Biochem Exam 2

What does 18:2(Δ9,12) indicate? A) 18 carbons, one double bond with a 12-carbon tail B) 18 carbons with double bonds at C9 and C12 counted from the methyl end C) 18 carbons with two C=C double bonds, at C9 and C12 counted from the carboxyl carbon D) 2 carbons with 18 double bonds

C) 18 carbons with two C=C double bonds, at C9 and C12 counted from the carboxyl carbon. The Δ number gives the position counted from the carboxyl carbon (C1); the number after the colon gives how many C=C bonds.

Which change would lower a fatty acid's melting point? A) Lengthening the chain B) Removing a double bond C) Increasing van der Waals contact D) Adding a cis double bond

D) Adding a cis double bond. Cis double bonds create kinks that pack poorly. Longer chains and more saturation raise melting point.

Which bond joins the head group X to glycerol 3-phosphate in a glycerophospholipid? A) Ester B) Phosphodiester C) Thioester D) Glycosidic

B) Phosphodiester. Tails are esters at C1 and C2, the head group is attached through a phosphodiester bond.

A phospholipid with only H as the head group has what net charge? A) +1 B) -1 C) -1.5 D) 0

C) -1.5. Oxygen partial charges on the phosphate give a net charge of about -1.5.

A protein is released from a membrane by high salt or high pH. It is: A) A lipid-anchored integral protein B) Peripheral C) Integral D) A β-barrel protein

B) Peripheral. Peripheral proteins release with mild treatments. Integral proteins need detergent.

Why do hydropathy plots miss β-barrel membrane proteins? A) They are too short to span the membrane B) They are covalently attached to lipids C) Alternating residues face the lipid, so there's no long hydrophobic stretch D) They have no hydrophobic residues

C) Alternating residues face the lipid, so there's no long hydrophobic stretch. β strands alternate residue orientation, unlike the ~20 consecutive hydrophobic residues of an α-helix TM segment.

A typical α-helical TM segment is about 20 residues because: A) It must fit in the ~3 Å selectivity filter B) It must reach the cytosol and extracellular space C) Each residue is 20 Å D) It spans the ~30 Å hydrophobic core

D) It spans the ~30 Å hydrophobic core. An α-helix rises about 1.5 Å per residue, so ~20 residues span ~30 Å.

Which is true of lipid movement in a bilayer? A) Both are fast B) Lateral diffusion is fast; flip-flop is slow unless catalyzed C) Both are slow without translocators D) Flip-flop is fast; lateral diffusion is slow

B) Lateral diffusion is fast; flip-flop is slow unless catalyzed. Translocators are needed to speed flip-flop.

Moving one Na⁺ into a cell ([Na⁺]in 12 mM, [Na⁺]out 145 mM, Δψ = -50 mV, 37 °C) has ΔG closest to: A) -6 kJ/mol B) +11 kJ/mol C) -11 kJ/mol D) -22 kJ/mol

C) -11 kJ/mol. RT ln(12/145) ≈ -6.4 plus (96.5)(-0.050) ≈ -4.8 gives about -11.2 kJ/mol.

The ZFΔψ term is dropped for glucose because: A) Glucose has no concentration gradient B) F is zero for neutral molecules C) Δψ is zero for sugars D) Glucose is uncharged (Z = 0)

D) Glucose is uncharged (Z = 0). Z is the charge of the solute.

A 100-fold concentration ratio is worth about how much free energy at 37 °C (rule of six)? A) 60 kJ/mol B) 12 kJ/mol C) 6 kJ/mol D) 24 kJ/mol

B) 12 kJ/mol. Each 10-fold is ≈6 kJ/mol, so 100-fold ≈ 12.

A transporter such as GLUT1 changes which quantity? A) The concentration at equilibrium B) ΔG of transport C) The direction of net flow D) ΔG‡ (lowered)

D) ΔG‡ (lowered). Like an enzyme, it lowers the activation barrier but does not change ΔG or equilibrium.

On a double-reciprocal plot for GLUT1, the x-intercept equals: A) -Vmax B) Kt C) -1/Kt D) 1/Vmax

C) -1/Kt. The y-intercept is 1/Vmax, the x-intercept is -1/Kt.

