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.

 

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