pH = pKa + log([base]/[acid]) pI = (pKa1 + pKa2)/2 If Isoelectric point (pI) < p

Question

pH = pKa + log([base]/[acid])

pI = (pKa1 + pKa2)/2

If Isoelectric point (pI) < pH, the amino acid loses proton and is negatively charged.

Negative Charge:

Aspartic acid – pI = 2.98

Methionine - pI = 5.74

Phenyl alanine – pI = 5.91

If Isoelectric point (pI) ~ pH, the amino acid is neutral.

Neutral Charge:

Isoleucine – pI = 6.04

Valine – pI = 6.02

Glycine – pI = 6.06

If Isoelectric point (pI) > pH, the amino acid gains proton and is positively charged.

Positive Charge:

Asparagine – pI = 10.76

Arginine – pI = 10.76

Lysine – pI = 9.47

Explanation / Answer

es previous | 3 of 21 next Part C A mixture of valine, glycine, isoleucine, aspartic acid. phenylalanine, arginine, lysine, asparagine, and methionine are put into an electrophoresis apparatus, with the buffer pH- 6.5 Sort each amino acid according to its charge in the buffer with a pH of 6.5 Drag each item to the appropriate bin. Hints hich is the e pH is n the pl Reset Help resis ids

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