Unit IV Homework Bio 12 Saddleback College Name______

For extra practice:

http://highered.mcgraw-hill.com/sites/0072351136/student_view0/chapter26/chapter_quiz.html

Homework: Osmotic Pressure Problem.

Neatly present all calculations requested below.

You know that plasma is a solution, composed of many different solutes in water (Table 25-2). Most of those solutes are sodium chloride (0.9g%). But there are also proteins in plasma (Total protein = 7g%), and albumin accounts for more than half of that protein. So, let’s assume albumin’s concentration in plasma is 4g%.

1) Compute the osmolalities for each substance, albumin and NaCl, in plasma.

Necessary information:

NaCl molecular weight = 58.5 g/mole. The molecule is formed by ionic bonds, and dissociates into 2 ions.

Albumin molecular weight = 70,000g/mole. Albumin is composed of atoms which are covalently bonded, and do not dissociate.

It is a fact that the osmotic pressure exerted by 1osmolal (1osmole/L) of solute is 19,300 mmHg; (that’s the same as 1mOsmolal (1mOsmole/L) solute exerting 19.3 mmHg pressure.)

(“mm Hg” is read as “millimeters of mercury” and it is the unit used to express pressures.)

2) Compute the osmotic pressures exerted by each of the substances, albumin and NaCl in plasma. Which substance is larger? There are more molecules of which substance in plasma? With this information, which substance exerts the greater osmotic pressure in plasma, NaCl or protein?

“Colloid osmotic pressure” is the term given to the osmotic pressure exerted by protein.

Reference: Guyton Hall, Pg 51-52 and 296-298


Renal Glucose Handling Homework Biology 12 Saddleback College

You know that renal handling of solutes is described by this equation:

Rate of Excretion = Rate of filtration + Rate of secretion – Rate of reabsorption.

(Review lecture material covered in class this week.)

The terminology is new to you, so remember that

• rate of excretion of any solute (s) is also called urinary output, and can be determined by the

equation:

Urinary output (s) = Urine concentration (s) X urine volume produced/minute

• rate of filtration of fluid is also called glomerular filtration rate (GFR)

• rate of filtration of any solute (s) is also called Tubular Load (s) determined by the equation:

Tubular Load (s) = GFR x Plasma concentration of (s)

Background:

As you know, glucose is freely filtered at the glomerulus, and is reabsorbed but not secreted.

Glucose is reabsorbed only in the proximal convoluted tubule, and the method for glucose reabsorption

involves carriers on the luminal membrane. The problem is, all of us only have a limited number of carriers

for glucose, and (as you saw in an early lab this semester on facilitated diffusion (refer to PhysioEx on

transport), if you have limited numbers of carriers, you limit the maximum rate or speed at which you can

move solutes across membranes. This type of transport, therefore, is called “transport maximum limited ”

or “Tm-limited reabsorption”. We see this type of reabsorption for other sugars and amino acids, but not

for other solutes like NaCl. We witness the effects of this rate-limited transport most often when a person

is diabetic, even though there’s absolutely nothing wrong with one’s kidney function! What’s going on?

Well, you need to understand the normal role of insulin first.

Insulin is the hormone released by the pancreatic beta cells when we eat a meal. It is the hormone

the signals our body cells to prepare to take up the glucose, which we’ve just ingested in the form of

carbohydrates, from the circulating blood. Except for neurons, most body cells cannot take up glucose until

they respond to insulin’s signal, because they first have to insert glucose carriers in their surface membranes for that specific purpose (remember, glucose is not lipid-soluble, so it requires a carrier for cellular uptake.)When we eat, plasma glucose levels rise, but with insulin, these levels are almost immediately lowered back to normal. If pancreatic function is normal, no matter how much carbohydrate you eat, your plasma glucose levels are always around 100mg%.

In diabetes mellitus, people either don’t make insulin or their insulin is ineffective. When these

individuals eat a meal, their plasma glucose levels rise, but do not decrease; consequently their plasma

glucose levels can hit very high levels, double, triple and sometimes seven or eight-fold increases from

normal. We will consider many of the consequences of this later. Right now, let’s consider what happens

in the proximal convoluted tubule.

Step 1: Calculate the tubular load of glucose for each of the following plasma glucose levels, using the

equation given above, and fill in the column of the table titled “Filtered Load of Glucose” with your results. Then plot each of these points on the graph. Draw a line through these points and label the line. Assume a GFR of 125 ml/min.

