Transcript DNA(Test - Mrs. DeNicola's Science Corner

DNA (Test 1)
3. When DNA replicates, each strand of the original
DNA molecule is used as a template for the synthesis
of a second, complementary strand. Which of the
following figures most accurately illustrates enzymemediated synthesis of new DNA at a replication fork?
DNA (Test 1)
7. Lactose digestions in E. coli begins with its hydrolysis by the
enzyme -galactosidase, lacZ, is part of a coordinatley regulated
operon containing the genes required for lactose utilization. Which of
the following figures correctly depicts the interactions at the lac
operon when lactose is NOT being utilized? (The legend defines the
shapes illustrated)
DNA(Test 1)
19. The first diagram below shows the levels of mRNA from two
different genes (bicoid and caudal) at different positions along the
anterior-positerior axis of a Drosophila (fruit fly) egg immediately
before fertilization. The second diagram shows the levels of the two
correspon ding proteins along the anterior—posterior axis shortly after
fertilization.
Which of the following conclusions is best supported by the data?
a. Bicoid protein inhibits translation of caudal mRNA.
b. Bicoid protein stabilizes caudal mRNA.
c. Translation of bicoid mRNA produces caudal protein.
d. Caudal protein stimulates development of anterior structures.
DNA(Test 1)
20. Sickle-cell anemia results from a point mutation in the HBB
gene. The mutation results in the replacement of an amino acid
that has a hydrophilic R-group with an amino acid that has a
hydrophobic R-group on the exterior of the hemoglobin
protein. Such a mutation would most likely result in altered:
a. properties of the molecule as a result of abnormal interactions
between adjacent hemoglobin molecules.
b. DNA structure as a result of abnormal hydrogen bonding
between nitrogenous bases.
c. Fatty acid structure as a result of changes in ionic interactions
between adjacent fatty acid chains.
d. Protein secondary structure as a result of abnormal hydrophobic
interactions between R-groups in the backbone of the protein.
DNA(Test 1) 24-28
In a transformation experiment, a sample of E. coli bacteria was mixed
with a plasmid containing the gene for resistance to the antibiotic
ampicillin (amp). Plasmid was not added to a second sample. Samples
were plated on nutrient agar plates, some of which were supplemented
with the antibiotic amp. The results of E. coli growth are summarized
below. The shaded area represents extensive growth of bacteria; dots
represent individual colonies.
24. Plates that have only amp-resistant bacteria growing
include which of the following?
a. I only
b. III only
c. IV only
d. I and II
25. Which of the following best explains why there is no
growth on plate II?
a. The initial E. coli culture was not amp resistant.
b. The transformation procedure killed the bacteria.
c. Nutrient agar inhibits E. coli growth.
d. The bacteria on the plate were transformed.
26. Plates I and II were included in the experimental design in order
to
a.
b.
c.
d.
demonstrate that E. coli cultures were viable
demonstrate that the plasmid can lose its amp gene
demonstrate that the plasmid is needed for E. coli growth
prepare the E. coli for transformation
27. Which of the following statements best explains why there are
fewer colonies on plate IV than on plate III?
a. Plate IV is the positive control.
b. Not all E. coli cells are successfully transformed.
c. The bacteria on plate III did not mutate.
d. The plasmid inhibits E. coli growth.
28. In a second experiment, the plasmid contained the gene
for human insulin as well as the amp gene. Which of the
following plates would have the highest percentage of bacteria
that are expected to produce insulin?
a. I only
b. III only
c. IV only
d. I and III
DNA (Test 1)
38. Based on your understanding of the ways in which signal
transmission mediates cell function, which of the following
predictions is most consistent with the information provided?
a. In an environment with low fixed nitrogen, treating the Anabaena
cells with a Ca-binding compound should prevent heterocyst
differentiation.
b. A strain that overexpresses the patS gene should develop many
more heterocysts in a low fixed nitrogen environment.
c. In an environment with abundant fixed nitrogen, free Ca levels
should be high in all cells so that no heterocysts develop.
d. In environments with abundant fixed nitrogen, loss of the hetR
gene should induce heterocyst development
DNA (Test 1)
39. The following figure 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?
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.
