Biology Exam 2

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In which compartments of the cell are secreted proteins, processed and sorted?

rough endoplasmic reticulum, (rER), Golgi (vesicle trafficking from rER to Golgi, and between the Golgi and the cell surface)

What is receptor mediated endocytosis and what role does the protein molecule calthrin play in this process?

Receptor mediated endocytosis: In general, endocytosis = internalization mechanism for bringing in material from outside the cell, accomplished through the pinching off of membrane vesicles from the cell surface. Receptor mediated endocytosis describes the internalization of molecules that are bound to membrane bound receptors. In other words, ligands bind membrane receptors -> internalization in vesicles.
Clathrin binds to adaptin molecules, which bind to the cytoplasmic domains of membrane receptors protein molecules. Clathrin mediates the invagination -> vesicle pinching off

What are the functions of lysosomes?

Lysosomes, which contain hydrolytic enzymes, fuse with incoming phagosomes or other vesicles leading to the degradation of the contents of these vesicles. Lysosomes are also responsible for the turn-over (recycling) of structures and molecules from within the cell (autophagy).

Tay-Sachs is one example of a large class of diseases that affect lysosomal function. What is the general name of this class of diseases? Explain why this name is an appropriate one based on the disease mechanism.

Lysosomal Storage Diseases
These diseases are characterized by a defect in lysosome function, frequently the lack of a functional hydrolytic enzyme. Such a defect leads to the inability to hydrolyze the target of lysosomal attack. Instead, the lysosome stores the compound(s) that is (are) normally targeted for turnover -> cellular damage and/or dysfunction

Explain the three different properties of signal transduction pathways that we discussed in class. Note that we are looking for general properties of signal transduction, not the Signal > Receptor > Transducer > Effect chain of events, nor the details of any particular signaling pathway.

Three concepts/principles:
1) Amplification of the signal
The strength of the signal can be amplified down the signal transduction chain. For example, 1 molecule of ligand can lead to the production of many molecules of an effector. The amplification is often achieved by catalytic enzymes, which can produce many molecules of a product when stimulated.
2) Distribution of the signal
The signal can be distributed in time and space throughout the cell, tissue, organ or organism. This distribution is often achieved by the use of second messenger molecules (e.g. Ca++, cAMP)
3) Specificity
A single signal molecule can have different effects in different cells due to the combination of different transducers and effectors.

6A) In the example we discussed in class, where in the cell do steroid hormone receptors reside?

cytoplasm

6B) What sequence of events occurs when a steroid hormone binds its receptor?

1) After ligand binding to the cytoplasmic receptor, the receptor's chaperone is released
2) The steroid hormone receptor and its ligand are transported into the nucleus
3) Changes in gene expression/activity occur

6C) Based on our discussion of eukaryotic cell biology, propose a mechanism that can explain the sequence of events that occurs when a steroid hormone binds its receptor.

1) Conformational change in receptor > chaperone release
2) Exposure of a previously masked nuclear localization signal (NLS) on the receptor; before ligand binding, the chaperone sequesters the receptor in the cytoplasm

Compare a ligand- receptor interaction to the interaction of an enzyme and its substrate

Similarities: reversible and specific binding
Differences; ligand is not changed by its interaction with the receptor, while the substrate is a reactant in the enzyme catalyzed reaction

G- proteins are sometimes viewed as 'self regulating' switches. Explain how G-proteins are activated and how they are inactivated.

Activated: exchanging a GDP molecule for a GTP molecule
Inactivated: GTP hydrolysis by the intrinsic GTPase activity of the G-protein

What is a protein kinase cascade?

Steps in a signal transduction chain that rely on sequential protein phosphorylation events. For example, activation of one protein kinase leads to the phosphorylation and activation of a second protein kinase, which in turn, phosphorylates a third protein kinase, etc.

How would a protein cascade be down-regulated (how can the signaling of a protein kinase be turned down or off)

Degradation or inactivation of the protein kinases
action fo a protein phosphatase, which removes the phosphate from the target protein

In class we talked about one drug that was a phosphodiesterase inhibitor. How would such drugs operation to alter signaling?

Phosphodiesterase (PDE) destroys cAMP (and/or cGMP) second messenger molecules. Therefore, PDE can shut off or disrupt a signal transduction cascade that relies on the cAMP (cGMP) second messengers. A PDE inhibitor would lead to sustained signaling due to persistence of the cyclic nucleotide second messenger.

Glycolysis (Inputs and Outputs)

Inputs: Glucose, ATP
Outputs: NADH (+ H+), ATP, Pyruvate

Citric Acid Cycle (Inputs and Outputs)

Inputs: Acetyl-CoA
Outputs: NADH (+ H+), GTP (ATP), FADH2

Mitochondrial electron transport chain + F0/F1 complex (Inputs and Outputs)

Inputs: high energy electrons donated from NADH and FADH2,
intermediary: proton-motive force
Outputs: ATP

Non-cyclic electron transport/Z-scheme in photosynthesis (Inputs and Outputs)

Inputs: electrons from H20, light energy
Intermediary: proton-motive force
Outputs: ATP, 02, NADPH (+ H+)

Cyclic electron transport in photosynthesis (Inputs and Outputs)

Inputs: Light energy
Intermediary: proton-motive force
Outputs: ATP

Any standard lab strain of E Coli will grow slowly under anaerobic conditions compared to aerobic conditions. Explain why.

Aerobic conditions support cellular respiration: the citric acid cycle and oxidative phosphorylation, which yield higher amounts of energy (#ATP)
Anaerobic conditions force the organism to generate energy by glycolysis, which yields far fewer ATP molecules per glucose molecule consumed.

What is the purpose of the Cori cycle?

Lactate formed by the muscle (by fermentation) is transported through the blood to the liver where it is converted by glucose through gluconeogenesis. Purpose: scavenging lactate; detoxifying muscle cells by removing lactate > converting to glucose; liver takes on part of the metabolic burden from the muscle

Explain either of the experiments discussed in the animations that demonstrated the importance of proton-motive force in generation of ATP by ATP synthase.

Expt 1: Manipulation of the pH on either side of the inner mitochondrial membrane, only when the pH was lower (more H+ ions) outside the inner mitochondrial membrane than in the matrix was ATP produced. Artificially produced proton gradient with higher [H+] outside

Expt 2: Artificial membrane vesicles with bacteriorhodopsin > light driven H+ pumping into vesicle. If ATP synthase is incorporated into the artificial membrane vesicle (oriented with the F1 subunit pointing outward)... ATP is produced

What is the role of the following proteins in the ATP synthase complex?

Gamma (y): Asymmetric protein that is attached to the F0 proton channel; movement of gamma subunit relative to the stationary beta catalytic subunit drives the conformational changes in beta > switch between O, L, T configuration and ATP formation
Subunit c: Transmembrane alpha helical protein that carries the proton on a protonated Asp residue from and to the b subunit split channel; comprises the spinning barrel of the F0 complex
Beta (B): ATPase or ATP synthase catalytic subunit; exists in different conformational states (O, L, T)

What specific subcellular compartment gets acidified during electron transport during the light reactions of photosynthesis in eukaryotes?

thylakoid interior

In eukaryotes, where do the light-independent reactions of photosynthesis take place?

stroma of the chloroplast

What are the inputs and outputs of the light-independent reactions of photosynthesis?

Inputs: ATP (energy) NADPH (+ H+) (reducing power), CO2
Outputs: reduced 3 carbon molecules: G3P (Glyceraldehyde 3-Phosphate)

What is the consequence of Rubisco's mixed carboxylase/oxygenase activities? Specifically, what are the immediate products of the two different reactions and what are the larger metabolic consequences?

