Experimental Design – Enzymes vs. Temperature

By: Keisha N, Yinling C

Purpose: To determine the effects of temperature on the reaction rate of the enzymes reactions

Hypothesis: The high temperature will cause a higher reaction rate causing more glucose to appear.

When heated above 45 degrees the protein will denature and there will be no glucose. The lower temperature will cause a slower reaction rate causing less glucose to appear.


  • Lactase enzyme drops
  • (lactose) Milk (100mL)
  • 1L of water
  • 5 Glucose test strips
  • 5 Test tubes
  • Test tube rack
  • 1 stir stick
  • Thermometer
  • Bowl of ice (ice bath)
  • Heat source (hot water bath)
  • Three beakers


  1. Fill one beaker with 500mL of water and place on the hotplate (medium – high heat)
  2. Fill the second beaker with 500mL of ice water (ice water bath)
  3. Fill each of the test tubes with 20mL of the lactose milk
  4. Add 2 drops of lactase enzyme drops to each of the test tubes
  5. Take one test tube and place it in the test tube rack to act as the control group, Make sure it is at room temp (23ºC).  Keep for 10 minutes.
  6. Take one test tube and place in the ice bath once it has dropped to 15ºC. Place in the bath for 10 minutes making sure that the temperature stays within close range.
  7. Take the third test tube and place it in the hot bath once it has reached 28ºC, leave the test tube in the hot bath for 10 minutes. Check temperature range regularly.
  8. Once 10 minutes has passed, take the test tubes out of the baths. Place a sample from each test tube on the test strips. Record colour. Repeat for beakers with body temperature and for 45ºC.
  9. Take the fourth test tube and place it in the hot bath, once it has reached body temperature (37ºC),  leave the test tube in the hot bath for 10 minutes.
  10. Take the last test tube and place it in the hot bath once it has reached 45ºC, leave the test tube and place in the bath for 10 minutes.

Data and Observations:

Beaker Temperature  Colour of the Test Strips Presence of Glucose (mmol/L) 
1 15ºC Green 14 mmol/L
2 Room Temperature (23ºC) brownOlive Green 56 mmol/L
3 28ºC  Light Brown 56 mmol/L
4 37ºC Light Dark Brown 111 or more mmol/L
45ºC Brown 111 or more mmol/L

 1. At what temperature was there the lowest count of glucose? – Why?

The lowest amount of glucose was present at 15ºC. This was because when you lower the temperature of the milk it causes the reaction rate of the enzymes to slow down, causing there to be less glucose present.

2. What would happen to the enzymes if it were placed into a test tube with a temperature higher than 45ºC?

If placed in a test tube above 45ºC there should be no glucose present in the milk. This is due to the fact that when an enzyme is heated to above 45ºC, the enzyme will start to denature, causing there to be no trace of glucose.

3. Why does 37ºC (body Temperature) produce the most optimal conditions for enzymes to work in?

As the temperature rises, reacting molecules have more kinetic energy; this increases the chances of successful collisions and thus, the reaction rate increases. At 37ºC, the enzyme’s catalytic activity is at its greatest and therefore making it the most optimal temperature for the human cell.


This lab allowed us to determine the effects of temperature on the reaction rate of the enzymes. We had hypothesized that there would be little to no glucose present when the temperature of the milk was lowered and that there would be high amounts of glucose when the temperature was raised, our hypothesis proved to be correct. We also predicted that no glucose would be present at 45ºC which was proven wrong. The incorrect hypothesis could be due to an error during the lab. At the lowest temperature (15ºC) there was little to no glucose present due to the low temperature and slow reaction rate. At he high temperatures we found that like we had hypothesized we found that there were high amounts of glucose present due to the high temp. and the increased reaction rate. When we had heated up the milk to 45ºC, the enzyme was supposed to denature however, due to error this was not the case. The most glucose was present at 37ºC and 45ºC. This lab gave us the ability to learn more about the relationship between temperature and how it effects enzymes.


To improve our lab design, we would have kept the amount of tests that we did however change the temperatures at which it happened. We tested too many temperatures that were in the same range and not much of a variety. We should have done at least another test at a very low temperature and one above 45ºC. It was also difficult for us to maintain a constant temperature causing there to be an error in our lab. The enzyme should have denatured once heated to 45ºC or higher.

