The Role of the Lungs and the Kidneys in Acid Base

Question 1

What is short term control of (H+) achieved by?

Answer 1

Dilution: the H+ generated in, say, exercising muscle are to some extent ‘shared’ with the rest of the body.
Buffering: a variety of chemical buffers complexes free H+ almost instantaneously.

Question 2

How is CO2 carried in plasma?

Answer 2

Hydration of carbon dioxide to carbonic acid occurs spontaneously, but relatively slowly; erythrocytes (diffuses in passively), and many tissues, contain carbonic anhydrase, an enzyme that catalyses this process, ensuring that equilibrium is rapidly established. The ionization of carbonic acid to form bicarbonate occurs spontaneously and very rapidly

Question 3

What is H+ buffered by?

Answer 3

haemoglobin (within erythrocytes) and by plasma proteins

Question 4

How does H+ affect the Bohr shift?

Answer 4

Protonated haemoglobin (HbH+) has a lower oxygen affinity than unprotonated Hb, so increased [H+] promotes ‘unloading’ of HbO2, while haemoglobin acts as an important buffer of H+ within erythrocytes.

Question 5

What happens to bicarbonate in the erythrocyte?

Answer 5

does not remain within the erythrocytes, but enters the plasma in exchange for chloride, through the operation of an antiporter in the erythrocyte membranes.

Question 6

How is the sequence reversed in the lungs?

Answer 6

CO2 is exhaled, causing the [H+] to drop (removal of the CO2 from the left hand side of the CO2/bicarbonate equilibrium pulls the equilibrium position to the left, removing [H+] from solution).
This drop in [H+] instantaneously causes the release of H+ from the haemoglobin (i.e. haemoglobin acts as a buffer of [H+]). This increases the oxygen affinity of Hb, promoting its oxygenation

Question 7

What conclusions can be made about the role of the lungs in acid base disturbances?

Answer 7

• Normal lung function makes a vital contribution to the maintenance of plasma [H+] within the narrow range compatible with health
• If the lungs are not functioning normally, there will be disruption of plasma [H+].
• Healthy lungs have the ability to correct any potentially dangerous change in plasma [H+].

Question 8

What is respiratory acidaemia?

Answer 8

If there is decreased ventilation, CO2 accumulates, the dissociation equilibrium will be pushed to the right and blood [H+] rises (pH falls)

Question 9

What is respiratory compensation?

Answer 9

In response to a rise in plasma [H+], the rate and depth of breathing increases and more CO2 is expired

Question 10

How does haemoglobin work?

Answer 10

As a standard buffer, its ionizing amino acid sidechains accepting or donating H+ as required, as happens with any protein.
By accepting an H+ when it becomes deoxygenated and vice versa according to the equilibrium shown in the figure.
Same effect as conventional buffering. Conversion of oxyhaemoglobin to deoxyhaemoglobin, the accompanying change in the protein conformation increases the pKa values of the free amino-groups at the N-termini, so that they bind H+ more readily.

Question 11

What is the Bohr effect?

Answer 11

Deoxy-haemoglobin is a weaker acid than oxy-haemoglobin; the change in pKa results in the binding of about 0.7H+, per O2 released.

Question 12

What is the equation for haemoglobin and oxyhaemoglobin?

Answer 12

HbO2 + H+ (reversible reaction) HbH+ + O2

Question 13

What happens in a working muscle?

Answer 13

The respiring tissue generates CO2 and/or lactic acid. Both will cause an increase in [H+] in the blood supplying the muscle.
This rise in [H+] will instantaneously cause a shift to the right in the position of the oxygen/haemoglobin equilibrium, releasing O2 from the haemoglobin and making it available for the respiring tissue
H+ effect additive to unloading of O2 caused by drop in ppO2

Question 14

How is the lungs the reverse of a working muscle?

Answer 14

The body expels CO2, so the PCO2 of the blood plasma drops, causing a fall in the [H+].
This fall causes a shift in the oxygen-binding equilibrium to the left, increasing the affinity for O2 and facilitating the oxygenation of haemoglobin.

Question 15

Describe the oxygen saturation curve for haemoglobin

Answer 15

Sigmoidal- reflecting cooperativity between the four oxygen-binding sites on the molecule.
• The lower the pH (i.e. the higher [H+ is), the more sigmoidal the curve, and the smaller the range of PO2 over which oxy-haemoglobin releases its bound oxygen. This is the Bohr effect.
• In the lungs haemoglobin is almost completely saturated with oxygen; when it circulates to tissues, oxygen is released because the local PO2 is lower. Notice that only a fraction of the bound oxygen is released, although a greater fraction is released at lower pH.
The binding curve for the oxygen-storage protein myoglobin (Mb) is not sigmoidal, but hyperbolic (there is only one oxygen-binding site, so there is no cooperativity in oxygen binding): Mb always has a higher oxygen affinity than Hb, whatever the pH, so haemoglobin always gives up its oxygen to myoglobin.