Oxygenase: converts the 5-carbon ribulose-bisphosphate to one 3 carbon and one 2 carbon compounds; oxidation of a 5 carbon sugar; leads to photorespiration
Carboxylase: converts the 5-carbon ribulose-bisphosphate to two molecules of 3 carbon compounds (3-phosphoglycerate, 3PG); carbon fixation
larger consequence: wasteful, reduces the efficiency of the carbon fixation

Hibernating mammals (and newborns) keep warm using a tissue called 'brown fat' which is rich in mitochondria. This tissue uses 'uncoupling' proteins to generate heat. What is being uncoupled and what is the mechanism by which these proteins work?

Uncoupling the oxidation of fuel molecules (and subsequent electron transport) from ATP production by dissipating the proton-motive force.
Uncoupling proteins allow protons to pass from the intermitochondiral membrane into teh matrix of the mitochondria, thereby bypassing the ATP synthase complex. The energy from this proton flow is released as heat.

The following DNA sequence is written in the convention 5' > 3' left to right
AAGTACCCTTGAGT
Write the base of the complimentary strand, following the same convention.

ACTCAAGGGTACTT

Chargaff's rules appear to apply to the sequence shown in 22A...should they?

In this particular case, it does, however this is a coincidence. There is no reason that a short base sequence for a single-strand should conform to Chargaff's rule. Chargaff's rules (#A = #T and #C = #G) is a reflection of Watson-Crick base pairing...because the base sequence shown in 22A is from a single-strand, there is no reason for the sequence to conform to Chargaff's rule.

What are the main structural features of B-form DNA? (The more detail the better)

-double stranded
-anti-parallel strands
-sugar-phosphate backbones on the outside of the molecule
-Watson-Crick base-pairs on the inside of the molecule
-right handed helix
-major and minor grooves
-10.5 bp per complete helical turn

-3.4 Angstroms per base-pair
- ~20 Angstroms wide

In class we discussed the structure and function of a heterotrimeric G-protein. In general, what is the function of this protein? Compare and contrast the structure of the proton in both its 'off' position and its 'on' position. If you did an SDS-PAGE electrophoresis experiment using the protein from its 'off' confirmation, how many bands would you see after Coomassie blue staining?

The function of the G-protein is to realy a signal form a receptor type protein in the membrane, to another effector type protein inside the cell. In the off-position, the G-protein consists of three subunits and is bound to GDP. In the on position, the PAGE, you would see three bands from the off confirmation protein.

What reaction does the enzyme pyruvate dehydrogenase catalyze? Your answer should include all the co-factors involved in this reaction and structures of the substrate and the product (but not the cofactors). Name one molecule that activates pyruvate dehydrogenase, and one molecule that inhibits the enzyme. For your two choices, explain the logic of the activation/inhibition.

PDH catalyzes the conversion of pyruvate to acetyl-CoA. Activators include pyruvate and ADP; inhibitors include acetyl-CoA, NADH, and ATP (equation below would be sufficient). A cell wants PDH working at its maximum capacity when it needs energy (ATP) and has the molecules necessary to use the CAC and oxidative phosphorylation. A high concentration of ADT generally indicates a lack of ATP, so PDH would be activated. A high concentration of pyruvate indicates that lots of the necessary compound to enter the CAC. In general, PDH will be inhibited when there is enough energy already, or molecules needed for CAC are lacking. So a high concentration of ATP will inhibit the enzyme. A high concentration of acetyl-CoA indicates that CAC is working at capacity, so there is no sense in converting more pyruvate to acetyl-CoA. NAD+ is required for the PDH reaction, and a lot of NADH indicates a lack of NAD+, so NADH inhibits the reaction until enough NAD+ is available.

A plasma membrane sodium/glucose symport is able to transport glucose into the cell against a concentration gradient, yet it is not an ATPase. Explain how this can happen.

There is sodium concentration gradient with its high end outside the cell. Sodium ions flow into the cell via this symport releasing energy to support the transport of glucose against concentration gradient. The sodium concentration gradient is established by a nearby sodium pump using energy from the hydrolysis of ATP.

A man heard an explosion nearby an decided to run away form the explosion as fast as he could. His body responded to this shock by releasing a hormone, epinephrine, to the blood stream. The following questions are related to this condition....
A) In response to epinephrine, liver cells released a lot of glucose to the blood stream. Describe the sequence of events from the site of perception of epinephrine to the production of glucose. Events taking place on membranes should be highlighted with an *. Your answer should include name of key enzymes, second messenger, and signaling intermediates. However, NO chemical structures are needed.

A) Epinephrine binds to its receptor (a 7TM receptor) > activation of heterotrimeric G-protein > activated alpha subunit of G-protein activates adenylyl cyclase* > (cAMP) increases > cAMP inactivates glycogen synthase, but activates protein kinase A > activated protein kinase A activates glycogen phosphorylase kinase > glycogen phosphorylase is activated > hydrolysis of glycogen to generate G-1-P that is converted to glucose > glucose secreted blood stream

34B) Glucose produced by liver was then transported to muscle to support the running this man was doing. Describe how glucose is metabolized to generate the ATP needed when there was plenty of oxygen. Your answer should include 1) names of pathways involved 2) number of carbon atoms in key intermediates and 3) cellular locations for each pathway. However, NO need for names of enzymes and chemical structures.

1) Glycolysis in cytosol C6 (glucose) >> C3 (glyceraldehydes) >> C3 Pyruvate
2) Pyruvate oxidation in matrix of mitochondria
C3 pyruvate > C2 acetyl CoA
3) Citric Acid Cycle (CAC) [Krebs cycle or TCA cycle OK] in matrix of mitochondria
C2 acetyle CoA + C4 > C6 citric acid > C5 + CO2 > C4 + CO2 (goes back to first step)
4) e transport and oxidative phosphorylation in the inner membrane of mitochondria to produce ATP

34C) After running for a while, the man felt exhausted and he has problems catching his breath. How was glucose metabolized under this condition when oxygen supply was severely limited? NO need for names of enzymes and chemical structures.

Glucose is metabolized to pyruvate via glycolysis generating ATP and NADH. When oxygen supply is limited, mitochondrial citric acid cycle and oxidative phosphorylation cannot occur. Pyruvate has to be reduced to lactate (lactic acid) so that NADH can be reoxidized to NAD+ in support of continuous glucose metabolism via glycolysis (NAD+ is needed in glycolysis). Therefore when oxygen supply is limited, glycolysis is the only process to produce small amounts of ATP and lactate is accumulated as an end product.

34D) The man finally reached safety when he felt fatigue in his muscles. Now he could rest and recover from the strenuous running. What happened to his muscle and liver cells at this stage? Your answer should include 1) names of pathways involved and 2) cellular locations for each pathway. However, no need for names of enzymes and chemical structures.

1) Lactate accumulated in the muscle is secreted to blood and eventually picked up by liver cells
2) Liver cells convert lactate to glucose via gluconeogenesis (in cytosol).
3) The regenerated glucose can be sent to muscle for future use.
4) The whole cycle between muscle (glucose > lactate) and liver (lactate > glucose) is called Cori cycle.

34E) During his running, what type of cytoskeleton was involved in the muscle contraction? What are the other types of cytoskeleton in typical eukaryotic cells? Arrange them (including the one involved in muscle contraction) in order of physical dimension (form largest to smallest). List at least one function for other types of cytoskeletons.

