What molecules are antibodies, and which cells make them?

Antibodies are immunoglobulins , made by B cells / plasma cells . They bind antigens specifically .

Basic antibody chain composition: how many heavy and light chains, and what holds them together?

2 heavy chains + 2 light chains , held by disulfide bonds (plus noncovalent interactions). Each chain folds into immunoglobulin domains .

What is the overall shape of an antibody, and where does it bind antigen?

Y-shaped . Antigen binds at the tips of the Y (the variable regions / Fab arms).

EXAM TRAP — True or false: "IgG appears earlier than IgM" in a primary immune response.

FALSE. In a primary response, IgM comes first , then IgG. Never reverse this.

In a PRIMARY immune response, which antibody isotype is produced FIRST?

IgM first. Naive B cells initially make IgM.

How does a B cell go from making IgM to making IgG/IgA/IgE?

After activation + T-cell help , the B cell undergoes class switching (isotype switching). IgM is the default; IgG is class-switched .

Why does the SECONDARY immune response produce IgG faster and stronger?

Memory B cells are already class-switched, so they make IgG rapidly and in larger amounts on re-exposure. IgG dominates the secondary response .

Is IgG a monomer, dimer, or pentamer? How many heavy and light chains?

IgG is a monomer : 2 identical heavy chains + 2 identical light chains .

Does IgG have a domain structure? (per the Q6 stem)

Yes — each chain folds into multiple immunoglobulin-fold domains (~100–200 residues each). IgG is the classic textbook example of domain structure.

EXAM TRAP — Does an antibody bind its antigen by COVALENT or NONCOVALENT interactions?

NONCOVALENT. Never say covalent. Binding is specific but reversible.

Name the four noncovalent forces that hold an antibody to its antigen.

Hydrogen bonds, ionic interactions, hydrophobic interactions, van der Waals forces. Specific but reversible.

In IgG, which region binds antigen and which region carries out effector functions?

Fab regions = the two arms = antigen binding . Fc region = the stem = effector functions (complement, phagocyte binding).

List four effector functions IgG can perform.

(1) Neutralize toxins/viruses, (2) opsonize microbes, (3) activate complement , (4) cross the placenta .

Which immunoglobulin can cross the placenta?

IgG (gives passive immunity to the fetus).

Where do the immunoglobulins migrate on serum protein electrophoresis (SPEP)?

In the gamma globulin fraction — the band that moves the least. A sharp monoclonal spike here = multiple myeloma .

Why is the gamma (antibody) band normally a broad smear rather than one sharp line?

Immunoglobulins are highly polymorphic — millions of distinct sequences from somatic recombination + hypermutation. A sharp spike instead = monoclonal ( myeloma ).

METHOD — "Structure tells function": what 3 questions do you ask on any structure card (IgG, TAG, LDL)?

(1) What are the components ? (2) What is the structure ? (3) What does that structure allow it to do ? For IgG: 2H+2L Y-shape → two Fab arms → bivalent antigen binding.

Q6 — Which statements were CORRECT about IgG? (appears earlier than IgM / has domain structure / has 2 heavy + 2 light chains / binds antigen noncovalently)

Correct = domain structure, 2 heavy + 2 light chains, binds antigen noncovalently . "Appears earlier than IgM" is FALSE (IgM is first).

MUST-MEMORIZE LINE #1 — the IgM/IgG timing rule.

IgM comes first; IgG is class-switched and dominates the secondary response.

MUST-MEMORIZE LINE #2 — the IgG one-liner (structure + binding).

IgG = monomeric, 2 heavy + 2 light chains, domain structure, binds antigen noncovalently (never covalent).

State the master pKa rule in one line: how does pH vs pK a decide whether an ionizable group is protonated or deprotonated?

When pH pKa → mostly DEPROTONATED . pK a is the pH where the group is half-and-half.

Translate "protonated" and "deprotonated" into actual charge for a carboxyl group vs an amino group.

Carboxyl : protonated = –COOH = neutral (0) ; deprotonated = –COO⁻ = negative (−1) . Amino : protonated = –NH₃⁺ = positive (+1) ; deprotonated = –NH₂ = neutral (0) . (Carboxyl loses a proton to go negative; amino loses a proton to go neutral.)

List the three pKa values for glutamic acid (as given on the exam figure) and name each group.

α- COOH pK a = 2.1; side-chain γ- COOH pK a = 4.1; α- NH₃⁺ pK a = 9.5. Two acidic carboxyls (2.1, 4.1) + one basic amino (9.5) .

Glutamate below pH 2.1 (all three groups): write each group's state and the net charge .

All below their pKa → all protonated: COOH (0), COOH (0), NH₃⁺ (+1). Net = +1 .

Glutamate between pH 2.1 and 4.1 : write each group's state and the net charge . What is this form called?

COO⁻ (−1), COOH (0), NH₃⁺ (+1). Net = 0 — this is the zwitterion / net-neutral form of glutamate.

Glutamate between pH 4.1 and 9.5 (the form shown on the Q29 figure): write each group's state and the net charge .

α- COO⁻ (−1), side-chain COO⁻ (−1), α- NH₃⁺ (+1). Net = −1 . (-1) + (-1) + (+1) = -1. This is the figure form.

Glutamate above pH 9.5 : write each group's state and the net charge .

All deprotonated: COO⁻ (−1), COO⁻ (−1), NH₂ (0). Net = −2 . (Amino group finally lost its proton → neutral.)

The Q29 form (α-COO⁻, side-chain COO⁻, α-NH₃⁺, net −1) occurs over what pH range , and why is the answer NOT a single pH?

It is the dominant form whenever pH is between 4.1 and 9.5 — i.e. above the side-chain COOH pKa (4.1, so it is deprotonated) but below the NH₃⁺ pKa (9.5, so it is still protonated). Physiological pH 7.4 sits inside this window.

Define the isoelectric point (pI) . Is it the pH where every group is neutral?

pI = the pH at which the molecule carries zero NET charge — the + and − charges exactly balance. NOT the pH where every group is uncharged (a named exam trap). At the pI the zwitterion form dominates.

Method: how do you pick which two pKa values to average for the pI?

Find the net-zero (zwitterion) species first, then average the two pKa values that flank it (the one just below and the one just above it). Neutral AA: (pKa₁+pKa₂)/2; acidic AA: (pKa₁+pK R )/2; basic AA: (pKa₂+pK R )/2.

Calculate the pI of glutamate using the exam pKa values, and state its net charge there.

Acidic AA → average the two acidic pKas flanking the neutral form: pI = (2.1 + 4.1)/2 = 3.1. At pH ≈ 3.1 glutamate's net charge = 0 . (Acidic AAs have pI ~3.)

Why is the pI of an acidic amino acid (Asp/Glu) low (~3), in one sentence of reasoning?

An acidic AA needs a low pH to suppress (re-protonate) the second carboxyl so its two negatives don't both ionize — only then do the charges balance to zero. Basic AAs (Lys/Arg) are the mirror image: pI ~9–11.

Zwitterion trap: the Q29 glutamate form has BOTH a + and a − charge. Can you call it "the zwitterion"?

Be careful. A zwitterion has both + and − charges, but "zwitterion" usually implies NET ZERO . The figure form has α- NH₃⁺ and two COO⁻ → net −1, NOT net zero . So do NOT call it the zwitterion if they mean the net-neutral species (that one is the pH 2.1–4.1 form, or pH = pI = 3.1).

Electrophoresis rule: which electrode does a net-negative molecule migrate toward? A net-positive one? A net-zero one?

Net negative → ANODE (+ electrode) . Net positive → CATHODE (− electrode) . Net zero → barely/does not migrate . Memory hook: opposites attract — go toward the opposite-sign electrode .

In terms of pH vs pI , when does an amino acid migrate to the ANODE vs the CATHODE?

If pH > pI → molecule is net negative → migrates to the anode (+) . If pH < pI → net positive → migrates to the cathode (−) . If pH = pI → no net charge → no migration . Each species sits still only at its own pI.

The shown glutamate (net −1) is placed in an electric field. Does it migrate, and toward which electrode?

Yes — it is net −1 , so it migrates toward the positive electrode (anode) . (Its pH window 4.1–9.5 is above its pI of 3.1, so pH > pI → net negative → anode.)

NAMED EXAM TRAP — True or false: "The predominant ionic form of glutamic acid at pH 7.4 has net charge = 0."

FALSE. At pH 7.4 both carboxyls are deprotonated (−1, −1; pKa 2.1 and 4.1 both below 7.4) and the amino is protonated (+1; pKa 9.5 above 7.4): (-1)+(-1)+(+1) = -1. Net = −1, not 0 (pH 7.4 is far above its pI of 3.1).

Common trap: at pH 7.4 an acidic amino acid is net −1, NOT −2. Why does the amino group save it from being −2?

Because the α- NH₃⁺ pKa (~9.5) is above 7.4, so it is still protonated and contributes +1 . Two COO⁻ (−1, −1) + one NH₃⁺ (+1) = net −1. You only reach −2 above pH 9.5, when the amino finally deprotonates.

Give the universal 6-step method for any "net charge at a given pH" amino-acid question.

(1) List every ionizable group. (2) Compare pH to each pK a . (3) pH pKa → deprotonated. (5) Assign each group's charge and SUM them for net charge . (6) Use net charge for electrophoresis (opposite-sign electrode). Works for single AAs and whole peptides (N-term + C-term + every side chain).

MUST-MEMORIZE LINE for glutamate (the never-miss for Q29): state it.

Glutamate over pH 4.1–9.5: two COO⁻ + one NH₃⁺ = net −1 ; pI = (2.1+4.1)/2 = 3.1; net −1 migrates to the anode (+) . Rule of thumb: pH pKa deprotonated .

A figure shows a glycerol backbone esterified to THREE fatty acids . What molecule is this?

A triacylglycerol (TAG) — glycerol + 3 fatty acids via 3 ester bonds . NOT a phospholipid (no phosphate, no polar head).

True or false: "This molecule (glycerol + 3 fatty acids) is an amphipathic phospholipid and a main constituent of cell membranes."

FALSE — this is the exact Q44 false statement (answer A). It is a TAG : no phosphate, not amphipathic , mostly hydrophobic, an energy-storage lipid — not a membrane lipid.

STRUCTURE TELLS FUNCTION: How many fatty acids and what head group distinguish a TAG from a phospholipid ?

TAG = glycerol + 3 fatty acids , NO phosphate → storage lipid. Phospholipid = glycerol + 2 fatty acids + a phosphate-containing polar head → amphipathic membrane lipid.

Why is a TAG NOT amphipathic but a phospholipid IS?

TAG has no charged/polar head — all three positions are nonpolar fatty acyl chains, so it is almost entirely hydrophobic . A phospholipid's phosphate head is hydrophilic while its 2 tails are hydrophobic → amphipathic → forms bilayers/membranes.

