Category: Biology 12

Purpose: To experiment and observe the changes in reaction rates when pH is manipulated with regards to enzymatic reactions

Hypothesis:   I  will be testing using glucose test strips , lactase enzyme ,test tubes ,milk , and pH standardized solutions ( pH 3,5,7,9,11).  I predict that lactase will function best at PH 6~7.4 given the fact that the PH of the small intestine is around 6-7 .

Materials:

6 Glucose test strips

Lactase enzyme

6 Test tubes

Test tube rack

Milk (pH 6.6)

Standardized solutions (pH 3, 5, 7, 9, 11)

Procedure

1. Label the test tubes with the corresponding pH.
2. Fill 1 test tube with 20ml of milk
3. Fill each of the remaining test tubes with 10ml of a different pH solution and 10ml of milk.
4. Add 2 drops of the lactase enzyme to each test tube.
5. Stir to each test tube to mix the solution with the lactase enzyme.
6. Check each solution with the glucose test strip to determine the numerical value of glucose detected.
7. Graph the data collected and draw the line of best fit.

Data:

pH	Milk (pH 6.6)	pH 3	pH 5	pH 7	pH 9	pH 11
g/dL	1/2	Negative	1/4	1/2	1/10	1/4
mmol/L	28	Negative	14	28	6	14

Observations:

The glucose strips changed colour according to the pH it was tested in. The glucose strips that were tested in the pH solutions closest to milk pH were the closest in colour. The glucose strips that were tested in the pH solutions that were the further away from milk pH ( more acidic pH) showed a negative result and no change occurred.

Analysis:

What our positive results are telling us is that lactase broke down the sugar lactose in  The milk. It shows that lactose digestion was most effective between pH 6.6 – pH 7. Our observations and results indicate that lactose digestion was not effective in a solution with a very low pH (pH 3) nor was it effective in higher pH either (pH 9 & pH 11). This tells us that the ideal pH for breaking down lactose is between 6.6 – 7.

Conclusion:

Our hypothesis proved to be correct in which pH lactase works best in.The milk in our control test tube and the third tube which contained pH 7 have similar results and proved lactase to be most effective in that range. Our results show that lactase is not effective In extreme pH whether it be very acidic or very basic. A happy medium was found to be at around pH 6.6~7.4 . The pH of the small intestine is around 7.3 giving rise to ideal pH conditions for lactase to break down the lactose found in milk and other diary products .

1- how is mRNA different from DNA ?

– DNA is made up of deoxyribose sugar while mRNA is made up of ribose sugar.DNA has thymine as its pyrimindine base while mRNA has uracil as its pyrimidine base. DNA is present in the nucleus while mRNA diffuses into the cytoplasm after synthesis. DNA has 2 strands that are anti-parallel that intertwine to form a double helix while mRNA is short and single stranded. mRNA is short lived , while DNA has a long lifespan.

 

2- describe the process of transcription 

the process of transcription begins with a specific section of the DNA unwinding which exposes a set of bases. Along the sense strand of DNA complementary RNA bases are brought in. In RNA , uracil binds to adenine . While in DNA cytosine binds to guanine . In the other strand of DNA ( missense strand) is not read read in eukaryotic cells.Adjacent RNA nucleotides form sugar – phosphate bonds and then the RNA strand is released from DNA . After that the DNA molecule rewinds and returns to its double helix form. This process occurs in the nucleus and is facilitated by RNA polymerase which is depicted in the form of a fuzzy peach in the photo below . The photo shows the RNA polymerase copying a particular segment of DNA into RNA .after this is  all done the red RNA strand separates from the DNA strands .

 

 

 RNA is shown as the red strand 

fuzzy peach – RNA Polymerase 

 

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

– The Pipe cleaner model is an excellent way to learn the steps of transcription because it is hands on and helps us better understand the process since we are building it ourselves . In addition to being hands on , the activity allows us to break down transcription into several steps and observe the process bit by bit and take our time to understand each step clearly. The different strand colours made it easier to distinguish between DNA and RNA and gave me a better understanding of the entire process rather than just reading the lesson of a board. The beads make the bases visually appealing and easier to distinguish , especially between single or double ringed bases.On the other hand , some discrepancies lie within this lab. For starters , the process of transcription is continuous meaning it happens without any stops or breaks. Another innaccuracy is with the size of the strands . Normally the DNA strands are much longer than the RNA but as you can see of our picture the red strand is almost as long as the DNA strand which in reality is false since RNA is much shorter.

