Author: nighinar2016

Computer Programming – Core Competency Reflection

by nighinar2016

Code:

CP11 – Rahimi – Snowmen Code.py

Result:

Reflection:

This project was to present how well we understood loops and how we were able to use them to simplify our code. Before starting, I had to imagine a picture that would use a lot of repetition and would allow me to use ‘for’ loops, ‘while’ loops, and counters. I decided to make snowmen because it uses a lot of repetition of shapes. The problem I faced when creating this code was that I didn’t fully understand the loops, so I wasn’t able to optimize my use of them. When I finished, I had only one loop and I didn’t know where else to fit the missing loops. It wasn’t until I presented my problem to my teacher, that I learned I had written sequences that weren’t necessary. The purpose of the loop is to avoid writing out the code for a shape every time it is used, but that was exactly what I did. I refined my code by making a definition of each shape and learning how to use the loops to make them repeat in different parts of the picture. I had a lot of thought go into my snowflakes and arms because the use of angles in Python is very different. After a bunch of trials and errors, I learned how to reset the angle for the shape in order to create the arms and snowflakes that I wanted. In the end, developing the design for each shape was a very interesting process that taught me a lot about drawing using Python, especially when using loops or angles.

The Effect of Temperature on Enzymes

by nighinar2016

Purpose: To see how changes in temperature affect the enzymes rate of reaction

Material:

-7 Glucose test strips

-6 test tubes

-10 ml Milk per test tube

-Lactase Enzyme (5 drops per test tube)

-Hot Plate

-Oven Mittens

-Kettle

-Ice

-Tap water

-6 Beakers

-A thermometer

Hypothesis: By increasing and decreasing the temperature, we will reduce the enzyme lactases ability to break down the lactose sugar found in milk (due to denaturing of the enzyme)

Procedure:

  1. Fill 6 test tubes with 10 ml of milk.
  2. Dip one of the glucose test strips into regular milk to have a control group. Record the temperature of the milk.
  3. Fill a beaker with ice, put one of the test tubes inside the ice for 10 minutes. After 10 minutes, dip a glucose test strip and record the temperature of the milk.
  4. Run the tap water at the coldest for less than a minute, fill a beaker with cold tap water, put one of the test tubes inside the water for 10 minutes. After ten minutes, dip a glucose test strip. Record the temperature of the milk.
  5. Boil the milk in a beaker on a hot plate. Once boiled, dip the glucose test strip. Record the temperature of the milk.
  6. Boil water in a kettle and pour into a beaker. Put one of the test tubes in the boiling water for 10 minutes. Once finished, dip one of the glucose test strips and record the temperature of the milk.
  7. Create a beaker with water of room temperature (about 23 degrees Celsius) by pouring hot and/or cold water, leave one of the test tubes of milk inside for 10 minutes. After, dip the glucose test strip in and record the temperature of the milk.
  8. Put one of the test tubes in a beaker of 39 degree Celsius water (body temperature), leave in for 10 minutes or until reached body temperature. Insert a glucose test strip, record the temperature of the milk.

Data:

Observations and Analysis: 

  • The hottest and room temperature had the same dark brown colour, therefore more glucose concentration. After that, body temperature and in ice had the closest colour.
  • The enzyme can function similar to inside the body at a high temperature of 93 degrees Celcius and of a similar temperature to body temperature of 25 degrees Celcius.

Questions:

  1. What are the observable effects that the changes in temperature had for each test tube?

There were no obvious observable effects. Using the glucose test strip, we could see that the glucose concentration had clearly changed depending on temperature. In certain test tubes, the glucose concentration was higher. This could be because the temperature caused the enzymes to no longer break down the sugar lactose. The temperature that had the least break down would have been room and boiling temp. The rest had a higher break down of sugar.

2. How did this lab help in understanding enzymes?

This lab helped us understand that enzymes are very sensitive and have very specific temperatures that they can fulfill their duties. When body temperature dropped only about ten degrees Celcius, the break down of enzymes changed mmol/L. That being said, our bodies need to hold extremely specific temperatures in order for functions to process.

