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a. DNA ligase
b. DNA polymerase
c. RNA polymerase
d. Restriction enzyme
e. Reverse transcriptase
f. Transformation
1. Enzyme found in retroviruses that produce
DNA from an RNA template. Answer: e
2. Enzyme used during replication to attach
Okazaki fragments to each other. Answer: a
3. Our bacteria that produced biolumescent
proteins.
Answer: f
Put the following phrases in order to form a
plasmid carrying recombinant DNA.
1. use restriction enzyme
2. use DNA ligase
3. remove plasmid from parent bacterium
4. introduce plasmid into new host bacterium
a. 1, 2, 3, 4
b. 4, 3, 2, 1
c. 3, 1, 2, 4
d. 2, 3, 1, 4
Answer: c
Restriction endonucleases (enzymes)
a. cut DNA.
b. are naturally found in prokaryotic cells.
c. are naturally found in eukaryotic cells.
d. both a and b
e. all of the above
Answer: d
The separation of DNA fragments by gel
electrophoresis is primarily achieved by
differential
a. sizes of fragments.
b. charges of fragments.
c. solubilities of fragments.
d. cleavage points of fragments.
e. radioactivity of fragments.
Answer: a
Which of the following is used for amplification?
a. Transformation
b. Polymerase chain reaction (PCR)
c. Restriction fragment length polymorphism
(RFLP)
d. Recombinant DNA technology
e. All of these
Answer: b
Which of the following is NOT needed to
make a recombinant DNA molecule?
a. foreign DNA
b. vector DNA
c. restriction enzymes
d. DNA ligase
e. DNA polymerase
Answer: e
Genetically engineered plants have or will
be used to
a. resist insects
b. resist herbicides
c. produce protein-enhanced beans, corn, and
wheat
d. produce animal neuropeptides, blood factors,
and growth hormones.
e. All of these are correct.
Answer: e
DNA that is made from mRNA is called
a. nonsense DNA (nDNA)
b. antisense DNA (aDNA)
c. complementary DNA (cDNA)
d. uncomplementary DNA (uDNA)
e. recombinant DNA (rDNA)
Answer: c
Foreign DNA can be inserted into vector DNA
because both DNA molecules
a. have the same genes.
b. have the same bases.
c. have “sticky ends”.
d. are not complementary to each other.
e. All of these are correct.
Answer: c
Which enzyme is used to seal breaks in a
DNA molecule?
a. DNA polymerase
b. RNA polymerase
c. restriction enzymes
d. DNA ligase
e. RNA ligase
Answer: d
1.
2.
3.
4.
Blastula stage Answer: E
Beginning of gastrulation Answer: F
Zygote Answer: A
Formation of germ layers Answer: G
1. Lane with smallest
band
Answer: A
2. Lane showing most
restriction sites
Answer: A
3. Lane with DNA
most closely related
to that in lane F
Answer: B
Which of the following describes the correct
sequence of stages during embryogenesis?
a.
b.
c.
d.
e.
cleavage, blastula formation, gastrulation
cleavage, gastrulation, blastula formation
blastula formation, gastrulation, cleavage
blastula formation, cleavage, gastrulation
gastrulation, cleavage, blastula formation
Answer: A
In animals, all of the following are associated
with embryonic development EXCEPT
a.
b.
c.
d.
e.
migration of cells to specific areas.
formation of germ layers.
activation of all the genes in each cell.
inductive tissue interactions.
cell division at a relatively rapid rate.
Answer: C
Cell differentiation
a. results from the loss of particular genes
from the nucleus of the differentiated cell.
b. results from the differential expression
of genes that are responsive to environmental
signals.
c. involves the persisting totipotency of early
embryonic cells in the mature organism.
d. results from the mutations in genes that
control the synthesis of DNA.
e. precedes cell determination.
Answer: B
Gel electrophoresis can be used for which of
the following purposes?
a. To group molecules based on their polarity
b. To measure the acidity of certain large
molecules
c. To measure the polarity of certain large
molecules
d. To separate out the proteins in a mixture
e. To measure the amount of protein in a
mixture of substances
Answer: D
The figure below shows several steps in the process of bacteriophage
transduction in bacteria. Which of the following explains how genetic variation
in a population of bacteria results from this process?
Answer: D
a. Bacterial proteins transferred from the donor bacterium by the phage to the
recipient bacterium recombine with genes on the recipient’s chromosome.
b. The recipient bacterium incorporates the transduced genetic material
coding for phage proteins into its chromosome and synthesizes the
corresponding proteins.
c. The phage infection of the recipient bacterium and the introduction of DNA
carried by the phage cause increased random point mutations of the
bacterial chromosome.
d. DNA of the recipient bacterial chromosome undergoes recombination with
DNA introduced by the phage from the donor bacterium, leading to a
change in the recipient’s genotype.
In a transformation experiment, a sample of E. coli bacteria was
mixed with a plasmid containing the gene for resistance to the
antibiotic ampicillin (ampr). Plasmid was not added to a second
sample. Samples were plated on nutrient agar plates, some of
which were supplemented with the antibiotic ampicillin. The
results of
E. coli growth are summarized below.
The shaded area represents extensive
growth of bacteria; dots represent
individual colonies of bacteria.
