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Determining the Branch Points and Reducing Ends of Amylopectin A 0.2-g sample of amylopectin was analyzed to determine the fraction of the total glucose residues, that are branch points in the structure. The sample was exhaustively methylated and then digested, yielding 50-μmol of 2,3-dimethylgluetose and 0.4 μmol of 1,2,3,6- letramethylglucose.
- What fraction of the total residues are branch points?
- I low many reducing ends does this sample of amylopectin have?
(a)
Interpretation:
The fraction of the total residues that are branch points in the structure of amylopectin is to be calculated.
Concept Introduction:
The simplest hydrolyzed form that is obtained from the carbohydrates is known as monosaccharide. Polysaccharides are those types of sugars which contain more than ten units of monosaccharides.
Amylopectin is a polysaccharide and is one of the two forms of starch. It is composed of long chain of glucose attached by the
Answer to Problem 8P
The fraction of the total residues that are branch points in the structure of amylopectin is
Explanation of Solution
The given mass of amylopectin sample is
The given amount of
The conversion of
Thus, the moles of
The given amount of
The molecular weight of glucose is
The molecular weight of water is
If glucose unit losses a water molecule while forming a glycosidic linkage in amylopectin, then, the molecular weight of glucose in amylopectin is
The moles of a substance is calculated by the equation as,
Substitute the value of mass and molar mass of the sample in the above equation.
The moles of
Substitute the values of moles of
Thus, the fraction of the total residues that are branch points in the structure of amylopectin is
(b)
Interpretation:
The total number of reducing ends possessed by the sample of amylopectin is to be calculated.
Concept Introduction:
The simplest hydrolyzed form that is obtained from the carbohydrates is known as monosaccharide. Polysaccharides are those types of sugars which contain more than ten units of monosaccharides.
Amylopectin is a polysaccharide and is one of the two forms of starch. It is composed of long chain of glucose attached by the
Answer to Problem 8P
The total number of reducing ends possessed by the sample of amylopectin is
Explanation of Solution
The given amount of
The moles of the reducing ends is calculated by the moles of
If one mole of compounds possesses
Thus, the total number of reducing ends possessed by the sample of amylopectin is
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Chapter 7 Solutions
Biochemistry
- Fura-2 Fluorescence (Arbitrary Unit) 4500 4000 3500 3000 2500 2000 1500 1000 500 [Ca2+]=2970nM, 25°C [Ca2+] 2970nM, 4°C [Ca2+]=0.9nM, 25°C [Ca2+] = 0.9nM, 4°C 0 260 280 300 340 360 380 400 420 440 Wavelength (nm) ← < The figure on the LHS shows the excitation spectra of Fura-2 (Em = 510 nm) in 2 solutions with two different Ca2+ ion concentration as indicated. Except for temperature, the setting for excitation & signal acquisition was identical.< ப a) The unit in Y-axis is arbitrary (unspecified). Why? < < b) Compare & contrast the excitation wavelength of the Isosbestic Point of Fura-2 at 25 °C & 4 °C. Give a possible reason for the discrepancy. < c) The fluorescence intensity at 25 °C & 4 °C are different. Explain why with the concept of electronic configuration. <arrow_forwarddraw in the structure of each amino acid (as L-amino acids) using the Fischer projection style. an example has been included. Draw the structure for glycine, alanine, valine, isoleucine, methionine, proline, phenylalanine, tryptophan, serine, threonine, asparagine, glutamine, lysine, arginine, aspartic acid, glutamic acid, histidine, tyrosine, cysteinearrow_forwarddraw in the structure of each amino acid (as L-amino acids) using the Fischer projection style. an example has been includedarrow_forward
- draw in the structure of each amino acid (as L-amino acids) using the Fischer projection style. an example has been includedarrow_forwardDraw out the following peptide H-R-K-E-D at physiological pH (~7.4). Make sure toreference table 3.1 for pKa values.arrow_forwardThe table provides the standard reduction potential, E', for relevant half-cell reactions. Half-reaction E'° (V) Oxaloacetate² + 2H+ + 2e malate²- -0.166 Pyruvate + 2H+ + 2e → lactate -0.185 Acetaldehyde + 2H+ + 2e¯ →→→ ethanol -0.197 NAD+ + H+ + 2e--> NADH -0.320 NADP+ + H+ + 2e →→ NADPH Acetoacetate + 2H+ + 2e¯ - -0.324 B-hydroxybutyrate -0.346 Which of the reactions listed would proceed in the direction shown, under standard conditions, in the presence of the appropriate enzymes? Malate + NAD+ oxaloacetate + NADH + H+ Malate + pyruvate oxaloacetate + lactate Pyruvate + NADH + H+ lactate + NAD+ Pyruvate + p-hydroxybutyrate lactate + acetoacetate Acetaldehyde + succinate ethanol + fumerate Acetoacetate + NADH + H+ → B-hydroxybutyrate + NAD+arrow_forward
- Arrange the four structures in order from most reduced to most oxidized. Most reduced R-CH2-CH3 R-CH2-CH₂-OH R-CH,-CHO R-CH₂-COO Most oxidizedarrow_forwardfor each pair of biomolecules, identify the type of reaction (oxidation-reduction, hydrolysis, isomerization, group transfer, or nternal rearrangement) required to convert the first molecule to the second. In each case, indicate the general type of enzyme and cofactor(s) c reactants required, and any other products that would result. R-CH-CH-CH-C-S-COA A(n) A(n) A(n) A(n) Palmitoyl-CoA R-CH-CH=CH-C-S-CoA ° trans-A-Enoyl-CoA reaction converts palmitoyl-CoA to trans-A2-enoyl-CoA. This reaction requires and also produces Coo HN-C-H CH₂ CH₂ CH CH CH, CH, L-Leucine CH, CH, D-Leucine 8/6881 COO HÌNH: reaction converts L-leucine to D-leucine. This reaction is catalyzed by a(n) H-C-OH H-C-OH C=0 HO-C-H HO-C-H H-C-OH H-C-OH H-C-OH CH,OH Glucose H-C-OH CH,OH Fructose OH OH OH CH-C-CH₂ reaction converts glucose to fructose. This reaction is catalyzed by a(n) OH OH OPO I CH-C-CH H Glycerol Glycerol 3-phosphate H reaction converts glycerol to glycerol 3-phosphate. This reaction requires H,N- H,N H…arrow_forwardAfter adding a small amount of ATP labeled with radioactive phosphorus in the terminal position, [7-32P]ATP, to a yeast extract, a researcher finds about half of the 32P activity in P; within a few minutes, but the concentration of ATP remains unchanged. She then carries out the same experiment using ATP labeled with 32P in the central position, [ẞ-³2P]ATP, but the 32P does not appear in P; within such a short time. Which statements explain these results? Yeast cells reincorporate P; released from [ß-³2P]ATP into ATP more quickly than P¡ released from [y-³2P]ATP. Only the terminal (y) phosphorous atom acts as an electrophilic target for nucleophilic attack. The terminal (y) phosphoryl group undergoes a more rapid turnover than the central (B) phosphate group. Yeast cells maintain ATP levels by regulating the synthesis and breakdown of ATP. Correct Answerarrow_forward
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BiochemistryBiochemistryISBN:9781305577206Author:Reginald H. Garrett, Charles M. GrishamPublisher:Cengage Learning