AP Bio Unit 5 College Board MCQ Test Questions and Answers Rated A
1. A model of crossing over during gamete formation A. Does synapsis of homol-
is shown in Figure 1.Based on Figure 1, which of the ogous chromosomes in
following questions could best be addressed? the parent cell contribute
A. Does synapsis of homologous chromosomes in the to an increase in genet-
parent cell contribute to an increase in genetic diversi- ic diversity in the daughter
ty in the daughter cells? cells?
B. Do sister chromatids separate and form diploid
daughter cells?
C. Do chromatids from nonhomologous chromosomes
rearrange to produce identical daughter cells?
D. Does synapsis of nonhomologous chromosomes
produce
2. Huntington's disease has been traced to the number C.An allele with more than
of CAG repeats in the HTT gene, which is located on 39 CAG repeats was inher-
chromosome 4. The phenotypic influence of individual ited by the attected per-
alleles with different numbers of repeats is shown in son.
Table 1.
Which of the following is most likely the immediate
cause of the first appearance of Huntington's disease
in a person?
Responses
A.A point mutation occurs in the HTT gene.
B.The first appearance of the CAG repeat occurs in the
HTT gene.
C.An allele with more than 39 CAG repeats was inher-
ited by the affected person.
D.The person inherited two alleles that each contained
20 CAG repeats.
3. In anaphase I of meiosis, cohesion between the cen- D. Is there a pattern to the
tromeres of sister chromatids is maintained while ho- movement of homologous
, AP Bio Unit 5 College Board MCQ Test Questions and Answers Rated A
mologous chromosomes migrate to opposite poles of chromosomes in the pres-
the cell along the meiotic spindle as represented in ence of this compound?
Figure 1.A compound that prevents the separation of
the homologous chromosomes in anaphase I is being
studied. Which of the following questions can be best
answered during this study?
Responses
A.Will the cells produced at the end of meiosis still be
genetically identical to each other in the presence of
this compound?
B.Will the long-term development of the individual be
affected by this meiotic error?
C.When do the centrosomes start to move apart during
meiosis I as compared to meiosis II?
D.Is there a pattern to the movement of homologous
chromosomes in the presence of this compound?
4. When a mustard plant seedling is transferred to an B.Leaf growth is promoted
environment with higher levels of carbon dioxide, the through increased pho-
new leaves have a lower stomata-to-surface-area ratio tosynthesis, but the ge-
than do the seedling's original leaves. netically regulated rate of
Which of the following best explains how the leaves stomatal production is not
from the same plant can have different stomatal den- altered, leading to a de-
sities when exposed to an elevated carbon dioxide crease in stomatal density.
level?
Responses
A. Increased photosynthesis leads to larger leaves that
need more stomata for photosynthesis, leading to an
increase in stomatal density.
B. Leaf growth is promoted through increased photo-
synthesis, but the genetically regulated rate of stom-
atal production is not altered, leading to a decrease in
, AP Bio Unit 5 College Board MCQ Test Questions and Answers Rated A
stomatal density.
C. Leaf growth is inhibited by decreased photosynthe-
sis, and the genetically regulated rate of stomatal pro-
duction remains the same, leading to an increase in
stomatal density.
D. Leaf growth is inhibited by decreased photosynthe-
sis, and the genet
5. Rubber rabbitbrush plants display heritable variation D.The critical value is 7.82,
in plant height and insect-induced gall formation.The and the null hypothesis
researchers calculated a chi-square value of 29.25. If can be rejected because
there are three degrees of freedom and the signifi- the calculated chi-square
cance level is p=0.05, which of the following state- value is greater than the
ments best completes the chi-square test? critical value.
A. The critical value is 0.05, and the null hypothesis
cannot be rejected because the calculated chi-square
value is greater than the critical value.
B. The critical value is 0.05, and the null hypothesis can
be rejected because the calculated chi-square value is
greater than the critical value.
C. The critical value is 7.82, and the null hypothesis
cannot be rejected because the calculated chi-square
value is greater than the critical value.
D. The critical value is 7.82, and the null hypothesis can
be rejected because the calculated chi-square value is
greater than the critical value.
