The proton gradient would not form because the electrons from NADH could not be obtained. No ATP synthesis would occur.
The proton gradient would be created more rapidly since NADH would donate its electrons more readily. ATP synthesis would increase.
The electrons would still pass from Complex III and Complex IV to create a smaller proton gradient. Not as much ATP could be made.
The protons from the matrix would have to diffuse through the membrane to maintain the proton gradient needed for ATP synthase.
Electrons from the proton gradient would be donated to Complex I so that it could function normally.
The light reactions would slow down since the supply of NADPH would decrease.
The light reactions would speed up since there would be an increase in electron availability.
NADPH would be reduced and the Calvin Cycle would occur more rapidly.
NADPH would be oxidized and the Calvin Cycle would cease to happen.
It is the oxidizing agent
It is the reducing agent
It will accept electrons from Complex I and NADH become reduced
It will donate electrons to Complex I and NADH become reduced.
A molecule of fat has oxygen-hydrogen bonds, which contain more energy than the carbon-carbon bonds in glucose
A molecule of fat has more non-polar covalent bonds that absorb more energy when broken than the polar covalent bonds in glucose
A molecule of fat has more polar covalent bonds that give off more energy when broken than the polar covalent bonds in glucose
A molecule of fat has a longer chain and therefore has more bonds to release energy, and a molecule of glucose has less available bonds to break
The final electron acceptor, oxygen, is very polar
The final electron acceptor, oxygen, is electropositive
The final electron acceptor, oxygen has a relatively high electronegativity, and therefore causes the driving force for the electrons to move down the chain
The final electron acceptor, oxygen, is electronegative, therefore, the electrons moving down the ETC will be attracted to it
A. Outside the cell (in the extracellular matrix)
B. Inside the cell (in the cytoplasm)
C. Passing through the lipid bilayer of the membrane
A and B
A. Matrix of a mitochondria
B. Smooth endoplasmic reticulum
E. A and B
The H+ ions would combine with bicarbonate (HCO3-) to form carbonic acid (H2CO3), which would dissociate to form CO2 and H20. CO2 would be exhaled.
Respiration would increase and subsequently increase CO2 in the blood, which would combine with the H+ to neutralize the H+ ions.
The H+ ions would combine with carbonic acid (H2CO3) to increase bicarbonate (HCO3-), which is a base, thereby neutralizing the acidity.
Have receptors that will bind the signaling molecule, and undergo a signal transduction pathway
Have receptors that bind to the signaling molecule, respond with the activation of the signal transduction pathway, and finally respond by ceasing cell division
Have ligands that bind to the signaling molecule, and immediately responds by ceasing further cell division
Have sugars protruding into the extracellular matrix to sense the signal and cause the activation of the signal transduction pathway, which in turn causes the cell to respond by ceasing further cell division
Only primary active transport can create an electrochemical gradient
Secondary active transport only involves the movement of solutes up their gradient
Primary Active Transporters are considered “pumps” and secondary active transporters are not
Primary active transport requires ATP and secondary active transport does not
There is a higher pH in the thylakoid so protons can be pumped out of the thylakoid via ATP synthase and synthesize ATP.
There is a lower pH in the thylakoid so protons can be pumped out of the thylakoid via ATP synthase and synthesize ATP.
The smaller proton gradient in the thylakoid would allow ATP synthase to use the protons in the gradient to make ATP.gradient in the stroma to move protons to the thylakoid and synthesize ATP in the thylakoid.
There is high hydroxyl concentration in the the thylakoid which can move through the ATP synthase and make ATP.
The separation of chromosomes
The bending of cilia and flagella
Movement of cargo such as a vesicle
Primary, Tertiary, Quaternary
The way in which they move.
The 9-2 arrangement of microtubules.
They both use the motor protein dynein
Both are found in eukaryotic cells.
Is dissolving a small, polar, or charged molecule
Is it dissolving a small, polar molecule
Is only forming hydrogen bonds, because water cannot dissolve anything without breaking bonds
Is dissolving the molecule by breaking the bonds in that molecule and replacing them with hydrogen bonds
This is because cellulose is polar and can allow the cell membrane of a cell to create a strong covalent bond to it
This is because cellulose is non-polar, and can allow for extensive hydrogen bonding between its chains
This is because the cellulose is polar, and can allow for extensive hydrogen bonding between its chains
This is because cellulose has many non-polar covalent bonds, which are, in general, stronger bonds than polar-covalent bonds, and therefore, will hold its structure well
The fatty acid tails are attracted to each other because the atoms involved are charged
The fatty acids are very polar and do not want to interact with the non-polar surroundings of the cell (the extracellular fluid and cytosol)
The fatty acid tails form strong hydrophillic interactions to keep them together
The atoms that make up the bonds in the fatty acid tails have similar electronegativities
NADPH contributes to the H+ gradient, which provides the energy to drive the formation of sugars
High energy electrons (which obtain energy from the sunlight) gain energy as they move down the ETC, producing the H+ gradient that drives ATP synthesis
The ATP that is produced in the Light reactions provides the energy that is required to form the bonds in sugars
The energy released from the breaking of bonds in NADPH at the Calvin Cycle provide the energy to form the bonds between sugars
B and C
The intermembrane space of the mitochondria
The lumen of the chloroplast thylakoids
The matrix of the mitochondria
A and B
B and C
Smooth Endoplasmic reticulum
Rough Endoplasmic Reticulum
The antioxidants are oxidizing agents and oxidize the damaged DNA.
The antioxidants are oxidizing agents and reduce the damaged DNA.
The antioxidants are reducing agents and reduce the damaged DNA.
The antioxidants are reducing agents and oxidize the damaged DNA.