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The Game of Dood

5/23/2016

For our final project, my partner, Maya and I created a board game called, The Game of Dood. We used the EK Statements (Essential Knowledge Statements):
1A: Change in the genetic makeup of a population over time is evolution.

more specifically

1A.3: Evolutionary change is also driven by random processes.
&
3E.1: Individuals can act on information and communicate it to others.

Below are instructions on how to play our game, what the game looks like, and how it pertains to AP Biology.

Instructions:

Pick an animal to play as and a game piece to use. Bears, Otters, Birds, or Koalas.
Draw a traits from the stack of the animal you choose to learn what traits you have. (Traits include long fur/feathers, short claws, dark or light fur/feathers, etc.)
Play rock paper scissors to decide who moves first.
Use one electronic die to decide how far forward each player/team moves.
1. Each player/team gets one turn each between the first player/team to go to the last player/team. Then it restarts. No do over rolls, trading, or take backs.
Once a player/team has moved choose a color card from the stack that matches the color of the placed landed on. Either green or blue.
Read the card and follow its directions.
Continue on until one player/team reaches the red tiles. From there have another player/team, not on a red tile or whose turn it is not, start drawing cards from the red cards and reading them to the player/team on the red tile whose turn it is. Each player/team has 15 seconds to guess the vocabulary word that is described on the card drawn. If the question is answered right you get to stay on the red tile you have moved to. If you get it wrong you must move back two tiles.
1. The red cards no longer pertain to the animals a player/team has chosen to play as.
Continue play until one player/team has reached the dood tile in the middle of the board.

*Note: The cards contain specific examples of the EK Statements. The Green deck holds statements from EK 1A and 1A.3 the blue deck holds examples of EK 3E.1. The deck of red cards are a mixture of all three statements and their main vocabulary.

Rules:

Once a card it drawn set it to the side of the stack. Then when all the cards for that stack are used reuse the card stack.
Use ONLY one die.
No changing your traits card.

Citations:
"Basic Facts about Sea Otters." Defenders of Wildlife. N.p., 19 Mar. 2012. Web. 23 May 2016.
Blake, Melissa. "E.K.1.A." Mrs. Blake at ECHS. Weebly, n.d. Web. 23 May 2016.
Blake, Melissa. "E.K. 3D-E." Mrs. Blake at ECHS. Weebly, n.d. Web. 23 May 2016.
"Bottlenecks and Founder Effects." Bottlenecks and Founder Effects. N.p., n.d. Web. 23 May 2016
"Koalas, Koala Pictures, Koala Facts - National Geographic." National Geographic. N.p., n.d. Web. 23 May 2016.

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Glowin That Bacteria

3/21/2016

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Post-Lab Questions:
1. Exogenous DNA does not passively enter E. coli cells that are not competent. What treatment do cells require to be competent?
The cells are given DNA in the presence of calcium, rubidium, or magnesium chloride and the suspension is moved from one temperature to another very quickly. That makes the cell wall and membrane permeable so the DNA can enter the cell.
2. Why doesn’t the recovery broth used in this experiment contain ampicillin?
Both the -DNA and the +DNA microfuge get recovery broth and if the the -DNA gets the Recovery broth it would kill the E. coli there which would not show how the DNA transforms the bacteria in the +DNA plates.
3. What evidence do you have that transformation was successful?
The cells were not killed by the ampicillin in the +DNA containers but in the -DNA/+AMP so the transformation was successful because the cells that were meant to be transformed showed the desired results of growth and the non-transformed cells did not grow but died.
4. What are some reasons why transformation may not be successful?
If the condition of changing the temperature to help the transformation along were not meant fully the cells would not become vulnerable to the DNA so that they could transform.
5. What is the source of the fluorescence? Why are some cells fluorescent and other cells not fluorescent?
The transformed cells with DNA added because they are made to contain Green Fluorescent Protein so it absorbs blue light and gives off green light as a response.

*For some of the things that might be harder to see in the infographic.*
Procedures
Conclusion
Pre-Lab Questions:
Transformation Energy:

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Mitosis and Meiosis Can Have as Many Babies as They Want.

