Back to the on-screen lesson ·
Acid-base speciation at zinc, carbonic anhydrase regeneration, and the limits of a two-state model.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
Calculate aqua and hydroxo populations and trace how a zinc enzyme regenerates its reactive ligand.
A Lewis acid accepts an electron pair, and zinc ions can accept donor pairs from water or protein groups. Recall that losing a proton converts water to hydroxide without transferring an electron to another chemical species. You have also distinguished equilibrium populations from reaction rates. That distinction will matter whenever a hydroxo fraction is used to discuss an enzyme's readiness to react.
| Term | What it means |
|---|---|
| Aqua ligand | A water molecule coordinated to a metal through oxygen. |
| Hydroxo ligand | A coordinated hydroxide ion, the conjugate base of a bound water molecule. |
| Lewis acid | An electron-pair acceptor; zinc can polarize a coordinated water molecule. |
| Proton shuttle | A group or connected pathway that transfers a proton between the active site and surrounding buffer. |
| Speciation | The distribution of a total amount among its chemical forms at stated conditions. |
| Carbonic anhydrase | An enzyme that accelerates the reversible interconversion of carbon dioxide and bicarbonate with water and protons. |
Free water is a weak acid. When its oxygen donates an electron pair to a positive metal center, the electronic environment of the water changes. A bound hydroxide can be stabilized relative to bound water, making proton loss easier than for an isolated water molecule. In acid-base language, the pKa of coordinated water can be much lower than the pKa of free water. The size of the change depends on the metal, its other ligands and the surrounding protein.
This gives zinc a catalytic role without requiring zinc to change oxidation state. In carbonic anhydrase, a familiar zinc enzyme, a metal-bound hydroxide acts as a nucleophile toward carbon dioxide. Zinc remains in its usual divalent state through this acid-base and ligand-exchange description. Treating every metal enzyme as a redox enzyme would miss the central chemistry here. Electron-pair donation and proton transfer need not be oxidation-reduction.
For a simple two-state site, label the protonated aqua form A and the hydroxo form B. Henderson-Hasselbalch gives $pH=pKa+\log_{10}([B]/[A])$. Let $r=[B]/[A]=10^{pH-pKa}$. The hydroxo fraction is $f=r/(1+r)$, and the aqua fraction is $1/(1+r)$. The denominator contains both forms. A ratio of ten does not mean a fraction of ten: ten hydroxo sites for every aqua site means ten out of eleven sites are hydroxo.
Our numerical models use invented pKa values and buffered conditions. They isolate the equilibrium between two specified forms. They do not include substrate binding, inhibitors, unfolding or every protonatable group in a real protein. Their value is that each assumption is explicit and the resulting population can be calculated and checked.
Another way: steps
Subtract bound-water pKa from the buffer pH. Raise ten to that difference to obtain hydroxo over aqua. Divide by one plus the ratio to obtain the hydroxo fraction. Multiply by one hundred for percent or by total sites for an amount. Keep that equilibrium amount separate from any statement about reaction speed.
Start with a zinc-bound hydroxide poised next to carbon dioxide. The hydroxide oxygen attacks the carbon atom, forming a new oxygen-carbon bond and producing bicarbonate. The metal and protein help position the reacting groups; writing a bulk hydroxide concentration alone would omit this local organization.
Bicarbonate then leaves the zinc coordination environment and water takes its place. The regenerated aqua complex is not yet the same state as the starting hydroxo complex. It still carries the proton that must be transferred away before the same nucleophilic form is restored. A proton-transfer pathway connects the metal-bound water to the surrounding buffer. In the well-studied human carbonic anhydrase II system, a histidine residue contributes to that proton-shuttling function.
Adding the steps cancels the metal-containing catalyst and gives carbon dioxide plus water forming bicarbonate plus a proton. At suitable conditions the reverse process also occurs. Catalysis accelerates approach to equilibrium; it does not create a different equilibrium constant for the overall reaction. A useful cycle drawing or ordered list should show both product release and regeneration. Leaving out proton transfer gives an incomplete cycle even if the carbon-containing product has already appeared.
An isolated zinc ion and a zinc ion inside a protein do not have identical chemical environments. The surrounding donor groups influence charge distribution and the stability of the aqua and hydroxo forms. Hydrogen bonds can further stabilize one state or orient a water network. Consequently, a pKa is a property of a specified site under specified conditions, not a universal number attached to the element zinc.
An altered amino acid can affect several quantities at once. It may change the metal's binding affinity, the bound-water pKa, access by carbon dioxide, the geometry of the reacting groups or the ease of proton transfer. Observing a slower enzyme after a substitution does not by itself identify which of these changed. Measurements of metal occupancy, equilibrium speciation and reaction rates answer different questions.
This is a reason to compare carefully controlled models. First keep the total sites and intrinsic chemical step fixed while changing only the pKa. Then examine a separate change to the proton-transfer rate. Combining all changes in one experiment may be biologically realistic, but it makes causal interpretation harder. The simple model is useful when its restricted question is stated: how many sites occupy each acid-base form before reaction under a given buffer pH?
