Question 1

In collagenous regions of proteins, what fraction of residues are glycine?

Accepted Answer

Introduction: Collagen is the most abundant protein in mammals and a hallmark of connective tissues. Its unique triple-helix architecture relies on a highly repetitive sequence pattern. This question targets recognition of the Gly-X-Y motif and the specific frequency of glycine needed to pack three chains tightly. Given Data / Assumptions: Collagen chains feature a repeating Gly-X-Y pattern. Glycine is the smallest amino acid with a single hydrogen side chain. Tight packing near the helix axis requires minimal steric bulk. Concept / Approach: Every third residue must be glycine to fit into the crowded center of the triple helix. The X and Y positions are often proline and hydroxyproline, which contribute to rigidity. Therefore, glycine constitutes one-third of residues in collagenous segments. Step-by-Step Solution: Recall the Gly-X-Y repeating sequence.Map frequency: glycine appears at positions 1, 4, 7, ...Count per triplet: 1 glycine in every 3 residues.Compute fraction: 1 out of 3 equals one-third. Verification / Alternative check: Biochemical and structural studies confirm glycine at every third position; substitution with larger residues destabilizes the triple helix and causes disease phenotypes (e.g., osteogenesis imperfecta). Why Other Options Are Wrong: One-fourth, one-tenth, two-thirds, half: do not match the strict Gly-X-Y pattern required for triple-helix packing. Common Pitfalls: Confusing overall glycine content of a whole protein with the strict periodicity in collagenous regions. Final Answer: one-third

Question 2

In protein chemistry, what does the term secondary structure specifically refer to?

Accepted Answer

Introduction: Proteins are described at multiple structural levels. This question distinguishes secondary structure from primary, tertiary, and quaternary arrangements by focusing on what is meant by regular local folding patterns such as alpha helices and beta sheets. Given Data / Assumptions: Primary structure: amino acid sequence. Secondary structure: repetitive, local backbone conformations stabilized mainly by hydrogen bonds. Tertiary structure: overall 3D fold of a single polypeptide chain. Quaternary structure: association of multiple polypeptides. Concept / Approach: Secondary structure refers to recurring patterns formed by backbone hydrogen bonds: alpha helices (i to i+4 hydrogen bonding) and beta sheets (inter- or intrachain). Turns and loops are also part of secondary structure. Side chains influence stability but the defining feature is backbone geometry and hydrogen bonding. Step-by-Step Solution: Identify the level: local, repetitive patterns, not the entire protein.Recognize main examples: alpha helices, beta sheets, beta turns.Stabilization: predominantly backbone hydrogen bonding.Conclude definition: regular local folding patterns of the backbone. Verification / Alternative check: Secondary structure assignments from crystallography and NMR (e.g., DSSP) classify residues as helix, sheet, turn, or coil based on backbone hydrogen bonding and geometry, not on side-chain arrangements alone. Why Other Options Are Wrong: Primary sequence: defines but is not itself secondary structure. Tertiary arrangement: describes the overall fold, not local motifs. Quaternary assembly and domain modules: involve multiple chains or higher-level organization. Common Pitfalls: Assuming side-chain interactions alone define secondary structure or conflating motifs (e.g., helix-turn-helix) with whole-protein tertiary arrangements. Final Answer: regular local folding patterns of the polypeptide backbone

Question 3

According to the Bohr effect, how does decreasing pH (more acidic conditions) affect the oxygen affinity of hemoglobin?

