Concept explainers
(a)
Interpretation: To identify whether lysine can or cannot be synthesized in the body in amounts adequate to meet the body’s needs.
Concept introduction: Amino acids are the main building blocks of proteins. Amino acids are classified as essential and non-essential depending upon whether they can be synthesized within the body itself or not.
Essential amino acids are those amino acids which cannot be synthesized by the body via biosynthesis and thus must be taken from the outside in form of dietary protein to meet the body’s need. Those amino acids which can be synthesized by biosynthesis within the liver are termed as non-essential amino acids.
The essential amino acids are arginine, histidine, isoleucine, leucine, lysine, methionine, phenylalanine, threonine, tryptophan, and valine.
The non-essential amino acids are alanine, asparagines, aspartic acid, cysteine, glutamic acid, glutamine, glycine, proline, serine, and tyrosine.
(b)
Interpretation: To identify whether cysteine can or cannot be synthesized in the body in amounts adequate to meet the body’s needs.
Concept introduction: Amino acids are the main building blocks of proteins. Amino acids are classified as essential and non-essential depending upon whether they can be synthesized within the body itself or not.
Essential amino acids are those amino acids which cannot be synthesized by the body via biosynthesis and thus must be taken from the outside in form of dietary protein to meet the body’s need. Those amino acids which can be synthesized by biosynthesis within the liver are termed as non-essential amino acids.
The essential amino acids are arginine, histidine, isoleucine, leucine, lysine, methionine, phenylalanine, threonine, tryptophan, and valine.
The non-essential amino acids are alanine, asparagines, aspartic acid, cysteine, glutamic acid, glutamine, glycine, proline, serine, and tyrosine.
(c)
Interpretation: To identify whether serine can or cannot be synthesized in the body in amounts adequate to meet the body’s needs.
Concept introduction: Amino acids are the main building blocks of proteins. Amino acids are classified as essential and non-essential depending upon whether they can be synthesized within the body itself or not.
Essential amino acids are those amino acids which cannot be synthesized by the body via biosynthesis and thus must be taken from the outside in form of dietary protein to meet the body’s need. Those amino acids which can be synthesized by biosynthesis within the liver are termed as non-essential amino acids.
The essential amino acids are arginine, histidine, isoleucine, leucine, lysine, methionine, phenylalanine, threonine, tryptophan, and valine.
The non-essential amino acids are alanine, asparagines, aspartic acid, cysteine, glutamic acid, glutamine, glycine, proline, serine, and tyrosine.
(d)
Interpretation: To identify whether tryptophan can or cannot be synthesized in the body in amounts adequate to meet the body’s needs.
Concept introduction: Amino acids are the main building blocks of proteins. Amino acids are classified as essential and non-essential depending upon whether they can be synthesized within the body itself or not.
Essential amino acids are those amino acids which cannot be synthesized by the body via biosynthesis and thus must be taken from the outside in form of dietary protein to meet the body’s need. Those amino acids which can be synthesized by biosynthesis within the liver are termed as non-essential amino acids.
The essential amino acids are arginine, histidine, isoleucine, leucine, lysine, methionine, phenylalanine, threonine, tryptophan, and valine.
The non-essential amino acids are alanine, asparagines, aspartic acid, cysteine, glutamic acid, glutamine, glycine, proline, serine, and tyrosine.
Want to see the full answer?
