Question

14.Which of the following amino acids have a net charge of +1 at pH 4 and a net charge of 0 at pH 8?

a) Glu
b) Arg
c) His
d) Tyr
e) Two of the above

Answer 1

the answer is a table like this
pK1 pK2 pKr pH Charge
Glu 2.19 9.67 4.25
-1 +1 * 4 Near 0
-1 +1 -1 8 -1
Arg 2.17 9.04 12.48
1 +1 +1 4 +1
-1 +1 +1 8 +1
His 1.82 9.17 6.00
-1 +1 +1 4 +1
-1 +1 0 8 0
Tyr 2.20 9.11 10.07
1 +1 +1 4 +1
-1 +1 +1 8 +1

Answer 2

can someone explain how to do the table to me?

Answer 3

use Henderson-Hasselbalch equation for every ionizable group to find the proportions of the species present at a given pH.

Answer 4

oh wait, that's already done for you on the right end of the table.

Answer 5

Just look at the 2 columns on the right

Answer 6

ok, I want to know how to find the charge using the henderson-hasselbach? can u go through Glu for me?

Answer 7

i guess i am having trouble finding the pk1,pk2 charge using the henderson-hasselbach eq.

Answer 8

you dont need to for this exercise (although we can if you want), everything you need is already calculated for you.

\(\color{red}{important}\)
pK1 pK2 pKr pH Charge
Glu 2.19 9.67 4.25
-1 +1 * 4 Near 0
-1 +1 -1 8 -1
Arg 2.17 9.04 12.48
1 +1 +1 4 +1
-1 +1 +1 8 +1
His 1.82 9.17 6.00
-1 +1 +1 \(\color{red}4\) \(\color{red}{+1}\)
-1 +1 0 \(\color{red}8\) \(\color{red}0\)
Tyr 2.20 9.11 10.07
1 +1 +1 4 +1
-1 +1 +1 8 +1

Answer 9

i would like to do it in case i have to fill out the table on an exam

Answer 10

okay, cool. so we use: \(pH=pKa+log\dfrac{[A^-]}{[HA]}\)

we need the pKa's Glu 2.19 4.25 9.67

at pH=4, \(8=2.19+log\dfrac{[A^-]}{[HA]}\)

\(\dfrac{[A^-]}{[HA]}=\dfrac{64.565}{1}\)

\([HA]=\dfrac{1}{1+64.565}*100\%=1.525\%\)

\([A^-]=\dfrac{64.565}{1+64.565}*100\%=98.475\%\)

so the \(\alpha -COOH\) exists 98.5% as \(COO^-\) and 2.5% as COOH.

Answer 11

You'd have to do this for every pH value.

note that this is given in the table

Glu 2.19 9.67 4.25
\(\color{red}{-1}\) +1 * 4 Near 0

Answer 12

uh how did u get 8 as the pH?

Answer 13

sorry i meant 4, i did the calculations with pH=4. it was a typographical error.

Answer 14

o ok that makes sense now

Answer 15

so what does the +1 or -1 mean when u find the ratio of [A-] to [HA]

Answer 16

that is the part i have trouble on finding how the amino acid in a chain is +1 -1 or 0. So if I can learn how to do it I can apply to other problems

Answer 17

first, you should know which of the pKa's corresponds to which ligand, because it's different if you have \(RNH_3^+\) or \(COOH\), one is charged and one isn't.

SO first, step would be to draw these things all protonated then draw all the ionizations as you work your way up the pH scale.

Now do the calculations, and you can extract the charge of each ligand from the percentages. For example, for \(\alpha\)-COOH, of Glu,

α−COOH exists 98.5% as COO− and 2.5% as COOH.

so we say that it has a -1 charge since most (98%) of it is COO−.

After you do this for each ligand, sum them up, for example,

Glu 2.19 9.67 4.25
-1 +1 0 pH = 4 charge=1+1+0=0

^ half is charged half is uncharged
so this would have a partial charge

Answer 18

by the way in the question they omitted pKa=4.25, i wrote zero (which is wrong) but disregard that.

Answer 19

uh is there a way to view the question marks?

Answer 20

refresh the page, i copied and pasted some of the stuff i wrote before and it's not recognizing it.

Answer 21

o ok

Answer 22

ok i think i understand now. appreciate the help

Answer 23

no problem

Answer 24

um so the second pkNH3+, it is [BH+]/[B]. So if we get something like [BH+]/[B]=.02137962, is that 0.02137962 as [BH+] and 1 as [B]?

Answer 25

@aaronq sorry idk if it shows a closed question if i respond again

Answer 26

\(\dfrac{[BH^+]}{[B]}=\dfrac{0.0214}{1}\)

\([BH^+]=\dfrac{0.0214}{1+0.0214}*100\%\)

\([B]=\dfrac{1}{1+0.0214}*100\%\)

Answer 27

so it is 2% in [BH+] form and 98% in [B]. what charge do u give this? is it 0 since it is in the B form and not BH+?

Answer 28

@aaronq

Answer 29

sorry, the site is not sending me notifications.

it would be zero charge for this group. i think you might've done this backwards, in the henderson hasselbalch you used the pKa not the pKb, so the ratio is \(\dfrac{base}{acid}=\dfrac{[B]}{[BH^+]}\)

Answer 30

o i c, they give it as pka, but i was using the henderson for pkb. so flip them since i am using pka. ok thx