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2 years ago

After which checkpoint is the cell first committed to continue the cell cycle through M?

Biology

1 answer:

Viktor [21]2 years ago

8 0

The correct answer is: B) G1

Normal cells (unlike cancer cells), move through the cell cycle in a regulated way in order to make sure that cells don't divide under conditions that are unfavorable for them. Information about their own internal state (nutrients, signal molecules, DNA integrity) is signal to go or not to go through the cell division. Because of that there are few checkpoints in the cell cycle at which the cell examines the signals and makes a “decision”. The major checkpoints are:

• The G1- the first point at which it must choose, once it passes the G1 checkpoint the cell enters S phase

• The G2-the cell checks DNA integrity and checks if replication is done well.

• The spindle checkpoint-at the transition from metaphase to anaphase.

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Read the description of evolution by natural selection in this section and describe the role that the environment plays in the t

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2 years ago

Iron-55 has a half-life of 3 years. How much of a 40 gram sample will remain after 12 years? A. 0 grams B. 4 grams C. 2.5 grams

Readme [11.4K]

Answer:

$\boxed {\boxed {\sf C. \ 2.5 \ grams }}$

Explanation:

We are asked to find how much of a 40 gram sample remains after 12 years.

Iron-55 has a half-life of 3 years. Therefore, after 12 years, 4 half-lives have been completed.

12 years/3 years = 4 half-lives

Every time a half-life is completed, half of the sample's mass decays. Remember we start with a 40 gram sample.

1 half- life: 40 g / 2 = 20 g
2 half-lives: 20 g / 2= 10 g
3 half-lives: 10 g / 2 = 5 g
4 half-lives: 5 g / 2 = 2.5 g

There is also a formula that can be used to solve this problem.

$A= A_o(\frac {1}{2})^{\frac{t}{hl}$

Where A₀ is the initial amount, t is the time, and hl is the half-life.

We know 40 grams is the inital amount, 12 years is the time, and 3 years is the halflife.

A₀= 40 g
t= 12
hl= 3

$A= 40 \ g (\frac{1}{2})^\frac{12}{3}$

$A= 40 \ g (\frac{1}{2})^4$

$A= 40 \ g * 0.0625$

$\bold {A= 2.5 \ g}$

After 12 years, <u>2.5 grams </u> of Iron-55 will remain.

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