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Excretion in Humans - Urea formation and dialysis

Grade 12IGCSEBiology

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Deamination: This occurs in the liver where excess amino acids are broken down. The nitrogen-containing amino group (−NH2-NH_2) is removed to form ammonia (NH3NH_3).

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Urea Synthesis: Ammonia (NH3NH_3) is highly toxic and alkaline. In the liver, it is combined with carbon dioxide (CO2CO_2) to form urea (CO(NH2)2CO(NH_2)_2), which is less toxic and soluble in water.

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Transport and Excretion: Urea is released into the bloodstream, transported to the kidneys, filtered out of the blood in the glomerulus, and excreted as a component of urine.

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Principles of Dialysis: Kidney dialysis is a treatment for kidney failure. It uses a partially permeable membrane to separate the patient's blood from dialysis fluid (dialysate).

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Dialysis Fluid Composition: The fluid contains a glucose and salt concentration equal to that of 'normal' blood to ensure no net loss of these essential substances. However, it contains 0%0\% urea to create a steep concentration gradient.

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Counter-current Flow: In the dialysis machine, blood and dialysis fluid flow in opposite directions. This maintains a concentration gradient for urea across the entire length of the membrane, maximizing the removal of waste products.

📐Formulae

2NH3+CO2→CO(NH2)2+H2O2NH_3 + CO_2 \rightarrow CO(NH_2)_2 + H_2O

Amino Acid→DeaminationAmmonia+Keto AcidAmino \ Acid \xrightarrow{Deamination} Ammonia + Keto \ Acid

Net Movement=P×A×(Cblood−Cdialysate)Net \ Movement = P \times A \times (C_{blood} - C_{dialysate})

💡Examples

Problem 1:

Explain why the dialysis fluid must be constantly replaced during the treatment session.

Solution:

As dialysis progresses, urea diffuses from the blood into the dialysis fluid. If the fluid is not replaced, the concentration of urea in the fluid (CdialysateC_{dialysate}) will eventually equal the concentration in the blood (CbloodC_{blood}).

Explanation:

According to the principles of diffusion, net movement stops when an equilibrium is reached (Cblood=CdialysateC_{blood} = C_{dialysate}). Constantly replacing the fluid ensures CdialysateC_{dialysate} remains near 00, maintaining a steep concentration gradient for maximum urea removal.

Problem 2:

Calculate the amount of urea produced if 1.7 g1.7 \ g of NH3NH_3 is processed in the liver (Molar masses: NH3=17 g/molNH_3 = 17 \ g/mol, CO(NH2)2=60 g/molCO(NH_2)_2 = 60 \ g/mol).

Solution:

n(NH3)=1.717=0.1 moln(NH_3) = \frac{1.7}{17} = 0.1 \ mol. From the equation 2NH3→1CO(NH2)22NH_3 \rightarrow 1CO(NH_2)_2, moles of urea = 0.05 mol0.05 \ mol. Mass of urea = 0.05×60=3.0 g0.05 \times 60 = 3.0 \ g.

Explanation:

Based on the stoichiometry of the urea cycle, two molecules of ammonia are required to produce one molecule of urea.

Urea formation and dialysis - Revision Notes & Key Diagrams | IGCSE Grade 12 Biology