CHAPTER 11
Application and Experimental Questions
E1.
Genetic material acts as a blueprint for an organism’s traits. Explain how the
experiments of Griffith indicated that genetic material was being transferred to the type R
bacteria.
Answer: A trait of pneumococci is the ability to synthesize a capsule. There needs to be a
blueprint for this ability. The blueprint for capsule formation was being transferred from
the type S to the type R bacteria. (Note: At the molecular level, the blueprint is a group of
genes that encode enzymes that can synthesize a capsule.)
E2.
With regard to the experiment described in Figure 11.2, answer the following:
A.
List several possible reasons why only a small percentage of the type R
bacteria was converted to type S.
B.
Explain why an antibody must be used to remove the bacteria that are not
transformed. What would the results look like, in all five cases, if the
antibody/centrifugation step had not been included in the experimental procedure?
C. The DNA extract was treated with DNase, RNase, or protease. Why was this
done? (In other words, what were the researchers trying to demonstrate?)
Answer:
A.
There are different possible reasons why most of the cells were not transformed.
1.
Most of the cells did not take up any of the type S DNA.
2.
The type S DNA was usually degraded after it entered the type R bacteria.
3.
The type S DNA was usually not expressed in the type R bacteria.
B.
The antibody/centrifugation steps were used to remove the bacteria that
had not been transformed. It enabled the researchers to determine the phenotype of the
bacteria that had been transformed. If this step was omitted, there would have been so
many colonies on the plate it would have been difficult to identify any transformed
bacterial colonies, because they would have represented a very small proportion of the
total number of bacterial colonies.
C.
They were trying to demonstrate that it was really the DNA in their DNA
extract that was the genetic material. It was possible that the extract was not entirely pure
and could contain contaminating RNA or protein. However, treatment with RNase and
protease did not prevent transformation, indicating that RNA and protein were not the
genetic material. In contrast, treatment with DNase blocked transformation, confirming
that DNA is the genetic material.
E3.
An interesting trait that some bacteria exhibit is resistance to killing by
antibiotics. For example, certain strains of bacteria are resistant to tetracycline, whereas
other strains are sensitive to tetracycline. Describe an experiment that you would carry
out to demonstrate that tetracycline resistance is an inherited trait encoded by the DNA of
the resistant strain.
Answer:
1.
Isolate and purify DNA from resistant bacteria.
2.
In three separate tubes, add DNase, RNase, or protease.
3.
Add sensitive bacteria to each tube. A small percentage may be
transformed.
4.
Plate on petri plates containing tetracycline.
Expected results: Tetracycline-resistant colonies should only grow when the DNA
has been exposed to RNase and protease, but not to DNase.
E4.
With regard to the experiment of Figure 11.5, answer the following:
A.
Provide possible explanations why some of the DNA is in the supernatant.
B.
Plot the results if the radioactivity in the pellet, rather than in the
supernatant, had been measured.
C.
Why were 32P and 35S chosen as radioisotopes to label the phages?
D.
List possible reasons why less than 100% of the phage protein was
removed from the bacterial cells during the shearing process.
Answer:
A.
There are several possible explanations why about 35% of the DNA is in the
supernatant. One possibility is that not all of the DNA was injected into the bacterial
cells. Alternatively, some of the cells may have been broken during the shearing
procedure, thereby releasing the DNA.
B.
If the radioactivity in the pellet had been counted instead of the supernatant, the
following figure would be produced:
32P and 35S were chosen as radioisotopes to label the phages because
C.
phosphorous is found in nucleic acids, while sulfur is found only in proteins.
D.
There are multiple reasons why less than 100% of the phage protein is removed
from the bacterial cells during the shearing process. For example, perhaps the shearing
just is not strong enough to remove all of the phages, or perhaps the tail fibers remain
embedded in the bacterium and only the head region is sheared off.
E5.
Does the experiment of Figure 11.5 rule out the possibility that RNA is the
genetic material of T2 phage? Explain your answer. If it does not, could you modify the
approach of Hershey and Chase to show that it is DNA and not RNA that is the genetic
material of T2 bacteriophage? Note: It is possible to specifically label DNA or RNA by
providing bacteria with radiolabeled thymine or uracil, respectively.
Answer: It does not rule out the possibility that RNA is the genetic material, because
RNA and DNA both contain phosphorus. One way to distinguish RNA and DNA is to
provide bacteria with radiolabeled uracil in order to label RNA or provide bacteria with
radiolabeled thymine to label DNA. (Note: Uracil is found only in RNA and thymine is
found only in DNA.) If they had propagated T2 phage in E. coli cells exposed to
radiolabeled uracil, the phages would not be radiolabeled. However, if they had
propagated phage in E. coli cells exposed to radiolabeled thymine, the T2 phages would
be radiolabeled. This would indicate that T2 phages contain DNA and not RNA, because
radiolabeled uracil would not label the genetic material of T2 bacteriophage.
E6.
In this chapter, we considered two experiments – one by Avery, MacLeod, and
McCarty and the second by Hershey and Chase – that indicated that DNA is the genetic
material. Discuss the strengths and weaknesses of the two approaches. Which
experimental approach did you find the most convincing? Why?
Answer: This is really a matter of opinion. The Avery, MacLeod, and McCarty
experiment seems to indicate directly that DNA is the genetic material, because DNase
prevented transformation and RNase and protease did not. However, one could argue that
the DNA is required for the rough bacteria to take up some other contaminant in the DNA
preparation. It would seem that the other contaminant would not be RNA or protein. The
Hershey and Chase experiments indicate that DNA is being injected into bacteria,
although quantitatively the results are not entirely convincing. Some 35S-labeled protein
was not sheared off, so the results do not definitely rule out the possibility that protein
could be the genetic material. But the results do indicate that DNA is the more likely
candidate.
E7.
The type of model building that was used by Pauling and Watson and Crick
involved the use of ball-and-stick units. Now we can do model building on a computer
screen. Even though you may not be familiar with this approach, discuss potential
advantages that computers might provide in molecular model building.
Answer:
1.
You can make lots of different shapes.
2.
You can move things around very quickly with a mouse.
3.
You can use mathematical formulas to fit things together in a systematic
way.
4.
Computers are very fast.
5.
You can store the information you have obtained from model building in a
computer file.
E8.
With regard to Chargaff’s data described in Table 11.1, answer the following:
A.
What is the purpose of paper chromatography?
B.
Explain why it is necessary to remove the bases in order to determine the
base composition of DNA.
C.
Would Chargaff’s experiments have been convincing if they had been
done on only one species? Discuss.
Answer:
A.
The purpose of chromatography was to separate the different types of bases.
B.
It was necessary to separate the bases and determine the total amount of
each type of base. In a DNA strand, all of the bases are found within a single molecule,
so it is difficult to measure the total amount of each type of base. When the bases are
removed from the strand, each type can be purified, and then the total amount of each
type of base can be measured by spectroscopy.
C.
Chargaff’s results would probably not be very convincing if done on a
single species. The strength of his data was that all species appeared to conform to the
AT/GC rule, suggesting that this is a consistent feature of DNA structure. In a single
species, the observation that A = T and G = C could occur as a matter of chance.
E9.
Gierer and Schramm exposed plant tissue to purified RNA from tobacco mosaic
virus and the plants developed the same types of lesions as if they were exposed to the
virus itself. What would be the results if the RNA was treated with DNase, RNase, or
protease prior to its exposure to the plant tissue?
Answer: If the RNA was treated with RNase, the plants would not be expected to develop
lesions. If treated with DNase or protease, lesions would still occur because RNA is the
genetic material, and DNase and protease do not destroy RNA.