AP Biology Practice Exam 2 ❮ 301
26. A
27. C
28. C
29. B
30. A
31. D
32. B
33. C
34. D
35. A
- B—It is not autosomal dominant because in
order for the second generation on the left to
have those two individuals with the condition,
one parent would need to display the condition
as well. It is probably not sex-linked because it
seems to appear as often in females as in males.
Autosomal recessive seems to be the best fit for
this disease.
- D—One first needs to determine the probability
that person C is heterozygous (Bb). We know that
person D is double recessive because she has the
condition. We know that the parents for person
C must be Bb and Bb because neither of them
has the condition, but they produced children
with the condition. The probability of person C
being heterozygous is 2 ⁄ 3, because a monohybrid
cross of his parents (Bb × Bb) gives the following
Punnett square:
Since you know that he doesn’t have the condi-
tion, he cannot be bb. This leaves just three pos-
sible outcomes, two of which are Bb. A cross
must then be done between the father (person C)
Bb and the mother (person D) bb. The chance of
their child being bb is 50 percent or 1 ⁄ 2. This
means that the chance of these two having a
child with the condition is 2 ⁄ 3 × 1 ⁄ 2 or 1 ⁄ 3.
- C—Albinism is the only autosomal recessive
condition on this list.
- C—It is 1 ⁄ 2, because finding out that one of their
children has the condition lets us know that the
father (person C) is definitelyBb. This changes
the probability of 2 ⁄ 3 to 1, meaning that the
probability of the two having another child with
this condition is simply the result of the Punnett
square of Bb ×bb, or 1⁄2.
- D—When you see a ratio like the one in this
problem—7:7:1:1 (approximately)—the genes are
probably linked. The reason the crumpled, gray,
and vestigial black flies exist at all is because
crossover must have occurred.
- A—To determine the crossover frequency in a
problem like this, simply add up the total
number of crossovers (75 + 45 =120) and divide
that sum by the total number of offspring (120 +
555 + 525 =1200). This results in 120/1200 or
10 percent.
- B—One map unit is equal to a 1 percent recom-
bination frequency.
- B—The data in the table show you that this
answer is the correct choice.
- D—The larger the value of Rf for a bunch of
pigments dissolved in a particular chromatogra-
phy solvent, the faster the pigments will migrate.
Beta carotene has the highest Rf value.
- B—Across the board it seems to have the lowest
rate of transpiration. You can make this leap
because, as mentioned on top of the larger chart,
all the leaves have the same surface area, allowing
you to compare their transpiration values.
- C—The average enamel thickness started at 10,
increased to 12, and then increased to 15. It is
therefore increasing overall.
- D—The average enamel thickness does not
describe the range of possible values; an individ-
ual with a thickness of 15 could reasonably come
from any of the three generations (if we took
intoaccount probability, we could say that the
individual most likely came from the 100th gen-
eration because this population has the highest
frequency of individuals with this thickness; how-
ever, the question does not ask for probabilities).
B b
B BB Bb
b Bb bb
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