Diffusion Lab

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

Hypothesis: The smaller the surface area and volume, the higher the rate of diffusion will occur.

1cm cube of Agar

2cm cube of Agar

3cm cube of Agar

Agar cubes after being in the base solution for 10 mins

Cut up cube of Agar

  1. In terms of maximizing diffusion, what was the most effective size cube that you tested?
    The most effective cube that we tested was the smallest cube that was 1cm cube of agar.
  2. Why was that size most effective at maximizing diffusion? What are the important factors that affect how materials diffuse into cells or tissues?
    Size was most effective at maximizing diffusion as its surface area is larger than its volume, which can be seen through the surface area to volume ratio, 6:1. The smaller surface area and volume, the easier it is for diffusion to occur. Some factors that affect diffusion are temperature, surface area, volume and the size/shape of the molecule itself.
  3. If a large surface area is helpful to cells, why do cells not grow to be very large?
    A large surface area is helpful to cells because it allows for more materials to enter the cells. A smaller surface area is also helpful as it allows for diffusion to occur at a faster rate, and move towards the center of the cell faster. When a cell grows its volume increases at a greater rate than its surface area, causing diffusion to happen at a slower rate; decreasing growth.
  4.  You have three cubes; A, B, and C. They have surface to volume ratios of 3:1, 5:2, and 4:1 respectively. Which of these cubes is going to be the most effective at maximizing diffusion? How do you know this?
    The most effective at maximizing diffusion would be Cube C as the surface areas to volume ratio (4:1) contains a surface area that is proportionately larger than the volume. There is more surface area to be diffused but, the smaller volume allows for diffusion to travel and reach the center of the cell quicker.
  5. How does your body adapt surface area-to-volume ratios to help exchange gases?
    Our lungs contains sacs known as Alveoli, they have a large surface area to volume ratio due to their folded shape, causing diffusion to happen more quickly. Each sac contains a thin wall which increases their surface area and makes it easier for them to deal with any inflation that may occur in the lungs. This also decrease the amount of volume that diffusion needs for it to occur.
  6. Why can’t certain cells, like bacteria, get to be the size of a small fish?
    Certain cells cannot get to the size of a small fish because it would not be practical for them. It would make it harder for cells to diffuse materials well. The cells are small to begin with as it allows for diffusion to occur at a much greater rate. Increasing the size of the cells would only slow down the rate of diffusion.
  7. What are the advantages of large organisms being multicellular?

    Some advantages of large organisms being multicellular is that they grow. Single celled organisms can only reproduce and make more copies of themselves as they are not capable of growing, another is that due to them being multicellular, they can have multiple functions making them more efficient.

Protein Synthesis

  1. How is mRNA different from than DNA?
    Both DNA and mRNA are polynucletides however, they have three distinct difference to one another.
    – Unlike DNA, which is double stranded, long and is shaped in a helix. Whereas mRNA differs as it is single stranded, short and is not in a shape of a double helix
    – DNA’s nucleotides contain contains the sugar deoxyribose but mRNA nucleotides contain the sugar; ribose.
    – While base pairing occurs in DNA, Adenine will always pair with Thymine. In mRNA, there is no Thymine but instead, Adenine pairs with Uracil in terms of all types of RNA including mRNA.

    DNA Backbone (Thymine is being represented by blue beads)

    RNA Backbone (Uracil is being represented by the brown beads)

    2. Describe the process of transcription
    Transcription is the process in which one gene’s DNA sequence is copied (transcribed) onto a strand of mRNA. It occurs in three phases: unwinding and unzipping of DNA, complimentary base pairing with DNA, and separation from DNA.
    All phases are in occurrence because of the enzyme RNA Polymerase.

    DNA being unzipped. A process that is facilitated by RNA Polymerase

    PHASE 1: Unwinding and Unzipping
    During the process of transcription, a portion of the DNA helix unwinds and unzips, exposing one gene.