At pH equal to pKa, the aqua and hydroxo concentrations are equal. Each is half of the total two-state population. One pH unit above pKa gives ten hydroxo sites per aqua site, approximately ninety-one percent hydroxo. One unit below gives the inverse ratio, approximately nine percent hydroxo. Increasing pH favors the deprotonated form, as acid-base reasoning predicts.
Under a restricted model in which only the hydroxo form reacts and every other kinetic factor remains constant, more hydroxo sites can increase the initial rate. That sentence contains assumptions. At high pH a real enzyme may change structure or protonation elsewhere, and proton transfer or product release may limit the cycle. An acid-base population curve therefore cannot prove that the measured rate will keep rising indefinitely.
Nor does an enzyme with all sites in the hydroxo form carry out infinitely many reactions at once. Each site must proceed through binding, chemical conversion, product release and regeneration. An equilibrium fraction is dimensionless, while a turnover frequency has units of inverse time. Equating the two is a dimensional error before it becomes a mechanistic error. If a problem asks for a rate, it must provide additional kinetic information or a clearly stated rate model.
The calculations treat pH as fixed by the surrounding buffer. This means the amount of proton exchanged by the modeled sites is too small to change the stated pH appreciably, or an external system maintains it. If a concentrated enzyme preparation releases a substantial proton amount into a weakly buffered solution, the initial pH may no longer apply. A full calculation then needs the buffer's composition and a proton balance.
The two forms in the denominator must account for all sites being counted. If some zinc sites bind an inhibitor, a substrate or an additional ligand, adding only aqua and hydroxo populations no longer gives the total active-site concentration. Either restrict the total to the unoccupied two-state pool or explicitly add the other species. Do not silently apply the same denominator to a chemically different inventory.
Activities rather than raw concentrations are the thermodynamic basis of acid-base constants. At this level the stated model absorbs or neglects activity effects so the concentration ratio can be used directly. Comparing values from different ionic strengths or temperatures requires more care than substituting two numbers into the same formula. The point of stating conditions is to make comparisons interpretable, not to complicate arithmetic unnecessarily.
Check the sign of the pH minus pKa difference before exponentiating. A negative difference should give a ratio smaller than one and a hydroxo percentage below fifty. A positive difference should do the opposite. Exactly at the midpoint the ratio is one, not zero, because ten to the zeroth power equals one.
Check the normalization next. Hydroxo percentage plus aqua percentage should equal one hundred apart from the final rounding. An answer above one hundred usually came from treating a species ratio as a fraction. For amounts, hydroxo sites cannot exceed the total zinc sites used in that calculation. Reconstruct the aqua amount by subtraction and divide hydroxo by aqua to recover the original ratio.
Finally label the result with what it actually measures. Micromoles of hydroxo sites is a population. Micromoles of bicarbonate per second is a rate. A statement about the first does not establish the second. A result can be numerically correct yet answer the wrong chemical question if the label is changed casually at the end. Keep unrounded fractions during multiplication and round only the reported amount.
A Wisconsin university group compares two invented model sites in the same pH-eight buffer. Both samples contain twenty-two micromoles of zinc sites. The first has bound-water pKa seven, so its hydroxo fraction is ten elevenths and its hydroxo amount is twenty micromoles. The second has pKa eight, so its fraction is one half and its amount is eleven micromoles. The first preparation has more hydroxo sites at equilibrium.
That result does not establish that it is the faster catalyst. The altered ligands might hinder carbon dioxide access or slow product release. A defensible report gives the speciation result first and calls for separate kinetic measurements. If both preparations had identical intrinsic reaction behavior and only hydroxo sites reacted, the equilibrium fractions could enter a simple initial-rate comparison, but those are extra assumptions to test rather than facts supplied by the pKa values.
An Illinois laboratory examines a proposed enzyme-assisted carbon dioxide capture step. In its simplified initial mixture, eleven micromoles of sites have bound-water pKa seven and the buffer is held at pH six. One eleventh of the sites are hydroxo, so one micromole occupies that form before substrate is introduced. Raising the buffered pH to eight would give ten micromoles in the same form if the two-state model and site inventory remain valid.
This demonstrates a large change in readiness for nucleophilic attack, but it does not calculate the capture plant's throughput or energy use. Those depend on gas transfer, reaction kinetics, product handling and maintenance of the buffer. The molecular calculation provides one input to an engineering assessment. Its clear boundary prevents a plausible microscopic explanation from becoming an unsupported claim about the entire process.
A metal can participate by polarizing a ligand and stabilizing an acid-base form. In this cycle zinc does not have to alternate between oxidation states to help water react. The proton transferred from coordinated water is not an electron transferred from zinc. Follow the chemical species rather than assuming a mechanism from the mere presence of a metal.
