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mote1985 [20]
3 years ago
9

You dilute your bacteriophage for plating 10^-6. You mix 0.1ml of the bacteriophage dilution with 0.3ml of bacteria and plate wi

th top agar. The next day you count your plaques. You get 46 pfu at this dilution. What is the final concentration (pfu/ml) of the phage stock?
Biology
1 answer:
Blizzard [7]3 years ago
7 0

Answer:

46 * 10^7 pfu/ml

Explanation:

Given -

Bacteriophage dilution = 0.1 ml

Volume of bacteria and plate with top agar  = 0.3 ml

Number of plaques = 46 pfu

Dilution (d) = 10^{-6}

Final concentration of phage stock is equal to

= \frac{pfu}{d*v}

Where pfu is the count of plaques

d is the dilution and v is the volume

Substituting the given values, we get -

\frac{46}{10^{-6} * 10^{-1}} \\= 46 * 10^7 pfu/ml

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Answer:

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La cromatina es un complejo altamente organizado de ADN y proteínas y es un componente principal del núcleo celular. Las proteínas histonas ayudan a organizar el ADN en unidades estructurales llamadas nucleosomas, que luego se ensamblan en una estructura compacta (cromatina) y, finalmente, en estructuras muy grandes de orden superior (cromosomas).

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3 years ago
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Alenkinab [10]
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Use the complementary rule: If we had a DNA strand of 20 nucleotides, and 6 of them
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Answer:

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