Genes and
Chromosomes
S-274
1. Packaging of DNA in a Virus Bacteriophage T2 has a DNA of molecular weight 120 10
6
con-
tained in a head about 210 nm long. Calculate the length of the DNA (assume the molecular weight of
a nucleotide pair is 650) and compare it with the length of the T2 head.
2. The DNA of Phage M13 The base composition of phage M13 DNA is A, 23%; T, 36%; G, 21%;
C, 20%. What does this tell you about the DNA of phage M13?
Answer The complementarity between A and T, and between G and C, in the two strands of
3. The Mycoplasma Genome The complete genome of the simplest bacterium known, Mycoplasma
genitalium, is a circular DNA molecule with 580,070 bp. Calculate the molecular weight and contour
length (when relaxed) of this molecule. What is Lk
0
for the Mycoplasma chromosome? If j0.06,
what is Lk?
chapter
24
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4. Size of Eukaryotic Genes An enzyme isolated from rat liver has 192 amino acid residues and is
coded for by a gene with 1,440 bp. Explain the relationship between the number of amino acid
residues in the enzyme and the number of nucleotide pairs in its gene.
5. Linking Number A closed-circular DNA molecule in its relaxed form has an Lk of 500. Approximately
how many base pairs are in this DNA? How is the linking number altered (increases, decreases,
doesn’t change, becomes undefined) when (a) a protein complex is bound to form a nucleosome,
(b) one DNA strand is broken, (c) DNA gyrase and ATP are added to the DNA solution, or (d) the
double helix is denatured by heat?
6. DNA Topology In the presence of a eukaryotic condensin and a type II topoisomerase, the Lk of a
relaxed closed-circular DNA molecule does not change. However, the DNA becomes highly knotted.
Chapter 24 Genes and Chromosomes S-275
type II topoisomerase
Condensin
Formation of the knots requires breakage of the DNA, passage of a segment of DNA through the
break, and religation by the topoisomerase. Given that every reaction of the topoisomerase would be
expected to result in a change in linking number, how can Lk remain the same?
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S-276 Chapter 24 Genes and Chromosomes
7. Superhelical Density Bacteriophage linfects E. coli by integrating its DNA into the bacterial chro-
mosome. The success of this recombination depends on the topology of the E. coli DNA. When the
superhelical density (j) of the E. coli DNA is greater than 0.045, the probability of integration is
20%; when jis less than 0.06, the probability is 70%. Plasmid DNA isolated from an E. coli cul-
ture is found to have a length of 13,800 bp and an Lk of 1,222. Calculate jfor this DNA and predict
the likelihood that bacteriophage lwill be able to infect this culture.
8. Altering Linking Number (a) What is the Lk of a 5,000 bp circular duplex DNA molecule with a
nick in one strand? (b) What is the Lk of the molecule in (a) when the nick is sealed (relaxed)? (c) How
would the Lk of the molecule in (b) be affected by the action of a single molecule of E. coli topoiso-
merase I? (d) What is the Lk of the molecule in (b) after eight enzymatic turnovers by a single
molecule of DNA gyrase in the presence of ATP? (e) What is the Lk of the molecule in (d) after four
enzymatic turnovers by a single molecule of bacterial type I topoisomerase? (f) What is the Lk of the
molecule in (d) after binding of one nucleosome?
9. Chromatin Early evidence that helped researchers define nucleosome structure is illustrated by the
agarose gel below, in which the thick bands represent DNA. It was generated by briefly treating chro-
matin with an enzyme that degrades DNA, then removing all protein and subjecting the purified DNA
to electrophoresis. Numbers at the side of the gel denote the position to which a linear DNA of the
indicated size would migrate. What does this gel tell you about chromatin structure? Why are the DNA
bands thick and spread out rather than sharply defined?
Answer The bands have a periodicity of about 200 bp (200, 400, 600 bp, etc.), showing that
the chromatin is protected from nuclease digestion at regular intervals of 200 bp. This sug-
10. DNA Structure Explain how the underwinding of a B-DNA helix might facilitate or stabilize the for-
mation of Z-DNA.
11. Maintaining DNA Structure (a) Describe two structural features required for a DNA molecule to
maintain a negatively supercoiled state. (b) List three structural changes that become more favorable
when a DNA molecule is negatively supercoiled. (c) What enzyme, with the aid of ATP, can generate
negative superhelicity in DNA? (d) Describe the physical mechanism by which this enzyme acts.
