Numerical on GFR

1. A person with a normal GFR of 120 mL/min has only 10 % functional nephrons left. Calculate the GFR per nephron.

Answer:

• Let N be the total number of nephrons in a healthy kidney.
• Since the person has only 10% functional nephrons left, the number of functional nephrons is 0.10 * N.
• The GFR per nephron can be calculated as:

$$ GFR\ per\ nephron = \frac{Total\ GFR}{Number\ of\ functional\ nephrons} = \frac{120\ mL/min}{0.10 * N}$$

$$ = 12\ mL/min/nephron $$

Therefore, the GFR per nephron in this person is 12 mL/min.

2. A healthy person has a creatinine concentration of 1 mg/dL in his blood. If the GFR is 100 mL/min, what is the amount of creatinine excreted in the urine per 24 hours?

Answer:

• The amount of creatinine excreted in the urine per minute can be calculated as:

$$ Creatinine\ clearance = GFR \times Creatinine\ concentration\ in\ blood $$ $$ = 100\ mL/min\ \times 1\ mg/dL $$

$$ = 100\ mg/min $$

• The amount of creatinine excreted in the urine per 24 hours can be calculated as:

$$ Creatinine\ excretion = Creatinine\ clearance\ \times 60\ min/hr \times 24\ hr/day $$ $$ = 100\ mg/min \times 60\ min/hr \times 24\ hr/day $$

$$ = 144,000\ mg/day$$

Therefore, the amount of creatinine excreted in the urine per 24 hours is 144,000 mg/day or 144 g/day.

Numerical on Tubular Reabsorption:

1. If the rate of GFR is 120 mL/min and the rate of tubular reabsorption is 115 mL/min, what is the volume of urine formed per minute?

Answer:

• The volume of urine formed per minute can be calculated as the difference between the GFR and the tubular reabsorption rate:

$$ Urine\ volume = GFR - Tubular\ reabsorption\ rate $$

$$ = 120\ mL/min - 115 mL/min $$ $$ = 5\ mL/min $$

Therefore, the volume of urine formed per minute is 5 mL.

2. If the concentration of glucose in the glomerular filtrate is 180 mg/dL and the concentration of glucose in the urine is 0 mg/dL, what is the percentage of glucose reabsorbed in the tubules?

Answer:

• The percentage of glucose reabsorbed in the tubules can be calculated using the following formula:

$$ Percentage\ Reabsorption\ = \frac{GF\ Glucose - Urine\ Glucose}{GF\ Glucose}\ \times 100%$$ $$ = \frac{180\ mg/dL - 0\ mg/dL}{180 mg/dL}\ \times 100%$$

$$ = 100%$$

Therefore, 100% of the glucose in the glomerular filtrate is reabsorbed in the tubules, indicating complete reabsorption of glucose in the proximal convoluted tubules.

Numerical on Tubular Secretion:

1. If the rate of GFR is 120 mL/min and the rate of tubular secretion of hydrogen ions is 45 mL/min, what is the net amount of hydrogen ions excreted in the urine per minute?

Answer:

• The net amount of hydrogen ions excreted in the urine per minute can be calculated as the difference between the tubular secretion rate and the GFR:

$$ Net\ hydrogen\ ion\ excretion = Tubular\ secretion\ rate - GFR$$

$$ = 45\ mL/min - 120\ mL/min $$ $$ = -75\ mL/min$$

Therefore, the net amount of hydrogen ions excreted in the urine per minute is -75 mL/min, indicating a net reabsorption of hydrogen ions.

2. If the concentration of penicillin in the glomerular filtrate is 100 mg/dL and the concentration of penicillin in the urine is 200 mg/dL, what is the percentage of penicillin secreted in the tubules?

Answer:

• The percentage of penicillin secreted in the tubules can be calculated using the following formula:

$$ Percentage\ Secretion = \frac{Urine\ Penicillin\ - GF\ Penicillin}{Urine\ Penicillin} \times 100% $$

$$ = \frac{200\ mg/dL - 100\ mg/dL}{200\ mg/dL} \times 100%$$

$$ = 100%$$

Therefore, 100% of the penicillin in the glomerular filtrate is secreted in the tubules, indicating complete tubular secretion of penicillin in the proximal convoluted tubules.