At $pH =2 \quad \stackrel{1}{ N } H _2 \text { and } \stackrel{2}{ N } H _2 \text { of Tyrosine and Lysine is +ve charged ( }+1 \text { each) }$

$+2\left|z_1\right|=2$

$\text { At } pH =6 \quad NH _2 \text { of Lysine ( }+1 \text {, }$

$COOH (-1) \text { of glutamic acid, }$

$\text { so because of dipolar ion exist }\left|z_2\right|=0$

$\text { At } pH =11 \quad COOH \text { of Glutamic acid (-1) }$

$COOH \text { of Lysine (-1) }$

$OH \text { of phenol ( }-1 \text { ) }$

$\left|z_3\right|=3$