Question

240 Human Phys Quantitative Questions-4 - Protected View - Saved - Search References Mailings Review View Help Quantitative Questions 4 - Urinary, Energy Metabolism & Exercise 1. (5 pts) If the plasma concentration of inulin is 1 mg inulin/mL plasma and the plasma concentration of substance X is 1 mg/mL and the GFR = 125 ml/min: 1.1 What is the filtration rate (mg/min) of inulin? and of substance X? 1.2 What is the excretion rate (mg/min) of inulin? Of substance X? 2. (5 pts) If 140 liters of plasma are filtered in a day and the filtration fraction is 20% 2.1 What is the renal blood flow 2.2 If the person has a hematocrit of 30%, what is the renal blood flow? 2.3 If renal blood flow is 20% of this person's cardiac output, what is her cardiac output in min? 3. (5 pts) Dinner - spaghetti and meatballs! HINT: p.697 of Silverthorn 3.1 What is the calorie content of a serving of spaghetti and meatballs that contain 6 g fat, 30 g carbohydrate and 8 g of protein? 3.2 What is the percentage of calories that comes from fat?

Answer 1

For inulin to = GFR x P. Conlonin Englmi) 1 Quantity filleted peenin = 125 1 Filtratoniate = 125 mg/min. = For Inulio I Quantity filleted peemin = Quantity excected pee min.. - GFR x P R xu - because inulin is a polyfuctose molecule with mol.ut 45500 Da The moleculae size of inulin is sufficiently small that it is freely filled at glomeentus. Howwe Inulin is too large to be reabsoebed passively by nephen di thee is no ait ve Transport peocessess that egeabsces a seceete inulin. So, Inillin exceited eate = 125 inglmin. And for Inubor plasma conch a veine conch And exceton cate = U XR-veine flow rate von deine coonn UXR = 125 2= 125/1 = 1252 Deine flow eate Thus X Filtration cate ha scanned withxeeton eate = 125 mgnin. & 125 mg/min CamScanner

2 Here 140 litres of plasma cue fitted in a day it Noinally 180 litree of plasma are fillard in a day - with CFR = 120 milonon So; for 180 120 - GFR 140 - ? > (93:33 mlmin GFR Filloceton flacton(%) = GFR x 100 RPF Renal plasma flow eate 2013 CF PX1003 RPF = 1900 9333 XIDO DPRD So, Renal blood flow L e fePF = 466.66 mlimin = RPF ta Hematoceit (Noerrolly Hematoce .466.66 1-0-45 So, [Renal blood flow = 84 8.48 milmin cs Scanish CamScanner

If a person has hematocent of 20% (03) Renal blood flow = RPF 1- Hematouit = 466-66 . I RBF = 666.66m/lmin of 1 If the patient Denal blood flow il 20% this person's Cardiac output Her caudac output is = 1.2L/min CS scanned with CamScanner