6. Students carry out a genetics experiment to inves- B.11.34
tigate the inheritance pattern of the white-eye trait
in fruit flies. In the experiment, the students cross a
red-eyed female with a white-eyed male to produce an
F1 generation. The students observe that all the flies
1. A model of crossing over during gamete formation A. Does synapsis of homol-
is shown in Figure 1.Based on Figure 1, which of the ogous chromosomes in
following questions could best be addressed? the parent cell contribute
A. Does synapsis of homologous chromosomes in the to an increase in genet-
parent cell contribute to an increase in genetic diversi- ic diversity in the daughter
ty in the daughter cells? cells?
B. Do sister chromatids separate and form diploid
daughter cells?
C. Do chromatids from nonhomologous chromosomes
rearrange to produce identical daughter cells?
D. Does synapsis of nonhomologous chromosomes
produce
2. Huntington's disease has been traced to the number C.An allele with more than
of CAG repeats in the HTT gene, which is located on 39 CAG repeats was inher-
chromosome 4. The phenotypic influence of individual ited by the attected per-
alleles with different numbers of repeats is shown in son.
Table 1.
Which of the following is most likely the immediate
cause of the first appearance of Huntington's disease
in a person?
Responses
A.A point mutation occurs in the HTT gene.
B.The first appearance of the CAG repeat occurs in the
HTT gene.
C.An allele with more than 39 CAG repeats was inher-
ited by the affected person.
D.The person inherited two alleles that each contained
20 CAG repeats.
3. In anaphase I of meiosis, cohesion between the cen- D. Is there a pattern to the
tromeres of sister chromatids is maintained while ho- movement of homologous
, AP Bio Unit 5 College Board MCQ Test Questions and Answers Rated A
mologous chromosomes migrate to opposite poles of chromosomes in the pres-
the cell along the meiotic spindle as represented in ence of this compound?
Figure 1.A compound that prevents the separation of
the homologous chromosomes in anaphase I is being
studied. Which of the following questions can be best
answered during this study?
Responses
A.Will the cells produced at the end of meiosis still be
genetically identical to each other in the presence of
this compound?
B.Will the long-term development of the individual be
affected by this meiotic error?
C.When do the centrosomes start to move apart during
meiosis I as compared to meiosis II?
D.Is there a pattern to the movement of homologous
chromosomes in the presence of this compound?
4. When a mustard plant seedling is transferred to an B.Leaf growth is promoted
environment with higher levels of carbon dioxide, the through increased pho-
new leaves have a lower stomata-to-surface-area ratio tosynthesis, but the ge-
than do the seedling's original leaves. netically regulated rate of
Which of the following best explains how the leaves stomatal production is not
from the same plant can have different stomatal den- altered, leading to a de-
sities when exposed to an elevated carbon dioxide crease in stomatal density.
level?
Responses
A. Increased photosynthesis leads to larger leaves that
need more stomata for photosynthesis, leading to an
increase in stomatal density.
B. Leaf growth is promoted through increased photo-
synthesis, but the genetically regulated rate of stom-
atal production is not altered, leading to a decrease in
, AP Bio Unit 5 College Board MCQ Test Questions and Answers Rated A
stomatal density.
C. Leaf growth is inhibited by decreased photosynthe-
sis, and the genetically regulated rate of stomatal pro-
duction remains the same, leading to an increase in
stomatal density.
D. Leaf growth is inhibited by decreased photosynthe-
sis, and the genet
5. Rubber rabbitbrush plants display heritable variation D.The critical value is 7.82,
in plant height and insect-induced gall formation.The and the null hypothesis
researchers calculated a chi-square value of 29.25. If can be rejected because
there are three degrees of freedom and the signifi- the calculated chi-square
cance level is p=0.05, which of the following state- value is greater than the
ments best completes the chi-square test? critical value.
A. The critical value is 0.05, and the null hypothesis
cannot be rejected because the calculated chi-square
value is greater than the critical value.
B. The critical value is 0.05, and the null hypothesis can
be rejected because the calculated chi-square value is
greater than the critical value.
C. The critical value is 7.82, and the null hypothesis
cannot be rejected because the calculated chi-square
value is greater than the critical value.
D. The critical value is 7.82, and the null hypothesis can
be rejected because the calculated chi-square value is
greater than the critical value.
6. Students carry out a genetics experiment to inves- B.11.34
tigate the inheritance pattern of the white-eye trait
in fruit flies. In the experiment, the students cross a
red-eyed female with a white-eyed male to produce an
F1 generation. The students observe that all the flies