    RNA Polymerase joining adjacent nucleotides

    mRNA Molecule

    PHASE 2: Complimentary Base Pairing
    Only one strand will be used as a template to produce an mRNA strand, this is know as the ‘sense strand’. Along the sense strand the complimentary RNA bases will bond together. Since RNA does not contain Thymine, therefore Adenine will pair with RNA’s equivalent base pair; Uracil.
    The sense strand partners with what is known as the nonsense strand, containing information that is not useful; not yielding a protein, if transcribed.
    RNA Polymerase will join the adjacent nucleotides to one another using H-Bonds. The nucleotides will bond, forming covalent bonds and build an mRNA backbone.

    mRNA is detached

    PHASE 3: Separation
    Until the entire gene has been transcribed the mRNA strand will detach from the DNA strand. The DNA molecule will rewind itself, reforming its double helix shape. This results in a strand of mRNA that gets modified before it moves out towards the nucleus.

    DNA reforms its double helix

    3. How did today’s activity do a good job of modelling the process of RNA transcription? In what ways was out model inaccurate?
    Some ways that this activity accurately modeled the process of RNA transcription. It accurately displayed how instead of pairing with Thymine, Adenine will pair with RNA’s Uracil by the use of a different coloured bead, as well as the difference in their backbones, modeled using a red backbone. The activity also gave a good representation of how the DNA and RNA bases paired with one another. The only inaccuracy in this activity I believe was that it did not demonstrate how the mRNA strand is much smaller to the DNA strand and it did not show what follows after transcription occurs and how the mRNA strand is processed which was shown as an important step.
    4. Describe the process of Translation: initiation, elongation, and termination.
    Translation is the process in which the code carried by mRNA transforms into a polypeptide. This process of translation occurs in three steps: Initiation, elongation and termination

    Ribosome reaches the start codon and is instructed to start translating and to code for an Amino Acid

    STEP 1: Initiation
    – In an mRNA, the instructions to build a polypeptide come in groups called codons (3 letter words located on mRNA). There are 3 “stop” codons that mark the polypeptide as finished; and one codon, AUG, is a “start” signal that starts of translation.
    – During this stage of translation, mRNA binds to the small ribosomal subunit in the area of the start codon (AUG). The first tRNA pairs with this codon. A large ribosomal subunit and small subunit will then join together. There are two sites in the ribosome in which bonding can occur, the “A-site” and “P-site”. The ribosome will then start to move along the mRNA until it reaches the start codon, AUG, in the “P-site”. This instructs it to start translating and code for an amino acid. A matching tRNA will bring in an amino acid that is represented by the mRNA codon.

    Amino Acids in both the “A” and “P” sites

    The tRNA in the “P-site” is released and leaves the ribosome;

    STEP 2: Elongation
    A ribosome is large enough where it is able to contain two tRNA molecules; the incoming and outgoing. The tRNA also contains three letter codes called anti-codons which are complimentary to mRNA codons. During this process, the ribosome will hold the mRNA and allow for the complimentary tRNA to attach to the binding sites. As the tRNA binds to the “P-site”, another will bond with the “A-site”. A peptide bond will form between the two amino acids. This bond causes the amino acids to let go of the tRNA in the “P-site” and start to bind to the neighbouring amino acids. The ’empty’ tRNA will then leave the ribosome and as a result, the ribosome will begin to move along the mRNA. This is done in order to make the tRNA is at the “P-site” and the new tRNA will start to bind with the mRNA codon at the “A-site”.

    tRNA in the “P-site” gets the stop codon and the chain stops growing

    STEP 3: Termination
    The cycle for elongation will continue until the mRNA reads a stop codon, terminating the process. This would result in a three letter word (codon) with no matching tRNA amino acid. This means that there are no new amino acids being added to the chain, causing the ribosome to separate into its two subunits and release the polypeptide.
    How did today’s activity do a good job of modelling the process of translation? In what ways was our model accurate?
    It was accurate that the model showed that it went through each of the stages carefully which allowed a better grasp on the concept. We got to see how the ribosome moves along the chain. It was also very beneficial how this activity showed how the codons and the anticodons paired with one another. However, there were some inaccuracies with the activity where we were not able to model the release factor that bonds to the stop codon and how the polypeptide is detached. Furthermore, the changing of the tRNA shape was not displayed not was the accurate shape of the amino acids displayed very well.

