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Griffith was trying to develop a vaccine for S. pneumonia when he was surprised to discover the phenomenon of bacterial transformation. Look at the second and third panel of Figure 16.2. Based on these results, what result was he expecting in the fourth panel? Explain.

Short Answer

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Griffith expected that the mouse would be alive when it was injected with a mixture of heat-killed S cells and living R cells. This is because the mouse was alive when the heat-killed S cells and living R cells were injected solely into the mouse.

Step by step solution

01

Bacterial transformation

Whena bacteria cell takes up a naked DNA (Deoxyribonucleic acid) molecule or fragment from the medium, it gets incorporated into the recipient chromosome. It becomes a part of the host chromosome. This process is called bacterial transformation.

Under natural conditions, when bacteria cell lyses, it releases its genetic material in the surrounding. When these DNA fragments come in contact with another bacterial cell, it results in bacteria transformation.

02

Griffith’s experiment with S. pneumonia

Griffith selected two strains ofS. pneumonia: the R type and the S type.The R strain was non-pathogenic, while the S strain was pathogenic.The presence of the outer capsule on the S type S. pneumoniaaccounts for the pathogenicity of this strain,which was absent in the R strain.

He injected the R and S strain into the mouse and studied the effect of these strains on the mouse.When he injected the mouse with living S cells in the first panel, he observed that the mouse died.

In the second panel, he injected the mouse with living R cells, and he observed that the mouse was alive. In the third panel, he injected the mouse with heat-killed S cells, and he observed that the mouse was alive.

In the fourth panel, he injected the mouse with a mixture of heat-killed S cell and living R cells; he observed that the mouse was dead.

03

Expected result versus observed result

When Griffith injected the mouse with the mixture of living R cells and heat-killed S cells, he was expecting that the mouse would be alive because when he injected heat-killed S cells alone in the mouse, the mouse was alive. When he injected with the living R cell alone, then also the mouse survived.

However, he was surprised to see the result because the mouse died in the fourth panel. To know the cause of the death, Griffith obtained the blood sample from the dead mouse. He found living R cells were present in the blood of the mouse.

He concluded that some heritable factors transformed from the dead S cells into the living R cells, making them pathogenic as they developed outer capsule-like S cells. Thus, the R cells became pathogenic due to the uptake and incorporation of the S cell DNA fragment into the R cell genome.

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Most popular questions from this chapter

Given a polynucleotide sequence such as GAATTC. Explain what further information you would need in order to identify which is the 5’ end. (See Figure 16.5)

Model building can be an important part of the scientific process. The illustration shown above is a computer-generated model of a DNA replication complex. The parental and newly synthesized DNA strands are colour coded differently, as are each of the following three proteins DNA pol III, the sliding clamp, and single-stranded binding protein.

  1. Using what you've learned in this chapter to clarify this model, label each DNA strand and protein.
  2. Draw an arrow to indicate the overall direction of DNA replication.

In his work with pneumonia-causing bacteria and mice, Griffith found that

  1. the protein coat from pathogenic cells was able to transform non-pathogenic cells.
  2. heat-killed pathogenic cells caused pneumonia.
  3. some substances from pathogenic cells were transferred to non-pathogenic cells, making them pathogenic.
  4. the polysaccharide coat of bacteria caused pneumonia.

The continuity of life is based on heritable information in the form of DNA, and structure and function are correlated at all levels of biological organization. In a short essay (100-150words), describe how the structure of DNA is correlated with its role as the molecular basis of inheritance.

Some bacteria may be able to respond to environmental stress by increasing the rate at which mutations occur during cell division. How might this be accomplished? Might there be an evolutionary advantage to this ability? Explain.

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