Cell Structure, Osmosis, and Diffusion

  

1. Answer the following questions:a. List four cell structures that are common to both plant and animal cells. (4 points)b. What structures are unique to plant cells? (2 points)c. What structures are unique to animal cells? (2 points)2. Below are five structures observed in cells. List the function of each structure. (5 points)a. Chloroplastsb. Golgi apparatusc. Ribosomesd. Central vacuolee. Rough endoplasmic reticulum3. William is observing a single-celled organism under a microscope and notices that it has a nucleus and iscovered in small, hair-like structures.a. Provide a probable name for this organism (1 point)b. Explain why William came to this conclusion. (2 points)4. Where in the cell are the chloroplasts located? (5 points)5. In the Spirogyra cells observed on the virtual microscope, about how many circular green chloroplasts wereseen in a single cell at 40X magnification? (2 points)6. What were the percent differences between the volumes of the potatoes in the osmosis experiment for eachsalt solution? (8 points)a. 0%b. 1.75%c. 3.5%d. 7%7. What extraneous variables might have affected how the results came out in the osmosis experiment? Describethree. (6 points)a.b.c.8. In osmosis, which direction does water move with respect to solute concentration? (2 points)9. Answer the following questions:a. Explain what would happen to a freshwater unicellular organism if it were suddenly released into asaltwater environment. Use the terms iissottoniicc, hyypottoniicc and hyypeerrttoniicc in the answer. (3 points)b. What would happen if a marine organism were placed in freshwater? (3 points)10. A student purchases and weighs 5 pounds of carrots from a local grocery store. She notices that the grocerystore constantly sprays its produce with distilled water. After returning home, she weighs the carrots again anddiscovers that they weigh only 4.2 lbs. They also no longer seem as crisp and taut. Provide a possibleexplanation for why the carrots weighed more at the store, based on the information learned in this lab. (5points)11. People always say that leeches can be removed from the body by pouring salt on them. Based on what thestudent learned about osmosis, provide an explanation that supports or refutes this information. (5 points)12. What is the surface-to-volume ratio and rate of diffusion for each potato cube from Procedure 3b? (6 points)a. Cube 1 surface-to-volume ratiob. Cube 1 rate of diffusionc. Cube 2 surface-to-volume ratiod. Cube 2 rate of diffusione. Cube 3 surface-to-volume ratiof. Cube 3 rate of diffusion13. Assume the potato cubes are cells. Which cube would be most efficient at moving materials into and out of thecube? Briefly explain the answer. (4 points)14. From what was observed in the potato procedure, how do the rate of diffusion and surface-to-volume ratio limitcell size? (5 points)15. One night, Hans decides to cook a hamburger and spaghetti with meatballs. To test ideas of surface-to-volumeratios, he makes a quarter pound hamburger and a quarter pound meatball and cooks them at the sametemperature. Which food item will cook the fastest and why? (5 points)16. While watching a special on animals, Brianna discovers that hares tend to lose heat through their ears. Based onthis and what is known about surface-to-volume ratios, propose an explanation as to why hares that live in hotclimates (such as the desert) have large, extended ears. (5 points)17. ((Applliiccaattiion)) How might the information gained from this lab pertaining to cell structures and diffusion beuseful to you in your everyday life or to a healthcare professional? (20 points)Key components of critical thinking and application include the following:1. Demonstrates application and comprehension of the scientific principles.2. Displays competence in applying scientific knowledge to your personal or professional life.3. Relevant content is supported by facts, data, and detailed examples.4. The application paragraph is organized and structured.CrriittiiccaallTThiinkkiinggaandApplliiccaattiionoffIInfforrmaattiion
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Lab 4
Cell Structure, Osmosis, and
Diffusion
Introduction: Connecting Your Learning
The basic building block of life is the cell. Each cell contains several structures, some of which are common to both
eukaryotic and prokaryotic cells and some that are unique to specific cell types. This lab will discuss cell structures
and how materials are moved in and out of the cell. Specifically, the principles of diffusion and osmosis will be
demonstrated by performing a scientific investigation that studies the effect of salt concentration on potato cells.