Chapter 24 Genes and Chromosomes S-277
200 bp
400 bp
600 bp
800 bp
1,000 bp
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S-278 Chapter 24 Genes and Chromosomes
(b) When a DNA molecule is negatively supercoiled, it is underwound. Any structural
12. Yeast Artificial Chromosomes (YACs) YACs are used to clone large pieces of DNA in yeast cells.
What three types of DNA sequences are required to ensure proper replication and propagation of a
YAC in a yeast cell?
13. Nucleoid Structure in Bacteria In bacteria, the transcription of a subset of genes is affected by
DNA topology, with expression increasing or (more often) decreasing when the DNA is relaxed. When a
bacterial chromosome is cleaved at a specific site by a restriction enzyme (one that cuts at a long, and thus
rare, sequence), only nearby genes (within 10,000 bp) exhibit either an increase or decrease in expression.
The transcription of genes elsewhere in the chromosome is unaffected. Explain. (Hint: See Fig. 24–36.)
14. DNA Topology When DNA is subjected to electrophoresis in an agarose gel, shorter molecules mi-
grate faster than longer ones. Closed-circular DNAs of the same size but with different linking numbers
also can be separated on an agarose gel: topoisomers that are more supercoiled, and thus more con-
densed, migrate faster through the gel. In the gel shown below, purified plasmid DNA has migrated
from top to bottom. There are two bands, with the faster band much more prominent.
(a) What are the DNA species in the two bands? (b) If topoisomerase I is added to a solution of
this DNA, what will happen to the upper and lower bands after electrophoresis? (c) If DNA ligase is
added to the DNA, will the appearance of the bands change? Explain your answer. (d) If DNA gyrase
plus ATP is added to the DNA after the addition of DNA ligase, how will the band pattern change?
Chapter 24 Genes and Chromosomes S-279
0.5 g/mL chloroquine
Average
Native
0
0.009
0.023
0.037
0.049
0.066
0.080
0.115
N
X
Runs as
Gel A
Runs as
Lk Lk
5 g/mL chloroquine
Average
Native
0
0.009
0.023
0.037
0.049
0.066
0.080
0.115
N
X
Runs as Runs as
Lk Lk
Gel B
15. DNA Topoisomers When DNA is subjected to electrophoresis in an agarose gel, shorter molecules
migrate faster than longer ones. Closed-circular DNAs of the same size but different linking number
also can be separated on an agarose gel: topoisomers that are more supercoiled, and thus more con-
densed, migrate faster through the gel—from top to bottom in the gels shown on page S-279. A dye,
chloroquine, was added to these gels. Chloroquine intercalates between base pairs and stabilizes a more
underwound DNA structure. When the dye binds to a relaxed, closed-circular DNA, the DNA is
underwound where the dye binds, and unbound regions take on positive supercoils to compensate. In
the experiment shown here, topoisomerases were used to make preparations of the same DNA circle
with different superhelical densities (). Completely relaxed DNA migrated to the position labeled N
(for nicked), and highly supercoiled DNA (above the limit where individual topoisomers can be distin-
guished) to the position labeled X.
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S-280 Chapter 24 Genes and Chromosomes
(a) In gel A, why does the 0 lane (i.e., DNA prepared so that 0, on average) have multiple
bands?
(b) In gel B, is the DNA from the 0 preparation positively or negatively supercoiled in the pres-
ence of the intercalating dye?
(c) In both gels, the 0.115 lane has two bands, one a highly supercoiled DNA and one relaxed.
Propose a reason for the presence of relaxed DNA in these lanes (and others).
(d) The native DNA (leftmost lane in each gel) is the same DNA circle isolated from bacterial cells
and untreated. What is the approximate superhelical density of this native DNA?
Answer
16. Nucleosomes The human genome contains just over 3.1 billion base pairs. Assuming it is covered with
nucleosomes that are spaced as described in this chapter, how many molecules of histone H2A are pre-
sent in one somatic human cell? (Do not consider reductions in H2A due to its replacement in some re-
gions by H2A variants.) How would the number change after DNA replication but before cell division?
Data Analysis Problem
17. Defining the Functional Elements of Yeast Chromosomes Figure 24–8 shows the major struc-
tural elements of a chromosome of baker’s yeast (Saccharomyces cerevisiae). Heiter, Mann, Snyder,
and Davis (1985) determined the properties of some of these elements. They based their study on the
finding that in yeast cells, plasmids (which have genes and an origin of replication) act differently from
chromosomes (which have these elements plus centromeres and telomeres) during mitosis. The plas-
mids are not manipulated by the mitotic apparatus and segregate randomly between daughter cells.