Question 16

What is meant by cooperativity in oxygen binding?

Answer 16

Tetramer- 4 subunits symmetrically arranged.
Exist in two possible conformations, called T (for taut) and R (for relaxed).
• In deoxy-Hb, all the subunits are in the T conformation, which has a relatively low oxygen affinity.
• On binding of the first molecule of O2 to a haem in one of the subunits, that subunit changes to the R conformation (caused by a small change in the size of the iron atom in the haem, which results in movement of the protein chains surrounding it). Note that the iron is always in the ferrous (Fe2+) state, whether oxygen is bound to it or not.
• Noncovalent inter-subunit bonds transmit the conformation change to neighbouring subunits, so the whole molecule flips to the R state, with increased oxygen affinity.
• The binding of the first O2 molecule thus increases the affinity of the haemoglobin for the next three molecules: the binding of O2 to Hb is cooperative.

Question 17

Why does cooperative binding produce a sigmoid saturation curve?

Answer 17

steep in the middle, ensuring that oxy-haemoglobin releases its bound oxygen over a narrow and physiologically appropriate range of partial pressure. The pKa values of several aminoacid side chains are different in the R and T states (higher in the T state): this means that the R/T equilibrium is pH-dependent, which is the basis of the Bohr effect.

Question 18

What happens to the H+ that are released by the ionisation of carbonic acid?

Answer 18

released H+ do acidify the plasma.
Biological membranes have very low passive permeability to H+.
H+/Na+ exchanger in the erythrocyte membrane.
H+-pumps are important in the stomach, the kidneys and in some endomembranes, but not in erythrocytes
H+-conducting channels in the erythrocytes occur as part of H+ pumps

Question 19

Where is H+ transmitted to if not buffered by haemoglobin?

Answer 19

plasma (otherwise there would be no such thing as respiratory acidaemia). This occurs through the action of the H+/Na+ antiporter

Question 20

How are erythrocytes like other cells (in terms of exchange)?

Answer 20

have an antiporter (exchange carrier protein) in the plasma membrane, which exchanges H+ for Na+. The action of this antiporter regulates the cytoplasmic H+ concentration.

Question 21

What is the plasma concentration of Na and why?

Answer 21

The concentration of Na+ is more than ten-fold greater in the plasma than it is within cells, because a ‘sodium pump’ (Na+,K+-ATPase) exports Na+, setting up a transmembrane concentration difference.

Question 22

How does Na+/H+ antiport use Na concentration difference?

Answer 22

drive import of Na+ into the cells and export H+, thus acidifying the plasma and making the cytoplasm more alkaline.

Question 23

What would happen if Na exchange ever reached equilibrium?

Answer 23

[H+] within the cells would be about 3 nM (pH~8.5)! However the activity of the H+/Na+ antiporter is tightly regulated, and equilibrium is never achieved; the cytoplasm becomes only slightly alkaline, relative to the plasma.

Question 24

How is the antiport inhibited?

Answer 24

The antiporter is inhibited by amiloride, a drug that has a diuretic effect because it lowers the rate of Na+ reabsorption of by the kidneys.

Question 25

Name non-volatile acids

Answer 25

• lactic acid (from glucose metabolism in exercising muscle),
• ‘ketone bodies’ acetoacetic and 3-hydroxybutyric acids (from fat degradation),
• sulphuric acid (from the oxidation of the sulphur-containing amino acids cysteine and methionine),
• phosphoric acid from the breakdown of nucleic acids
• acids derived from ingested substances, such as salicylic acid from aspirin

Question 26

What are the equations to work out (H+) and (CO2)?

Answer 26

(H+)= 800 x (CO2)/(HCO3-)
(CO2)= (H+)(HCO3-)/800

Question 27

What systems are effective in limiting the increase in (H+)?

Answer 27

combination of the bicarbonate buffering system and respiratory compensation

Question 28

What line of defence is the kidneys?

Answer 28

fourth line of defence against a change in [H+] in body fluids (the first three lines being dilution, buffering and the lungs).

Question 29

Why does the renal system response take so long?

Answer 29

hours to develop and may continue for days, because it requires the synthesis and insertion of membrane proteins. Although it is slow to act it is an extremely powerful correcting force.

Question 30

What are the renal responses (in order) to high plasma (H+)?

Answer 30

-Increased H+ secretion to urine
-HCO3- re-absorption from urine
-Addition of new HCO3- to plasma
Overall result= reduced plasma (H+), increased plasma (HCO3-)

Question 31

What are the renal responses (in order) to low plasma (H+)?

Answer 31

-Decreased H+ secretion into urine
-Increased HCO3- excretion into urine
Overall result increased plasma (H+), reduced plasma (HCO3-)

Question 32

What is the free H+ concentration?

Answer 32

30 μmol/L; the remainder are bound to urinary buffers, of which phosphate is the most important