Accepted Answer

Introduction: The Bohr effect describes how hemoglobin's oxygen-binding affinity is modulated by pH and carbon dioxide. Physiologically, tissues produce acid and CO2 during metabolism, signaling hemoglobin to release oxygen where it is needed. This question checks your understanding of the direction of the affinity change when pH falls. Given Data / Assumptions: Lower pH corresponds to more protons in solution. CO2 hydration in blood generates carbonic acid, lowering pH. Hemoglobin exhibits cooperative oxygen binding regulated by allosteric effectors. Concept / Approach: Protonation of specific hemoglobin residues stabilizes the T (tense) state relative to the R (relaxed) state. Stabilizing T reduces oxygen affinity, shifting the oxygen dissociation curve to the right, thereby facilitating oxygen release in acidic, metabolically active tissues. Step-by-Step Solution: Relate decreased pH to increased protonation of hemoglobin.Protonation stabilizes salt bridges favoring the T state.T-state stabilization reduces oxygen affinity.Result: rightward shift of the oxygen dissociation curve and enhanced oxygen unloading. Verification / Alternative check: Clinical and physiological data show that acidosis (lower pH) reduces saturation at a given partial pressure of oxygen, consistent with the Bohr effect; alkalosis produces the opposite. Why Other Options Are Wrong: Increase affinity or no effect: contradicts the Bohr effect. Tissue-specific increases: the effect is systemic and context-driven by pH and CO2, not an exception limited to muscles or lungs. Common Pitfalls: Confusing Bohr effect (pH and CO2 influence) with the Haldane effect (oxygenation of blood affecting CO2 transport). Final Answer: decrease in oxygen affinity

Question 4

Disulfide bonds most commonly stabilize the native structures of which class of proteins?

Accepted Answer

Introduction: Disulfide bonds are covalent links formed by oxidation of cysteine thiols. Their prevalence varies across cellular compartments and protein classes. This question tests the association between disulfide bonding and the environment in which proteins fold and function. Given Data / Assumptions: Intracellular cytosol is generally a reducing environment. Extracellular and certain organellar spaces are more oxidizing. Disulfide bond formation requires oxidation of two cysteine residues. Concept / Approach: Because disulfides are stabilized in oxidizing settings, they are far more common in secreted and extracellular proteins (e.g., antibodies, many hormones). In the reducing cytosol, disulfides are less stable unless formed in specific compartments (e.g., endoplasmic reticulum) before secretion. Step-by-Step Solution: Identify the redox status of cellular compartments.Connect oxidizing environments with stable disulfide bonds.Recognize extracellular proteins often transit through the ER where disulfides form.Conclude extracellular proteins are most commonly stabilized by disulfide bonds. Verification / Alternative check: Proteomic surveys reveal a high density of disulfides in secreted proteins and low frequency in cytosolic proteins, consistent with redox chemistry and protein trafficking pathways. Why Other Options Are Wrong: Dimeric or hydrophobic proteins: dimerization or hydrophobicity does not guarantee disulfides. Intracellular proteins: reducing cytosol disfavors disulfide stability. Membrane proteins: many lack disulfides unless extracellular loops require them. Common Pitfalls: Assuming that multimeric state implies disulfides or overlooking the role of cellular redox conditions. Final Answer: extracellular proteins

Question 5

Glycine has the molecular formula C2H5O2N. If ten glycines are linked linearly by peptide (condensation) reactions to form an oligomer, what is the correct overall molecular formula?

Accepted Answer

Introduction: Forming peptide bonds between amino acids is a condensation process that eliminates water. Accounting correctly for the number of water molecules lost is essential for deriving the empirical formula of a peptide from its monomer building blocks. Given Data / Assumptions: Monomer formula for glycine: C2H5O2N. Ten glycines linked linearly create nine peptide bonds. Each peptide bond formation eliminates one H2O. Concept / Approach: The total formula is the sum of monomers minus the water molecules removed during condensation. For n residues in a linear chain, there are n − 1 peptide bonds and thus n − 1 waters lost. Step-by-Step Solution: Compute monomer sum: 10 * C2H5O2N = C20H50O20N10.Count peptide bonds: 10 residues → 9 bonds → 9 waters removed.Account for water loss: 9 * H2O = H18O9 removed.Subtract: H50 − 18 = H32; O20 − 9 = O11; C and N unchanged.Resulting formula: C20H32O11N10. Verification / Alternative check: General formula check: for n glycines, overall becomes C(2n) H(5n − 2(n − 1)) O(2n − (n − 1)) N(n) which simplifies to C(2n) H(3n + 2) O(n + 1) N(n). For n = 10, this yields C20H32O11N10, confirming the calculation. Why Other Options Are Wrong: C20H50O20N10: ignores water loss entirely. C20H40O10N10 and C20H42O12N10: remove the wrong number of hydrogens and oxygens. C20H68O29N10: chemically implausible for a condensation product. Common Pitfalls: Forgetting that the number of waters lost equals n − 1 for a linear chain, or mistakenly subtracting one water per residue. Final Answer: C20H32O11N10

