Protein Structure and Function Questions and Answers

Q: The amino acid that interrupts α-helix conformation in protein by developing kinks in the structure :

The amino acid is proline.In proline the side chain is bonded back to the backbone nitrogen, forming a ring. Its nitrogen therefore has no hydrogen to donate to the hydrogen bond that holds the α-helix together, and the ring fixes the phi angle. Where proline occurs inside a helix, it breaks the pattern and puts a kink in the chain; it is a well-known helix breaker.Leucine is a strong helix former, and valine and serine weaken helices somewhat but do not cause kinks the way proline does.

Q: A Ramchandran plot describes, for a particular amino acid in a polypeptide chain, stearically allowed angles for :

The plot describes the allowed angles for the Cα-C and Cα-N bonds.The Ramachandran plot maps the two backbone torsion angles of each residue: phi (rotation about the N-Cα bond) and psi (rotation about the Cα-C bond). The peptide bond itself is planar, so these two rotations fix the backbone conformation, and the plot shows which combinations avoid steric clashes, the regions of alpha-helix and beta-sheet.The Cα-CB bond leads into the side chain, and Cα-H bonds carry a hydrogen; rotations about neither are plotted, so every option naming them is wrong.

Q: Which of the following bonds in proteins have a partial double bond character ?

The bond with partial double bond character is C – N, the peptide bond.In the peptide bond between the carbonyl carbon of one residue and the amide nitrogen of the next, the nitrogen's lone pair is delocalised into the carbonyl. Resonance gives the C-N bond about 40% double bond character, so it is shorter than a normal C-N single bond, planar and does not rotate freely. This planarity is why backbone conformation is described only by the phi and psi angles.The Cα-C, Cα-Cβ and Cα-N bonds are ordinary single bonds. Rotation about Cα-N (phi) and Cα-C (psi) is what lets the chain fold.

Q: How many different linear tripeptides can be made from three different L-α neutral amino acids, using each amino acid only once in the chain ?

6 different linear tripeptides can be made.A linear peptide has direction, from N-terminus to C-terminus, so Ala-Gly-Ser and Ser-Gly-Ala are different molecules. Using three different amino acids once each, the number of tripeptides is the number of orderings of three items:\[ 3! = 3 \times 2 \times 1 = 6 \]3 counts only which amino acid is first, or treats the chain as a ring. 27 is \(3^{3}\), the count if each position could use any of the three with repetition, which the stem excludes. 12 has no matching rule.

Q: NMR is a very important technique for determination of biopolymer structure. NMR experiments are not limited to the study of protons only. Which of the following represents a correct group of isotopes used to obtain resonance signals ?

The correct group is \(^{1}\)H, \(^{13}\)C, \(^{15}\)N, \(^{19}\)F.An NMR signal needs a nucleus with non-zero spin, and sharp lines in biopolymer work come from spin-1/2 nuclei. \(^{1}\)H, \(^{13}\)C, \(^{15}\)N and \(^{19}\)F all have spin 1/2, which is why proteins are labelled with \(^{13}\)C and \(^{15}\)N for multidimensional NMR.\(^{12}\)C has spin 0 and gives no signal, which rules out the options containing it. \(^{14}\)N, the common nitrogen isotope, has spin 1 and a quadrupole moment that broadens its lines, so the biomolecular work uses \(^{15}\)N instead.

Q: Indicate the ionic species that predominates at pH 4, 8 and 11 for ammonia :

The species are \(\mathrm{NH}_{4}^{+}\) at pH 4, \(\mathrm{NH}_{4}^{+}\) at pH 8 and \(\mathrm{NH}_{3}\) at pH 11.The ammonium ion has a pKa of about 9.25. By the Henderson-Hasselbalch relation, the protonated form dominates below the pKa and the base form above it:\[ \begin{aligned} \mathrm{pH} &= \mathrm{p}K_a \\ &+ \log\frac{[\mathrm{NH_3}]}{[\mathrm{NH_4^+}]} \end{aligned} \]At pH 4 and pH 8 the pH is below 9.25, so \(\mathrm{NH}_{4}^{+}\) predominates; at pH 8 the ratio is still about 1 to 18 in its favour. At pH 11, above the pKa, \(\mathrm{NH}_{3}\) predominates.The other options put \(\mathrm{NH}_{3}\) at low pH or \(\mathrm{NH}_{4}^{+}\) at pH 11, the reverse of this rule.