1/29/2016

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Investigation 1:

Part 1: Modeling Mitosis

Hypothesis: If a model of mitosis is built out of oreo cookies then each phase can be correctly displayed using sprinkles as chromosomes, the cream as cytoplasm, and the cookie itself as the cell.

Materials:

Four oreos
Lots of sprinkles
Data table

Procedures:
1. Take a picture of one of the oreo cookies. Then take another cookie and place it on top of the original to show how the cell grows during the first stages of interphase. Take a picture.
2. Gently twist one of the cookies apart so that you have one of the cookie sides with cream and the half side without cream.
3. Each cookie represents a cell; ice cream sprinkles represents chromosome pairs.
4. Put sprinkles (chromosomes) all scrambled in the middle of the cookie side with cream to show the first part of interphase where the DNA is tangled and coiled. Take a picture. Spread out the sprinkles on the cookie to show unraveling DNA still in interphase. Take a picture.
5. Match up the sprinkles (chromosomes) with one another sprinkle of the same color to show replication which is prophase. Take a picture.
6. Aline the sprinkles in middle of the oreo(cell) to demonstrate metaphase. Take a picture.
7. Separate the paired sprinkles to opposite sides of the oreo to demonstrate pulling of chromosomes by spindle fibers in anaphase. Take a picture.
8. Break the oreo in half with half of the sprinkles on one side and the rest on the other. This stage shows telophase as the cells splits to form a daughter cell. Take a picture.
9. Gently twist two other cookies apart so that you have one of the cookie sides with cream and the half side without cream.
10. Spread the amount of sprinkles from the split cookie in Step 8 on each new cookie with cream side. (If your original cookie had 9 pairs then the two new cookies should have one sprinkle from each pair.)

Analysis Questions:

Identify 1 step in your model where a mistake could occur. Describe the consequences of that mistake on the cell and on the organism.

During anaphase should a pair chromatids not be pulled apart from one another then one cell will be without that specific chromosome and the other will a more than the necessary amount needed. The absence of a chromosome or the extra chromosome can result in severe damage to the cell and it will either die or be defective.

Describe the limitations of your model versus observing this in the real world.

There were no spindle fibers. We did not have a another small object to symbolize spindle fibers and making them out of sprinkles might have been confusing. Whereas in the real world spindle fibers are large branches that connect to the centromere and pull apart the sister chromatids.

Conclusion: Mitosis is the process made to create the necessary cells when others die of or need replacement. It has six phases with four checkpoints to make sure that mistakes weren’t made. However in my diagram makes could have been made when not displaying spindle fibers to show how they connect to the chromatids. Also to show that the chromatids do not simply just split apart but are forced that way. In the end my hypothesis was supported mitosis was fully diagrammed using oreos cookies.

Part 2: Modeling Meiosis

Hypothesis: If a model of meiosis is built out of oreo cookies then each phase can be correctly displayed including the first rounds of prophase, metaphase, anaphase, telophase, and cytokinesis as well as the second round.

Materials:

Six oreos
Lots of sprinkles
Data table

Analysis Questions:

Identify 1 step in your model where a mistake could occur. Describe the consequences of that mistake on the organism's possible progeny.

During prophase, crossing over could be skipped or missed and that could be a cause for error in the model. If crossing over does not happen diversity is lacked and certain genes like recessive genes might not be expressed in later generations because crossing over did not occur. Valuable traits and advantageous genes might be lost because of not mixing and matching DNA.

Compare and Contrast the process of mitosis and meiosis.

Mitosis goes through one full cycle of cell division that includes interphase. Meiosis goes through two cell division processes and only has interphase during the very first. However both do have the same process and steps during all processes of cell division not including the lack of an interphase two during meiosis and crossing over during prophase I of meiosis.
Conclusion: Mitosis and meiosis are much the same processes. Both create more cells in the body through cell division. However in meiosis it is for reproduction and mitosis is for the good of the body. If new skins cells are needed mitosis will occur. On the other hand in reproduction in sperm or eggs meiosis is needed to create genetic diversity. Both processes have the same steps but meiosis is doubled and therefore longer.

Investigation 2:

Hypothesis: If it takes onion root tips twenty four hours to complete the cell cycle slowly then it will take each phase at least sixty minutes.

