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3 years ago

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Give at least three examples of types of mutations that would disrupt the process of mitotic chromosome segregation. That is, ex

plain in what DNA structures or in genes encoding what kinds of proteins you would find these segregation-disrupting mutations.

1 answer:

Nutka1998 [239]3 years ago

5 0

Answer: Insertions and Deletions. ...

Large-scale mutations. ...

Sickle Cell Disease and Malaria

Explanation:

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The 3 can only be produced in the Mediterranean biome. This biome is known to have dry and wet weather. However, the summer seasons in this biome is very dry and usually lasts longer. Hence, the vegetation is found to be drought resistant. The 3 plants mentioned are known to be naturally drought resistant, hence it is in this biome that it thrives and survives best.

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In open ocean waters, it can be subdivided even further. What are those divisions?

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In the traditional phylogeny (A), the phylum Platyhelminthes is depicted as a sister taxon to the rest of the protostome phyla,

Vadim26 [7]

Answer:

In the traditional phylogeny platyhelminthes is depicted as a lophotrochozoan phylum. What probably led to this change?

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Explanation:

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3 years ago

g If a person takes a prescribed dose of 10 milligrams of Valium, the amount of Valium in that person's bloodstream at any time

Elenna [48]

Answer:

a. 8.1 milligrams

b. 40.07 hours

c. 8.859 milligrams

Explanation:

If a person takes a prescribed dose of 10 milligrams of Valium, the amount of Valium in that person's bloodstream at any time can be modeled by

$A_{t}=10e^{-0.0173t}$

Where A(t) = amount of Valium remaining in the blood after t hours

t = time or duration after the drug is taken

a. we have to calculate the amount of drug remaining in the bloodstream after 12 hours

$A_{12}=10e^{-0.0173\times12}$

$A_{12}=10e^{-0.2076}$

= 10×0.81253

= 8.1 milligrams

b. In this part we have to calculate the time when A(t) = 5 milligrams

$5=10e^{-0.0173\timest}$

$\frac{5}{10}=e^{-0.0173t}$

0.5 = $e^{-0.0173t}$

Now we take natural log on both the sides of the equation.

ln(0.5) = ln( $e^{-0.0173t})$

-0.69314 = -0.0173t

t = $\frac{0.69314}{0.0173}$

t = 40.0658

≈ 40.07 hours

c. In this part we have to calculate the rate, by which amount of drug will decay in the bloodstream after 7 hours.

$A_{7}=10e^{-0.0173\times7}$

$A_{7}=10e^{-0.1211}$

= 10×0.8859

= 8.859 milligrams

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3 years ago