3. What was the difference between boiled milk and hot milk?

The boiled milk had a much higher concentration of glucose than that of the hot milk. This means the boiling milk must’ve had its enzymes denatured by the temperature, causing a higher concentration of sugars that haven’t been broken down by the enzyme lactase.

4. How does this lab show how enzymes work in the body?

If the body temperature test tube was accurate, having seen that it produced a glucose test strip of about 56 mmol/L, it would have a glucose concentration in between all the different variations of temperature. This means that the enzymes of the test tube were not denatured and could not not break any sugars down nor were they sped up and started breaking more sugars than usual. The enzymes in the body have a specific temperature that helps them keep a specific sugar level that was clearly different than all the other temperatures when you look at the graph.

5. How could this lab have been improved? 

This lab could have been improved with specific temperatures rather than conditions. Also, more variation of temperature closer to body temperature to see more how an initial higher and lower temperature can affect enzymes.

Also, the reactions didn’t model definitions. By definition, the colder temperature should render the speed of the rate of enzymes breaking down sugars and high heat should do the opposite. As we see in the picture of the glucose test strips, ice, room temp, and boiling temp are of different temperatures but have very similar effects on the enzymes. So, this experiment might have factors that affected its accuracy.

Conclusion: This lab was an okay representation of the break down of sugars in different temperatures. An interesting thing that we didn’t predict is that, seen in the graph and the glucose test strips, the highest temperature and room temperature had the same effect on the enzymes.

Diffusion in Agar Cubes

by nighinar2016

What determines the efficiency of diffusion throughout the model ‘cell’?

Hypothesis: The diffusion is affected by size and volume. The further to the cubes centre and the wider the surface means that the material has a wider and longer space to travel until it reaches full diffusion, therefore rendering the larger surface area and volume cube to be less efficient.

The Agar cubes before diffusion.
The Agar cubes after soaking in the 0.1M sodium hydroxide solution for about 10 minutes. The pink colouring is caused by the pH indicator mixed in with the Agar cubes.
These are the cubes cut in half. You can see how much of the base (pink) was able to penetrate the cube.

DATA TABLE:                                                                                                                                      

Cube Size Total cube volume (cm3) Total volume that was not pink

(cm3)

 Volume of the diffused cube

(total volume – volume that was not pink)

Percent

Diffusion

 Surface area of cube (cm2) Surface area to volume ratio
1cm  

1cm3

  

0.25cm3

  

0.75cm3

  

75%

  

6cm2

  

6:1
2cm  

8cm3

 

  

6.84cm3

  

1.16cm3

  

14.5%

  

24cm2

  

3:1
3cm  

18cm3

 

  

14.3cm3

  

3.75cm3

  

20.8%

  

42cm2

  

7:3

Question

  1. In terms of maximizing diffusion, what was the most effective size cube that you tested?
    • The most effective cube was the 1cm cube, in the photo it shows that the 1cm cube was more pink than the others, and had very littel agar left that was not pink. The hydroxide solution was able to extend through this cube and cover more of it’s area.
  2. Why was that size most effective at maximizing diffusion? What are the important factors that affect how materials diffuse into cells or tissues
    • The important factors that helpedmaximize diffusion was surface area and volume. The smaller cube had more hydroxide in it because it had less space that need hydroxide solution absorbed. The smaller cube has a smaller surface area to a lower volume, meaning that the diffusion reaches a wider surface and has less material to penetrate. This helps for it to be the highest diffused percentage.
  3. If a large surface area is helpful to cells, why do cells not grow to be very large?
    • Cells need to have materials reach the center. High surface area is helpful, but must also have a low volume. Cells don’t want to get too big to lose their low volume aspect. The large surface area allows more areas for materials to enter and the low volume helps them travel a short period until the centre. Therefore, a smaller cell comes with a larger surface area compared to volume which helps cell functions such as materials entering the cell.
  4. You have three cubes, A, B, and C. They have surface to volume ratios 6:1, 3:1, 7:3 respectively. Which of these cubes is going to be the most effective at maximizing diffusion, how do you know this?
    • Cube C (1cm cube, 6:1 ratio). It has a large surface area and a low volume, the surface area is much larger than its volume. This ratio helps gain the highest percent diffusion.
  5. How does your body adapt surface area-to-volume ratios to help exchange gases?
    • Our body adapts the ratio for Alevoli, an air-filled sac inside our lungs. They need a large surface area to volume ratio to allow gas exchnage to occur more rapidly in our body.
  6. Why can’t certain cells, like bacteria, get to be the size of small fish?
    • Bacteria are single-celled organisms. As seen in the lab, cells prefer to have a lower volume, so being the size of a fish would not help the functions of the cell. A cell needs to stay small for diffusion of materials.
  7. What are the advantages of large organisms being multicellular?
    • Our body has multiple cells with unique purposes. This helps human functions, so having organisms be created of one cell would render us useless. In addition, multi-cellular organisms have the advantage and the ability that they can grow, whereas a single-celled organisms like bacertia can only reproduce and multiple itself. Multi-ceullar organisms have the potential to do so much more than a single-celled organism because of the help of many cells.

