Team+Flowers

flat =Members= Brooke Marissa Keeley Elissa

=Characteristics of Woody Stems Lab= Part 1: External structure of a woody twig Examine one of your twigs closely with a hand lens. The buds are the most conspicuous structures on a dormant stem. Locate the terminal bud at the top of the twig. Notice the bud scales covering the growing point within the bud. Locate the series of rings encircling the twig some distance below the terminal bud. These are bud-scale scars left where the scales of the terminal buds of the previous year were attached. The portion of the twig between the terminal bud and the first bud scale scar marks the growth of the previous season. The growth of the season before that and other seasons can be observed by measuring between the bud-scale scars. 1. Count the sections between bud-scale scars and determine the age of your twig. One of our group’s twig was four years old, and the other group’s twig was two years old. 2. Has growth in length occurred the same rate each year? The growth in length has not occurred at the same rate each year. 3. Why? There could have been a drought or something similar that caused the twig to grow less that year.

At intervals along the twig, you will find circular, oval, or shield shaped leaf scars, which mark the point of attachment of leaf petioles from the previous season. Leaf scars are located at the nodes. 4. Examine your twig and determine how many leaf scars are located at a node. There are two leaf scars located at each node. 5. Classify the leaf arrangement as opposite, alternate, or whorled. The leaf arrangement is opposite. 6. Examine the twig and determine the number of nodes produced for each growing season (the space between two nodes is called an internode.) There is one node produced for each growing season. 7. Is the same number of nodes produced each growing season? The same number of nodes are produced each growing season.

Examine a leaf scar and notice minute dots called bundle scars, which show the location of the xylem and phloem that carried water and minerals from stem to leaf.

All of the leaf scars have the same number of bundle scars, and the arrangement is the same.

Look for lateral buds located along the sides of the stem. These are smaller than the terminal bud and are usually different in shape. Lateral buds situated above the leaf scars are axial buds. Examine the internodes for tiny pores called lenticels. These are especially common in younger bark. 9. Describe the form and location of the lenticels. The lenticels are located in the internode. They are very small bumps on the bark of the twig. 10. What is the function of the lenticels? The function of the lenticels is to allow gas exchange, generally in a younger branch.

=Thumb War Infographic=



=Monocot Dicot Lab=

We can find the young leaves. There is one. The outside is the seed coat. Monocot
 * Corn Seed**


 * Bean Seed**

You cannot open a corn seed like a bean seed because it is a monocot. Dicot


 * Split Pea Seed**

The Split Pea Seed is a monocot because it only has one cotyledon. Monocot

**Stem Slides**

Monocot

Dicot

Dicot stems have a pith, while monocot stems do not. Dicot vascular tissues are in a ring-like pattern, while monocot vascular tissues are scattered throughout the ground tissue. These tissues link all the parts of the plant, allowing water, nutrients, and other compounds to be carried throughout the plant.

The difference between a herbaceous and a woody stem is that herbaceous stems die down to soil level at the end of the growing season, while woody stems have a wooden stalk above ground at all times. This is because woody stems have secondary growth, while herbaceous stems do not.
 * Herbaceous and Woody Stems **



= Characteristics of Living Things Lab = Beginning Picture Beginning Observations:
 * Flask A (water and yeast) is murky, almost like a cloudy pool.
 * Flask B (water and molasses) seems to be stagnant.
 * Flask C (water, molasses, and yeast) is foamy on the top and looks like beer.

Later Picture Later Observations:
 * Flask A (water and yeast) stayed the same.
 * Flask B (water and molasses) stayed the same.
 * Flask C (water, molasses, and yeast) turned its test tube yellow, and bubbles began to visibly rise in the flask.

Yeast Budding Picture Observation: The yeast is budding, which is when a cell produces another set of chromosomes, splits into two nuclei, and then eventually splits into two cells.