The cytoskeleton involved was actin (micro) filament.
Cytoskeletons in cell (from large to small in dimension):
Microtubules (chromosome movement during cell division) > intermediate filaments (nuclear lamina for maintaining nuclear structure) > microfilaments.

Draw a diagram and use it to describe how the F0 complex of a mitochondrial ATP synthase rotates in response to the flow of H+? This F0 complex has 10 to 14 subunits, and use these numbers to calculate how many H+ need to go through F0 in generating one ATP. Also, use these numbers to calculate how many ATP molecules are expected to be synthesized when one mitochondrial NADH is oxidized by the electron transport system.

F0 of ATPase (as shown in the diagram) is actually a proton channel surrounded by 10-14 c subunits embedded in membrane. Amino acid 61 in each subunit c is an aspartic acid (Asp) that becomes negatively charged when it is not protonated, but neutral when it is protonated.
The unprotonated subunit c (Asp is negatively charged) cannot be in contact with the hydrophobic interior of membrane, and has to be 'protected' by subunit a which has two half proton channels. The protonated (with H+) Asp can move out of the cover of subunit a, thus the whole complex of F0 turns when this happens. Subunit a covers two unprotonated c subunits. The half proton channel on the left allows one proton to flow in converting the left subunit c to neutral, which then moves out of the cover of subunit a and is intact with the hydrophobic interior of membrane. During the same rotation of F0, a protonated subunit moves under the cover of subunit a, and releases a proton to the other side of membrane via the half channel on the right.
The above actions are repeated to allow the F0 to rotate. For the F0 complex to rotate 360 takes 10 protons if there are 10 c subunits, or 14 protons if there are 14 c subunits. Each turn allows 3 ATP formed in the attached F1 complex (the ATP synthase itself). Thus it takes 10/3 = 3.3 or 14/3 = 4.6 protons to generate one ATP. Since electrons from one NADH in mitochondria release enough energy to pump 10 proton across the mitochondrial inner membrane, these 10 electrons can generate 2.2 (10/4.6) to 3 ATP (10/3.3)

What are the forces/atomic interactions holding two single strands of DNA together in a double helical configuration?

1) hydrogen bonds between bases from the two strands of DNA
2) hydrophobic interactions due to base-stacking in the helical structure

Both NA+ and K+ are positively charged ions. Describe mechanisms that allow Na and K channels to distinguish these two ions.

Specificity for Na channel: Na+ is smaller than K+. The sodium channel is large enough for Na to go through, but too small for the larger K to go through.
Specificity for K channel: One third of the total length of K channel is too narrow for hydrated Na+ or K+ ions to go through. THus, both ions have to remove the hydration shell, causing their energy level to increase (desolvation energy). Once inside the narrow part of the channel, both ions interact with oxygen atoms on the r-groups of amino acids lining the narrow passage. with its large size, K + interacts well with those oxygen atoms, thus reducing its energy level (resolvation energy). On the other hand, the smaller Na+ cannot interact with oxygen atoms well due to its smaller size, hence much smaller resolvation energy drop compared to K+. Therefore, K+ going through the K channel is thermodynamically favorable, but not favorable for Na+.

what is the key enzyme in Calvin cycle in photosynthesis? Describe the quaternary structure of this enzyme (including the genomes encoding these subunits). In what compartment is this enzyme localized in chloroplast? What are the substrates and products for the reaction catalyzed by this enzyme in typical C3 photosynthesis (no need for structures)? Where is NADPH needed in Calvin Cycle?

RuBP carboxylase (rubisco): 8 lage subunits encoded by chloroplast genome and 8 small subunits encoded by nuclear genome.
Located in the stroma (matrix) of chloroplast
Substrates: RuBP (ribulose 1,5 biphosphate), CO2 (or O2 at high temp)
Products: two molecules of 3PG (3-phosphoglyceric acid) or one molecule of 3PH and a C2 compound (when O2 is the substrate)
NADPH is needed to reduce 3PG to glyceraldehydes-3-phostphate (G3P, the smallest sugar)

What causes DNA supercoiling? What is the enzyme regulating DNA supercoiling?

Two causes:
1) Deviation in the overall helical twist per bp (ie the helical structure is either over or under wound/twisted)
2) Topological constraints such as formation of circular DNA or both ends of a linear DNA are 'fixed'
Enzymes regulating DNA supercoiling: topoisomerases

Describe the experiment demonstrating that a H+ concentration gradient across a biological membrane is sufficient to produce ATP. Antimycin A is an electron transport inhibitor. Would the addition of antimycin A in this experiment affect the outcome? What would happen if DNP (dinitrophenol) is added in this experiment?

Antimycin A should not affect this process because there is no electron transport involved.
DNP is an uncoupler causing dissipation of the pH gradient, thus the [H+] concentration gradient driven ATP synthesis will be inhibited by DNP

Draw a diagram of a typical chloroplast and label all membranes and other compartments. Label the following locations/events, using a to g on the diagram
a- site where energy from photons is absorbed
b- site of Calvin cycle
c- site of electron transport
d- location where high [H+] is expected
e- location where newly synthesized ATP is located
f- location where newly synthesized NADPH is located
g- site of oxygen production

diagram

what is the most important enzyme in the Calvin cycle? Name and diagram the structure of its substrates. How are the reactino products of this enzyme eventually converted to a six-carbon sugar? Where are ATp and NADPH needed in the Calvin cycle?

Rubisco (RuBP carboxylase) is the most important enzyme in the Calvin cycle. It catalyze the following rxn with ribulose 1,5 biphospate (RuBP) and CO2 as substrates. The products are reduced to generate glyceraldehyde 3 phosphate (G3P or GAP), a step requiring ATP and NADPH. G3P (GAP) are then converted to a six-carbon sugar via gluconeogenesis.

Assume you have just consumed a high protein meal with proteins enriched with a simple amino acid alanine (R group is CH3-). Describe how this alanine becomes the storage lipid, triaglyceride (TAG). In your answer you need to include the pathways, key intermediates (no chemical structures needed) and cellular location of biochemical events.

1) Alanine is converted to the a-keto acid pyruvate in cytosol.
2) Pyruvate is converted to acetyl CoA via pyruvate oxidation in mitochondria
3) Acetyl CoA is converted to fatty acid via fatty acid synthesis in cytosol
4) Pyruvate is converted to glyceraldehyde (phosphate) via gluconeogenesis (halfway in the pathway) in cytosol
5) Glyceraldehyde (phosphate) is reduced to glycerol (phosphate) in cytosol
6) Glycerol (phosphate) and fatty acids form TAG on ER

34F) Now that the shocking experience to this man is over and the epinephrine level starts to decline, how does his body turn of the signaling transduction pathway that has been activated by epinephrine?

1) Activated 7TM receptor can be turned off by interacting with 'arrestin'
2) Heterotrimeric G-protein cycles from 'on' state (a-subunit with GTP) to 'off' state (GTP is hydrolyzed to give GDP)
3) cAMP can be hydrolyzed (destroyed) by phosphodiesterase
4) protein kinases can be dephosphorylated by protein phosphatases

What is a nucleosome structure?

A nucleosome is a protein/DNA complex with about 150-200 nt of dsDNA wrapped around small basic (positively charged) nuclear histone protein core (2 each of H2A, H2B, H3 and H4) (One more histone (H1) binds on the outside of nuclesome)

The following diagram is related to mitochondrial electron transport discussed in class. In oxidative phosphorylation, the so-called P:O ratio (number of ATP molecules produced per oxygen atom consumed) is an important parameter. Assuming that the F0 region (embedded in the membrane) of ATP synthase in mitochondria has 10 identical subunits (not counting subunit a), what is the P:O ratio for NADH and for FADH2?