The Na⁺-glucose symporter can theoretically concentrate glucose about: A) 60-fold B) 6-fold C) 6000-fold D) 600-fold

C) 6000-fold. Two Na⁺ give ΔG ≈ -22.4 kJ/mol, so e^(22.4/2.58) ≈ 5.9 × 10³.

The Na⁺K⁺ ATPase moves per ATP: A) 2 Na⁺ out, 3 K⁺ in B) 1 Na⁺ out, 1 K⁺ in C) 3 Na⁺ in, 2 K⁺ out D) 3 Na⁺ out, 2 K⁺ in

D) 3 Na⁺ out, 2 K⁺ in. This net +1 charge leaving makes it electrogenic.

How does the K⁺ channel pass K⁺ quickly yet selectively? A) Carbonyl oxygens replace K⁺'s hydration shell at no energetic cost B) K⁺ binds very tightly in the filter C) The filter is wide enough for hydrated K⁺ D) Na⁺ is too large to enter

A) Carbonyl oxygens replace K⁺'s hydration shell at no energetic cost. This resolves the paradox: selective but without tight binding. Na⁺ is smaller (0.95 Å) and holds its waters closer, so carbonyls are too far apart to replace them.

Which describes amplification in the β-adrenergic pathway? A) One receptor activates many Gs, producing many cAMP and active PKA B) GTP hydrolysis turns Gsα off C) One hormone binds one receptor with one effect D) cAMP is degraded to 5'-AMP

A) One receptor activates many Gs, producing many cAMP and active PKA. Enzymes activating enzymes increases the number of affected molecules geometrically. The others describe specificity or termination.

A GEF activates a G protein by: A) Binding the γ-phosphate B) Speeding GDP release so GTP loads C) Speeding GTP hydrolysis D) Phosphorylating the G protein

B) Speeding GDP release so GTP loads. GAPs do the opposite by speeding GTP hydrolysis.

PKA is activated by: A) Insulin receptor phosphorylation B) cAMP binding to the regulatory subunits C) GTP binding D) Phosphorylation by phosphorylase kinase

B) cAMP binding to the regulatory subunits. R2C2 + 4 cAMP → R2(cAMP)4 + 2 free active C subunits.

Which enzyme is the Ras GEF in insulin signaling? A) MEK B) Grb2 C) Raf-1 D) Sos

D) Sos. Grb2 docks to phosphorylated IRS-1 and brings in Sos, which causes Ras to release GDP and bind GTP.

In the insulin MAP kinase cascade, the correct order is: A) Ras → Raf-1 → MEK → ERK B) Raf-1 → Ras → ERK → MEK C) Ras → MEK → Raf-1 → ERK D) ERK → MEK → Raf-1 → Ras

A) Ras → Raf-1 → MEK → ERK. Raf-1 (MAPKKK) phosphorylates MEK (MAPKK), which phosphorylates ERK (MAPK).

What makes the insulin receptor different from the β-adrenergic receptor? A) It's a GPCR with seven helices B) It's a receptor tyrosine kinase, not a GPCR C) It lacks a signaling cascade D) It signals only through cAMP

B) It's a receptor tyrosine kinase, not a GPCR. The insulin receptor autophosphorylates Tyr residues.

Which are the two off switches in the β-adrenergic pathway? A) Isoproterenol and β-blockers B) Adenylyl cyclase and PKA C) GEF and Sos D) Gsα GTPase and cAMP phosphodiesterase

D) Gsα GTPase and cAMP phosphodiesterase. The G protein inactivates itself via intrinsic GTPase, and phosphodiesterase degrades cAMP to AMP.

A β-blocker is: A) An inhibitor of PKA B) An antagonist of the β-adrenergic receptor C) A GAP for Gsα D) An agonist with tighter binding than epinephrine

B) An antagonist of the β-adrenergic receptor. Isoproterenol is an agonist; β-blockers are antagonists.

During starvation (days 2-3), the brain increasingly uses: A) Fatty acids directly B) Ketone bodies C) Muscle glycogen D) Lactate from red blood cells

B) Ketone bodies. Ketone bodies, made by the liver from fatty acids, spare glucose and thus muscle protein.

Muscle ATP + ADP + AMP stays at 5.6 mM during exercise, but ATP:AMP changes from 50 to 8. This illustrates: A) Tissue diversity B) Compartmentalization C) Fuel sparing D) Constant totals, variable ratios

D) Constant totals, variable ratios. This is the sixth metabolic principle.