What is the primary biological FUNCTION of a triacylglycerol?

The body's main energy-storage lipid . Stored in adipocytes ; transported in chylomicrons and VLDL . (Phospholipids build membranes; TAG stores energy.)

MUST-MEMORIZE LINE: Complete both — TAG = ___ ; phospholipid = ___

TAG = glycerol + 3 fatty acids = storage lipid . Phospholipid = glycerol + 2 fatty acids + phosphate = membrane lipid.

Name the THREE lipases that hydrolyze TAG and the ONE-LINE job of each.

Pancreatic lipase — digests dietary TAG in the gut lumen. Lipoprotein lipase (LPL) — hydrolyzes lipoprotein TAG (chylomicron/VLDL) at capillaries. Hormone-sensitive lipase (HSL) — mobilizes stored adipocyte TAG during fasting.

Where does PANCREATIC lipase act, and what does it produce from dietary TAG?

Acts in the intestinal lumen . Hydrolyzes dietary TAG → free fatty acids + 2-monoacylglycerol . This is the digestion step.

Where is LIPOPROTEIN lipase (LPL) located, and what does it act on?

On the luminal surface of capillary endothelium (adipose & muscle), anchored by heparan sulfate. It cleaves TAG inside chylomicrons and VLDL → free fatty acids + glycerol for tissue uptake.

What obligate cofactor/activator does LPL require, and from which particle is it donated?

ApoC-II — without it LPL is essentially inactive. ApoC-II is donated to chylomicrons/VLDL from circulating HDL , then returned. (Deficiency of LPL or apoC-II = Type I hyperlipoproteinemia, massive hypertriglyceridemia.)

Which hormone activates ADIPOSE LPL, and what does this accomplish in the fed state?

Insulin stimulates adipose LPL. In the fed state this lets adipose tissue store fat — fatty acids freed by LPL are taken up and re-esterified into stored TAG.

Public flashcard deck · 259 cards

Polish II · Exam-Specific Weaknesses

Round 2 — your brutally-honest weakness questions (Q6, Q29, Q44, Q45, Q49, Q51, Q52, Q57, Q62, Q63, Q68, Q76): IgG/Ig timing, amino-acid pKa/charge/pI, TAG vs phospholipid + lipases, DNA repair, buffers/H-H, bilirubin/jaundice, PKU, dilutions & %, polymerases, Vmax/Km, lipoproteins, + the reasoning method. Colored to the book; math in LaTeX.

