Why is the net charge 0 for cooh? Shouldn't it be +1?
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Why is the net charge 0 for cooh? Shouldn't it be +1?
![Glutamate, Glu, E
H
HƠN - C co
T
CH₂
pha 4,5
PH 2.1 +NH3 +1
4.1 *NH3
+1
CH₂
4,5 ENH2JINH 23 H2 có
-
pka 4.1
PH 2.1 COOH
4.1 [co] = [CODA] -1/2
9.5 CDD
Overall charge pH 2.3 +1/2
2
4.1
-1/₂ >
9.5 - 11/2
5
N
3
pka 2.1
at PH 2.1 [co] = [cook] - 1/₂
4.1
Coo
-1
9.5
4
All
)?
PI = 2₁1 + 4.1 = 3.1
2
COO
8
Sep 16
1:26 +](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fe9a96325-e1c3-4b19-b1b4-8ad61c01db3d%2F91276cd1-9b94-47e6-bab9-6ba14c8bf275%2Fwjhi1jn_processed.png&w=3840&q=75)
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- Lefer to the figure showing the molecular structure of dimethylmercury. H–C-Hg C-H H. Exposure of human skin to a single drop of dimethylmercury can lead to death, as it is highly poisonous and passes easily cell membranes. Based on its structure, why is it able to pass so easily through cell membranes? I-0-I IICIHConsider the positively charged amino acid lysine Lys2+ 21 COOH I H&N-C-H I pH 14 12 10 8 6 4 2 0 CH₂ I CH₂ I CH₂ I CH₂ T NH₂+ 0 Nelson p85 2.18 = 2.18 PK₁ Lys+ COO™ I H₂N-C-H H₂N-C-H ī I -----) 8.95 Lysº 8.95 pK₂ pka carboxyl = 2.19 pkaamino = 9.67 pka sidechain = 4.25 COO™ I CH₂ I CH₂ I CH₂ I CH₂ I NH₂¹ 1.0 2.0 Equivalents of OH added- COO™ I H₂N-C-H I 10.79 1 10.79 pk Isoelectric point Lys CH₂2 I CH₂ I CH₂ I CH₂ T NH₂ 3.0 +H3N NH3+ T CH₂ T CH₂ CH₂ CH₂ -COO™ H Lysine (Lys, K) Physiological pH = 7.4 < pl → Amino acid is positively charged at physiological pH 1. Consider glutamate in its fully protonated form (e.g. in a pH = 1 solution) 1) Draw all the forms of glutamate at various pH 2) Calculate the pl of this amino acid 3) Sketch a titration curve showing pH as a function of added [OH-] and locate the predominant forms of histidine in the curve STEPS: 1. Find the H atoms that can be removed on the molecule 2. Associated a pka value to each removable H. 3. Draw the Aa structure at:…NAZO NHZ Ala-Cys-Glu -Tyr - Trp - Lys - Arg - His -Pro-G ly Glu pka 4.15 SH Tyr 10.10 Draw Charges Lys 10.67 Olt A3 12.10 +NH₂ Ntrm 2) Calculate net charge 3) write out I letter code 300 Ctim 3 juli of peptich (above) Ⓒ pH; 1,7,12
- Based on the following titration table of Glycine, would you be able to recreate the same for Histidine? Glicina Increasing pH H-N-CH,-COOH -H-N-CH,-co = H-N-CH,-C00- Net charge: +1 pH COOH-pKa = 2.3 NH3-pka = 9.6 CN AEq +1 +1 2.3 (0.5°-1)+(0.5°0)=-0.5 +0.5 0.5 9.6 -1 (0.5*+1)+(0.5°0)=+0.5 -0.5 1.5 14 Given that the pKas for Histidine are the following: pka1 = 1.82 pKa2 = 9.17 pКаз %3 6.00 pL = 7.59 2.The skeletal structures of the two amino acids, glycine and lysine, are given below along with the values of the relevant acid dissociation constants (pKa). NH,* PK, - 10.79 - HạN*CH2CO,¯ S pK¸=2.35 (CH2)4 - H3N*CHCO,- pK,=9,78 lysine (Lys) glycine (Gly) pK, = 