Which group is a nucleophile only when unprotonated? A) Lys –NH2 B) Carbonyl carbon C) Protonated imine carbon D) Phosphorus of a phosphate

A) Lys –NH2. Lys and His are nucleophiles only when unprotonated; the others are electrophiles.

Which is an electrophile? A) Carbon of a protonated imine (Schiff base) B) Uncharged His imidazole C) Hydroxide ion D) Carbanion

A) Carbon of a protonated imine (Schiff base). The protonated imine carbon is a strong electrophile and electron sink.

In aldol addition, the nucleophile and electrophile are: A) Water attacks a phosphate B) Carbonyl oxygen attacks a carbanion C) Amine attacks a proton D) Enol/enolate carbon attacks a carbonyl carbon

D) Enol/enolate carbon attacks a carbonyl carbon. The nucleophilic carbon is made so by deprotonation next to a carbonyl.

Which has the MOST negative ΔG′° of hydrolysis? A) 1,3-BPG (-49.3) B) PEP (-61.9) C) Acetyl-CoA (-31.4) D) ATP (-30.5)

B) PEP (-61.9). PEP > 1,3-BPG > acetyl-CoA > ATP, in kJ/mol of free energy released.

Why is PEP's hydrolysis so exergonic? A) Charge repulsion in PEP's phosphate B) PEP has two phosphoanhydride bonds C) PEP is a thioester D) The product pyruvate tautomerizes from enol to the more stable keto form

D) The product pyruvate tautomerizes from enol to the more stable keto form. Enol-to-keto tautomerization of the product is the key.

Why is acetyl-CoA a high-energy compound? A) Thioesters are less resonance-stabilized than oxygen esters, while the acetate product is stabilized B) It is an acyl phosphate C) It contains a phosphoanhydride D) It is a Schiff base

A) Thioesters are less resonance-stabilized than oxygen esters, while the acetate product is stabilized. Hydrolysis is more favorable than for an ester.

What does the ATP hydrolysis example give for ΔG in an erythrocyte ([ATP] 2.25 mM, [ADP] 0.25 mM, [Pi] 1.65 mM)? A) About +22 kJ/mol B) About -30.5 kJ/mol C) About -52 kJ/mol D) About -22 kJ/mol

C) About -52 kJ/mol. Q ≈ 1.8 × 10⁻⁴, RT ln Q ≈ -22, so ΔG = -30.5 + (-22) ≈ -52 kJ/mol.

ΔG = 0 occurs when: A) Q = 1 B) The reaction is at standard conditions C) Q = K′eq D) ΔG′° = 0

C) Q = K′eq. At equilibrium, Q equals K′eq and no net flux occurs.

Positive ΔE′° corresponds to: A) Positive ΔG′° B) ΔG′° = 0 C) Negative ΔG′° D) No electron transfer

C) Negative ΔG′°. ΔG′° = -nFΔE′°.

Ethanol + NAD⁺ → acetaldehyde + NADH (E′° NAD⁺/NADH = -0.320 V; acetaldehyde/ethanol = -0.197 V) has ΔG′° of: A) -23.7 kJ/mol B) -47.4 kJ/mol C) +23.7 kJ/mol D) +11.9 kJ/mol

C) +23.7 kJ/mol. ΔE′° = -0.123 V; ΔG′° = -(2)(96.5)(-0.123) = +23.7.

NADH can be tracked spectrophotometrically at: A) 340 nm B) 260 nm C) 280 nm D) 540 nm

A) 340 nm. NADH (not NAD⁺) absorbs at 340 nm.

NADPH is mainly used for: A) Substrate-level phosphorylation B) Collecting electrons from catabolism C) Carrying electrons in flavoproteins D) Reductive biosynthesis and keeping glutathione reduced

D) Reductive biosynthesis and keeping glutathione reduced. NADH is for catabolism; NADPH for anabolism, largely made by the PPP.

Which glycolytic step is the committed step? A) PFK-1 (step 3) B) Hexokinase (step 1) C) Aldolase (step 4) D) Pyruvate kinase (step 10)

A) PFK-1 (step 3). G6P has other fates (glycogen, PPP), so hexokinase can't commit.

