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DNA and Protein Synthesis Hands-On Labs, Inc. Version 42-0051-00-02

Review the safety materials and wear goggles when working with chemicals. Read the entire exercise before you begin. Take time to organize the materials you will need and set aside a safe work space in which to complete the exercise.

Experiment Summary:

You will learn the structure and function of DNA and RNA. You will learn the similarities and differences between DNA and RNA. You will learn the process of protein synthesis and create and use models to demonstrate both transcription and translation.

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EXPERIMENT

Learning Objectives Upon completion of this laboratory, you will be able to:

● Review the structure and function of DNA.

● Identify the codons that code for amino acids in DNA and RNA.

● Explain the purpose of start and stop codons in protein synthesis.

● Summarize the steps involved in protein synthesis and define a ribosome and its three sites.

● Summarize the steps of transcription, including: initiation, elongation, and termination.

● Summarize the steps of translation, including; initiation, elongation, and termination.

● Illustrate and model the processes of transcription and translation.

● Construct a series of tRNA molecules and write the anti-codons and amino acids each tRNA carries.

● Explain the difference in the number of amino acids that were present at the start and at the end of the translation model.

Time Allocation: 3 hours

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Experiment DNA and Protein Synthesis

Materials Student Supplied Materials

Quantity Item Description 1 Camera, digital or Smartphone 1 Pair of scissors 1 Printer

10 Sheets of printer paper 1 Pen or pencil 1 Tape

HOL Supplied Materials

Quantity Item Description 1 DNA Nucleotide Template 1 RNA Nucleotide Template 1 tRNA Template

Note: To fully and accurately complete all lab exercises, you will need access to:

1. A computer to upload digital camera images.

2. Basic photo editing software such as Microsoft Word® or PowerPoint®, to add labels, leader lines, or text to digital photos.

3. Subject-specific textbook or appropriate reference resources from lecture content or other suggested resources.

Note: The packaging and/or materials in this LabPaq kit may differ slightly from that which is listed above. For an exact listing of materials, refer to the Contents List included in your LabPaq kit.

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Experiment DNA and Protein Synthesis

Background DNA, Codons, and Proteins

Deoxyribonucleic acid (DNA), the genetic material of all living organisms, is composed of two chains of nucleotides wound together in a double-helical formation. Nucleotides, the molecules responsible for the structural units of DNA, are composed of three sections: a phosphate group (PO4), a sugar (deoxyribose) group, and a nitrogenous base. There are four different DNA nucleotides: adenine (A), thymine (T), cytosine (C), and guanine (G), which are identical in their phosphate and sugar groups, but vary in their nitrogenous bases. The bonds between the sugar and phosphate groups of each nucleotide form the sugar-phosphate backbone of DNA and the two strands wind together as a result of base pairing: AT (Adenine-Thymine) or GC (Guanine- Cytosine).

The arrangement of the four DNA nucleotides creates the genetic code, the blueprint for all living things. The genetic code is composed of codons, triplets of nucleotides that contain the code for the production of amino acids, which are strung together to create proteins (polypeptide chains). Proteins are highly complex, organic substances that provide a vast number of functions in living organisms, including maintenance of cells and growth. Thus, proteins are essential components of living tissues including: skin, bones, and muscle. The four different nucleotides provide 64 different codons (four options for each of three positions = 43 = 64 options), which code for one of twenty amino acids or a stop codon. See Table 1.

Table 1. Codon Chart (DNA)

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Experiment DNA and Protein Synthesis

Protein Biosynthesis and Transcription

Protein biosynthesis is the process where cells use the genetic code to build proteins. The process differs slightly between prokaryotes (single-celled organism with no organelles or distinct nucleus) and eukaryotes (single or multi-celled organisms with organelles and DNA contained in a distinct nucleus). In the context of this experiment, the focus will be on the main steps and commonalities between the prokaryotic and eukaryotic protein biosynthesis steps. There are two main steps in protein biosynthesis: transcription and translation.

Transcription is the process by which single stranded RNA is synthesized from DNA. RNA (ribonucleic acid), like DNA, is composed of nucleotides with a phosphate, a sugar (ribose), and one of four nitrogenous bases. The nucleotides adenine, cytosine, and guanine exist in both DNA and RNA, however; in RNA, the nucleotide uracil (U) replaces thymine (T), and binds with adenine (A). See Figure 1.