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Front
What molecules are antibodies, and which cells make them?
Back
Antibodies are immunoglobulins , made by B cells / plasma cells . They bind antigens specifically .
Front
Basic antibody chain composition: how many heavy and light chains, and what holds them together?
Back
2 heavy chains + 2 light chains , held by disulfide bonds (plus noncovalent interactions). Each chain folds into immunoglobulin domains .
Front
What is the overall shape of an antibody, and where does it bind antigen?
Back
Y-shaped . Antigen binds at the tips of the Y (the variable regions / Fab arms).
Front
EXAM TRAP — True or false: "IgG appears earlier than IgM" in a primary immune response.
Back
FALSE. In a primary response, IgM comes first , then IgG. Never reverse this.
Front
In a PRIMARY immune response, which antibody isotype is produced FIRST?
Back
IgM first. Naive B cells initially make IgM.
Front
How does a B cell go from making IgM to making IgG/IgA/IgE?
Back
After activation + T-cell help , the B cell undergoes class switching (isotype switching). IgM is the default; IgG is class-switched .
Front
Why does the SECONDARY immune response produce IgG faster and stronger?
Back
Memory B cells are already class-switched, so they make IgG rapidly and in larger amounts on re-exposure. IgG dominates the secondary response .
Front
Is IgG a monomer, dimer, or pentamer? How many heavy and light chains?
Back
IgG is a monomer : 2 identical heavy chains + 2 identical light chains .
Front
Does IgG have a domain structure? (per the Q6 stem)
Back
Yes — each chain folds into multiple immunoglobulin-fold domains (~100–200 residues each). IgG is the classic textbook example of domain structure.
Front
EXAM TRAP — Does an antibody bind its antigen by COVALENT or NONCOVALENT interactions?
Back
NONCOVALENT. Never say covalent. Binding is specific but reversible.
Front
Name the four noncovalent forces that hold an antibody to its antigen.
Back
Hydrogen bonds, ionic interactions, hydrophobic interactions, van der Waals forces. Specific but reversible.
Front
In IgG, which region binds antigen and which region carries out effector functions?
Back
Fab regions = the two arms = antigen binding . Fc region = the stem = effector functions (complement, phagocyte binding).
Front
List four effector functions IgG can perform.
Back
(1) Neutralize toxins/viruses, (2) opsonize microbes, (3) activate complement , (4) cross the placenta .
Front
Which immunoglobulin can cross the placenta?
Back
IgG (gives passive immunity to the fetus).
Front
Where do the immunoglobulins migrate on serum protein electrophoresis (SPEP)?
Back
In the gamma globulin fraction — the band that moves the least. A sharp monoclonal spike here = multiple myeloma .
Front
Why is the gamma (antibody) band normally a broad smear rather than one sharp line?
Back
Immunoglobulins are highly polymorphic — millions of distinct sequences from somatic recombination + hypermutation. A sharp spike instead = monoclonal ( myeloma ).
Front
METHOD — "Structure tells function": what 3 questions do you ask on any structure card (IgG, TAG, LDL)?
Back
(1) What are the components ? (2) What is the structure ? (3) What does that structure allow it to do ? For IgG: 2H+2L Y-shape → two Fab arms → bivalent antigen binding.
Front
Q6 — Which statements were CORRECT about IgG? (appears earlier than IgM / has domain structure / has 2 heavy + 2 light chains / binds antigen noncovalently)
Back
Correct = domain structure, 2 heavy + 2 light chains, binds antigen noncovalently . "Appears earlier than IgM" is FALSE (IgM is first).
Front
MUST-MEMORIZE LINE #1 — the IgM/IgG timing rule.
Back
IgM comes first; IgG is class-switched and dominates the secondary response.
Front
MUST-MEMORIZE LINE #2 — the IgG one-liner (structure + binding).
Back
IgG = monomeric, 2 heavy + 2 light chains, domain structure, binds antigen noncovalently (never covalent).
Front
State the master pKa rule in one line: how does pH vs pK a decide whether an ionizable group is protonated or deprotonated?
Back
When pH < pKa → group is mostly PROTONATED; pH > pKa → mostly DEPROTONATED . pK a is the pH where the group is half-and-half.
Front
Translate "protonated" and "deprotonated" into actual charge for a carboxyl group vs an amino group.
Back
Carboxyl : protonated = –COOH = neutral (0) ; deprotonated = –COO⁻ = negative (−1) . Amino : protonated = –NH₃⁺ = positive (+1) ; deprotonated = –NH₂ = neutral (0) . (Carboxyl loses a proton to go negative; amino loses a proton to go neutral.)
Front
List the three pKa values for glutamic acid (as given on the exam figure) and name each group.
Back
α- COOH pK a = 2.1; side-chain γ- COOH pK a = 4.1; α- NH₃⁺ pK a = 9.5. Two acidic carboxyls (2.1, 4.1) + one basic amino (9.5) .
Front
Glutamate below pH 2.1 (all three groups): write each group's state and the net charge .
Back
All below their pKa → all protonated: COOH (0), COOH (0), NH₃⁺ (+1). Net = +1 .
Front
Glutamate between pH 2.1 and 4.1 : write each group's state and the net charge . What is this form called?
Back
COO⁻ (−1), COOH (0), NH₃⁺ (+1). Net = 0 — this is the zwitterion / net-neutral form of glutamate.
Front
Glutamate between pH 4.1 and 9.5 (the form shown on the Q29 figure): write each group's state and the net charge .
Back
α- COO⁻ (−1), side-chain COO⁻ (−1), α- NH₃⁺ (+1). Net = −1 . (-1) + (-1) + (+1) = -1. This is the figure form.
Front
Glutamate above pH 9.5 : write each group's state and the net charge .
Back
All deprotonated: COO⁻ (−1), COO⁻ (−1), NH₂ (0). Net = −2 . (Amino group finally lost its proton → neutral.)
Front
The Q29 form (α-COO⁻, side-chain COO⁻, α-NH₃⁺, net −1) occurs over what pH range , and why is the answer NOT a single pH?
Back
It is the dominant form whenever pH is between 4.1 and 9.5 — i.e. above the side-chain COOH pKa (4.1, so it is deprotonated) but below the NH₃⁺ pKa (9.5, so it is still protonated). Physiological pH 7.4 sits inside this window.
Front
Define the isoelectric point (pI) . Is it the pH where every group is neutral?
Back
pI = the pH at which the molecule carries zero NET charge — the + and − charges exactly balance. NOT the pH where every group is uncharged (a named exam trap). At the pI the zwitterion form dominates.
Front
Method: how do you pick which two pKa values to average for the pI?
Back
Find the net-zero (zwitterion) species first, then average the two pKa values that flank it (the one just below and the one just above it). Neutral AA: (pKa₁+pKa₂)/2; acidic AA: (pKa₁+pK R )/2; basic AA: (pKa₂+pK R )/2.
Front
Calculate the pI of glutamate using the exam pKa values, and state its net charge there.
Back
Acidic AA → average the two acidic pKas flanking the neutral form: pI = (2.1 + 4.1)/2 = 3.1. At pH ≈ 3.1 glutamate's net charge = 0 . (Acidic AAs have pI ~3.)
Front
Why is the pI of an acidic amino acid (Asp/Glu) low (~3), in one sentence of reasoning?
Back
An acidic AA needs a low pH to suppress (re-protonate) the second carboxyl so its two negatives don't both ionize — only then do the charges balance to zero. Basic AAs (Lys/Arg) are the mirror image: pI ~9–11.
Front
Zwitterion trap: the Q29 glutamate form has BOTH a + and a − charge. Can you call it "the zwitterion"?
Back
Be careful. A zwitterion has both + and − charges, but "zwitterion" usually implies NET ZERO . The figure form has α- NH₃⁺ and two COO⁻ → net −1, NOT net zero . So do NOT call it the zwitterion if they mean the net-neutral species (that one is the pH 2.1–4.1 form, or pH = pI = 3.1).
Front
Electrophoresis rule: which electrode does a net-negative molecule migrate toward? A net-positive one? A net-zero one?
Back
Net negative → ANODE (+ electrode) . Net positive → CATHODE (− electrode) . Net zero → barely/does not migrate . Memory hook: opposites attract — go toward the opposite-sign electrode .
Front
In terms of pH vs pI , when does an amino acid migrate to the ANODE vs the CATHODE?
Back
If pH > pI → molecule is net negative → migrates to the anode (+) . If pH < pI → net positive → migrates to the cathode (−) . If pH = pI → no net charge → no migration . Each species sits still only at its own pI.
Front
The shown glutamate (net −1) is placed in an electric field. Does it migrate, and toward which electrode?
Back
Yes — it is net −1 , so it migrates toward the positive electrode (anode) . (Its pH window 4.1–9.5 is above its pI of 3.1, so pH > pI → net negative → anode.)
Front
NAMED EXAM TRAP — True or false: "The predominant ionic form of glutamic acid at pH 7.4 has net charge = 0."
Back
FALSE. At pH 7.4 both carboxyls are deprotonated (−1, −1; pKa 2.1 and 4.1 both below 7.4) and the amino is protonated (+1; pKa 9.5 above 7.4): (-1)+(-1)+(+1) = -1. Net = −1, not 0 (pH 7.4 is far above its pI of 3.1).
Front
Common trap: at pH 7.4 an acidic amino acid is net −1, NOT −2. Why does the amino group save it from being −2?
Back
Because the α- NH₃⁺ pKa (~9.5) is above 7.4, so it is still protonated and contributes +1 . Two COO⁻ (−1, −1) + one NH₃⁺ (+1) = net −1. You only reach −2 above pH 9.5, when the amino finally deprotonates.
Front
Give the universal 6-step method for any "net charge at a given pH" amino-acid question.
Back
(1) List every ionizable group. (2) Compare pH to each pK a . (3) pH < pKa → protonated. (4) pH > pKa → deprotonated. (5) Assign each group's charge and SUM them for net charge . (6) Use net charge for electrophoresis (opposite-sign electrode). Works for single AAs and whole peptides (N-term + C-term + every side chain).
Front
MUST-MEMORIZE LINE for glutamate (the never-miss for Q29): state it.
Back
Glutamate over pH 4.1–9.5: two COO⁻ + one NH₃⁺ = net −1 ; pI = (2.1+4.1)/2 = 3.1; net −1 migrates to the anode (+) . Rule of thumb: pH < pKa protonated, pH > pKa deprotonated .
Front
A figure shows a glycerol backbone esterified to THREE fatty acids . What molecule is this?
Back
A triacylglycerol (TAG) — glycerol + 3 fatty acids via 3 ester bonds . NOT a phospholipid (no phosphate, no polar head).
Front
True or false: "This molecule (glycerol + 3 fatty acids) is an amphipathic phospholipid and a main constituent of cell membranes."
Back
FALSE — this is the exact Q44 false statement (answer A). It is a TAG : no phosphate, not amphipathic , mostly hydrophobic, an energy-storage lipid — not a membrane lipid.
Front
STRUCTURE TELLS FUNCTION: How many fatty acids and what head group distinguish a TAG from a phospholipid ?
Back
TAG = glycerol + 3 fatty acids , NO phosphate → storage lipid. Phospholipid = glycerol + 2 fatty acids + a phosphate-containing polar head → amphipathic membrane lipid.
Front
Why is a TAG NOT amphipathic but a phospholipid IS?
Back
TAG has no charged/polar head — all three positions are nonpolar fatty acyl chains, so it is almost entirely hydrophobic . A phospholipid's phosphate head is hydrophilic while its 2 tails are hydrophobic → amphipathic → forms bilayers/membranes.
Front
What is the primary biological FUNCTION of a triacylglycerol?
Back
The body's main energy-storage lipid . Stored in adipocytes ; transported in chylomicrons and VLDL . (Phospholipids build membranes; TAG stores energy.)
Front