Materials:

Link for the online lab
Calculater

Procedures: *Note: The actual lab failed to produce any results so an online lab was used for this Investigation.*

Click on the given link. Time Spent in the Phases of Mitosis
Put the given cells into the phase they belong in as they are given to you.
Count the amount of cells in each phase and record it in the data table.
Calculate the number percent of cells in each phase using the number of cells in that phase and the amount of cells total. Record in the data table.
Calculate the time spent in each phase by doing: % of cells stage X 1440 minutes. Record in the data table.

Analysis:

To calculate the amount of time spent in each phase of the cell cycle from the percentage of cells in that stage. On the average, it take 1,440 minutes (24 hours) for onion root tip cells to complete the cell cycle.

% of cells in stage x 1440 minutes = _____ minutes of cell cycle spent in stage.

Conclusion: In conclusion, although my hypothesis was correct it was also incorrect and vague. It was incorrect because telophase did not take sixty minutes, it only needed 38.8 minutes, but was correct because all other phases took longer than sixty minutes to be completed. My hypothesis supported all but one phase. However, errors could have been made when calculating the percent total of cells counted because when I rounded off I did not increase the last digit only kept it the same. So when calculating the time spent in the phase my numbers might have been slightly different than what is fully correct.

Investigation 3:

Based on your knowledge about human chromosomal disorders and nondisjunction due to loss of control during the cell cycle, identify the name of the syndromes and karyotypes of the following patients.

*Note: The picture on the right is of normal chromosomes from a male and female captured in metaphase and is used as a point of reference.*

Conclusion:
Any variations missing or extra chromosomes significantly affects a person. In all of the patient's chromosomes see above the addition of one single chromosome resulted in a defect in their genes. Each have a specific syndrome determined by whichever chromosome they have an extra of. Having an extra 21 chromosome resulted in Down Syndrome. An extra copy of chromosome 18 creates Edwards Syndrome. If everything in the chromosomes are not exactly right as a result syndromes can be seen.

Investigation 4: Meiosis & Crossing Over
Study the pictures of Sordia in the picture below by counting at least 50 asci and scoring them as either parental or recombinant.

If the ascospores are arranged 4 dark/4 light, count the ascus as “No crossing over.”
If the arrangement of ascospores is in any other combination, count it as “Crossing over.”
Record your results in the data table provided.

Analysis Questions:

Once you have determined if crossing over has occurred in at least 50 hybrid asci, record your data in the table below.
Based on your counts, determine the percentage of asci showing crossover. Record in the table below.
Divide the percent showing crossover by 2. This is your gene to centromere distance. (The percentage of crossover asci is divided by 2 because only half of the spores in each ascus are the result of a crossover event.)

Conclusion: In conclusion it is shown that about fifty percent of Asci crossover. Majority of the Asci had the four dark four light combination that were parental. Many others however were a mixture of of light and dark. Mistakes happened when counting. It was assumed that any presence of four light and four dark ascospores were parental and did not cross over. The correct show of no crossing over must be arranged four light and four dark or vice versa. That was not counted in that way during the experiment so my numbers showing how many crossing over and how many are not are incorrect.

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Taking, Saving, Using Free Energy

12/4/2015

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E.K.: 2A2

Statement: Organisms capture and store free energy for use in biological processes.

My project was done on a google presentation.

E.K. 2A2 PROJECT
(Click Me)

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Beans the Magical O2 Consuming Fruit

11/20/2015

Objective: The objective of this experiment is to apply the gas laws to the function of the respirometer. Students will observe cell respiration of germinating and nongerminating seeds and describe the effects of temperature on the rate of cell respiration.

Hypothesis: If lower temperature slows down cellular respiration and in turn decreases oxygen consumption then the vials held in room temperature will perform better cellular respiration and consume more oxygen.

