Question1 3.33/5 pts 1.The correct formula for getting the isoelectric point for the tripeptide shown in question #2 is [ Select] 2. For this tripeptide, (shown at a pH of 7) 1 CH-C-N-CHcOO CH CH2 H CH2 H2C CH3 H2C 3 ?? i. The amino terminal residue is [Select ] caboxyl terminal end is Tyrosine ii. The one-letter code formula for this tripeptide would be and the [ Select ] ii. How many ionizable protons (thus pKa values) does this tripeptide contain. [Select ] iv. As pictured above at a pH of 7, what is the net charge on this tripeptide [Select ]
Solution
1.The tripeptide is Val-Glu-Tyr. The isoelectric point is an average of pka values of protonation and deprotonation of neutral form of the peptide. In this case it should be the avergae of pka of alpha COOH group of terminal Tyrosine and side chain COOH group of glutamic acid.
pka of alpha COOH of tyrosine= 2.20
pka of side chain COOH of glutamic acid =4.25
Therefore formula is I.P. = (pka 1 + pka)/2 = (2.20 + 4.25)/2= 3.225
2. i) The amino terminal residue is Valine.
ii) One letter code formula is V-E-Y
iii) Three ionizable protons- for the two carboxylic acids each and one for the alpha amino group.
iv) The net charge at pH 7 is -1.
.
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Question1 3-33-5 pts 1-The correct formula for getting the isoelectric.docx
1. Question1 3.33/5 pts 1.The correct formula for getting the isoelectric point for the tripeptide
shown in question #2 is [ Select] 2. For this tripeptide, (shown at a pH of 7) 1 CH-C-N-CHcOO
CH CH2 H CH2 H2C CH3 H2C 3 ?? i. The amino terminal residue is [Select ] caboxyl terminal
end is Tyrosine ii. The one-letter code formula for this tripeptide would be and the [ Select ] ii.
How many ionizable protons (thus pKa values) does this tripeptide contain. [Select ] iv. As
pictured above at a pH of 7, what is the net charge on this tripeptide [Select ]
Solution
1.The tripeptide is Val-Glu-Tyr. The isoelectric point is an average of pka values of protonation
and deprotonation of neutral form of the peptide. In this case it should be the avergae of pka of
alpha COOH group of terminal Tyrosine and side chain COOH group of glutamic acid.
pka of alpha COOH of tyrosine= 2.20
pka of side chain COOH of glutamic acid =4.25
Therefore formula is I.P. = (pka 1 + pka)/2 = (2.20 + 4.25)/2= 3.225
2. i) The amino terminal residue is Valine.
ii) One letter code formula is V-E-Y
iii) Three ionizable protons- for the two carboxylic acids each and one for the alpha amino
group.
iv) The net charge at pH 7 is -1.
![Question1 3.33/5 pts 1.The correct formula for getting the isoelectric point for the tripeptide
shown in question #2 is [ Select] 2. For this tripeptide, (shown at a pH of 7) 1 CH-C-N-CHcOO
CH CH2 H CH2 H2C CH3 H2C 3 ?? i. The amino terminal residue is [Select ] caboxyl terminal
end is Tyrosine ii. The one-letter code formula for this tripeptide would be and the [ Select ] ii.
How many ionizable protons (thus pKa values) does this tripeptide contain. [Select ] iv. As
pictured above at a pH of 7, what is the net charge on this tripeptide [Select ]
Solution
1.The tripeptide is Val-Glu-Tyr. The isoelectric point is an average of pka values of protonation
and deprotonation of neutral form of the peptide. In this case it should be the avergae of pka of
alpha COOH group of terminal Tyrosine and side chain COOH group of glutamic acid.
pka of alpha COOH of tyrosine= 2.20
pka of side chain COOH of glutamic acid =4.25
Therefore formula is I.P. = (pka 1 + pka)/2 = (2.20 + 4.25)/2= 3.225
2. i) The amino terminal residue is Valine.
ii) One letter code formula is V-E-Y
iii) Three ionizable protons- for the two carboxylic acids each and one for the alpha amino
group.
iv) The net charge at pH 7 is -1.](https://image.slidesharecdn.com/question13-33-5pts1-thecorrectformulaforgettingtheisoelectric-230131184949-b2619830/85/Question1-3-33-5-pts-1-The-correct-formula-for-getting-the-isoelectric-docx-1-320.jpg)
