Biochem Exam 3

Question 1

What are the products of glycolysis?

Answer 1

2 Pyruvate, 2 ATP (net), 2 NADH, 2 H₂O

Question 2

Where does glycolysis occur?

Answer 2

The cytoplasm

Question 3

Energy Investment Phase

Answer 3

In this phase, the cell uses up 2 ATP to “activate” the glucose intro 2 triose phosphates.

Question 4

Energy Generation Phase

Answer 4

In this phase, the 6C molecule is split into two 3C molecules (G3P), and each goes through the same reactions — so everything here happens twice per glucose.

Each G3P yields:
2 ATP (so 2 × 2 = 4 total ATP)
1 NADH (so 2 × 1 = 2 total NADH)

Question 5

Name the 10 steps of glycolysis and the enzymes

Answer 5

1. Phosphorylation of Glucose
   (Hexokinase)

Low Km (high affinity); liver uses glucokinase (hexokinase IV) with high Km so it responds to high blood glucose.

2. Isomerization of Glucose-6-Phosphate (Phosphoglucose isomerase)
3. Phosphorylation of Fructose-6-Phosphate (Phosphofructokinase-1 (PFK-1)

Allosterically regulated; main control point. Rate-limiting.

4. Cleavage of Fructose-1,6-bisphosphate (Aldolase)

Breaks 6C sugar into 2 3C sugars.

5. Isomerization of Dihydroxyacetone Phosphate (Triose phosphate isomerase)
6. Oxidative Phosphorylation of Glyceraldehyde-3-phosphate (GAPDH)

Forms NADH + high-energy intermediate (1,3-BPG).

7. Phosphorylation Transfer to ATP (Phosphoglycerate kinase)
8. Phosphorolization of 3-phosphoglycerate (Phosphoglycerate mutase)
9. Dehydration of 2-phosphoglycerate (Enolase)
10. Phosphorylation Transfer to ATP (Pyruvate kinase)

Question 6

How many NADH molecules are made per glucose in glycolysis?

Answer 6

2

Question 7

Which enzyme catalyzes the first step of glycolysis?

Answer 7

Hexokinase

Question 8

What molecule is produced at the end of glycolysis?

Answer 8

Pyruvate

Question 9

What are the total (not net) ATP molecules produced in glycolysis?

Answer 9

4

Question 10

Which molecule is oxidized to produce NADH?

Answer 10

G3P

Question 11

The fourth reaction involving Gly-4 ( aldolase) is an especially important reaction in glycolysis. Why is this?

Answer 11

The fourth reaction of glycolysis is especially important because it splits a six-carbon sugar into two three-carbon molecules: dihydroxyacetone phosphate (DHAP) and glyceraldehyde-3-phosphate (G3P).

Question 12

PYRUVATE OXIDATION enzyme

Answer 12

Pyruvate dehydrogenase complex

Pyruvate (3C) → Acetyl-CoA (2C)

Produces:
1 CO₂
1 NADH
1 Acetyl-CoA

Question 13

Pyruvate dehydrogenase catalyzes three reactions

Answer 13

1) Oxidation of pyruvate’s carboxyl group, releasing the first CO2 of cellular respiration

2) Reduction of NAD+ to NADH

3) Combination of the remaining two-carbon fragment with coenzyme A to form acetyl CoA

Question 14

What are the 8 steps and enzymes of TCA?

Answer 14

1) Acetyl CoA (2C) + Oxaloacetate (4C) → Citrate (6C)

Citrate synthase

2) Citrate is converted to → Isocitrate

Aconitase

3) Isocitrate → α-Ketoglutarate

Isocitrate dehydrogenase

4) α-Ketoglutarate → Succinyl-CoA

α-Ketoglutarate dehydrogenase

5) Succinyl-CoA → Succinate + ATP (or GTP)

Succinyl-CoA Synthetase

6) Succinate oxidized → Fumarate + FADH₂

Succinate Dehydrogenase

7) Fumarate hydrated → Malate

Fumarase

8) Malate → Oxaloacetate

Malate Dehydrogenase

Question 15

Overall Products (per 1 Acetyl CoA)

Answer 15

3 NADH
1 FADH₂
1 ATP (or GTP)
2 CO₂ (byproduct)

Question 16

Since 1 glucose gives 2 pyruvates, everything doubles per glucose molecule in TCA

Answer 16

6 NADH
2 FADH₂
2 ATP (or GTP)
4 CO₂

Question 17

Among the products of glycolysis, which compounds contain energy that can be used by other biological reactions?

Answer 17

ATP
NADH
Pyruvate

Question 18

Enzymes

Answer 18

Biological catalysts that serve to speed up the rate of a biochemical reaction without being affected by the reaction.

Question 19

Induced fit model

Answer 19

The enzyme is induced to undergo a slight alteration to achieve optimum fit for the substrates.

Question 20

Oxidoreductase

Answer 20

Catalyzes the transfer of electrons between molecules.

Oxidases, reductases, peroxidases, and oxygenases are all oxidoreductases. These enzymes catalyze redox reactions (transfer of electrons).

Question 21

Hydrolase

Answer 21

Catalyzes the breaking of a molecule by adding water. Ex: An enzyme that breaks a bond between two nucleotides by adding water.

Question 22

Lyase

Answer 22

Catalyzes the breaking of a molecule without the use of water. Ex: An enzyme that breaks a bond between two nucleotides without using water.

Question 23

Ligase

Answer 23

Catalyzes the joining of two molecules. Ex: An enzyme that seals the gap between two adjacent Okazaki fragments.

