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Developmental Biology Assignment

Submitted by hamacdonald on Tue, 11/21/2017 - 16:19

Hailey MacDonald November 19th 2017 Writing Assignment 4

Mittens or Gloves?
There is no denying the appreciation and usefulness humans have for digits, fingers. It is

a main separating factor between humans and other mammals. What we might not think about is how important the timing of the production of digits are. Although it is a complex process, in theory has been explained more simply by Alan Turings using his mathematical model, A Turing- Type Mechanism.

The basic idea behind this model is there are two chemicals. One of the two chemicals is called the activator meaning it turns things on. Specifically, the activator found in mice embryos is fgf. The other of the two chemicals is the inhibitor known as Gli3 in mice, which turns things off. In this case, it stops the production of digits. At first, these two chemicals fluctuate randomly at similar levels. The activator is turned on, it then turns on the inhibitor which quickly turns the activator off. As seen in the figure below, the activator builds up which also causes the inhibitor to build up. Eventually the inhibitor has a high enough concentration to severely limits the amount of activator and turning digit development off. The two chemicals together work to create a wave like pattern. This pattern is what gives us the spaces in-between our fingers. The activator triggers digit development, and the inhibitor is what gives us these negative spaces

between each finger.

What controls this pattern is the Hox genes, or more specifically the distal Hox genes (distal meaning far away, our digits are furthest away from our bodies). Hox genes are what control the body plan of the embryo on the anterior-posterior (head- tail) axis. The amount of distal hox genes being expressed controls the digit formation by starting the self-regulating loop between the activator and inhibitor.

In order to determine what controls the normal digit formation, mutants, or mice with more and skinner fingers than normal mice were observed. This is known as polydactyl. Something went wrong in the timing or development of the digits causing an abnormal amount of digits to form.

The mutants in which the inhibitor, Gli3 was not expressed showed increased number of digits until the digits began to fuse together in a

mitten- like structure. This can be seen in the figure to the left. The less Gli3 that inhibits the activator that produces digit development, the more digits form and the skinner these digits are

Reference:

K. (2011, January 13). Alan Turing’s Reaction-Diffusion Model – Simplification of the

Complex. Retrieved November 19, 2017, from

Ecology Assignment 5

Submitted by hamacdonald on Sat, 11/18/2017 - 21:19

Hailey MacDonald

Ecology 287

November 18th, 2017

Assignment A5

 

Part A (3 pts): Fig. 1 shows the relationship between local and regional species richness in a hypothetical region. Do you expect to observe this relationship between regional and local species richness in nature? Explain. (That is, if no, why not? If yes, why and under what circumstances?)

 

            I would not expect to observe this relationship between local and regional species richness in nature. The slope of the line produced in Figure 1 is higher than one meaning the local richness is higher than regional richness. It would not be possible to observe a higher local species richness than regional species richness because the total amount of species found in the local area also factors into the number of species in the regional area meaning the regional area would have to have a higher number of species richness.

 

Part B (6 pts): You are investigating patterns of species richness on two different continents to determine how local and regional species richness varies between them. Your two study areas are study area 1 and study area 2. For each study site, you sample the local species richness in a small area that is a subset of the entire site, and then find the regional species richness for the whole site. Your data are shown in Fig. 2. Which process, regional or local, is the dominant driver of the pattern in each study area? Explain your answer

 

            For Study area 1, regional processes appear to be the driving limiting local diversity. This is because the line forms a linear slope which means dispersal is limiting local richness growth. The regional richness is increasing at a steady rate and the local richness is slightly less. This means dispersal must be the issue because the species are not going to enough local areas to increase the local richness at the same rate as the regional richness.

            In the Study area 2, the driving limiting factor is local processes. This is what forms the horizontal line. The local processes most likely abiotic factors such as limited resources. These limit the local richness even with the addition of more species to the region.

 

Part C (3 pts): Now imagine you’re studying species richness on islands. Do you think the regional species richness on the mainland will affect how many species are predicted to be found on an island based on the equilibrium theory of island biogeography? Why or why not?

