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Alja [10]
1 year ago
12

g In the absence of allosteric effectors, the enzyme phosphofructokinase displays Michaelis–Menten kinetics (see Fig. 7.15). The

v0/Vmax ratio is 0.90 when the concentration of the substrate, fructose-6-phosphate, is 0.10 mM. Calculate the KM for phosphofructokinase under these conditions (in units of mM).
Chemistry
1 answer:
Furkat [3]1 year ago
6 0

Answer:

The value of the Michaelis–Menten constant is 0.0111 mM.

Explanation:

Michaelis–Menten 's equation:

v_o=V_{max}\times \frac{[S]}{(K_m+[S])}=k_{cat}[E_o]\times \frac{[S]}{(K_m+[S])}

V_{max}=k_{cat}[E_o]

Where:

v_o = rate of formation of products

[S] = Concatenation of substrate

[K_m] =  Michaelis constant

V_{max}  = Maximum rate achieved

k_{cat} = Catalytic rate of the system

[E_o] = Initial concentration of enzyme

On substituting all the given values

We have :

\frac{v_o}{V_{max}}=0.90

[S] = 0.10 mM

\frac{v_o}{V_{max}}=\frac{[S]}{(K_m+[S])}

0.90=\frac{0.10 mM}{K_m+0.10 M}

K_m=0.0111 mM

The value of the Michaelis–Menten constant is 0.0111 mM.

You might be interested in
The pKs of succinic acid are 4.21 and 5.64. How many grams of monosodium succinate (FW = 140 g/mol) and disodium succinate (FW =
Varvara68 [4.7K]

Answer:

9.744g of monosodium succinate.

4.925g of disodium succinate.

Explanation:

To find pH of the buffer produced by the mixture of monosodium succinate-Disodium succinate is obtained from H-H equation:

pH = pKa + log ([Na₂Suc] / [NaHSuc])

As you want a pH of 5.28 and pKa is 5.64:

5.28 = 5.64 + log ([Na₂Suc] / [NaHSuc])

-0.36 = log ([Na₂Suc] / [NaHSuc])

0.4365 = ([Na₂Suc] / [NaHSuc]) <em>(1)</em>

<em />

As total concentration of the buffer is 100mM = 0.100M:

0.100M = [Na₂Suc] + [NaHSuc] <em>(2)</em>

Replacing (2) in (1):

0.4365 = (0.100M - [NaHSuc] / [NaHSuc])

0.4365 = (0.100M - [NaHSuc] / [NaHSuc])

0.4365 [NaHSuc] = 0.100M - [NaHSuc]

1.4365 [NaHSuc] = 0.100M

[NaHSuc] = 0.0696M

And:

[Na₂Suc] = 0.0304M

As volume of the buffer is 1L:

[NaHSuc] = 0.0696 moles

[Na₂Suc] = 0.0304 moles

Using molar mass of both substances:

Mass of monosodium succinate:

0.0696moles * (140g / 1mol) =<em> 9.744g of monosodium succinate.</em>

Mass of disodium succinate:

0.0304moles * (162g / 1mol) =<em> 4.925g of disodium succinate.</em>

<em></em>

5 0
2 years ago
A 0.1153-gram sample of a pure hydrocarbon was burned in a C-H combustion train to produce 0.3986 gram of CO2and 0.0578 gram of
Mars2501 [29]

<u>Answer:</u> The mass of carbon and hydrogen in the sample is 0.1087 g and 0.0066 g respectively and the percentage composition of carbon and hydrogen in the sample is 94.27 % and 5.72 % respectively.

<u>Explanation:</u>

The chemical equation for the combustion of hydrocarbon having carbon and hydrogen follows:

C_xH_y+O_2\rightarrow CO_2+H_2O

where, 'x' and 'y' are the subscripts of carbon and hydrogen respectively.

We are given:

Mass of CO_2=0.3986g

Mass of H_2O=0.0578g

Mass of sample = 0.1153 g

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

  • <u>For calculating the mass of carbon:</u>

In 44g of carbon dioxide, 12 g of carbon is contained.

So, in 0.3986 g of carbon dioxide, \frac{12}{44}\times 0.3986=0.1087g of carbon will be contained.

  • <u>For calculating the mass of hydrogen:</u>

In 18g of water, 2 g of hydrogen is contained.

So, in 0.0578 g of water, \frac{2}{18}\times 0.0578=0.0066g of hydrogen will be contained.

To calculate the percentage composition of a substance in sample, we use the equation:

\%\text{ composition of substance}=\frac{\text{Mass of substance}}{\text{Mass of sample}}\times 100      ......(1)

  • <u>For Carbon:</u>

Mass of sample = 0.1153 g

Mass of carbon = 0.1087 g

Putting values in equation 1, we get:

\%\text{ composition of carbon}=\frac{0.1087g}{0.1153g}\times 100=94.27\%

  • <u>For Hydrogen:</u>

Mass of sample = 0.1153 g

Mass of hydrogen = 0.0066 g

Putting values in equation 1, we get:

\%\text{ composition of hydrogen}=\frac{0.0066g}{0.1153g}\times 100=5.72\%

Hence, the mass of carbon and hydrogen in the sample is 0.1087 g and 0.0066 g respectively and the percentage composition of carbon and hydrogen in the sample is 94.27 % and 5.72 % respectively.

8 0
1 year ago
Explain why a propane torch is lit inside a hot air balloon during preflight preparations. Which gas law applies?
Olegator [25]

Propane torch is lit inside a hot air balloon during pre-flight preparation because the heat from the touch is needed to heat the cold air inside the balloon, so that the air will expand and become less dense and rise, thus providing a lift for the balloon. This is line with charle's law, which states that, the volume of a fixed mass of ideal gas is directly proportional to the absolute temperature. This law implies that, as the temperature of the air inside the balloon increase, the volume of the balloon also increases.

5 0
2 years ago
Read 2 more answers
How many moles of lead (ii) chromate are in 51 grams of this substance? answer in units of mol?
maria [59]

Mass of lead (II) chromate is 51 g. The molecular formula is PbCrO_{4} and its molar mass is 323.2 g/mol

Number of moles can be calculated using the following formula:

n=\frac{m}{M}

Here, m is mass and M is molar mass.

Putting the values,

n=\frac{(51 g}{323.1937 g/mol}=0.1578 mol

Therefore, number of moles of lead (II) chromate will be 0.1578 mol.

5 0
1 year ago
Rank the compounds by the ease with which they ionize under sn1 conditions. rank the compounds from easiest to hardest. to rank
marysya [2.9K]
The rate of Formation of Carbocation mainly depends on two factors'

                    1)  Stability of Carbocation:
                                                              The ease of formation of Carbocation mainly depends upon the ionization of substrate. If the forming carbocation id tertiary then it is more stable and hence readily formed as compared to secondary and primary.

                     2) Ease of detaching of Leaving Group:
                                                                                   The more readily and easily the leaving group leaves the more readily the carbocation is formed and vice versa. In given scenario the carbocation formed is tertiary in all three cases, the difference comes in the leaving group. So, among these three substrates the one containing Iodo group will easily dissociate to form tertiary carbocation because due to its large size Iodine easily leaves the substrate, secondly Chlorine is a good leaving group compared to Fluoride. Hence the order of rate of formation of carbocation is,

                                            R-I > R-Cl > R-F

                                               B   >  C  >  A

3 0
1 year ago
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