MUST-MEMORIZE LINE: Complete both — TAG = ___ ; phospholipid = ___
Back
TAG = glycerol + 3 fatty acids = storage lipid . Phospholipid = glycerol + 2 fatty acids + phosphate = membrane lipid.
Front
Name the THREE lipases that hydrolyze TAG and the ONE-LINE job of each.
Back
Pancreatic lipase — digests dietary TAG in the gut lumen. Lipoprotein lipase (LPL) — hydrolyzes lipoprotein TAG (chylomicron/VLDL) at capillaries. Hormone-sensitive lipase (HSL) — mobilizes stored adipocyte TAG during fasting.
Front
Where does PANCREATIC lipase act, and what does it produce from dietary TAG?
Back
Acts in the intestinal lumen . Hydrolyzes dietary TAG → free fatty acids + 2-monoacylglycerol . This is the digestion step.
Front
Where is LIPOPROTEIN lipase (LPL) located, and what does it act on?
Back
On the luminal surface of capillary endothelium (adipose & muscle), anchored by heparan sulfate. It cleaves TAG inside chylomicrons and VLDL → free fatty acids + glycerol for tissue uptake.
Front
What obligate cofactor/activator does LPL require, and from which particle is it donated?
Back
ApoC-II — without it LPL is essentially inactive. ApoC-II is donated to chylomicrons/VLDL from circulating HDL , then returned. (Deficiency of LPL or apoC-II = Type I hyperlipoproteinemia, massive hypertriglyceridemia.)
Front
Which hormone activates ADIPOSE LPL, and what does this accomplish in the fed state?
Back
Insulin stimulates adipose LPL. In the fed state this lets adipose tissue store fat — fatty acids freed by LPL are taken up and re-esterified into stored TAG.
Front
Two LPL isoforms: which tissue's LPL has the LOW Km (high affinity), and why does that matter?
Back
Muscle LPL has the low Km (high affinity) → active even when plasma TAG is low, so muscle gets fatty acids first at rest. Adipose LPL has a high Km → works mainly after a meal (post-prandial storage).
Front
Where is HORMONE-SENSITIVE lipase (HSL) located, and what does it do?
Back
Inside adipocytes . It hydrolyzes stored TAG → free fatty acids + glycerol , releasing them into blood for other tissues to burn ( fat mobilization ).
Front
Which hormonal conditions ACTIVATE HSL, and which condition turns it OFF?
Back
ON during fasting/stress: glucagon , adrenaline/epinephrine , and low insulin (cAMP → PKA). OFF when insulin is high (fed state).
Front
DISCRIMINATION TRAP — Which lipase mobilizes STORED adipocyte fat during fasting: LPL or HSL?
Back
HSL (hormone-sensitive lipase) mobilizes stored fat. NOT LPL — LPL works on circulating lipoprotein TAG at capillaries; HSL works on intracellular stored TAG.
Front
DISCRIMINATION TRAP — Which lipase digests DIETARY TAG in the intestine: pancreatic lipase or LPL?
Back
Pancreatic lipase digests dietary TAG in the gut. NOT LPL — LPL acts on TAG already packaged in chylomicrons/VLDL out in the capillaries, not on food in the lumen.
Front
REASONING METHOD: When a question shows a glycerol-based lipid, how do you classify it in one step?
Back
Count fatty acids and look for phosphate. 3 fatty acids , no phosphate → TAG (storage). 2 fatty acids + phosphate head → phospholipid (membrane, amphipathic).
Front
REASONING METHOD: Given a lipase, how do you instantly recall its job? (3-part decision)
Back
Match by LOCATION : gut lumen = pancreatic lipase (dietary TAG). Capillary endothelium = LPL (lipoprotein TAG, needs apoC-II ). Inside adipocyte = HSL (stored TAG, driven by glucagon/adrenaline ). Location = function .
Front
MUST-MEMORIZE NEVER-MISS LINE: state the one-line role of pancreatic lipase, LPL, and HSL.
Back
Pancreatic lipase digests dietary TAG; LPL hydrolyzes lipoprotein TAG (chylomicron/VLDL, apoC-II-activated); HSL mobilizes adipocyte stored TAG.
Front
Both products of LPL and HSL acting on TAG are the same two molecules — name them. (And which tissue cannot re-use the glycerol?)
Back
TAG hydrolysis yields free fatty acids + glycerol . Adipocytes lack glycerol kinase , so they cannot re-phosphorylate that glycerol — it goes to the liver (which has glycerol kinase) for gluconeogenesis.
Front
Match the lesion to its repair pathway: base alkylation (e.g. O6-methylguanine) is fixed by which system?
Back
Direct repair by the enzyme MGMT — the base is not removed; the methyl group is transferred off directly.
Front
Match the lesion to its repair pathway: uracil in DNA is fixed by which system?
Back
Base excision repair (BER) — uracil is a small, non-bulky base lesion; a DNA glycosylase removes it.
Front
Match the lesion to its repair pathway: bulky DNA adducts and UV thymine dimers are fixed by which system?
Back
Nucleotide excision repair (NER) — these are bulky, helix-distorting lesions; a short oligonucleotide is excised.
Front
Match the lesion to its repair pathway: post-replication errors (wrong base, insertion/deletion loops, slippage) are fixed by which system?
Back
Mismatch repair (MMR) — corrects mismatches that slip past proofreading after synthesis.
Front
The 4-question SHORTCUT: how do you assign the right repair pathway? (Method card — say all four.)
Back
(1) One damaged single base , not bulky? BER . (2) Bulky helix distortion ? NER . (3) Wrong base AFTER replication ? MMR . (4) Methyl removed directly , base kept? MGMT / direct repair .
Front
What is the defining feature of direct repair that separates it from BER and NER?
Back
It fixes the damage WITHOUT removing the base — no strand is excised. The lesion is chemically reversed in place (e.g. MGMT transfers the methyl group off).
Front
Exactly how does MGMT repair O6-methylguanine ?
Back
It transfers the methyl group from O6-methylguanine onto a cysteine in its own active site , restoring normal guanine. It is a suicide enzyme — permanently inactivated after one repair event.
Front
Why does uracil appear in DNA, and why is BER needed to remove it?
Back
Cytosine spontaneously deaminates to uracil (~70–200/cell/day). DNA should NOT contain uracil ; if left, the next polymerase reads it as thymine, locking in a C→T transition .
Front
List the 4 steps of BER in order (and the enzyme at each step).
Back
(1) DNA glycosylase removes the bad base → AP site. (2) AP endonuclease cuts the backbone. (3) DNA polymerase fills the gap. (4) DNA ligase seals the nick. Glycosylase → endonuclease → polymerase → ligase .
Front
Besides uracil, name other lesions handled by BER .
Back
Small, non-bulky base damage : deaminated bases, oxidized bases (e.g. 8-oxoguanine , removed by OGG1), and small alkylated bases . BER is the workhorse for tiny, non-distorting damage.
Front
List the steps of NER in order.
Back
(1) Recognize the helix distortion . (2) Cut on both sides of the lesion. (3) Remove a short oligonucleotide (~ 25–30 nt in eukaryotes ; ~12–13 in bacteria). (4) DNA polymerase fills. (5) DNA ligase seals.
Front
Which lesions specifically trigger NER ?
Back
Bulky, helix-distorting lesions : UV-induced cyclobutane pyrimidine (thymine) dimers and bulky chemical/DNA adducts (PAHs, aflatoxin). BER cannot handle these.
Front
Defective NER causes which clinical disease?
Back
Xeroderma pigmentosum (XP) — autosomal recessive loss of any XPA–XPG gene; patients cannot repair UV pyrimidine dimers → extreme sun sensitivity and early skin cancers .
Front
Which errors does MMR specifically correct?
Back
Replication errors after synthesis that escaped proofreading: a wrong base (e.g. G:T), and insertion/deletion loops from strand slippage (especially at microsatellites).
Front
Defective MMR produces what fingerprint and which inherited disease?
Back
Fingerprint = microsatellite instability . Disease = Lynch syndrome (HNPCC) — ~3–5% of colorectal cancers; human genes MSH2, MSH6, MLH1, PMS2 .
Front
PATHWAY-DIAGNOSIS reasoning: a question says " uracil is present in DNA ." Which pathway, and what is the one-line reasoning?
Back
BER . Reasoning: uracil is a single, small, non-bulky base error from cytosine deamination → glycosylase plucks one base → BER, never NER.
Front
PATHWAY-DIAGNOSIS reasoning: a question says " UV thymine dimer / bulky adduct distorting the helix ." Which pathway, and why NOT BER?
Back
NER . Reasoning: the lesion distorts the double helix ; BER only handles small non-distorting bases, so a bulky distortion forces NER (excise ~25–30 nt).
Front
TRAP — True or false: UV thymine dimers are repaired by base excision repair (BER) .
Back
FALSE . Thymine dimers are bulky helix-distorting lesions repaired by NER , not BER. (BER = small non-bulky bases only.)
Front
TRAP — True or false: O6-methylguanine pairs with cytosine and is repaired by base excision repair .
Back
FALSE on both counts . O6-methylguanine mispairs with thymine (→ GC→AT transition) and is fixed by MGMT direct repair , which removes the methyl without excising the base.
Front
TRAP — True or false: Humans use the enzyme photolyase to repair UV pyrimidine dimers .
Back
FALSE . Photolyase (photoreactivation) is NOT found in placental mammals . Humans rely on NER for dimers; photolyase works in bacteria/yeast/non-mammals.
Front
MUST-MEMORIZE never-miss line: Uracil in DNA = ?
Back
BER (base excision repair). Cytosine deaminated → uracil → glycosylase removes it.
Front
MUST-MEMORIZE never-miss line: Bulky adduct / thymine dimer = ?
Back
NER (nucleotide excision repair). Defective NER = Xeroderma pigmentosum .
Front
MUST-MEMORIZE never-miss line: Replication mismatch (after synthesis) = ?
Back
MMR (mismatch repair). Defective MMR = Lynch syndrome + microsatellite instability.
Front
MUST-MEMORIZE never-miss line: O6-methylguanine = ?
Back
MGMT direct repair — transfers the methyl off, keeps the base, no excision; suicide enzyme.
Front
State the Henderson-Hasselbalch equation for a buffer.
Back
pH = pK a + log([base]/[acid]) i.e. pH = pK a + log([A − ]/[HA]) — base (conjugate base) on top , acid on bottom.
Front
Henderson-Hasselbalch: when [base] = [acid] (equimolar buffer), what does pH equal?
Back
pH = pKₐ Because log([base]/[acid]) = log(1) = 0, so pH = pK a + 0. This is the most tested HH fact.
Front
Q49 SETUP — Identify the acid and base forms: a phosphate buffer is made from NaH₂PO₄ and Na₂HPO₄ . Which is the acid and which is the conjugate base?
Back
Acid form = NaH₂PO₄ → H₂PO₄⁻ Base form = Na₂HPO₄ → HPO₄²⁻ Equilibrium: H 2 PO 4 − → H + + HPO 4 2− , with pK₂ = 7.2 . Trap: mixing these up flips the ratio.
Front
Q49 — Why must you compare MOLES (not concentrations) in this Henderson-Hasselbalch problem?
Back
Because the two solutions have different volumes (100 mL acid vs 200 mL base). HH uses the ratio of amounts in the SAME final mixed volume, so the volume cancels only if you use moles = volume × concentration for each form.
Front
Q49 — Compute it: 100 mL of 0.12 M NaH₂PO₄ (acid) + 200 mL of 0.06 M Na₂HPO₄ (base), pK₂ = 7.2 . Find the pH.
Back
Acid: 0.100 L × 0.12 M = 0.012 mol Base: 0.200 L × 0.06 M = 0.012 mol pH = 7.2 + log(0.012/0.012) = 7.2 + log(1) = 7.2 + 0 pH = 7.2
Front
SHORTCUT — In any Henderson-Hasselbalch problem, if the moles of acid = moles of base , what is the pH instantly?
Back
pH = pKₐ (no calculation needed). Spot equal moles and write down the pKₐ — that is the entire Q49 answer ( pH = 7.2 ).
Front
TRAP #1 (Q49) — A student uses only the volumes (200 mL base vs 100 mL acid) and concludes base > acid, so pH > 7.2. Why is this WRONG?