Question 6

Ramachandran plot: what does it illustrate about permissible backbone dihedral angles in proteins?  In protein structure analysis, the Ramachandran plot maps the sterically allowed regions for the backbone dihedral angles phi (φ) and psi (ψ) of pept…

Accepted Answer

Introduction: Proteins adopt specific three-dimensional shapes constrained by their backbone geometry. The Ramachandran plot is a foundational tool that visualizes the sterically allowed combinations of the backbone dihedral angles phi (φ) and psi (ψ). Given Data / Assumptions: Backbone dihedrals considered: phi (N–Cα) and psi (Cα–C’). Peptide bond itself is planar due to resonance. Steric clashes between atoms restrict many φ, ψ combinations. Concept / Approach: The plot reveals allowed basins corresponding to secondary structures such as right-handed α-helix, β-sheet, and left-handed helix. Disallowed regions arise from steric overlap among backbone and side-chain atoms. Step-by-Step Solution: 1) Define φ, ψ as the two rotatable backbone angles per residue (excluding peptide bond).2) Evaluate steric constraints from van der Waals radii; many φ, ψ pairs are forbidden.3) Observed protein structures populate a few dense regions (α, β, sometimes left-handed α).4) Therefore, φ, ψ do not take any value nor a single value; they cluster into a few allowed regions. Verification / Alternative check: Inspection of known structures shows residues in α-helices around φ ≈ -60°, ψ ≈ -45°, and in β-sheets around φ ≈ -120°, ψ ≈ 120° (approximate ranges), confirming clustering. Why Other Options Are Wrong: “Peptide bond planar” is true but not what the plot primarily illustrates. “Any value” ignores sterics. “Only a single value” contradicts observed diversity. “Depends only on charge” ignores steric packing. Common Pitfalls: Confusing peptide bond planarity (ω) with φ, ψ flexibility; ignoring glycine/proline outliers. Final Answer: The phi (φ) and psi (ψ) angles cluster into a few allowed regions (e.g., α-helix, β-sheet, left-handed helix).

Question 7

CNBr cleavage mapping: Val-Lys-Glu-Met-Ser-Trp-Arg-Ala  Cyanogen bromide (CNBr) specifically cleaves on the C-terminal side of methionine residues. What peptide fragments are produced from the sequence Val-Lys-Glu-Met-Ser-Trp-Arg-Ala?

Accepted Answer

Introduction: Cyanogen bromide (CNBr) is a classic reagent for peptide mapping because it cleaves specifically at methionine residues, enabling predictable fragmentation for sequencing strategies. Given Data / Assumptions: Sequence: Val-Lys-Glu-Met-Ser-Trp-Arg-Ala. CNBr cleaves at the C-terminus of Met, generating a new fragment ending with Met. No Met at the extreme C-terminus beyond the single internal Met. Concept / Approach: Identify all Met positions, then cut immediately after each Met to form fragments. The presence of only one Met means two fragments will form. Step-by-Step Solution: 1) Locate Met: Val-Lys-Glu-Met-Ser-Trp-Arg-Ala.2) CNBr cleavage rule: cut after Met.3) Fragments obtained: Val-Lys-Glu-Met and Ser-Trp-Arg-Ala. Verification / Alternative check: Counting residues pre- and post-cleavage preserves total length (8 residues = 4 + 4), confirming consistency. Why Other Options Are Wrong: Other answers either split at non-Met positions or place the cut before Met, contradicting CNBr specificity. Common Pitfalls: Confusing CNBr (cleaves after Met) with trypsin (after Lys/Arg) or chymotrypsin (aromatics). Final Answer: Val-Lys-Glu-Met + Ser-Trp-Arg-Ala

Question 8

If the backbone dihedral angles phi (φ) and psi (ψ) for every residue in a protein are known, what global structural feature can be determined?