Analysis & Conclusion
 * 1) Why did the bromothymol blue change colors when a classmate exhaled into the test tube? The bromothymol blue changed colors when a classmate exhaled into the test tube because his respiration released CO2, which interacts with the water to make Carbonic Acid (H2CO3), an acid. Bromotymol Blue turns yellow when it senses an acid.
 * 2) What does the production of carbon dioxide gas in the one flask indicate? The production of carbon dioxide gas in the Flask C indicates that the yeast is living, because it “breathes” air (respirates), and that it uses the oxygen and gives off the carbon dioxide.
 * 3) How can you **be sure** that the carbon dioxide gas was produced by the yeast? We can be sure that the carbon dioxide gas was produced by the yeast because of respiration (when you turn glucose and oxygen into carbon dioxide and energy.)
 * 4) What does the presence of buds indicate about the yeast? The presence of buds indicates that the yeast is reproducing.
 * 5) Why were more buds present in one of the mixtures? More buds were present in Flask C because the yeast was reproducing more quickly. This is because Flask C contained water and sugar, the two things that yeast needs to reproduce. Flask A only contained water and yeast, so there was no food for the yeast to use to reproduce.

=Comparing Plant and Animal Cells Lab=

Onion Cell

Questions:

 1. Describe the shape and arrangement of the onion cells. The onion cells are rectangular and arranged very close together.

 2. What happened to the cells when concentrated salt solution was added to the cells? Why do you think this happened? When the concentrated salt solution was added to the cells, the cell membrane shriveled up. This happened because the salt caused the cell membrane to dry out, much like salt on a slug.

Onion Cell After Salt

Cheek Cell 

Questions:

 1. Describe the shape and arrangement of the cheek cells. The cheek cells were circular and scattered about the slide.

<span style="font-family: Arial,Helvetica,sans-serif; font-size: 90%;"> 2.What was the purpose of adding the Lugol's iodine or Methylene blue to the slide? The purpose of adding the Lugol’s iodine or Methylene blue to the slide is so that we could see the parts of the cells.

=<span style="font-family: Arial,Helvetica,sans-serif; font-size: 90%;">Cell Transport Lab =

<span style="font-family: Arial,Helvetica,sans-serif; font-size: 90%; line-height: 0px; overflow: hidden;"> The iodine water and starch solution demonstrates diffusion. In the beginning, the solution outside the dialysis bag was hypertonic because it contained a higher concentration of iodine than the starch solution inside the dialysis bag, which was hypotonic. Overnight, the two solutions became isotonic as the iodine diffused through the selectively permeable membrane (the dialysis bag). This is diffusion because it involves movement of the iodine, not the water.
 * Iodine Water with Starch Solution Dialysis Bag**


 * Tap Water with Potato**


 * Salt Water with Potato**

The tap water demonstrates osmosis. In the beginning, the tap water was hypertonic because it contained a higher concentration of water molecules than inside the potato, which was hypotonic. Overnight, the two solutions became isotonic as the water molecules went through the selectively permeable membrane into the potato cells. This caused the potato cells to expand. This is osmosis because it involves movement of the water molecules. The salt solution also demonstrates osmosis. In the beginning, the potato was hypertonic because it contained a higher concentration of water molecules than the salt solution, which was hypotonic. Overnight, the solutions became isotonic as the water molecules diffused into the salt solution. This is also osmosis because it involves movement of the water molecules.

=Cell Model=

=Protists Lab=


 * Stentor**


 * Spirostomum**


 * Daphnia**

=Cell Size Lab=


 * Agar Before**


 * Agar After**

<span style="font-family: Arial,Helvetica,sans-serif;">1. Compare and contrast the three cubes after they were sliced in half. The three cubes all experienced the same rate of diffusion, but the Phenolphthalein did not reach the center of the larger cubes because they were thicker. Unlike the 1cm cube, the centers of the 2cm and 3cm cubes were still white.
 * Analysis Questions**

<span style="font-family: Arial,Helvetica,sans-serif;"> 2. Which "cell" seemed to be most and least efficient at getting outside substances into the cell? Explain. The 3cm “cell” seemed to be least efficient in getting outside substances into the cell because its larger volume caused the Phenolphthalein to take longer to spread throughout the entire cell (to the center). The 1cm “cell” seemed to be the most efficient because the Phenolphthalein was able to spread throughout the entire cell.