9.18 pK, - 2.16 For an aqueous solution of glycine alone, calculate the value of pH at which the ratio of the concentration of neutral glycine zwitterions to the concentration of protonated cation is 102. On your under each of the following conditions. In the blank, provide the total charge of the dipeptide. (Example: If the charge is two plus write the answer simply as 2, if the charge is negative two write the answer as -2). draw the skeletal structure of the dipeptide, Lys-Gly, when it is solvated in an aque us solution i. pH= 1 ii. pH= 12OOC H₂C 1 H₂C H₂C H₂C-C HC H₂C NH N CON HN C COO™® 1 CH₂ CH₂ C-CH₂ CH C-CH, G=CH₂ "ooc H₂C H₂C H₂C H₂C-C HC 0-N H₂C Fe(ll) COO 1 CH₂ CH₂ C-CH₂ C-CH₂ G=CH₂ The structure below is Fe- protoporphyrin IX. It is linked by bridging groups to form porphyrin ring. Addition of Fe to the porphyrin rings produces the heme group. Identify the group of atoms that can for H-bond interaction with water.
- The amino acid arginine ionizes according to the following scheme: NH, NH2 NH2 H NH, C=N C=N C=N C=N-H pk = 2.17 -H pK. = 8.99 -H pK = 125 -H* H. NH NH NH NH H (CH,)a H (CH), +H* (CH)a (CH,)a H,N-C-coo- +H* +H* H-N*-C-COOH H-N-C-Co- H,N-C-co- H. нн H. H II II IV (a) Calculate the isoelectric point of arginine. You can neglect contributions from form I. Why? (b) Calculate the average charge on arginine when pH = 9.20. (Hìnt: Find the average charge for each ionizable group and sum these together.) (c) Is the value of average charge you calculated in part b reasonable, given the pl you calculated in part a? Explain your answer.T At pH=2.2, which of the following is true ÇOOH ÇO0 Ç00 HN-CH H,N-CH H&N-CH HN-CH CH2 CH2 CH, pK2 CH2 CH2 pK, CH2 pKR CH2 CH2 COOH СООН COO COO- II IV O 50 % of carboxylic acid group in the side chain is ionized O Most of the molecules are in the form II None 50% of carboxylic acid group attached to alpha carbon is ionized 50% of amino group is ionizedWhich of the following shows the reaction by which valine and leucine form a peptide bond H. H. H O -COO H2N-C-C-NH- C- C-OH + HO CH2 CH,SH ČH3 А. C. CH(CH)2 H. HN-C-C-NH-C- C-OH H,N-CH-C- + H,O HC-C-H CH2 CH2 H CH3 CH H,C-C-H CH(CH3)2 CH В. D.
- Based on the following titration table of Glycine, would you be able to recreate the same for Histidine? Increasing pH – Glicina H H-N-CH,-COOH = H-N-CH,-CO0- = H-N-CH,-coo- H H Net charge: +1 -1 pH COOH-pKa = 2.3 NH3-pKa = 9.6 CN ΔΕ +1 +1 2.3 (0.5*-1)+(0.5*0)=-0.5 +1 +0.5 0.5 9.6 -1 (0.5*+1)+(0.5*0)=+0.5 -0.5 1.5 14 -1 -1 2 Given that the pKas for Histidine are the following: pka1 = 1.82 pKa2 = 9.17 pКа3 %3D 6.00 pl = 7.59AMINO ACIDS 1. Write the tripeptide structure for: val-ser-cys. 2. Write the reactions for: phe + serthe st 6. velv, ) At pH 7.0, the amino acid lysine has the structure shown here. Draw the structure of this molecule at the pH of 2.0 and 13.0 respectively. H₂N-CH- CH₂ CH₂ CH₂ CH₂ ΘΝΗ, Lysine -00