Back
Because the base concentration is half the acid's (0.06 vs 0.12 M). The 2× volume exactly cancels the ½× concentration , so moles are equal and pH = pK a = 7.2. Use moles, never volume alone.
Front
TRAP #2 (Q49) — A student uses only the concentrations (0.12 M acid vs 0.06 M base) and concludes acid > base, so pH < 7.2. Why is this WRONG?
Back
Because the base volume is twice the acid's (200 vs 100 mL). The ½× concentration is offset by the 2× volume , so moles are equal and pH = 7.2. Always convert both forms to moles first.
Front
METHOD — Give the step-by-step recipe to solve ANY buffer-pH (Henderson-Hasselbalch) problem.
Back
1) Identify the acid and its conjugate base . 2) Convert both to moles (volume × concentration). 3) pH = pK a + log([base]/[acid]). 4) If base moles = acid moles → pH = pKₐ .
Front
True or false: In the phosphate buffer equilibrium H 2 PO 4 − ⇌ H + + HPO 4 2− , HPO₄²⁻ is the acid form.
Back
FALSE. H₂PO₄⁻ is the ACID (it donates the H⁺); HPO₄²⁻ is the conjugate BASE . More protons = acid. H₂PO₄⁻ = acid, HPO₄²⁻ = base.
Front
What is the buffer range — the pH window over which a buffer actually works?
Back
pKₐ ± 1 pH unit Outside this the ratio [base]/[acid] exceeds 10:1 or 1:10, one component is essentially exhausted , and the buffer fails.
Front
When is buffer capacity (β) maximized — i.e., when does the buffer best resist pH change?
Back
When [acid] = [base] , i.e. when pH = pKₐ . β = (Δ n)/(Δ pH) — the moles of strong acid/base needed to shift pH by 1 unit. Maximal capacity = minimal pH change.
Front
DISCRIMINATION — Two acetate buffers (pKₐ 4.77): one 1.0 M, one 0.1 M, both equimolar. Compare their pH and their buffer capacity .
Back
Same pH = 4.77 (both equimolar → pH = pKₐ). Different capacity : the 1.0 M buffer has ~10× greater capacity (capacity depends on absolute concentration). pH and capacity are independent properties.
Front
Write the Henderson-Hasselbalch equation for the bicarbonate buffer and the ratio that gives normal plasma pH.
Back
pH = 6.1 + log([HCO 3 − ]/[CO 2 ]) Normal: [HCO 3 − ] = 24 mM, [CO 2 ] = 1.2 mM → ratio 20:1 → 6.1 + log(20) = 6.1 + 1.30 = 7.40.
Front
Bicarbonate buffer: how do you convert pCO₂ (mmHg) to [CO₂] (mM) ?
Back
[CO₂] (mM) = pCO₂ (mmHg) × 0.03 e.g. 40 mmHg × 0.03 = 1.2 mM. Memorize the solubility coefficient α = 0.03 mM/mmHg for every blood-gas calculation.
Front
Why is the bicarbonate buffer effective at pH 7.4 even though its pKₐ is only 6.1 (1.3 units outside the ±1 range)?
Back
Because blood is an open system : the lungs continuously exhale CO₂ (the acid component), so it never accumulates. Added strong acid converts HCO₃⁻ → CO₂ , which is simply blown off . The open-system coupling beats the pKₐ mismatch.
Front
Log shortcut: what is log(20) , and why is that the number behind plasma pH 7.40?
Back
log(20) = log(2 × 10 1 ) = log 2 + 1 = 0.30 + 1 = 1.30 So bicarbonate: pH = 6.1 + 1.30 = 7.40. log 2 ≈ 0.30 is worth memorizing for fast pH math.
Front
Which body buffer dominates in blood plasma , and which dominate inside cells / red blood cells ?
Back
Plasma: the bicarbonate buffer . Inside cells/RBCs: proteins (especially hemoglobin) and phosphate (HPO₄²⁻/H₂PO₄⁻) . Location drives which buffer matters.
Front
MUST-MEMORIZE LINES — recite the two never-miss buffer rules for Q49-type questions.
Back
(1) Buffer pH = pKₐ + log(base/acid) — use moles when volumes differ. (2) Acid moles = base moles → pH = pKₐ (Q49: pH = 7.2 ). Bonus: H₂PO₄⁻ = acid, HPO₄²⁻ = base; capacity max at [acid]=[base]; range = pKₐ ± 1.
Front
In the Q51 stem, what two lab findings were given, and what is the correct diagnosis ?
Back
Increased conjugated (direct) bilirubin in blood + decreased urobilinogen in urine → obstructive (post-hepatic) jaundice .
Front
Trace the normal bilirubin pathway from old RBC to gut in 5 steps.
Back
(1) Old RBC destroyed: hemoglobin → heme → biliverdin → unconjugated bilirubin . (2) Unconjugated bilirubin (hydrophobic) travels bound to albumin . (3) Liver conjugates it with glucuronic acid via UDP-glucuronosyltransferase (UGT) → conjugated (water-soluble) bilirubin . (4) Secreted into bile → intestine . (5) Gut bacteria → urobilinogen → stercobilin (feces) / some reabsorbed → urine as urobilin.
Front
Which enzyme opens the heme ring, and what are its byproduct and immediate product?
Back
Heme oxygenase (uses O 2 + NADPH) cleaves the ring → releases carbon monoxide (CO) as a byproduct (a rare endogenous CO source) + releases iron + yields green biliverdin .
Front
Which enzyme/cofactor converts biliverdin → bilirubin ?
Back
Biliverdin reductase , using NADPH → reduces green biliverdin to yellow-orange bilirubin (the bruise color change: green → yellow).
Front
What enzyme conjugates bilirubin, what does it add, and what does conjugation accomplish?
Back
UDP-glucuronosyltransferase (UGT) adds two molecules of glucuronic acid → forms bilirubin diglucuronide = conjugated (direct) bilirubin. This makes bilirubin water-soluble so it can be secreted into bile and excreted.
Front
Unconjugated vs conjugated bilirubin — solubility, what it binds, and whether it appears in urine?
Back
Unconjugated (indirect) : water-insoluble , travels bound to albumin , NOT filtered → never in urine . Conjugated (direct) : water-soluble , NOT albumin-bound, CAN be filtered → appears in urine when it backs up (dark urine).
Front
Pre-hepatic (hemolytic) jaundice : cause, which bilirubin rises, urine bilirubin, urine urobilinogen?
Back
Cause = too much RBC breakdown (hereditary spherocytosis, G6PD deficiency, sickle cell, malaria, autoimmune). Unconjugated bilirubin ↑ . Urine bilirubin = ABSENT (unconjugated is albumin-bound, not filtered). Urine urobilinogen = INCREASED (more bilirubin reaches gut).
Front
Hepatocellular (hepatic) jaundice : cause, bilirubin pattern, and which liver enzymes are markedly elevated?
Back
Cause = damaged liver cells (viral hepatitis, alcoholic liver disease, cirrhosis, drug toxins). Pattern = BOTH conjugated and unconjugated may rise (mixed). Transaminases (ALT/AST) markedly elevated ; may see bilirubinuria. (Neonatal physiologic jaundice is hepatic in mechanism — low UGT at birth.)
Front
Post-hepatic (obstructive) jaundice : cause, which bilirubin rises, urine bilirubin, urine urobilinogen, stool color?
Back
Cause = bile cannot reach intestine (gallstone, pancreatic head cancer, cholangiocarcinoma). Conjugated bilirubin ↑ in blood. Urine bilirubin = PRESENT (dark urine). Urine urobilinogen = ABSENT/low (little reaches gut). Stool = clay-colored (acholic) . Also: ALP markedly elevated, itching from bile salts.
Front
METHOD — How do you reason from labs to obstructive jaundice in Q51?
Back
Conjugated high + urine urobilinogen low = the liver successfully conjugated it (so conjugation works) BUT bile couldn't reach the intestine (so no urobilinogen forms). Therefore the block is AFTER conjugation = obstruction .
Front
METHOD — General 3-step approach to ANY jaundice question?
Back
(1) Which bilirubin is up — unconjugated (problem BEFORE/at uptake-conjugation: hemolysis or liver) vs conjugated (problem secreting/draining: liver or obstruction). (2) Is bilirubin in urine? No → unconjugated (hemolytic); Yes → conjugated present. (3) Urobilinogen ↑ = hemolytic; ↓/absent = obstruction .
Front
Trap (urine bilirubin): True or false — in hemolytic jaundice, bilirubin appears in the urine.
Back
FALSE. In hemolytic jaundice the excess is unconjugated bilirubin, which is albumin-bound and not filtered by the glomerulus → NO bilirubin in urine . (Urobilinogen IS increased — but that is not bilirubin itself.)
Front
Trap (urobilinogen direction): Which way does urine urobilinogen move in obstructive vs hemolytic jaundice?
Back
Obstructive: urobilinogen DOWN/absent (bile blocked, little bilirubin reaches gut). Hemolytic: urobilinogen UP (more bilirubin reaches gut). Opposite directions — do not mix them up.
Front
Trap (stool & urine color): In complete biliary obstruction, what color are the stool and urine, and why?
Back
Stool = clay-colored (acholic) — no bilirubin reaches gut, so no stercobilin (the brown pigment). Urine = dark — water-soluble conjugated bilirubin accumulates in blood and is filtered by the kidney.
Front
Why does unconjugated bilirubin require albumin, and what neonatal danger does it pose?
Back
It is lipid-soluble / water-insoluble , so it binds albumin non-covalently to stay soluble in blood AND to keep it from crossing membranes freely. If it crosses into a neonate's brain it causes kernicterus . (Phototherapy makes water-soluble photoisomers excretable without conjugation.)
Front
A UGT deficiency underlies which two named syndromes, and how do they differ?
Back
UDP-glucuronosyltransferase deficiency → Crigler-Najjar (complete absence, type I — lethal without liver transplant) vs Gilbert syndrome (partial reduction — benign, mild unconjugated jaundice during fasting/illness). Both raise unconjugated bilirubin.
Front
What do gut bacteria convert conjugated bilirubin into, and what gives stool its brown and urine its yellow color?
Back
Gut bacteria reduce conjugated bilirubin to colorless urobilinogen . Most is excreted in stool as stercobilin ( brown stool ); a small reabsorbed fraction spills into urine as urobilin (yellow urine).
Front
Discrimination MCQ: A patient has elevated direct (conjugated) bilirubin, dark urine, clay-colored stools, absent urine urobilinogen, and elevated ALP. Type of jaundice?
Back
Post-hepatic (obstructive) jaundice. Conjugated bilirubin is water-soluble → backs up into blood → urine (dark). No bilirubin reaches gut → no urobilinogen/stercobilin → acholic stool. ALP rises from biliary pressure.
Front
Discrimination MCQ: Which jaundice type shows elevated UNCONJUGATED bilirubin, NO urine bilirubin, and INCREASED urine urobilinogen?
Back
Pre-hepatic (hemolytic) jaundice. Excess RBC breakdown overwhelms conjugation → unconjugated ↑ (albumin-bound, not in urine ); more bilirubin reaches gut → urobilinogen ↑ .
Front
MUST-MEMORIZE LINE (Q51 one-liner): Summarize obstructive jaundice in one sentence.
Back
Obstruction = conjugated bilirubin backs up (↑ in blood, dark urine), the intestine gets less bilirubin, so urobilinogen falls and stool turns clay-colored (acholic). ALP markedly elevated.
Front
In phenylketonuria (PKU) , what is the deficient enzyme in the classical form?
Back
Phenylalanine hydroxylase (PAH) — the most common enzyme deficiency in PKU.
Front
Write the NORMAL reaction catalyzed by phenylalanine hydroxylase (substrates, cofactor, products).
Back
Phenylalanine + O 2 + BH4 (tetrahydrobiopterin) → tyrosine + H 2 O + BH2 .
Front
What is the cofactor required by phenylalanine hydroxylase ?
Back
Tetrahydrobiopterin (BH4) — oxidized to BH2 in the reaction (regenerated by dihydropteridine reductase using NADH).
Front
MUST-MEMORIZE one-liner: PKU = ___ deficiency = phenylalanine cannot become ___.
Back
PKU = phenylalanine hydroxylase deficiency = phenylalanine cannot become tyrosine .
Front
In PKU, phenylalanine accumulates and is shunted into which three "phenylketones"?
Back
Phenylpyruvate , phenyllactate , and phenylacetate — all increased in PKU.