Plates that have only ampicillinresistant bacteria growing include
which of the following?
a. I only
b. III only
c. IV only
Answer: C
d. I and II
In a transformation experiment, a sample of E. coli bacteria was mixed
with a plasmid containing the gene for resistance to the antibiotic
ampicillin (ampr). Plasmid was not added to a second sample. Samples
were plated on nutrient agar plates, some of which were supplemented
with the antibiotic ampicillin. The results of E. coli growth are
summarized
below. The shaded area represents extensive
growth of bacteria; dots represent
individual colonies of bacteria.
Which of the following best explains why
there is no growth on plate II?
A. The initial E. coli culture was not ampicillin
resistant.
B. The transformation procedure killed the
bacteria.
C. Nutrient agar inhibits E. coli growth.
D. The bacteria on the plate were
transformed.
Answer: A
In a transformation experiment, a sample of E. coli bacteria was
mixed with a plasmid containing the gene for resistance to the
antibiotic ampicillin (ampr). Plasmid was not added to a second
sample. Samples were plated on nutrient agar plates, some of
which were supplemented with the antibiotic ampicillin. The
results of
E. coli growth are summarized below.
The shaded area represents extensive
growth of bacteria; dots represent
individual colonies of bacteria.
Plates I and III were included in the
experimental design in order to
a. demonstrate that the E. coli cultures were
viable.
b. demonstrate that the plasmid can lose its ampr
gene
c. demonstrate that the plasmid is needed for E.
coli growth
d. prepare the E. coli for transformation
Answer: A
In a transformation experiment, a sample of E. coli bacteria was
mixed with a plasmid containing the gene for resistance to the
antibiotic ampicillin (ampr). Plasmid was not added to a second
sample. Samples were plated on nutrient agar plates, some of
which were supplemented with the antibiotic ampicillin. The
results of
E. coli growth are summarized below.
The shaded area represents extensive
growth of bacteria; dots represent
individual colonies of bacteria.
Which of the following statements best
explains why there are fewer colonies
on plate IV than on plate III?
A.
B.
C.
D.
Plate IV is the positive control.
Not all E. coli cells are successfully transformed.
The bacteria on plate III did not mutate.
The plasmid inhibits E. coli growth.
Answer: B
A student placed 20 tobacco seeds of the same species on moist paper towels in each of two petri
dishes. Dish A was wrapped completely in an opaque cover to exclude all light. Dish B was not
wrapped. The dishes were placed equidistant from a light source set to a cycle of 14 hours of light
and 10 hours of dark. All other conditions were the same for both dishes. The dishes were
examined after 7 days and the opaque cover was permanently removed from dish A. Both dishes
were returned to the light and examined again at 14 days. The following data were obtained.
Dish A
Day 7 Covered Day 14 Uncovered
Dish B
Day 7 Uncovered Day 14
Uncovered
Germinated seeds
12
20
20
20
Green leaved seedlings
0
14
15
15
Yellow leaved seedlings
12
6
5
5
Mean stem length
below first set of leaves
8 mm
9 mm
3 mm
3 mm
Which of the following best supports the hypothesis that the difference in leaf color is genetically
controlled?
a.
b.
c.
d.
The number of yellow leaved seedlings in dish A on day 7
The number of germinated seeds in dish A on days 7 and 14
The death of all the yellow leaved seedlings
The existence of yellow leaved seedlings as well as green leaved ones on
day 14 in dish B
Answer: D
Homeotic genes
A. encode transcription factors that control
the expression of genes responsible for
specific anatomical structures.
B. are found only in Drosophila and other
arthropods.
C. encode proteins that form anatomical
structures in the fly.
D. are responsible for patterning during plant
development.
Answer: A
Genes that regulate development
are highly conserved. This means
that
A. large difference have evolved
among multicellular organisms.
B. they have changed very little over
the course of evolution.
C. they are always turned on.
D. they have been lost in some
lineages.
Answer: B
The diagram below shows a developing worm embryo at the four-cell stage.
Experiments have shown that when cell 3 divides, the anterior daughter cell
gives rise to muscle and gonads and the posterior daughter cell gives rise to
the intestine. However, if the cells of the embryo are separated from one
another early during the four-cell stage, no intestine will form. Other
experiments have shown that if cell 3 and cell 4 are recombined after the initial
separation, the posterior daughter cell of cell 3 will once again give rise to
normal intestine. Which of the following is the most plausible explanation for
these findings?
a. A cell surface protein on cell 4 signals
cell 3 to induce formation of the worm’s
intestine.
b. The plasma membrane of cell 4 interacts
with the plasma membrane of the
posterior portion of cell 3, causing
invaginations that become microvilli.
c. Cell 3 passes an electrical signal to cell
4, which induces differentiation in cell 4.
d. Cell 4 transfers genetic material to cell 3,
which directs the development of
Answer: A
intestinal cells.
Hox genes
A. encode protein domains that are
important in development and have
been highly conserved over
evolutionary time.
B. Help determine cell fate within each
segment of a developing Drosophila
embryo.
C. can produce the wrong structure in the
wrong place when mutated.
D. All of the above.
Answer: D