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Masteriza [31]
3 years ago
6

A purified protein has a molecular mass of 360 kDa when measured by size exclusion chromatography. When analyzed by gel electrop

horesis in the presence of SDS, three bands are observed, with molecular masses of 160, 140, and 60 kDa. When gel electrophoresis is carried out in the presence of SDS and dithiothreitol, three bands are once again observed, with molecular masses of 140, 80, and 60 kDa. What is the subunit composition of the protein?
Chemistry
1 answer:
Dimas [21]3 years ago
4 0

Answer:

A protein has four subunits whose molecular masses are 140, 80, and 60 kDa.

A disulfide bond links the two 80 kDa subunits (possibly identical).

Explanation:

Given that:

A protein has four subunits whose molecular masses are 140, 80, and 60 kDa.

A disulfide bond links the two 80 kDa subunits (possibly identical).

As a result of SDS and dithiothreitol analysis treatment, the molecular masses can not be 360 in total. They are 280, which implies that they are in short of 80 kDa. This means that there are possibilities that two groups with a molecular mass of 80 kDa which are joined by a disulfide bond.

The presence of SDS and dithiothreitol acts as a reducing agent, and they can break disulfide bonds whose pH is greater than 7, i.e. those in basic condition.

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daser333 [38]

Answer:

0.077M is the concentration of the hydroxyl ion

Explanation:

Dilution factor is the ratio between the aliquot that is taken of a solution and the total volume of the diluted solution.

For the problem, dilution factor is:

7.53cm³ / 147cm³ =<em> 0.05122</em>

To obtain molarity of a diluted solution you must multiply dilution factor and initial molarity of the solution, thus:

1.5 M × 0.05122 = <em>0.077M is the concentration of the hydroxyl ion</em>

4 0
4 years ago
Which of the following measurements is expressed to three significant figures?
tiny-mole [99]
D. is the answer
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8 0
3 years ago
In preparation for a demonstration, your professor brings a 1.50−L bottle of sulfur dioxide into the lecture hall before class t
solmaris [256]

Answer:

4.81 moles

Explanation:

The total pressure of the gas = Pressure at which gauge reads zero + pressure read by it.

Pressure at which gauge reads zero = 14.7 psi

Pressure read by the gauge = 988 psi

Total pressure = 14.7 + 988 psi = 1002.7 psi

Also, P (psi) = P (atm) / 14.696

Pressure = 1002.7 / 14.696  = 68.2297 atm

Temperature = 25 °C

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15  

So,  

T = (25 + 273.15) K = 298.15 K  

Volume = 1.50 L

Using ideal gas equation as:

PV=nRT

where,  

P is the pressure

V is the volume

n is the number of moles

T is the temperature  

R is Gas constant having value = 0.0821 L.atm/K.mol

Applying the equation as:

68.2297 atm × 1.5 L = n × 0.0821 L.atm/K.mol × 298.15 K  

⇒n = 4.81 moles

4 0
3 years ago
Sample Response: The change in her position (from beginning to end), along with the direction, is considered her displacement. T
kykrilka [37]

The sample response given in the question is right.

To find the answer, we need to know more about the distance and displacement.

<h3>How distance differ from displacement?</h3>
  • Displacement is the shortest distance between the initial and final points of a body.
  • It is the change in position with a fixed direction.
  • Displacement is a vector quantity and can be positive, negative or zero values.
  • Distance is the length of actual path of the body between initial and final positions.
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Thus, we can conclude that the given sample response is right.

Learn more about distance here:

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3 0
2 years ago
A sample of nitrogen is initially at a pressure of 1.7 kPa, a temperature of -10 C and a volume of 7.5 m3. Then the volume is de
zhannawk [14.2K]

Answer:

\boxed{\text{2.6 kPa}}

Explanation:

To solve this problem, we can use the Combined Gas Laws:

\dfrac{p_{1}V_{1} }{T_{1}} = \dfrac{p_{2}V_{2} }{T_{2}}

Data:

p₁ = 1.7 kPa; V₁ = 7.5 m³;  T₁ =   -10 °C

p₂ = ?;          V₂ = 3.8 m³; T₂ = 200  K

Calculations:

(a) Convert temperature to kelvins

T₁ = (-10   + 273.15) K = 263.15 K

(b) Calculate the pressure

\begin{array}{rcl}\dfrac{1.7 \times 7.5 }{263.15} & = & \dfrac{p_{2} \times 3.8}{200}\\\\0.0485 & = & 0.0190p_{2}\\p_{2} & = & \textbf{2.6 kPa}\\\end{array}\\\text{The new pressure of the gas is \boxed{\textbf{2.6 kPa}}}

7 0
3 years ago
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