Download Ch 14- 17 Unit Test - Akron Central Schools

Ch 14- 17 Unit Test
Genetics
• When Thomas Hunt Morgan crossed his red-eyed F1
generation flies to each other, the F2 generation
included both red- and white-eyed flies. Remarkably,
all the white-eyed flies were male. What was the
explanation for this result?
• A) The gene involved is on the Y chromosome.
• B) The gene involved is on the X chromosome.
• C) The gene involved is on an autosome, but only in
males.
• D) Other male-specific factors influence eye color in
flies.
• B) The gene involved is on the X chromosome.
• Which of the following is the meaning of the
chromosome theory of inheritance as expressed in the
early twentieth century?
• A) Individuals inherit particular chromosomes
attached to genes.
• B) Mendelian genes are at specific loci on the
chromosome and, in turn, segregate during meiosis.
• C) No more than a single pair of chromosomes can be
found in a healthy normal cell.
• D) Natural selection acts on certain chromosome
arrays rather than on genes.
• B) Mendelian genes are at specific loci on the
chromosome and, in turn, segregate during
meiosis.
• SRY is best described as ____.
• A) a gene present on the X chromosome that
triggers female development
• B) an autosomal gene that is required for the
expression of genes on the Y chromosome
• C) a gene region present on the Y chromosome
that triggers male development
• D) an autosomal gene that is required for the
expression of genes on the X chromosome
• C) a gene region present on the Y
chromosome that triggers male development
• Cinnabar eyes is a sex-linked, recessive
characteristic in fruit flies. If a female having
cinnabar eyes is crossed with a wild-type
male, what percentage of the F1 males will
have cinnabar eyes?
• A) 0%
• B) 25%
• C) 50%
• D) 100%
• D) 100%
• Sex determination in mammals is due to the SRY
gene. Which of the following could allow a person
with an XX karyotype to develop a male
phenotype?
• A) the loss of the SRY gene from an autosome
• B) translocation of SRY to a X chromosome
• C) a person with an extra autosomal
chromosome
• D) a person with one normal and one shortened
(deleted) X
• B) translocation of SRY to a X chromosome
• In humans, clear gender differentiation occurs, not at
fertilization, but after the second month of gestation.
What is the first event of this differentiation?
• A) formation of testosterone in male embryos
• B) formation of estrogens in female embryos
• C) activation of SRY in male embryos and
masculinization of the gonads
• D) activation of SRY in females and feminization of the
gonads
• B) formation of estrogens in female embryos
• All female mammals have one active X
chromosome per cell instead of two. What causes
this?
• A) activation of the XIST gene on the X
chromosome that will become the Barr body
• B) activation of the BARR gene on one X
chromosome, which then becomes inactive
• C) inactivation of the XIST gene on the X
chromosome derived from the male parent
• D) attachment of methyl (CH3) groups to the X
chromosome that will remain active
• A) activation of the XIST gene on the X
chromosome that will become the Barr body
• What is an adaptive advantage of recombination
between linked genes?
• A) Recombination is required for independent
assortment.
• B) Recombination must occur or genes will not
assort independently.
• C) New allele combinations are acted upon by
natural selection.
• D) The forces on the cell during meiosis II results
in recombination.
• C) New allele combinations are acted upon by
natural selection.
• In a series of mapping experiments, the recombination
frequencies for four different linked genes of
Drosophila were determined as shown in the figure
above. What is the order of these genes on a
chromosome map?
• A) rb-cn-vg-b
• B) cn-rb-b-vg
• C) b-rb-cn-vg
• D) vg-cn-b-rb
• c) b-rb-cn-vg
• A phenotypically normal prospective couple seeks genetic
counseling because the man knows that he has a translocation of a
portion of his chromosome 4 that has been exchanged with a
portion of his chromosome 12. Although his translocation is
balanced, he and his wife want to know the probability that his
sperm will be abnormal. What is your prognosis regarding his
sperm?
• A) 1/4 will carry the two normal chromosomes, 4 and 12, 1/4 will
have only the two translocation chromosomes and no normal
chromosomes 4 and 12, and 1/2 will have both normal and
translocated chromosomes.
• B) All will carry the same translocation as the father.
• C) None will carry the translocation.
• D) 1/2 will be normal and the rest will have the father's
translocation.
• A) 1/4 will carry the two normal
chromosomes, 4 and 12, 1/4 will have only the
two translocation chromosomes and no
normal chromosomes 4 and 12, and 1/2 will
have both normal and translocated
chromosomes.
• What is a syndrome?
