Δ G 0 value should be determined for the excitation of P700 and its comparison with the energy in an Einstein of 700nm photons should be explained. Concept introduction: P700 is primary donor for photosystem I where P denotes the pigment and since it absorbs at 700nm hence named as P700. In the process of photo excitation, when light is absorbed by the photosystem, one electron of chlorophyll P700 gets excited to the higher energy level. An einstein energy is defined as the energy associated with one mole of photons of certain wavelength.

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Answer 1

Question

Chapter 15, Problem 1P

Interpretation Introduction

Interpretation:

$ΔG0$ value should be determined for the excitation of P700 and its comparison with the energy in an Einstein of 700nm photons should be explained.

Concept introduction:

P700 is primary donor for photosystem I where P denotes the pigment and since it absorbs at 700nm hence named as P700.

In the process of photo excitation, when light is absorbed by the photosystem, one electron of chlorophyll P700 gets excited to the higher energy level.

An einstein energy is defined as the energy associated with one mole of photons of certain wavelength.

Expert Solution & Answer

Answer to Problem 1P

Solution:

Total $ΔG0=164.0245kJ/mol$

$ΔG0$ obtained is almost $95.92%$ of the Einstein energy of 700nm photons.

Given:

The raise in potential for the P700 reaction center is +0.4 to −1.3 volts.

Explanation of Solution

Step1:

Let us find change in potential first as:

$ΔE0=Efinal0−EInitial0=−1.3V−(+0.4V)=−1.7V$

Now, the free energy change is calculated as follows:

$ΔG0=−nFΔE0$

Where, n is number of electrons involved, F is Faraday's constant (96485C/mol) and E⁰ is cell potential.

Here,

$ΔE0=−1.7V=−1.7J/C(1V=1J/C)$

$n=1$

Since, the P700 gets excited through only one electron, so value of "n" will be 1.

Putting all the values we get:

$ΔG0=−nFΔE0=−1×96485C/mol×−1.7J/C=164024.5J/mol=164.0245kJ/mol$

Step2:

Einstein is energy of one mole of photons.

Let us find the energy of one photon as:

$E=h×cλ$

Where, h is Planck's constant, c is speed of light and λ is wavelength.

Putting the values:

$h=6.626×10−34J.sc=3×108m/sλ=700nm=7×10−7m$

Hence,

$E=6.626×10−34J.s×3×108m/s7×10−7m=2.84×10−19J$

Here, energy of one photon is $2.84×10−19J$ .

Hence, an einstein energy will be energy of one mole of photon, which can be calculated as:

$Einstein=E×NA=2.84×10−19(J/photon)×6.022×1023(photons/mol)=1.71×105J/mol=171kJ/mol$

Step3:

Comparison of the two energies ( $ΔG0$ and Einstein) is done as follows:

In the given photoexcitation, the percentage of energy conserved by the P700 is calculated as:

$Percent of energy conserved in P700= Energy conserved by P700Einstein energy of photon×100= 164.0245kJ/mol171kJ/mol×100=95.92%$

Hence, $ΔG0$ obtained is almost $95.92%$ of the Einstein energy of 700nm photons.

Conclusion

The total obtained, $ΔG0=164.0245kJ/mol$ .

It is almost $95.92%$ of the Einstein energy of 700nm photons.

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You’ve isolated a protein and determined that the Native molecular weight of the holoenzyme is 160 kD using size exclusion chromatography. Analysis of this protein using SDS-PAGE revealed 2 bands, one at 100 kD and one at 30 kD. The enzyme was found to be 0.829% NAD (by weight). What further can be said regarding the structure of the polypeptide?

Answer 2

Question

Chapter 15, Problem 1P

Interpretation Introduction

Interpretation:

$ΔG0$ value should be determined for the excitation of P700 and its comparison with the energy in an Einstein of 700nm photons should be explained.

Concept introduction:

P700 is primary donor for photosystem I where P denotes the pigment and since it absorbs at 700nm hence named as P700.

In the process of photo excitation, when light is absorbed by the photosystem, one electron of chlorophyll P700 gets excited to the higher energy level.

An einstein energy is defined as the energy associated with one mole of photons of certain wavelength.

