Unit 1: Biochemistry — Practice Quiz (AS Learning)

Question 1

Which monomer pairs with which polymer correctly?

Amino acid → starch
Glucose → glycogen
Nucleotide → triglyceride
Glycerol → DNA

Solution: Glucose monomers join via alpha-1,4 (and alpha-1,6 branching) glycosidic bonds to form glycogen, the animal storage polysaccharide. Amino acids form proteins; nucleotides form nucleic acids; glycerol + 3 fatty acids form a triglyceride.

Question 2

A saturated fatty acid differs from an unsaturated fatty acid because it:

Contains nitrogen
Has no C=C double bonds in its hydrocarbon tail
Is hydrophilic
Is a polymer

Solution: Saturated = no C=C; tails are straight, pack tightly, solid at room temperature (e.g., butter). Unsaturated has one or more C=C, creating kinks; usually liquid (oils).

Question 3

Which level of protein structure is stabilized primarily by hydrogen bonding between backbone atoms?

Primary
Secondary
Tertiary
Quaternary

Solution: Secondary structure (alpha-helices, beta-sheets) is held by H-bonds between backbone N-H and C=O groups. Primary is the amino-acid sequence; tertiary involves R-group interactions (disulfide, ionic, hydrophobic); quaternary involves multiple subunits.

Question 4

In a Michaelis–Menten plot of reaction rate (v) vs [substrate], V_max represents:

The substrate concentration at half-maximum rate
The maximum rate when all enzyme active sites are saturated
The activation energy
The optimum pH

Solution: V_max is the asymptotic maximum rate when every active site is occupied — adding more substrate cannot speed it up further. K_m = [S] at half V_max and is inversely related to enzyme–substrate affinity.

Question 5

A competitive inhibitor of an enzyme will:

Decrease V_max; K_m unchanged
Increase apparent K_m; V_max unchanged
Decrease both V_max and K_m
Bind irreversibly to the substrate

Solution: A competitive inhibitor binds the active site, competing with substrate. With enough substrate, V_max can still be reached, so V_max is unchanged but apparent K_m rises (more substrate is needed for half V_max). Non-competitive inhibitors lower V_max.

Question 6

Calculate K_m: an enzyme reaches V_max = 80 µmol/min. At [S] = 5 mM, v = 40 µmol/min.

Solution: K_m = [S] at half V_max. Half V_max = 40 µmol/min, observed at [S] = 5 mM. Therefore K_m = 5 mM.

Question 7

The induced-fit model of enzyme action proposes that:

The active site is rigid and substrate must match perfectly
The active site changes shape upon substrate binding to enhance catalysis
Enzymes are consumed during the reaction
All enzymes work at pH 7

Solution: Koshland's induced-fit model: the active site is flexible. Substrate binding induces a conformational change that strains substrate bonds, lowering activation energy. Replaces the older rigid lock-and-key view.

Question 8

Which transport mechanism requires ATP and moves a solute against its gradient?

Simple diffusion
Osmosis
Facilitated diffusion
Primary active transport

Solution: Primary active transport (e.g., Na+/K+ ATPase: 3 Na+ out, 2 K+ in per ATP) hydrolyzes ATP directly to move solutes uphill. Facilitated diffusion uses channel/carrier proteins but follows the gradient (no ATP).

Question 9

A red blood cell placed in a hypotonic solution will:

Shrivel (crenate)
Swell and possibly lyse
Remain unchanged
Become turgid like a plant cell

Solution: In a hypotonic solution (lower solute outside), water enters the cell by osmosis. Animal cells lack a cell wall, so they swell and may lyse (hemolysis). Plant cells become turgid because the wall resists.

Question 10

In the fluid-mosaic model, which feature most affects membrane fluidity at low temperatures?

Higher cholesterol content keeps the membrane fluid
More saturated fatty acids increase fluidity
Glycoproteins increase fluidity
Phosphate heads make the membrane rigid

Solution: Cholesterol is a fluidity buffer: at low T it prevents tails packing too tightly (maintains fluidity); at high T it restrains motion (prevents excess fluidity). Unsaturated fatty acids (kinks) increase fluidity; saturated decrease it.

Question 11

Calculate water potential. A plant cell has solute potential ψ_s = −0.7 MPa and pressure potential ψ_p = +0.4 MPa. What is ψ?

Solution: ψ = ψ_s + ψ_p = −0.7 + 0.4 = −0.3 MPa. Water moves from higher (less negative) ψ to lower ψ.

Question 12

Endocytosis differs from facilitated diffusion because endocytosis:

Brings large particles or fluid into the cell using vesicles, requiring ATP
Moves only ions
Is a passive process
Uses aquaporins

Solution: Endocytosis (phagocytosis = solids; pinocytosis = fluids; receptor-mediated = specific) engulfs material in membrane vesicles, requires ATP and cytoskeletal rearrangement. Facilitated diffusion is passive, single-molecule, channel/carrier-mediated.