278 ❯ STEP 5. Build Your Test-Taking Confidence
- B—The expected heterozygous probability does
not match up with the actual. This population is
not in Hardy–Weinberg equilibrium. - B—The homozygous frequency is higher than
expected; one explanation for this is that the
homozygotes are being selected for. - A—Water will flow outof the bag because the
solute concentration of the beaker is hypertonic
compared to the dialysis bag. Osmosis passively
drives water from a hypotonic region to a hyper-
tonic region. - B—Water would now flow intothe bag because
the solute gradient has been reversed. Now the
beaker is hypotonic compared to the dialysis bag.
Water thus moves into the bag. - C—Water potential =pressure potential +solute
potential. Water passively moves from regions
with high water potential toward those with lower
water potential. - D—The mother (person B) must be heterozy-
gous Aa because she and her husband (aa) have
produced children that have the double recessive
condition. This means that person B (the mother)
must have contributed an a and that the cross is
Aa×aa—and the probability is 1 ⁄ 2. - D—To answer this question, we must first
determine the probability that person D is het-
erozygous. We know she is not aa because she
does not have the condition. Since we know that
the father hasthe condition, we know for certain
that his genotype is aa. Both of mother D’s par-
ents must be heterozygous since neither of them
have the condition, but they have produced a child
with the condition. The probability that mother D
is heterozygous Aa is 2 ⁄3. The probability that a
couple with the genotypes Aa ×aa have a double
recessive child is 1 ⁄ 2.The probability that these
two will have a child with the condition is 1 ⁄ 2 ×
2⁄3 =1⁄3 =0.333.
54. E—If the couple has a child (person E) with the
recessive condition, then we know for certain
that mother D must be heterozygous. It is defi-
nitely an aa × Aa cross, leaving a 50 percent
chance that their child will be aa.
55. B
56. B
57. A—The total solute potential for side A is 1.0
MPa (remember that for NaCl, i =2), and the
total solute potential for side B is 0.8 MPa.
Therefore, side A has a higher concentration of
solute (hypertonic).
58. D—Water will move from a hypotonic solution
(side B) toward a hypertonic solution (side A).
Sodium will diffuse from a region of more sodium
(side B) to a region of less sodium (side A).
59. A—Heterozygous individuals carry the recessive
gene but are themselves healthy.
60. D—Low numbers of stomata help to reduce
water loss, helpful in hot and dry regions.
61. A—Prokaryotic and eukaryotic cells do have
similar structures, the organelles in eukaryotic
cells took care of having a smaller surface
area–to–volume ratio, and eukaryotic cells are
not able to multiply faster.
62. B—Changing envelope proteins are created
because of genetic variation in the genes that
code for these proteins.
63. D—This cladogram shows a closer relationship
between X and Y.
A a
A AA Aa
a Aa aa