Accepted Answer

Introduction: Backbone dihedral angles phi (φ) and psi (ψ) define the local geometry of each residue. Knowing them for all residues largely specifies the polypeptide backbone conformation, enabling reconstruction of the three-dimensional fold. Given Data / Assumptions: All φ, ψ angles are known for each residue. Bond lengths and bond angles are near standard values. No consideration of absolute position (translation/rotation) needed to define shape. Concept / Approach: With φ, ψ for each residue and standard peptide geometry (including planar peptide bonds), one can sequentially place each backbone atom in 3D space, yielding the protein’s tertiary structure up to overall rigid-body transformations. Step-by-Step Solution: 1) Use known bond lengths/angles to define local frames at Cα.2) Apply φ around N–Cα and ψ around Cα–C’ to position the next residue.3) Iterate along the chain to build the entire backbone trace.4) Side-chain placement then follows using rotamer libraries (not required for backbone shape). Verification / Alternative check: Backbone reconstruction from φ, ψ is standard in computational modeling; the resulting coordinates align with experimentally determined tertiary folds. Why Other Options Are Wrong: Secondary structure is a subset of tertiary structure; quaternary assembly requires inter-subunit information; thermodynamic stability cannot be deduced solely from geometry; primary sequence cannot be inferred from angles. Common Pitfalls: Equating backbone reconstruction with side-chain packing accuracy; ignoring cis peptide bonds and proline exceptions. Final Answer: Complete tertiary (3D) structure of the polypeptide chain

Question 9

Deoxyhemoglobin coordination: which ligands bind Fe(II) in the heme pocket of deoxy Hb?

Accepted Answer

Introduction: Hemoglobin’s function depends on how the Fe(II) center is coordinated in the heme pocket. In deoxyhemoglobin, Fe(II) is high-spin and five-coordinate; in oxyhemoglobin, binding O2 gives a sixth ligand and low-spin character. Given Data / Assumptions: We are considering the deoxygenated state (deoxy Hb). Heme contributes four in-plane nitrogen ligands. One histidine (proximal His F8) binds axially to Fe(II). Concept / Approach: Count coordination number and identify axial ligands. In deoxy Hb, there is no bound O2 or water at the sixth site; the distal His does not coordinate directly but stabilizes ligands via H-bonding when present. Step-by-Step Solution: 1) Porphyrin macrocycle provides 4 N donors.2) Proximal His provides the 5th coordination to Fe(II).3) Sixth site is vacant in deoxy Hb; upon oxygenation, O2 occupies it as the sixth ligand. Verification / Alternative check: Spectral and structural data show a five-coordinate, high-spin Fe(II) in deoxy Hb and six-coordinate, low-spin Fe(II) in oxy Hb. Why Other Options Are Wrong: Options with water or multiple histidines misrepresent deoxy coordination. A water ligand occurs in some met or aquo states, not in deoxy Hb. Common Pitfalls: Confusing distal His (non-bonding in deoxy) with a coordinating ligand; assuming O2/water present in the deoxygenated state. Final Answer: Four nitrogens of heme and the proximal histidine (five-coordinate)

Question 10

During embryonic development, which globin chain functionally replaces the adult β chain in embryonic hemoglobins?

Accepted Answer

Introduction: Hemoglobin expression is developmentally regulated. Distinct globin chains are expressed in embryonic, fetal, and adult stages to optimize oxygen delivery across changing physiological conditions. Given Data / Assumptions: Embryonic hemoglobins (e.g., Hb Gower-1, Hb Gower-2) are under consideration, not fetal HbF or adult HbA. Adult β chains are minimal or absent during early embryogenesis. Concept / Approach: Embryonic hemoglobins pair ζ (zeta, α-like) with ε (epsilon, β-like) chains. Later, fetal hemoglobin uses γ (gamma) instead of β; finally, adult hemoglobin uses β with α. Step-by-Step Solution: 1) Embryo: ζ and ε chains dominate (β is not yet expressed).2) Fetus: γ replaces ε in β-like role (HbF: α2γ2).3) Adult: β chains replace γ (HbA: α2β2). Verification / Alternative check: Globin gene clusters show temporal switches: ε (embryo) → γ (fetus) → β (adult). Why Other Options Are Wrong: γ is fetal, not embryonic; δ is a minor adult chain; α is present but does not replace β in the β-like role; ζ is α-like, not β-like. Common Pitfalls: Confusing fetal (γ) with embryonic (ε) hemoglobins. Final Answer: Epsilon (ε)

Question 11

Dominant secondary structure element in myoglobin and hemoglobin: which motif predominates?

