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When cell has stalled DNA replication fork, which checkpoint should be predominantly activated?

Option: 1

G1/S

 

Option: 2

G2/M

 

 

 

Option: 3

M

 

 

Option: 4

Both G2 M and M

 

G2/M should be activated as the cell has stalled DNA replication fork.

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Posted by

Ajit Kumar Dubey

At 300 K and 1 atm, 15 mL of a gaseous hydrocarbon requires 375 mL air containing 20% O2 by volume for complete combustion.  After combustion, the gases occupy 330 mL.  Assuming that the water formed is in liquid form and the volumes were measured at the same temperature and pressure, the formula of the hydrocarbon is :
Option: 1  C4H8  
Option: 2  C4H10
Option: 3  C3H6
Option: 4  C3H8
 

Volume of N2  in air = 375 × 0.8 = 300 ml

Volume of O2 in air = 375 × 0.2 = 75 ml
 

C_{x}H_{y} +\left ( x +\frac{y}{4} \right )O_{2} \; \rightarrow \; xCO_{2}(g) + \frac{y}{2} H_{2}O(l)

15ml                15\left ( x +\frac{y}{4} \right )

  0                         0                            15x                 -

 

After combustion total volume

330 =V_{N_{2}} + V_{CO_{2}}

330 = 300 + 15x 

x = 2 

Volume of O2 used

15\left ( x +\frac{y}{4} \right ) = 75

\left ( x +\frac{y}{4} \right ) = 5

y = 12 

So hydrocarbon is = C2H12

None of the options matches it therefore it is a BONUS.

----------------------------------------------------------------------

Alternatively  Solution


 C_{x}H_{y} +\left ( x +\frac{y}{4} \right )O_{2} \; \rightarrow \; xCO_{2}(g) + \frac{y}{2} H_{2}O(l)

15ml              15\left ( x +\frac{y}{4} \right )

  0                         0                            15x                 -

Volume of O2 used

15\left ( x +\frac{y}{4} \right ) = 75

\left ( x +\frac{y}{4} \right ) = 5

If further information (i.e., 330 ml) is neglected, option (C3H8 ) only satisfy the above equation.

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Posted by

Ritika Jonwal

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Which of the following is an anionic detergent?
Option: 1 Sodium stearate
Option: 2 Sodium lauryl sulphate
Option: 3 Cetyltrimethyl ammonium bromide
Option: 4 Glyceryl oleate

As we have learned

Phenol formation from Benzenesulphonic acid -

Acidification of sodium salt gives phenol.

- wherein

C_{6}H_{5}SO_{3}Na+NaOH\xrightarrow[HCl]{fuse}C_{6}H_{5}-OH+Na_{2}SO_{3}

 

 

 Sodium lauryl sulphate has anionic charge 

 

 

 

 

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Posted by

Ritika Jonwal

The absolute configuration of is :
Option: 1 (2R, 3S)
Option: 2 (2S, 3R)
Option: 3 (2S, 3S)
Option: 4 (2R, 3R)
 

As we learnt ,

 

Chiral Carbon -

Those carbon on which four different groups are present.

- wherein

 

 

For C2 \rightarrow 

\Rightarrow This rotation suggests R but the least prior group is at horizontal position so the configuration is R.

For C3 \rightarrow 

\Rightarrow This rotation suggests S but the least prior group is at horizontal position so the configuration is S.

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Posted by

Ritika Jonwal

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The equilibrium constant at 298 K for a reaction A+BC+D is 100.  If the initial concentration of all the four species were 1 M each, then equilibrium concentration of D (in mol L−1) will be :
Option: 1  0.182  
Option: 2  0.818  
Option: 3  0.818  
Option: 4  1.182  
 

As we dicussed in the concept

Law of Chemical equilibrium -

At a given temperature, the product of concentration of the reaction products raised to the respective stoichiometric coefficient in the balanced chemical equation divided by the product of concentration of the reactants raised to their individual stoichiometric coefficients has a constant value.