Why is G6P isomerized to F6P in step 2? A) So aldolase can cleave C3-C4 into two equal phosphorylated 3-C pieces B) To release ATP C) To trap glucose in the cell D) To make the molecule more water soluble

A) So aldolase can cleave C3-C4 into two equal phosphorylated 3-C pieces. Moving the C=O to C2 places it next to C3.

Which glycolytic step directly consumes NAD⁺? A) GAP dehydrogenase B) Enolase C) Phosphoglycerate kinase D) Pyruvate kinase

A) GAP dehydrogenase. Step 6 oxidizes GAP's aldehyde and makes NADH.

Which two steps perform substrate-level phosphorylation? A) GAPDH and aldolase B) Enolase and mutase C) Phosphoglycerate kinase and pyruvate kinase D) Hexokinase and PFK-1

C) Phosphoglycerate kinase and pyruvate kinase. Both transfer a phosphoryl from a high-energy compound (1,3-BPG, PEP) to ADP.

How much net ATP does glycolysis yield per glucose? A) 38 B) 0 C) 2 D) 4

C) 2. 4 ATP are made minus 2 invested.

GAP dehydrogenase uses which catalytic residue? A) Active-site Glu as a general base B) Zn²⁺ C) Active-site Cys forming a thioester intermediate D) Active-site Lys forming a Schiff base

C) Active-site Cys forming a thioester intermediate. The thioester is high-energy and lets Pi form the acyl phosphate.

Human (class I) aldolase differs from bacterial aldolase in that it: A) Forms a Schiff base with an active-site Lys, giving an enamine B) Uses TPP C) Doesn't need a carbonyl D) Uses Zn²⁺ to stabilize an enolate

A) Forms a Schiff base with an active-site Lys, giving an enamine. Enamine is more stable than the enolate, lowering the transition state.

Pyruvate kinase becomes exergonic because: A) NAD⁺ is regenerated B) ATP is hydrolyzed C) A Schiff base forms D) Enol pyruvate tautomerizes to keto pyruvate

D) Enol pyruvate tautomerizes to keto pyruvate. The enol product is trapped by tautomerization to the stable keto form.

Under anaerobic conditions, fermentation exists to: A) Regenerate NAD⁺ so GAP dehydrogenase can continue B) Make additional ATP C) Dispose of NADPH D) Regenerate ADP

A) Regenerate NAD⁺ so GAP dehydrogenase can continue. Without NAD⁺, glycolysis stalls.

Net glucose → 2 lactate gives: A) 38 ATP B) 2 ATP and no net NADH C) 4 ATP D) 2 ATP and 2 NADH

B) 2 ATP and no net NADH. NADH from step 6 is reoxidized by LDH.

Which enzyme and cofactor are used by yeast to convert pyruvate to acetaldehyde? A) Pyruvate decarboxylase with TPP B) Alcohol dehydrogenase with NADH C) Pyruvate carboxylase with biotin D) LDH with NAD⁺

A) Pyruvate decarboxylase with TPP. Alcohol dehydrogenase then reduces acetaldehyde to ethanol.

Why does pyruvate decarboxylase need TPP while aldolase does not? A) It breaks the bond α to the carbonyl, so an electron sink is needed B) It makes C–C bonds C) It breaks a bond β to the carbonyl D) It hydrolyzes a Schiff base

A) It breaks the bond α to the carbonyl, so an electron sink is needed. Aldolase's bond is β to the carbonyl so the carbonyl absorbs the electrons.

Which cofactor vitamin gives TPP? A) B6 B) B3 (niacin) C) B2 (riboflavin) D) B1 (thiamine)

D) B1 (thiamine). The slide says TPP is derived from vitamin B1.

Where does gluconeogenesis occur in humans? A) Muscle B) Brain C) Red blood cells D) Liver

D) Liver. Muscle and brain don't perform it.

Which is NOT a bypass reaction in gluconeogenesis? A) FBPase-1 B) Phosphoglycerate kinase running in reverse C) PEP carboxykinase D) Glucose 6-phosphatase

B) Phosphoglycerate kinase running in reverse. PGK is one of the seven reversible steps shared with glycolysis.

Why can't hexokinase simply run backward in gluconeogenesis? A) It's in a different compartment B) It has no active site for G6P C) It lacks a cofactor D) G6P + ADP → glucose + ATP is highly endergonic

D) G6P + ADP → glucose + ATP is highly endergonic. Hydrolysis of the phosphate ester is the favorable alternative.

