Biochemistry

Step 1: Recall formula for dissociation constant .
Step 2: Substitute values.
Step 3: Convert to nM.
Step 4: Conclusion.
The dissociation constant is nM.

Step 1: Recall aldol cleavage in glycolysis.
In glycolysis, aldolase cleaves fructose-1,6-bisphosphate into two 3-carbon molecules: DHAP and GAP. The same logic applies when considering cleavage of a 6-carbon molecule like glucose-6-phosphate.
Step 2: Analyze possible products.
A 6-carbon compound split via aldol cleavage forms two 3-carbon fragments → equal chain length.
Step 3: Conclusion.
Thus, aldol cleavage yields products of equal carbon number (C3 + C3).

Step 1: Understanding the process.
The oxidation of aldehydes to acids is a crucial step in the metabolism of 1,3-bisphosphoglycerate, where energy is released. This energy is used to form the high-energy mixed anhydride bond.
Step 2: Analyzing the options.
(A) an aldehyde to acid: Correct — The oxidation of aldehydes to acids generates the necessary energy for the bond formation.
(B) an alcohol to acid: Incorrect — This does not apply to the generation of the mixed anhydride bond in the case of 1,3-bisphosphoglycerate.
(C) an alcohol to aldehyde: Incorrect — This is not related to the generation of the anhydride bond.
(D) NADH to NAD : Incorrect — While NADH oxidation is essential in metabolism, it does not directly generate the mixed anhydride bond.
Step 3: Conclusion.
The correct answer is (A) an aldehyde to acid, as this oxidation reaction releases energy used to form the mixed anhydride bond.

Step 1: Identify the structure.
Oleic acid is an 18-carbon monounsaturated fatty acid, denoted as C H O , with one cis double bond at the 9th carbon.
Step 2: Determine saturation.
Because of the presence of a C=C double bond, it is unsaturated. Saturated fatty acids contain no double bonds between carbons.
Step 3: Conclusion.
Hence, oleic acid is an unsaturated fatty acid.

Step 1: Recall ubiquinone’s properties.
Ubiquinone (Coenzyme Q) is a small, lipid-soluble electron carrier in the mitochondrial inner membrane.
It shuttles electrons between complexes I/II and III.
Step 2: Analyze statements.
(A) True — It can accept two electrons one by one, functioning as a bridge between 2e donors (NADH) and 1e acceptors (cytochromes).
(B) True — Its hydrophobic nature allows diffusion within the lipid bilayer.
(C) False — Ubiquinone is hydrophobic, not hydrophilic.
(D) Partially true but incomplete. Direct electron transfer to cytochromes occurs via cytochrome b and c , not directly by ubiquinone.
Step 3: Conclusion.
Hence, statements (A) and (B) are correct.

Step 1: Recall the step in glycolysis.
In glycolysis, glucose-6-phosphate (an aldose) is isomerized to fructose-6-phosphate (a ketose) by the enzyme phosphoglucose isomerase.
Step 2: Role of isomerization.
This conversion moves the carbonyl group from C-1 to C-2, forming a ketose. This change enables the molecule to later form a symmetrical 6-carbon intermediate that can be split evenly into two 3-carbon compounds.
Step 3: Importance for cleavage.
During the aldolase step, fructose 1,6-bisphosphate splits into dihydroxyacetone phosphate and glyceraldehyde 3-phosphate — a reaction that is possible only because the carbonyl group is positioned at C-2.
Step 4: Conclusion.
Therefore, the isomerization to fructose 6-phosphate ensures proper bond cleavage between C-3 and C-4.

Step 1: Concept.
Each peptide bond links two amino acids.
Therefore, an n-residue linear peptide contains peptide bonds.
Step 2: Substitute values.
Step 3: Conclusion.
Hence, a 20-residue peptide has 19 peptide bonds.

Step 1: Write the relation between and equilibrium constant .}
Step 2: Find .}
Reaction: At equilibrium, Step 3: Substitute values.
Approximate literature value: ≈ –5010 J/mol (due to rounding & standard conditions).
Step 4: Conclusion.