Resources and Assignments
Multimedia
Resources
Virtual Microscope
Required
Assignments
Lesson 4 Lab 4
From the Lab Kit
Vernier caliper
Metric ruler
100 mL Graduated cylinder
2 teaspoons Salt
Measuring spoons (1 teaspoon and ½ teaspoon)
Student Supplied
Required Materials
2 cups Distilled (preferred), deionized, or purified drinking water
2 Potatoes
Knife or single-edge razor
Cutting board
Tape
Pen or marker
5 Plastic cups (8 oz)
A vial of food dye (blue, red or green)
Focusing Your Learning
Lab Objectives
By the end of this lesson, you should be able to:
1. Identify the structures and their respective functions of idealized plant and animal cells.
2. Compare and contrast the processes of diffusion and osmosis.
3. Define and calculate surface-to-volume ratio.
4. Explain the factors that limit cell size.
Background Information
In 1662, Robert Hooke investigated the properties of cork when he discovered cells. He named them after small
rooms in a monastery because they reminded him of them. Years later, in 1837, Schleiden and Schwann were
attributed with developing the cell theory. While their original theory was modified, the fundamental ideas behind the
theory held true. Three general postulates are included in the cell theory: 1) All organisms are composed of cells. 2)
The cell is the unit of life. 3) All cells arise from pre-existing cells.
Because a cell is the basic building block of living things, it is important to become familiar with its characteristics.
Several structures comprise a cell. Many of these structures are visible with the use of a standard compound
microscope. Below are pictures of idealized plant and animal cells, illustrating the important structures.
Click on image to enlarge.
The c e l l m e m b ra n e encloses all cells and is responsible for separating the internal environment from the
extracellular space (the space between cells). Because other structures within the cell are also surrounded by a
membrane, the outer membrane is often called the p l a s m a m e m b ra n e .
Click on image to enlarge.
The cell membrane is semi-permeable, allowing certain molecules to enter into the cell freely, while others are
prohibited from entering the cell. It is composed of p h o s p h o l i p i d s , which have a head consisting of a phosphate
group and a tail of two fatty acid chains. The phosphate group is attracted to water ( h y d ro p h i l i c ) while the fatty acid
chains are repulsed by water ( h y d ro p h o b i c ). When in water, the properties of the phospholipids cause them to form
two layers: The hydrophobic tails face the inside of the double layer (away from the water), and the hydrophilic heads
face out (toward the water). Because two layers are formed, the membrane is made up of a p h o s p h o l i p i d b i l a y eer ,
as seen in the image.
The c e l l w a l l surrounds the cell membrane in plant cells, bacteria, and some fungi. In plant cells, the cell wall is
composed of cellulose. In bacteria, the wall is made mostly of polypeptides (protein) or polysaccharides
(carbohydrates). The cell wall provides support and protection and is responsible for giving plant cells their shape.
Another important structure found only in eukaryotic cells is the n u c l e u s . This structure contains the genetic
information and is the control center of the cell. Protecting the nucleus is a double-membrane called the n u c l e a r
e n v e l o p e , which, like the plasma membrane, is semi-permeable. It is important to note that although prokaryotes
lack a nucleus, they still contain genetic information.
Within the nucleus is the n u c l e o l u s . This is the site where ribosomes are formed. Ribosomes function to assemble
proteins. Many cells have multiple nucleoli, which contain concentrated areas of DNA and RNA.
F l a g e l l a (singular is flagellum) is Latin for whip. Flagella are whip-like projections often found in prokaryotes,
eukaryotic single-celled organisms, and some specific cells (like human sperm). These structures extend beyond the
cell membrane and cell wall and are used for locomotion (movement). Although flagella are found in both eukaryotes
and prokaryotes, the structure of the flagella is different for each cell type.
C i l i a (singular is cilium) are structurally similar to eukaryotic flagella but are smaller and more hair-like. Cilia are
found in some eukaryotic organisms. Some cilia are used for locomotion, as in the single-celled paramecium. In other
organisms, the cilia act as a filter. Sometimes, cilia are used not to move the cell itself, but to move objects through a
cell (akin to a conveyer belt).
Va c u o l e s are specialized organelles that are responsible for storing starch, water, and pigments. They also act as a
repository for metabolic wastes. Some plant cells contain a large, central vacuole, which occupies almost the entire
cell. Central vacuoles are responsible for providing support, which is based on the amount of water or pressure
against the cell wall. If too much water is lost in the central vacuole, a plant will lose its support and appear to droop.
