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fiasKO [112]
2 years ago
11

What is the advantage to having two different protein-digesting enzymes, rather than just one?

Biology
1 answer:
Alex Ar [27]2 years ago
5 0
Increased efficiency
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During ________ the chromosomes, which were distinct and condensed, begin to spread out into a tangle of chromatin. a nuclear en
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During telophase in mitosis or meiosis II, where chromosomes which were previously distinct and condensed, de-condense and spread out into a tangle of chromatin. At the same time, the nuclear envelope re-forms around each cluster of chromosomes in each of the daughter cells, and spindle fibres disintegrate.

Chromosomes are unlikely to de-condense and spread out into a tangle of chromatin during telophase I (i.e. Telophase of Meiosis I), since Meiosis is still ongoing, and the genetic material should be in its condensed form during Meiosis II as well. 

Hope this helps! :)
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What do you call the place where two branches split apart
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It’s called a branch point, represents where a single lineage evolved into a distinct new one.
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Which of the following would you least likely find living in the chaparral biome?
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Calculate the final concentration of BSA in the problems below using the formula
Rudiy27

Answer:

A. C_2=1.5\frac{mg}{mL}

B. C_2=0.075\frac{mg}{mL}

C. C_2=0.01\frac{mg}{mL}

D. C_2=0.001\frac{mg}{mL}

Explanation:

Hello.

In this case, we must compute the final concentration in all the cases so we solve for it in the given equation:

C_2=\frac{C_1V_1}{V_2}

Thus, we proceed as follows:

A. Here, the final diluted solution includes the 300 μL of the 5 mg/ml-BSA and the 700 μL of TBS.

C_2=\frac{300\mu L*5\frac{mg}{mL} }{(300+700)\mu L}\\\\C_2=1.5\frac{mg}{mL}

B. Here, the final diluted solution includes the 50 μL of the 1.5 mg/ml-BSA, the 450 μL of water and the 500 μL of TBS.

C_2=\frac{50\mu L*1.5\frac{mg}{mL} }{(50+450+500)\mu L}\\\\C_2=0.075\frac{mg}{mL}

C. Here, the final diluted solution includes the 10 μL of the 1 mg/ml-BSA and the 990 μL of TBS.

C_2=\frac{10\mu L*1\frac{mg}{mL} }{(10+990)\mu L}\\\\C_2=0.01\frac{mg}{mL}

D. Here, the final diluted solution includes the 10 μL of the 0.1 mg/ml-BSA and the 990 μL of TBS.

C_2=\frac{10\mu L*0.1\frac{mg}{mL} }{(10+990)\mu L}\\\\C_2=0.001\frac{mg}{mL}

Best regards.

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Believe it or not, the tundra is like the desert in one very important way. The similarity results in plants with a like adaptat
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Answer:

Low rainfall and available water; waxy coating on leaves to prevent drying.

Explanation:

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