1. 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.
2. Enzyme used during replication to attach
Okazaki fragments to each other.
3. Our bacteria that produced biolumescent
proteins.
Answer: e
Answer: a
Answer: f

2. 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
1, 2, 3, 4
4, 3, 2, 1
3, 1, 2, 4
2, 3, 1, 4
Answer: c

3. Restriction endonucleases (enzymes) cut DNA.
are naturally found in prokaryotic cells.
are naturally found in eukaryotic cells.
both a and b
all of the above
Mrs. Ohlsen is the best Biology teacher in HISTORY! Eat Mor Chikin
Answer: d

4. The separation of DNA fragments by gel
electrophoresis is primarily achieved by
differential
sizes of fragments.
charges of fragments.
solubilities of fragments.
cleavage points of fragments.
radioactivity of fragments.
Answer: a

5. Which of the following is used for amplification?
Transformation
Polymerase chain reaction (PCR)
Restriction fragment length polymorphism
(RFLP)
d. Recombinant DNA technology
e. All of these
Answer: b

6. Which of the following is NOT needed to
make a recombinant DNA molecule?
foreign DNA
vector DNA
restriction enzymes
DNA ligase
DNA polymerase
Answer: e

7. Genetically engineered plants have or will
be used to
resist insects
resist herbicides
produce protein-enhanced beans, corn, and
wheat
d. produce animal neuropeptides, blood factors,
and growth hormones.
e. All of these are correct.
Answer: e

8. DNA that is made from mRNA is called
nonsense DNA (nDNA)
antisense DNA (aDNA)
complementary DNA (cDNA)
uncomplementary DNA (uDNA)
recombinant DNA (rDNA)
Answer: c

9. Foreign DNA can be inserted into vector DNA
because both DNA molecules
have the same genes.
have the same bases.
have “sticky ends”.
are not complementary to each other.
All of these are correct.
Answer: c

10. Which enzyme is used to seal breaks in a
DNA molecule?
DNA polymerase
RNA polymerase
restriction enzymes
DNA ligase
RNA ligase
Answer: d

11. Answer: E
Blastula stage
Beginning of gastrulation
Zygote
Formation of germ layers
Answer: F
Answer: A
Answer: G

12. Lane with smallest
band
2. Lane showing most
restriction sites
3. Lane with DNA
most closely related
to that in lane F
Answer: A
Answer: A
Answer: B

13. Which of the following describes the correct
sequence of stages during embryogenesis?
cleavage, blastula formation, gastrulation
cleavage, gastrulation, blastula formation
blastula formation, gastrulation, cleavage
blastula formation, cleavage, gastrulation
gastrulation, cleavage, blastula formation
Answer: A

14. In animals, all of the following are associated
with embryonic development EXCEPT
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

15. Cell differentiation
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

16. Gel electrophoresis can be used for which of
the following purposes?
To group molecules based on their polarity
To measure the acidity of certain large
molecules
To measure the polarity of certain large
To separate out the proteins in a mixture
To measure the amount of protein in a
mixture of substances
Answer: D

17. 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
Bacterial proteins transferred from the donor bacterium by the phage to the recipient bacterium recombine with genes on the recipient’s chromosome.
The recipient bacterium incorporates the transduced genetic material coding for phage proteins into its chromosome and synthesizes the corresponding proteins.
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.
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.
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18. In a transformation experiment, a sample of E
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 ampicillin-resistant bacteria growing include which of the following?
I only
III only
IV only
I and II
Answer: C
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19. Answer: A Nutrient agar inhibits E. coli growth.
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?
The initial E. coli culture was not ampicillin resistant.
The transformation procedure killed the bacteria.
Nutrient agar inhibits E. coli growth.
The bacteria on the plate were transformed.
Answer: A
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20. In a transformation experiment, a sample of E
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
demonstrate that the E. coli cultures were viable.
demonstrate that the plasmid can lose its ampr gene
demonstrate that the plasmid is needed for E. coli growth
prepare the E. coli for transformation
Answer: A
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21. In a transformation experiment, a sample of E
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?
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
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22. 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 Dish B Day 7 Covered Day 14 Uncovered Day 7 Uncovered Day 14 Uncovered Germinated seeds Green leaved seedlings Yellow leaved seedlings Mean stem length below first set of leaves mm 9 mm 3 mm mm
Which of the following best supports the hypothesis that the difference in leaf color is genetically controlled?
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
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23. Homeotic genes Answer: A
encode transcription factors that control the expression of genes responsible for specific anatomical structures.
are found only in Drosophila and other arthropods.
encode proteins that form anatomical structures in the fly.
are responsible for patterning during plant development.
Answer: A
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24. Genes that regulate development are highly conserved. This means that
large difference have evolved among multicellular organisms.
they have changed very little over the course of evolution.
they are always turned on.
they have been lost in some lineages.
Answer: B
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25. 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 cell surface protein on cell 4 signals cell 3 to induce formation of the worm’s intestine.
The plasma membrane of cell 4 interacts with the plasma membrane of the posterior portion of cell 3, causing invaginations that become microvilli.
Cell 3 passes an electrical signal to cell 4, which induces differentiation in cell 4.
Cell 4 transfers genetic material to cell 3, which directs the development of intestinal cells.
Answer: A
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26. Hox genes
encode protein domains that are important in development and have been highly conserved over evolutionary time.
Help determine cell fate within each segment of a developing Drosophila embryo.
can produce the wrong structure in the wrong place when mutated.
All of the above.
Answer: D
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