• A) a characteristic facial appearance
• B) a trait that leads to cancer at some stage in
life
• C) a group of traits typically found in
conjunction with a particular chromosomal
aberration or gene mutation
• D) a specific characteristic that appears in
conjunction with one specific aneuploidy
• C) a group of traits typically found in
conjunction with a particular chromosomal
aberration or gene mutation
• Which of the following is an example of
monosomy?
• A) Turner's syndrome
• B) Klinefelter's syndrome
• C) Down syndrome
• D) trisomy X
• A) Turner's syndrome
• In his transformation experiments, what did Griffith
observe?
• A) Mixing a heat-killed pathogenic strain of bacteria with
a living nonpathogenic strain can convert some of the living
cells into the pathogenic form.
• B) Mixing a heat-killed nonpathogenic strain of bacteria
with a living pathogenic strain makes the pathogenic strain
nonpathogenic.
• C) Infecting mice with nonpathogenic strains of bacteria
makes them resistant to pathogenic strains.
• D) Mice infected with a pathogenic strain of bacteria can
spread the infection to other mice.
• A) Mixing a heat-killed pathogenic strain of
bacteria with a living nonpathogenic strain can
convert some of the living cells into the
pathogenic form.
• Which of the following investigators was (were)
responsible for the following discovery? In DNA
from any species, the amount of adenine equals
the amount of thymine, and the amount of
guanine equals the amount of cytosine.
• A) Alfred Hershey and Martha Chase
• B) Oswald Avery, Maclyn McCarty, and Colin
MacLeod
• C) Erwin Chargaff
• D) Matthew Meselson and Franklin Stahl
• C) Erwin Chargaff
• It became apparent to Watson and Crick after
completion of their model that the DNA
molecule could carry a vast amount of
hereditary information in which of the
following?
• A) sequence of bases
• B) phosphate-sugar backbones
• C) complementary pairing of bases
• D) side groups of nitrogenous bases
• A) sequence of bases
• What is meant by the description "antiparallel"
regarding the strands that make up DNA?
• A) The twisting nature of DNA creates
nonparallel strands.
• B) The 5' to 3' direction of one strand runs
counter to the 5' to 3' direction of the other
strand.
• C) Base pairings create unequal spacing between
the two DNA strands.
• D) One strand contains only purines and the
other contains only pyrimidines.
• B) The 5' to 3' direction of one strand runs
counter to the 5' to 3' direction of the other
strand.
• Suppose you are provided with an actively dividing
culture of E. coli bacteria to which radioactive thymine
has been added. What would happen if a cell replicates
once in the presence of this radioactive base?
• A) One of the daughter cells, but not the other, would
have radioactive DNA.
• B) Neither of the two daughter cells would be
radioactive.
• C) All four bases of the DNA would be radioactive.
• D) DNA in both daughter cells would be radioactive.
• D) DNA in both daughter cells would be
radioactive.
• Eukaryotic telomeres replicate differently than
the rest of the chromosome. This is a
consequence of which of the following?
• A) the evolution of telomerase enzyme
• B) DNA polymerase that cannot replicate the
leading strand template to its 5' end
• C) gaps left at the 5' end of the lagging strand
• D) gaps left at the 3' end of the lagging strand
because of the need for a primer
• C) gaps left at the 5' end of the lagging strand
• The leading and the lagging strands differ in that ____.
• A) the leading strand is synthesized in the same direction
as the movement of the replication fork, and the lagging
strand is synthesized in the opposite direction
• B) the leading strand is synthesized by adding nucleotides
to the 3' end of the growing strand, and the lagging strand
is synthesized by adding nucleotides to the 5' end
• C) the lagging strand is synthesized continuously, whereas
the leading strand is synthesized in short fragments that are
ultimately stitched together
• D) the leading strand is synthesized at twice the rate of
the lagging strand
• A) the leading strand is synthesized in the
same direction as the movement of the
replication fork, and the lagging strand is
synthesized in the opposite direction
• What is the function of topoisomerase?
• A) relieving strain in the DNA ahead of the
replication fork
• B) elongating new DNA at a replication fork
by adding nucleotides to the existing chain
• C) unwinding of the double helix
• D) stabilizing single-stranded DNA at the
replication fork
• A) relieving strain in the DNA ahead of the
replication fork
• Which of the following help(s) to hold the
DNA strands apart while they are being
replicated?
• A) primase
• B) ligase
• C) DNA polymerase
• D) single-strand DNA binding proteins
• D) single-strand DNA binding proteins
• Which of the following would you expect of a
eukaryote lacking telomerase?
