Biomolecules Set-3

Q1. Glycine has $pK_a^{\mathrm{COOH}}=2.34$ and $pK_a^{\mathrm{NH_3^+}}=9.60$. The isoelectric point $pI$ of glycine is

Q2. Consider the peptide $\mathrm{H_2N\!-\!Ala\!-\!Lys\!-\!Glu\!-\!COOH}$. Given $pK_a$ values: $pK_a(\mathrm{N\!-\!term})=9.0$, $pK_a(\mathrm{C\!-\!term})=2.0$, $pK_a(\mathrm{Lys\ side\ chain})=10.5$, $pK_a(\mathrm{Glu\ side\ chain})=4.1$. The net charge of this peptide at pH $7.0$ is

Q3. A double-stranded DNA fragment contains $100$ base pairs and has GC content $40\%$. If each A–T pair forms $2$ H‑bonds and each G–C pair forms $3$ H‑bonds, the total number of hydrogen bonds in the duplex is

Q4. Assertion (A): A lysine residue buried in the hydrophobic core of a folded protein has a lower $pK_a$ than the same lysine exposed to aqueous solvent.
Reason (R): Burial in a nonpolar environment destabilizes the positively charged (protonated) form because of the large desolvation penalty and lack of stabilizing solvent interactions; therefore the neutral (deprotonated) form is favoured and its $pK_a$ decreases.

Q5. The peptide $\mathrm{H_2N\!-\!Arg\!-\!Lys\!-\!Glu\!-\!Asp\!-\!COOH}$ has $pK_a$ values: $pK_a(\mathrm{N\!-\!term})=9.0$, $pK_a(\mathrm{C\!-\!term})=2.0$, $pK_a(\mathrm{Arg})=12.5$, $pK_a(\mathrm{Lys})=10.5$, $pK_a(\mathrm{Glu})=4.3$, $pK_a(\mathrm{Asp})=3.9$. The isoelectric point $pI$ of this peptide is approximately

Q6. An enzyme follows Michaelis–Menten kinetics with $K_m=2\ \text{mM}$ and $V_{\max}=100\ \mu\text{mol min}^{-1}$. Using $v=\dfrac{V_{\max}[S]}{K_m+[S]}$, calculate the initial rate $v$ when $[S]=2\ \text{mM}$.

Q7. A mixture contains glycine, lysine and glutamic acid at pH $7.0$. Given $pK_a$ values: glycine ($pK_a^{\alpha\text{-COOH}}=2.34,\ pK_a^{\alpha\text{-NH}_3^+}=9.60$), lysine ($pK_a^{\alpha\text{-COOH}}=2.18,\ pK_a^{\alpha\text{-NH}_3^+}=8.95,\ pK_a^{\text{side-NH}_3^+}=10.53$), glutamic acid ($pK_a^{\alpha\text{-COOH}}=2.19,\ pK_a^{\text{side-COOH}}=4.25,\ pK_a^{\alpha\text{-NH}_3^+}=9.67$). Which amino acid will migrate towards the anode (positive electrode) in electrophoresis at pH $7.0$?

Q8. An enzyme has $K_m=2\ \text{mM}$ and $V_{\max}=100\ \mu\text{mol min}^{-1}$. In presence of a competitive inhibitor the apparent $K_m$ increases to $4\ \text{mM}$ while $V_{\max}$ remains unchanged. If the initial rate at $[S]=2\ \text{mM}$ without inhibitor is to be restored in the presence of the inhibitor, what substrate concentration $[S]$ is required? (Use $v=\dfrac{V_{\max}[S]}{K_m+[S]}$.)

Q9. Consider the dipeptide H–Glu–Lys–OH (N‑terminus = Glu, C‑terminus = Lys). Use these $pK_a$ values: N‑terminal $\alpha$–NH$_3^+$ $pK_a=9.5$, C‑terminal $\alpha$–COOH $pK_a=2.1$, Glu side‑chain COOH $pK_a=4.3$, Lys side‑chain NH$_3^+$ $pK_a=10.5$. What is the net charge of this dipeptide at pH $7.0$?

Q10. A catalytic histidine in an enzyme acts as a general base; its $pK_a$ in the wild‑type enzyme is $6.0$. A mutation changes the microenvironment and raises the histidine $pK_a$ to $7.0$. At pH $7.0$, the fraction of deprotonated histidine is given by $\displaystyle f=\frac{1}{1+10^{pK_a-\text{pH}}}$. Assuming the catalytic rate is directly proportional to the fraction of deprotonated histidine, what approximate fraction of the original catalytic activity remains after the mutation?