH = 2.5 ATP: Total 3 NADH for 1 Acetyl COA = 3*2.5 = 7.5 ATPOxidation of Succinate to fumarate. (FAD) isreduced and forms FADH2 . Enzyme: succinate dehydrogenase.1 FADH2 = 1.5 ATP succinyl CoA converted to Succinate byremoving the COA using GTP and generate ATP in process Enzyme: succinyl-CoA synthetase.

1 ATP  Total ATP for 1Acetyl COA = 3 NADH + 1 FADH2 + 1 ATP = 7.5+ 1.5+ 1= 10 ATP. For 2 Acetyl COA = 20 ATP.Answer 5:Glucose 1 phosphate conversion to lactate yields 3 ATPequivalents1 ATP from Phosphofructokinase reaction2 ATP form Phosphoglycerate Kinase reaction Converting twomolecules of lactate to one molecule of glucose 1 phosphate needs 6 ATP.

2 ATP for Pyruvatecarboxylase reaction 2 ATP in the PEPcarbokinase reaction 2 ATP forPhosphoglycerate Kinase reaction.  Answer 6: rRNA are processedand assembled into their ribosomal subunits within nucleus and exported so itis resistant to nucleases.tRNA are processedfrom a primary transcript and heavy modification in nucleoside was seen andhave an extensive secondary structure which makes themresistant to ribonuclease degradation.Cappingof mRNA was conducted during the transcription by an enzyme complex.

Mainreason for capping is protection of mRNA from 5′ exonuclease enzymes. Removalof one nucleotide from mRNAsynthesizes an inefficient protein so that preservation of the mRNA translationis very important.Main functions of 5′ capping isregulating transport of mRNA from nucleus and protection from exonucleases andpromotion of translation, 5′ proximal intron excision. Answer 7:mRNA: It has codons for peptide synthesis. It makes up to 3-5 % oftotal RNA.

Show base relationship to DNA.tRNA: It has anticodons that can base pair or link the exact requiredamino acid corresponding to mRNA codon. Also called as Adapter molecules.

Itmakes up to 15-20 % of total RNA.rRNA: It is a molecule in cell that form a part of ribosome and thenexported to cytoplasm and help in translation process. Catalyzes the formation of thepeptide bond. It makes up 80% of total RNA. I have no base relationship to DNA.

Synthesis of mRNA, tRNA, rRNA:mRNA: It isformed by transcription process from DNA with the help of RNA polymerase whichmake a copy of gene from DNA form to mRNA. It is transferred from Nucleus to Cytoplasm.tRNA: It isalso formed by transcription process from DNA with help if RNA polymerase threeenzyme in the nucleus. It is formed by nuclear processing of precursormolecule.rRNA: It isformed by transcription from DNA using RNA polymerase one into large RNAmolecule. Later addition of sequences is added by many other polymerasessimultaneously to form giant RNA molecules.

Structure and function of mRNA, tRNA, rRNA:mRNA: It islinear shape molecule. IT carries genetic information from DNA to ribosome in Cytosolwhich serves as a template for proteins synthesis and unpaired bases are bindedto mRNA and tRNA. 5’end terminal is capped by the 7 – methyl guanosinetriphosphate cap. It helps in recognizing the mRNA by the translationmachinery. Capping prevents cleavage by 5′ exonucleases. 3′ end have a polymerof adenylate residues which protect from 3′ exonucleases.

tRNA: Ithas a primary and secondary structure. Primary structure the nucleotidesequence of all tRNA molecules allow intrastand complementary that forms asecondary structure. Each tRNA have extensive internal base pairing and forms aclover like structure.

The hydrogen bonding stabilizes the structure. Cloverleaf structure have 5 arms1. Acceptor arm – It is 3′ end, thehydroxyl of Adenine binds with carboxyl groups of Amino acid.2. Anticodon arm- Opposite ends ofAcceptor arms, it binds specifically with mRNA by hydrogen bonding.3. DHU arm – Serves as site torecognize enzymes that helps to add amino acid to acceptor arm.