            I do think if the mainland has a higher species richness that the island will also have a high species richness and if the mainland has a lower species richness the island would similarly show lower species richness. The more species that exist on the mainland increases the chances that these species will immigrate to the island. The more species on the mainland competing for resources would likely encourage more species to immigrate to the island.

            

Orgo Procedure

Submitted by hamacdonald on Fri, 11/17/2017 - 14:40

      In a 250 mL round bottom flask, nutmeg (1.010 g, 2.29 mmol) and tert-butyl methyl ether (3 mL) were added and boiled for ten minutes. Once the solution had boiled and cooled, the liquid layer was filtered into a 25- mL Erlenmeyer flask via microscale filtration techniques. Tert-butyl methyl ether (2 mL, 22.69) was added to the round bottom flask, heated, and filtered again (percent recovery, 24.578- 24.107). After the second filtration, the 25- mL Erlenmeyer flask was warmed and air was blown over the solution until all of the solvent evaporated and the solid yellow product remained. The product was dried for five minutes and the crude weight was obtained (0.471 g, 0.65 mmol) and acetone (9.4 mL) was added and warmed until the solid dissolved. The solution was cooled to room temperature for 5 minutes and then cooled in an ice bath for 15 minutes. The solid product was filtered via suction filtration and washed with cold acetone. After filtration and drying the crystals were collected, weighed (0.189 g, 0.261 mmol) and the melting point range (49-51 °C) of the product was obtained.

Orgo Lab

Submitted by hamacdonald on Fri, 11/17/2017 - 14:39

 

 

Purpose:

 The purpose of this experiment is to isolate and recrystallize trimyristin from nutmeg and then hydrolyze the trimyristin in a mixture containing 6M NaOH, 95% ethanol, and HCL to form myristic acid.

Reaction Scheme:

 

Procedure:

            In a 250 mL round bottom flask, nutmeg (1.010 g, 2.29 mmol) and tert-butyl methyl ether (3 mL) were added and boiled for ten minutes. Once the solution had boiled and cooled, the liquid layer was filtered into a 25- mL Erlenmeyer flask via microscale filtration techniques. Tert-butyl methyl ether (2 mL, 22.69) was added to the round bottom flask, heated, and filtered again (percent recovery, 24.578- 24.107). After the second filtration, the 25- mL Erlenmeyer flask was warmed and air was blown over the solution until all of the solvent evaporated and the solid yellow product remained. The product was dried for five minutes and the crude weight was obtained (0.471 g, 0.65 mmol) and acetone (9.4 mL) was added and warmed until the solid dissolved. The solution was cooled to room temperature for 5 minutes and then cooled in an ice bath for 15 minutes. The solid product was filtered via suction filtration and washed with cold acetone. After filtration and drying the crystals were collected, weighed (0.189 g, 0.261 mmol) and the melting point range (49-51 °C) of the product was obtained.

            A sample from the product (0.59 g, 0.816 mmol) was added to a new round boiling flask with 6 M NaOH (2 mL, 50 mmol) and 95% ethanol (2 mL, 43.41 mmol). The solution was brought to a boil and refluxed for 45 minutes. During this time, the remaining product (0.130 g, 0.18 mmol) was dissolved in boiling acetone (1 mL), cooled to room temperature for 10 minutes, cooled in an ice bath for 10 minutes and recrystallized a second time. The weight of the second recrystallization (.061 g, 0.084 mmol) and the melting point range (51-53°C) was observed. After the 45- minute hydrolysis, the flask was cooled to room temperature and the solution was poured into a 50 mL beaker containing water (8 mL, 444.44 mmol). HCl (2 mL, 65.38) was added dropwise to the beaker. The beaker was then cooled on ice for ten minutes with stirring and then filter via suction filtration. The product was dried overnight then the weight (0.077 g, 14.7%) and melting point (52-53 °C) were measured.