C e n t r i o l e s are found in all animal cells and some plant cells. These structures, which occur in pairs, are responsible
for the cytoskeleton. The cytoskeleton is composed of m i c ro t u b u l e s , microfilaments, and intermediate filaments. It
is with these long structures that the cytoskeleton provides support, maintains the cell shape, and anchors the
organelles. The cytoskeleton is also used for moving structures or products.
Within eukaryotes is an e n d o m e m b ra n e system. In this system, the endoplasmic reticulum, which consists of a
membrane that forms folds and pockets, connects the nuclear envelope, the G o l g i a p p a ra t u s (or Golgi complex),
and cell membranes. This system is often called the factory of the cell because each of the individual organelles
contributes to the production and delivery of proteins, lipids, and other molecules.
The nucleus contains the blueprints for proteins. These plans are then passed to the ro u g h e n d o p l a s m i c re t i c u l u m
( R E R ) . This structure is composed of several folds of a membrane and is covered with r i b o s o m e s (these bumps are
why it is called rough endoplasmic reticulum). Once the ribosomes receive the plans, the protein is built. Some
proteins will move to the G o l g i c o m p l ex . Other proteins will move to the smooth endoplasmic reticulum (it is called
smooth because it lacks ribosomes). These proteins instruct the organelle to build other molecules, such as lipids and
carbohydrates. Like some proteins from the RER, some of these molecules will move to the Golgi complex.
The G o l g i a p p a ra t u s is the central post office area of the cell. It receives the products of the rough endoplasmic
reticulum and smooth endoplasmic reticulum, packages them, and ships them to their intended destination.
Another structure found only in photosynthetic cells is the c h l o ro p l a s t . This specialized structure belongs to a class
of membrane-lined sacs called p l a s t i d s (like the vacuole). The chloroplast contains pigments and is responsible for
creating food through photosynthesis.
Eukaryotic cells contain an organelle called the m i t o c h o n d r i a , which is the site of energy production. This structure
is often referred to as the powerhouse of the cell. Cellular energy is stored in the form of a d e n o s i n e t r i p h o s p h a t e
( AT P ) .
The ability of a cell to absorb water and nutrients is an important aspect of its survival. D i f f u s i o n is the movement of
solutes (dissolved molecules) in a solution or matrix from an area of high concentration to an area of lower
concentration. Molecules move down the concentration gradient: from an area of high concentration to an area of low
concentration. The greater the concentration differential, the faster the rate of diffusion. The size, shape, and
composition of the solute also affect the ability of a substance to diffuse. These factors become increasingly
important when considering the diffusion of substances across the cell membrane. Diffusion, being a passive process,
is quite efficient across small distances. However, as distances become longer, the efficiency of diffusion decreases.
O s m o s i s is the movement of water across a selectively permeable membrane from an area of lower concentration (of
solute) to an area of higher concentration (of solute). Remember that everything in the universe is constantly moving
toward a state of equilibrium. Living cells contain a small amount of salt. For example, human cells contain 0.85%
NaCl. If the solution outside the cell has this same concentration, the solution is said to be i s o t o n i c . Because there is
no net difference in solutes between the inside and outside of the cell, there is no net movement of water. Higher
concentrations of solutes outside of the cell are termed h y p e r t o n i c , while lower concentrations are termed
hypotonic.
An important concept that affects how well a cell can absorb and pass material through the membrane is the surfaceto-volume ratio. This formula for calculating this ratio is:
Surface area ÷ Volume
Because cells constantly interact with their external environments to obtain nutrients and remove wastes, it is critical
that they maintain a proper surface-to-volume ratio.
As objects of the same shape increase in size, the surface-to-volume ratio decreases. For example, suppose there are
two cubes. Cube 1 is 1 cm x 1 cm x 1 cm, and Cube 2 is 10 cm x 10 cm x 10 cm. To calculate the surface-to-volume
ratio, the formula for determining the surface area (SA) of a cube (length x width x number of sides) and the formula
for the volume (V) of a cube (length x width x height) must be known. Once the formulas for calculating surface area
and volume of a cube are known, the surface area to volume ratios can be calculated, as seen below.