Without a selectable marker to force the host cells to retain them (see Fig. 9–4), these plasmids are
rapidly lost. In contrast, chromosomes, even without a selectable marker, are manipulated by the mi-
totic apparatus and are lost at a very low rate (about 10
5
per cell division).
(c) Based on these data, what can you conclude about the size of the centromere required for nor-
mal mitotic segregation? Explain your reasoning.
(d) Interestingly, all the synthetic chromosomes created in these experiments were circular and
lacked telomeres. Explain how they could be replicated more-or-less properly.
Heiter and colleagues next constructed a series of linear synthetic chromosomes that in-
cluded the functional centromeric sequence and telomeres, and measured the total mitotic error
rate (% loss % nondisjunction) as a function of size:
Heiter and colleagues set out to determine the important components of yeast chromosomes by
constructing plasmids with various parts of chromosomes and observing whether these “synthetic
chromosomes” segregated properly during mitosis. To measure the rates of different types of failed
chromosome segregation, the researchers needed a rapid assay to determine the number of copies of
synthetic chromosomes present in different cells. This assay took advantage of the fact that wild-type
yeast colonies are white whereas certain adenine-requiring (ade
) mutants yield red colonies on
nutrient media. Specifically, ade2
cells lack functional AIR carboxylase (the enzyme of step 6a in
Figure 22–35) and accumulate AIR (5-aminoimidazole ribonucleotide) in their cytoplasm. This excess
AIR is converted to a conspicuous red pigment. The other part of the assay involved the gene SUP11,
which encodes an ochre suppressor (a type of nonsense suppressor; see Box 27–4) that suppresses
the phenotype of some ade2
mutants.
Heiter and coworkers started with a diploid strain of yeast homozygous for ade2
; these cells are
red. When the mutant cells contain one copy of SUP11, the metabolic defect is partly suppressed and
the cells are pink. When the cells contain two or more copies of SUP11, the defect is completely sup-
pressed and the cells are white.
The researchers inserted one copy of SUP11 into synthetic chromosomes containing various ele-
ments thought to be important in chromosome function, and then observed how well these chromo-
somes were passed from one generation to the next. These pink cells were plated on nonselective me-
dia, and the behavior of the synthetic chromosomes was observed. Specifically, Heiter and coworkers
looked for colonies in which the synthetic chromosomes segregated improperly at the first division
after plating, giving rise to a colony that is half one genotype and half the other. Because yeast cells
are nonmotile, this will be a sectored colony, with one half one color and the other half another color.
(a) One way for the mitotic process to fail is nondisjunction: the chromosome replicates but the
sister chromatids fail to separate, so both copies of the chromosome end up in the same daugh-
ter cell. Explain how nondisjunction of the synthetic chromosome would give rise to a colony
that is half red and half white.
(b) Another way for the mitotic process to fail is chromosome loss: the chromosome does not enter
the daughter nucleus or is not replicated. Explain how loss of the synthetic chromosome would
give rise to a colony that is half red and half pink.
By counting the frequency of the different colony types, Heiter and colleagues could estimate
the frequency of these aberrant mitotic events with different types of synthetic chromosome.
First, they explored the requirement for centromeric sequences by constructing synthetic chromo-
somes with different-sized DNA fragments containing a known centromere. Their results are
shown below.
Chapter 24 Genes and Chromosomes S-281
Size of
centromere-
Synthetic containing Chromosome Nondisjunction
chromosome fragment (kbp) loss (%) (%)
1 none — 50
2 0.63 1.6 1.1
3 1.6 1.9 0.4
4 3.0 1.7 0.35
5 6.0 1.6 0.35
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(e) Based on these data, what can you conclude about the chromosome size required for normal mi-
totic segregation? Explain your reasoning.
(f) Normal yeast chromosomes are linear, range from 250 kbp to 2,000 kbp in length, and have a
mitotic error rate of about 10
5
per cell division. Extrapolating the results from (e), do the
centromeric and telomeric sequences used in these experiments explain the mitotic stability of
normal yeast chromosomes, or must other elements be involved? Explain your reasoning. (Hint:
A plot of log (error rate) vs. length will be helpful.)
Answer
Synthetic Size Total error
chromosome (kbp) rate (%)
6 15 11.0
7 55 1.5
8 95 0.44
9 137 0.14