Question

To determine the mechanism of reabsorption of compound Y, scientists evaluated ion concentrations in the peritubular capillaries. Here is what they found: K* decrease increase Nat [ion]peritubular capillary ionltubular lumen increase decrease When they pre-treated the patient with rotenone, an inhibitor of the electron transport chain in the mitochondria, the rate of absorption of compound Y looked like Time Based on this information, propose a mechanism of transport for the reabsorption of Compound Y

Answer 1

q1. There are three type od transport mechanism

A) Passive transport-> Dependent on concentration of the solute. And the movement of solute is along the concentration gradient ( ie from high concentration to low concentration, if the solute is allowed to cross the membrane). It does not need any ATP. Rate of trans port depedns on the concentration gradient.

b) Facilitated or carrier mediated transport: This type of trasnport is done via carrier protein. when all the carrier protein is occupied( saturated) the rate of transport rate becomes constant. This depends on concentration gradient plus availability of carrier proteins.

C) Active transport: Primary active transport. This happens against the concentration gradient, uses ATP energy for transport mechanism. Do es not depends on the concentration gradient. If ATP depleted then trasnport will stop.

Now the answer: You see the graph started from zero to maximum then to zero, information given is that electron transport chain is blocked by rotenone. Electron transport chain is related to the ATP production in crebs cycle. So new ATP production stopped by rotenone. Now the with stored ATP inside cell reabsorption of the compund Y is reached maximum-> then ATP store of the cell started to diminish as there is no new ATP production due to blocked electron transport chain [ though few ATP produced by glycolysis, that can not match the transport speed, glycolysis eventually stops due ATP=0, because the initial step of glycolysis need 2 molecule ATP] --> as a result reabsorption of the compound Y started to decrease and when all the ATP depleted reabsorption od copound Y completely stopped.

So the reabsorption mechanism is ACTIVE TRANSPORT.

Again you can assume that Na+ is co-absorbed with the Compound Y via same active transporter, becuase data is given that tubular lumen has decreased Na+ concentration. And how to explain K+, as you know that in kidney there is na+ k+ co transporter which drives K+ out to lumen in exchange of Na+

Please see the picture:

Transpont Transport [concentration gradient is maintained at a constant value] Time Concentration grdient Passive transport All the carrier protein is saturated and concentration gradient is maintained at aonstant value Active transport [ provided if ATP supply continues] no new ATP production, cell ATP store starting to deplete ATP availab here for tran port Active transport ATP depleted state Cell ATP amount reached to 0 [ATP-0] so transport stopped transport time>

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Q2.

We know that if there is no absorption, no secretion and no protein binding of a drug , then GFR is equal to clearance.

We know clerance ( CL) = rate of drug elimination by kidney / Plasma concentration

                               Rate of drug elimation     = ( Drug concentration in urine x Urine output / time ).

So we can write CL =( urine drug concentration x uine output/time ) / Plasma concentration of drug.

So for Drug X clearance CL_X= ( 56 ng ml^-1 x 5/1.5 ml Hr ^-1) / 44 ng ml^-1

                                                       = ( 186.67 )/44 ml /Hour

                                         $\approx$ 4.24 ml /Hour

As for drug X all the drug amount is excreted, CL_x = GFR = 4.24 ml/ Hour

Q2.

GFR = 4.24 ml/ Hr

For drug Y amount of drug Y filtered out via Glomerulus : CL_GFRY= GFR x Plasma concentration of Y

                                                                                  CL_GFRY = 4.24 x 250 ng /ml / Hour

But CL_GFRY x 65/100 ng reabsorbed

      CL_GFRY x 35/100 ng secreted by proximal tubule

      CL_GFRY x 4/100 ng secreted by distal tubule

So final excretion:

glomerular filter amount - reabsorbed amount + secreted amount

= CL_GFRY - CL_GFRY65 /100 + CL_GFRY35 /100 + CL_GFRY4/100

= CL_GFRY [ 1 - 65/100 + 35/100 + 4/100 ]

=CL_GFRY[ (100 - 65 +35 +4 )/100 ]

=CL_GFRY[ ( 74)/100]

=CL_GFRY (74 /100)

=(4.24 x 250 ) ( 74 /100 ) ng/hr

= 784.4 $\approx$ 784 ng/hr

So Drug Y is excretion rate is 784 ng/Hour.

So clearance of CL_Y= (784 ng/hr) /(250 ng/ml ) = 3.14 ml /hr