Parts C and F are wrong. Please help answer Part C and Part F. Thank you! Answer
ID: 1083809 • Letter: P
Question
Parts C and F are wrong. Please help answer Part C and Part F. Thank you!
Answer the following seven questions (a through g) based upon a tripeptide sequence with the following amino acids (one-letter symbols are given): S A D a) What are the full names of the amino acids? Spelling counts, but capitalization does not. For an acidic amino acid, assume it is ionized and use the - ate ending instead of-ic acid. S: Serine A: Alanine D Aspartate b) What are the three-letter abbreviations for the amino acids? S Ser A: Ala D: Asp c) What is the overall charge of the most abundant tripeptide species at neutral pH (pH 7.0)? Number Enter your answer as a whole number. d) At physiological pH, what is the best description of the chemical properties of the side chain (R group)? polar, un aged-D E R A hydrophobic D negatively chargedExplanation / Answer
Peptide sequence of SAD has three amino acids namely serine, alanine and aspartate. There are three ionizable groups in this peptide sequence as given below:
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A functional group (amino or carboxylic acid) is in protonated form if pH is less than its pKa. Thus, amino group of serine will be in protonate form (NH3+ form) at pH less than 9.2 and it will have +1 charge. At pH above 9.2 serine amino group will be in deprotonated form (NH2 form) and will have charge zero.
At pH less than 2.1, alpha carboxylic acid functional group willl be in protonated form (COOH form) and it will have charge zero. Above pH = 2.1, it will be in deprotonate form (COO - form) and it will have charge of -1.
At pH less than 3.9, side chain carboxylic acid functional group willl be in protonated form (COOH form) and it will have charge zero. Above pH = 3.9, it will be in deprotonate form (COO - form) and it will have charge of -1.
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Thus, at pH less than 1, all the three functional groups will be in protonated form (NH3+ form and COOH form).Therefore, charge on peptide will be +1 (+1+0+0= +1)
As pH is increased to 7, only amino group will be protonated (NH3+ form) and other two carboxylic acid groups will be deprotonated form(COO - form). Therefore, at pH = 7 total charge on peptide will be -1 (+1-1-1 = -1).
As pH is further increased to 11.5, all the functional groups will be in deprotonated form (NH2 form and COO – form). Therfore, at pH = 11.5, total charge on peptide will be -2 (0-1-1 = -2)
Thus, correct answer of option C is -1 and correct answer of option F is -2.