Plasma Glucose concentration (mg/ml) / Filtered Load of Glucose (mg/min) / Glucose reabsorption rate (mg/min) / Glucose excretion rate (mg/min)
1
2
3
4
5
6
7
8
9
10

Step 2: The Tm-limited reabsorption rate for glucose is 320 mg glucose / minute. Any amount of

glucose, up to that maximum, can be completely reabsorbed, but nothing more. Enter the amount of

glucose reabsorbed each minute (glucose reabsorption rate) in the next column. Plot the data points for

reabsorption on the same graph (in another color) and draw a line through these points. Label the line

“reabsorption.” What do you notice about the line? Can you identify the values on the x and y axis where

this line turns? What do these two values represent?

Step 3: Calculate the rate of glucose excretion from the equation provided above. Enter each

value in the column labeled Glucose Excretion Rate. Plot the data points for excretion on the same graph

(in another color) and draw a line through these points. Label the line “excretion.” What do you notice

about the line? Can you identify the values on the x and y axis where this line turns? What do these two

values represent?

MEMBRANE TRANSPORT

1.  Use the following information to answer questions A – F. Assume

the cell is permeable to all solutes except protein.

CELL SOLUTION

10% K+ 10% Cl-

10% Na+

10% Protein 60% Water

A.  Which direction will potassium diffuse? ______

B.  Which direction will chloride diffuse? ______

C.  Which direction will sodium diffuse? ______

D.  Which direction will protein diffuse? ______

E.  What is the concentration of water inside the cell? ______

F.  What direction will water diffuse? ______

2.

S0LUTION A SOLUTION B SOLUTION C

0.3% NaCl 0.9% NaCl 15% NaCl

__% H20 __% H20 __% H20

A.  For each solution, determine the % of water.

Solution A = ______% H20

Solution B = ______% H20

Solution C = ______% H20

B.  Indicate if the solution is iso-, hypo-, or hyperosmotic to the cell.

Solution A ______

Solution B ______

Solution C ______

C.  State what changes (crenation or lysis) will occur, if any, to the cells when put into the above solutions

Cell A ______

Cell B ______

Cell C ______

D.  Patients are often administered normal saline solutions. Which of the above solutions is considered to be a normal saline solution? Why is this concentration the safest solution to give a patient? (pg 298-299 G&H)

3. In patients with renal disease, filtrate formation decreases and wastes accumulate in the blood. Dialysis is a process in which the blood is cleaned and its chemical composition is adjusted to normal levels. During dialysis, the patient’s blood is passed through dialysis tubing which is permeable only to selected substances. The tubing is immersed in a solution that differs from the composition of the patients blood (plasma). As the blood passes through the tubing, wastes and excess ions can diffuse out of the blood into the surrounding solution. Likewise, substances can also be added to the patient’s blood.

Normal Plasma Values of Selected Solutes

Urea 7-26 mg/dl

Potassium 3.5-5.1 mEq/l

Sodium 136-145 mEq/l

Albumin (protein) 3.2-5.0 g/dl

Bicarbonate 22-26 mEq/l

Hypothetical plasma values in a patient with renal disease

Urea > 30 mg/dl

Potassium > 5.5 mEq/l

Sodium < 130 mEq/l

Albumin 4.5 g/dl

Bicarbonate < 20 mEq/l

Determine the solute concentration of the solution needed to adjust the patient’s blood to normal values by determining if the concentration in the solution will be greater than, lesser than, or equal to the concentration of solutes in the patient’s blood. Remember at the end, you want your patient’s blood to have the same plasma values as a normal person.

BLOOD IN DIALYSIS TUBING SOLUTION

Urea ______

Potassium ______

Sodium ______

Albumin ______

Bicarbonate ______

For each solute, indicate the

direction of diffusion (into solution, into blood, or not at all).

5

Urea ______

Potassium ______

Sodium ______

Albumin ______

Bicarbonate ______

5

Effect of Adding Saline Solution to the Extracellular fluid: calculation of fluid shifts and osmolarities after infusion of hyperosmotic saline. (G&H pgs 299-301).

Objective: Calculate the sequential effects of infusing different solutions on extracellular and intracellular fluid volumes and osmolarities.