1. Flow of H+ drives F0 region of ATP synthase to turn in the membrane. The passing of each H+ allows one subunit to turn one notch. Thus, it takes 10 H+ to make a 360 turn for F0.
2. F0 is connected with gamma subunit of the 'head' F1 where ATP is synthesized. Each full turn of F1 makes thee ATP molecules.
3. Therefore, every ten H+ drives the synthesis of 3 ATP molecules
4. NADH delivers 2 electrons to complex I in the diagram and electrons pass through complexes II, III, and IV and eventually are accepted by one oxygen to form water. Along the way, ten H+ are pumped to the intermembrane space. Thus, the P:O ratio for NADH is three ATP per O consumed.
5. FADH2 delivers 2 electrons to complex II in the diagram and electrons pass through complexes II and IV and are eventually accepted by one oxygen to form water. Along the way, 6 H+ are pumped to the intermembrane space. THus the P:O ratio for NADH is 3*6/10 = 1.8

Draw a diagram to illustrate the epinephrine triggered signaling process leading to an increase in blood sugar level. Your diagram should include information about steps occurring in and around plasma membrane. How does caffeine affect this signaling process? Use the diagram to explain steps in this process that are reversible. Also, circle the steps that amplify the signal. Explain why certain steps but not others, have an amplification effect.

Caffeine inhibits the cAMP degrading enzyme, phosphodiesterase (PDE) causing cAMP to accumulate, thus enhancing blood sugar levels.
Four reversible steps:
1) ligand (epinephrine) being associated (binding) or dissociated (not binding) with receptor
2) G-protein switching between 'on' (GTP bound form) and 'off' (GDP bound form) states
3) cAMP being synthesized by adenylyl cylase and destroyed by PDE
4) protein kinases and phosphatases phosphorylating and dephosphorylating target proteins
Any step involving an enzyme can amplify the signal because an enzyme molecule can catalyze the formation of numerous products

Second messengers can be dervied from special phospholipid molecules in the membrane. Draw a diagram how this can happen. Your answer should include the enzyme involved, names of second messengers, and unique properties of these second messengers.

diagram
IP3 is hydrophillic thus can move around in cytosol to affect processes in other cellular compartments. DAG is hydrophobic and moves in the membrane to affect other membrane-associated processes.

Based on what we have discussed in Bio2960, figure out how many ATPs are generated if a molecule of palmitate (16 carbons), a fatty acid enriched in palm oil, is fully metabolized.

1) palmitate with 16 carbons is converted to palmityl (acyl) CoA couple with the hydrolysis of one ATP to AMP -2 ATPs
2) Palmityl CoA forms 8 acetyl CoAs via 7 runs of Beta oxidation. Each run of B-oxidation generates 1 FADH2 and 1 NADH so the total is 7 FADH2 and 7 NADH. Each FADH2 produces 1.5 ATPs in oxidative phosphorylation and each NADH is worth 2.5 ATPs via the same process. Thus the total ATPs produced by Beta-oxidation is 7*(1.5 + 2.5) -2(or -1) = 26 (or 27)
3) Each acetyl CoA enters citric acid cycle to produce 3 NADHs, 1 FADH2 and 1 GTP (ATP). Thus, the 8 acetyl CoAs generated from palmityl CoA produce a total of 80 ATPs = 8[(2.53) + 1.5 + 1]
4) Total ATPs produced from the complete degradation of an palmitate is 26 (or 27) + 80 = 106 (107)

The following series of questions are related to acetyl CoA, D-glyceraldehyde-3-phosphate, pyruvate, and lactate.
A) Draw the structures of these four compounds (the detailed structure of CoA is not needed, however the complete structure of phosphate needs to be presented) and link (order) them, relative to glucose (i.e., from glucose to compound 1, compound 2, and other compounds, but no other information is needed).
B) B) If C3 (#3 carbon) of the glucose molecule is labeled with 14C, circle the C on each these four compounds which are likely to be labeled. (Use the structures you draw for (A) above. No explanation is necessary.)
C) Describe the important features in each link in (A) above (total of four links, each
could be multiple enzymatic steps) among these four compounds and glucose in terms of formation, consumption and changes of coenzymes and ATP (no structures and enzymes are needed, but your answer needs to include the quantitative information, e.g., how many ATP consumed/generated per molecule of compound X)

A) diagram
The acid oxygen can be either protonated, i.e., with a H, or dissociated as shown. 2 pts for each correct structure
Methyl group can be presented as CH3-Detailed phosphate structure is needed Acetyl CoA without S is OK
B) diagram
C) Glucose to D-glyceraldehyde 3 phosphate: Two ATPs are consumed per glucose metabolized.
-D-Glyceraldehyde 3-phosphate to pyruvate: Two ATPs and one NADH are formed per D-glyceraldehyde 3-phosphate metabolized to pyruvate.
-Pyruvate to lactate: One NADH is consumed (re-oxidized) per molecule of pyruvate reduced to lactate.
-Puruvate to acetyl CoA: One NADH is formed and one CoA-SH (or HS-CoA) is consumed per molecule of pyruvate converted to acetyl CoA.

All of these four compounds are also linked with other pathways/processes in cellular metabolism. List name of these pathways/processes that are linked to each of these compounds (details of each pathway are not needed)

-D-Glyceraldehyde-3-phosphate is linked to glycolysis, gluconeogenesis, glycerol (lipid) metabolism and Calvin Cycle (photosynthetic CO2 fixation; photosynthesis)
-Pyruvate is linked to glycolysis, gluconeogenesis, fermentation, amino acid (alanine) synthesis, and citric acid cycle (krebs or TCA cycle)
-Lactate is linked to glycolysis, gluconeogenesis, Cori cycle
-Acetyl CoA is linked to glycolysis (or pyruvate oxidation/decarboxylation), citric acid cycle, fatty acid synthesis, and fatty acid degradation oxidation

RuBP carboxylase is the most abundant protein on eart. Where in a chloroplast is this enzyme localized? Use chemical structures to illustrate the reaction it catalyzes when CO2 concentration is high. If CO2(14) is used, which C on teh glucose molecule will be labeled?

-Enzyme localized in stroma portion of chloroplasts
-if CO2(14) is used, the carboxylic C of 3 phosphoglyceric acid will be labeled, after reduction to form glyceraldehyde, two of which will be converted to a glucose with the aldehyde carbon being #3 and #4 on the glucose molecule. Thus #3 and #4 of glucose will be labeled

Describe the experiment in which scientists showed that a proton concentration gradient is sufficient for ATP synthesis. Cyanide is an inhibitor to the last component in the electron transport system. What would happen if cyanide is added in this experimental system? What would happen if DNP (dinitrophenol) is added to the system?

-Experiment: Equilibrate isolated mitochondria with pH 8 buffer, then very quickly place them in pH 4 buffer to generate an artificial [H+] gradient, leading to synthesis of ATP.
-Cyanide will not affect this experiment because no electron transport takes place (proton concentration gradient in this experiment is not formed via electron transport system).
-DNP is an uncoupler causing dissipation of the pH gradient, thus the [H+] concentration gradient driven ATP synthesis will still be inhibited by DNP

Location of pyruvate decarboxylation

matrix of mitochondria

Location of citric acid cycle

matrix of mitochondria

Location of electron transport system

inner membrane of mitochondria

Location of High [H+]

intermembrane space of mitochondria

ATP (formed by oxidative phosphorylation)

matrix of mitochondria

fatty acid degradation

matrix of mitochondria

fatty acid synthesis

the environment right outside the mitochondria

What is the role of helicase? Why is its activity very much dependent on ATP?

