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General Biology — Practice Questions flashcards

6 of 55 questions free to study here · 23 topics · the whole deck is built into Willys

Multiple-choice practice questions, one best answer each, on membranes, energy, signaling, cell division, genetics, gene regulation, evolution, organism diversity, physiology and ecology.

Practice deck · part of General Biology I/II · written against: Standard two-semester majors sequence; OpenStax Biology 2e, consulted as a reference; no text copied. Updated 2026-10-10.

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  1. Pancreatic cells get a brief pulse of radioactive amino acids followed by a long chase with unlabeled ones. The cells were first treated with a drug that blocks vesicle transport from the ER to the Golgi apparatus. Where will most of the labeled protein be at the end of the chase?

    A) In the Golgi apparatus
    B) In the medium outside the cell
    C) In the rough endoplasmic reticulum
    D) In secretory vesicles near the membrane
    C) In the rough endoplasmic reticulum

    Secretory proteins are made on the rough ER and then move to the Golgi, to secretory vesicles and out of the cell, the order Palade's pulse-chase work revealed. Blocking the ER-to-Golgi step traps the labeled protein at its first stop.
    Cell Structure and Membranes
  2. Liver cells are stimulated with epinephrine in the presence of a drug that blocks phosphodiesterase. Compared with cells given epinephrine alone, what is expected?

    A) cAMP is never made, so glycogen breakdown does not start
    B) cAMP rises normally, but protein kinase A cannot be activated
    C) Epinephrine enters the cells and acts on genes directly
    D) cAMP stays high longer, prolonging glycogen breakdown
    D) cAMP stays high longer, prolonging glycogen breakdown

    Phosphodiesterase switches the signal off by converting cAMP to AMP. Blocking it leaves cAMP elevated, so the kinase cascade that activates glycogen phosphorylase keeps running and the response lasts longer.
    Cell Signaling
  3. The template strand of a gene reads 3′-TAC TTC AAA-5′. A mutation changes the T at the fourth position of this strand to an A. What kind of mutation is this?

    A) A missense mutation that replaces lysine with another amino acid
    B) A silent mutation that leaves the protein unchanged
    C) A frameshift mutation that changes every later codon
    D) A nonsense mutation that creates a stop codon
    D) A nonsense mutation that creates a stop codon

    The original mRNA is 5′-AUG AAG UUU-3′ (Met–Lys–Phe). After the change the second codon is UAG, a stop codon, so translation ends after methionine. A one-base substitution does not shift the reading frame.
    Gene Expression and Regulation
  4. A mutant strain of the cyanobacterium Anabaena forms heterocysts that still carry out the oxygen-releasing reactions of photosystem II. What would you predict?

    A) Nitrogen fixation rises, because heterocysts gain more ATP
    B) Photosynthesis fails, because heterocysts lack chlorophyll
    C) Growth is unaffected, because heterocysts do not fix nitrogen
    D) Nitrogen fixation fails, because O₂ inactivates nitrogenase
    D) Nitrogen fixation fails, because O₂ inactivates nitrogenase

    Heterocysts protect nitrogenase, which oxygen destroys, by lacking photosystem II and so releasing no O₂. A heterocyst that splits water would poison its own nitrogen-fixing enzyme.
    Prokaryote and Protist Diversity
  5. Helping a full sibling (r = 1/2) raise offspring costs a helper 1 offspring of its own. By Hamilton's rule, what is the threshold that the sibling's extra offspring from the help must exceed for an allele for helping to spread?

    A) 1/2
    B) 1
    C) 2
    D) 4
    C) 2

    Hamilton's rule says helping spreads when rB > C; with r = 1/2 and C = 1, B must exceed 1 ÷ 1/2 = 2. A benefit just over 1 would be enough only if the relative shared all of the helper's genes.
    Animal Behavior
  6. A captive-breeding population has 20 breeding males and 80 breeding females. Using Ne=4NmNfNm+NfN_e = \frac{4 N_m N_f}{N_m + N_f}⁠, what is its effective population size?

    A) 40
    B) 64
    C) 80
    D) 100
    B) 64

    Ne=4×20×80100=64N_e = \frac{4 \times 20 \times 80}{100} = 64⁠. Because so few males father the offspring, the population loses genetic diversity by drift as fast as an ideal population of 64, not 100.
    Conservation Biology

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Review status: Second-reader reviewed, with corrections applied.

Checked against:

  • OpenStax Biology 2e, consulted as a reference; no text copied.
  • Original questions written in the style of an introductory college biology final exam; no official, textbook or question-bank items were copied.

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