Materials:
- 3 Cork/pipet assemblies
- 3 Vials
- Absorbent cotton
- KOH solution
- Nonabsorbent cotton
- 100 ml graduated cylinder
- 25 non-germinated beans
- beads
- Tape
- Thermometer
- 25 germinated beans
- Tray
- Silicone glue
- Timer
- Parafilm
- Ice

Procedures: *Note: In the procedure you will notice there are six vials labeled A, B, C, D, E, F, and G. However, in my experiment vials A, B, and C were used for both the 20oC water bath and 10oC ice bath. They fill the spots where ever D, E, and/or F are seen ormentioned.*

1. Set up an ice bath at 10° C and a room temperature bath (24° C) as assigned by the instructor. This should be done prior to beginning the other steps in order to allow enough time for the water baths to equilibrate to the required temperature. To attain and keep the 10° C temperature, add ice.
2. Label 6 vials (respirometers) A, B, C, D, E and F.
• Place a circle of absorbent cotton (approximately the size of a nickel) into the bottom of each vial.
• Carefully soak with 1- 2 ml of 15% KOH solution, not allowing any KOH solution to touch the side walls of the vials.
• Place a circle of nonabsorbent cotton into the vials directly on top of the KOH/cotton circle in the vial. This will keep the KOH solution from touching peas during the experiment.
3. Determine the volume of germinated peas.
• Fill a 100 ml graduated cylinder with 50 ml of water.
• Add 25 germinated peas to the cylinder and measure the increase in water volume. The difference represents the volume of the peas.
Pea volume = __________ ml.
• Place the peas on a paper towel. These germinated peas will be used in Vial A.
• Repeat steps for Vial D.
4. Determine the volume of non-germinated peas.
• Fill a 100 ml graduated cylinder with 50 ml water.
• Add 25 non-germinating peas.
• Add glass beads to raise the volume to equal that obtained with swollen germinated peas.
• Remove the peas and beads. Place them on a paper towel. The non-germinated peas and beads will be used in respirometer B.
• Repeat steps for Vial E.
5. Determine the volume of glass beads.
• Fill the 100 ml graduated cylinder with 50 ml water.
• Add glass beads to raise the volume so it equals the volume of germinating beans as determined previously in Step 3.
• Place glass beads on a paper towel, to be used in Vial C.
• Repeat steps for Vial F.
6. Place one set of germinating peas into Vial A. Insert the cork/pipet assembly into the vial. Repeat with the second set of germinating peas for Vial D.
7. Place one set of non-germinated peas + glass beads into Vial B. Insert the cork/pipet assembly into the vial. Repeat with the second set of non-germinated peas + glass beads for Vial E.
8. Place one set of glass beads into Vial C. Insert the cork/pipet assembly into the vial. Repeat with the second set of glass beads for Vial F.
9. Wrap Parafi lm® or plastic wrap tightly around the seams (cork and tube) to seal any potential leak.
10. Make sure the ice water bath has equilibrated to 10° C. Record the temperature of the room temperature bath.
11. Place a piece of masking or lab tape over the water bath to suspend pipet tips out of the water during the equilibration phase.
12. Place Vials A, B & C into the 10° C water bath and Vials D, E & F into the 24° C water bath. Place them with the calibrated side of the pipet facing up to allow measurements to be taken.
13. Allow the respirometers to equilibrate in the water baths for 8 minutes.
14. After the equilibration phase, immediately submerge each respirometer. Water in the water bath will enter the pipets and travel a short distance. As respiration occurs inside the vials, oxygen is consumed and the pressure drops. Over time, as pressure drops, additional water from the water bath enters the pipets. Make sure the vials do not fill up with water. If it does, there is a leak which must be corrected. Reassemble the faulty respirometer.
15. Arrange the vials so you can read the volume markings on each pipet. Place lead donuts or other weighted objects on the vials to keep vials submerged.
16. Record the starting temperature. Maintain temperature by adding ice or water as necessary during the experiment.
17. Record the starting point (“Time 0”) volume of each pipet. Take readings of the volume of water in each pipet every 5 minutes for 20 minutes. Record these values in the tables on the following page.
18. Collect class data for Vials A, B, C, D, E and F for both temperature water baths.
19. Correct volumes measured for changes in environmental variables.

Data Tables:Note: The following data tables were results provided by the instructor. In my experiment an error occurred which prevented the result of adequate data.

1. Graph your corrected data (difference) for Vials A, B, and C. Place time (in minutes) on the x-axis and volume (ml O2 consumed) on the y-axis. Data from both temperatures should be plotted on the same graph. Draw the best straight line through the data points.