- wherein

aA+bB\rightleftharpoons cC+dD


K_{c}=\frac{[C]^{c\:[D]^{d}}}{[A]^{a}\:[B]^{b}}

[A],\:[B],\:[C]\:[D]

are equilibrium concentration

 

 For reaction              A+B\rightleftharpoons C+D

Initial concentration IM      IM       IM       IM

At equilibarium if degree of dissociation is A+B\rightleftharpoons C+D\alpha then

A+B\rightleftharpoons C+D

1-\alpha   1-\alpha      H\alpha   1+\alpha

K_{c}=\frac{(1+\alpha)^{2}}{(1-\alpha)^{2}}=100

10=\frac{1+\alpha}{1-\alpha }

10-10\alpha =1+\alpha

9=11\alpha

\alpha =\frac{9}{11}

Concentration of D is 1+\alpha=I+\frac{9}{11}

=1.818

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Posted by

Ritika Jonwal

The heats of combustion of carbon and carbon monoxide are −393.5 and −283.5 kJ mol−1, respectively.  The heat of formation (in kJ) of carbon monoxide per mole is:
Option: 1 110.5
Option: 2 676.5
Option: 3 -676.5
Option: 4 -110
 

\mathrm{C_s+O_2_g \rightarrow CO_2_g \: \: \: \: \Delta H= -393.5 KJmol^{-1}}

CO_{\left ( g \right )}+\frac{1}{2}O_{2} \: _{\left ( g \right )}\rightarrow CO_{2} \:_{(g)} \: \: \: \: \: \Delta H=-283.5 kJmol^{-1}

C_{\left ( s \right )}+\frac{1}{2}O_{2} \:_{\left ( g \right )}\rightarrow CO_{\left ( g \right )}

\therefore \Delta H= -393.5+283.5

            = -110.0\ kJmol^{-1}

Therefore, Option(4) is correct

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Posted by

Ritika Jonwal

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The concentration of fluoride, lead, nitrate, and iron in a water sample from an underground lake was found to be 1 ppm, 40 ppm, 100 ppm and 0.2 ppm, respectively.  This water is unsuitable for drinking due to high concentration of :  
Option: 1 Fluoride
Option: 10  Lead
Option: 19 Nitrate
Option: 36 Iron

As we learned in concept

Hard Water -

It contains calcium and Magnesium salt in the form of hydrogen carbonate , chloride and sulphate

- wherein

Hard water does not give Lathers with soap.

 

Maximum permissible concentration of constituents

lead : 50 ppm

nitrate : 50 ppm

Iron : 0.2 ppm

fluoride: 1 ppm

 

Therefore, option (3) is correct.

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Posted by

Ritika Jonwal

Which one of the following statements about water is FALSE?
Option: 1 Water is oxidized to oxygen during photosynthesis..
Option: 2 Water can act both as an acid and as a base
Option: 3 There is extensive intramolecular hydrogen bonding in the condensed phase
Option: 4 Ice formed by heavy water sinks in normal water.

There is intermolecular hydrogen bounding in condensed phase of water and not intramolecular hydrogen bonding.

During Photosynthesis, Water is oxidised to Oxygen according to the reaction

6CO_2 + 6 H_2O \quad \longrightarrow \quad C_6H_{12}O_6 + 6O_2

Water can react with both acid as well as base

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Posted by

Ritika Jonwal

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The correct sequence of reagents for the following conversion will be :
Option: 1 CH_{3}MgBr, \; \left [ Ag\left ( NH_{3} \right )_{2} \right ]^{+}OH^{-} , H^{+}/CH_{3}OH  

Option: 3 \left [ Ag\left ( NH_{3} \right )_{2} \right ]^{+}OH^{-},   CH_{3}MgBr,    H^{+}/CH_{3}OH

Option: 5 \left [ Ag\left ( NH_{3} \right )_{2} \right ]^{+}OH^{-},  H^{+}/CH_{3}OH , CH_{3}MgBr,

Option: 7 CH_{3}MgBr,  H^{+}/CH_{3}OH ,  \left [ Ag\left ( NH_{3} \right )_{2} \right ]^{+}OH^{-},
 

?

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Posted by

vishal kumar

 Among the following compounds, the increasing order of their basic strength is
Option: 1  (I) < (II) < (IV) < (III)
Option: 2  (I) < (II) < (III) < (IV)
Option: 3  (II) < (I) < (IV) < (III)
Option: 4  (II) < (I) < (III) < (IV)
 

Among the following compounds, the increasing order of their basic strength is (II)<(I)<(IV)<(III)
(III) is most basic due to +I effect of −CH3 group attached to N. This increases electron density on N so that N can easily donate its lone pair of electrons to suitable acid.
(II) is least basic as the lone pair of electrons on N is part of pi electrons of aromatic ring. Donation of this lone pair of electrons will break the stability of the aromatic ring.
(I) is less basic than (IV) as the lone pair of electrons on N is in conjugation with aromatic ring.

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Posted by

vishal kumar

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