Expert Solution & Answer

Answer to Problem 1P

Solution:

Total $ΔG0=164.0245kJ/mol$

$ΔG0$ obtained is almost $95.92%$ of the Einstein energy of 700nm photons.

Given:

The raise in potential for the P700 reaction center is +0.4 to −1.3 volts.

Explanation of Solution

Step1:

Let us find change in potential first as:

$ΔE0=Efinal0−EInitial0=−1.3V−(+0.4V)=−1.7V$

Now, the free energy change is calculated as follows:

$ΔG0=−nFΔE0$

Where, n is number of electrons involved, F is Faraday's constant (96485C/mol) and E⁰ is cell potential.

Here,

$ΔE0=−1.7V=−1.7J/C(1V=1J/C)$

$n=1$

Since, the P700 gets excited through only one electron, so value of "n" will be 1.

Putting all the values we get:

$ΔG0=−nFΔE0=−1×96485C/mol×−1.7J/C=164024.5J/mol=164.0245kJ/mol$

Step2:

Einstein is energy of one mole of photons.

Let us find the energy of one photon as:

$E=h×cλ$

Where, h is Planck's constant, c is speed of light and λ is wavelength.

Putting the values:

$h=6.626×10−34J.sc=3×108m/sλ=700nm=7×10−7m$

Hence,

$E=6.626×10−34J.s×3×108m/s7×10−7m=2.84×10−19J$

Here, energy of one photon is $2.84×10−19J$ .

Hence, an einstein energy will be energy of one mole of photon, which can be calculated as:

$Einstein=E×NA=2.84×10−19(J/photon)×6.022×1023(photons/mol)=1.71×105J/mol=171kJ/mol$

Step3:

Comparison of the two energies ( $ΔG0$ and Einstein) is done as follows:

In the given photoexcitation, the percentage of energy conserved by the P700 is calculated as:

$Percent of energy conserved in P700= Energy conserved by P700Einstein energy of photon×100= 164.0245kJ/mol171kJ/mol×100=95.92%$

Hence, $ΔG0$ obtained is almost $95.92%$ of the Einstein energy of 700nm photons.

Conclusion

The total obtained, $ΔG0=164.0245kJ/mol$ .

It is almost $95.92%$ of the Einstein energy of 700nm photons.

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Answer 3

Question

Chapter 15, Problem 1P

Interpretation Introduction

Interpretation:

$ΔG0$ value should be determined for the excitation of P700 and its comparison with the energy in an Einstein of 700nm photons should be explained.

Concept introduction:

P700 is primary donor for photosystem I where P denotes the pigment and since it absorbs at 700nm hence named as P700.

In the process of photo excitation, when light is absorbed by the photosystem, one electron of chlorophyll P700 gets excited to the higher energy level.

An einstein energy is defined as the energy associated with one mole of photons of certain wavelength.

Expert Solution & Answer

Answer to Problem 1P

Solution:

Total $ΔG0=164.0245kJ/mol$

$ΔG0$ obtained is almost $95.92%$ of the Einstein energy of 700nm photons.

Given:

The raise in potential for the P700 reaction center is +0.4 to −1.3 volts.

Explanation of Solution

Step1:

Let us find change in potential first as:

$ΔE0=Efinal0−EInitial0=−1.3V−(+0.4V)=−1.7V$

Now, the free energy change is calculated as follows:

$ΔG0=−nFΔE0$

Where, n is number of electrons involved, F is Faraday's constant (96485C/mol) and E⁰ is cell potential.

Here,

$ΔE0=−1.7V=−1.7J/C(1V=1J/C)$

$n=1$

Since, the P700 gets excited through only one electron, so value of "n" will be 1.

Putting all the values we get:

$ΔG0=−nFΔE0=−1×96485C/mol×−1.7J/C=164024.5J/mol=164.0245kJ/mol$

Step2:

Einstein is energy of one mole of photons.

Let us find the energy of one photon as:

$E=h×cλ$

Where, h is Planck's constant, c is speed of light and λ is wavelength.

Putting the values:

$h=6.626×10−34J.sc=3×108m/sλ=700nm=7×10−7m$

Hence,

$E=6.626×10−34J.s×3×108m/s7×10−7m=2.84×10−19J$

Here, energy of one photon is $2.84×10−19J$ .