Problem: A nurse, obtains a saline solution and administers 0.5L of 9% sodium chloride solution into the EC fluid compartment of a 70kg patient whose initial plasma osmolarity is 280mOsm/L. The nurse assumes with blind faith that someone else knows the chemistry and that this is what the patient needs. What would be the intracellular and extracellular fluid volumes and osmolarities after osmotic equilibrium?

Step 1: calculate the initial conditions including the volume, concentration, and total mOsmoles in each compartment. Assume ECF volume is 20% of body weight and ICF volume is 40% body weight. (You can go back to your first lab unit when you dealt with body fluid compartments). Remember that one kg has the volume of one liter.

Volume (L) / Concentration (mOsm/L) / Total (mOsm)
EC fluid / 280
IC fluid / 280
Total body fluid / 280

Next calculate the total mOsmoles added to the ECF in 0.5L of 9% sodium chloride.

o  What does 9% represent?

o  How many grams would you have in one liter?

o  The molecular weight of NaCl is 58.5 grams, so determine the molarity (M/L).

o  Knowing the nurse administered only 0.5L, how many moles were administered?

o  1 mole of NaCl is equivalent to 2 osmoles due to dissociation. How many osmoles did the nurse administer to the patient?

o  How many milliosmoles is this?

Step 2: Calculate the instantaneous effect of adding your ______milliosmoles of sodium chloride to your patient. There would be no change in the ICF concentration or volume, and there would be no osmotic equilibrium. In the ECF, however, there would be an additional ______mosmoles of total solute. Keep in mind your ECF now has an additional 0.5L volume as well. Determine the concentration by dividing your new total mosmoles in the ECF by your new volume.

Volume (L) / Concentration (mOsm/L) / Total (mOsm)
EC fluid
IC fluid / 280
Total body fluid / b) / No equilibrium / a)

Step 3: calculate the volumes and concentrations that would occur within a few minutes after osmotic equilibrium develops. The concentrations in the ICF and ECF compartments would be equal and can be calculated by dividing the total mOsmoles in the body ______(see “a” in your chart above) by the total volume ______of the body (see “b” from your chart above). Your new concentration ______will be in all body fluid compartments. Assume no solute or water has been lost form the body and that there is no movement of NaCl into or out of cells, now calculate the volumes of the ICF and ECF compartments.

Divide the total mOsmoles of the ICF by the concentration to yield a volume ______.

Divide the total mOsmoles of the ECF by the concentration to yield a volume ______.

Fill in your table below:

Volume (L) / Concentration (mOsm/L) / Total (mOsm)
EC fluid
IC fluid
Total body fluid

What has happened to the volume of the body compartments? Which gained? Which lost volume?

So, the question now is, do you have an understanding about osmolarity? What do you think this solution has done to the patient’s tissues? Lysis? Crenation?

Please repeat this process but now the patient is an extreme athlete who is being encouraged to chug 2L of pure water to re-hydrate. (remember that the transcellular compartment is part of the ECF volume). Table one stays the same as the previous example, but tables 2 and 3 are different and are included below for you to fill-in.

Step 2: Calculate the instantaneous effect of adding 2L of pure water to the athlee. There would be no change in the ICF concentration or volume, and there would be no osmotic equilibrium.

Volume (L) / Concentration (mOsm/L) / Total (mOsm)
EC fluid
IC fluid / 280
Total body fluid / b) / No equilibrium / a)
Volume (L) / Concentration (mOsm/L) / Total (mOsm)
EC fluid
IC fluid
Total body fluid

What has happened to the volume of the body compartments? Which gained? Which lost volume?


Below is the picture of a nephron that will be used on your upcoming lecture exam. Make sure you understand what each arrow is pointing to. Be ready to match statements to the picture.

Matching: Below is a picture of a nephron. Choose the best portion of the nephron that matches the descriptions below and mark them on your scantron. Answers may be used more than once.

First: Identify what each label is pointing to and then answer the following descriptions:

This site is only permeable to ions.

This site has the Na+/K+/2Cl- transporter.

Site of water reabsorption, ions, glucose and amino acids.

Site of highest capillary hydrostatic pressure.

Site of highest capillary colloid osmotic pressure.

This is a functional contact between the Distal convoluted tubule and the afferent and efferent arterioles.

This is just a sample, you should expect MANY more of these types of questions.