2. What accounts for the difference in oxygen consumption seen between the germinating and non-germinating beans?Germinating beans are maturing and needs to perform cellular respiration in order to keep growing so the beans consume oxygen. Non-germinating beans are dormant and do not consume as much oxygen because they do not need to.
3. List some of the constant controls in this experiment.

Volume of the vials
8 minute equilibrium period

4. Why do the beads seem to be using oxygen?
There is a change in pressure in the vial and the air has to get out. Also, there is oxygen already in the vial and because oxygen is less dense than water it has to come out of the submerged vial. So oxygen bubbles flow out of the tube.
5. Why are the readings corrected using the bead values?
The values are corrected because the beads are controlled to measure change in water volume because of temperature and pressure. The correct values show those outside influences on the respirometer.
6. What is the function KOH in this experiment?
KOH removes CO2 from the system.
7. From the slope of the lines, determine the rate of oxygen consumption at 10° C and room temperature for the germinating and non-germinating beans. Determine the slope of the lines over a middle section of each line by dividing the difference in volume reading by the difference in time. Volume (ml O2 consumed) values are determined from the line.

8. Compare the rate of oxygen consumption at 10°C and room temperature. Why are they different?
They are different because cellular respiration works better in a regular or higher temperature. Anything below what is normal slows down the process and oxygen cannot be consumed as quickly or in large amounts.
9. How do you think the rates of respiration would change in peas that have been germinating for 0, 24, 48, 72, and 96 hours. Why?
Over that period of time I think the rates would increase significantly and then decrease rapidly because after the beans have germinated they would have grown leaves or buds. When they plant part of the seed appears it will no longer need respiration but photosynthesis.
10. Write a hypothesis using the same experimental design to compare the rates of respiration in a mouse at both room temperature 24° C and at 10° C.
If cellular respiration and oxygen consumption is higher in average or room temperature then placing a mouse in both room temperature and 10oC the higher rate will be in the room temperature setting.
11. Using the same experimental design, write a hypothesis to test the respiration rate of a 15 g reptile and a 15 g mammal at 10° C.
If a 15g reptile is more comfortable in 10oC then it will perform less cellular respiration and have a lower oxygen consumption rate than a 15g mammal at 10oC because the mammal needs to keep homeostasis and the oxygen intake will help.
12. What basic cellular process is responsible for the oxygen consumption?
The ETC is responsible for oxygen consumption because oxygen is used at the end of that process to carry away low energy electrons from the cell.

14. What is the independent variable?
Time is the independent variable.
15. What is the dependent variable?
ml O2 consumed is the dependent variable.
16. What two hypotheses are being tested in this experiment?
The two tested hypotheses are:

If cellular respiration works better in warmer temperatures then there will be more oxygen consumption in the room temperature vials.
If germinated beans perform more cellular respiration than non-germinated beans then there will be more oxygen consumption in the germinated beans vial.

13. Graph the results from the Corrected Difference column for the germinated peas and dry peas at both room temperature and at 10° C. Label the horizontal x-axis "Time in Minutes.” Label the vertical y-axis "ml O2 Consumed.”

Conclusion:
While performing this experiment two errors were made. The first being not all vials were held underwater during all timed intervals. The second was a hole found in the cork of the A vial. Both errors resulted in the use of given data. However with errors included and using the given data my hypothesis was proven correct. All three vials held in a room temperature water bath consumed more oxygen. The increase consumption of oxygen is only possible with a faster performing cellular respiration process. The vials in the low temperature water bath had a lower rate of oxygen consumption which in turn means the cellular respiration process was more slow than the warmer bath. If repetition of the lab was necessary more pictures would have been taken of the process to more thoroughly explain the steps. The cork with the hole would also be replaced to obtain the correct data and take away one of the sources of error.

Blast Blab

10/4/2015

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Investigation 1:

Exercise 1:
Chimpanzees and humans share 96% of their DNA which would place them closely on a cladogram. Humans and fruit flies are placed farther apart on a cladogram since they share only approximately 60% of their DNA. Draw a cladogram that shows the evolutionary relationship between humans, chimpanzees, and fruit flies.

Exercise 2:
Using the given cladogram (found below), answer the following questions:

According to the cladogram, what organisms have hair?
According to the cladogram, what four structures do tigers possess?
According to the cladogram, which structure evolved first between lungs and dry skin?