Q: Which region of CD spectrum gives details of secondary structure of a protein ?

Secondary structure is read from the Far-UV region of the CD spectrum.In the far UV, roughly 190 to 250 nm, the absorbing groups are the peptide bonds of the backbone, and their circular dichroism depends on the backbone conformation. An alpha helix gives negative bands at 222 and 208 nm and a positive band near 192 nm; beta sheet and random coil give their own characteristic shapes, so the proportions of each can be estimated.The near UV, about 250 to 320 nm, is dominated by aromatic side chains and disulphide bonds and reports on tertiary structure. The visible region shows CD only for chromophores such as haem or bound cofactors. The whole UV range is not used as one region for this purpose.

Q: Paradox related to protein folding is :

The protein folding paradox is Levinthal's paradox.Cyrus Levinthal pointed out in 1969 that a polypeptide of even 100 residues has so many possible conformations (of the order of \(3^{100}\)) that sampling them one by one would take longer than the age of the universe, yet real proteins fold in milliseconds to seconds. The resolution is that folding follows pathways down an energy landscape (the folding funnel) rather than a random search.The C-value paradox concerns genome size: DNA content does not track the complexity of organisms. 'Sherman paradox' and 'Lombard's paradox' are not about protein folding; Lombard's paradox describes the simultaneous contraction of the hamstrings and quadriceps when standing up.

Q: Ubiquitin ligases are associated with degradation of :

Ubiquitin ligases are associated with degradation of Protein.Ubiquitin is attached to target proteins in three steps: E1 activates it, E2 carries it, and the E3 ubiquitin ligase recognises the specific substrate and transfers ubiquitin to a lysine on it. A chain of four or more ubiquitins linked through lysine 48 marks the protein for destruction by the 26S proteasome. This pathway removes cyclins, misfolded proteins and many regulators (Nobel Prize in Chemistry, 2004).DNA is broken down by nucleases, RNA by ribonucleases and the exosome, and carbohydrates by glycosidases. None of these depends on ubiquitin tagging.

11.

The amino acid that interrupts α-helix conformation in protein by developing kinks in the structure :

A.Serine
B.Valine
C.Proline
D.Leucine

Exam Year(s)

MH SET - 2019, SET Exam - 2019,

Answer: C

The amino acid is proline.

In proline the side chain is bonded back to the backbone nitrogen, forming a ring. Its nitrogen therefore has no hydrogen to donate to the hydrogen bond that holds the α-helix together, and the ring fixes the phi angle. Where proline occurs inside a helix, it breaks the pattern and puts a kink in the chain; it is a well-known helix breaker.

Leucine is a strong helix former, and valine and serine weaken helices somewhat but do not cause kinks the way proline does.

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12.

A Ramchandran plot describes, for a particular amino acid in a polypeptide chain, stearically allowed angles for :

A.Cα-C and Cα-CB bonds
B.Cα-C and Cα-N bonds
C.Cα-C and Cα-H bonds
D.Cα-N and Cα-H bonds

Exam Year(s)

MH SET - 2020, SET Exam - 2020,

Answer: B

The plot describes the allowed angles for the Cα-C and Cα-N bonds.

The Ramachandran plot maps the two backbone torsion angles of each residue: phi (rotation about the N-Cα bond) and psi (rotation about the Cα-C bond). The peptide bond itself is planar, so these two rotations fix the backbone conformation, and the plot shows which combinations avoid steric clashes, the regions of alpha-helix and beta-sheet.

The Cα-CB bond leads into the side chain, and Cα-H bonds carry a hydrogen; rotations about neither are plotted, so every option naming them is wrong.

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13.

Which of the following bonds in proteins have a partial double bond character ?

A.Cα – C
B.Cα – Cβ
C.Cα – N
D.C – N

Exam Year(s)

MH SET - 2021, SET Exam - 2021,

Answer: D

The bond with partial double bond character is C – N, the peptide bond.

