Please explain how to do this. I’ve watched videos and read the chapter on this

Question

Please explain how to do this. I’ve watched videos and read the chapter on this and am confused about how to find the PI for the peptide below. Thank you! Val Glu Ile Lys Met -NH, 9.60 9.69 9.62 9.68 9.68 9.15 10.43 9.21 9.13 9.39 8.80 9.13 10.6 9.82 9.67 9.17 10.78 9.11 8.95 9.04 Acid ?-COOH RH or RHt Gly Ala Val Leu Ile Ser Thr Met Phe Trp Asn Gln Pro Asp Glu His Cys Tyr Lys Arg For these amino acids, the R group ionization occurs before the a-NH, ionization. 2.34 2.34 2.32 2.36 2.36 2.21 2.63 2.28 1.83 2.38 2.02 2.17 1.99 2.09 2.19 1.82 1.71 2.20 2.18 2.17 3.86* 4.25** 6.0* 8.33* 10.07 10.53 12.48

Explanation / Answer

You had been already given the value of pK1 which is alpha-COOH value and pK2 value is the value of alpha NH3+ and value of ionizahble group is RH or RH+ value.

If the side chain does not have an ionizable group then the pI is simply the average of the alpha-NH3 and alpha-COOH pKa values.

If the side chain has ionizable groups then all three pKa must be considered :

1. If the side chain is acidic (asp,glu) then average the side chain pKa with alpha-COOH pKa .

2. If the side chain is basic (his,arg,lys) then average the side chain pKa with alpha-NH3 pKa .

3. For other ionizable groups (tyr, cys) determine which is the middle pKa and average it with the alpha-COOH pKa .

now u need to calculate pI value which can be calculated by using the formula :

pI= (pK1 + pK2 )/2

But for ionizable group follow the above rules.

for Val it is

pI = (2.32 + 9.62)/2 = 5.97 or you can write it down as 6.0

For Glu

pI = (2.19+4.25)/2 = 3.22

For Ile

pI = ( 2.36 + 9.68)/2 = 6.02

For Lys

pI =( 8.95 + 10.53)/2 = 9.74

For Met

pI = (2.28 + 9.21)/2 = 5.74

Related Questions

Navigate

• Browse (All)
• Browse P
• Subjects

---