Helicase is the enzyme responsible for unwinding dsDNA to form ssdNA template during DNA replication. Since the two DNA strands are held tightly together by numerous H-bonds and hydrophobic interactions among bases, the energy released by the hydrolysis of ATP is needed for this reaction.

What type of chemical bond is formed by the action of DNA ligase?

Phospho(di)ester linkage (between 5' C of one deoxyribose and 3'C of the next deoxyribose

The Michaelis-Menton equation below is important in analyzing the kinetic properties of enzymatic reactions. v = Vmax[(S)/Km + (S)]
A) What are the two key parameters in this equation? What do they mean?
B) To make the estimation of these parameters more accurate, the Lineweaver-Burk double reciprocal plot has been devised baed on this equation. Draw a well-labled graph with solid line to show how you would use the double reciprocal plot to determine the value of these two key parameters. Include how you derive the equation to make it suitable for this double reciprocal plot analysis.
C) Use the graph above, but with a dashed line, to show the kinetics of the same enzymatic reaction in the presence o a competitive inhibitor.

A)
-Vmax is the maximal velocity for the enzymatic reaction (when substrate is in plenty supply)
-Km is the concentration of substrate when the reaction velocity reaches 50% (or one half) of Vmax. Km measures the affinity between enzyme and substrate.

B)
-Original equation v = Vmax [(S)/Km + (S)]
-take reciprocal for both sides:
1/v = (Km + (S))/(Vmax*(S))
-regroup right side:
1/v = Km/(Vmax x [S]) + [S]/(Vmax x [S])
-further rearrangement:
1/v = (Km/Vmax) x (1/[S]) + (1/Vmax)

-When the data are plotted (1/[S] vs 1/v), -a typical graph is shown below: When 1/[S] reaches zero (the [S] is infinite), the intercept on the 1/v axis equals to 1/Vmax.
-When 1/v reaches zero, the intercept on the 1/[S] axis is -1/Km.
-Therefore, the values of Vmax and Km can be more accurately estimated by the double reciprocal analysis.

C)
-A competitive inhibitor competes with the substrate for the same active site. Thus, when the [S] becomes very high, i.e., 1/[S] reaching zero, the maximal velocity is unchanged. (The intercept on the 1/v axis remains the same)

Compare and contrast E Coli polymerase I and III. Your answer should include the function, protein structure, associated activities, and processivity. Explain why the unique properties of polymerase I are ideal for the function of this enzyme.

chart
The function of DNA pol I is to destroy the RNA primer using 5' to 3' exonuclease act and replace it with DNA (DNA pol act). Since its processivity is low, it will work on short stretch of nucleic acid, which is ideal for replacing RNA primer with DNA without affect DNA beyond the primer.

Compare the differences between chloroplast non-cyclic phosphorylation and mitochondrial oxidative phosphorylation

chart

Draw a diagram of chloroplast and mitochondrion and label the important membranes and sub-organelle compartments. In the chloroplast diagram, indicate locations for chloroplast DNA and ribulose 1,5 biphosphate carboxylase. In the mitochondrial diagram indicate locations for ATP synthase and TCA cycle.

diagrams

Draw the chemical structure of pyruvate at pH7 in the box below. Pyruvate is an important metabolic crossroad in the cell. Use the arrows to highlight reactions using pyruvate as the substrate. Your answer should include cofactors, products, and significance of each reaction.

diagrams

Glyceraldehyde-3-phosphate (GAP and G3P) is also an important metabolic crossroads. Draw its structure within the box below and use it to explain how GAP is
A) further metabolized to produce ATP and NADH.
B) related to photosynthetic CO2 fixation in plants (when CO2 supply is not a problem) and
C) converted to a part of major lipids
Your answer should include
1) chemical structures of all intermediates involved and
2) names of co-factors needed

diagrams

The following sets of data are the activities of photophofuctokinase (PFK 1) under different fructose 6-phosphate or ATP concentrations in the presence of various concentrations of fructose 2,6 biphosphate. (F2, 6 BP)
A) Besides PFK 1, what is the other enzyme regulated by F2, 6BP synthesis? What pathways does F2, 6BP regulate?
B) What are the unique features of enzymes responsible for F2,6BP synthesis and degradation?
C) For fig (A): Explain the effect F2,6BP on the kinetic properties of PFK 1
D) For Fig (B): Explain
1) why the PFK 1 activity increases when ATP concentration increases form 0-0.5 mM
2) Why the PFK I activity decreases dramatically when ATP concentration is higher than 1 mM
3) What the significance is that F2,6BP can somehow alleviate the inhibition of PFK 1 by high concentrations of ATP

A) FBPase 1 (Fructose 1,6 biphosphatase) is inhibited by F2,6BP which inhibits gluconeogenesis but enhances glycolysis
B) The two enzymes of F2,6BP synthesis/degradation are PFK2 and FBPase 2. They are two different domains on the same polypeptide. The differential activities of these two enymes are regulated by protein phosphorylation/dephosphorylation
C) In Fig (A) F2,6BP appears to lower the Km, but not the Vmax of PFK 1; the Km decreases in the presence of F2,6BP, which means the enzyme has higher affinity toward its substrate (F6P) in the presence of F2,6BP.
D) In Fig (B)
1) the enzyme activity increases when ATP increases from 1 to 0.5 mM because ATP is is another substrate for PFK 1 (it is a kinase)
2) when ATP becomes very high, it signals the high energy state of the cell. HEnce there is a feedback allosteric inhibition of PFK 1 to reduce the rate of glycolysis so that less ATp is produced
3) THe level of F2, 6BP reflects the level of F6P the substrate of PFK 1. The activity of PFK 1 is a balance between the inhibitory effect by high ATP and the promotive effect by F2,6BP. Thus even when ATP is high, its inhibitory effect is reduced by the high F2,6BP

For a typical enzymatic reaction, double reciprocal plots were determined for three different enzyme concentrations.
A) Which of the following three families of curves would you expect to observe?
B) For figure A in question 73 which of the following families of curves fit best.

A) The middle set of curve fits because the rate of an enzymatic reaction is proportional to the amount of active enzyme as long as there is a sufficient supply of substrate. Thus higher amounts of enzyme lead to higher Vmax without affecting the affinity between enzyme and substrate (Km remains the same).
B) In Fig 4 in question 73 the three conditions have the same Vmax but different Km, thus the one of the right fits the data best.

We have learned long time ago that lipids contain much more energy than carbohydrates. For examp,e, stearic acid is a C(18) saturated fatty acid and glucose as you know has 6 carbons. However when a molecule of stearic acid and three molecules of glucose (both cases have a total of 18 carbons) are fully metabolized to generate CO2 the stearic acid case produces almost 40% more ATP than the glucose case. Without going into detailed calculations, explain how this could be the case baed on what you have learned in class.

Although glycolysis for sugar degradation and B-oxidation for fatty acid degradation each yields some energy, the bulk of energy for stearic acid and glucose degradation comes from acetyl CoA which enters TCA cycle to yield energy and CO2. One molecule of stearic acid (C18) generates 9 molecules of acetyl CoA via B-oxidation, but 3 molecules of glucose (3C(6)) only generate 6 molecules of acetyl CoA (32 acetyl CoA per glucose) via glycolysis/pyruvate oxidation. Thus stearic acid degradation produces much more energy than degradation of 3 glucose molecules.

Epinephrine enhances glucose release from liver cells. What would happen if the liver cells are mutated or treated with the following chemicals?
A) cAMP in the absence of epinephrine

cAMP is produced as a second messenger in epinephrine signaling process. Thus, treatment with cAMP even in the absence of epinephrine will still elicit the same downstream effect-enhancing glucose release from liver cells.

