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marin [14]
3 years ago
15

Mitotic cyclins control the progression of S phase and mitosis. G1 phase and G2 phase. S phase and G2 phase. G2 phase and mitosi

s​
Chemistry
2 answers:
kotegsom [21]3 years ago
7 0

Answer:

The cell cycle involves different stages such as interphase, the Mitotic phase followed by cytokinesis.

The interphase involves G0 phase, G1 phase, Synthesis or S phase and G2 phase. The G stands for Gap phase.

The interpase involves the preparation of cellular contents, duplication of DNA. Then during mitosis karyokinesis occurs which is the division of duplicated chromosomes. And then through cytokinesis, the cytoplasmic contents also divides yielding two daughter cells.

Most of the cells normally enter into resting phase I.e., G0 phase and then start the process of cell division again. But some cells enter into G0 phase after the division and remain for longer duration. And this stage is called as quiescent or inactive stage. The mature heart muscle cell and nerve cell are the typical examples for this.

There are also some checkpoints in the cell cycle to check for any damage in the DNA from one phase to another phase and is mediated by Cyclins.

Apart from cyclins, the cyclin-dependent kinase also control the cell division.

These kinases are the enzymatic proteins which were activated by the binding of cyclins to it.

As the levels of different cyclins were different at each phase the amount of kinase activity is also regulated.

The kinases phosphorylates the target proteins which inturn helps in cell cycle such as break down of nuclear membrane during mitotic phase and promoting DNA replication in interphase.

These kinases were also down regulated by phosphorylating the sites where cyclin binds.

Therefore, cyclin dependent kinases can control cell cycle activities.

So, the correct answer is 'Cell cycle activities'

Umnica [9.8K]3 years ago
7 0

Answer:

The orderly series of events that lead to duplication of cell content followed by division of a cell into daughter cells is termed as the cell cycle. It consists of two phases, namely interphase and M phase. The M phase of cell cycle is marked by the division of nucleus and cytokinesis to produce daughter cells. Interphase is the period between two subsequent M phases and consists of three phases:

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Determine the molality of ions in a solution formed by dissolving 0.187 moles of NaCl in 456 grams of water. The density of the
PSYCHO15rus [73]

Answer:

Molality Ions Na⁺ = 0.410 m

Molality Ions Cl⁻ = 0.410 m

Total ion molality = 0.820 m

Explanation:

Molality (m) is defined as moles of solute that are dissolved in 1000g (1kg) of solvent. In this case, the molality of the ions is requested, so it should be borne in mind that when the NaCl salt dissolves in the water, it dissociates forming ions as follows:

NaCl ⇒Na⁺ (aq) + Cl⁻ (aq)

Therefore, <em>for each mole of salt that dissolves, you will have 1 mole of Na⁺ and 1 mole of Cl⁻ </em>

Knowing the amount of initial solute moles and the mass of water, we proceed to perform the calculations for each ion:

<em>Ions Na⁺</em>

456 g of H₂O ____ 0.187 mol Na⁺

1000 g of H₂O ____ X = 0.410 m

Calculation: 1000 g x 0.187 mol / 456 g = 0.410 mol

<em>Ions Cl⁻</em>

456 g of H₂O ____ 0.187 mol Cl⁻

1000 g of H₂O ____ X = 0.410 m

Calculation: 1000 g x 0.187 mol / 456 g = 0.410 m

If both results are added, the total molality of the ions will be obeyed:

0.410 m + 0.410 m = 0.820 m

This means that <em>for every kilogram of water, 0.820 moles of dissociated ions from NaCl are dissolved.</em>

8 0
3 years ago
What mass of oxygen (O2) forms in a reaction that forms 15.90 g C6H12O6? (Molar mass of O2 = 32.00 g/mol; molar mass of C6H12O6
Dafna11 [192]

The answer is: the mass of oxygen is 16.95 grams.

The overall balanced photosynthesis reaction:  

6CO₂ + 6H₂O → C₆H₁₂O₆ + 6O₂.  

m(C₆H₁₂O₆) = 15.90 g; mass of glucose.

n(C₆H₁₂O₆) = m(C₆H₁₂O₆) ÷ M(C₆H₁₂O₆).

n(C₆H₁₂O₆) = 15.9 g ÷ 180.18 g/mol.

n(C₆H₁₂O₆) = 0.088 mol; amount of glucose.

From chemical reaction: n(C₆H₁₂O₆) : n(O₂) = 1 : 6.

n(O₂) = 6 · 0.088 mol.

n(O₂) = 0.53 mol; amount of oxygen.

m(O₂) = 0.53 mol · 32.00 g/mol.

m(O₂) = 16.95 g; mass of oxygen.

5 0
3 years ago
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Reaction 1: C3H8(g)+5O2(g)→3CO2(g)+4H2O(g), ΔH1=−2043 kJ
Leokris [45]
Given teh equation adn the heat of reaction, reaction 2's heat of reaction can be obtained by simply multiplying teh heat of reaction of 1 by 3. The final answer is -6129 kJ. 
6 0
3 years ago
An unsaturated solution ________.
Greeley [361]

Answer:

I think the answer is D.

Explanation:

Because if it is unsaturated then it can dissolve more solutes.

8 0
3 years ago
Wet steam at 1100 kPa expands at constant enthalpy (as in a throttling process) to 101.33 kPa, where its temperature is 105°C. W
nikklg [1K]

Answer:

There are 3 steps of this problem.

Explanation:

Step 1.

Wet steam at 1100 kPa expands at constant enthalpy to 101.33 kPa, where its temperature is 105°C.

Step 2.

Enthalpy of saturated liquid Haq = 781.124 J/g

Enthalpy of saturated vapour Hvap = 2779.7 J/g

Enthalpy of steam at 101.33 kPa and 105°C is H2= 2686.1 J/g

Step 3.

In constant enthalpy process, H1=H2 which means inlet enthalpy is equal to outlet enthalpy

So, H1=H2

     H2= (1-x)Haq+XHvap.........1

    Putting the values in 1

    2686.1(J/g) = {(1-x)x 781.124(J/g)} + {X x 2779.7 (J/g)}

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1904.976 (J/g) = x1998.576 (J/g)

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So, the quality of the wet steam is 0.953

                   

7 0
3 years ago
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