The Physics of Brazilian Jiu-Jitsu

by nighinar2016

Brazilian Jiu-jitsu is torque in action. Brazilian Jiu-Jitsu is a martial art that uses leverage and force to apply joint locks and chokeholds to defeat the opponent. It has the concept of a smaller, weaker person being able to defend themselves against a much larger, heavier person. Seeing as a smaller person may not have as much strength against the larger person, they can use BJJ techniques such as joint locks to still win. Joint locks and Choke holds use torque to inflict pain. If a smaller person wishes to win, they can use the idea of torque and create levers using the human body. By apply force to these levers, a submission will be created. It is important for BJJ fighters to understand torque and use it to their advantage; without understanding levers and force, a heavier person could easily defeat a weaker person.

Why does Jiu-Jitsu follow this progression? 

  1. Get your opponent to the ground
  2. Get past your opponent’s legs
  3. Control their hips and shoulders
  4. Add a submission

This progression is important because it neutralizes the opponent’s ability to throw the fighter’s balance off and use a submission on the fighter. It is especially important if the fighter is smaller, this progression gives the smaller opponent advantage seeing as the control is given to the fighter and not the opponent. Once control is established, the fighter is able to create locks (fulcrums) and apply force to create a submission and win the match.

Why does Brazilian Jiu-Jitsu want control?

Control gives the fighter the ability to use the opponent’s centre of gravity and limbs against them. They have the ability to choose where the force is applied and how to create a submission.

Determine the Lever, Fulcrum, Force, and Torque of Joint Locks

Kimura  In this joint lock, the fight uses the opponents forearm to create a lever. The fulcrum is created at the elbow with the fighters arm under the opponents elbow so it does not move. The force is being applied by the fighters another hand. This submission shows the control of the fighter over the opponent and how he uses this torque to inflict pain and have the opponent tap out. The calculation of torque is F x d
Armbar In this joint lock, the fulcrum is created at the shoulder with the hip keeping the fulcrum intact. The whole arm becomes a lever, it is not allowed to bend because the arm is rotated in a certain direction. The force applied at the hand can break the elbow. This submission uses torque to have the opponent to tap out. The calculation of torque is F x d

Determine the Lever, Fulcrum, Force,  and Torque of a Choke

Bow and Arrow The fulcrum here is created at the hip, the leg of the fighter holding it in place to not move. There are two forces being applied, F1 causing a clockwise rotation and F2 causing a counter-clockwise rotation. In this submission, F1 causes force on the throat which therefore causes a choking submission. F2 is responsible on balancing F1 so F1 has enough impact to cause a submission. The calculation of torque is (F1 x r1) + (F2 x r2) if we assume that the forces applied are at 90 degree angles (perpendicular to the lever).

