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Georgia [21]
2 years ago
14

When NEM is added to a purified solution of creatine kinase, Cys 278 is alkylated, but no other Cys residues in the protein are

modified. What can you infer about the Cys 278 residue based on this observation
Chemistry
1 answer:
USPshnik [31]2 years ago
5 0

Answer:

The answer is given below

Explanation:

Cys 278 residue is the only available cysteine which is alkylated  by the addition of N-Ethylmaleimide or NEM (alkylating agent). It works by only alkylating the sulfhydryls. In this case, Cys 278 residue is the only one which has exposed cysteine residue.

While the other residues have their sulfhydryls group either involved in the synthesis of disulfide bonds of proteins or their Cys residues are intrinsically placed in the proteins and cannot be alkylated with NEM.

NEM cannot alkylate if its protein is not available in the free form or it is in bounded form. For NEM to alkylate Cys 278, it should be free and should have sulfhydryls available for alkylation.

Alkylation: it is the transfer of alkyl groups. Alkyl groups contain Hydrogen and Carbon in their structure.

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A cup of gold colored metal beads was measured to have a mass 425 grams. By water displacement, the volume of the beads was calc
Butoxors [25]
If the volume of 425 grams was 48.0 cm³, simply divide

g/cm³ = 425 g/48 cm³ = 8.85 g/cm³

If using water in water displacement, 1 mL = 1 cm³

8.85 g/cm³ = 8.85 g/mL

This density is most closely aligned with that of B. Copper

Hope I helped!
4 0
2 years ago
What volume of beaker contains exactly 2.23x10^-2 mol of nitrogen gas at STP?
nikklg [1K]

Answer:

V = 0.5 L

Explanation:

Given data:

Moles of nitrogen = 2.23×10⁻² mol (0.0223 mol)

Temperature = 273 K

Pressure = 1 atm

Volume = ?

Solution:

PV = nRT

V = nRT / P

V = 0.0223 mol × 0.0821 atm. mol⁻¹. L . k⁻¹ × 273 K / 1 atm

V = 0.5 L

4 0
2 years ago
A 150 W electric heater operates for 12.0 min to heat an ideal gas in a cylinder. During this time, the gas expands from 3.00 L
Brut [27]

<u>Answer:</u> The change in internal energy of the gas is 108.835 kJ

<u>Explanation:</u>

To calculate the work done for reversible expansion process, we use the equation:

W=P\Delta V=-P(V_2-V_1)

where,

W = work done

P = pressure = 1.03 atm

V_1 = initial volume = 3.00 L

V_2 = final volume = 11.0 L

Putting values in above equation, we get:

W=-(1.03)\times (11.0-3.00)=8.24L.atm=834.9J=0.835kJ     (Conversion factor:  1 L. atm = 101.325 J)

Calculating the heat from power:

Q=P\times t

where,

Q = heat required

P = power = 150 W

t =  time = 12 min = 720 s       (Conversion factor:  1 min = 60 s)

Putting values in above equation:

Q=150\times 720=108000J=108kJ

The equation for first law of thermodynamics follows:

Q=dU+W

where,

Q = total amount of heat required = 108 kJ

dU = Change in internal energy = ?

W = work done  = -0.835 kJ

Putting values in above equation, we get:

108kJ=dU+(-0.835)\\\\dU=(108+0.835)=108.835kJ

Hence, the change in internal energy of the gas is 108.835 kJ

7 0
2 years ago
A 7.591-9 gaseous mixture contains methane (CH4) and butane
mestny [16]

Answer:

65.71%

Explanation:

First, we can write the mass of the mixture, thus:

7.519g = X + Y <em>(1)</em>

<em>Where X is the mass of methane and Y the mass of butane</em>

<em />

Also, the reactions of combustion are:

CH₄ + 2O₂ → CO₂ + 2H₂O

<em>2 moles of oxygen react per mole of methane</em>

C₄H₁₀ + 13/2 O₂ → 4CO₂ + 5H₂O

<em>13/2 moles of oxygen react per mole of methane</em>

<em />

That means, in therms of moles of oxygen we can write:

0.9050 moles = 2X/16.04 + 13/2Y/ 58.12

0.9050 = 0.12469X + 0.11184Y <em>(2)</em>

<em>Where 16.04 and 58.12 are molar masses of methane and butane</em>

That is because if X is the mass of methane:

X g Methane * (1mol / 16.04g) = Moles methane

Moles methane * (2 moles Oxygen / mole methane) = Moles oxygen

Replacing (1) in (2):

0.9050 = 0.12469X + 0.11184 (7.519 - X)

0.9050 = 0.12469X + 0.841 - 0.11184X

0.0641 = 0.01285X

X = 4.988g = Mass of methane.

And mass percent of methane is:

4.988g / 7.591g * 100

<h3>65.71%</h3>

7 0
2 years ago
(ii) When shale gas is burned, the hydrogen sulfide reacts with oxygen.
Elenna [48]

Answer:

See explanation

Explanation:

Now , we have the equation of the reaction as;

2H2S(g) + 302(g)------->2SO2(g) + 2H2O(g)

This equation shows that SO2 gas is produced in the process. Let us recall that this same SO2 gas is the anhydride of H2SO4. This means that it can dissolve in water to form H2SO4

So, when SO2 dissolve in rain droplets, then H2SO4 is formed thereby lowering the pH of rain water. This is acid rain.

4 0
2 years ago
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