1. Based of the given cladogram it is shown that tigers and gorillas have hair.
2. According to the diagram tigers have jaws, lungs, dry skin, and hair.
3. Lungs evolved first between lungs and dry skin.

Investigation 2:

Exercise 3:
Make a cladogram based off of the data table below.

Exercise 4:
After answering question (a) complete the cladogram based on the gene similarity for part (b).

a) Why is the percentage of similarity in the protein always higher than the percentage of similarity in the gene for each of the species?
Protein similarity is always higher because there are more ways for genes to order themselves to make the protein than there are types of proteins that do the same job. So genes can differ a little more and still make the same protein.
(b)

Investigation 3: Uncovering Fossil Specimen using BLAST

Part A: MORPHOLOGICAL OBSERVATION OF THE FOSSIL SPECIMEN AND FORMING THE HYPOTHESIS

A team of scientists have uncovered the following fossil specimen near Hells Creek, Montana.

1. Make some general observations about the morphology (physical structure) of the fossil and record your observations in the space provided.

My Observation:

long tail
half circle head
two legs
small, short thin neck
bumps (perhaps scales) down the back
thick body
vertebrae down what appear to be the spine and through the tail
one eye on the visible side of the fossil

2. Little is known about the fossil other than it appears to be a new species. Upon careful examination of the fossil, small amounts of soft tissue have been discovered. The scientists were able to extract proteins from the tissue and use the information to sequence several genes. The task is to use BLAST to analyze these genes and determine the most likely placement of the fossil species on the following fossil cladogram:

3. Hypothesis: If the observation made on the fossil specimen prove to be true then it will accurately fit on the branch housing the birds and the crocodilian.
The hypothesis graphed on a cladogram is pictured below.

Part B: USING BLAST TO ANALYZE GENES AND DETERMINE THE MOST LIKELY PLACEMENT OF THE FOSSIL SPECIES

Materials:

Laptop
http://blogging4biology.edublogs.org/2010/08/28/college-board-lab-files/
http://blast.ncbi.nlm.nih.gov/Blast.cgi
Gene Squences

Procedure:
Step 1: Instruction for Downloading Gene Files
Before starting this investigation, students need to download four gene files (Gene 1- Gene 4) to their flash drive or computer. These files are located at the following web address: http://blogging4biology.edublogs.org/2010/08/28/college-board-lab-files/
Note that these files will not open on your computer. They only work when opened on the BLAST website.

Step 2: Instructions for BLAST Queries
Upload the gene sequences into BLAST by following the instructions below:
1. Go to the BLAST homepage: http://blast.ncbi.nlm.nih.gov/Blast.cgi
2. Click on “Saved Strategies” on the top of the page.
3. Under “Upload Search Strategy,” click on “Browse” and locate one of the gene files you saved onto your computer.
4. Click “View.”
5. A screen will appear with the parameters for your query already configured.
NOTE: Do not alter any of the parameters. Scroll down the page and click on the “BLAST” button at the bottom.
6. After collecting and analyzing all of the data for that particular gene (see instructions below), repeat this procedure for the other three gene sequences.

Step 3: Instructions for Analyzing BLAST Queries
1. Scroll down to the section titled, “Sequences producing signifi cant alignments”. The list of organisms that appear below this section are those with sequences identical to or most similar to the gene of interest. The most similar sequences are listed fi rst and as you move down the list, the sequences become less similar to your gene of interest.
2. Try clicking on a particular species listed to get a full report that includes the species’ classification scheme, the research journal the gene was first reported in, and the sequence of bases that appear to align with your gene of interest.
3. Click “Distance tree of results,” a cladogram of the species with similar sequences to your gene of interest placed on the cladogram will be shown according to how closely their matched gene aligns with your gene of interest.

Step 4: Analysis of Results
Species share similar genes because of common ancestry. The more similar genes two species have in common, the more recent their common ancestor. Thus, the two species will be located closer on a cladogram. As you collect information from BLAST for each of the gene fi les explain whether the data supports your original hypothesis and your original placement of the fossil species on the cladogram.
For each BLAST query, consider the following:

What species has the most similar gene sequence as your gene of interest?
Where is that species located on the cladogram?
How similar is that gene sequence?
What species has the least similar gene sequence as your gene interest?

