How many ATP will be yielded in 2 moles of an 17:0 and 18:0 carbon chain?
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How many ATP will be yielded in 2 moles of an 17:0 and 18:0 carbon chain?
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- If 100% of the free energy from the metabolism of glucose is used for the conversion of ADP to ATP, how manymolecules of ATP can be produced from each gram of glucose (mm = 180.16)?The oxidation of 1 mol of glucose supplies enough metabolic energy to form 36 mol of ATP. Oxidation of 1 mol of a typical dietary fat like tristearin (C57H116O6) yields enough energy toform 458 mol of ATP. How many molecules of ATP can form per gram of (a) glucose; (b) tristearin?An amylose chain is 5000 glucose units long. At how many places must it be cleaved to reduce the average length to 2500 units? To 1000 units? To 200 units? What percentage of the glycosidic links are hydrolyzed in each case? (Even partial hydrolysis can drastically alter the physical properties of polysaccharides and thus affect their structural role in organisms.)
- Answer the following questions regarding the trisaccharide composed of a glucose, mannose and galactose a) What is the configuration on the anomeric carbon for the each of the sugars b) Which sugar end(s) are reducing? c) What is the numbering on the linkage connecting the glucose to the mannose? d) What is the numbering on the linkage connecting the mannose to the galactose? OH CH,OH Galactose но. но, CH,OH но но. он Glucose но. он MannoseAn amylose chain is 5000 glucose units long. At how many places must it be cleaved to reduce the average length to 2500 units? To 1000 units? To 200 units? What percentage of the glycosidic links are hydrolyzed in each case? (Even partial hydrolysis can drastically alter the physical properties of polysaccharides and thus affect their structural role in )Although the first two carbons of fructose and glucose are identical in structure to DHAP and GADP (from glycolysis), DHAP and GADP equilibriate on their in solution to favor the ketone over the aldehyde, while fructose and glucose do not. Why? a)The larger size of the molecule sterically hinders the isomerization b)The larger sugars have more OH groups which hydrogen bond and disrupt isomerization c)The larger sugars cyclize, and there is no carbonyl to isomerize in the cyclic form d)The larger sugars cyclize, and in the cyclic form the hydrogen bonding is very strong e)The larger sugars are less soluble in water than the smaller sugars
- Acid phosphatases are an important group of enzymes that can be detected in human blood serum. Under slightly acidic conditions (pH 5.0), this group of enzymes can hydrolyze biological phosphate esters as follows: R-O-P-O;² + H 20 → R-OH + HO-P-O;² Acid phosphatases are produced and can be detected in erythrocytes, kidney, spleen, the liver, and prostrate gland. The enzyme from the prostrate gland is clinically important because an increased activity in the blood is frequently an indication of cancer of the prostrate gland. Tartrate ion can strongly inhibit the phosphatase from the prostrate gland, but not acid phosphatases from other tissues. How can you use the information above to develop a specificH OH CH2OH D H ОН H الحزن ОН CH2OH ОН H ОН I Which of the following statements correctly describes this structure? (A) The monomer units are bonded by beta 1-2 glycosidic linkage. B The monomer units are bonded by alpha 1-2 glycosidic linkage. The monomer units are bonded by alpha 1-4 glycosidic linkage. The monomer units are bonded by beta 1-4 glycosidic linkage. H CH2OH H ОН H ОН H ОНA camel hump contains 12 kg of triacylgylcerols. (a) Given that there are 0.491 moles of ATP per gram of fat, how many moles of ATP could be produced by the fat in the camel hump?(b) If the hydrolysis of ATP releases 7.3 kcal/mole, how many kilocalories are produced by the utilization of the fat?
- With alanine AA, and using citric acid cycee and glycolysis, which Carbon atom would be labeled 1st with 14C n succinate? Why?Consider the structure of the tripeptide (in its fully protonated form) below. H H H + I || H₂N-C-C-N-C-C-N-C-C-OH 1 I | H CH₂ H CH₂ CH₂ T C=O OH AA1 0=0 || HC-CH3 CH3 AA2 AA3 0=C 1. Give the sequence of the tripeptide using the ONE-LETTER DESIGNATION (UPPERCASE LETTER) with NO spaces and symbols between each letter. 2. How many ionizable groups are there in the tripeptide? Give the numerical value (e.g., 10 not ten). • pH 10: {Choices: -2, -1, 0, +1, +2} 3. Which amino acid residue has one ionizable group left upon forming the tripeptide? {Choices: AA1, AA2, AA3, none, all} 4. Give the net charge of the dominant structure of the tripeptide at the given pH values. The pK, values of the amino acids are given in Table 1. • pH 4: {Choices: -2, -1, 0, +1, +2}The free energy of hydrolysis of an α(1→4) glycosidic bond is −15.5 kJ ⋅mol−1, whereas that of an α(1→6) glycosidic bond is −7.1 kJ ⋅ mol−1. Use these data to explain why glycogen debranching includes three reactions [breaking and re-forming α(1→4) bonds and hydrolyzing α(1→6) bonds], whereas glycogen branching requires only two reactions [breaking α(1→4) bonds and forming α(1→6) bonds].