 

 

 

Translation Activity 

1-  describe the process of translation- initiation , elongation , and termination. ( photos illustrate the process in order)

– initiation: The process starts with the mRNA reading the start codon (AUG) and attaches to the R site of the ribosome . The AUG codon always initiates translation and codes for the amino acid : methionine . tRNA binds to the start codon of mRNA . The tRNA has a binding site of 3 bases called anticodons that are complementary to the mRNA codon. The  methionly-tRNA is in the P site of the ribosome . The A site next to it is available to the tRNA carrying the next Amino Acid.

 

– elongation :  more amino acids are added and connected together to form a polypeptide. A Peptide bond is formed between the new amino acid and the growing polypeptide chain. Then , the amino acid is removed from the first tRNA  ( bonds break).

The first tRNA that was in the P site is released , and the tRNA in the A site is moved over to the P site.

The ribosome moves over one codon along the mRNA . This movement shifts the second tRNA to the P site . The third tRNA with the third amino acid can move into the A site and bind with the next codon on mRNA.

the process repeats and the chain elongates .

 

termination :the elongation process repeats until the stop codon is reached. ( UAA , UAG , UGA)

the stop codons do not code for amino acids but act as signals to stop translation

A protein called release factor binds directly to the stop codon in the A site. This causes a water molecule to be added to the end of the polypeptide chain , and the chain then separates from the last tRNA.

The protein is complete now. The mRNA breaks down and the ribosome splits into large and small subunits .

 

The pictures below illustrate the three steps of translation : initiation, elongation , and termination .

 

 

 

2- how did today’s activity do a good job of modelling the process of translation? In what ways was it inaccurate?

 

The activity did a good job of modelling translation by breaking it down into several steps and showing each small step for us to fully grasp the concept of the A and P sites , tRNA anticodons , and how translation begins. the different paper figures and colours made the process visually easier to grasp and understand. The fact the activity was hands on made it more meaningful and easier to understand. On the other hand , it was not possible to show the 3 D components of the process as this process is usually happening continuously and at a steady pace. In the end , I believe this activity gave me a better understanding of translation and definitely helped when I went back and read the steps of translation. Since we worked hands on , I better understood the process and re reading the lesson helped reinforce my understanding p.

 

 

 

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DNA Structure questions :

1. DNA  is made out of deoxyribose, phosphates and nitrogenous bases. DNA is a large polymer made out of nucleotide monomers and has 2 backbones.The structure of DNA consists of Two strands that intertwine to form a double helix and are anti-parallel.Meaning they are read in the opposite directions. If the first strand is read in the 3′ to 5′ direction that means the second strand will be read in the 5′ to 3′ direction. The Nucleotides ( A, C , G ,T)  join up and form hydrogen bonds between their complementary bases. Bonds with T , and C bonds with G , This is called complementary base pairing.

 

2. This activity helps model the DNA structure clearly because it gives us a better visual of the double helix structure. The fact that we are creating a 3D model allows us to better observe the complementary base pairing and better understand the concept behind the anti-parallel strands. The color-coded beads that symbolized nitrogen bases help us further our understanding of the purine/pyrimidine bases. Since the purine bases were always shown by 2 beads rather than 1. A possible change that can be implemented to improve this project would be to improve the modeling of the anti-parallel strands. instead of beads that may slip off or move out of place, we could use a white pipe cleaner and color the sides to represent a sugar or phosphate instead of using beads.

 

 

 

 

 

 

DNA REPLICATION:

 

1. DNA replication occurs before a cell divides to ensure all the cell’s genetic information is copied before it can proceed to divide.

 

2. The first step in DNA replication is the unwinding of the double helix. DNA Helicase is responsible for the unwinding of the strand that leaves the nitrogen bases exposed and breaks the hydrogen bonds between them. The second step is Complementary Base pairing,  in which the nucleotides move into place and form H – bonds with their partner nucleotides on the strand. This process is done by DNA polymerase. The third and final step is the joining of adjacent nucleotides by forming sugar-phosphate bonds between them. The leading strand is continuous as the DNA unzips. Meanwhile, the lagging strand forms fragments as the DNA continues to unzip. DNA ligase proceeds to glue any of the fragments that are present. The reason why the process is different for the lagging strand is because of the fragments that form because it is read much slower than the leading strand.

 

 

3. For the complementary base pairing, we used candy to model the DNA polymerase, similarly with the Joining of adjacent nucleotides we also used candy to model the DNA Ligase.