<span style="font-family: Arial,Helvetica,sans-serif;"> 3. Which of your calculations seems to explain what you observed in your cell models? Why do you think so? The surface area to volume ratio seems to explain what we observed in our cell models. This is because when there is a higher surface area to a lower volume, the Phenolphthalein is able to reach more of the cell in the allotted time. When there is more surface area, there is more opportunity for the Phenolphthalein to diffuse into the agar. Since there is less volume, it takes less time for the Phenolphthalein to reach the entire cell.

<span style="font-family: Arial,Helvetica,sans-serif;"> 4. Speculate on a relationship between cell size and efficiency. Your statement should resemble a hypothesis. (Remember: Use an If....., then..... statement.) If a cell is smaller, then it is more efficient.


 * Data Table**
 * <span style="font-family: Arial,Helvetica,sans-serif;">Cube Size || <span style="font-family: Arial,Helvetica,sans-serif;">Surface Area of Cube || <span style="font-family: Arial,Helvetica,sans-serif;">Volume of Cube || <span style="font-family: Arial,Helvetica,sans-serif;">Surface Area to Volume Ratio || <span style="font-family: Arial,Helvetica,sans-serif;">Diffusion Distance || <span style="font-family: Arial,Helvetica,sans-serif;">Rate of Diffusion ||
 * <span style="font-family: Arial,Helvetica,sans-serif;">3cm || <span style="font-family: Arial,Helvetica,sans-serif;">54 sq cm || <span style="font-family: Arial,Helvetica,sans-serif;">27 cubic cm || <span style="font-family: Arial,Helvetica,sans-serif;">2:1 || <span style="font-family: Arial,Helvetica,sans-serif;">.5cm || <span style="font-family: Arial,Helvetica,sans-serif;">.05cm/min ||
 * <span style="font-family: Arial,Helvetica,sans-serif;">2cm || <span style="font-family: Arial,Helvetica,sans-serif;">24 sq cm || <span style="font-family: Arial,Helvetica,sans-serif;">8 cubic cm || <span style="font-family: Arial,Helvetica,sans-serif;">3:1 || <span style="font-family: Arial,Helvetica,sans-serif;">.5cm || <span style="font-family: Arial,Helvetica,sans-serif;">.05cm/min ||
 * <span style="font-family: Arial,Helvetica,sans-serif;">1cm || <span style="font-family: Arial,Helvetica,sans-serif;">6 sq cm || <span style="font-family: Arial,Helvetica,sans-serif;">1 cubic cm || <span style="font-family: Arial,Helvetica,sans-serif;">6:1 || <span style="font-family: Arial,Helvetica,sans-serif;">.5cm || <span style="font-family: Arial,Helvetica,sans-serif;">.05cm/min ||

=Catalase Lab=
 * ** Temperature ** || **Liver** || **Ground** || **Bean** || **Potato** ||
 * **Cold** || **Bubbles through entire tube** || **Little bit of bubbling** || **Fizzy** || **Fizzy** ||
 * ** Room ** || **Bubbles through entire tube** || **Small bubbling (Fizzy)** || **Fizzy** || **Fizzy** ||
 * **Warm** || ** Bubbles flowing out of tube ** || ** Fizzy ** || ** Fizzy ** || ** Fizzy ** ||
 * Room Temperature Hydrogen Peroxide**