Hence, an einstein energy will be energy of one mole of photon, which can be calculated as:

$Einstein=E×NA=2.84×10−19(J/photon)×6.022×1023(photons/mol)=1.71×105J/mol=171kJ/mol$

Step3:

Comparison of the two energies ( $ΔG0$ and Einstein) is done as follows:

In the given photoexcitation, the percentage of energy conserved by the P700 is calculated as:

$Percent of energy conserved in P700= Energy conserved by P700Einstein energy of photon×100= 164.0245kJ/mol171kJ/mol×100=95.92%$

Hence, $ΔG0$ obtained is almost $95.92%$ of the Einstein energy of 700nm photons.

Conclusion

The total obtained, $ΔG0=164.0245kJ/mol$ .

It is almost $95.92%$ of the Einstein energy of 700nm photons.

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Answer 4

Question

Chapter 15, Problem 1P

Interpretation Introduction

Interpretation:

$ΔG0$ value should be determined for the excitation of P700 and its comparison with the energy in an Einstein of 700nm photons should be explained.

Concept introduction:

P700 is primary donor for photosystem I where P denotes the pigment and since it absorbs at 700nm hence named as P700.

In the process of photo excitation, when light is absorbed by the photosystem, one electron of chlorophyll P700 gets excited to the higher energy level.

An einstein energy is defined as the energy associated with one mole of photons of certain wavelength.

Expert Solution & Answer

Answer to Problem 1P

Solution:

Total $ΔG0=164.0245kJ/mol$

$ΔG0$ obtained is almost $95.92%$ of the Einstein energy of 700nm photons.

Given:

The raise in potential for the P700 reaction center is +0.4 to −1.3 volts.

Explanation of Solution

Step1:

Let us find change in potential first as:

$ΔE0=Efinal0−EInitial0=−1.3V−(+0.4V)=−1.7V$

Now, the free energy change is calculated as follows:

$ΔG0=−nFΔE0$

Where, n is number of electrons involved, F is Faraday's constant (96485C/mol) and E⁰ is cell potential.

Here,

$ΔE0=−1.7V=−1.7J/C(1V=1J/C)$

$n=1$

Since, the P700 gets excited through only one electron, so value of "n" will be 1.

Putting all the values we get:

$ΔG0=−nFΔE0=−1×96485C/mol×−1.7J/C=164024.5J/mol=164.0245kJ/mol$

Step2:

Einstein is energy of one mole of photons.

Let us find the energy of one photon as:

$E=h×cλ$

Where, h is Planck's constant, c is speed of light and λ is wavelength.

Putting the values:

$h=6.626×10−34J.sc=3×108m/sλ=700nm=7×10−7m$

Hence,

$E=6.626×10−34J.s×3×108m/s7×10−7m=2.84×10−19J$

Here, energy of one photon is $2.84×10−19J$ .

Hence, an einstein energy will be energy of one mole of photon, which can be calculated as:

$Einstein=E×NA=2.84×10−19(J/photon)×6.022×1023(photons/mol)=1.71×105J/mol=171kJ/mol$

Step3:

Comparison of the two energies ( $ΔG0$ and Einstein) is done as follows:

In the given photoexcitation, the percentage of energy conserved by the P700 is calculated as:

$Percent of energy conserved in P700= Energy conserved by P700Einstein energy of photon×100= 164.0245kJ/mol171kJ/mol×100=95.92%$

Hence, $ΔG0$ obtained is almost $95.92%$ of the Einstein energy of 700nm photons.

Conclusion

The total obtained, $ΔG0=164.0245kJ/mol$ .

It is almost $95.92%$ of the Einstein energy of 700nm photons.

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Answer 5

Chapter 15, Problem 1P

Interpretation Introduction

Interpretation:

$ΔG0$ value should be determined for the excitation of P700 and its comparison with the energy in an Einstein of 700nm photons should be explained.

Concept introduction:

P700 is primary donor for photosystem I where P denotes the pigment and since it absorbs at 700nm hence named as P700.

In the process of photo excitation, when light is absorbed by the photosystem, one electron of chlorophyll P700 gets excited to the higher energy level.

An einstein energy is defined as the energy associated with one mole of photons of certain wavelength.

