The substituents R_1 and R_2 for nine peptide | vedclass

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Question

(A) The given structure is a tetrapeptide with a terminal $-NH_3^+$ group and a terminal $-COO^-$ group. At $pH = 7.0$,the net charge depends on the s

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(A) The given structure is a tetrapeptide with a terminal $-NH_3^+$ group and a terminal $-COO^-$ group. At $pH = 7.0$,the net charge depends on the side chains $R_1$ and $R_2$.For a peptide to be positively charged at $pH = 7.0$,it must contain basic side chains (like $-(CH_2)_4NH_2$) that remain protonated,while acidic side chains (like $-CH_2COOH$) must be absent or balanced.Let us analyze the side chains:- $I$: $R_1=H, R_2=H$ (Neutral)- $II$: $R_1=H, R_2=CH_3$ (Neutral)- $III$: $R_1=CH_2COOH$ (Acidic),$R_2=H$ (Net negative charge)- $IV$: $R_1=CH_2CONH_2$ (Neutral),$R_2=(CH_2)_4NH_2$ (Basic) $\rightarrow$ Positively charged- $V$: $R_1=CH_2CONH_2, R_2=CH_2CONH_2$ (Neutral)- $VI$: $R_1=(CH_2)_4NH_2, R_2=(CH_2)_4NH_2$ (Basic) $\rightarrow$ Positively charged- $VII$: $R_1=CH_2COOH$ (Acidic),$R_2=CH_2CONH_2$ (Neutral) $\rightarrow$ Negatively charged- $VIII$: $R_1=CH_2OH$ (Neutral),$R_2=(CH_2)_4NH_2$ (Basic) $\rightarrow$ Positively charged- $IX$: $R_1=(CH_2)_4NH_2$ (Basic),$R_2=CH_3$ (Neutral) $\rightarrow$ Positively chargedThe peptides $IV, VI, VIII,$ and $IX$ are positively charged at $pH = 7.0$. Thus,there are $4$ such peptides.

Peptide	$R_1$	$R_2$
$I$	$H$	$H$
$II$	$H$	$CH_3$
$III$	$CH_2COOH$	$H$
$IV$	$CH_2CONH_2$	$(CH_2)_4NH_2$
$V$	$CH_2CONH_2$	$CH_2CONH_2$
$VI$	$(CH_2)_4NH_2$	$(CH_2)_4NH_2$
$VII$	$CH_2COOH$	$CH_2CONH_2$
$VIII$	$CH_2OH$	$(CH_2)_4NH_2$
$IX$	$(CH_2)_4NH_2$	$CH_3$

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