A mixture of three protiens, (A) (pepsin), (B) (haemo-globin), and (C ) (lysozyme) was seperated by elctrophoresis method at pH=7. The pH at isoelectric point (pI) of the proteins are pI of (A), (B), and (C ) which are 1.1,6.7, and 11.0, respectively. which of the statement are correct?

A mixture of three protiens, (A) (pepsin), (B) (haemo-globin), and (C

1.	A mixture of three protiens, (A) (pepsin), (B) (haemo-globin), and (C ) (lysozyme) was seperated by elctrophoresis method at pH=7. The pH at isoelectric point (pI) of the proteins are pI of (A), (B), and (C ) which are 1.1,6.7, and 11.0, respectively. which of the statement are correct?
Answer» Pepsin (A) will migrate to the CATHODE. Lysozyme (C ) will migrate to the anode. Haemoglobon will not migrate. At pH=7, (A) and (C ) would PRECIPITATE out while (B) would remain in the solution. Solution :Is in a very acidic range, has more anionic groups at `pH=7`, and will migrate to the positive electrode (cathode). `(B)(PI=6.7)` is present mostlywith net zero charge and moves very littel or will not migrate. `(C )(pH=11.0)`, the very BASIC protein, EXISTS mainly in the cationic form and migrates to the negative electrode (anode). At `pI`, amino acids have least solubities, so `(B)` `(pH=6.7)`, `[pH` very close to `pH` of the mixture `(pH=7)]`, is least soluble and would precipitate out, while `(A)` and `(C )` would remain in the solution.