Check out a sample textbook solutionChapter 26 Solutions
General, Organic, and Biological Chemistry
- Don't used Ai solutionarrow_forwardDon't used Ai solutionarrow_forward5. A solution of sucrose is fermented in a vessel until the evolution of CO2 ceases. Then, the product solution is analyzed and found to contain, 45% ethanol; 5% acetic acid; and 15% glycerin by weight. If the original charge is 500 kg, evaluate; e. The ratio of sucrose to water in the original charge (wt/wt). f. Moles of CO2 evolved. g. Maximum possible amount of ethanol that could be formed. h. Conversion efficiency. i. Per cent excess of excess reactant. Reactions: Inversion reaction: C12H22O11 + H2O →2C6H12O6 Fermentation reaction: C6H12O6 →→2C2H5OH + 2CO2 Formation of acetic acid and glycerin: C6H12O6 + C2H5OH + H₂O→ CH3COOH + 2C3H8O3arrow_forward
- Show work. don't give Ai generated solution. How many carbons and hydrogens are in the structure?arrow_forward13. (11pts total) Consider the arrows pointing at three different carbon-carbon bonds in the molecule depicted below. Bond B 2°C. +2°C. cleavage Bond A •CH3 + 26.← Cleavage 2°C. + Bond C +3°C• CH3 2C Cleavage E 2°C. 26. weakest bond Intact molecule Strongest 3°C 20. Gund Largest argest a. (2pts) Which bond between A-C is weakest? Which is strongest? Place answers in appropriate boxes. C Weakest bond A Produces Most Bond Strongest Bond Strongest Gund produces least stable radicals Weakest Stable radical b. (4pts) Consider the relative stability of all cleavage products that form when bonds A, B, AND C are homolytically cleaved/broken. Hint: cleavage products of bonds A, B, and C are all carbon radicals. i. Which ONE cleavage product is the most stable? A condensed or bond line representation is fine. 13°C. formed in bound C cleavage ii. Which ONE cleavage product is the least stable? A condensed or bond line representation is fine. • CH3 methyl radical Formed in Gund A Cleavage c.…arrow_forwardBr. COOH Br, FCH COOH E FeBr ASOCI B NH (CH,CO),OD Br₂ 2 C alcKOHarrow_forward
- Find A to F (all)arrow_forwardShow work. don't give Ai generated solutionarrow_forwardHi!! Please provide a solution that is handwritten. Ensure all figures, reaction mechanisms (with arrows and lone pairs please!!), and structures are clearly drawn to illustrate the synthesis of the product as per the standards of a third year organic chemistry course. ****the solution must include all steps, mechanisms, and intermediate structures as required. Please hand-draw the mechanisms and structures to support your explanation. Don’t give me AI-generated diagrams or text-based explanations, no wordy explanations on how to draw the structures I need help with the exact mechanism hand drawn by you!!! I am reposting this—ensure all parts of the question are straightforward and clear or please let another expert handle it thanks!!arrow_forward
- Hi!! Please provide a solution that is handwritten. Ensure all figures, reaction mechanisms (with arrows and lone pairs please!!), and structures are clearly drawn to illustrate the synthesis of the product as per the standards of a third year organic chemistry course. ****the solution must include all steps, mechanisms, and intermediate structures as required. Please hand-draw the mechanisms and structures to support your explanation. Don’t give me AI-generated diagrams or text-based explanations, no wordy explanations on how to draw the structures I need help with the exact mechanism hand drawn by you!!! I am reposting this—ensure all parts of the question are straightforward and clear or please let another expert handle it thanks!!arrow_forward. (11pts total) Consider the arrows pointing at three different carbon-carbon bonds in the molecule depicted below. Bond B 2°C. +2°C. < cleavage Bond A • CH3 + 26. t cleavage 2°C• +3°C• Bond C Cleavage CH3 ZC '2°C. 26. E Strongest 3°C. 2C. Gund Largest BDE weakest bond In that molecule a. (2pts) Which bond between A-C is weakest? Which is strongest? Place answers in appropriate boxes. Weakest C bond Produces A Weakest Bond Most Strongest Bond Stable radical Strongest Gund produces least stable radicals b. (4pts) Consider the relative stability of all cleavage products that form when bonds A, B, AND C are homolytically cleaved/broken. Hint: cleavage products of bonds A, B, and C are all carbon radicals. i. Which ONE cleavage product is the most stable? A condensed or bond line representation is fine. 人 8°C. formed in bound C cleavage ii. Which ONE cleavage product is the least stable? A condensed or bond line representation is fine. methyl radical •CH3 formed in bund A Cleavagearrow_forwardWhich carbocation is more stable?arrow_forward
- General, Organic, and Biological ChemistryChemistryISBN:9781285853918Author:H. Stephen StokerPublisher:Cengage LearningOrganic And Biological ChemistryChemistryISBN:9781305081079Author:STOKER, H. Stephen (howard Stephen)Publisher:Cengage Learning,Chemistry for Today: General, Organic, and Bioche...ChemistryISBN:9781305960060Author:Spencer L. Seager, Michael R. Slabaugh, Maren S. HansenPublisher:Cengage Learning
- Introduction to General, Organic and BiochemistryChemistryISBN:9781285869759Author:Frederick A. Bettelheim, William H. Brown, Mary K. Campbell, Shawn O. Farrell, Omar TorresPublisher:Cengage LearningWorld of Chemistry, 3rd editionChemistryISBN:9781133109655Author:Steven S. Zumdahl, Susan L. Zumdahl, Donald J. DeCostePublisher:Brooks / Cole / Cengage LearningWorld of ChemistryChemistryISBN:9780618562763Author:Steven S. ZumdahlPublisher:Houghton Mifflin College Div