Figure 1. Nitrogenous bases in DNA and RNA. Note that while DNA is double-stranded, RNA exists as a single-strand. © udaix

There are three steps in transcription: initiation, elongation, and termination. A general depiction of transcription is shown in Figure 2.

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Experiment DNA and Protein Synthesis

Figure 2. General schematic of transcription. © The National Human Genome Research Institute

In initiation, an enzyme called RNA polymerase binds to the DNA promoter, which is the DNA sequence that initiates transcription. The RNA polymerase causes the two strands of DNA to begin unwinding and separate from one another. In elongation, the RNA polymerase travels downstream (3’ to 5’) along the DNA antisense strand, elongating the mRNA transcript in the 5’ to 3’ direction. The DNA antisense strand is the template strand from which the mRNA is transcribed. Figure 2 illustrates how transcription creates an mRNA copy of the DNA sense, or coding, strand, with uracil replacing thymine in the newly constructed mRNA. As the RNA polymerase continues to move downstream, the two strand of DNA re-wind into a double-helix formation. In termination, the RNA polymerase detaches from the DNA and releases the transcribed mRNA. In a prokaryote, the released mRNA is complete and ready to move into translation, while in a eukaryote the released RNA undergoes a series of steps where it is processed before moving into translation as mRNA.

Translation

Translation, the second main step of protein synthesis, is the process by which the mRNA (created in transcription) is converted into a protein. In a prokaryote, translation occurs in the cytoplasm, the same site as transcription; while in eukaryotes, translation occurs in the cytoplasm, where it is carried after transcription has completed in the nucleus. There are two major players in translation; transfer RNA (tRNA) and ribosomes. tRNA is a clover-shaped molecule that acts as the interpreter between mRNA and the protein it will help to synthesize. A ribosome is an organelle which functions as the site of protein synthesis. Ribosomes are made of ribosomal RNA (rRNA) and protein molecules, and are divided into two subunits: large and small. The small ribosomal

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Experiment DNA and Protein Synthesis

subunit binds the mRNA and reads the information contained in the mRNA nucleotide sequence. The large ribosomal subunit contains three binding sites: the peptidyl-tRNA site (P site), the aminoacyl-tRNA site (A site), and the exit site (E site). See Figure 3.

Figure 3. The ribosome.

There are three steps in translation: activation and initiation, elongation, and termination. A general depiction of translation is shown in Figure 4.

Figure 4. General schematic of translation.

In the first step of translation, activation and initiation, the mRNA is threaded between the

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Experiment DNA and Protein Synthesis

small and large subunits of the ribosome. The ribosome signals the start of translation when it encounters and binds to the first start codon (AUG, which codes for the amino acid methionine) at the A site. The tRNA carrying the anticodon (UTC) and the methionine binds to the mRNA codon at the ribosomal A site, creating the initiation complex, signaling translation to begin. The tRNA bound mRNA then moves into the P site, which brings in the next tRNA carrying the complementary anticodon and amino acid to the mRNA now in the A site.

The amino acid in the P site then forms a peptide bond with the amino acid in the A site, releasing the amino acid from the tRNA in the P site and moving the empty tRNA into the E site. Simultaneously, the tRNA in the A site, holding the two peptide-bonded amino acids, then moves into the P site, signaling the next tRNA to bind to the mRNA in the A site. Using Figure 4 as an example, as the empty tRNA (which had been carrying valine) exits the E site, a peptide bond is formed between lysine (in the P site) and cysteine (in the A site). The lysine (peptide bound to valine, glutamate, serine, and glycine) then detaches from the tRNA in the P site and attaches to the tRNA in the A site. The tRNA in the A site (now carrying the cysteine, lysine, valine, glutamate, serine, and glycine) then moves into the P site, releasing the tRNA that had carried the lysine from the E site. As the tRNA is released from the E site, tRNA carrying the anticodon AUA and the amino acid tyrosine (yellow Tyr) binds to the mRNA in the A site, continuing the process. This continuous process, called elongation, builds the polypeptide chain (protein) one amino acid at a time until the mRNA reads a stop codon (UAA, UAG, or UGA).

When a stop codon is encountered in the mRNA at the A site, termination is signaled. In termination the stop codon signals the end of elongation, which cleaves the protein from the tRNA, allowing it to exit the ribosome. The two ribosome subunits and the mRNA then dissociate from one another, completing the translation process. The protein then undergoes a series of steps including post-translational modifications and protein folding to assume its new shape.