DNA (Test 2)
5. Which of the following correctly matches the class of RNA with its
function?
a. mRNA transfers a message from DNA in the nucleus to the
tRNA in the cytoplasm.
b. tRNA transfers a message mRNA in the nucleus to the rRNA in
the cytoplasm.
c. mRNA transfers a message from DNA in the nucleus to the
ribosomes in the cytoplasm.
d. tRNA transfers a message from rRNA in the nucleus to mRNA in
the cytoplasm.
DNA (Test 2)
7. During which scenario can a viral infection introduce
genetic variation into the host?
a. The combination of human flu virus & bird flu virus to
create the H5N1 subtype of flu virus
b. Integration of viral DNA into bacterial DNA during the
lysogenic cycle
c. Maturation and assembly of viral components during the
lytic cycle
d. Bacterial infections in humans such as botulism, tetanus,
and leprosy
DNA (Test 2)
19. Bacteria live in an environment that is ever changing. In
order to be efficient there is no need to produce all of the same
enzymes and proteins all of the time. One way to control the
production of enzymes and other proteins is to
a. use a posttranscriptional control such as a delay of the
mRNA exiting the nucleus.
b. use internal cues to trigger gene regulation by proteins that
bind to their DNA.
c. use a promoter gene to signal where transcription is to
begin.
d. use an active repressor to cleave and splice the DNA
sequences removing unneeded portions.
DNA (Test 2) 46 - 50
Each cell in multicellular eukaryotes,
including humans, has a copy of all
genes; however, different genes are
actively expressed in different cells.
Muscle cells, for example, have a
different set of genes that are turned on
in the nucleus and a different set of
proteins that are active in the cytoplasm
than do nerve cells. Like prokaryotic
cells, a variety of mechanisms regulate
gene expression in eukaryotic cells.
These mechanisms regulate gene
expression in eukaryotic cells. These
mechanisms can be grouped under 5
primary levels of control; 3 of them
pertain to the nucleus, and 2 pertain to
the cytoplasm. In other words, control
of gene activity in eukaryotes extends
from transcription to protein activity.
DNA (Test 2)
46. Because their environment is constantly changing, bacteria
do not need the same enzymes all of the time. As a way to
regulate the expression of its genes a portion of the DNA codes
for a repressor that binds to the operator and stops the
production of mRNA. This description is most accurate for
a. a regulator gene
b. a promoter
c. an operator
d. an active repressor
DNA(Test 2)
47. The most critical level of eukaryotic genetic controls is the control
of transcription in the nucleus. Transcription is controlled by DNA
binding proteins called a transcription factor. Which description most
accurately describes the actions of a transcription factor?
a. TF’s work during DNA unpacking, physically moving nucleosomes
aside, exposing promoters so that transcription may begin.
b. TF’s bind to the promoter of the DNA, and with the help of
mediator proteins and transcription activators forms a hairpin loop
with the enhancer on the DNA. This complex attracts and bind
RNA pol so that transcription may begin.
c. TF’s are active in the eukaryotic nucleus and control genetic
expression by removing introns from pre-mRNA prior to its exit
from the nucleus.
d. TF’s are active outside of the eukaryotic nucelus and control
genetic expression by removing exons from post-mRNA after it has
exited the nucleus.
DNA (Test 2)
48. E. coli prefers to break down glucose and has a way to ensure that
the lactose operon is maximally turned on only when glucose is absent.
A molecule called cyclic AMP (cAMP) accumulates when glucose is
absent. What is the effect of cAMP as it binds to an active CAP and
the CAP binding site on the DNA?
a. cAMP is an enzyme which promotes the breakdown of lactose into
its constituent parts
b. cAMP acts as an inactive repressor, acting so that RNA polymerase
cannot attach to the DNA
c. The binding site of cAMP to the DNA causes the DNA to bend
exposing the promoter to RNA polymerase
d. The binding of cAMP to the DNA causes the DNA to form a
hairpin loop, allowing transcription to begin.
DNA(Test 2)
49. There is a group of proteins that assist bacteria in the
controls of genetic expression. These proteins bind to DNA
near the structural genes, essentially stopping RNA
polymerase from attaching to DNA, halting the production of
mRNA. Which protein is best described this way?
a. enzymes
b. RNA polymerase
c. Lactose
d. Active and inactive repressors