In the peptide bond between the carbonyl carbon of one residue and the amide nitrogen of the next, the nitrogen's lone pair is delocalised into the carbonyl. Resonance gives the C-N bond about 40% double bond character, so it is shorter than a normal C-N single bond, planar and does not rotate freely. This planarity is why backbone conformation is described only by the phi and psi angles.

The Cα-C, Cα-Cβ and Cα-N bonds are ordinary single bonds. Rotation about Cα-N (phi) and Cα-C (psi) is what lets the chain fold.

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14.

How many different linear tripeptides can be made from three different L-α neutral amino acids, using each amino acid only once in the chain ?

A.3
B.6
C.12
D.27

Exam Year(s)

MH SET - 2021, SET Exam - 2021,

Answer: B

6 different linear tripeptides can be made.

A linear peptide has direction, from N-terminus to C-terminus, so Ala-Gly-Ser and Ser-Gly-Ala are different molecules. Using three different amino acids once each, the number of tripeptides is the number of orderings of three items:

\[ 3! = 3 \times 2 \times 1 = 6 \]

3 counts only which amino acid is first, or treats the chain as a ring. 27 is \(3^{3}\), the count if each position could use any of the three with repetition, which the stem excludes. 12 has no matching rule.

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15.

NMR is a very important technique for determination of biopolymer structure. NMR experiments are not limited to the study of protons only. Which of the following represents a correct group of isotopes used to obtain resonance signals ?

A.\(^{1}\)H, \(^{12}\)C, \(^{15}\)N, \(^{19}\)F
B.\(^{1}\)H, \(^{13}\)C, \(^{14}\)N, \(^{19}\)F
C.\(^{1}\)H, \(^{12}\)C, \(^{14}\)N, \(^{19}\)F
D.\(^{1}\)H, \(^{13}\)C, \(^{15}\)N, \(^{19}\)F

Exam Year(s)

MH SET - 2021, SET Exam - 2021,

Answer: D

The correct group is \(^{1}\)H, \(^{13}\)C, \(^{15}\)N, \(^{19}\)F.

An NMR signal needs a nucleus with non-zero spin, and sharp lines in biopolymer work come from spin-1/2 nuclei. \(^{1}\)H, \(^{13}\)C, \(^{15}\)N and \(^{19}\)F all have spin 1/2, which is why proteins are labelled with \(^{13}\)C and \(^{15}\)N for multidimensional NMR.

\(^{12}\)C has spin 0 and gives no signal, which rules out the options containing it. \(^{14}\)N, the common nitrogen isotope, has spin 1 and a quadrupole moment that broadens its lines, so the biomolecular work uses \(^{15}\)N instead.

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16.

Indicate the ionic species that predominates at pH 4, 8 and 11 for ammonia :

A.\(\mathrm{NH}_{4}^{+}\) at pH 4, \(\mathrm{NH}_{4}^{+}\) at pH 8 and \(\mathrm{NH}_{3}\) at pH 11
B.\(\mathrm{NH}_{4}^{+}\) at pH 4, \(\mathrm{NH}_{3}\) at pH 8 and \(\mathrm{NH}_{4}^{+}\) at pH 11
C.\(\mathrm{NH}_{3}\) at pH 4, \(\mathrm{NH}_{4}^{+}\) at pH 8 and \(\mathrm{NH}_{4}^{+}\) at pH 11
D.\(\mathrm{NH}_{3}\) at pH 4, \(\mathrm{NH}_{4}^{+}\) at pH 8 and \(\mathrm{NH}_{3}\) at pH 11

Exam Year(s)

MH SET - 2021, SET Exam - 2021,

Answer: A

The species are \(\mathrm{NH}_{4}^{+}\) at pH 4, \(\mathrm{NH}_{4}^{+}\) at pH 8 and \(\mathrm{NH}_{3}\) at pH 11.