Epinephrine enhances glucose release from liver cells. What would happen if the liver cells are mutated or treated with the following chemicals?
B) GTP-y-S, a non-hydrolyzable GTP, after epinephrine is withdrawn

After epinephrine is withdrawn the GTP bound to the Galpha subunit is supposed to be hydrolyzed to generate GDP, which will turn off the signaling. However, treatment with the non-hydrolyzable GTP will keep Galpha active, thus glucose continues to be released by liver cells.

Epinephrine enhances glucose release from liver cells. What would happen if the liver cells are mutated or treated with the following chemicals?
C) A protein kinase A inhibitor in the presence of epinephrine

A protein kinase A inhibitor will break the signaling pathway. THus, glucose will not be released by the liver cells even under the influence of epinephrine.

Epinephrine enhances glucose release from liver cells. What would happen if the liver cells are mutated or treated with the following chemicals?
D) IP3 (inisitol triphosphate) in the absence of epinephrine

No effort, ie the release of glucose from liver cells will be reduced after epinephrine withdrawal because IP3 is not involved in this signaling pathway

Epinephrine enhances glucose release from liver cells. What would happen if the liver cells are mutated or treated with the following chemicals?
E) caffeine after epinephrine is withdrawn

caffeine is an inhibitor of phosphodiesterase that degrades cAMP. THus, treatment of caffeine will keep cAMP level high even after epinephrine is withdrawn, and liver cells will keep releasing glucose.

Epinephrine enhances glucose release from liver cells. What would happen if the liver cells are mutated or treated with the following chemicals?
F) A mutation in the catalytic subunit of protein kinase A rendering it incapable of binding to the regulatory subunit.

The catalytic subunits of protein kinase A are inactive if bound to the regulatory subunits. Normally, the binding of cAMP to the regulatory subunits will dissociate the catalytic subunits, allowing them to be active. Thus, a mutation of catalytic subunits making it not bound to regulatory subunits will keep the catalytic subunits active all the time. Hence, liver cells, of this mutant still release glucose even when there is no stimulation by epinephrine.

Draw a diagram to explain how the medicine Viagra induces blood flow. INclude information about physiology, cell types, various signaling molecules and specific effect of Viagra.

diagram
Viagra is an inhibitor of phosphodiesterase which degrades cGMp to GMP. Therefore, Viagra enhances the level of cGMP which in turn relaxes smooth muscle lining the blood vessel. The effect leads to higher blood flow.

2',3' dideoxynucleotide triphosphates have been used as DNA synthesis inhibitors. Draw the chemical structure of 2'3' dideoxyribose to explain how it could inhibit DNA synthesis

Since 3'-OH is needed to form an ester linkage with phosphate which is also linked to the next nucleotide the lack of ) in the 3' position of 2'3' dideoxyribose will terminate DNA synthesis because no more new deoxynucleotide can be added to the existing DNA

What types of molecular interactions are involved in stabilizing DNA double helix?

H-bonds between complementary bases and hydrophobic interactions among stacked bases.

A) Determine the percentage of all the bases in a double-stranded DNA that is 20% thymine.
B) The composition (in mole-fraction) of one of the strands of a double helical DNA molecule is [A] = 0.30 and [G] = 0.24. What can you say about [T] and [C] in the same strand? What can you say about the base composition in the complementary strand?
C) Assuming that the two double-stranded DNA molecules in parts A and B are the same length, which one of them has higher melting temperature? Why?
D) The bacterial virus M13 has a DNA genome that does not obey 'Chargaff's rules'. Can you provide an explanation for this observation?

A) [T] = 20% so [A] = 20%. The remaining 60% is equally divided by [G] and [C], so each is 30%.
B) same strand: [T] + [C] = .46 (mole-fraction)
complementary strand: [T]=.30, [C]=.24 and [A] + [G] = .46
C) The DNA in (A) is 40% in A/T pairing, and A/T pairing in the DNA in (B) is >60%. Since A/T pairing has only two H-bonds (vs 3 H-bonds for G/C pairing), it takes lower temperature to dissociate the A DNA than B DNA. Hence the A DNA has higher melting temperature.
D) The M13 genome is a single stranded DNA. It is not a complementary double-stranded DNA, thus no A/T and G/C pairings, hence does not follow the 'Chargaff's rules'

G-proteins are sometimes viewed as 'self regulating' switches. Explain how G-proteins are activated and how they are inactivated.

the G-protein cycle has two important state, the ON and OFF states. The On state G-protein has a GTP bound to it. After triggering downstream events, GTP is hydrolyzed to GDP and Pi, converting the G-protein to the OFF state. The OFF state can be turn back on by exchanging GDP with a new GTP. Thus, each time the ON state performs its function, it turns itself OFF at the same time due to the hydrolysis of GTP.

GTP-y-S is an analog of GTP, which cannot be hydrolyzed to GDp and Pi. What is the effect of GTP-y-S on the function of heterotrimeric G-protein?

The G-protein cycle has two important states, the ON state and the OFF state. The On state G-protein has a GTP bound to it. AFter triggering downstream events, GTP is hydrolyzed to form GDP and Pi, converting the G-protein to its off state. If GTP is replaced with non-hydrolyzable GTP-y-S, this On to Off conversion will not take place, and the system will remain on all the time.

G-proteins usually have the GTPase activity capable of hydrolyzing GTP to GDP +Pi. What would happen if this activity is lost?

This will keep the G-protein at the on state and the downstream event will continue even after the original signal is no longer present.

Lipid soluble hormones, such as testosterone, cross plasma membranes of all cells but affect only certain target cells. Why?

Although testosterone can cross plasma membrane and enter many different cell types, only cells with specific (cytosolic) receptors capable of recognizing this hormone can respond to its presence.

cAMP is a second messenger for many functions in many cells. How can the same messenger act in different ways in different cells?

the cAMP responding system is different in different cells, such as different types of kinases responding to cAMP in different cells, thus cAMP can elicit different responses in difference cells. Even in the same cell, there could be different responses to cAMP. For example, cAMP in liver cells decreases the activity of glycogen synthase, but at the same time enhances the activity of protein kinase A, eventually leading ot glycogen degradation. In other cells, cAMP is known to regulate ion channel activities.

What are the mechanisms of caffeine in heightening alertness?

1) structure of caffeine similar to adenosine (adenine linked to ribose), thus it blocks the adenosine effect on reducing arousal.
2) Caffeine is an inhibitor of phosphodiesterase (PDE) thus causing cGMP levels to increase, leading to smooth muscle relaxation and dilation of blood vessels.
3) Caffeine is an inhibitor of phosphodiesterase (PDE), thus also causing cAMP level to increase, leading to increase in blood sugar level.

Both IP3 and DAG are derived from the same original phospholipid molecule but why IP3travels fast in the cytosol and DAG travels fast in the membrane?

IP3 is highly hydrophillic thus can travel fast in aqueous environment of cytosol. DAG is hydrophobic thus moves preferentially in the membrane.

Some protein kinases are inactive unless they are phosphorylated on key serine and threonine residues. In some cases, active enzymes can be generated by mutating these serine or threonine residues to aspartate or glutamate. Propose and explanation.

When serine is phosphorylated, its R-group resembles that of an aspartic acid. Similarly, a phosphorylated threonine is similar to glutamic acid. Thus, by replacing serine or threonine to aspartic acid or glutamic acid through mutation, respectively, one can make some protein kinases active all the time.