Front
What is the metabolic SEQUENCE that produces the phenylketones from accumulated phenylalanine in PKU?
Back
Phenylalanine → transamination to phenylpyruvate → reduction to phenyllactate → oxidation to phenylacetate .
Front
Which phenylketone produces the characteristic odor in PKU, and what does that odor smell like?
Back
Phenylacetate — gives urine/sweat a musty / mousey odor .
Front
Why does untreated PKU cause hypopigmentation (decreased skin/hair color)?
Back
Tyrosine is the precursor of melanin (via tyrosinase ). With PAH blocked, tyrosine is relatively deficient , so melanin synthesis falls.
Front
List the major CLINICAL features of UNTREATED classical PKU.
Back
Intellectual disability , seizures , musty/mousey odor , hypopigmentation , eczema (eczematoid rash) , vomiting, and developmental/motor delay.
Front
In PKU, tyrosine status: essential or non-essential? Explain.
Back
Tyrosine becomes conditionally essential — it is normally made from phenylalanine, but that route is blocked , so it must be supplied in the diet.
Front
How is the brain damage in untreated PKU thought to arise (mechanism)?
Back
High phenylalanine competes with other large neutral amino acids for transport across the blood–brain barrier , disrupting neurotransmitter synthesis in the brain .
Front
What is the TREATMENT of classical PKU?
Back
Low-phenylalanine diet (restrict dietary Phe), supplement tyrosine , BH4 if responsive, and avoid aspartame (it contains phenylalanine).
Front
When must the low-phenylalanine diet be started in classical PKU, and why?
Back
Within the first two weeks of life — to prevent irreversible neurological damage; continued at least through childhood. (Neonatal screening detects it early.)
Front
Why must PKU patients avoid aspartame ?
Back
Aspartame contains phenylalanine , which they cannot metabolize.
Front
A MINORITY of PKU cases are NOT due to PAH deficiency — what is the alternative cause?
Back
BH4 (tetrahydrobiopterin) deficiency — defective BH4 synthesis or deficiency of dihydropteridine reductase .
Front
Why does BH4 -deficient PKU NOT respond to a low-phenylalanine diet alone?
Back
BH4 is ALSO the cofactor for tyrosine hydroxylase and tryptophan hydroxylase , so its loss impairs catecholamine AND serotonin synthesis — which diet restriction cannot fix.
Front
EXAM TRAP — Why is phenylalanine aminotransferase a WRONG answer for "the deficient enzyme in PKU"?
Back
It is NOT deficient — it is the enzyme that performs the alternative metabolism of accumulated Phe (forming phenylpyruvate). The deficient enzyme is phenylalanine hydroxylase .
Front
EXAM TRAP — In a PKU stem, why is DHFR (dihydrofolate reductase) a distractor, not the answer?
Back
DHFR is a folate -pathway enzyme (regenerates THF) — unrelated to phenylalanine. The PKU enzyme is phenylalanine hydroxylase .
Front
EXAM DISCRIMINATION — A PKU question offers these enzymes. Pick the deficient one and reject the others: phenylalanine aminotransferase, DHFR, tyrosine deaminase, phenylalanine decarboxylase, phenylalanine hydroxylase.
Back
Correct = phenylalanine hydroxylase . Rejects: phenylalanine aminotransferase (makes phenylpyruvate, alt route), DHFR (folate), tyrosine deaminase (tyrosine catabolism), phenylalanine decarboxylase (not the disease enzyme).
Front
HOW TO REASON a PKU stem: newborn, mousey urine, decreased pigmentation, plasma phenylalanine HIGH, tyrosine LOW — what is the defect and why is tyrosine low?
Back
Defect = phenylalanine hydroxylase deficiency (classical PKU). Tyrosine is low because it is normally made FROM phenylalanine by PAH (using BH4 ); the blocked reaction makes tyrosine conditionally essential. Mousey odor = phenylacetate .
Front
Q57 (the exact tested item): A dye must be diluted 20× ; you need to prepare 2 mL of the diluted dye. How much starting (stock) dye do you use?
Back
100 µL (0.1 mL) of stock dye. Method: stock volume = final volume ÷ dilution factor. V stock = V final /DF = (2 mL)/20 = 0.1 mL = 100 µL Then add 1.9 mL (1900 µL) of solvent to reach 2 mL total.
Front
METHOD CARD: To make a fixed final volume at a given fold-dilution (e.g. 20×, 10×), what is the master rearrangement for the stock volume ?
Back
Stock volume = final volume ÷ dilution factor V stock = V final /DF A 20× dilution means the final volume is 20× the stock, so stock = 1/20 of final. Solvent added = V final - V stock .
Front
True or false: "Diluted 20×" means you take 1 part stock and add 20 parts solvent (so final volume = 21 parts).
Back
FALSE — classic trap. 20× dilution means the final volume is 20× the stock , i.e. 1 part stock + 19 parts solvent = 20 parts total. Stock = 1/20 of final, NOT 1/21. For 2 mL final: 0.1 mL stock + 1.9 mL solvent.
Front
Q76 (the exact tested item): You dissolve 8 g of sugar in 200 mL of coffee. What is the percent concentration (% w/v) ?
Back
4% (w/v). Method: % w/v = grams per 100 mL. % w/v = (8 g)/(200 mL) = (4 g)/(100 mL) = 4%
Front
METHOD CARD: What is the definition of percent w/v , and what does " 5% solution " literally mean in grams and mL?
Back
% w/v = grams of solute per 100 mL of solution So 5% = 5 g per 100 mL ; 0.9% saline = 0.9 g per 100 mL; 4% = 4 g/100 mL. General: % w/v = (g solute)/(mL solution) × 100
Front
Using density ≈ 1 g/mL for dilute aqueous solutions, why does the Q76 answer come out the same whether you compute % w/v or % w/w ?
Back
Because 1 mL ≈ 1 g , so 200 mL of coffee ≈ 200 g. (8 g)/(200 g) = 0.04 = 4% — same as the w/v result. The lectures explicitly permit density ≈ 1 g/mL for dilute aqueous solutions, which makes mass and volume percent interchangeable here.
Front
THE TRAP CARD: You add 8000 µL of water to 2 mL of solution. How many times (×) is it diluted?
Back
5× diluted. Trap: "how many times diluted" = final volume ÷ original volume , NOT solvent ÷ original. 8000 µL = 8 mL water + 2 mL original = 10 mL final . DF = V final /V original = (10 mL)/(2 mL) = 5×
Front
UNIT CONVERSION (must be automatic): Convert 1 mL, 0.1 mL, and 2 mL to microliters (µL).
Back
1 mL = 1000 µL 0.1 mL = 100 µL 2 mL = 2000 µL (8 mL = 8000 µL, 1.9 mL = 1900 µL). The factor is always ×1000 from mL → µL.
Front
METHOD CARD: What is the dilution factor (DF) formula, both from volumes and as a way to find the stock volume?
Back
From volumes: DF = V final /V stock Rearranged for stock: V stock = V final /DF The dilution factor is the same number as the fold-dilution (a 20× dilution → DF = 20).
Front
METHOD CARD (concentration dilutions): What is the master equation for diluting a solution by adding solvent , and what physical quantity does it hold constant?
Back
C 1 V 1 = C 2 V 2 It works because the number of moles of solute does NOT change when you add solvent — only the volume changes . Solve for whichever of C 1 , V 1 , C 2 , V 2 is unknown.
Front
Using C 1 V 1 = C 2 V 2 : to dilute 20 mL of 0.2 M HCl down to 0.01 M , what final volume do you need and how much water do you add?
Back
Final volume = (C 1 V 1 )/C 2 = (0.2 × 20)/0.01 = 400 mL Water added = 400 − 20 = 380 mL . Remember: the answer is the final volume ; subtract the starting volume to get the water to add.
Front
SERIAL / CHAIN DILUTION method: A solution is diluted 5×, then an aliquot is diluted 4× more. What is the overall dilution factor , and how do you find the original concentration?
Back
Multiply all the dilution factors together : overall DF = 5 × 4 = 20× . To get the original concentration, work backwards from the final: C original = C final × DF total . Serial dilutions are multiplicative , never additive.
Front
UNIT-CHAIN card: Going from M → mM → µM → nM , what is the factor at each step, and what is 1 M in each unit?
Back
Each step is ×10³ (×1000) . 1 M = 10 3 mM = 10 6 µM = 10 9 nM So 0.1 M = 100 mM = 10⁵ µM .
Front
RULE OF THE CROSS (alligation) method: To mix a 1% and a 0.1% solution into a 0.5% target, what ratio of volumes do you use?
Back
Volumes are inversely proportional to each component's distance from the target. 1% provides |0.5 - 0.1| = 0.4 parts; 0.1% provides |1.0 - 0.5| = 0.5 parts. Ratio (1% : 0.1%) = 0.4 : 0.5 = 4 : 5 . For 180 g total: 80 g of 1% + 100 g of 0.1% . Always verify the back-calculation.
Front
When you mix an acid and a base (e.g. KOH + HCl), what MUST you do before computing the final concentration?
Back
Account for the neutralization reaction first . Strategy: (1) compute moles of each reactant , (2) subtract stoichiometrically (what's left over), (3) divide the remaining moles by the TOTAL final volume . You cannot just dilute — H⁺ and OH⁻ cancel first.
Front
SANITY-CHECK trap: For Q57, a student writes "stock = 2 mL × 20 = 40 mL." Why is this wrong on its face?
Back
You can't need 40 mL of stock to make only 2 mL of final product — the stock must be smaller than the final volume when you're diluting. Diluting always means: divide by the fold (2 ÷ 20 = 0.1 mL), never multiply. If your stock volume > final volume, you flipped the operation.
Front
MUST-MEMORIZE LINE 1 (Dilutions): Fill in the three core formulas.
Back
DF = final volume ÷ stock volume V stock = V final /DF C 1 V 1 = C 2 V 2 Diluting = divide by the fold; moles are conserved, only volume changes; serial dilutions multiply .
Front
MUST-MEMORIZE LINE 2 (Percent): State the % w/v rule and the two worked answers you must know cold.
Back
% w/v = grams per 100 mL (5% = 5 g/100 mL). % w/v = (8 g)/(200 mL) = 4% and V stock = (2 mL)/20 = 100 µL. Use density ≈ 1 g/mL so 1 mL ≈ 1 g; 1 mL = 1000 µL.
Front
In which direction do both DNA polymerase and RNA polymerase synthesize a new strand?
Back
Always 5'→3' — each new nucleotide is added to the 3'-OH of the growing chain. There is no 3'→5' polymerization.
Front
In which direction do DNA/RNA polymerases read the template strand?
Back
The template is read 3'→5' — the opposite direction to synthesis (which is 5'→3'). Synthesis 5'→3', read 3'→5' — they are mirror directions.
Front
The must-memorize one-liner for polymerase directionality (read / synthesize / add to which end):
Back
Polymerases read 3'→5' , synthesize 5'→3' , and add nucleotides to the 3'-OH of the growing strand.
Front
What bond do DNA and RNA polymerases form between nucleotides?
Back
A phosphodiester bond — between the 3'-OH of one sugar and the 5'-phosphate of the next nucleotide.
Front
Mechanistically, how is the phosphodiester bond formed (what attacks what)?
Back
The growing chain's 3'-OH attacks the α-phosphate of the incoming nucleotide triphosphate , forming the bond and releasing pyrophosphate ( PPi ) .
Front
What is the energy source that drives polymerase bond formation?
Back
Hydrolysis of the high-energy nucleotide phosphate bonds — specifically, subsequent hydrolysis of the released pyrophosphate (PPi) makes the reaction thermodynamically irreversible. (No separate ATP is needed; the NTP substrate itself carries the energy.)
Front
What substrate (building block) does DNA polymerase use?
Back
Deoxyribonucleoside triphosphates (dNTPs) : dATP, dGTP, dCTP, dTTP — makes DNA (contains thymine, no 2'-OH).
Front
What substrate (building block) does RNA polymerase use?
Back
Ribonucleoside triphosphates (NTPs) : ATP, GTP, CTP, UTP — makes RNA (contains uracil, has 2'-OH).
