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NNADVOKAT [17]
3 years ago
7

What would happen if a cell replicated once in the presence of this radioactive base?Suppose one were provided with an actively

dividing culture of E. coli bacteria to which radioactive thymine had been added.A. One of the daughter cells, but not the other, would have radioactive DNA.B. Neither of the two daughter cells would be radioactive.C. All four bases of the DNA would be radioactive.D. Radioactive thymine would pair with nonradioactive guanine.E. DNA in both daughter cells would be radioactive.
Biology
1 answer:
Ket [755]3 years ago
8 0

Answer:

The correct answer is option E. "DNA in both daughter cells would be radioactive".

Explanation:

Radioactive thymidine incorporation has irremediable effects in thymine and cytosine in bacteria such as E. coli. Although there are some reports of recovery of E. coli from UV-irradiation DNA damage, the presence of radioactive bases such as thymine is often irremediable. Therefore, after a dividing culture of E. coli exposed to radioactive thymine divides, DNA in both daughter cells would be radioactive.

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2 years ago
Yeast cells are recovered from a fermentation broth by using a tubular centrifuge. Sixty percent of the cells are recovered at a
Serga [27]

Answer:

\\ \omega = 5033.22 \ rpm

Explanation:

Given that:

R_1 = 0.6\\\\R_2 = 0.95\\\\\omega_1 = 4000 \ rpm\\\\\omega_2 = ???\\\\Q_1 = 12\ l/min\\\\Q_2 =  12\ l/min

For a continuous centrifuge ; solid recovery R ∝ \frac{\omega^2}{Q}

where;

ω = angular velocity of the centrifuge in rpm

Q = flow rate

SO;

R_1 \alpha \frac{4000^2}{12}

R_2 \alpha  \frac{\omega^2}{12}

\frac{R_1}{R_2} = \frac{4000^2}{12}  ÷  \frac{\omega^2 }{12}

\frac{R_1}{R_2} = \frac{4000^2}{12} * \frac{12}{\omega^2}

\frac{R_1}{R_2} = \frac{4000^2}{\omega^2}\\\\\frac{0.6}{0.95} = \frac{4000^2}{\omega^2}

{\omega^2} = \frac{4000^2*0.95}{0.6}\\\\\\{\omega} = \sqrt{ \frac{4000^2*0.95}{0.6}}\\\\\\ \omega = 5033.22 \ rpm

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