Expert Solution & Answer

Answer to Problem 1P

Solution:

Total $ΔG0=164.0245kJ/mol$

$ΔG0$ obtained is almost $95.92%$ of the Einstein energy of 700nm photons.

Given:

The raise in potential for the P700 reaction center is +0.4 to −1.3 volts.

Explanation of Solution

Step1:

Let us find change in potential first as:

$ΔE0=Efinal0−EInitial0=−1.3V−(+0.4V)=−1.7V$

Now, the free energy change is calculated as follows:

$ΔG0=−nFΔE0$

Where, n is number of electrons involved, F is Faraday's constant (96485C/mol) and E⁰ is cell potential.

Here,

$ΔE0=−1.7V=−1.7J/C(1V=1J/C)$

$n=1$

Since, the P700 gets excited through only one electron, so value of "n" will be 1.

Putting all the values we get:

$ΔG0=−nFΔE0=−1×96485C/mol×−1.7J/C=164024.5J/mol=164.0245kJ/mol$

Step2:

Einstein is energy of one mole of photons.

Let us find the energy of one photon as:

$E=h×cλ$

Where, h is Planck's constant, c is speed of light and λ is wavelength.

Putting the values:

$h=6.626×10−34J.sc=3×108m/sλ=700nm=7×10−7m$

Hence,

$E=6.626×10−34J.s×3×108m/s7×10−7m=2.84×10−19J$

Here, energy of one photon is $2.84×10−19J$ .

Hence, an einstein energy will be energy of one mole of photon, which can be calculated as:

$Einstein=E×NA=2.84×10−19(J/photon)×6.022×1023(photons/mol)=1.71×105J/mol=171kJ/mol$

Step3:

Comparison of the two energies ( $ΔG0$ and Einstein) is done as follows:

In the given photoexcitation, the percentage of energy conserved by the P700 is calculated as:

$Percent of energy conserved in P700= Energy conserved by P700Einstein energy of photon×100= 164.0245kJ/mol171kJ/mol×100=95.92%$

Hence, $ΔG0$ obtained is almost $95.92%$ of the Einstein energy of 700nm photons.

Conclusion

The total obtained, $ΔG0=164.0245kJ/mol$ .

It is almost $95.92%$ of the Einstein energy of 700nm photons.

Answer 6

Chapter 15, Problem 1P

Interpretation Introduction

Interpretation:

$ΔG0$ value should be determined for the excitation of P700 and its comparison with the energy in an Einstein of 700nm photons should be explained.

Concept introduction:

P700 is primary donor for photosystem I where P denotes the pigment and since it absorbs at 700nm hence named as P700.

In the process of photo excitation, when light is absorbed by the photosystem, one electron of chlorophyll P700 gets excited to the higher energy level.

An einstein energy is defined as the energy associated with one mole of photons of certain wavelength.

Expert Solution & Answer

Answer to Problem 1P

Solution:

Total $ΔG0=164.0245kJ/mol$

$ΔG0$ obtained is almost $95.92%$ of the Einstein energy of 700nm photons.

Given:

The raise in potential for the P700 reaction center is +0.4 to −1.3 volts.

Explanation of Solution

Step1:

Let us find change in potential first as:

$ΔE0=Efinal0−EInitial0=−1.3V−(+0.4V)=−1.7V$

Now, the free energy change is calculated as follows:

$ΔG0=−nFΔE0$

Where, n is number of electrons involved, F is Faraday's constant (96485C/mol) and E⁰ is cell potential.

Here,

$ΔE0=−1.7V=−1.7J/C(1V=1J/C)$

$n=1$

Since, the P700 gets excited through only one electron, so value of "n" will be 1.

Putting all the values we get:

$ΔG0=−nFΔE0=−1×96485C/mol×−1.7J/C=164024.5J/mol=164.0245kJ/mol$

Step2:

Einstein is energy of one mole of photons.

Let us find the energy of one photon as:

$E=h×cλ$

Where, h is Planck's constant, c is speed of light and λ is wavelength.

Putting the values:

$h=6.626×10−34J.sc=3×108m/sλ=700nm=7×10−7m$

Hence,

$E=6.626×10−34J.s×3×108m/s7×10−7m=2.84×10−19J$

Here, energy of one photon is $2.84×10−19J$ .