Proteins are thermodynamically unstable and deltaG for protein hydrolysis is very negative, yet proteins can be quite stable. How?

The activation energy for protein hydrolysis is high, thus this hydrolysis reaction cannot take place in the absence of an enzyme.

An allosteric enzyme usually has two forms, active and inactive. If a competitive inhibitor (NOT an allosteric inhibitor) is added to a solution containing this enzyme, the ratio of active form of this enzyme to inactive form actually increases. Explain how this can happen.

An allosteric enzyme has two forms: an active form is maintained when the active site is occupied, and an inactive form is maintained when the regulatory site is occupied by an allosteric inhibitor. A non-allosteric competitive inhibitor competes against the substrate for the same active site, thus it actually helps maintain the 'active' form.

What are the two most important parameters in enzyme kinetics? Which of these two is likely affected by competitive inhibition?

Vmax and Km are the two most important parameters in enzyme kinetics. Km is likely affected by competitive inhibition while Vmax is NOT likely to be affected (Vmax is the V0 when [S] is very high, thus S can out compete the inhibitor)

Each of the following molecules is processed by glycolysis to pyruvate. How many ATP molecules of ATP are generated from each of them?
A) Glucose 6 phosphate
B) glyceraldehyde 3-phosphate
C) sucrose

A) G6P will be converted to F1,6 BP consuming one ATP, but the two GAP produced from splitting F1,6BP will each produce two ATPs when they are converted to pyruvate. Thus, the total ATP produced is 2 X 2 -1 = 3.
B) One glyceraldehyde 3-phosphate produces two ATP when it is converted to pyruvate.
C) A sucrose molecule is composed of one glucose and one fructose, and each of them produces 2 ATPs when converted to pyruvate (in glycolysis). Thus, the total number of ATP produced is 2 X 2 = 4.

Explain how a biochemical compound could lose or gain 1 C (+ or - one C). Mention an example in your answer.

- one C: alpha-keot acids tend to lose the carboxylic group as CO2 (decarboxylation)
+ one C: the carboxylation reaction converting CO2 to a new carboxylic group, a process usually dependent on the vitamin, biotin

Gluconeogenesis can be considered as the reversal of glycolysis except two important steps. What are these two steps?

1) the conversion between F6P and F1,6BP (F6P to F1,6BP catalyzed by PFK for glycolysis and F1,6BP to F6P catalyzed by FBPase for gluconeogenesis
2) the conversion between PEP (phosphoenolpyruvate) and pyruvate (PEP to pyruvate catalyzed by pyruvate kinase and pyruvate to PEP in two steps

The following questions are related to the sequence of biochemical reactions: pyruvate>oxaloacetate (OAA)>phosphoenolpyruvate (PEP)
A) In what process does this sequence take place?

A) This is the beginning of gluconeogenesis. Alternatively, the pyruvate to OAA conversion is a way for cell to replenish TCA cycle intermediates if for whatever reason they are in short supply.

The following questions are related to the sequence of biochemical reactions: pyruvate>oxaloacetate (OAA)>phosphoenolpyruvate (PEP)
B) Where in the cell does this sequence take place?

B) Pyruvate to OAA takes palce in the matrix of mitochondria. OAA to PEP takes place in cytosol.

The following questions are related to the sequence of biochemical reactions: pyruvate>oxaloacetate (OAA)>phosphoenolpyruvate (PEP)
C) How many ATP equivalents are needed for this sequence?

C) Two, one for each step

The following questions are related to the sequence of biochemical reactions: pyruvate>oxaloacetate (OAA)>phosphoenolpyruvate (PEP)
D) Which of these two steps require the vitamin biotin?

The carboxylation between pyruvate and OAA

How many ATP molecules and how many NADH molecules are needed to convert two pyruvate molecules to one glucose molecule?

A) Pyruvate to OAA to PEP needs two ATPs for each pyruvate.
B) 3PGA to GAP needs one NADH each and one ATP each.
C) F1,6 bisP to F6P does NOT generate ATP.
D) F6P to G6P to glucose does NOT generate ATP.
For ATP: a) + b) + c) + d) = 2 X 2 + 2 X 1 + 0 + 0 = 6 ATPs
For NADH: 2 NADHs are needed.

What is the purpose of the Cori cycle?

The Cori cycle is needed to
1) coordinate the supply of glucose from liver with the utilization of glucose in muscle, and
2) recycle the lactic acid produced in muscle back to glucose/glycogen in liver. Basically, the Cori cycle has two halves, one in liver and the other in muscle with blood circulation connecting these two halves. In liver, glycogen is broken down to generate glucose, which is then transported to muscle to provide energy for muscle contractions. Due to the anaerobic condition generated after prolonged exercise, muscle cells accumulate a large amount of lactic acid that is transported back to liver where lactic acid is converted to glucose and eventually to glycogen via gluconeogenesis.

In the 1930's, the compound 2,4-dinitrophenol (DNP) was promoted and used as a diet aid. The fat-soluble molecule acts a proton ionophore that binds protons on one side of a membrane and drifts to the opposite side where it loses the protons. DNP very effectively leads to increased metabolic rate and can be fatal. Deaths from DNP overdoses are associated with rises in body temperature. The US government banned the use of DNP as a drug in 1938. Explain the mechanism for DNP action - in other words, why does the drug cause an increased metabolic rate, body temperature and death?

DNP is an "uncoupling" agent that disconnects the mitochondrial electron transport system from ATP production. DNP dissipates the electron-motive force generated by the electron transport chain of respiration in the mitochondria; consequently, the rate of ATP production is reduced. The free energy released by diffusion of protons back into the matrix due to the presence of DNP is released as heat, causing the spike in body temperature. Death by DNP overdoses are likely to be caused by hyperthermia, but a severe reduction of ATP production will also cause problems at the cellular level. Metabolic rates rise because of the mobilization of energy stores - sugars and fats are oxidized at a higher rate to try to compensate for the dissipation of the proton-motive force in the mitochondria.

Fast growing cancer cells with limited blood vessel development could have glycolysis rate as much as 200 times higher than normal cells. Explain why.

In order to provide the energy needed for fast growth, cancer cells have to produce ATP even at the time oxygen supplies because blood vessel development sometimes cannot catch up with the growth of cancer cells (local hypoxia). The elevated glycolysis allows cancer cells to produce ATP even when there is not sufficient supply of oxygen to support mitochondrial activities.

In which sub-mitochondrial compartment do you expect to find enzymes for TCA (citric acid) cycle, high [H+], and DNA?

TCA (citric acid) cycle and DNA are in matrix, the inner most compartment in mitochondria.
High [H+] is present in the intermembrane space between outer and inner membranes.

If you feed the cell with a 14C labeled glucose labeled with 14C at the fourth carbon. When do you expect this 14C is released as 14CO2? Do you expect any of the Citric Acid Cycle intermediates to be labeled? Explain your answer.

The 4th C on a glucose molecule will become the 1st C on a glyceraldehyde-3- phosphate molecule when the six-C fructose 1,6 BP is split into two C3 molecules (the lower half of the fructose, i.e., #4, #5 and #6 C, becomes glyceraldehyde 3-phosphate). Thus, the 4th C on glucose becomes the 1st C on the pyruvate molecule, the ending point of glycolysis. This 1st C is released as CO2 when pyruvate is oxidized to become acetyl CoA. Thus, acetyl CoA is not labeled, hence no citric acid cycle intermediates will be labeled.

Hibernating mammals (and newborns) keep warm using a tissue called "brown fat," which is rich in mitochondria. This tissue uses "uncoupling" proteins to generate heat. What is being uncoupled and what is the mechanism by which these proteins work?

