What z Net charge of a dipeptide Glu-Lys at physiological pH?

Answer 1

The net charge is 0. First, lets look at the two amino acids separately. For Lys, you have:

pka1=2.2 (coo-)

pka2=9.1(NH2+)

pka3=10.5 (lateral chain)

For glu, you have:

pka1=2.1(COO-)

pka2=4.1 (lateral chain)

pka3=9.5(NH2+)

First, below 2.1:

100% of : NH2+---------CH----CO-NH---CH-----COOH (CHARGE=2+)

COOH NH+

At 2.1:

50% of :NH2+---------CH----CO-NH---CH-----COOH

COOH NH+

50% of :NH2+---------CH----CO-NH---CH-----COOH (here, we see that there was no

COOH NH+ change, WHY: the carboxyl of glu should be ionized at 2.1, but it is attached to the amino group of lys, so there is no change

At 2.2

...of NH2+---------CH----CO-NH---CH-----COO- (charge+1)

COOH NH+

At 4.1

....of NH2+---------CH----CO-NH---CH-----COO- (CHARGE=0)

COO- NH+

At pI = 6.8

100 % of NH2+---------CH----CO-NH---CH-----COO- (CHARGE=0)

COO- NH+

Since the next pka is at 9.1, we don't have any change yet at the physical pH 7.4. Thus, the charge is zero. I didn't put the pI in between b4, it would just take too much time, but I hope that you do understand the concept now.

Good luck