• A) a high probability of somatic cells
becoming cancerous
• B) an inability to produce Okazaki fragments
• C) an inability to repair thymine dimers
• D) a reduction in chromosome length in
gametes
• D) a reduction in chromosome length in
gametes
• You briefly expose bacteria undergoing DNA replication to
radioactively labeled nucleotides. When you centrifuge the
DNA isolated from the bacteria, the DNA separates into two
classes. One class of labeled DNA includes very large
molecules (thousands or even millions of nucleotides long),
and the other includes short stretches of DNA (several
hundred to a few thousand nucleotides in length). These
two classes of DNA probably represent ____.
• A) leading strands and Okazaki fragments
• B) lagging strands and Okazaki fragments
• C) Okazaki fragments and RNA primers
• D) leading strands and RNA primers
• A) leading strands and Okazaki fragments
• DNA contains the template needed to copy itself,
but it has no catalytic activity in cells. What
catalyzes the formation of phosphodiester bonds
between adjacent nucleotides in the DNA
polymer being formed?
• A) ribozymes
• B) DNA polymerase
• C) ATP
• D) deoxyribonucleotide triphosphates
• B) DNA polymerase
• What provides the energy for the
polymerization reactions in DNA synthesis?
• A) ATP
• B) DNA polymerase
• C) breaking the hydrogen bonds between
complementary DNA strands
• D) the deoxyribonucleotide triphosphate
substrates
• D) the deoxyribonucleotide triphosphate
substrates
• What is the difference between the leading strand and the
lagging strand in DNA replication?
• A) The leading strand is synthesized in the 3' ® 5' direction
in a discontinuous fashion, while the lagging strand is
synthesized in the 5' ® 3' direction in a continuous fashion.
• B) The leading strand is synthesized continuously in the 5'
® 3' direction, while the lagging strand is synthesized
discontinuously in the 5' ® 3' direction.
• C) The leading strand requires an RNA primer, whereas
the lagging strand does not.
• D) There are different DNA polymerases involved in
elongation of the leading strand and the lagging strand.
• B) The leading strand is synthesized
continuously in the 5' ® 3' direction, while the
lagging strand is synthesized discontinuously
in the 5' ® 3' direction.
• Which of the following represents the order of
increasingly higher levels of organization of
chromatin?
• A) nucleosome, 30-nm chromatin fiber, looped
domain
• B) looped domain, 30-nm chromatin fiber,
nucleosome
• C) nucleosome, looped domain, 30-nm
chromatin fiber
• D) 30-nm chromatin fiber, nucleosome, looped
domain
• A) nucleosome, 30-nm chromatin fiber,
looped domain
• Which of the following is most critical for the
association between histones and DNA?
• A) Histones are small proteins.
• B) Histones are highly conserved (that is,
histones are very similar in every eukaryote).
• C) There are at least five different histone
proteins in every eukaryote.
• D) Histones are positively charged.
• D) Histones are positively charged.
• What does it mean when we say the genetic code
is redundant?
• A) A single codon can specify the addition of
more than one amino acid.
• B) The genetic code is different for different
domains of organisms.
• C) The genetic code is universal (the same for all
organisms).
• D) More than one codon can specify the addition
of the same amino acid.
•
More than one codon can specify the
addition of the same amino acid.
• Which of the following does not occur in
prokaryotic gene expression, but does occur in
eukaryotic gene expression?
• A) mRNA, tRNA, and rRNA are transcribed.
• B) RNA polymerase binds to the promoter.
• C) A cap is added to the 5' end of the mRNA.
• D) RNA polymerase requires a primer to
elongate the molecule.
• C) A cap is added to the 5' end of the mRNA.
•
•
•
•
A ribozyme is ____.
A) a catalyst that uses RNA as a substrate
B) an RNA with catalytic activity
C) an enzyme that catalyzes the association
between the large and small ribosomal
subunits
• D) an enzyme that synthesizes RNA as part of
the transcription process
• B) an RNA with catalytic activity
• A primary transcript in the nucleus of a
eukaryotic cell is ____ the functional mRNA,
while a primary transcript in a prokaryotic cell
is ____ the functional mRNA.
• A) the same size as; smaller than
• B) larger than; the same size as
• C) larger than; smaller than
• D) the same size as; larger than
• B) larger than; the same size as
• A signal peptide ____.
• A) directs an mRNA molecule into the
cisternal space of the ER
• B) terminates translation of messenger RNA
• C) helps target a protein to the ER
• D) signals the initiation of transcription
• C) helps target a protein to the ER
• What type of bonding is responsible for
maintaining the shape of the tRNA molecule
shown in the figure above?
• A) ionic bonding between phosphates
• B) hydrogen bonding between base pairs
• C) van der Waals interactions between
hydrogen atoms
• D) peptide bonding between amino acids
• B) hydrogen bonding between base pairs
• What must occur before a newly made
polypeptide is secreted from a cell?
