A synthetic peptide contains 4 negatively charged and 8 positively charged amino acids at pH 6.1. When the peptides are treated with an alkaline solution having a pH of 8.9, 4 positively charged groups become deprotonated. At pH 8.9, what is the net charge on that peptide? What will be the direction of movement (towards cathode or anode) if the alkali treated peptides are now placed in an electric filed? 0; Movement of the peptide will not happen -8; Movement of the peptide will be towards anode +12; Movement of the peptide will be towards anode -4; Movement of the peptide will be towards cathode

The net charge of a peptide can be calculated by subtracting the number of negatively charged amino acids from the number of positively charged ones.

Initially, the peptide has 8 positively charged and 4 negatively charged amino acids. So, the net charge at pH 6.1 is 8 - 4 = +4.

When the pH is increased to 8.9, 4 of the positively charged groups become deprotonated, meaning they lose a proton and therefore lose their positive charge. So, the number of positively charged amino acids decreases from 8 to 4.

Now, the net charge at pH 8.9 is 4 (positively charged) - 4 (negatively charged) = 0.

So, the net charge on the peptide at pH 8.9 is 0.

In an electric field, particles move according to their charge. Positively charged particles move towards the cathode (negative electrode), and negatively charged particles move towards the anode (positive electrode).

Since the peptide has a net charge of 0, it will not move in the electric field.

So, the correct answer is "0; Movement of the peptide will not happen".