Biochem Exam 2 (second half)`

How do you calculate fractional saturation and partial pressure for a protein like hemoglobin?

Use the Hill equation: Y = pO₂ⁿ/(p50ⁿ + pO₂ⁿ). Rearranged (extra): pO₂ = p50 · [Y/(1−Y)]^(1/n)

What is linear hill plot?

The Hill plot turns a curved binding curve into a straight line, so you can read two numbers directly: how cooperative the binding is (n) and how tightly it binds (p50).

Linear hill plot equation

log(Yo2/(1-Yo2)) = nlogpo2 - nlogp50

Axes, slopes, and intercepts of Linear Hill plot

y-axis: log(Yo2/(1-Yo2))

x-axis = nlogpo2

y-intercept=-nlogp50

x-intercept=logp50

slope=n

How do you obtain the Hill coefficient and partial pressure values from a Hill plot?

Find n: pick two points on the straight part of the line and compute rise over run, Δy/Δx.

1. Find log p50: read where the line crosses y = 0 (the x-intercept).

2. Convert: p50 = 10^(x-intercept). The axis is a log, so you have to undo it.

3. Check with the y-intercept (optional): −n·log p50 should match.

what slope and horizontal poistions tells you about about

slope is cooperativity: a steeper slop means more cooperative binding

horizontal position is affinity: a line further left has a smaller p50, so it binds tighter

Why is hemoglobin’s Hill plot is not one straight line?

Hb’s plot has three regions

Region	Y range	y-axis range	Slope (n)	p50	Meaning
① Low	< 0.1	below about −0.95	1	about 30 torr	Non-cooperative. The 1st O₂ binds independently, with Hb in its low-affinity T-state.
② Middle	0.1 to 0.9	−0.95 to +0.95	2.8	26 torr	Cooperative. After the 1st O₂ binds, the next ones bind with greater affinity (T → R).
③ High	> 0.9	above about +0.95	1	about 0.3 torr	Non-cooperative. Hb is in the high-affinity R-state, and the last O₂ binds independently and very tightly.

myoglobin hill plot intercepts and slope

slope: 1

x-int: 0.45

y-int: -0.45

x and y intercepts of Hb

region 1: x-int=1.5, y-int=-.15

region 2: x-int=1.4, y-int=-3.9

region 3: x-int=-0.5, y-int=0.5

why is Hb plot so complicated?

• Hb can exist in two forms: a low-affinity T-state and a high-affinity R-state.

• Each has its own n = 1 line. The R-state line sits far to the left (p50 about 0.3 torr) and the T-state line sits far to the right (p50 about 30 torr).

• Hb’s real data starts on the T-state line (①), switches steeply to the R-state line (②), and ends on it (③).

• The steep middle section is the T → R transition, and its slope is the Hill coefficient. The maximum slope (about 3) is as steep as it gets.

Overall: Hb goes from a deoxy state with low O₂ affinity (the “T-state”) to an oxy state with high O₂ affinity (the “R-state”).

What are the differences between the T and R states of hemoglobin?

	T-state (tense)	R-state (relaxed)
Form	Deoxy-Hb	Oxy-Hb
O₂ affinity	Low	High
When it dominates	Low pO₂	High pO₂

what is the molecular basis of cooperativity and why

R: each O2 that binds pushes the equilibirum toward R

ion pairs and hydrogen bonds in the T-state

extra interactions in deoxyhemoglobin (T-state). ion-pair interactions in the T-state break in the R-state.

• The N-terminal –NH₃⁺ of Val 1 (blue “+”) forms an ion pair (with a Cl⁻ bridging in the α chain).

• The C-terminal –COO⁻ (circled; the red “−”) of Arg 141 (α) or His 146 (β) pairs with a Lys (+) from the other subunit and with other residues.

• Asp 126 (α) or Asp 94 (β) (−) pairs with the positive side chain of the C-terminal residue.

• Tyr side chains H-bond to backbone C=O groups.

These salt bridges and H-bonds link the subunits together, holding Hb in the T-state and making O₂ binding harder. Moving to the R-state breaks them, so the R-state is looser (relaxed) and binds O₂ more easily.

what is the Bohr effec

pH affects O₂ binding: pH ↑ → O₂ binding ↑ (shifts toward R); pH ↓ → binding ↓ (toward T). Higher pH deprotonates –NH₃⁺ to –NH₂, which destabilizes the T-state. In muscle, CO₂ ↑ → H⁺ ↑ → pH ↓ → O₂ released; in the lungs, the reverse.

why does pH increasing shift toward r

because –NH₃⁺ is deprotonated to –NH₂ and the T-state is destabilized. The T-state depends on positive charges (like the N-terminal –NH₃⁺) forming ion pairs. Remove the protons and those ion pairs can’t form.

it’s the reverse for low pH: more protons keep the groups charged, ion pairs persist, the T-state is stabilized, and Hb releases O₂ more readily.

How does carbon dioxide affect the oxygen binding properties of hemoglobin? bohr effect

In muscle:
CO₂ ↑ → HCO₃⁻ ↑ and H⁺ ↑ → pH ↓ → T ⇌ R shifts toward T → O₂ affinity ↓ → favors O₂ release.

In the lungs:
CO₂ ↓ (exhaled) → the reaction runs backwards, HCO₃⁻ is converted to CO₂ and H⁺ is consumed → pH ↑ → shifts toward R → O₂ affinity ↑ → favors O₂ binding.

How does carbon dioxide affect the oxygen binding properties of hemoglobin?covalent modifictation

The N-terminal amine of Hb (–NH₂) can be covalently modified by CO₂, forming a carbamate: –NH–COO⁻. CO₂ ↑ → carbamate formation → favors T → O₂ affinity ↓ → O₂ released.