Discussion:

Crude trimyristin was extracted from nutmeg (1.010 g, 2.29 mmol) and recrystallized using acetone. Following the first recrystallization, majority of the sample underwent hydrolysis and acidification to form myristic acid. The remainder of the sample was recrystallized a second time. After the original extraction from nutmeg the crude trimyristin was recovered (0.471 g, 46.63%). This low yield makes sense because nutmeg is not composed solely of trimyristin. This trimyristin (0.471, 0.65 mmol) was then recrystallization and recovered (0.189 g, 18.71%). A low yield after recrystallization is common because impurities are ridden from the starting sample. However this very low percent recovery could be due to the left over crystals that remained in the flask after transfer to the suction filtration. Some crystals were also not able to be scraped off the side of the funnel from the suction filtration. The melting point of the trimyristin after the first recrystallization was 49-51°C. The true melting point of trimyristin is 56-57°C so the low recorded melting point range indicates the sample was not yet pure. A sample of the trimyristin product (0.130 g, 0.18) was recrystallized a second time (0.061 g, 46.92%). This percent yield makes sense because the sample had more impurities needed to be filtered out of the product. The melting point of the trimyristin after the second recrystallization was 51-53 °C. This increased melting point range indicated there were less impurities left in the product then the first because compounds melt at lower temperatures when they contain contaminants. However, even after the second recrystallization the sample is still not considered pure as it is still significantly lower than the true range of 56-57 °C. The main product of the trimyristin (0.059 g, 0.816 mmol) underwent hydrolysis and acidification to form the target product of myristic acid (0.077 g, 14.7%) The low percent yield is due to the loss of product during hydrolysis. The solution was too close to the heat source causing it to overheat and it went up the distillation column sticking to the sides. A rinse with 95% ethanol was done but not all of the product was removed from the insides of the column. The melting point range of this target sample was 52-53 °C. This melting point range confirmed the product identity because the melting point of myristic acid is actually 54.5 °C. The target product was not 100% pure because the measure melting point range was a few degrees lower than the true melting point range. However the accuracy of the thermometer is within 2 °C which allows the measured result to be a reliable confirmation that the final product was the targeted myristic acid.

Boltwood 6 Reflection

Submitted by hamacdonald on Wed, 11/15/2017 - 16:02

This week at Boltwood we started out again with some crafts. We had everyone make turkeys from tracing their hands. Everyone seemed to enjoy the activity. Barbara came a little later to the craft so everyone else finished before her. Nicky finished very quickly so I helped her make a crown she seemed really excited about this. Next the group made crescent roles and per usual ended with a game, this week we played bingo. This week the girls were kind of upset because one of their in house nurses was leaving. We talked about this and also talked a little bit about death of family members because Rosie brought up missing her mom. We talked about this briefly and were able to change the topic back to positive things. Overall it was a good week! 

Boltwood 6 Reflection

Submitted by hamacdonald on Wed, 11/15/2017 - 16:02

This week at Boltwood we started out again with some crafts. We had everyone make turkeys from tracing their hands. Everyone seemed to enjoy the activity. Barbara came a little later to the craft so everyone else finished before her. Nicky finished very quickly so I helped her make a crown she seemed really excited about this. Next the group made crescent roles and per usual ended with a game, this week we played bingo. This week the girls were kind of upset because one of their in house nurses was leaving. We talked about this and also talked a little bit about death of family members because Rosie brought up missing her mom. We talked about this briefly and were able to change the topic back to positive things. Overall it was a good week! 