In 2009, Dr. Ada Yonath won the Nobel Prize in Chemistry. She was the first woman to win the Nobel Prize in Chemistry

in 45 years, since Dorothy Crowfoot Hodgkin in 1964, and the first woman in the Middle East to ever win the Chemistry Nobel Prize. Her award,

shared with Dr. Thomas Steitz and Dr. Venkatraman Ramakrishnam, was the result of her work on the

structural determination of the ribosome, determining the structure of both the small and large ribosomal

subunits. Her work lead to the conclusion that a ribosome is a ribozyme (ribonucleic acid enzyme), that

organizes its substrates in the stereochemistry necessary for the formation of peptide bonds. From her work came the new and exciting crystallization technique called cryo bio-crystallography, which allows for the crystallization of large biological macromolecules at cryogenic temperatures (approximately -320°F)

allowing the macromolecules to maintain their solution state.

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Experiment DNA and Protein Synthesis

Exercise 1: Protein Synthesis In this exercise, you will model the steps of protein synthesis, starting with a single strand of nucleotides and ending with a protein.

1. Print 6 copies of the DNA Nucleotide Template, 4 copies of the RNA Nucleotide Template, and 1 copy of the tRNA Template. It is preferable, but not necessary, to print them in color. The templates are located in the “Supplemental Documents” folder of your digital courseware.

2. Review the coding strand of DNA (5’ to 3’) in Data Table 1 of your Lab Report Assistant.

3. Create the template strand of DNA (3’ to 5’) and record in Data Table 1.

4. Gather the scissors, tape, and the 6 printed copies of the DNA Nucleotide Template. Cut out the nucleotides from the template. It is not necessary to cut out the entire nucleotide; rather, cut the nucleotide in a rectangular shape, only cutting out the details of the nitrogenous bases. See Figure 5.

Figure 5. Cutting out DNA nitrogenous bases.

5. Using the DNA nucleotides, create the entire double strand of DNA by matching up and taping together the base pairs. See Figure 6 as an example.

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Experiment DNA and Protein Synthesis

Figure 6. Pairing of DNA nucleotides.

6. Take a photograph of the completed double strand of DNA with your name and the data showing in the photograph. Resize and insert the photograph into Data Table 2 of your Lab Report Assistant. Refer to the appendix entitled “Resizing an Image” for guidance with resizing an image.

7. Determine the mRNA strand that transcription would produce from the DNA template strand and record the mRNA strand in Data Table 1.

8. Gather the 4 printed copies of the RNA Nucleotide Template. Cut out the nucleotides from the template. It is not necessary to cut out the entire nucleotide; rather, cut the nucleotide in a rectangular shape, only cutting out the details of the nitrogenous bases.

9. Using the RNA nucleotides, create the mRNA strand by matching up and taping together the base pairs.

10. Take a photograph of the mRNA strand with your name and the date showing in the photograph. Resize and insert the photograph in Data Table 2.

11. Starting with the first mRNA nucleotide, determine what amino acids the codons in the mRNA are coding for and record in Data Table 1.

Note: Use Table 2 to determine the amino acids coded by RNA codons.

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Experiment DNA and Protein Synthesis

Table 2. Codon Chart (RNA)

12. Gather the printed copy of the tRNA Template and cut out the tRNAs.

13. Build the line of tRNAs that would flow into the A site during translation. Write the anti- codons into each tRNA and the amino acid the mRNA codes for. See Figure 7 as an example of the tRNA that would be created from the mRNA codons CCU.

Note: Use Figure 4 in the Background section as needed to help organize your thoughts and identify where in the mRNA strand translation would begin.

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Experiment DNA and Protein Synthesis

Figure 7. tRNA created from mRNA (CCU). Note that the anti-codons (GGA) and the name of the amino acid (gly = glycine) are written into the tRNA.

14. Take a photograph of the tRNAs (in order) with your name and the date showing in the photograph. Resize and insert the photograph in Data Table 2.

15. Write the name of the each amino acid in the final protein created from translation and record in Data Table 1.

16. When you are finished uploading photos and data into your Lab Report Assistant, save your file correctly and zip the file so you can send it to your instructor as a smaller file. Refer to the appendix entitled “Saving Correctly” and the appendix entitled “Zipping Files” for guidance with saving the Lab Report Assistant correctly and zipping the file.