A second mistake is claiming that a catalyst drives the net equilibrium toward product. If initial conditions favor the reverse reaction, the enzyme can accelerate that direction too. Its cycle gives a faster route between the same overall reactants and products. Finally, the hydroxo fraction is not the fraction of enzymes permanently active: individual sites interconvert, and the fraction describes a population under the stated conditions.
State the buffered conditions.
$pH=7,\quad pKa=7$
The model contains only aqua and hydroxo forms.
Subtract the acid constant.
$pH-pKa=7-7=0$
This is the logarithm of the species ratio.
Exponentiate the difference.
$r=10^0=1$
Equal concentrations give a ratio of one.
Normalize the hydroxo population.
$f=1/(1+1)=0.5$
Both forms belong in the total.
Express both percentages.
$100f=50\%,\quad100(1-f)=50\%$
The two percentages account for all sites.
Identify the acid-base offset.
$pH-pKa=8-7=1$
The solution favors the conjugate base.
Calculate hydroxo over aqua.
$r=10^1=10$
Ten hydroxo sites correspond to one aqua site.
Normalize the site ratio.
$f=10/(1+10)=10/11$
Eleven parts make up the total.
Convert the fraction to percent.
$100(10/11)\approx90.91\%$
Keep the exact fraction until reporting.
Check the complementary fraction.
$100(1/11)\approx9.09\%$
The two rounded percentages add to one hundred.
State the total site inventory.
$C=33\ \mathrm{micromol},\ pH=6,\ pKa=7$
All counted sites belong to the two-state pool.
Calculate the species ratio.
$r=10^{6-7}=0.1$
Aqua is favored below the midpoint.
Normalize the hydroxo fraction.
$f=0.1/(1+0.1)=1/11$
A ratio to aqua must be converted to a fraction of all sites.
Multiply by available sites.
$33(1/11)=3\ \mathrm{micromol}$
This is an equilibrium amount of hydroxo sites.
Reconstruct the remaining population.
$33-3=30\ \mathrm{micromol},\quad3/30=0.1$
The reconstructed ratio matches the acid-base equation.
Distinguish population from speed.
$3\ \mathrm{micromol}\ne3\ \mathrm{micromol/s}$
No time or kinetic constant was supplied.
Find the base-to-acid ratio.
$r=10^{9-8}=10$
The pH is one unit above pKa.
Normalize the hydroxo fraction.
$f=10/(1+10)$
Add both forms in the denominator.
Report the hydroxo percentage.
Order one idealized carbonic anhydrase hydration cycle, beginning with zinc-bound water. The last step must regenerate that same ligand state.
Number the steps in order (write the number in the box):
Complete a two-state zinc-water calculation at pH $6$ with bound-water pKa $7$. Round the hydroxo percentage to two decimal places.
Calculate hydroxo over aqua.
$r=$ r
Exponentiate the pH minus pKa difference.
Normalize and convert to percent.
$100f=$ f
The denominator includes both species.
Check the population bound.
$0<f<1$
A fraction cannot exceed the available sites.
A two-state zinc-site model has a metal-bound-water pKa of $7$ at pH $8$. Find the ratio of hydroxo to aqua sites, then the hydroxo and aqua percentages. Use $r=10^{pH-pKa}$ and round percentages to two decimal places.
| hydroxo / aqua | hydroxo (%) | aqua (%) | |
|---|---|---|---|
| zinc sites |
At pH $6$, a two-state zinc site has hydroxo-to-aqua ratio $0.1$. Its base-ten logarithm is $-1$. Infer the bound-water pKa.
| inferred pKa | |
|---|---|
| bound water |
Order one idealized carbonic anhydrase hydration cycle, beginning with zinc-bound bicarbonate. The last step must regenerate that same ligand state.
Number the steps in order (write the number in the box):
A two-state zinc-site model has a metal-bound-water pKa of $7$ at pH $6$. Find the ratio of hydroxo to aqua sites, then the hydroxo and aqua percentages. Use $r=10^{pH-pKa}$ and round percentages to two decimal places.
| hydroxo / aqua | hydroxo (%) | aqua (%) | |
|---|---|---|---|
| zinc sites |
A Wisconsin research lab models $101$ micromol of zinc sites using only aqua and hydroxo forms. Bound-water pKa is $8$ and buffer pH is $6$. Find the hydroxo percentage and micromoles of hydroxo sites before substrate is added. Round final answers to two decimal places.
| hydroxo (%) | hydroxo sites (micromol) | |
|---|---|---|
| before substrate |
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
A two-state zinc-site model has a metal-bound-water pKa of $8$ at pH $9$. Find the ratio of hydroxo to aqua sites, then the hydroxo and aqua percentages. Use $r=10^{pH-pKa}$ and round percentages to two decimal places.
| hydroxo / aqua | hydroxo (%) | aqua (%) | |
|---|---|---|---|
| zinc sites |
Can you distinguish a hydroxo-to-aqua ratio from a hydroxo fraction, and explain why neither alone specifies turnover rate?
16. A model site has bound-water pKa eight at pH nine. Find its hydroxo percentage., step 3
$100f\approx90.91\%$
Round at the end.