Hence, an einstein energy will be energy of one mole of photon, which can be calculated as:

$Einstein=E×NA=2.84×10−19(J/photon)×6.022×1023(photons/mol)=1.71×105J/mol=171kJ/mol$

Step3:

Comparison of the two energies ( $ΔG0$ and Einstein) is done as follows:

In the given photoexcitation, the percentage of energy conserved by the P700 is calculated as:

$Percent of energy conserved in P700= Energy conserved by P700Einstein energy of photon×100= 164.0245kJ/mol171kJ/mol×100=95.92%$

Hence, $ΔG0$ obtained is almost $95.92%$ of the Einstein energy of 700nm photons.

Conclusion

The total obtained, $ΔG0=164.0245kJ/mol$ .

It is almost $95.92%$ of the Einstein energy of 700nm photons.

Answer 7

Chapter 15, Problem 1P

Interpretation Introduction

Interpretation:

$ΔG0$ value should be determined for the excitation of P700 and its comparison with the energy in an Einstein of 700nm photons should be explained.

Concept introduction:

P700 is primary donor for photosystem I where P denotes the pigment and since it absorbs at 700nm hence named as P700.

In the process of photo excitation, when light is absorbed by the photosystem, one electron of chlorophyll P700 gets excited to the higher energy level.

An einstein energy is defined as the energy associated with one mole of photons of certain wavelength.

Answer 8

Expert Solution & Answer

Answer to Problem 1P

Solution:

Total $ΔG0=164.0245kJ/mol$

$ΔG0$ obtained is almost $95.92%$ of the Einstein energy of 700nm photons.

Given:

The raise in potential for the P700 reaction center is +0.4 to −1.3 volts.

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Explanation of Solution

Step1:

Let us find change in potential first as:

$ΔE0=Efinal0−EInitial0=−1.3V−(+0.4V)=−1.7V$

Now, the free energy change is calculated as follows:

$ΔG0=−nFΔE0$

Where, n is number of electrons involved, F is Faraday's constant (96485C/mol) and E⁰ is cell potential.

Here,

$ΔE0=−1.7V=−1.7J/C(1V=1J/C)$

$n=1$

Since, the P700 gets excited through only one electron, so value of "n" will be 1.

Putting all the values we get:

$ΔG0=−nFΔE0=−1×96485C/mol×−1.7J/C=164024.5J/mol=164.0245kJ/mol$

Step2:

Einstein is energy of one mole of photons.

Let us find the energy of one photon as:

$E=h×cλ$

Where, h is Planck's constant, c is speed of light and λ is wavelength.

Putting the values:

$h=6.626×10−34J.sc=3×108m/sλ=700nm=7×10−7m$

Hence,

$E=6.626×10−34J.s×3×108m/s7×10−7m=2.84×10−19J$

Here, energy of one photon is $2.84×10−19J$ .

Hence, an einstein energy will be energy of one mole of photon, which can be calculated as:

$Einstein=E×NA=2.84×10−19(J/photon)×6.022×1023(photons/mol)=1.71×105J/mol=171kJ/mol$

Step3:

Comparison of the two energies ( $ΔG0$ and Einstein) is done as follows:

In the given photoexcitation, the percentage of energy conserved by the P700 is calculated as:

$Percent of energy conserved in P700= Energy conserved by P700Einstein energy of photon×100= 164.0245kJ/mol171kJ/mol×100=95.92%$

Hence, $ΔG0$ obtained is almost $95.92%$ of the Einstein energy of 700nm photons.

Answer 12

Conclusion

The total obtained, $ΔG0=164.0245kJ/mol$ .

It is almost $95.92%$ of the Einstein energy of 700nm photons.

Answer 13

Ch. 15 - Prob. 1P Ch. 15 - Prob. 2P Ch. 15 - Prob. 3P Ch. 15 - Prob. 4P Ch. 15 - The flux of solar energy reaching the Earth's...Ch. 15 - The substance dichlorophenyldimethylurea (DCMU) is...Ch. 15 - Prob. 7P Ch. 15 - Prob. 8P Ch. 15 - Prob. 9P Ch. 15 - Prob. 10P

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Answer to Problem 1P

Explanation of Solution

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