SBI4U Biology Practice Exam — Free

Separate the two jobs of the inner mitochondrial membrane, because the whole question rests on their being separate. The electron transport chain (ETC), embedded in the cristae, takes the high-energy electrons carried by $\mathrm{NADH}$ and $\mathrm{FADH_2}$, passes them along a series of carriers to oxygen (the final electron acceptor, which is reduced to water) and uses the energy released to pump $\mathrm{H^+}$ from the matrix into the intermembrane space. That builds a proton gradient. ATP synthase then lets those protons flow back into the matrix and uses the flow to join ADP and inorganic phosphate; this is chemiosmosis, and it supplies the great majority of the roughly 30–32 ATP made per glucose (older textbooks say 36–38). The two processes are coupled only because the membrane is otherwise impermeable to $\mathrm{H^+}$: the one way back is through ATP synthase.

The added chemical gives protons a second way back, so the gradient drains away without turning ATP synthase. Electrons still flow to oxygen, so oxygen is still consumed; in practice consumption rises, because the ETC runs faster once it is no longer pumping against a steep gradient. The energy that would have been captured in ATP is released as heat. Such chemicals are called uncouplers. 2,4-dinitrophenol was once sold (dangerously) as a diet drug, and a natural uncoupling protein in brown fat warms newborn babies and hibernating mammals by exactly this mechanism.

Option B is the classic trap: it reverses cause and effect. The ETC builds the gradient; the gradient does not drive the ETC. Option C assumes that any proton entering the matrix makes ATP, but only protons passing through ATP synthase do; a leak bypasses the enzyme. Option D describes an ETC inhibitor such as cyanide, which blocks electron flow so that $\mathrm{NADH}$ cannot be re-oxidized to $\mathrm{NAD^+}$ and the Krebs cycle backs up. Here electron flow continues, so $\mathrm{NAD^+}$ is regenerated normally and the Krebs cycle keeps running.

Oxygen is still consumed, but ATP production falls sharply and the energy is lost as heat (option A).

Start from what produces each phase of a neuron's electrical activity, because the toxin removes only one kind of channel.

(a) The resting potential (about −70 mV, inside negative) is set by two things that TTX does not touch. The sodium–potassium pump moves three $\mathrm{Na^+}$ out of the cell for every two $\mathrm{K^+}$ it brings in, and $\mathrm{K^+}$ leak channels let potassium diffuse back out down its concentration gradient, leaving large negatively charged proteins behind. Voltage-gated sodium channels are closed at the resting potential; they open only when the membrane depolarizes to threshold. Blocking a channel that is already shut changes nothing, so the resting potential stays at about −70 mV.

(b) An action potential begins when a stimulus depolarizes the membrane to threshold (about −55 mV). Voltage-gated $\mathrm{Na^+}$ channels then open, $\mathrm{Na^+}$ floods in and the inside swings positive to about +40 mV. That rapid depolarization is the rising phase, and it is also what brings the next patch of membrane to threshold, so the impulse travels along the axon. With the sodium channels blocked, a stimulus can still produce a small local (graded) depolarization through other channels, but the regenerative, all-or-none rise cannot occur, the membrane never approaches +40 mV and nothing is passed to the neighbouring membrane. The voltage-gated $\mathrm{K^+}$ channels are intact but irrelevant: there is no depolarization to repolarize from. Motor neurons therefore stop signalling to muscle, which is why pufferfish poisoning causes paralysis, including of the muscles used for breathing.

The common trap is to predict that the resting potential collapses, because sodium is the ion students associate with neurons. The resting state is maintained by potassium movement and the pump; sodium channels matter only during the action potential itself.

(a) Resting potential unchanged, about −70 mV; (b) no action potentials can be generated or conducted, so the neuron cannot transmit impulses (paralysis).

(a) Transcription builds mRNA by complementary base pairing with the template strand: template A pairs with U, T with A, G with C and C with G. RNA polymerase reads the template $3' \rightarrow 5'$ and assembles the mRNA $5' \rightarrow 3'$, so the two strands are antiparallel and the mRNA's 5′ end sits opposite the template's 3′ end.

• Template: 3′-TAC GAC ATG AAA ACT-5′
• mRNA: 5′-AUG CUG UAC UUU UGA-3′

(b) Read the mRNA $5' \rightarrow 3'$ in non-overlapping triplets from the start codon: AUG | CUG | UAC | UUU | UGA. From the excerpt, AUG = Met (start), CUG = Leu, UAC = Tyr, UUU = Phe and UGA = stop. No tRNA matches a stop codon; a release factor binds instead and the chain is released, so the stop codon adds nothing. The polypeptide is Met–Leu–Tyr–Phe: four amino acids. The common error is to copy the template with only T swapped for U, giving UAC GAC AUG AAA ACU (Tyr–Asp–Met–Lys–Thr); the fact that this sequence does not begin with the start codon AUG is the clue that something went wrong. Counting the stop codon as a fifth amino acid is the other frequent slip.

(c) Only the third triplet changes, so only the third codon needs re-checking.

Allele 1: template ATG becomes ATC, transcribed as UAG. UAG is a stop codon, so translation ends after Met–Leu. This is a nonsense mutation (a base substitution that creates a premature stop codon). The polypeptide is truncated to two amino acids and is almost certainly non-functional, since it lacks half of its residues and cannot fold into the normal shape.

Allele 2: template ATG becomes ATA, transcribed as UAU. UAU also codes for tyrosine, so the polypeptide is still Met–Leu–Tyr–Phe. This is a silent mutation: the genetic code is degenerate (most amino acids have several codons, usually differing only in the third base), so the amino acid sequence, and therefore the protein, is unchanged.

The same base substituted two different ways gives one harmless allele and one devastating one; what matters is not the size of the change in the DNA but what it does to the reading of the codons. Neither is a frameshift, because no bases were added or removed.

(a) 5′-AUG CUG UAC UUU UGA-3′; (b) Met–Leu–Tyr–Phe, four amino acids; (c) allele 1: UAG (stop), nonsense mutation, truncated non-functional Met–Leu; allele 2: UAU (Tyr), silent mutation, protein unchanged.