Accepted Answer

Introduction: Protein secondary structure comprises repeating motifs like α-helices and β-sheets. Globins (myoglobin and hemoglobin) are classic α-helical proteins. Given Data / Assumptions: Considering globin fold proteins. Myoglobin and hemoglobin share a conserved fold composed largely of helices A–H. Concept / Approach: The globin fold is dominated by α-helices connected by short loops; β-structure is minimal to absent. Step-by-Step Solution: 1) Identify structural class: all-α proteins.2) Count elements: 7–8 helices per subunit, loops connect helices, negligible β-strand content.3) Conclude α-helix predominance. Verification / Alternative check: Crystal structures and secondary-structure assignments (e.g., DSSP) show majority α-helix content in globins. Why Other Options Are Wrong: β-strands and β-turns are present in many proteins but not dominant in globins; “All equally” is incorrect; collagen triple helix is not present in globins. Common Pitfalls: Assuming all proteins contain balanced α and β content. Final Answer: α-helix (alpha helix)

Question 12

How is the heme group held in the globin pocket? Identify the key bond that anchors the porphyrin to the protein.

Accepted Answer

Introduction: Heme proteins position the porphyrin cofactor precisely to enable reversible ligand binding. In globins, a specific axial coordination anchors heme within the pocket. Given Data / Assumptions: Globin family (myoglobin/hemoglobin), not c-type cytochromes. Iron is Fe2+ in the physiologically active state. Concept / Approach: The proximal histidine (His F8) coordinates Fe2+ axially, anchoring heme and transmitting conformational changes upon ligand binding. The distal histidine modulates ligand binding through hydrogen bonding, not direct coordination in deoxy state. Step-by-Step Solution: 1) Identify axial ligands in globins: proximal His binds Fe2+ directly.2) Distinguish from cytochrome c, where covalent thioether bonds to cysteines anchor heme.3) Conclude Fe2+–His as the anchoring bond. Verification / Alternative check: Structural data show Fe–N(His) bond length consistent with axial coordination in globins. Why Other Options Are Wrong: Fe3+ states reflect met forms; cysteine ligation and thioether attachments are characteristic of other heme proteins, not globins. Common Pitfalls: Confusing globins with c-type cytochromes or P450 enzymes. Final Answer: Fe2+ to histidine (proximal His) coordination

Question 13

Heterotrimeric G proteins: which subunit binds and hydrolyzes guanine nucleotides (GDP/GTP)?

Accepted Answer

Introduction: Heterotrimeric G proteins transmit signals from G protein–coupled receptors (GPCRs) by cycling between GDP-bound inactive and GTP-bound active states. Given Data / Assumptions: Components: Gα, Gβ, Gγ. Ligand-activated GPCRs act as GEFs (guanine nucleotide exchange factors) for Gα. Concept / Approach: Only the Gα subunit binds and hydrolyzes GDP/GTP. Upon GTP loading, Gα dissociates from Gβγ to engage effectors; intrinsic GTPase activity resets the cycle. Step-by-Step Solution: 1) GPCR activation promotes GDP release from Gα.2) GTP binds Gα, activating it.3) Gα-GTP and Gβγ signal to downstream effectors.4) Gα hydrolyzes GTP to GDP, reassociating with Gβγ. Verification / Alternative check: Biochemical assays show nucleotide binding/hydrolysis centered on Gα; Gβγ modulates effectors but does not bind GTP. Why Other Options Are Wrong: Gβ and Gγ do not bind/hydrolyze GTP; “delta” is not a heterotrimeric subunit; Gβγ dimer does not directly bind GTP. Common Pitfalls: Assuming Gβγ holds nucleotides; mixing up small GTPases (Ras-like) with heterotrimeric G proteins. Final Answer: Alpha (Gα)

Structure and Properties of Peptides Questions