CUBE 1
CUBE 2
S u r f a c e A re a : 1cm x 1cm x 6 sides = 6cm 2 10cm x 10cm x 6 sides = 600cm 2
Vo l u m e :
1cm x 1cm x 1cm = 1cm 3
10cm x 10cm x 10cm = 1000cm 3
SA/V:
6cm 2 /1cm 3 = 6.0 cm 2 /cm 3
600cm 2 /1000cm 3 = .6cm 2 /cm 3
As shown in the calculations above, the ratio for Cube 2 is significantly smaller than the ratio for Cube 1. The same
trend holds true for cells. As a cell gets larger, the SA/V ratio decreases, meaning that it is not as efficient in moving
material in and out of the cell. In other words, the size of the cell membrane relative to the contents of the cell
decreases as the cell size increases.
An illustration of the importance in maintaining a high surface-to-volume ratio can be found in the human digestive
system. Cells in the human digestive system contain villi, which are finger-like projections. Because of their shape,
they have a large surface area for a small volume.
Procedures
1. C e l l S t r u c t u re a n d F u n c t i o n
a. L a b e l t h e f o l l ow i n g i d e a l i z e d p l a n t a n d a n i m a l c e l l s .
b. O b s e r v i n g C e l l S t r u c t u re s U n d e r a M i c ro s c o p e
i. Utilize the Virtual Microscope to view several cell structures. When using the virtual microscope,
complete the following steps in the order provided below. Failure to properly perform the steps in
the correct order will result in failure to complete subsequent steps. Click to view optional detailed
instructions  .
ii. Drag and drop the desired slide onto the microscope.
iii. Click on the stage clip knob on the left of the microscope stage.
iv. Adjust the interpupillary distance. First click on the title interpupillary distance. Next, place the
pointer on the images and adjust them until the two images are observed as one image.
v. Adjust the slide position. Place the pointer on the positioner and adjust the slide so there is a clear
view of the specimen.
vi. Adjust the iris diaphragm until a comfortable light is obtained.
vii. Adjust the diopter until a clear image is obtained. Use the line on the slide and move it up or down.
viii. Adjust the coarse focus. Use the line and move it up or down until a clear image is obtained.
ix. Adjust the fine focus. Use the line and move it up or down until a clear image is obtained.
x. Adjust the magnification by clicking on the objective numbers on the microscope.
xi. Using the virtual microscope, view Spirogyra. Identify and draw an image of the chloroplasts.
xii. Using the virtual microscope, view the slide of a paramecium. Identify and draw an image showing
the cilia.
xiii. Using the virtual microscope, view the slide of the Euglena. Identify and draw an image of the
flagella.
2. D e m o n s t ra t i o n o f O s m o s i s i n a P o t a t o
a. L e a r n t o u s e t h e c a l i p e r .
1. Take the Vernier caliper out of the lab kit. Examine the scale on the tool and try to measure the
length of an object. Look closely at the scale. The metric scale will be used for measurements in this
lab.
2. Read the scale by measuring exactly 2 cm (20 mm). Next, measure 4.5 cm (45 mm).
3. This caliper is accurate enough to measure to the nearest tenth of a millimeter (measured by the
small, scored lines in the window). With the caliper in hand, go to this instructional Web site, which
describes how to use a Vernier caliper.
Watch the scale move as in an actual measurement.
b. S e t u p t h e ex p e r i m e n t
1. Get four 8 oz. cups from the lab kit. Place a piece of tape on each cup or glass. Using a pen or
marker, label the tape on each cup with one of the following percentages: 0%, 1.75%, 3.5%, and 7%.
2. Using the graduated cylinder, measure out 100 mL of distilled water. Pour the water into the fifth,
unlabeled cup.
3. Measure out 1.5 level teaspoons of salt and add it to the unlabeled cup containing 100 mL of
distilled water. Mix completely. This is the 7% salt solution.
4. Using the graduated cylinder, measure out 50 mL of this mixture and pour it from the graduated
cylinder into the cup labeled 7%.
5. Add distilled water up to the 100 mL mark of the graduated cylinder to make the next dilution.
Adding 50 mL distilled water to 50 mL of a 7% solution will result in 100 mL of a 3.5% solution.
6. Using the graduated cylinder, measure out 50 mL of the 3.5% solution and pour it from the
graduated cylinder into the cup labeled 3.5%.
7. Add distilled water up to the 100 mL mark of the solution in the graduated cylinder to make the next
dilution. Adding 50 mL of distilled water to the 50 mL of the 3.5% solution will result in 100 mL of a
1.75% solution.
8. Using the graduated cylinder, measure out 50 mL of the 1.75% solution and pour it from the
graduated cylinder into the cup labeled 1.75%.