Key to understanding early embryo development: dorsal Amount and Location

Submitted by hamacdonald on Sun, 11/05/2017 - 23:48

Key to understanding early embryo development: dorsal Amount and Location

Siegfried Roth, David Stein, and Christiane Nüsslein-Volhard discovered the importance of genes and morphogens in the development of Drosophila melanogaster, fruit flies, body plan and body segments. They were able to identify what causes the fertilized eggs to specialize and give rise to differing positions and body parts. They discovered dorsal regulates ventral development in embryo structures. And the concentration of dorsal in the nuclei determines whether or not ventral or dorsal development pursues. The dorsal proteins were found in the nuclei of all the cells of both ventral and dorsal but based on the gradient high dorsal levels trigger ventral development and low dorsal levels correlate to dorsal development. This paper proves dorsal getting into the nuclei is critical for development this can be seen in Figure 4. It shows the dorsalized mutants and the protein is not present in the nuclei creating embryos with no ventral development. Then in Figure 5, the dorsal null is seen, in which Western Blots were performed on dorsal mutants and it showed dorsal protein was present in every one meaning the location of dorsal is critical as well as the amount present. The importance of their findings applies to more than just the model system of fruit flies. Analogues genes have been found in man and are crucial when looking at the early development of human embryos. These findings can be used to help understand congenital malformations which are the leading cause of infant death in the United States.

Outside Resource:

 

"Physiology or Medicine 1995 - Press Release". Nobelprize.org. Nobel Media AB 2014. Web. 5

Boltwood 5

Submitted by hamacdonald on Thu, 11/02/2017 - 15:31

This week at Boltwood it was Halloween. The girls were dressed up again and we all dressed up as well for this week. Everyone was in good spirits and enjoyed the craft we did. We cut out ghosts and everyone (aside Rosie who wanted one made for her) decorated their own ghosts. Even Judy did the craft which is nice because she usually doesn’t enjoy this kind of activity. We brought ice cream which was also a big hit with everyone. We talked about how later this month is Rosie’s birthday and she seems to be really looking forward to us being there on her birthday. The week before her birthday we are going to have her help us pick out somethings to do and a snack she would like. Judy also appreciated the card we got her and she said her surgery went well which was good to hear!

Boltwood 4

Submitted by hamacdonald on Thu, 11/02/2017 - 15:31

This week in Boltwood, three fourths of the group went dancing again and Judy stayed behind with us. Everyone was already dressed in there Halloween costumes for the dance, Nicky was a cheerleader, Barbara and Rosie were both witches! They all seemed very excited for the costume dance. We did a big group stretch and Nicky showed off some of her dance moves.

Once everyone left, it was a pretty relaxing week because Judy does not require much special attention like the other girls, she can follow conversation easily. This week came at a goodtime for Judy because she was pretty stressed for the following day. She confided in Kate about her fears for her surgery tomorrow and then the nurse filled us all in about what she was going to have to do tomorrow. The night before any surgery is very stressful for anyone so I was glad she got all of our attention and we tried to talk to her a little bit about it to ease her mind. We ended with the dice game and all wished Judy good luck tomorrow. I’m excited to see her next week and hope she is doing better. Since she is very religious, I was thinking sending her a prayer card for speedy recovery might cheer her up! 

Persuasion Activity

Submitted by hamacdonald on Thu, 11/02/2017 - 15:24

The vaccine should be given to German Shepard’s. This elite specie of dog is well known for its use in the police force, military, and assistance in guiding the blind. It is one of the most popular breeds in the United States for various reasons. They are very intelligent dogs, they top the list for the most intelligent dogs. They are loyal and heroic by nature due to their keen sense of smell and ability to detect weapons. They have been known to save owners and victims of trauma on many occasions. Fox news accounts the story from May 2015 in which a police dog saved its partners life post ambush attack. The German Shepard was able to chase away the assailants before they could murder his partner.

As a breed, German Shepard’s are very healthy and have very few major health issues. This breed is made to protect and love it’s owner. It makes for a very effective guard dog as it is loyal to its owners but wary of intruders. If someone were to break into your home a German Shepard would alert you of this attack and defend your children. Although they can be aggressive at times, the German Shepard gets along well with other pets and have an all-around agreeable personality. They’re personalities make them easy dogs to train, as seen by their choice for police dogs, they learn quickly. They are also very active dogs making them susceptible to learn new things and training for long periods of time without getting tired.

These dogs are very durable to varying weather conditions due to their thick double coat they can live in varying climates and withstand seasonable change.  They also have a relatively long life span for dogs of 14 years. German Shepard’s would be the natural choice to save some this new retrovirus.

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