Front
THE exam trap: Does RNA polymerase require a primer ?
Back
NO. RNA polymerase can initiate de novo — it needs only a DNA template, no primer. (This is the key difference from DNA polymerase.)
Front
Does DNA polymerase require a primer ? Why?
Back
YES. DNA polymerase cannot start a strand from nothing — it can only extend an existing strand with a free 3'-OH . The primer is a short RNA made by primase .
Front
Quick contrast: which polymerase needs a primer and which does not?
Back
DNA polymerase needs a primer (free 3'-OH); RNA polymerase does NOT need a primer. Both still synthesize 5'→3' .
Front
What enzyme makes the RNA primer that DNA polymerase requires, and what is it functionally?
Back
Primase — which is itself an RNA polymerase . It lays down a short (~10 nt) RNA primer on the template, providing the free 3'-OH that DNA polymerase extends.
Front
Are the common features shared by DNA and RNA polymerases? (1) synthesize 5'→3' (2) read template 3'→5' (3) form phosphodiester bonds (4) use nucleotide triphosphates (5) energy from phosphate hydrolysis
Back
ALL TRUE. Both enzymes share every one of these five features — the only differences are substrate (dNTP vs NTP) and the primer requirement (DNA pol needs one, RNA pol does not).
Front
Method to reason about any polymerase question — the 4-step checklist:
Back
(1) Direction: synthesize 5'→3' , read template 3'→5' . (2) Add to the 3'-OH . (3) Substrate = NTP/dNTP , energy from PPi hydrolysis . (4) Primer? DNA pol yes / RNA pol no .
Front
What is the proofreading activity of DNA polymerase , and which direction does it move?
Back
A 3'→5' exonuclease activity intrinsic to the polymerase: it backs up, excises a mispaired base, and reinserts the correct one. (Note: this is exonuclease proofreading, NOT the 5'→3' synthesis direction.)
Front
Which eukaryotic DNA polymerase lacks 3'→5' proofreading activity?
Back
DNA polymerase alpha (α) — it makes the primer + short initiator segment then hands off, so it has no proofreading . All other replicative polymerases (δ, ε, γ) DO proofread.
Front
Match the eukaryotic DNA polymerases to their roles: alpha (α) , delta (δ) , gamma (γ) .
Back
Pol α = primase + short initiator (no proofreading); Pol δ = main replicative polymerase (bulk synthesis); Pol γ = replicates the mitochondrial genome. (Pol ε also aids replication/repair.)
Front
Does DNA ligase synthesize DNA? What does it actually do?
Back
No — ligase cannot synthesize DNA de novo . It only seals nicks , forming a phosphodiester bond between two pre-existing strands (e.g., joining Okazaki fragments).
Front
What does K_m equal, defined in terms of velocity?
Back
K_m = the substrate concentration [S] at which velocity is half of V_max . v = ½V max when [S] = K m
Front
K_m and affinity: does a LOW K_m mean high or low affinity? HIGH K_m?
Back
Low K_m = HIGH affinity (enzyme reaches ½V_max at low [S]). High K_m = LOW affinity (needs a lot of substrate to get moving). Trap: K_m and affinity move in opposite directions.
Front
When velocity v = V max , four things are true. Name them.
Back
1. The enzyme is saturated (all active sites occupied). 2. Adding more substrate does NOT increase rate . 3. [S] ≫ K m . 4. Velocity depends on enzyme concentration (not substrate). All four are TRUE.
Front
MUST-MEMORIZE (one breath): at V_max , what matters and what no longer matters?
Back
At V_max the enzyme is saturated : substrate no longer matters ; enzyme amount does . The ONLY way to raise V_max is more enzyme .
Front
Does V_max depend on enzyme concentration? Does K_m ?
Back
V_max DOES depend on enzyme concentration — more enzyme = higher V_max (V max = k cat ·[E] total ). K_m does NOT depend on enzyme concentration — it is a property of the enzyme-substrate pair.
Front
The Michaelis-Menten equation — write it.
Back
v = V max [S]/(K m + [S]) Plots as a hyperbola of v vs [S] ; asymptotes toward V_max (never reached at finite [S]).
Front
On the M-M hyperbola: at LOW [S] vs HIGH [S], how does rate behave?
Back
Low [S]: rate depends strongly on [S] (curve is nearly linear, first-order). High [S]: enzyme is saturated , rate plateaus at V_max . At V_max, [S] ≫ K m .
Front
The Lineweaver-Burk (double-reciprocal) equation — write it.
Back
1/v = K m /V max ·1/[S] + 1/V max A straight line y=mx+b: plot 1/v (y-axis) vs 1/[S] (x-axis); slope =K m /V max .
Front
Lineweaver-Burk: what do the y-intercept and x-intercept give you?
Back
y-intercept = 1/V max (so V max =1/y-int). x-intercept = (-1)/K m (so K m = -1/x-int). y → V_max, x → K_m (both reciprocals).
Front
THE reciprocal TRAP: a HIGHER y-intercept on Lineweaver-Burk means V_max is __? A more NEGATIVE x-intercept means K_m is __?
Back
Higher y-intercept = LOWER V_max (since y-int = 1/V_max). More negative x-intercept = SMALLER K_m (since x-int = −1/K_m). Trap: they are reciprocals — read them backwards and you call the inhibition type wrong.
Front
COMPETITIVE inhibition: what happens to K_m and V_max, and can excess substrate rescue it?
Back
K_m INCREASES (apparent), V_max UNCHANGED . YES — rescuable by excess substrate (substrate outcompetes the inhibitor for the active site). Inhibitor resembles the substrate , competes for the same active site.
Front
NONCOMPETITIVE (pure) inhibition: K_m and V_max changes, and rescuable?
Back
K_m UNCHANGED , V_max DECREASES . NOT rescuable by excess substrate. Inhibitor binds an allosteric site (not the active site), so flooding with substrate cannot outcompete it.
Front
UNCOMPETITIVE inhibition: K_m and V_max changes, and where does the inhibitor bind?
Back
BOTH K_m and V_max DECREASE . Inhibitor binds only the ES complex (not free enzyme) — it cannot bind until substrate has bound. Trapping ES as ESI lowers apparent K_m but those complexes are dead-end, so V_max falls. Not rescuable.
Front
Lineweaver-Burk SIGNATURES: match each inhibition type to its intercept change.
Back
Competitive : same y-intercept , x-intercept shifts right toward 0 (V_max same, K_m up). Noncompetitive : same x-intercept , y-intercept rises (K_m same, V_max down). Uncompetitive : parallel lines (slope unchanged, both down).
Front
EXAM TRAP: An enzyme + inhibitor shows K_m unchanged and V_max decreased (e.g. K_m stays 0.5 mmol/L, V_max drops 0.15 → 0.03 mmol/L/min). Which inhibition type? Rescuable?
Back
Noncompetitive (or irreversible) . V_max down, K_m unchanged → NOT rescuable by adding substrate. On LB: x-intercept (−1/K_m) stays, y-intercept (1/V_max) rises.
Front
EXAM TRAP: Inhibitor A → same x-intercept, higher y-intercept . Inhibitor B → same y-intercept, less-negative x-intercept . Which is which?
Back
A = noncompetitive (same K_m → same x-int; lower V_max → higher y-int). B = competitive (same V_max → same y-int; higher apparent K_m → x-int shifted toward 0). y-intercept tracks V_max; x-intercept tracks K_m.
Front
TRAP: How can you tell irreversible inhibition apart from noncompetitive on a kinetics plot alone?
Back
You can't — they look identical (both: V_max down, K_m unchanged, same x-intercept). The distinction is functional: washing out the inhibitor restores activity for reversible noncompetitive, but NOT for irreversible (covalent). An exam may mark BOTH as correct in the same scenario.
Front
Distinguish V_max from k_cat (turnover number). Which is the intrinsic enzyme property?
Back
V max = k cat ×[E] total k_cat = molecules of substrate one enzyme converts per second when saturated — intrinsic , independent of [E]. V_max scales with how much enzyme you have (double enzyme → double V_max). Trap: don't confuse the two.
Front
Clinical: Methotrexate competitively inhibits which enzyme , and what is the downstream block?
Back
Methotrexate competitively inhibits dihydrofolate reductase (DHFR) → blocks THF → blocks nucleotide synthesis → blocks rapidly dividing cancer cells . Binds ~1000× tighter than substrate (practically irreversible, mechanistically competitive ). ACE inhibitors (captopril) are the other competitive example.
Front
Clinical isozyme map: which enzyme/subunit marks the heart (MI) , and which mark liver and muscle?
Back
CK-MB → myocardium (MI marker) ; CK-MM → skeletal muscle; CK-BB → brain. ALT/AST → liver damage . LDH1 rises before LDH5 in MI. Saturating-substrate velocity reads enzyme concentration → enzyme leakage = tissue damage signal.
Front
Q68 tested fact: What is the main lipid carried by LDL ?
Back
Cholesterol — both esterified ( cholesteryl esters in the core) and free. LDL is the cholesterol-delivery particle.
Front
True or false: LDL is smaller than HDL.
Back
FALSE — size order largest→smallest is chylomicron > VLDL > IDL > LDL > HDL . HDL is the SMALLEST (and densest). LDL is larger than HDL.
Front
What is the general architecture of any lipoprotein particle (core vs surface)?
Back
Hydrophobic CORE : TAGs + cholesteryl esters . SURFACE monolayer : phospholipids, free cholesterol, apoproteins . Amphipathic shell → makes nonpolar fat water-soluble for transport.
Front
Memory line — match each particle to the lipid it carries: chylomicron / VLDL / LDL / HDL.
Back
Chylomicron + VLDL carry TAG; LDL carries cholesterol; HDL cleans cholesterol (reverse transport). Memorize this one line — it answers most lipoprotein MCQs.
Front
Chylomicron : origin, cargo, and main apoprotein ?
Back
Made by the intestine ; transports dietary TAG to tissues; main apo = apoB-48 (also picks up apoC-II + apoE from HDL). Remnants cleared by liver via apoE .
Front
VLDL : origin, cargo, and main apoprotein ?
Back
Made by the liver ; transports endogenous TAG ; main apo = apoB-100 (also apoC-II + apoE ). VLDL is TAG-rich, apoB-100.
Front
LDL : where does it come from, what does it deliver, and which single apoprotein stays on it?
Back
Derived from VLDL → IDL metabolism; delivers cholesterol to tissues; the ONLY apo remaining is apoB-100 (apoC-II and apoE are lost). LDL = cholesterol-rich, apoB-100 only.
Front
HDL : function, and main apoprotein ?
Back
Made by liver/intestine ; runs reverse cholesterol transport (scavenges free cholesterol from tissues → liver); main apo = apoA-I , which activates LCAT . HDL = reverse transport, apoA-I.
Front
Trap — apoB-48 vs apoB-100: which binds the LDL receptor , and where is each made?
Back
apoB-48 = intestine only, chylomicrons only, CANNOT bind the LDL receptor . apoB-100 = liver only, on VLDL/IDL/LDL, IS the LDL-receptor ligand . Do not assume both use the same receptor.
Front
Through what receptor and apoprotein is LDL taken up by cells — and is it only the liver?
Back
Via the LDL receptor (B/E receptor) , which recognizes apoB-100 , by receptor-mediated endocytosis. Taken up by MANY tissues, not only liver (cells need cholesterol for membranes, steroids, bile acids).
Front
What is the cofactor/activator of lipoprotein lipase (LPL) , and where does LPL sit?
Back
Activator = apoC-II . LPL is on the luminal surface of capillary endothelium (adipose/muscle); it cleaves TAG in chylomicrons/VLDL → fatty acids + glycerol . Insulin stimulates adipose LPL (fed state).
Front
LPL isoforms — why does muscle get fatty acids before adipose when supply is low?
Back