Note: Use a textbook or internet source, as necessary, for a list of the three-letter amino acid abbreviations and the full amino acid names.

Questions A. How many amino acids were coded for by the mRNA? How many amino acids were present

in the final protein chain created in translation? In detail, explain the differences in the two numbers; why were some amino acids coded for by the mRNA but not present in the final protein chain? What amino acids were omitted from the final protein chain? Explain your answers.

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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.

Focusing Your Learning

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 be- hind 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.

The cell membrane encloses all cells and is responsible for separating the internal en- vironment from the extracellular space (the space between cells). Because other struc- tures within the cell are also surrounded by a membrane, the outer membrane is of- ten called the plasma membrane.

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The cell membrane is semi-permeable, allowing cer- tain molecules to enter into the cell freely, while oth- ers are prohibited from entering the cell. It is com- posed of phospholipids, which have a head consist- ing of a phosphate group and a tail of two fatty acid chains. The phosphate group is attracted to water

(hydrophilic) while the fatty acid chains are repulsed by water (hydrophobic). When in water, the properties of the phospholipids cause them to form two layers: The hy- drophobic tails face the inside of the double layer (away from the water), and the hy- drophilic heads face out (toward the water). Because two layers are formed, the mem- brane is made up of a phospholipid bilayer, as seen in the image.

The cell wall 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 nucleus. This struc- ture contains the genetic information and is the control center of the cell. Protecting the nucleus is a double-membrane called the nuclear envelope, 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 nucleolus. This is the site where ribosomes are formed. Ri- bosomes function to assemble proteins. Many cells have multiple nucleoli, which con- tain concentrated areas of DNA and RNA.

Flagella (singular is flagellum) is Latin for whip. Flagella are whip-like projections of- ten 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 eukary- otes and prokaryotes, the structure of the flagella is different for each cell type.

Cilia (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

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a filter. Sometimes, cilia are used not to move the cell itself, but to move objects through a cell (akin to a conveyer belt).

Vacuoles 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 re- sponsible 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.

Centrioles 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 microtubules, microfilaments, and intermediate filaments. It is with these long struc- tures 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 endomembrane system. In this system, the endoplasmic retic- ulum, which consists of a membrane that forms folds and pockets, connects the nu- clear envelope, the Golgi apparatus (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 rough endoplasmic reticulum (RER). This structure is composed of several folds of a membrane and is covered with ribosomes (these bumps are why it is called rough en- doplasmic reticulum). Once the ribosomes receive the plans, the protein is built. Some proteins will move to the Golgi complex. Other proteins will move to the smooth en- doplasmic 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 com- plex.

The Golgi apparatus 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 chloroplast. This special- ized structure belongs to a class of membrane-lined sacs called plastids (like the vac-

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uole). The chloroplast contains pigments and is responsible for creating food through photosynthesis.

Eukaryotic cells contain an organelle called the mitochondria, which is the site of en- ergy production. This structure is often referred to as the powerhouse of the cell. Cel- lular energy is stored in the form of adenosine triphosphate (ATP).

The ability of a cell to absorb water and nutrients is an important aspect of its sur- vival. Diffusion is the movement of solutes (dissolved molecules) in a solution or ma- trix 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.

Osmosis 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 isotonic. Because there is no net difference in solutes between the inside and outside of the cell, there is no net movement of water. Higher concen- trations of solutes outside of the cell are termed hypertonic, while lower concentra- tions are termed hypotonic.

An important concept that affects how well a cell can absorb and pass material through the membrane is the surface-to-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.

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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 de- termining 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

Surface Area: 1cm x 1cm x 6 sides = 6cm2 10cm x 10cm x 6 sides = 600cm2

Volume: 1cm x 1cm x 1cm = 1cm3 10cm x 10cm x 10cm = 1000cm3

SA/V: 6cm2/1cm3 = 6.0 cm2/cm3 600cm2/1000cm3 = .6cm2/cm3

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 sur- face area for a small volume.

Procedures

1. Cell Structure and Function a. Label the following idealized plant and animal cells.

b. Observing Cell Structures Under a Microscope 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

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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 ad- just 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. Demonstration of Osmosis in a Potato

a. Learn to use the caliper. 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.

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b. Set up the experiment 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 grad- uated 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% solu- tion 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 exact- ly 10 mm x 10 mm x 40 mm (1 cm x 1 cm x 4 cm). It is critically impor- tant 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

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ending volumes.

