CHAPTER 13
Application and Experimental Questions
E1.
Answer the following questions that pertain to the experiment of Figure 13.3.
A.
What would be the expected results if the Meselson and Stahl experiment
were carried out for four or five generations?
B.
What would be the expected results of the Meselson and Stahl experiment
after three generations if the mechanism of DNA replication was dispersive?
C.
As shown in the data, explain why three different bands (i.e., light, halfheavy, and heavy) can be observed in the CsCl gradient.
Answer:
A.
Four generations: 7/8 light, 1/8 half-heavy; five generations: 15/16 light,
1/16 half-heavy
B.
All of the DNA double helices would be 1/8 heavy.
C.
The CsCl gradient separates molecules according to their densities. 14Ncontaining compounds have a lighter density compared to 15N-containing compounds.
The bases of DNA contain nitrogen. If the bases contain only 15N, the DNA will be
heavy; it will sediment at a higher density. If the bases contain only 14N, the DNA will
be light; it will sediment at a lower density. If the bases in one DNA strand contain 14N
and the bases in the opposite strand contain 15N, the DNA will be half-heavy; it will
sediment at an intermediate density.
E2.
An absentminded researcher follows the steps of Figure 13.3 and when the
gradient is viewed under UV light, the researcher does not see any bands at all. Which of
the following mistakes could account for this observation? Explain how.
A.
The researcher forgot to add 14N-containing compounds.
B.
C.
The researcher forgot to add lysozyme.
The researcher forgot to turn on the UV lamp.
Answer:
A.
You would probably still see a band of DNA, but you would only see a heavy
band.
B.
You would probably not see a band because the DNA would not be released from
the bacteria. The bacteria would sediment to the bottom of the tube.
C.
You would not see a band. UV light is needed to see the DNA, which absorbs
light in the UV region.
E3.
Figure 13.4b shows an autoradiograph of a replicating bacterial chromosome. If
you analyzed many replicating chromosomes, what types of information could you learn
about the mechanism of DNA replication?
Answer: You might be able to determine the number of replication forks and their
approximate locations. For chromosomes with a single origin, you could determine that
replication is bidirectional. However, you would not get any molecular information about
the DNA replication process.
E4.
The technique of dideoxy sequencing of DNA is described in Chapter 19. The
technique relies on the use of dideoxyribonucleotides (shown in Figures 19.16 and
19.17). A dideoxyribonucleotide has a hydrogen atom attached to the 3’ carbon atom
instead of an –OH group. When a dideoxyribonucleotide is incorporated into a newly
made strand, the strand cannot grow any longer. Explain why.
Answer: For a DNA strand to grow, a phosphoester bond is formed between the 3 –OH
group on one nucleotide and the innermost 5 phosphate group on the incoming
nucleotide (see Figure 13.15). If the –OH group is missing, a phosphoester bond cannot
form.
E5.
Another technique described in Chapter 19 is the polymerase chain reaction
(PCR) (see Figure 19.5). This method is based on our understanding of DNA replication.
In this method, a small of amount of double-stranded template DNA is mixed with a high
concentration of primers. Nucleotides and DNA polymerase are also added. The template
DNA strands are separated by heat treatment, and when the temperature is lowered, the
primers bind to the single-stranded DNA, and then DNA polymerase replicates the DNA.
This increases the amount of DNA that is made from the primers. This cycle of steps (i.e.,
heat treatment, lower temperature, allow DNA replication to occur) is repeated again and
again and again. Because the cycles are repeated many times, this method is called a
chain reaction. It is called a polymerase chain reaction because DNA polymerase is the
enzyme that is needed to increase the amount of DNA with each cycle. In a PCR
experiment, the template DNA is placed in a tube and the primers, nucleotides, and DNA
polymerase are added to the tube. The tube is then placed in a machine called a
thermocycler, which raises and lowers the temperature. During one cycle, the temperature
is raised (e.g., to 95°C) for a brief period and then lowered (e.g., to 60°C) to allow the
primers to bind. The sample is incubated at a slightly higher temperature for a few
minutes to allow DNA replication to proceed. In a typical PCR experiment, the tube may
be left in the thermocycler for 25 to 30 cycles. The total time for a PCR experiment is a
few hours.
A.
Why is helicase not needed in a PCR experiment?
B.
How is the sequence of each primer important in a PCR experiment? Do
the two primers recognize the same strand or opposite strands?
C.
The DNA polymerase used in PCR experiments is a DNA polymerase
isolated from thermophilic bacteria. Why is this kind of polymerase used?
D.
If a tube initially contained 10 copies of double-stranded DNA, how many
copies of double-stranded DNA (in the region flanked by the two primers) would be
obtained after 27 cycles?
Answer:
A.
Heat is used to separate the DNA strands, so you do not need helicase.
B.
Each primer must be a sequence that is complementary to one of the DNA
strands. There are two types of primers, and each type binds to one of the two
complementary strands.
C.
A thermophilic DNA polymerase is used because DNA polymerases
isolated from nonthermophilic species would be permanently inactivated during the
heating phase of the PCR cycle. Remember that DNA polymerase is a protein, and most
proteins are denatured by heating. However, proteins from thermophilic organisms have
evolved to withstand heat, which is how thermophilic organisms survive at high
temperatures.
With each cycle, the amount of DNA is doubled. Because there are initially 10 copies of
the DNA, there will be 227 copies after 27 cycles. 10227 = 1.34109 = 1.34
billion copies of DNA. As you can see, PCR can amplify the amount of DNA by a
staggering amount!