Numerical Problems/ISC/Class 11
Created by Titas Mallick
Biology Teacher • M.Sc. Botany • B.Ed. • CTET (CBSE) • CISCE Examiner
Created by Titas Mallick
Biology Teacher • M.Sc. Botany • B.Ed. • CTET (CBSE) • CISCE Examiner
Numerical Problems - Excretory System
In a healthy individual, the concentration of creatinine in plasma is 0.01 mg/mL, and its concentration in urine is 1.2 mg/mL. If the urine flow rate is 1 mL/minute, calculate the Glomerular Filtration Rate (GFR) in mL/minute, assuming creatinine is freely filtered and not reabsorbed or secreted.
Solution:
The GFR can be calculated using the clearance formula, which for a substance like creatinine (freely filtered, not reabsorbed or secreted) is:
GFR = (Urine Concentration of Substance * Urine Flow Rate) / Plasma Concentration of Substance
Given:
GFR = (U_creat * V) / P_creat GFR = (1.2 mg/mL * 1 mL/minute) / 0.01 mg/mL GFR = 1.2 / 0.01 mL/minute GFR = 120 mL/minute
Therefore, the Glomerular Filtration Rate (GFR) is 120 mL/minute.
If a person's GFR is 125 mL/minute, and 99% of the filtrate is reabsorbed, what is the approximate daily urine output in liters?
Solution:
First, calculate the total volume of filtrate formed per day:
Now, calculate the volume of urine formed, given that 99% of the filtrate is reabsorbed:
Convert milliliters to liters:
Therefore, the approximate daily urine output is 1.8 liters.
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