MOLECULAR BIOLOGY OF THE CELL, SIXTH EDITION
CHAPTER 4: DNA, CHROMOSOMES, AND GENOMES
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.
( T ) The other chain is heavier, i.e. it has a greater mass.
( T ) There are no C residues in the other chain.
( F ) The 5′-terminal residue of the other chain is G.
( F ) The other chain is pyrimidine-rich.
The sequence of the other chain is 5′-TAGAATGGG-3′, which makes it slightly heavier because
it is mainly composed of the bulkier purine bases.
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.
( 4 ) Hydrogen-bonding
( 3 ) Covalent linkage
( 2 ) Phosphate group
( 1 ) Nitrogen-containing base
1
2
5
3
4
( 5 ) Deoxyribose sugar
In the double-stranded DNA, the sugar-phosphate backbones form two covalently
continuous chains, while the nitrogenous bases from one chain are hydrogen-bonded to
those of the other chain according to the Watson–Crick model.
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
T
C
T
A
A
A
G
-3′
The final sequence should be 5′-CTTTAGATCTAAAG-3′. The complementary strand would
then have the exact same sequence. This is an example of a palindromic sequence.
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.
7pg
There are about 6.4 × 109 nucleotide pairs (np) in a diploid nucleus. The average mass of 660
daltons is equivalent to 660 grams per mole (6 × 1023) of the nucleotide pairs. Thus, the total
mass of nuclear DNA is:
(6.4 × 109 np) × (660 g/mole) / (6 × 1023 np/mole) = ~7 × 10–12 g = ~7 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
In principle, replication would have been conceptually as simple without any of the other
features.
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)
Biological complexity correlates better with the number of genes than it does with genome size,
number of chromosomes, or number of genes per chromosome.
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.
( F ) Only about 1.5% of the human genome is highly conserved.
( T ) Almost half of our genome is composed of repetitive sequences.
( T ) Genes occupy almost a quarter of the genome.
( T ) There are roughly as many pseudogenes in the human genome as functional
genes.
( F ) Transposable elements occupy almost 10% of our genome.
( F ) On average, exons comprise 1.5% of our genes.
Only about 5% of our genome is highly conserved. Our relatively long genes (including their
exons, introns, and some regulatory sequences) cover almost a quarter of the genome, but only
about 1.5% of our genome is composed of exonic sequences. These exons constitute roughly 6%
of our genes. In contrast, a whopping 50% of our genome is made of various repeated sequences,
most notably the transposable DNA elements. Please refer to Table 4–1 for the data.
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
Multiplying the size of the chromosome by the distance between consecutive bases gives the
end-to-end distance as follows: (200 × 106 nucleotide pairs [np]) × (0.34 × 10–9 m/np) = ~0.068
m = ~7 cm.
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.
centromere, origin of replication, telomere
These three elements are required in a functional chromosome.
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.
( T ) It contains repeated sequences at the ends of the chromosomes.
( T ) It is NOT generally longer in higher organisms compared to yeast.
( O ) Each eukaryotic chromosome has many such sequences.
( T ) There are normally two such sequences in each eukaryotic chromosomal DNA
molecule.
( C ) There is normally one such sequence per eukaryotic chromosomal DNA
molecule.
( O ) It is where DNA duplication starts in S phase.
( C ) It attaches the chromosome to the mitotic spindle via the kinetochore structure.
There is normally one centromere per chromosomal DNA molecule, and two per mitotic
chromosome. The mitotic kinetochore structure forms at the centromere. In contrast, the
telomeres are at the two ends of each linear chromosome. There are usually many replication
origins per eukaryotic chromosome. Replication origins and centromeres are both generally
much longer in higher eukaryotes compared to yeast.
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
The various chromatin proteins, including histone and non-histone proteins, wrap and fold the
DNA to achieve an astonishing compaction ratio. The nuclear envelope and the nuclear matrix
are dispensable for this effect, as evident by the high compaction seen in mitotic chromosomes
when the nucleus is disassembled.
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.
The homologs are derived from both parents, normally have the same set of loci, and show the
same banding pattern. However, the sequences are not expected to be identical.
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.
The yeast has a concise genome with a much higher ratio of coding to noncoding 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.
( F ) The histones are highly acidic proteins.
( T ) The histone fold consists of three α helices.
( T ) The core histones are much more conserved than the H1 histone.
( T ) The N-terminal tails of the core histones undergo a variety of reversible post–
translational modifications.
( F ) Every nucleosome core is made up of three polypeptide chains.
( F ) The H1 histone is absent in the 30-nm fibers.
The highly basic histone proteins are made of the histone fold that consists of three α helices
connected by two loops. Additionally, each core histone has an N-terminal tail which, along with
the rest of the protein, can be modified post-translationally. Each nucleosome core particle
contains eight histone proteins, two copies of each type. The less well conserved histone H1 is
not part of the nucleosome core, but is present 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.
( 2 ) Nucleosome core particle
( 4 ) Linker DNA
( 3 ) Histone octamer
( 1 ) Non-histone protein
The nucleosome core particle is composed of a histone octamer wrapped by 147 nucleotide pairs
of DNA, and is connected to its neighbors via a linker DNA of variable length. Non-histone
proteins are also abundant in the chromatin.
1
2
4
3
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
The H3–H4 and the H2A–H2B dimers appear to be stable intermediates in histone assembly and
exchange. The H3–H4 tetramers are also stable, and are thought to be assembled (and inherited)
mostly as a single unit.
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.