Table 1. Potato core measurements

0% saltsolution 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 measure- ments for the two cores and record in the data table above. The cores can now be discarded.

15. Repeat Steps 11 – 14 three more times: one time for each cup. 3. Illustration of the Importance of Surface-to-Volume Ratios

a. Calculate the surface-to-volume ratio of the following potato cubes: 1. CUBE 1: Length, width, and height are all 5 mm 2. CUBE 2: Length, width, and height are all 3 mm

b. Effect of cell size on diffusion rate 1. With clean hands, cutting board, and knife, cut the skin off of the pota-

to. 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.

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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

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.

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LAB 4

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. List five structures observed in the cell images and provide the function of each structure. (5 points)

a. Structure 1 and function b. Structure 2 and function c. Structure 3 and function d. Structure 4 and function e. Structure 5 and function

3. William is observing a single-celled organism under a microscope and notices that it has a nucleus and is covered 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 cir-

cular green chloroplasts were seen in a single cell? (2 points) 6. What were the percent differences between the volumes of the potatoes in the

osmosis experiment for each salt 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? Describe three. (6 points)

a. b. c.

8. In osmosis, which direction does water move with respect to solute concentra- tion? (2 points)

9. Answer the following questions: a. Explain what would happen to a freshwater unicellular organism if it were

suddenly released into a saltwater environment. Use the terms isotonic,

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hypotonic and hypertonic 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 grocery store constantly sprays its produce with distilled water. After returning home, she weighs the carrots again and discovers that they weigh only 4.2 lbs. They also no longer seem as crisp and taut. Provide a possible explanation for why the carrots weighed more at the store, based on the information learned in this lab. (5 points)

11. People always say that leeches can be removed from the body by pouring salt on them. Based on what the student learned about osmosis, provide an explanation that supports or refutes this information. (5 points)

12. What is the rate of diffusion for the potato cubes from the surface-to-volume ex- periment (procedure 3b)? (6 points)

a. Cube 1 b. Cube 2 c. Cube 3

13. Assume the potato cubes are cells. Which cube would be most efficient at mov- ing materials into and out of the cube? 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 limit cell size? (5 points)

15. One night, Hans decides to cook a hamburger and spaghetti with meatballs. To test ideas of surface-to-volume ratios, he makes a quarter pound hamburger and a quarter pound meatball and cooks them at the same temperature. 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 on this and what is known about surface-to-vol- ume ratios, propose an explanation as to why hares that live in hot climates (such as the desert) have large, extended ears. (5 points)

17. (Application) How might the information gained from this lab pertaining to cell structures and diffusion be useful to a student employed in a healthcare related profession? (20 points)

Page 1“Antibiotic Resistance” by Maureen Leonard

by Maureen Leonard Biology Department Mount Mary College, Milwaukee, WI

Antibiotic Resistance: Can We Ever Win?

Part I – Measuring Resistance Katelyn was excited to start her summer job in her microbiology professor’s research laboratory. She had enjoyed Dr. Johnson’s class, and when she saw the ” yer recruiting undergraduate lab assistants for the summer, she had jumped at the opportunity. She was looking forward to making new discoveries in the lab.

On her # rst day, she was supposed to meet with Dr. Johnson to talk about what she would be doing. She knew the lab focused on antibiotic resistance in Staphylococcus aureus, espe- cially MRSA (methicillin-resistant S. aureus ).

She still remembered the scare her family had last year when her little brother, Jimmy, got so sick. He’d been playing in the neighborhood playground and cut his lip when he fell o$ the jungle gym. Of course he always had cuts and scrapes—he was a # ve-year-old boy! % is time though his lip swelled up and he developed a fever. When her mother took him to the doctor, the pediatrician said the cut was infected and had prescribed cephalothin, an antibiotic related to penicillin, and recommended ” ushing the cut regularly to help clear up the infection.

Two days later, Jimmy was in the hospital with a fever of 103°F, coughing up blood and having trouble breathing. % e emergency room doctors told the family that Jimmy had developed pneumonia. % ey started him on IV antibiotics, including ceftriaxone and nafcillin, both also relatives of penicillin.

