(a)
Interpretation:
The given set of elements has to be arranged in increasing order of atomic radius.
Concept Introduction:
The atomic radius of a chemical element is a measure of the size of its atoms, the atomic radius is distance from the centre of the nucleus to the boundary of the surrounding cloud of electrons.
The atomic radius decreases from left side to right side in the periodic table whereas atomic radius increases from top to bottom in the periodic table.
(b)
Interpretation:
The increasing atomic radius of the element should be identified.
Concept introduction:
The atomic radius of a chemical element is a measure of the size of its atoms, the atomic radius is distance from the center of the nucleus to the boundary of the surrounding cloud of electrons.
The atomic radius decreases from left side to right side in the periodic table whereas atomic radius increases from top to bottom in the periodic table.
(c)
Interpretation:
The increasing atomic radius of the element should be identified.
Concept Introduction:
The atomic radius of a chemical element is a measure of the size of its atoms, the atomic radius is distance from the center of the nucleus to the boundary of the surrounding cloud of electrons.
The atomic radius decreases from left side to right side in the periodic table whereas atomic radius increases from top to bottom in the periodic table.
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EBK FUNDAMENTALS OF GENERAL, ORGANIC, A
- draw 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_forwarddraw in the structure of each amino acid (as L-amino acids) using the Fischer projection style. an example has been includedarrow_forward
- Draw 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_forwardArrange 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_forward
- for 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_forwardCompare the structure of the nucleoside triphosphate CTP with the structure of ATP. NH₂ 0- 0- 0- ·P—O—P—O—P—O—CH₂ H H H H OH OH Cytidine triphosphate (CTP) Consider the reaction: ATP + CDP ADP + CTP NH 0- 0- 0- ¯0— P—O— P—O—P-O-CH₂ H Η о H H OH OH Adenosine triphosphate (ATP) NH₂ Now predict the approximate K'eq for this reaction. Now predict the approximate AG for this reaction. Narrow_forward
- The standard free energy, AGO, of hydrolysis of inorganic polyphosphate, polyP, is about −20 kJ/mol for each P; released. In a cell, it takes about 50 kJ/mol of energy to synthesize ATP from ADP and Pi. ○ P O Inorganic polyphosphate (polyP) Is it feasible for a cell to use polyP to synthesize ATP from ADP? Why or why not? No. The reaction is unidirectional and always proceeds in the direction of polyP synthesis from ATP. Yes. If [ADP] and [polyP] are kept high, and [ATP] is kept low, the actual free-energy change would be negative. No. The synthesis of ATP from ADP and P; has a large positive G'o compared to polyP hydrolysis. Yes. The hydrolysis of polyP has a sufficiently negative AG to overcome the positive AGO of ATP synthesis. Correct Answerarrow_forwardIn the glycolytic pathway, a six-carbon sugar (fructose 1,6-bisphosphate) is cleaved to form two three-carbon sugars, which undergo further metabolism. In this pathway, an isomerization of glucose 6-phosphate to fructose 6-phosphate (as shown in the diagram) occurs two steps before the cleavage reaction. The intervening step is phosphorylation of fructose 6-phosphate to fructose 1,6-bisphosphate. H H | H-C-OH H-C-OH C=0 HO-C-H HO-C-H phosphohexose isomerase H-C-OH H-C-OH H-C-OH H-C-OH CH₂OPO CH₂OPO Glucose 6-phosphate Fructose 6-phosphate What does the isomerization step accomplish from a chemical perspective? Isomerization alters the molecular formula of the compound, allowing for subsequent phosphorylation. Isomerization moves the carbonyl group, setting up a cleavage between the central carbons. Isomerization causes the gain of electrons, allowing for the eventual release of NADH. Isomerization reactions cause the direct production of energy in the form of ATP.arrow_forwardFrom data in the table, calculate the AG value for the reactions. Reaction AG' (kJ/mol) Phosphocreatine + H₂O →>> creatine + P -43.0 ADP + Pi → ATP + H₂O +30.5 Fructose +P → fructose 6-phosphate + H₂O +15.9 Phosphocreatine + ADP creatine + ATP AG'O ATP + fructose → ADP + fructose 6-phosphate AG'° kJ/mol kJ/molarrow_forward
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