 * Cold Hydrogen Peroxide**


 * Warm Hydrogen Peroxide**



=Penny Lab= Observations: The water formed a bubble over the penny as drops were added. For the soapy water, not as many drops could be added to the penny. Once the surface tension was broken, all of the water spilled off the penny.
 * Analysis**
 * 1) <span style="font-family: Arial,Helvetica,sans-serif; font-size: 110%;">Write the equation for the breakdown of hydrogen peroxide. H2O2 → H2O + O2 --- Hydrogen Peroxide breaks down into water and oxygen.
 * 2) <span style="font-family: Arial,Helvetica,sans-serif; font-size: 110%;">What large group of biomolecules (carbohydrate, lipid, or protein) does catalase belong to? What do members of this group have in common?Catalase belongs to the protein group. Members of the protein group “do stuff.”
 * 3) <span style="font-family: Arial,Helvetica,sans-serif; font-size: 110%;">Is it possible to reuse biomolecules like catalase? Explain. It is possible to reuse biomolecules like catalase. This is because it can bond or break down multiple sets of sugars.
 * 4) <span style="font-family: Arial,Helvetica,sans-serif; font-size: 110%;">How did the amount of reaction change between room temperature, warm, and cold hydrogen peroxide? Room temperature and cold hydrogen peroxide had about the same reaction. Warm hydrogen peroxide had a greater reaction than the other two.
 * 5) <span style="font-family: Arial,Helvetica,sans-serif; font-size: 110%;">What happens to an organism if biomolecules like catalase become useless? Explain. If biomolecules like catalase become useless, then the organism would have no way of breaking down hydrogen peroxide, and it would become poisoned and eventually die.
 * || Hypothesis || Trial 1 || Trial 2 || Trial 3 || Trial 4 || Average ||
 * Water || 20 Drops || 29 Drops || 16 Drops || 37 Drops || 33 Drops || 29 Drops ||
 * Soapy Water || 20 Drops || 16 Drops || 15 Drops || 21 Drops || 21 Drops || 18 Drops ||

=Light Intensity Activity=

What are the best possible conditions for making the maximum of ATP? High Light Intensity, Mid-Wavelength

Hypothesize what values you believe are the best conditions (what wavelength and light intensity creates the most ATP.) 200 Light Intensity, 700 Wavelength


 * Wavelength || 400 || 450 || 500 || 550 || 600 || 650 || 700 || 750 ||  || 400 || 450 || 500 || 550 || 600 || 650 || 700 || 750 ||
 * Light Intensity |||||||||||||||| Percent ||  |||||||||||||||| ATP Number ||
 * 40 || 13 || 3 || 1.5 || 3 || 4 || 17 || 1 || 0.6 ||  || 2 || 1 || 1 || 1 || 1 || 2 || 1 || 1 ||
 * 80 || 26 || 6 || 3 || 6 || 8 || 34 || 2 || 1.2 ||  || 3 || 1 || 1 || 1 || 1 || 4 || 1 || 1 ||
 * 120 || 39 || 9 || 4.5 || 9 || 12 || 51 || 3 || 1.8 ||  || 4 || 1 || 1 || 1 || 1 || 5 || 1 || 1 ||
 * 160 || 52 || 12 || 6 || 12 || 16 || 68 || 4 || 2.4 ||  || 5 || 2 || 1 || 2 || 2 || 9 || 1 || 1 ||

When the light intensity increases, the ATP number and percent increases. When wavelength is at 650, there is the most ATP production. There is also high ATP production when the wavelength is low.

=Chromatography Lab=

Right Strip--Brooke & Elissa Solvent Front—5cm Pigment Front Yellow—4cm Rf: .8 Pigment Front Green—4.5cm Rf: .9 Pigment Front Yellow/Green—5cm Rf: 1

Left Strip--Marissa & Keeley Solvent Front--4cm Pigment Front Yellow--2.5cm Rf: .625 Pigment Front Green--4cm Rf: 1

Because these measurements are not necessarily exact, it is possible that the green in both of our groups' leaves is the same pigment. However, our yellow pigments seem to be different. This could mean that we had different types of chlorophyll in our leaves. Because other groups had different results than we did, their leaves probably had different types of chlorophyll.