It was lucky for Jimmy that one of the doctors decided to check for MRSA, because that’s what it was! MRSA is resistant to most of the penicillin derivatives. Most cases of MRSA are hospital-acquired from patients who are already susceptible to infection, but the ER doctor explained that community-acquired MRSA was becoming more common. % e doctor then switched the treatment to vancomycin, a completely di$ erent kind of antibiotic, and Jimmy got better quickly after that.

Katelyn had dropped Jimmy o$ at swimming lessons just before coming to work at the lab. As she waited in the hallway for Dr. Johnson, she hoped that she would be at least a small part of helping other people like Jimmy deal with these scary resistant microbes. She was surprised when the professor burst out of the lab, almost running into her.

“Hi Katelyn, I’m really sorry but I have to run to a meeting right now—they sprung it on me last minute. % ere are a bunch of plates in the incubator right now that need their zones of inhibition measured. I’ll be back in a few hours,” Dr. Johnson said as he rushed down the hallway with a stack of folders.

Katelyn dug out her old lab notebook to look up what she was supposed to do. She found the lab where she and her fellow students had examined the antimicrobial properties of antibiotics using the Kirby-Bauer disk di$ usion tech- nique. Looking at the plates Dr. Johnson had told her about, she saw they had all been “lawned,” or completely coated with microbes to make a thick hazy layer over the agar surface. She could also see paper disks with letters on them, and some of the disks had clear zones around them where the microbe had been inhibited (Fig. 1). Her notebook explained how to measure the zone of inhibition around the disks (Fig. 2).

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ttle brother, Jimmy, got hi li h h f ll $ h j l

1

We’re looking for undergraduate lab assistants! If yes, e-mail Dr. Johnson to grab a spot today!

(You must have taken Biology 200 Microbiology to apply)

Interested in studying microbial antibiotic resistance?

Do you want to work in a research lab? Are you interested in bacteria?

Have you heard of antibiotic resistance?

Page 2“Antibiotic Resistance” by Maureen Leonard

Plate 1. Plate 2. Plate 3.

S. aureus

PE

CE ME

VA

S. aureus

PE

CE ME

VA

S. aureus

PE

CE ME

VA

PE

CE ME

MRSA

VA

PE

CE ME

MRSA

VA

PE

CE ME

MRSA

VA

Figure 1. Agar plates of S. aureus or MRSA lawns with antibiotic disks placed on them.

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Page 3“Antibiotic Resistance” by Maureen Leonard

Figure 2. Katelyn’s diagram of how to measure a zone of inhibition from her microbiology lab notebook.

x xi

i 1

n

n

Exercise1 Measure the zones of inhibition for each antibiotic on the plates shown in Figure 1 and note the measurements in the spaces in Table 1 below. (Note: % e Kirby-Bauer method is standardized so that no zone of inhibition is scored as a 0, and all others include the disk as part of the zone.)

Key: PE = penicillin, ME = methicillin, CE = cephalothin, and VA = vancomycin

Plate S. aureus MRSA

1

PE

ME

CE

VA

2

PE

ME

CE

VA

3

PE

ME

CE

VA

An average, or mean (x ), is a measure of central tendency in the data, or what value occurs in the middle of the data set. % e mean is calculated by adding up all the values for a given set of data, then dividing by the sample size (n).

Average

Standard deviation measures the spread of the data—as in how variable the data set is. % e standard deviation (s ) is calculated by the following:

Inhibition (clear) zone

Measure in mm

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Page 4“Antibiotic Resistance” by Maureen Leonard

Standard error measures the di$ erence between the sample you have taken and the whole population of values. % e standard error (SE) is calculated as follows:

s (x x) 2

n 1

SE s n

Exercise 2 In Table 2 below calculate and record the averages and standard errors for each antibiotic in S. aureus and MRSA.

S. aureus MRSA

Average SE Average SE

PE

ME

CE

VA

Exercise 3 Now, redraw Tables 1 and 2 into a single, more organized table. Be sure to label the table appropriately.

Standard deviation

Standard error

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Page 5“Antibiotic Resistance” by Maureen Leonard

Exercise 4 Graph the results from Table 2. Be sure to label the # gure and the axes correctly.

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Page 6“Antibiotic Resistance” by Maureen Leonard

Questions 1. What do you think the experimental question is? 2. What hypotheses can you come up with to answer the experimental question? 3. If your hypothesis is correct, what would the plates look like (i.e., what predictions would you make for each

hypothesis)? 4. Is the experiment you just collected data for an appropriate test of the experimental question you came up with

in your answer to Question 1? 5. Which antibiotics where most e$ ective against S. aureus? Against MRSA? 6. When comparing the antibiotics e$ ective against both, were there di$ erences in e$ ectiveness? 7. What other questions do the data shown in Figure 1 make you think of?