9. Empty the remaining 1.75% solution down the drain of the sink and rinse out the graduated cylinder
with tap water.
10. Using a sharp steak or kitchen knife, slice eight pieces of potato exactly 10 mm x 10 mm x 40 mm (1
cm x 1 cm x 4 cm). It is critically important that these potato core pieces are cut as precisely as
possible; they need to all start out having the same volume. A single- edge razor blade may work
better than a knife.
11. Determine the volume of the potato cores. The volume, is calculated by multiplying the width x
height x length. Therefore, each core starts out with a volume of 4,000 cubic millimeters or 4 cubic
centimeters. Measure the cores with both the mm ruler and the calipers. Measuring with the calipers
to the nearest millimeter will be good enough for this lab. Create a data table like the one below to
record the beginning and ending volumes.
Table 1. Potato core measurements
0% salt
solution
1.75% salt
solution
3.5% salt
solution
7% salt
solution
Beginning average volume (cu
mm)
Ending average volume (cu
mm)
Percent difference
12. Place two measured cores into each solution overnight, or for at least 8 hours. That time period is
not critical to the results; it can be longer.
13. Remove the cores from one of the cups and pat them dry with a paper towel. The solution may now
be discarded down the drain of a sink.
14. Using the caliper, measure the height, width, and length of the cores, and then determine the
volume of each core. Average the measurements for the two cores and record in the data table
above. The cores can now be discarded.
15. Repeat Steps 13 – 14 three more times: one time for each cup.
3. I l l u s t ra t i o n o f t h e I m p o r t a n c e o f S u r f a c e – t o – Vo l u m e R a t i o s
a. C a l c u l a t e t h e s u r f a c e – t o – v o l u m e ra t i o o f t h e f o l l ow i n g p o t a t o c u b e s :
1. CUBE 1: Length, width, and height are all 5 mm
2. CUBE 2: Length, width, and height are all 3 mm
b. E f f e c t o f c e l l s i z e o n d i f f u s i o n ra t e
1. With clean hands, cutting board, and knife, cut the skin off of the potato.
2. Using the knife, cut two cubes of potato with dimensions of 1 cm x 1 cm x 1 cm.
3. Using the knife, cut two cubes of potato with dimensions 1.5 cm x 1.5 cm x 1.5 cm.
4. Using the knife, cut two cubes of potato with dimension of 2 cm x 2 cm x 2 cm.
5. Place distilled water into a cup or glass. Add the vial of food coloring to the water until a dark color is
achieved.
6. Carefully place the potato cubes in the solution. The cubes must be completely submerged in the
water. Let them stand in the solution for 2 to 4 hours.
7. After 2 to 4 hours, remove the cubes. Using the knife, cut each cube in half.
8. Using the ruler, measure how far the solution has diffused into each potato cube.
9. Record the results. A sample data table is included below that may be used to organize and record
the results.
10. Complete the following calculations to determine the rate of diffusion and record the results.
Rate of Diffusion (cm/min)= Distance of diffusion ÷ time.
Cube
Distance of Diffusion Rate of Diffusion
1 cm cubed
1 cm cubed
1.5 cm cubed
1.5 cm cubed
2 cm cubed
2 cm cubed
Average Rate of Diff.
Assessing Your Learning
Warning: You are expected to submit your own, individual
work. Using work completed by anyone other than yourself is
plagiarism. This includes resources found on internet sites.
Posting assessments on an unauthorized web site, soliciting
assessment answers or the acquisition of assessments,
assessment answers, and other academic material is
cheating. Cheating and/or plagiarism will result in a failing
grade for the course.
Compose answers to the questions below in Microsoft Word and save the file as a backup copy in the event that a
technical problem is encountered while attempting to submit the assignment. Make sure to run a spell check. Copy
the answer for the first question from Microsoft Word by simultaneously holding down the Ctrl and A keys to select
the text, and then simultaneously holding down the Ctrl and C keys to copy it. Then, click the link on the Lab Preview
Page to open up the online submit form for the laboratory. Paste the answer for the first question into the online
dialog boxes by inserting the cursor in the dialog box and simultaneously holding down the Ctrl and V keys. The
answer should now appear in the box. Repeat for each question. Review all work to make sure that all of the
questions have been completely answered and then click on the Submit button at the bottom of the page.
LAB 4
1. Answer the following questions:
a. List four …
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