Step 5: Drawing the Cladogram
Redraw the original cladogram and include your final placement of the fossil species.

The redrawn cladogram based on the collected information.

Conclusion: Based off of genes 1-4 my hypothesis was supported in placing the fossil specimen on the same branch of the given cladogram as the birds and the crocodilian. When examining the cladograms that show genes 1-3 each the fossil specimen is shown to be on the closest branch to birds. In the last cladogram the fossil specimen is on the closest branch to alligators which are very close relatives to crocodiles. The branch on the given cladogram containing crocodilians and birds makes for a reasonable spot to place the unidentified fossil specimen based simply off of the four genes. Therefore my placement of the fossil specimen was a very accurate spot on the given cladogram.

Investigation 4: DIY Gene Comparisons
The next step is to learn how to find and BLAST your own genes of interest. To locate a gene, go to the Entrez Gene website (http://www.ncbi.nlm.nih.gov/gene) and search for the gene. Once you have found the gene on the site, copy the gene sequence and input it into a BLAST query.

Materials:

Laptop
Entrez Gene website (http://www.ncbi.nlm.nih.gov/gene)
http://blast.ncbi.nlm.nih.gov/Blast.cgi)

Procedure:
1. On the Entrez Gene website search the term “human actin”.
2. Click on the first link that appears and scroll down to “NCBI Reference Sequences”.
3. Click on the fi rst fi le name “NM 001100.3” under “mRNA and Proteins”.
4. Click on “FASTA” just below the gene title.
5. The nucleotide sequence displayed is that of the actin gene in humans.
6. Copy the gene sequence and go to the BLAST homepage (http://blast.ncbi.nlm.nih.gov/Blast.cgi).
7. Click on “nucleotide blast” under the Basic BLAST menu.
8. Paste the sequence into the box where it says “Enter Query Sequence.”
9. Give the query a title in the box provided if you plan on saving it for later.
10. Under “Choose Search Set” select the type or genome you want to search (human genome, mouse genome, or all genomes available).
11. Under “Program Selection” choose whether or not you want highly similar sequences or somewhat similar sequences. Choosing somewhat similar sequences will provide you with more results.
12. Click BLAST.

Gene of Focus: GAPDH (glyceraldehyde 3-phosphate dehydrogenase)

Hypothesis: If the gene is vital in the process of cellular respiration and the breakdown of glucose then it will be a common gene throughout many species because all life needs energy and many have to break it down it fully utilize it.
In humans, what is the importance of the gene you chose?
It helps break down glucose and carbon molecules in the body. Without the breakdown of glucose and carbon the body would not have energy to perform any tasks. It is an important part of cellular respiration.
Would you expect to find that gene is all organisms? Why or why not?
I would expect to find this gene in all organism because all organism need to break down the carbon they intake and glucose they make. All organisms need energy and breaking down that glucose and carbon to properly use them later on.

Conclusion: After completing this I can say that GAPDH is found in many organisms do to its involvement in cellular respiration. All living organisms need energy to survive and that energy comes from the breakdown glucose and carbon dioxide. However, some organisms need very little glucose and don’t perform cellular respiration GAPDH is still very much common in other organism requiring more energy. The constant appearance GAPDH shows its importance in the survival of many organisms. It is an unskippable step to create energy and vital to life.

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The Game of Dood

Glowin That Bacteria

Mitosis and Meiosis Can Have as Many Babies as They Want.

Taking, Saving, Using Free Energy

Beans the Magical O2 Consuming Fruit

Data Tables:Note: The following data tables were results provided by the instructor. In my experiment an error occurred which prevented the result of adequate data.

Blast Blab

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Mitosis and Meiosis Can Have as Many Babies as They Want.

Taking, Saving, Using Free Energy

Beans the Magical O2 Consuming Fruit﻿

Data Tables:*Note: The following data tables were results provided by the instructor. In my experiment an error occurred which prevented the result of adequate data.*

Blast Blab﻿

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Beans the Magical O2 Consuming Fruit

Data Tables:Note: The following data tables were results provided by the instructor. In my experiment an error occurred which prevented the result of adequate data.

Blast Blab