• A) It must be translated by a ribosome that
remains free within the cytosol.
• B) Its signal sequence must target it to the ER,
after which it goes to the Golgi.
• C) Its signal sequence must be cleaved off before
the polypeptide can enter the ER.
• D) Its signal sequence must target it to the
plasma membrane, where it causes exocytosis.
• B) Its signal sequence must target it to the ER,
after which it goes to the Golgi.
• Which one of the following, if missing, would
usually prevent translation from starting?
• A) exon
• B) 5' cap
• C) AUG codon
• D) poly-A tail
• C) AUG codon
• How does termination of translation take
place?
• A) The end of the mRNA molecule is reached.
• B) A stop codon is reached.
• C) The 5' cap is reached.
• D) The poly-A tail is reached.
• B) A stop codon is reached.
• Which of the following statements is true about
protein synthesis in prokaryotes?
• A) Extensive RNA processing is required before
prokaryotic transcripts can be translated.
• B) Translation can begin while transcription is
still in progress.
• C) Prokaryotic cells have complicated
mechanisms for targeting proteins to the
appropriate cellular organelles.
• D) Unlike eukaryotes, prokaryotes require no
initiation or elongation factors.
• B) Translation can begin while transcription is
still in progress.
• A nonsense mutation in a gene ____.
• A) changes an amino acid in the encoded
protein
• B) has no effect on the amino acid sequence
of the encoded protein
• C) introduces a premature stop codon into
the mRNA
• D) alters the reading frame of the mRNA
• C) introduces a premature stop codon into
the mRNA
• During meiosis, a defect occurs in a cell that results in the failure of
microtubules, spindle fibers, to bind at the kinetochores, a protein
structure on chromatids where the spindle fibers attach during cell
division to pull sister chromatids apart. Which of the following is the most
likely result of such a defect?
• A) New microtubules with more effective binding capabilities to
kinetochores will be synthesized to compensate for the defect.
• B) Excessive cell divisions will occur resulting in cancerous tumors and
an increase in the chromosome numbers known as polyploidy.
• C) The defect will be bypassed in order to and ensure normal
chromosome distribution in the new cells.
• D) The resulting cells will not receive the correct number of
chromosomes in the gametes, a condition known as aneuploidy.
• D) The resulting cells will not receive the
correct number of chromosomes in the
gametes, a condition known as aneuploidy.
• Inheritance patterns cannot always be explained by
Mendel’s models of inheritance. If a pair of
homologous chromosomes fails to separate during
meiosis I, select the choice that shows the
chromosome number of the four resulting gametes
with respect to the normal haploid number (n)?
•
•
•
•
A)
B)
C)
D)
n+1; n+1; n-1; n-1
n+1; n-1; n; n
n+1; n-1; n-1; n-1
n+1; n+1; n; n
• A) n+1; n+1; n-1; n-1
• In E. coli replication the enzyme primase is used to attach a 5 to 10
base ribonucleotide strand complementary to the parental DNA
strand. The RNA strand serves as a starting point for the DNA
• polymerase that replicates the DNA. If a mutation occurred in the
primase gene, which of the following would you expect?
• A) Replication would only occur on the leading strand.
• B) Replication would only occur on the lagging strand.
• C) Replication would not occur on either the leading or lagging
strand.
• D) Replication would not be affected as the enzyme primase in
involved with RNA synthesis.
• C) Replication would not occur on either the
leading or lagging strand.
• Hershey and Chase used a DNA-based virus for their
work. What would the results have been if they had
used an RNA virus?
• A) With an RNA virus radioactive protein would have
been in the final pellet.
• B) With an RNA virus radioactive RNA would have
been in the final pellet.
• C) With an RNA virus neither sample would have had
a radioactive pellet.
• D) With an RNA virus the protein shell would have
been radioactive in both samples.
• B) With an RNA virus radioactive RNA would
have been in the final pellet.
• The lagging strand is characterized by a series of short segments of
DNA (Okazaki fragments) that will be joined together to form a
finished lagging strand. The experiments that led to the discovery of
Okazaki fragments gave evidence for which of the following ideas?
• A) DNA polymerase is a directional enzyme that synthesizes
leading and lagging strands during replication.
• B) DNA is a polymer consisting of four monomers: adenine,
thymine, guanine, and cytosine.
• C) DNA is the genetic material.
• D) Bacterial replication is fundamentally different from eukaryotic
replication. The key shouldn’t be way longer than the distractors.
• A) DNA polymerase is a directional enzyme
that synthesizes leading and lagging strands
during replication.