CHAPTER 12
Application and Experimental Questions
E1.
Two circular DNA molecules, which we will call molecule A and molecule B, are
topoisomers of each other. When viewed under the electron microscope, molecule A
appears more compact compared to molecule B. The level of gene transcription is much
lower for molecule A. Which of the following three possibilities could account for these
observations?
First possibility: Molecule A has 3 positive supercoils and molecule B has 3 negative
supercoils.
Second possibility: Molecule A has 4 positive supercoils and molecule B has 1 negative
supercoil.
Third possibility: Molecule A has 0 supercoils and molecule B has 3 negative supercoils.
Answer: The second possibility, in which molecule A has 4 positive supercoils and
molecule B has 1 negative supercoil fits these data. Molecule A would be more
compacted because it has more supercoils. Also, molecule B would be more
transcriptionally active, because it is more negatively supercoiled. The first possibility
does not fit the data because both molecules have the same level of supercoiling so
molecule A and molecule B would have the same level of compaction. The third
possibility does not fit the data because molecule B would be more compact.
E2.
Let’s suppose that you have isolated DNA from a cell and have viewed it under a
microscope. It looks supercoiled. What experiment would you perform to determine if it
is positively or negatively supercoiled? In your answer, describe your expected results.
You may assume that you have purified topoisomerases at your disposal.
Answer: Supercoiled DNA would look curled up into a relatively compact structure. You
could add different purified topoisomerases and see how they affect the structure via
microscopy. For example, DNA gyrase relaxes positive supercoils, while topoisomerase I
relaxes negative supercoils. If we added topoisomerase I to a DNA preparation and it
became less compacted, then the DNA was negatively supercoiled.
E3.
We seem to know more about the structure of eukaryotic chromosomal DNA than
bacterial DNA. Discuss why you think this is so, and list several experimental procedures
that have yielded important information concerning the compaction of eukaryotic
chromatin.
Answer:
1.
The repeating nucleosome structure was revealed from DNase I digestion studies.
2.
Purification studies showed that the biochemical composition is an
octamer of histones.
3.
More recently, crystallography has shown the precise structure of the
nucleosome.
4.
Microscopy has revealed information about the 30 nm fiber and the
attachment of chromatin to the nuclear matrix.
In general, it is easier to understand the molecular structure of something when it forms a
regular repeating pattern. The eukaryotic chromosome has a repeating pattern of
nucleosomes. The bacterial chromosome seems to be more irregular in its biochemical
composition.
E4.
When chromatin is treated with a moderate salt concentration, the linker histone
H1 is removed (see Figure 12.12a). Higher salt concentration removes the rest of the
histone proteins (see Figure 12.18b). If the experiment of Figure 12.11 were carried out
after the DNA was treated with moderate or high salt, what would be the expected
results?
Answer: With a moderate salt concentration, the nucleosome structure is still preserved
so the same pattern of results would be observed. DNase I would cut the linker region
and produce fragments of DNA that would be in multiples of 200 bp. However, with a
high salt concentration, the core histones would be lost, and DNase I could cut anywhere.
On the gel, you would see fragments of almost any size. Because there would be a
continuum of fragments of many different sizes, the lane on the gel would probably look
like a smear rather than having a few prominent bands of DNA.
E5.
Let’s suppose you have isolated chromatin from some bizarre eukaryote with a
linker region that is usually 300 to 350 bp in length. The nucleosome structure is the
same as in other eukaryotes. If you digested this eukaryotic organism’s chromatin with a
high concentration of DNase I, what would be your expected results?
Answer: You would get DNA fragments of about 446 to 496 bp (i.e., 146 bp plus 300 to
350).
E6.
If you were given a sample of chromosomal DNA and asked to determine if it is
bacterial or eukaryotic, what experiment would you perform and what would be your
expected results?
Answer: Lots of possibilities. You could digest it with DNase I and see if it gives
multiples of 200 bp or so. You could try to purify proteins from the sample and see if
eukaryotic proteins or bacterial proteins are present.
E7.
Consider how histone proteins bind to DNA and then explain why a high salt
concentration can remove histones from DNA (as shown in Figure 12.18b).
Answer: Histones are positively charged and DNA is negatively charged. They bind to
each other by these ionic interactions. Salt is composed of positively charged ions and
negatively charged ions. For example, when dissolved in water, NaCl becomes individual
ions of Na+ and Cl–. When chromatin is exposed to a salt such as NaCl, the positively
charged Na+ ions could bind to the DNA and the negatively charged Cl– ions could bind
to the histones. This would prevent the histones and DNA from binding to each other.
E8.
In Chapter 21, the technique of fluorescence in situ hybridization (FISH) is
described. This is another method that can be used to examine sequence complexity
within a genome. In this method, a particular DNA sequence, such as a particular gene
sequence, can be detected within an intact chromosome by using a DNA probe that is
complementary to the sequence. For example, let’s consider the β-globin gene, which is
found on human chromosome 11. A probe that is complementary to the β-globin gene
will bind to the β-globin gene and show up as a brightly colored spot on human
chromosome 11. In this way, researchers can detect where the β-globin gene is located
within a set of chromosomes. Because the β-globin gene is unique, and because human
cells are diploid (i.e., have two copies of each chromosome), a FISH experiment would
show two bright spots per cell; the probe would bind to each copy of chromosome 11.
What would you expect to see if you used the following types of probes?
A.
A probe that is complementary to the AluI sequence
B.
A probe that is complementary to a tandemly repeated sequence near the
centromere of the X chromosome
Answer:
A.
Because the AluI sequence is interspersed throughout all of the
chromosomes, there would be many brightly colored spots along all chromosomes.
B.
Only the centromeric region of the X chromosome would be brightly
colored.