 

DNA and Protein Synthesis

by nighinar2016

DNA Model

We began with our 5′ backbone, its nucleotides, and its base pairing. The blue pipe cleaner represents the 5-carbon sugar backbone. The pink is the phosphates. Yellow is Adenine, blue is Thymine, purple is Guanine, and green is Cytosine.
We finished the hydrogen bonding, the complementary base pairings were in and the backbones were hydrogen bonded together.
The hydrogen bonds cause a slight twist which gives DNA the double helix form.

Explain the structure of DNA

DNA has two strands of the sugar-phosphate: deoxyribose. These strands are anti-parallel, the leading strand is 5′ and the complementary 3′. There are nucleotides that are hydrogen bonded to the sugar-phosphate strand. These bases are bonded to their complementary base, the only other nucleotide they can bond with, Adenine with Thymine and Cytosine with Guanine, they are paired together by hydrogen bonds. The hydrogen bonds have a slight charge, causing an attraction between other bonds, forming the twist in the double helix shape.

How does this activity help model the structure of DNA? What changes could we make to improve the accuracy of this model?

This activity helped model DNA by representing the sugar and its phosphate, and the hydrogen bonds with a blue pipe cleaner for the sugar, pink bead for the phosphate, and a white pipe cleaner for the hydrogen bond. This really represented where everything is, clearly showing the order of phosphate and hydrogen bond. It also had different colours for the bases, which really showed how certain nucleotides have specific bonding pairs. It also represented how it was a 1 pyrimidine to 2 purines.

The accuracy of this model could be helped with measurements. It was hard to know what the distance of the backbones were or how far apart the phosphates were. It would have helped to get a visual of a DNA enlarged slightly, this way you could see the bond between the sugar, phosphate, and the hydrogen bond and how they are spread out along the backbone.

 

DNA Replication

DNA helicase comes in and untwists and unzips the DNA.

DNA polymerase synthesizes the complementary base pairs. It works upwards on the 5′ strand (the strand that starts with the phosphate) and down on the 3′ strand (the strand anti-parallel to the 5′)
The DNA polymerase is attaching the compliment base as DNA lisage comes in and glues fragments together.
The replication process ends with two identical DNA strands.

When does DNA replication occur?

During cell division, DNA replication occurs in order to have DNA in every new cell. This allows every cell to have the instructions to create protein.

Name and describe the 3 steps involved in DNA replication. Why does the process occur differently on the “leading” and “lagging” strands?

The three steps are: unwinding and unzipping, complementary base pairing, and joining. The enzymes that help in this process are DNA Helicase, DNA Polymerase, and DNA Ligase. DNA Helicase unwinds and unzips the DNA, hydrogen bonds are unbonded and the nucleotide pairs are separated. DNA Polymerase comes in with the paired nucleotide for the lone nucleotides. Finally, DNA Ligase “glues” everything together and there are two identical DNA strands left. The nucleotides being put in place must follow the 3′ to 5′ pattern, and being anti-parallel, there is a “leading” and “lagging” strand. This means that the polymerase must follow two different ways to add the nucleotides. The leading is easy and the polymerase follows the helicase. The polymerase on the lagging strand starts from the top and must go back and forth as the new DNA is being formed. This results in a lot of pairing being attached later or being fixed by the ligase.

The model today wasn’t a great fit for the process we were exploring. What did you do to model the complementary base pairing and joining of adjacent nucleotides steps of DNA replication? In what ways was this activity well suited to showing this process? In what ways was in inaccurate? 

This process is well suited to give a visual representation of how the enzymes move and how the new DNA is truly replicated. The only problem was with the DNA polymerase, we couldn’t fully see it acting differently with the “leading” and “lagging” strands as we moved the candies out of the way and just attached the new strands ourselves. We couldn’t really see how the polymerase had to bring the nucleotides, or how it had to work almost harder on the lagging strand. Also, the ligase didn’t really show much of a difference. The enzyme passed and everything got attached, we couldn’t see it fix errors or bring the sugar-phosphate.

 

RNA Transcription

When mRNA is formed, the DNA separates.
The sense strand (5′) is transcripted in the process that the 5-carbon sugar backbone, ribose, is created with the help of the RNA polymerase that attaches the compliment base.
The mRNA detaches itself after the transcription.
In the end, the DNA reforms and the single strand RNA has the instructions and can leave the nucleus.