How does BPG affect the oxygen binding properties of hemoglobin?

• BPG binds tightly and non-covalently to the T-state (deoxyHb).

• It sits in the central cavity between the β subunits, which is lined with positively charged side chains (the blue “+” residues in your figure, such as Lys, His, and the N-terminal amine). The BPG’s negative charges (red) pair with them.

• In the R-state, the binding pocket is too narrow to fit BPG, since the β subunits have moved closer together.

So BPG stabilizes the T-state:

T + BPG ⇌ T·BPG → equilibrium pulled toward T → O₂ affinity ↓ → helps release O₂.

BPG structure

see notes. it’s a small, very negatively charged molecule

How does BPG function in the high-altitude adaptation?

At altitude, arterial pO₂ falls (100 → 55 torr), so O₂ release drops (38% → 30%). The body makes more BPG, so p50 rises from 26 to 31 torr and release recovers to about 37%.

What are the differences between fetal and adult hemoglobin? How does this different relate with Hb function?

• Adult Hb: α₂β₂. Fetal Hb: α₂γ₂ (the β chains are replaced by γ chains).

• In the γ chain, one His that is in the β chain becomes Ser. That His is one of the residues that bind BPG

• With the positive His gone, fetal Hb doesn’t bind BPG well.

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What does fetal Hb help the fetus?

1. During pregnancy, maternal BPG ↑, so in the mother’s blood O₂ affinity ↓ (T-state favored).

2. Fetal Hb isn’t affected by BPG, so it keeps its higher O₂ affinity.

3. The affinity difference favors the transfer of O₂ from the mother to the fetus across the placenta.

Skeletal muscle organization

Muscle → fascicle (bundle of muscle fibers) → individual muscle fiber (a cell, with many nuclei) → myofibril → sarcomere

what composes muscle fibers?

A muscle fiber (a cell) is a parallel array of myofibrils, each made of repeating sarcomeres containing thick and thin filaments.

What proteins comprise the thick and thin filaments of muscle?

Thick = myosin (2 heavy + 4 light chains; N-terminal heads, C-terminal coiled-coil tails).

Thin = actin (G-actin monomers polymerize non-covalently into F-actin).

Muscle contraction cycle

1. ATP binds to myosin heads; actin binding site opens; myosin head releases actin

2. Active site closes and ATP is hydrolyzed (to ADP + Pi), causing cocking of myosin head, a conformational change that moves it away from actin.

3. The myosin head binds weakly to the actin

4. Pi is released, resulting in strong binding of myosin head to actin

5. The head moves (swivels) and the thin filament moves with it. This is where the stored energy becomes motion. ADP is still bound (the “transient intermediate”).

6. ADP is released, and the head is back where it started, bound to actin. A new ATP binds and the cycle repeats.

Myosin structure

• 6 polypeptide chains: 2 heavy chains and 4 light chains.

• The N-terminal ends of the heavy chains form the myosin head, a globular domain (with the light chains attached, circled in red).

  • The C-terminal ends of the heavy chains form a long coiled-coil, where the two heavy chains wind around each other. Your note reminds you of the a-b-c-d-e-f-g heptad repeat with nonpolar residues at a and d, the same pattern as keratin

Thick filament structure

• A thick filament contains hundreds of myosin units bundled together.

• The tails pack together to form the filament’s core, and the heads stick out around the surface (the circled loops in the figure).

• The middle has a bare zone with no heads

Two forms of actin

	G-actin	F-actin
Shape	Globular (a single monomer)	Fibrous (a polymer)
Role	Building block	Forms the thin filament

F-actin forms when many G-actin monomers aggregate with one another non-covalently.

Big idea of myosin-actin interaction

Motor proteins convert chemical energy (from ATP hydrolysis) into mechanical energy (to move the thick and thin filaments).

What is an antibody?

B-cells make it, they detect bacteria/viruses outside cells and tag them for deconstruction

what is antigen

any molecule—usually a protein, polysaccharide (sugar), lipid, or nucleic acid—that your immune system can recognize and use to identify whether a substance belongs in your body

Describe the structure of an antibody (chains; constant/variable regions and disulfide bond formation, Fab and Fc domain), and identify the antigen binding site.

Y-shaped, 2 heavy + 2 light chains, joined by disulfide bonds. Constant regions (stem and lower section) are the same across antibodies of a type; variable regions (V_L and V_H) differ in every antibody and recognize the antigen. Papain cleaves at the hinge into Fc (stem) and two Fab (antigen-binding) fragments. The antigen-binding site is at the tips, formed by the hypervariable loops of V_L and V_H, binding non-covalently (K_alpha in the very low range, so tight).

Describe how our immune system generates a large diversity of antibodies

Gene recombination: light chains combine V, J, C segments and heavy chains combine V, D, J, C. Each B cell makes a unique combination, giving about 10¹⁸ different antibodies.

What is Kalpha in antibodies

K_d is the [antigen] required to occupy half of the Ab. A smaller K_d = a stronger interaction.

What are the characteristic features of a carbohydrate?

Carbohydrates (sugars, glycans, saccharides) are polyhydroxy aldehydes or ketones with several chiral centers.

What is the difference between an aldose and a ketose?

An aldose has an aldehyde (C1 C=O, as in glucose); a ketose has a ketone (C=O inside connected to two R groups

D-glucose fisher

see notes

D-glucose Hayworth

see notes

D-glucose chair confromation

see notes

What is the anomeric carbon and alpha and beta anomers of a carbohydrate?

C1 is the anomeric carbon, a new stereocenter formed on cyclization. α: the C1 –OH points down; β: it points up. They interconvert by bond breaking (opening to the linear form and re-closing from the other face), while chair conformers interconvert without bond breaking.

enatiomer

all centers inverted (G-glucose and L-glucose)

epimer

one center inverted