The ammonium ion has a pKa of about 9.25. By the Henderson-Hasselbalch relation, the protonated form dominates below the pKa and the base form above it:

\[ \begin{aligned} \mathrm{pH} &= \mathrm{p}K_a \\ &+ \log\frac{[\mathrm{NH_3}]}{[\mathrm{NH_4^+}]} \end{aligned} \]

At pH 4 and pH 8 the pH is below 9.25, so \(\mathrm{NH}_{4}^{+}\) predominates; at pH 8 the ratio is still about 1 to 18 in its favour. At pH 11, above the pKa, \(\mathrm{NH}_{3}\) predominates.

The other options put \(\mathrm{NH}_{3}\) at low pH or \(\mathrm{NH}_{4}^{+}\) at pH 11, the reverse of this rule.

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17.

Which region of CD spectrum gives details of secondary structure of a protein ?

A.Far-UV region
B.Near-UV region
C.Entire-UV region
D.Both UV and Visible region

Exam Year(s)

MH SET - 2023, SET Exam - 2023,

Answer: A

Secondary structure is read from the Far-UV region of the CD spectrum.

In the far UV, roughly 190 to 250 nm, the absorbing groups are the peptide bonds of the backbone, and their circular dichroism depends on the backbone conformation. An alpha helix gives negative bands at 222 and 208 nm and a positive band near 192 nm; beta sheet and random coil give their own characteristic shapes, so the proportions of each can be estimated.

The near UV, about 250 to 320 nm, is dominated by aromatic side chains and disulphide bonds and reports on tertiary structure. The visible region shows CD only for chromophores such as haem or bound cofactors. The whole UV range is not used as one region for this purpose.

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18.

Paradox related to protein folding is :

A.C. Value paradox
B.Sherman paradox
C.Lombard's paradox
D.Levinthal's paradox

Exam Year(s)

MH SET - 2023, SET Exam - 2023,

Answer: D

The protein folding paradox is Levinthal's paradox.

Cyrus Levinthal pointed out in 1969 that a polypeptide of even 100 residues has so many possible conformations (of the order of \(3^{100}\)) that sampling them one by one would take longer than the age of the universe, yet real proteins fold in milliseconds to seconds. The resolution is that folding follows pathways down an energy landscape (the folding funnel) rather than a random search.

The C-value paradox concerns genome size: DNA content does not track the complexity of organisms. 'Sherman paradox' and 'Lombard's paradox' are not about protein folding; Lombard's paradox describes the simultaneous contraction of the hamstrings and quadriceps when standing up.

Discuss in Forum

19.

Ubiquitin ligases are associated with degradation of :

A.Protein
B.DNA
C.RNA
D.Carbohydrates

Exam Year(s)

MH SET - 2023, SET Exam - 2023,

Answer: A

Ubiquitin ligases are associated with degradation of Protein.

Ubiquitin is attached to target proteins in three steps: E1 activates it, E2 carries it, and the E3 ubiquitin ligase recognises the specific substrate and transfers ubiquitin to a lysine on it. A chain of four or more ubiquitins linked through lysine 48 marks the protein for destruction by the 26S proteasome. This pathway removes cyclins, misfolded proteins and many regulators (Nobel Prize in Chemistry, 2004).

DNA is broken down by nucleases, RNA by ribonucleases and the exosome, and carbohydrates by glycosidases. None of these depends on ubiquitin tagging.

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20.

Which of the following heat shock proteins is present in almost every compartment of the Eukaryotic cell ?

A.Hsp60
B.Hsp70
C.Hsp90
D.Hsp100

Exam Year(s)

MH SET - 2023, SET Exam - 2023,

Answer: B

The heat shock protein found in almost every compartment is Hsp70.

The Hsp70 family has members in the cytosol and nucleus (Hsc70 and Hsp70), in the lumen of the endoplasmic reticulum (BiP, also called Grp78), in the mitochondrial matrix (mtHsp70) and in chloroplasts. Wherever proteins are being folded, imported or refolded after stress, an Hsp70 binds exposed hydrophobic segments in an ATP-dependent cycle.

Hsp60 (the chaperonin) is mainly mitochondrial, with its relative in the chloroplast; Hsp90 works mainly in the cytosol and nucleus, with an ER form (Grp94); and the Hsp100 disaggregases (ClpB, Hsp104) occur in bacteria, fungi and plants, in the cytosol and some organelles, but not throughout the cell's compartments as Hsp70 does.

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