How is mRNA different from DNA? 

The differences are that RNA has a 5-carbon sugar called Ribose as its backbone, whereas DNA has a 5-carbon sugar called Deoxyribose. Also, in DNA Adenine pairs with Thymine. In RNA, Uracil is Adenines complimentary base pair. Finally, DNA is a double strand and very long. mRNA is shorter, it only reads certain sections, and is single-stranded, all of this allowing it to be able to leave the nucleus.

Describe the process of transcription. 

The DNA unwinds and unzips, the leading strand has nucleotides of RNA form with it. In this case, Uracil will pair with Adenine as it is from the RNA. These form hydrogen bonds. The RNA polymerase enzyme forms this bond and the backbone. Once this strand is created, they split off again resulting in the reattachment of a DNA strand and a single backbone mRNA.

How did today’s activity do a good job of modelling the process of RNA transcription? In what ways was our model inaccurate?

This activity represented the mRNA transcription well in that is shows how the DNA gets copied only from the leading strand and how mRNA is a single strand at the end of the process. What this activity forget to show is how mRNA is shorter than DNA. Also, what the DNA strand that is not being copied does. Lastly, this activity didn’t really explain how mRNA polymerase does the whole job.

 

RNA Transcription and Translation

In the process of Transcription, the RNA Polymerase transcribes the DNA message to the mRNA strand.
After the RNA Polymerase is finished, there is a DNA strand and a single strand mRNA (messanger RNA)  that have the DNA’s message and is ready to leave the nucleus.
This is where the process of Initiation begins. The ribosome holds onto the mRNA and begins by reading the start codon (three nucleotides that describe an amino acid). The first codon starts at the P-site.
It begins with a tRNA (transfer RNA) bringing in the correct amino acid to the codon and filling the P-site.
The codon in the A-site gets filled with the tRNA and its respective amino-acid.
The process of elongation begins as the amino acid from the P-site attaches itself to the amino acid at the A-site. The tRNA leaves the P-site.
Ribosomes prefer their P-site being filled, so it will shift down the mRNA to have the codon in the P-site.
The new codon in the A-site gets filled.
The amino acids from the P-site will bond with the new amino acid and the tRNA will leave the P-site.
The Ribosome moves over. This process continues until the ribosome reaches the STOP codon.
At the STOP codon, the ribosome will release the mRNA, the tRNA, and the polypeptide chain. This is termination. The final result is a small portion of a protein, a partial amino acid sequence (30 out of 1255 total).

Describe the process of translation: initiation, elongation, and termination.

The process of translation begins with initiation. A messenger RNA (mRNA) that has left the nucleus after transcribing DNA’s message gets attached to a ribosome. This ribosome has two subunits that bind together, P and A site, this is why the red cut out has two openings, the first being the P-site and the second being the A-site. The start codon (three letter sequence) is found and begins at the P-site. The codon gets read in the P-site with a transfer RNA that brings the anticodon and the amino acid (as seen in the photo, the green cut out holds the anticodon towards the codon and the amino acid on the other end). The A-site then gets filled with a tRNA with the anticodon and the amino acid. Elongation begins when the amino acid in the P-site attaches itself to the amino acid in the A-site and the P-site loses its tRNA. The ribosome shifts to fill its P-site with what was originally in the A-site. A new tRNA fills the A-site and the amino acid of the P-site attaches itself to the A-site amino acid. This process continues until the ribosome reads the stop codon, this is the termination stage. The stop codon does not have a matching amino acid so the chain of polypeptides is released. The last photo is a small portion of what the ribosome created.

How did today’s activity do a good job of modelling the process of translation? In what ways was our model inaccurate?

This activity accurately represented all the stages of translation. The model cut-outs helped give a visualization of how things moved. The only things that could’ve helped the understanding are if the ribosome was shown how it became P-site and A-site. Also, in translation, there are many ribosomes for one mRNA which was missing in this activity. They could’ve shown the dissociation of the ribosomal subunits in a complete sequence.