Question 24

Isomerase

Answer 24

Catalyzes an isomerization reaction, which is an intramolecular rearrangement of bonds in a molecule. Ex: An enzyme that converts a cis double bond into a trans double bond.

Question 25

Prosthetic group

Answer 25

The cofactor or coenzyme is bound extremely tightly to the enzyme.

Question 26

Apoenzymes

Answer 26

Enzymes are alone

Question 27

Enzyme functions

Answer 27

1) Lower activation energy of reaction 2) Affect the kinetics of the reaction, but not the thermodynamics or equilibrium constant. 3) Regeneration

Question 28

Vmax and Km

Answer 28

Vmax: The maximum velocity of the reaction when the enzyme is saturated with substrate. Km: The substrate concentration at which the reaction rate is half of Vmax. Low Km = high affinity (enzyme grabs onto substrate easily). High Km = low affinity (substrate needs to be at a higher concentration to get the enzyme to work).

Question 29

Competitive inhibition

Answer 29

Inhibitor competes for the active site → substrate needs to compete harder = higher Km. But if you add enough substrate, you still reach the same Vmax. Graph clue: Lines cross on the Y-axis (Vmax same).

Question 30

Noncompetitive inhibition

Answer 30

The inhibitor does not bind at the active site, but at an allosteric (allo- means other) site. A change in shape initiated by an inhibitor binding to the allosteric site changes the shape of the active site, making it unable to bind to its substrate. Km: Unchanged (affinity of enzyme for substrate stays the same) Vmax: Decreased (maximum rate drops because fewer working enzymes) Like turning off a blender (enzyme) regardless of whether fruit (substrate) is inside — it can’t blend either way. Graph clue: Lines cross on the X-axis (Km same).

Question 31

Uncompetitive inhibition

Answer 31

Similar to non-competitive inhibitors but, the inhibitor only bind to the enzyme when substrate is bound to the enzyme. Both, the apparent KM and apparent Vmax, are decreased. ↓Km, ↓Vmax → inhibitor binds only to ES complex. Inhibitor only binds after substrate is bound → locks enzyme in inactive form = both Vmax and Km go down. Graph clue: Lines are parallel on Lineweaver-Burk plots.

Question 32

In the Lineweaver-Burk plot of an enzyme reaction, the Km is given by the ________. In the Lineweaver-Burk plot (also called a double reciprocal plot) of an enzyme reaction, the KMKM​ is given by the:

Answer 32

Negative reciprocal of the x-intercept.

Question 33

Describes the complexes

Answer 33

Complex I (NADH–CoQ Reductase): Electrons from NADH → FMN → Fe-S clusters → CoQ ✅ Pumps protons Complex II (Succinate–CoQ Reductase): Electrons from succinate → FAD → Fe-S → CoQ ❌ No protons pumped CoQ (Ubiquinone): Carries electrons from I & II → Complex III Complex III (CoQ–Cytochrome c Reductase): Transfers electrons from CoQH₂ → cytochrome c ✅ Pumps protons Cytochrome c: Carries electrons from III → IV Complex IV (Cytochrome c Oxidase): Transfers electrons from cytochrome c → O₂ → H₂O ✅ Pumps protons

Question 34

pH Gradient

Answer 34

Intermembrane space: acidic (low pH), high H+ Matrix: basic (high pH) ~1 pH unit difference = 10× difference in [H⁺] H⁺ flows back into the matrix through ATP synthase. This flow powers the conversion of ADP + Pi → ATP.

Question 35

Oxidative Phosphorylation Overview

Answer 35

Where: Inner mitochondrial membrane Purpose: Use electrons from NADH and FADH₂ to pump protons and generate ATP

Question 36

Electron Donors

Answer 36

NADH → Complex I (pumps protons) FADH₂ → Complex II (does not pump protons)

Question 37

ATP Yield

Answer 37

Each NADH → ~2.5 ATP Each FADH₂ → ~1.5 ATP Total from 1 glucose: Glycolysis: 2 ATP TCA: 2 ATP ETC: ~34 ATP → ~38 ATP total

Question 38

What is the difference in pH levels between the mitochondrial membrane and the inter membrane space during oxidative phosphorylation? Please briefly explain why.

Answer 38

The intermembrane space has a lower pH (more acidic). The matrix has a higher pH (more basic). As electrons pass through the electron transport chain (ETC) in the inner mitochondrial membrane, protons (H⁺) are pumped from the matrix into the intermembrane space by complexes I, III, and IV. This creates a proton gradient.

Question 39

The final electron acceptor in the ETC is __________.

Answer 39

Oxygen

Question 40

The electrochemical gradient generated by proton pumping is called the ____________.

Answer 40

Proton motive force

Question 41

Why does FADH₂ yield less ATP than NADH?

Answer 41

FADH₂ donates its electrons to Complex II, which does not pump protons. This means fewer protons are pumped across the membrane → less proton motive force → less ATP

Question 42

What is the role of ATP synthase in oxidative phosphorylation?

Answer 42

ATP synthase (Complex V) uses the energy of the proton gradient to catalyze the formation of ATP from ADP and Pi. As protons flow down their gradient through F₀, the γ subunit rotates, causing conformational changes in F₁ that drive ATP synthesis.

Question 43

What are the three conformational states of the αβ dimers in the F₁ portion of ATP synthase, and what does each one do?

Answer 43

L (Loose): Binds ADP + Pi T (Tight): Catalyzes the formation of ATP O (Open): Releases ATP

Question 44

What would happen to oxidative phosphorylation if there were no oxygen available?

Answer 44

Without oxygen, electrons can't be passed to the final acceptor (O₂) → the ETC backs up → no proton gradient is generated → ATP synthase stops → no ATP from oxidative phosphorylation.