94. You have two tubes containing bacteriophage labelled with radioactive phosphorous (tube A) and
radioactive Sulphur (tube B) that are devoid of bacteria. You use these bacteriophage to infect separate E.
coli cultures. After infection you separate bacteria from the virus and check them for radioactivity. You will
find:
(1) Radioactivity in both bacterial samples.
(2) Radioactivity in none of them as bacteria have been totally separated from the iruses.
(3) Radioactivity in bacteria infected with viruses from tube A.
(4) Radioactivity in bacteria infected with viruses from tube B.
Introduction:
In the 1950s, Martha Chase and Alfred Hershey conducted an experiment that would change our understanding of genetics and molecular biology. The Hershey-Chase experiment used bacteriophages (viruses that infect bacteria) to determine whether DNA or protein carries genetic information. By labeling the DNA with radioactive phosphorus and the protein with radioactive sulfur, they were able to trace which component entered bacterial cells during infection. Let’s break down the experiment and its results in detail.
The Hershey-Chase Experiment: Key Concepts
In the Hershey-Chase experiment, bacteriophages were used to infect E. coli bacteria. The experiment involved labeling two different components of the bacteriophage with radioactive isotopes:
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Radioactive phosphorus (P-32) was used to label the DNA of the bacteriophage, as DNA contains phosphorus in its phosphate backbone.
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Radioactive sulfur (S-35) was used to label the protein coat of the bacteriophage, as proteins contain sulfur in the amino acid methionine and cysteine but do not contain phosphorus.
The idea was to track which component of the virus (either DNA or protein) was injected into the bacteria during infection, thereby determining which one carries the genetic information required for viral replication.
The Procedure and Results:
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Tube A (Radioactive Phosphorus): Bacteriophages with labeled phosphorus were used to infect E. coli.
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Tube B (Radioactive Sulfur): Bacteriophages with labeled sulfur were used to infect a separate E. coli culture.
After infection, the virus particles were removed, and the bacterial cells were separated from the virus. The next step was to check the bacterial samples for radioactivity.
Results Interpretation:
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Radioactive phosphorus (P-32) is incorporated into the DNA of the bacteriophage, so when these bacteriophages infect E. coli, the radioactive DNA enters the bacterial cell.
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Radioactive sulfur (S-35) is incorporated into the protein coat of the bacteriophage, which remains outside the bacterial cell during infection. Only the DNA enters the bacterial cell and carries the genetic information.
After the infection, bacteria from the Tube A (labeled with radioactive phosphorus) showed radioactivity, meaning the DNA from the virus was injected into the bacteria. However, bacteria from Tube B (labeled with radioactive sulfur) did not show any radioactivity, because the sulfur-labeled protein coat did not enter the bacterial cell.
Conclusion:
The experiment concluded that DNA is the genetic material responsible for viral replication because only the radioactive DNA entered the bacterial cells and was detected, while the radioactive protein did not. This finding was crucial in establishing DNA as the molecule that carries genetic information in living organisms.
The Correct Answer:
Given the experiment’s design and results, the correct answer is:
(3) Radioactivity in bacteria infected with viruses from tube A.
This result indicates that the phosphorus-labeled DNA entered the bacterial cells, carrying the genetic information for the bacteriophage to reproduce.
