### Hexapeptide Charge and pK Analysis**Step 1**: Determine the total positive charge on the hexapeptide when all acidic and basic groups are fully protonated. Enter your answer without the sign.- **Charge**: [ _______ ]**Step 2**: Determine the pK value of the group that **loses** a proton to form the zwitterionic species with a net charge of zero.- **pK**: [ _______ ]**Step 3**: Determine the pK value of the group that **gains** a proton to form the zwitterionic species with a net charge of zero.- **pK**: [ _______ ]**Step 4**: Estimate the pI value of the hexapeptide.- **pI**: [ _______ ] ### Understanding Isoelectric Focusing and pI Calculation of PeptidesIsoelectric focusing can separate peptides based on their relative contents of acidic and basic residues. When voltage is applied to an electrophoretic gel with a pH gradient, each peptide migrates until it reaches a point where it has no net charge. This pH at this point is referred to as the isoelectric point (pI) for the peptide.One can estimate the pI fairly accurately (within 0.1 or 0.2 pH units) using the pK values of all the proton dissociable groups in the peptide.#### Example HexapeptideConsider the hexapeptide **Met-Asp-Lys-Arg-Ala-Tyr**.#### Reference Table of pK Values| Amino acid | pK || ------------------ | ---- || Arg | 12.5 || Asp | 3.7 || Cys | 8.2 || Glu | 4.3 || His | 6.0 || Lys | 10.5 || Tyr | 10.5 || *peptide-NH₃⁺ | 8.0 || *peptide-COOH | 3.4 |**Note:** Amino and carboxyl terminal values differ from the α amino and carboxy values of a single amino acid.### Calculating the Total Positive Charge**Step 1:** Determine the total positive charge on the hexapeptide when all acidic and basic groups are fully protonated. Enter your answer without the sign.**Charge:** [Input Box]### Detailed Explanation of the TableThe table provided lists the pK values for the amino acids and terminal groups. Each amino acid has specific pK values that need to be considered for calculating the net charge at a given pH, and ultimately the pI of the peptide. For instance:- **Arg** has a high pK value of 12.5, indicating it will contribute to the positive charge until the pH exceeds this value.- **Asp** and **Glu** have low pK values (3.7 and 4.3, respectively) indicating they will lose their proton and contribute to the negative charge at relatively low pH values.Using this table, you can sequentially determine which groups

Step 1: Determine the total positive charge on the hexapeptide when all acidic and basic groups are fully protonated. Enter your answer without the sign. charge: Step 2: Determine the pK value of the group that loses a proton to form the zwitterionic species with a net charge pK = of zero. Step 3: Determine the pK value of the group that gains a proton to form the zwitterionic species with a net charge pK = of zero. Step 4: Estimate the pl value of the hexapeptide. pI =

Step 1: Determine the total positive charge on the hexapeptide when all acidic and basic groups are fully protonated. Enter your answer without the sign. charge: Step 2: Determine the pK value of the group that loses a proton to form the zwitterionic species with a net charge pK = of zero. Step 3: Determine the pK value of the group that gains a proton to form the zwitterionic species with a net charge pK = of zero. Step 4: Estimate the pl value of the hexapeptide. pI =

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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Question 10.d of 11 Submit using Cr/Cr3* and Zn/Zn²* at 25 °C. The following half-reactions are provided for each metal: Cr3* (aq) + 3 e → Cr(s) E°red = -0.744 V Zn* (aq) + 2 e → Zn(s) E°red = -0.763 V Given that the standard cell potential is 0.019 V and the overall equation is 2 Cr³* (aq) + 3 Zn (s) → 2 Cr (s) + 3 Zn²* (aq), what is the cell potential for this cell at 25 °C when [Zn²*] = O.0243 M and [Cr³*] = 0.1656 M? %3D | v 1 2 4 6. C 7 8 +/- x 10 0 Tap here or pull up for additional resources 3. LO
a) If the Kb of a weak base is 4.7E-9, what is the Ka of its conjugate acid? Do not include any units with your answer. Your answer should have two significant figures. For scientific notation use the same as that in the problem (i.e. 3.3E-2 is the same as 3.3 x 10-2). b) Diethylamine, (CH3CH2)2NH, has a Kb of 8.6 x 10-4. What is the pH of a 0.22 M solution of diethylammonium chloride, [(CH3CH2)2NH2]Cl? Your answer should be to two places past the decimal point.
A student added a total of 75 mL sodium hydroxide to titrate an amino acid solution with 0.5M NaOH. They also calculated that they added a total of 14 mL of sodium hydroxide to arrive at pH 2.6 and a total of 32 mL of sodium hydroxide to arrvive at pH 4.8.What was the initial concentration of the amino acid solution in milimolar (mM)? The calculations must be presented such that those can be recapitulated by the marker. Provide the answer with no decimal places. Show all details of your working out. State the answer in a full sentence.
Will 3-methylhexanoic acid be positive or neutral with an pH of 6.38?
Abbreviating malonic acid, CH2(CO2H)2, as H2A (pK1 = 2.847, pK2 = 5.696), find the pH of 0.100 M solution of NaHA.
ball & stick This amino acid is Not started Submit Answer - + labels to access important values if needed for this question. Retry Entire Group 9 more group attempts remaining < ENG 6
An unknown amino acid labeled KEMISTRI has an IpH value equal to 5.41. If its pKa1 is equal to 2.03, calculate the pKa2 of KEMISTRI. Show complete solutions. Based on the table below, what is the probable identity of KEMISTRI?
The pka's of the ascorbic acid are pka1 = 4.10 and pka2 = 11.80. If the net charge of ascorbic acid is exactly -1, what is the pH of the solution? Use 2 decimal places.
Consider the following five weak acids: Succinic acid: pKa = 4.16HOBr: Ka = 2.06 x 10-9Dinicotinic: pKa = 2.08Nitrobenzoic: Ka = 6.95 x 10-3Saccharin: pKa = 11.68 Order these acids from the strongest (1) to the weakest (5)
An unknown compound, X, is thought to have a carboxyl group with a pK₂ of 2.5 and a second ionizable group with a pKa between 5.0 and 8.0. When 50.0 mL of 0.2 M NaOH is added to 75.0 mL of a 0.2 M solution of X at pH 2.5, the pH increases to 7.13. Calculate the pK₂ of the second ionizable group of X. pK₂ = 6.97 Incorrect