Muscle LPL has a LOW Km (high affinity) , so it works even at low plasma TAG → muscle has first priority. Adipose LPL has a HIGH Km → only efficient after a meal (high TAG) → storage is post-prandial.
Front
What enzyme matures HDL, what activates it, and what does it do?
Back
LCAT (lecithin:cholesterol acyltransferase), activated by apoA-I ; it esterifies free cholesterol (fatty acid from lecithin) → cholesteryl ester sinks into the HDL core. Drives reverse cholesterol transport.
Front
How does free cholesterol delivered by LDL regulate the cell, in three coordinated ways?
Back
(1) Suppresses HMG-CoA reductase (stops endogenous synthesis); (2) activates ACAT (re-esterifies cholesterol for storage); (3) downregulates the LDL receptor (no more uptake than needed). This triple feedback breaks in familial hypercholesterolemia.
Front
Trap — impaired LDL uptake causes which blood lipid pattern?
Back
High LDL + high cholesterol = hypercholesterolemia — NOT isolated hypertriglyceridemia . Familial hypercholesterolemia (Type IIa) = defective LDL receptor OR apoB-100 → atherosclerosis.
Front
Type I hyperlipoproteinemia (LPL or apoC-II deficiency) : lab pattern and the key trap?
Back
Massive hypertriglyceridemia (chylomicrons + VLDL not cleared), milky plasma, pancreatitis. LDL/HDL paradoxically LOW. Trap: NO increased coronary risk in the classic form.
Front
Familial hypercholesterolemia (Type IIa) : cause, and why do statins fail in homozygotes ?
Back
Cause = defective LDL receptor or defective apoB-100 → LDL not cleared → foam cells → atherosclerosis . Statins upregulate LDL-receptor expression, so they help heterozygotes but NOT homozygotes (no functional receptor to upregulate).
Front
Method card — to answer ANY lipoprotein MCQ, what 4 things do you ask about the particle?
Back
(1) Origin (gut vs liver)? (2) Main cargo (TAG vs cholesterol)? (3) Main apoprotein (B-48 / B-100 / A-I)? (4) Which enzyme/receptor handles it (LPL/apoC-II, LDL-R/apoB-100, LCAT/apoA-I)? Structure → function.
Front
Method card — "work backward" pathway logic: chylomicrons absent in fasting plasma — except when?
Back
Chylomicrons/remnants should be virtually absent fasting . If present → suspect a defect in LPL or its activator apoC-II (i.e. Type I ). Persistent TAG-rich particles = unloading enzyme failure.
Front
Never-miss summary: give the cargo + signature apoprotein for all four particles in one line.
Back
Chylomicron = dietary TAG, apoB-48 · VLDL = endogenous TAG, apoB-100 · LDL = cholesterol, apoB-100 · HDL = reverse cholesterol transport, apoA-I .
Front
The 4 weakness categories on this exam — name all four (the cross-cutting reasoning map).
Back
1) Structure tells function · 2) Charge, pH, and math · 3) Pathway diagnosis · 4) Kinetic reasoning. Every bolded miss falls into one of these — name the category first, then apply its method.
Front
Category 1 " Structure tells function ": which questions belong, and what 3 questions do you ask?
Back
Belongs: IgG structure , TAG vs phospholipid, LDL structure, DNA/RNA polymerases . Ask: What are the components? What is the structure? What does that structure ALLOW it to do?
Front
Category 2 " Charge, pH, and math ": which questions belong, and what is the one-word method?
Back
Belongs: glutamate pKa, buffer pH, dilutions, percent. Method: SLOW DOWN and calculate — these punish rushing. Don't eyeball; write the numbers.
Front
Category 3 " Pathway diagnosis ": which questions belong, and what is the reasoning direction?
Back
Belongs: bilirubin/jaundice , PKU , DNA repair. Method: identify what ACCUMULATES, then work backward (e.g. phenylpyruvate high → PAH deficiency; conjugated bilirubin high + urobilinogen low → obstruction; uracil in DNA → BER).
Front
Category 4 " Kinetic reasoning ": which questions belong, and what is the method?
Back
Belongs: Vmax /Km, polymerase directionality, enzyme saturation. Method: memorize the few core rules and apply them every single time — no derivation, just recall the rule.
Front
Pathway-diagnosis worked examples — give the three "what accumulates → cause" inferences the method produces.
Back
Phenylpyruvate high → PAH deficiency. Conjugated bilirubin high + urobilinogen low → obstruction . Uracil in DNA → BER . (Accumulation points to the broken step just upstream.)
Front
Never-miss #1: In a primary immune response, which antibody appears FIRST?
Back
IgM before IgG — naive B cells make IgM first, then class-switch to IgG. NOT the reverse.
Front
TRUE or FALSE (exam trap): "IgG appears earlier than IgM."
Back
FALSE . IgM comes first ; IgG is the class-switched antibody that comes later.
Front
Never-miss #2: Describe IgG structure and how it binds antigen.
Back
2 heavy + 2 light chains , folded immunoglobulin domain structure, antigen binding is noncovalent (H-bonds, ionic, hydrophobic, van der Waals).
Front
TRUE or FALSE (exam trap): "IgG binds antigen covalently."
Back
FALSE . Antigen binding is noncovalent — specific but reversible. Never say covalent.
Front
Never-miss #3: State the pKa rule for protonation state.
Back
pH protonated ; pH > pKa → group is deprotonated . (Low pH = lots of H+ = stays protonated.)
Front
FILL IN (trap): "When pH > pKa, the group is mostly ____."
Back
deprotonated . (pH ABOVE pKa = lost its proton. Below pKa = protonated.)
Front
Never-miss #4: Glutamate at pH 4.1–9.5 — list the group charges and the net charge.
Back
Two COO⁻ (α-carboxyl + side-chain carboxyl) and one NH3⁺ → net −1 . (−1 −1 +1 = −1.) A net −1 species migrates toward the anode.
Front
Never-miss #5: Distinguish a triacylglycerol from a phospholipid by structure.
Back
TAG = glycerol + 3 fatty acids (storage). Phospholipid = glycerol + 2 fatty acids + phosphate head (amphipathic membrane lipid). Count the fatty acids: 3 = TAG, 2 = phospholipid.
Front
Glycerol backbone esterified to THREE fatty acids, no phosphate — is it amphipathic and a main membrane constituent?
Back
No — that's a triacylglycerol (mostly hydrophobic, energy storage). Amphipathic membrane lipid = phospholipid (2 FA + phosphate head).
Front
Never-miss #6: Match the three lipases to what they hydrolyze and where.
Back
Pancreatic lipase → digests dietary TAG in the intestinal lumen . LPL → hydrolyzes lipoprotein TAG (chylomicron/VLDL) at capillaries . HSL → mobilizes stored TAG in adipocytes (fasting).
Front
Never-miss #7: Uracil found in DNA is repaired by which system?
Back
Base excision repair (BER) — glycosylase removes the base → AP site → AP endonuclease → polymerase → ligase. (Cytosine deaminates to uracil.)
Front
Never-miss #8: Bulky adducts and thymine dimers are repaired by which system?
Back
Nucleotide excision repair (NER) — removes a short oligonucleotide around the helix-distorting lesion. (Defective NER = Xeroderma pigmentosum.)
Front
Never-miss #9: A post-replication base mismatch / slippage loop is fixed by which system?
Back
Mismatch repair (MMR) — corrects wrong bases and insertion/deletion loops AFTER synthesis.
Front
Never-miss #10: O6-methylguanine (an alkylated base) is repaired by what?
Back
MGMT — direct repair . MGMT transfers the methyl group onto itself; the base is fixed WITHOUT removal.
Front
DNA-repair discrimination: a UV-induced thymine dimer is repaired by ____ (NOT BER).
Back
NER (bulky/helix-distorting). BER = small non-bulky base damage like uracil; MMR = replication mismatches; MGMT = direct alkyl removal.
Front
Repair shortcut (one line each): one damaged base? bulky distortion? wrong base after replication? methyl removed directly?
Back
One damaged base → BER . Bulky distortion → NER . Wrong base after replication → MMR . Methyl removed directly → MGMT / direct repair .
Front
Never-miss #11: State the Henderson-Hasselbalch buffer equation.
Back
pH = pK a + log([base]/[acid]) — i.e. pH = pKa + log(base/acid) . Compare MOLES, not concentrations, when volumes differ.
Front
Never-miss #12: If acid moles equal base moles in a buffer, what is the pH?
Back
pH = pKa — because log(1) = 0. (Equal moles ⇒ the log term vanishes.)
Front
Buffer trap: 100 mL of 0.12 M acid vs 200 mL of 0.06 M base, pKa 7.2 — does the volume/conc difference change the pH?
Back
No — convert to MOLES: 0.012 mol each → base/acid = 1 → pH = 7.2 . Using volume-only or concentration-only is the trap.
Front
Never-miss #13: Obstructive (posthepatic) jaundice — give the conjugated bilirubin, urobilinogen, stool, and urine findings.
Back
Conjugated bilirubin HIGH, urobilinogen LOW , pale stool , dark urine . Bile can't reach the intestine, so little urobilinogen forms.
Front
Jaundice MCQ: conjugated bilirubin ↑ in blood + urobilinogen ↓ in urine = which jaundice?
Back
Obstructive (posthepatic) — liver conjugated it, but bile couldn't reach the intestine, so urobilinogen falls.
Front
Never-miss #14: Phenylketonuria (PKU) is a deficiency of which enzyme?
Back
Phenylalanine hydroxylase (PAH) — phenylalanine can't be converted to tyrosine , so it accumulates.
Front
PKU enzyme MCQ: increased phenylpyruvate/phenyllactate/phenylacetate — which enzyme is deficient?
Back
Phenylalanine hydroxylase . Phe builds up and overflows into the phenylketones.
Front
Never-miss #15: The Phe → Tyr reaction requires which enzyme and which cofactor?
Back
Phenylalanine hydroxylase + cofactor tetrahydrobiopterin (BH4) . (Phe + O₂ + BH4 → Tyr + H₂O + BH2.)
Front
Never-miss #16: A 20× dilution to a final 2 mL — how much stock do you start with?
Back
100 µL stock (= 0.1 mL), plus 1.9 mL solvent. Stock = final volume ÷ dilution factor = 2 mL ÷ 20.
Front
Never-miss #17: Define % w/v.
Back
grams per 100 mL (e.g. 5% w/v = 5 g/100 mL; 8 g in 200 mL = 4 g/100 mL = 4%).
Front
Never-miss #18: State the directionality of BOTH DNA and RNA polymerase.
Back
Both read template 3'→5', synthesize 5'→3' , adding nucleotides to the free 3' OH end.
Front
Polymerase trap: which polymerase requires a primer, and which does not?
Back
DNA polymerase NEEDS a primer (free 3' OH). RNA polymerase does NOT need a primer . Both still synthesize 5'→3'.
Front
Never-miss #19: At v = Vmax, what is true about the enzyme and about [S] relative to Km?
Back
Enzyme is saturated and [S] >> Km. Adding more substrate does nothing more.
Front
Never-miss #20: Does Vmax depend on enzyme concentration? Does Km?
Back
Vmax depends on enzyme concentration (more enzyme → higher Vmax). Km does NOT depend on enzyme concentration.
Front
Never-miss #21: Competitive inhibition — what happens to Km and Vmax?
Back
Km UP, Vmax SAME . Substrate can outcompete the inhibitor, so max rate is still reachable.
Front
Never-miss #22: Pure noncompetitive inhibition — what happens to Km and Vmax?
Back
Km SAME, Vmax DOWN . Affinity unchanged; max rate falls.
Front
Inhibitor MCQ trap: an inhibitor causes Km ↓ AND Vmax ↓ — which type is it?
Back
Uncompetitive — binds the ES complex; both Km and Vmax decrease (Lineweaver-Burk = parallel lines). Competitive = Km↑/Vmax same; noncompetitive = Km same/Vmax↓.
Front
Never-miss #23–26: One line each for LDL, VLDL, chylomicron, HDL (main lipid/role + main apoprotein).
Back
LDL = cholesterol-rich , apoB-100. VLDL = TAG-rich , apoB-100. Chylomicron = dietary TAG , apoB-48. HDL = reverse cholesterol transport , apoA-I.

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