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Page 7“Antibiotic Resistance” by Maureen Leonard

Part II – Resistance Among the # rst antibiotics used on a large scale was penicillin, which was discovered in 1929 by Alexander Fleming. It was # nally isolated and synthesized in large quantities in 1943. Penicillin works by interfering with the bacterial cell wall synthesis. Without a cell wall, the bacterial cells cannot maintain their shape in changing osmotic conditions. % is puts signi# cant selective pressure on the microbes to evolve, as they cannot survive the osmotic stress. Any microbe that can resist these drugs will survive and reproduce more, making the population of microbes antibiotic resistant.

% e speci# c mechanism of penicillin is the prevention of cell wall synthesis by the -lactam ring of the antibiotic (Fig. 3), which binds and inhibits an enzyme required by the bacterium in this process.

% e enzyme is called penicillin-binding protein (PBP), even though it is an enzyme involved in cell wall synthesis. Normally enzymes have names that indicate what they do and end in the su, x -ase, like lactase, the enzyme that breaks down lactose. Figure 4 is a representation of PBP and its active site.

Bacterial cell walls are layered structures, where each layer is made of peptidoglycan, a sugar and protein polymer. Each layer is cross-linked to the next, strengthening the wall and allowing the cell to resist osmotic pressure. % e way the enzyme PBP works is to form those cross-bridges by joining strings of amino acids together in the active site, which is a groove in the protein (Fig. 5).

Active site

Figure 4. PBP (penicillin-binding protein) active site is a groove allowing formation of cross-links in the bacterial cell wall.

Figure 5. Cross-link formation in bacterial cell walls by PBP (penicillin-binding protein).

PBP

Peptidoglycan layers

Amino acids

Cross-bridge

Figure 3. % e -lactam ring common to the penicillin family of antibiotics.

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Page 8“Antibiotic Resistance” by Maureen Leonard

% e PBP takes amino acid residues attached to peptidoglycan layers and forms bridges between them within the active site groove. % is cross-linking, or cross-bridging, stabilizes and strengthens the cell wall. -lactam antibiotics interfere with the PBP enzyme by binding to the active site, blocking the site from the amino acids (Fig. 6).

% ere are over 80 natural and semi-synthetic forms of -lactam antibiotics, including cephalothin and methicillin. Vancomycin also interferes with cell wall synthesis, but its mechanism of action is to bind directly to the cell wall components (Figs. 7 and 8).

Figure 6. Inhibition of PBP (penicillin-binding protein) by

-lactam blocking the active site.

NH

O

Figure 7. PBP (penicillin-binding protein), the enzyme that allows the bacterial cell wall to form cross-bridges, is inhibited by the -lactam family of antibiotics. % is prevents proper cell wall synthesis and the bacterium will succumb to osmotic stress.

a. Normal PBP binding and cross-bridge formation

b. PBP inhibited by -lactam antibiotic

c. Cell wall does not form properly

+ =

PBP

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Page 9“Antibiotic Resistance” by Maureen Leonard

% e # rst MRSA case was discovered in 1961 in a British hospital, and was the result of a mutation in the enzyme normally inhibited by the -lactam ring of methicillin. % e site where the antibiotic would bind no longer allowed access to the ring, so the enzyme continued to function normally. % e microbe acquired a new gene that, when made into protein, was a di$ erent version of PBP, one that couldn’t be inhibited by penicillin.

Questions 1. Describe what is happening in Figures 7 and 8 in a complete sentence of your own words. 2. What are the di$ erences in how -lactam antibiotics and vancomycin work? 3. What other mechanisms might arise to allow resistance to the -lactam antibiotics? 4. Could resistance arise to vancomycin? Why or why not?

Figure 8. Vancomycin inhibits cell wall synthesis a di$ erent way by binding PBP’s substrates and preventing cross-bridging. % is prevents proper cell wall synthesis and the bacterium will succumb to osmotic stress.

a. Normal PBP binding and cross-bridge formation

b. Vancomycin binds PBP substrate

c. Cell wall does not form properly

PBP

+

Vancomycin