5 Steps to a 5 AP Biology, 2014-2015 Edition

(Marvins-Underground-K-12) #1

300 ❯ STEP 5. Build Your Test-Taking Confidence



  1. D—Habituation is the loss of responsiveness to
    unimportant stimuli or stimuli that do not provide
    appropriate feedback. This is a prime example of
    habituation.

  2. A—Directional selection occurs when members
    of a population at one end of a spectrum are
    selected against, while those at the other end are
    selected for. Taller giraffes are being selected for;
    shorter giraffes are being selected against.

  3. A—When interbreeding ceases because some
    sort of barrier separates a single population into
    two (an area with no food, a mountain, etc.), the
    two populations evolve independently, and if
    they change enough, then, even if the barrier is
    removed, they cannot interbreed. This is allopatric
    speciation.

  4. D—Chapter 19, despite being last, is a very
    important chapter. The experiments are very well
    represented on the AP Biology exam, and you
    should read this chapter carefully and learn how to
    design and interpret experiments.

  5. C

  6. A

  7. D

  8. B—The endosymbiotic theory proposes that
    mitochondria and chloroplasts evolved through
    the symbiotic relationship between prokaryotic
    organisms.

  9. D—Fibroblast growth factoris said to be involved,
    but fibroblast cells are not.

  10. D—Posttranscriptional modification actually
    occurs in the nucleus.

  11. C—Genetic drift is a change in allele frequen-
    cies that is due to chance events. When drift
    dramaticallyreduces population size, it is called
    a “bottleneck.”

  12. A—Neutrophils are phagocytic cells of the
    immune system. They roam the body looking
    for rubbish to clear.

  13. C—Only photosystem I is involved in the cyclic
    reactions. Photosystem II is not.

  14. A—Genetics has no memory... it will be 1 ⁄ 2
    forever.

  15. A—DNA migrates from a negative charge to a
    positive charge. The rest are true.

  16. B—0.04 = q^2. Therefore, the square root of
    0.04=q=0.20 and p+q=1. So p+0.20=1.
    Therefore,p=0.80, and 2pqis the frequency
    of the heterozygote condition: 2(0.20)(0.80) =
    0.320=32 percent.

  17. C—Phytochrome is an important pigment to
    the process of flowering. Of its two forms, the
    active form, Pfr, is responsible for the production
    of the hormone florigen, which is thought to
    assist in the blooming of flowers.

  18. D—Secondary carnivores > primary carnivores >
    primary consumers = herbivores > primary
    producers.

  19. C—The example to know is the cattle egrets that
    feast on insects aroused into flight by cattle graz-
    ing in the insects’ habitat. The birds benefit
    because they get food, but the cattle do not appear
    to benefit at all.

  20. B—Conjugation is the sexual reproduction of
    bacteria.

  21. B—In competitive inhibition, an inhibitor mol-
    ecule resembling the substrate binds to the active
    site and physically blocks the substrate from
    attaching.

  22. C—The other four are the four main factors that
    can affect enzyme efficiency.

  23. B—The activation energy of a reaction is the
    amount of energy needed for the reaction to
    occur. Notice that the activation energy for the
    enzymatic reaction is much lower than the
    nonenzymatic reaction.

  24. D

  25. B


❯ Answers and Explanations for Practice Exam 2


PART A: MULTIPLE-CHOICE QUESTIONS

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