Blog Post 1: DNA, Replication


  1. Explain the structure of DNA – Use the terms nucleotides,  Anti-parallel strands and complimentary base paring.
    DNA (Deoxyribosenucleic acid) is a large polymer that consists 4 nucleotides, each of them made up of a phosphate group, a sugar group (deoxyribose), and also a nitrogenous base:
    PYRIMIDINE: Adenine and Guanine (Single ringed)
    PURINE: Thymine and Cytosine (Double ringed)

    DNA Backbone

    DNA Structure and Base Pairing

    The two backbones of DNA are formed through the bonding of the sugar-phosphate pieces of adjacent nucleotides. The complimentary bases faces inwards and are formed through Hydrogen bonds between the base pairs. The bases attaches together always with the same partner; complimentary base pairing (Adenine -> Thymine & Guanine -> Cytosine). The base pairs then form two long strands that spiral, creating a double helix shape. The two DNA strands are anti-parallel, they lie parallel to one another and but read in opposite directions.

    Double Helix

    forms a Double Helix

    2. How does this activity help model the structure of DNA? What changes could we make to improve the accuracy of this model? Be detailed and constructive. 
    This activity helped model the basic structure of DNA through the use of pipe-cleaners and beads to show its different parts. The pipe-cleaners were used to show the backbones as well for the Hydrogen bonds that were formed. The beads were used to represent the four complimentary bases and also the number of rings each base contained. We used two beads to show the double rings in the Purine bases and we used one bead to show the single ring in Pyrimidines. The different bead colours used for the bases helped make a clear visual on how complimentary base pairing worked.
    The model we made also demonstrated how the strands are anti-parallel to one another.
    Some changes that could be made to this modelling activity to make it more accurate would be to have more accurate measurements included in order to show the spacing between everything on the DNA strands. Also, the activity does not accurately demonstrate the number of hydrogen bonds that are formed between the base pairs.
    3. When does DNA Replication occur?
    DNA replication occurs prior to cell division. It is a  semi-conservative process as each new molecule must contain one backbone from the original DNA strand, allowing each molecule to have the same instructions to make new proteins.
    4. Name and describe the 3 steps involved in DNA replication. Why does the process
    There are 3 stages that occur in DNA replication: Unwinding, Complimentary Base Pairing, and Joining.

    DNA Helicase ‘unzips’ the DNA molecule

    The DNA’s double helix shape begins to unwind, the enzyme known as the helicase “breaks” the hydrogen bonds that are holding the complimentary base pairs together. The separation of two single strands of DNA creates a shape referred to as the replication fork.

    DNA Polymerase begins working on the strands

    Nucleotides presents in the nucleus move and form a Hydrogen bond with “partners” on the template strands of DNA. This process is facilitated by DNA polymerase. One strand is in the 3′ to 5′ direction (towards the replication fork), this is the leading strand. The other strand is in the 5’to 3′ direction (which is away from the replication fork),this is the lagging strand. The replication that occurs on the leading strand is continuous. A primer binds to the end of the leading strand ad acts as a starting point for DNA synthesis. DNA polymerase binds to the leading strand and ‘walks’ along it, adding new complimentary bases to the strand along the way in a 5′ to 3′ direction. Replication that occurs on the lagging strand is discontinuous. A number of RNA primers bind to the lagging strand at various points. Parts of DNA (Okazaki fragments), are then added to the lagging strand at various points. The reason the lagging strands replication is discontinuous is due to the fact that the Okazaki fragments needs to join up later.

    Covalent bonds form between nucleotides on the new strand. The leading strand will remain continuous as the DNA “unzips”. On the lagging strand, fragments will begin to form as the DNA “unzips”. The fragments will be glued together by DNA ligase. As a result, two daughter strands are formed.

    Daughter Strands

    5. The model today wasn’t a great fit for the process we were exploring. What did you do to model the complimentary 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 it inaccurate?
    The way we modeled the complementary base pairing, we paired some of the ‘free’ nucleotides to the complimentary base on the template strand. To model the joining of adjacent nucleotides, we attached the newly based pairs to one of the backbones. The activity provided a good visual on how the leading and lagging strands are read. It also showed how the bases were attache to the leading strand. Although, there were some inaccuracy in the usage of the candies as the candies were only placed in the areas where it was thought to be placed and function. We did not get to see how they interacted and worked. It was easy to comprehend how DNA Polymerase interacted with the leading strand but was harder to grasp and show for the lagging strand.

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