4. T, C arm- Involves binding oftRNA to ribosomes.5. Extra arm – Only 75% of tRNA hasextra arm.The tertiary structure is alsoformed by internal bonding of hydrogen in clover leaf between T and D arms.

rRNA: large,small rRNA combine along ribosomal proteins to form large, small subunit of ribosome. These complex structures,which physically move along an mRNA molecule. Also help in binding tRNAs andaccessory molecules that are required for synthesis of proteins.Answer 8:After DNA strands are separated twostrands were formed one is Leading and other is called as lagging strands.Leading strands always lead from 5′ to 3′ and lagging strand reads from 3′ to 5′. As DNAstrands are antiparallel only one continuous strand can synthesis at 3′ end ofthe leading strand because of DNA polymerase property to start synthesis from5′ to 3′. DNA polymerase is highly specific for 3′- OH terminal of new strand. DNA polymerase attacks by nucleophilic by the 3′-OH of thenucleotide at the 3′ end of the strand on the 5′-?-phosphorus of the deoxynucleoside 5′-triphosphate.

A primer (segment of new strand) is needed opposite to leadingstrand to which nucleotide are attached. The primer should be in place beforeDNA polymerase start to act. The polymerase can only add nucleotides to apreexisting strand. So, the lagging end is unavailablefor the DNA polymerase to interact.

Lagging strand forms a short section of DNAa result of discontinuation replication. Many RNA primers are made by primaseand bind to many sites of lagging strand and forms chunks of DNA called asOkazaki fragments and then added to lagging strand in 5′ to 3′.   Answer 9: Step 2: Isomerization of glucose-6-phosphateto fructose 6- phosphate.Enzyme: Phosphoglucomutase; ?G= +2.8 KJ.Phosphoglucomutase belongs to Isomerases. Isomerase catalyzes the shifting of a functional group fromone carbon to other within a molecule. Step 4: Fructose-1,6-bisphosphate is breakdown to: dihydroxyacetone phosphate(DHAP) and glyceraldehyde 3-phosphate.

Enzyme: Aldolase; ?G= +24.6KJ.Aldolasebelongs to class Lyases. Aldolase catalyze an aldol cleavage reaction. Step 5: DHAP and GAP are isomersof are readily inter-converted.  GAPis a substrate for the next step in glycolysis so all of the DHAP is eventuallydepleted.

Enzyme:Triose phosphate Isomerase; ?G=+7.6 KJ.Triosephosphate Isomerase belongs to Isomerasesclass. Interconverting of aldolases and ketoses are Involved.

 Step 6: GAP isdehydrogenated to form 1,3-bisphosphoglycerate.Enzyme:Glyceraldehyde 3-phosphate dehydrogenase (GAPDH); ?G= +2.6 KJ.Glyceraldehyde3-phosphate dehydrogenase belongs to classOxidoreductases. Step 8: Conversion of 3-phosphoglycerate to 2-phosphoglycerate. Thephosphate shifts from C3 to C2 to form 2-phosphoglycerate.

Enzyme:Phosphoglycerate mutase; ?G=+6.4 KJ.Phosphoglyceratemutase belongs to class Isomerases. Answer 10:  Synonymous codonsthat instruct ribosome complex to add arginine are:                                                                        CGU, CGC, CGA, CGG, AGA, AGGSynonymous codonsfor Methionine:  AUGSynonymous codonsfor termination of proteins synthesis: UAA,UGA, UAG Synonymouscodons that signalthe initiation of synthesis: AUG.  BonusQuestion:Tumor cells grow under limited oxygen supply initialstages as they devoid of capillary supply.

So, the cells depend on glycolysisby converting glucose to pyruvate and lactate for ATP production but the ATPproduced is only 2 ATP per glucose, to compensate that tumor cells absorb moreglucose than normal cells. More lactic acid is formed there will be change in pHto acidic. Increase in glycolysis is achieved by increased synthesis ofglycolytic enzymes and plasma membrane transporters. With high rate ofglycolysis, the tumor cells can survive anaerobic conditions.High rate of glucose uptake used in pinpointlocation of tumors. In positron emission tomography isotope labelled glucoseanalog is taken up and not metabolized by tissue. The decay of isotope yieldspositrons that will be detected by a detector and help in pinpointing thelocation of tumor precisely.

The intensity of the positrons emitted is detectedin the pet scan is translated from green to red.

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