Uncoupling the oxidation of fuel molecules (and subsequent electron transport) from ATP production by dissipating the proton-motive force. Uncoupling proteins allow protons to pass from the mitochondrial intermembrane space into the matrix of the mitochondria, thereby bypassing the ATP synthase complex. The energy from this proton flow is released as heat.

Explain why most fatty acids contain even-numbered carbon chain.

Because the building block of a fatty acid is the two-carbon acetyl CoA.

What is energy charge? Is glycolysis favored at high or low energy charge?

Energy charge the ratio of ([ATP] + 1⁄2 [ADP])/ [ATP] + [ADP] + [AMP] Glycolysis will be favored when energy charge is low signaling a need to increase ATP formation.

What is the function of fructose 2,6 bisphosphate in regulating glycolysis

Epinephrine triggers the synthesis of cAMP, which activates PKA (see signaling pathway of epinephrine). As shown in the diagram above, PKA phosphorylates PFK2-FBPase2 knocking out PFK2 activity, hence reducing the level of F2,6BP, leading to slowdown in glycolysis and enhancement in gluconeogenesis for the synthesis of glucose.
F2,6BP is a regulatory molecule activating PFK, thus speeding up glycolysis, and also reducing the inhibitory effect of ATP on PFK. F2,6BP also inhibits the activity of fructose 1,6-bisphosphatase, a key enzyme in gluconeogenesis.

What are the unique features of phosphofructokinase 2 (PFK2) and fructose 2,6 bisphosphate phosphatase 2 (FBPase2). [Please note these enzymes are for fructose 2,6-bisphosphate metabolism, while the substrate/product for glycolysis/gluconeogenesis is fructose 6-phosphate and fructose 1,6- bisphosphate]

These are two opposing enzymatic activities located in two different domains in the same polypeptide chain. PFK2 is responsible for the synthesis of F2,6BP, while FBPase2 catalyzes the hydrolysis of F2,6BP to F6P. Phosphorylation and de-phosphorylation regulates the differential activities of these two enzymes (as shown in the diagram above).

explain why higher activity of FBPase2, an enzyme converting fructose 2,6 bisphosphate to fructose 6 phosphate, reduces glycolysis.

Higher activity of FBPase2 will reduce the level of F2,6BP, which in turn cannot enhance PFK activity needed for the production for F1,6BP, a committal step for glycolysis.

Explain how glucagon, whose physiological role is similar to that of epinephrine, leads to higher level of glucose even not counting on glycogen being hydrolyzed.

Glucagon triggers the synthesis of cAMP, which activates PKA (see signaling pathway of epinephrine). As shown in the diagram above, PKA phosphorylates PFK2-FBPase2 knocking out PFK2 activity, hence reducing the level of F2,6BP, leading to slowdown in glycolysis and enhancement in gluconeogenesis for the synthesis of glucose.

Explain how high levels of fructose 6 phosphate inhibit glyconeogenesis.

High levels of F6P lead to the formation of the regulator F2,6BP which in turns inhibits F1,6 BPase, a key enzyme in gluconeogenesis.

What are the likely consequences of a genetic disorder rendering fructose 1, 6 bisphosphatase less sensitive to regulation by fructose 2,6 bisphosphate?

If fructose 1, 6 bisphosphatase is less sensitive to the inhibition by F2,6BP but PFK is still activated by F2,6BP, fructose 1,6 bisphosphate tends to be converted back to F6P even when F2,6BP level is high. This generates a futile cycle to burn up ATP, not a very healthy situation (See diagram below—the inter- conversion between F6P and F1,6BP takes place all the time [back and forth between them and causes the hydrolysis of ATP each time] ).

If water labeled with 18O is added to a suspension of photosynthesizing chloroplasts, which of the following compounds will first become labeled with 18O: ATP, NADPH, O2, or 3PGA (3-phosphoglyceric acid which can be reduced to form G3P [GAP])? If water labeled with 3H is added to a suspension of photosynthesizing chloroplasts, which of the same compounds will first become labeled with 3H? If CO2 labeled with 14C is added, which of the same compounds will first become labeled with 14C?

18O first shows up in O2. 3H first shows up in NADPH. 14C first shows up in 3 PGA (3-phosphoglyceric acid or 3-phosphoglycerate or glycerate 3 phosphate [though the last version is rarely used].

What is the expected reaction catalyzed by rubisco (RuBP carboxylase)? What are its problems? What modifications of photosynthesis could alleviate most of these problems?

Rubisco (RuBP carboxylase) catalyzes the carboxylation of ribulose 1,5- bisphosphate (RuBP) converting it to two molecules of 3 PG (3- phosphoglyceric acid) that are then reduced to form the simplest sugar, glyceralhehyde 3 phosphate (G3P). However, rubisco can also reacts with oxygen that is quite abundant in air (20%) causing the breakdown of RuBP to form a C2 compound plus a 3PG. Two of these C2 compounds are then condensed to form a C3 compound and release a CO2. This wasteful process is more apparent when plants are grown in warm climate. To reduce this waste, some plants have evolved to carry out C4 photosynthesis in which two neighboring plant cells collaborate in
1) concentrating CO2 by one cell, and
2) carrying out conventional Calvin cycle CO2 fixation in another cell. The first cell uses another enzyme, PEP carboxylase, to convert CO2 and a C3 compound, PEP, to form a C4 compound that is then transported to the second cell where C4 is converted back to C3 , releasing CO2. This efficient CO2 concentrating mechanism allows the second cell to accumulate enough CO2 for rubisco to function as a carboxylase and for the conventional Calvin cycle to take place (see diagram below).

In which sub-mitochondrial compartment do you expect to find enzymes for TCA (citric acid) cycle, pyruvate dehydrogenase, β-oxidation of fatty acid, high [H+], and DNA?

TCA (citric acid) cycle, pyruvae dehydrogenase, and β-oxidation and DNA are in matrix, the inner most compartment in mitochondria.
High [H+] is present in the intermembrane space between outer and inner membranes.

Why can't animals convert fats to sugars while plants can?

The bulk of fats are triacylglycerides that can be hydrolyzed to glycerol and fatty acids. Glycerol can easily enter glycolysis after one step of oxidation (to glyceraldehyde), and fatty acids have to be degraded to acetyl CoA first. Acetyl CoA can enter TCA cycle to be oxidized to CO2 and the process generates a lot of energy, but no net gain of C atoms. Thus, animals cannot covert acetyl CoA (derived from fatty acid degradation) to sugars. In plants and some microorganisms, a different pathway, the glyoxylate pathway, allows acetyl CoA to be condensed with OAA (C4) to form the C6 citrate, which then bypasses the two CO2 generating steps, and is converted to a C4 intermediate and the C2 glyoxylate. Glyoxylate is then condensed to form a C4 compound that is eventually converted to the C4 OAA. The output of this glyoxylate cycle, a C4 compound, is then converted to sugar via gluconeogenesis.

After receiving the DNA you have designed, you put these two DNA samples in the same solution, heat them up, and then gradually let the solution cool off to allow the two complementary strands to find each other so that double-stranded DNA is formed (the "annealing" process). What would happen to these DNAs if their concentrations in the solution are very low?

If the concentrations of these DNAs are very low, they tend to anneal (forming complementary double strands) with itself (intramolecular hybrid) like the following diagram.

Suppose that you want to radioactively label DNA but not RNA in dividing and growing bacterial cells, which radioactive molecule would you add to the culture medium?

Labeled thymidine or thymine. {Deoxyribose is not the best choice because it could be converted to ribose}

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