Amino acids are bifunctional having - COOH and - NH 2 groups. Depending upon the position, they may by α amino acids, β, γ, or δ - amino acids. There are about 20 α amino acids, commonly found in proteins. They are soluble in water, but insoluble in organic solvents. It is due to formation of zwitterions. Thee show the properties of - NH 2 group as well as - COOH group. Every amino acid has its own characteristic isoelectric point. What is the structure of the a bove amino acid in a solution of pH =9.87 ?

Question

Amino acids are bifunctional having - COOH and - NH 2 groups. Depending upon the position, they may by α amino acids, β, &gamma;, or δ - amino acids. There are about 20 α amino acids, commonly found in proteins. They are soluble in water, but insoluble in organic solvents. It is due to formation of zwitterions. Thee show the properties of - NH 2 group as well as - COOH group. Every amino acid has its own characteristic isoelectric point. What is the structure of the a bove amino acid in a solution of pH =9.87 ? (A) H 2 N - CH 2- CH 2- CH 2- CH 2- underset overset|⊕NH3CH- COOH (B) H 3 overset⊕N - CH 2- CH 2- CH 2- CH 2- underset overset|⊕NH3CH - COOH (C) H 3 overset⊕N - CH 2- CH 2- CH 2- CH 2- underset overset|⊕NH3CH - COO - (D) H 3 overset⊕N - CH 2- CH 2- CH 2- CH 2- underset overset|⊕NH3CH - COO -

Tardigrade Admin · Accepted Answer

Since the given amino acid has two - NH 2 group and one COOH group, the isoelectric point is obtained as the average value of pK a of both - NH 2 groups only. Thus pI =(8.95+10.79/2)=(19.47/2)=9.87

Q. What is the structure of the a bove amino acid in a solution of $p H = 9.87 ?$

$H_{2} N - C H_{2} - C H_{2} - C H_{2} - C H_{2} - \oplus N H_{3} ∣ C H - COO H$

$H_{3} N \oplus - C H_{2} - C H_{2} - C H_{2} - C H_{2} - \oplus N H_{3} ∣ C H - COO H$

$H_{3} N \oplus - C H_{2} - C H_{2} - C H_{2} - C H_{2} - \oplus N H_{3} ∣ C H - CO O^{-}$

$H_{3} N \oplus - C H_{2} - C H_{2} - C H_{2} - C H_{2} - \oplus N H_{3} ∣ C H - CO O^{-}$

Since the given amino acid has two $- N H_{2}$ group and one $COO H$ group,
the isoelectric point is obtained as the average value of $p K_{a}$ of both $- N H_{2}$ groups only.
Thus $p I = \frac{8.95 + 10.79}{2} = \frac{19.47}{2} = 9.87$

Q. What is the structure of the a bove amino acid in a solution of pH=9.87?

H2​N−CH2​−CH2​−CH2​−CH2​−⊕NH3​∣​CH​−COOH

H3​N⊕−CH2​−CH2​−CH2​−CH2​−⊕NH3​∣​CH​−COOH

H3​N⊕−CH2​−CH2​−CH2​−CH2​−⊕NH3​∣​CH​−COO−

H3​N⊕−CH2​−CH2​−CH2​−CH2​−⊕NH3​∣​CH​−COO−

Since the given amino acid has two −NH2​ group and one COOH group, the isoelectric point is obtained as the average value of pKa​ of both −NH2​ groups only. Thus pI=28.95+10.79​=219.47​=9.87

Q. What is the structure of the a bove amino acid in a solution of $p H = 9.87 ?$

$H_{2} N - C H_{2} - C H_{2} - C H_{2} - C H_{2} - \oplus N H_{3} ∣ C H - COO H$

$H_{3} N \oplus - C H_{2} - C H_{2} - C H_{2} - C H_{2} - \oplus N H_{3} ∣ C H - COO H$

$H_{3} N \oplus - C H_{2} - C H_{2} - C H_{2} - C H_{2} - \oplus N H_{3} ∣ C H - CO O^{-}$

$H_{3} N \oplus - C H_{2} - C H_{2} - C H_{2} - C H_{2} - \oplus N H_{3} ∣ C H - CO O^{-}$

Since the given amino acid has two $- N H_{2}$ group and one $COO H$ group,
the isoelectric point is obtained as the average value of $p K_{a}$ of both $- N H_{2}$ groups only.
Thus $p I = \frac{8.95 + 10.79}{2} = \frac{19.47}{2} = 9.87$