Reference no: EM131325435
Part 1.
Suppose you are given a new temperature sensitive bacterial mutant that grows normally at 37 degrees C but cannot replicate its chromosomes properly at 42 C. To investigate the nature of the underlying defect, you incubate the cells at 42 C with radioactive substrates required for DNA synthesis. After one hour you find that the cell population has doubled its DNA content, suggesting that DNA replication can still occur at 42 C. Moreover, centrifugation reveals that all of this DNA has the same large molecular weight as does the original DNA present in the cells. When the DNA is denatured, however, you discover that 75% of the resulting single stranded DNA has a molecular weight that is half that of the original double-stranded DNA, and the remaining 25% of the DNA has a much lower molecular weight.
Based on the preceding results, what gene do you think is defective in these cells?
A. The cells have a defective gene for DNA polymerase III, causing the enzyme to lose its function at 42°C. DNA polymerase III is required during DNA replication for removing RNA primers and filling the resulting gaps that are created as part of lagging strand synthesis.
B. The cells have a defective gene for DNA polymerase II, causing the enzyme to lose its function at 42°C. DNA polymerase II is required during DNA replication for removing RNA primers and filling the resulting gaps that are created as part of lagging strand synthesis.
C. The cells have a defective gene for DNA gyrase, causing the enzyme to lose its function at 42°C. DNA gyrase is required during DNA replication for creating swivel points in the DNA molecule that prevents overwinding (positive supercoiling) of the DNA ahead of the replication fork.
D. The cells have a defective gene for DNA ligase, causing the enzyme to lose its function at 42°C. DNA ligase is required during DNA replication for joining together the short Okazaki fragments that are created as part of lagging strand synthesis.
Part 2
How such a defect would account for the experimental results that you observed?
Select the three correct statements.
A. The DNA denaturation step in the experiment released the two full-length strands of the original DNA, the new full-length leading strand, and the unjoined Okazaki fragments of the lagging strand.
B. The full-length strands together account for 75% of the DNA.
C. The full-length strands together account for 50% of the DNA.
D. 25% of the DNA have a low molecular weight because it represents the unjoined Okazaki fragments of the lagging strand.
E. The DNA denaturation step in the experiment released the one full-length strands of the original DNA and the unjoined Okazaki fragments of the lagging strand.
F. 50% of the DNA have a low molecular weight because it represents the unjoined Okazaki fragments of the lagging strand.
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