Pyruvate carboxylase is located in: A) Cytosol B) Nucleus C) Plasma membrane D) Mitochondria

D) Mitochondria. Oxaloacetate can't cross the inner membrane, which affects how NADH is exported.

Which provides the phosphate for PEP formation by PEP carboxykinase? A) Pi B) 1,3-BPG C) ATP D) GTP

D) GTP. GTP ≡ ATP energetically (ΔG′° = 0 for nucleoside diphosphate kinase).

The total cost of making glucose from 2 pyruvate is: A) 2 ATP + 2 GTP B) 2 ATP C) 6 ATP D) 4 ATP + 2 GTP

D) 4 ATP + 2 GTP. Glycolysis yields only 2 ATP, which is why both directions are favorable.

What does a flux control coefficient of 1 mean? A) A single step is rate-limiting B) All steps contribute equally C) The step has no effect on flux D) The step is at equilibrium

A) A single step is rate-limiting. C = 1 there and 0 elsewhere; the C values sum to 1.

If C = 0.5, a 20% rise in enzyme activity produces about what change in flux? A) +10% B) +20% C) 0% D) +40%

A) +10%. ΔJ/J = C · ΔE/E = 0.5 × 20%.

A step with Vf/Vr ≈ 1.02 is: A) Far from equilibrium with a high C B) Irreversible C) Near equilibrium with ΔG ≈ 0 and C ≈ 0 D) Rate-limiting

C) Near equilibrium with ΔG ≈ 0 and C ≈ 0. Near-equilibrium steps can run either direction and carry little control.

Which glycolytic enzymes have ΔG ≪ 0 in cells? A) Hexokinase, PFK-1, pyruvate kinase B) GAPDH, PGK, mutase C) Aldolase, TPI, enolase D) Only PFK-1

A) Hexokinase, PFK-1, pyruvate kinase. These three, with ΔG of ≈ -33, -22, -17 kJ/mol, are the nonequilibrium steps.

In exercising muscle, which rises most to relieve PFK-1's ATP inhibition? A) F2,6BP B) AMP (~600%) C) ADP D) Acetyl-CoA

B) AMP (~600%). ATP falls only ~10%; adenylate kinase (2 ADP ⇌ ATP + AMP) amplifies the change in AMP.

Pyruvate kinase is activated by: A) Acetyl-CoA B) G6P C) F1,6BP (feed-forward) D) ATP

C) F1,6BP (feed-forward). ATP and acetyl-CoA inhibit; F1,6BP activates.

Which describes F2,6BP? A) Made by PFK-1 B) Activates PFK-1 and inhibits FBPase-1 C) Activates both PFK-1 and FBPase-1 D) Inhibits PFK-1 and activates FBPase-1

B) Activates PFK-1 and inhibits FBPase-1. It ensures glycolysis and gluconeogenesis don't run simultaneously.

The bifunctional PFK-2/FBPase-2 is phosphorylated by PKA. Result: A) Both off B) PFK-2 on, FBPase-2 off, [F2,6BP] rises C) Both on D) PFK-2 off, FBPase-2 on, [F2,6BP] falls

D) PFK-2 off, FBPase-2 on, [F2,6BP] falls. Phosphorylated = FBPase on; dephosphorylated = PFK on.

After a carbohydrate-rich meal, insulin leads to: A) Gluconeogenesis up B) Active PP1, glycogen synthesis up, glycolysis up C) High cAMP D) Active PKA, glycogen breakdown up

B) Active PP1, glycogen synthesis up, glycolysis up. PP1 dephosphorylates phosphorylase kinase, phosphorylase, and glycogen synthase.

Glycogen phosphorylase cleaves α1→4 bonds by: A) Hydrolysis, giving free glucose B) Phosphorolysis, giving glucose 1-phosphate C) Oxidation D) Transglycosylation

B) Phosphorolysis, giving glucose 1-phosphate. It uses Pi, not water, and works at nonreducing ends.

What is the benefit of phosphorolysis? A) It skips the hexokinase ATP, so G1P yields 3 ATP vs. 2 for free glucose B) It makes NADH C) It produces fructose D) It releases free glucose

A) It skips the hexokinase ATP, so G1P yields 3 ATP vs. 2 for free glucose. G1P → G6P by phosphoglucomutase.

