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MOLECULAR BIOLOGY OF THE CELL, SIXTH EDITION
CHAPTER 4: DNA, CHROMOSOMES, AND GENOMES
© Garland Science 2015
1.
In a double-stranded DNA molecule, one of the chains has the sequence CCCATTCTA
when read from the 5′ to the 3′ end. Indicate true (T) and false (F) statements below regarding
this chain. Your answer would be a four-letter string composed of letters T and F only, e.g.
TTFT.
( ) The other chain is heavier, i.e. it has a greater mass.
( ) There are no C residues in the other chain.
( ) The 5′-terminal residue of the other chain is G.
( ) The other chain is pyrimidine-rich.
2.
Indicate which numbered feature (1 to 5) in the schematic drawing below of the DNA
double helix corresponds to each of the following. Your answer would be a five-digit number
composed of digits 1 to 5 only, e.g. 52431.
( ) Hydrogen-bonding
( ) Covalent linkage
( ) Phosphate group
( ) Nitrogen-containing base
( ) Deoxyribose sugar
1
2
5
3
4
3.
Complete the DNA sequence below such that the final sequence is identical to that of the
complementary strand. Your answer would be a seven-letter string composed of letters A, C, T,
and G only, e.g. TTCTCAG.
5′-
C
T
T
T
A
G
A
-3′
4.
A DNA nucleotide pair has an average mass of approximately 660 daltons. Knowing the
number of nucleotides in the human genome, how many picograms of DNA are there in a diploid
human nucleus? Avogadro’s number is 6 × 1023. Write down the picogram amount without
decimals (round the number to the closest integer), e.g. 23 pg.
5.
Which of the following features of DNA underlies its simple replication procedure?
A. The fact that it is composed of only four different types of bases
B. The antiparallel arrangement of the double helix
C. The complementary relationship in the double helix
D. The fact that there is a major groove and a minor groove in the double helix
6.
Which of the following correlates the best with biological complexity in eukaryotes?
A. Number of genes per chromosome
B. Number of chromosomes
C. Number of genes
D. Genome size (number of nucleotide pairs)
7.
Indicate true (T) and false (F) statements below about the human genome. Your answer
would be a six-letter string composed of letters T and F only, e.g. FTFFFT.
( ) Only about 1.5% of the human genome is highly conserved.
( ) Almost half of our genome is composed of repetitive sequences.
( ) Genes occupy almost a quarter of the genome.
( ) There are roughly as many pseudogenes in the human genome as functional genes.
( ) Transposable elements occupy almost 10% of our genome.
( ) On average, exons comprise 1.5% of our genes.
8.
Chromosome 3 contains nearly 200 million nucleotide pairs of our genome. If this DNA
molecule could be laid end to end, how long would it be? The distance between neighboring base
pairs in DNA is typically around 0.34 nm.
A. About 7 mm
B. About 7 cm
C. About 70 cm
D. About 7 m
E. None of the above
9.
For the Human Genome Project, cloning of large segments of our genome was first made
possible by the development of yeast artificial chromosomes, which are capable of propagating
in the yeast Saccharomyces cerevisiae just like any of the organism’s 16 natural chromosomes.
In addition to the cloned human DNA, these artificial vectors were made to contain three
elements that are necessary for them to function as a chromosome. What are these elements?
Write down the names of the elements in alphabetical order, and separate them with commas,
e.g. gene, histone, nucleosome.
10.
Indicate whether each of the following descriptions better applies to a centromere (C), a
telomere (T), or an origin of replication (O). Your answer would be a seven-letter string
composed of letters C, T, and O only, e.g. TTTCCTO.
( ) It contains repeated sequences at the ends of the chromosomes.
( ) It is NOT generally longer in higher organisms compared to yeast.
( ) Each eukaryotic chromosome has many such sequences.
( ) There are normally two such sequences in each eukaryotic chromosomal DNA
molecule.
( ) There is normally one such sequence per eukaryotic chromosomal DNA molecule.
( ) It is where DNA duplication starts in S phase.
( ) It attaches the chromosome to the mitotic spindle via the kinetochore structure.
11.
The eukaryotic chromosomes are organized inside the nucleus with a huge compaction
ratio of several-thousand-fold. What is responsible for such a tight packaging?
A. The various chromatin proteins that wrap and fold the DNA
B. The nuclear envelope which encapsulates the chromosomes
C. The nuclear matrix that provides a firm scaffold
D. All of the above
12.
The two chromosomes in each of the 22 homologous pairs in our cells ...
A. have the exact same DNA sequence.
B. are derived from one of our parents.
C. show identical banding patterns after Giemsa staining.
D. usually bear different sets of genes.
E. All of the above.
13.
Compared to the human genome, the genome of yeast typically has …
A. more repetitive DNA.
B. longer genes.
C. more introns.
D. longer chromosomes.
E. a higher fraction of coding DNA.
14.
Indicate true (T) and false (F) statements below regarding histones. Your answer would
be a six-letter string composed of letters T and F only, e.g. TTFFFF.