For a 10-glucose glycogen chain with 1 branch, glycolysis yields: A) 30 ATP B) 28 ATP C) 18 ATP D) 20 ATP

B) 28 ATP. 8 G1P × 3 + 2 glucose × 2 = 28.

Muscle can't release glucose from glycogen to the blood because: A) It lacks phosphorylase B) It lacks glucose 6-phosphatase C) GLUTs can't transport glucose D) It lacks phosphoglucomutase

B) It lacks glucose 6-phosphatase. G6P isn't recognized by GLUTs and is trapped.

Branch points in glycogen are linked by: A) β1→4 bonds B) Ester bonds C) α1→6 bonds D) α1→4 bonds

C) α1→6 bonds. Chains are α1→4, with branches every 12-14 residues on a glycogenin core.

Which pair shows the correct fed-state liver regulation? A) Both active B) Both inactive C) Glycogen synthase active, phosphorylase kinase inactive D) Glycogen synthase inactive, phosphorylase active

C) Glycogen synthase active, phosphorylase kinase inactive. The fed state (insulin → PP1) favors synthesis and lowers breakdown.

In muscle, glucagon causes: A) Glycogen breakdown B) No effect, because muscle has no glucagon receptor C) Glycogen synthesis D) Gluconeogenesis

B) No effect, because muscle has no glucagon receptor. Epinephrine, not glucagon, signals in muscle.

In muscle, which does epinephrine promote? A) Release of blood glucose B) Glycogen synthesis C) Gluconeogenesis D) Glycogen breakdown and glycolysis

D) Glycogen breakdown and glycolysis. Muscle has no gluconeogenesis and uses glucose for itself.

Which proteins in the table are phosphorylated by PKA and lose activity? A) MEK, ERK B) Phosphorylase kinase, FBPase-2 C) Glycogen synthase, PFK-2, pyruvate kinase L D) Insulin receptor, IRS-1

C) Glycogen synthase, PFK-2, pyruvate kinase L. PKA activates phosphorylase kinase and FBPase-2 but inactivates synthase, PFK-2, and liver PK.

Which is a main role of the PPP? A) NADPH and ribose 5-phosphate production B) ATP production C) Gluconeogenesis D) Fermentation

A) NADPH and ribose 5-phosphate production. NADPH for biosynthesis and glutathione reduction; ribose 5-P for nucleotides.

In the oxidative PPP, which carbon of glucose is lost as CO2? A) C6 B) C2 C) C1 D) C3

C) C1. 6 C in → 5 C out + 1 CO2.

The oxidative PPP converts G6P to: A) Fructose 6-P + NADPH B) Ribose 5-P + 2 NADH + 2 CO2 C) Pyruvate + ATP D) Ribose 5-P + 2 NADPH + CO2

D) Ribose 5-P + 2 NADPH + CO2. Equation: G6P + 2 NADP⁺ + H2O → ribose 5-P + 2 NADPH + 2 H⁺ + CO2.

Which enzyme moves a 2-carbon unit using TPP? A) Aldolase B) Transketolase C) Pyruvate kinase D) Transaldolase

B) Transketolase. Transaldolase moves 3 carbons using a Schiff base.

Transaldolase and human aldolase share which strategy? A) Zn²⁺ stabilization B) Active-site Lys Schiff base forming an enamine C) TPP D) Thioester intermediate

B) Active-site Lys Schiff base forming an enamine. Same chemical strategy.

Sedoheptulose 7-P + GAP ⇌ ? A) Erythrose 4-P + F6P B) Ribose 5-P + xylulose 5-P C) DHAP + F6P D) Ribulose 5-P + G6P

A) Erythrose 4-P + F6P. This is the transaldolase reaction.

If a cell needs NADPH but not ribose 5-P, the pentoses are: A) Excreted B) Stored as glycogen C) Converted to pyruvate only D) Converted to F6P and GAP by TK/TA, then F6P → G6P to re-enter the oxidative phase

D) Converted to F6P and GAP by TK/TA, then F6P → G6P to re-enter the oxidative phase. Six pentoses → five hexoses.

Red blood cells depend on the PPP because: A) They need ribose for DNA B) They need NADPH to keep glutathione reduced against oxidative damage C) They need ATP from the PPP D) They have mitochondria

B) They need NADPH to keep glutathione reduced against oxidative damage. The PPP is critical in RBCs for NADPH.