( ) The histones are highly acidic proteins.
( ) The histone fold consists of three α helices.
( ) The core histones are much more conserved than the H1 histone.
( ) The N-terminal tails of the core histones undergo a variety of reversible post-
translational modifications.
( ) Every nucleosome core is made up of three polypeptide chains.
( ) The H1 histone is absent in the 30-nm fibers.
15.
Indicate which feature (1 to 4) in the schematic drawing below of a chromatin fiber
corresponds to each of the following. Your answer would be a four-digit number composed of
digits 1 to 4 only, e.g. 2431.
1
2
4
3
( ) Nucleosome core particle
( ) Linker DNA
( ) Histone octamer
( ) Non-histone protein
16.
In assembling a nucleosome, normally the …(1) histone dimers first combine to form a
tetramer, which then further combines with two … (2) histone dimers to form the octamer.
A. 1: H1–H3; 2: H2A–H2B
B. 1: H3–H4; 2: H2A–H2B
C. 1: H2A–H2B; 2: H1–H3
D. 1: H2A–H2B; 2: H3–H4
E. 1: H1–H2; 2: H3–H4
17.
The chromatin remodeling complexes play an important role in chromatin regulation in
the nucleus. They …
A. can slide nucleosomes on DNA.
B. have ATPase activity.
C. interact with histone chaperones.
D. can remove or exchange core histone subunits.
E. All of the above.
18.
Which of the following is true regarding heterochromatin in a typical mammalian cell?
A. About 1% of the nuclear genome is packaged in heterochromatin.
B. The DNA in heterochromatin contains all of the inactive genes in a cell.
C. Genes that are packaged in heterochromatin are permanently turned off.
D. The different types of heterochromatin share an especially high degree of compaction.
E. Heterochromatin is highly concentrated in the centromeres but not the telomeres.
19.
The position effect variegation (PEV) phenotype described in this chapter can be used to
identify new genes that regulate heterochromatin formation. For instance, strains of Drosophila
melanogaster with the White variegation phenotype have been subjected to mutagenesis to
screen for dominant mutations (in other genes) that either enhance or suppress PEV, meaning the
mutations result in either lower or higher red pigment production, respectively. Which of the
following mutations is expected to be an enhancer of variegation?
A. A mutation that results in the loss of function of the fly’s HP1 (heterochromatin
protein 1) gene.
B. A loss-of-function mutation in a gene encoding a histone deacetylase that
deacetylates lysine 9 on histone H3.
C. A gain-of-function mutation in a gene encoding a histone methyl transferase that
trimethylates lysine 9 on histone H3, resulting in a hyperactive form of the enzyme.
D. A gain-of-function mutation in a gene encoding a histone acetyl transferase that
normally acetylates lysine 9 on histone H3, resulting in higher expression of the
protein.
20.
The acetylation of lysines on the histone tails …
A. loosens the chromatin structure because it adds positive charges to the histone.
B. recruits the heterochromatin protein HP1, resulting in the establishment of
heterochromatin.
C. can be performed on methylated lysines only after they are first demethylated.
D. is sufficient for the formation of an open chromatin structure.
E. is a covalent modification and is thus irreversible.
21.
Nucleosomes that are positioned like beads on a string over a region of DNA can interact
to form higher orders of chromatin structure. Which of the following factors can contribute to the
formation of the 30-nm chromatin fiber from these nucleosomes?
A. Interactions that involve the histone tails of neighboring nucleosomes
B. Interaction of the linker histone H1 with each nucleosome
C. Binding of proteins to DNA or the histones
D. ATP-dependent function of chromatin remodeling complexes
E. All of the above
22.
Indicate whether each of the following histone modifications is generally associated with
active genes (A) or silenced genes (S). Your answer would be a four-letter string composed of
letters A and S only, e.g. SSAS.
( ) H3 lysine 9 acetylation
( ) H3 serine 10 phosphorylation
( ) H3 lysine 4 trimethylation
( ) H3 lysine 9 trimethylation
23.
Indicate whether each of the following histone modifications adds a negative charge to
the histone (A), removes a positive charge from the histone (B), or does neither of these (C).
Your answer would be a four-letter string composed of letters A, B, and C only, e.g. CABA.
( ) H3 lysine 9 acetylation
( ) H3 serine 10 phosphorylation
( ) H3 lysine 4 trimethylation
( ) H3 lysine 9 trimethylation
24.
To study the chromatin remodeling complex SWR1, a researcher has prepared arrays of
nucleosomes on long DNA strands that have been immobilized on magnetic beads. These
nucleosomes are then incubated with an excess of the H2AZ–H2B dimer (which contains the
histone variant H2AZ) in the presence or absence of SWR1 with or without ATP. She then
separates the bead-bound nucleosomes (bound fraction) from the rest of the mix (unbound
fraction) using a magnet, elutes the bound fraction from the beads, and performs SDS-PAGE on
the samples. This is followed by a Western blot using an antibody specific to the H2AZ protein
used in this experiment. The results are shown below, with the presence (+) or absence (–) of
