Anatomy and Physiology 12 – Blog post

Today is October 5th and my Anatomy and Physiology 12 class learned about the process of DNA replication. Below are review questions from the process we learned.

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1. Explain the structure of DNA

DNA is known as one of the most complex molecules and its structure is very fascinating. The shape of DNA is a double helix, which looks like a twisted ladder. DNA is a polymer of nucleic acid while the monomer for this structure would be the nucleotides. Nucleotides are made up of three parts: a sugar (deoxyribose in DNA), a phosphate and a nitrogen containing base. The bonds connecting the phosphate and the nitrogen base to the sugar are covalent bonds. There are 4 different kinds of nitrogen containing bases and they are all complementary, meaning they will always pair up the same way.  There are 4 different base types – 2 purines and 2 pyrimidines. The 2 purine bases are Adenine (A) and Guanine (G). The two pyrimidines are Cytosine and Thymine (T). A way to remember which bases bond with which is remembering (A)pples go with (T)rees, and (C)ars go with (G)arages. This is complementary base pairing that is essential to know in the structure of DNA.

Another thing to note about the structure of DNA is that the backbones are anti-parallel which means they travel in different directions. This is indicated by the orientation of the sugar. The leading side goes from 3’ to 5’ and the lagging strand goes from 5’ to 3’. These numbers are indicators of which carbon on the sugar is holding onto which specific bond. This will show us the direction the carbon is oriented and which phosphate and sugar are apart of the same nucleotide. A way I like to remember which sides the 3’ and 5’ are, is by looking and the direction the oxygen in the sugar is pointing. The oxygen in the sugar will point to the 5’ end.

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2. When does DNA replication occur?

DNA replication will occur during the s-stage of the cell’s life cycle in preparation for mitosis. When a cell undergoes replication, it is essential that all genetic information is perfectly passed on to daughter cells. Because a parent cell must split into two, there must be a perfect replication of the DNA to pass on all the genetic information to all cells. This process of DNA replication occurs specifically during the S-Stage of interphase.

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3. 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 involved in DNA replication are:

• Unwinding and unzipping

The step of unwinding and unzipping is very straight forward. An enzyme is used to unwind the double helix shape and unzip the strands. This enzyme is called DNA helicase and it also breaks the H-bonds between the nitrogen base pairs.

• Complementary base pairing

Now that we have individual DNA strands, new nucleotides will be moved into place and bond with their complementary base. This process is done through DNA polymerase. The DNA polymerase can only read from the 3’ to the 5’ side and because of DNA’s anti-parallel shape, it must read the bases in two different directions. This causes two different DNA polymerases to travel in different directions. One traveling on the leading strand and another traveling on the lagging strand.

• Joining of adjacent nucleotides within each backbone

The joining process of nucleotides is when  new nucleotides start to form covalent bonds together. The leading strand is the one where the DNA polymerase can travel in one direction the entire time and follow behind the DNA helicase. This strand is easy and will all form bonds together after the DNA polymerase sequences it through. With the lagging strand, the DNA polymerase must travel backwards in a series of segments. Another enzyme called RNA primase will lay a small (primer) start for the polymerase to go backwards in individual segments. Each segment starting with a primer is called an Okazaki fragment. Then another DNA polymerase will replace the primers and another enzyme called DNA Ligase will connect all the fragments together. This complicated process is why Hank Green calls the lagging strand a “scumbag”. A visual way of this process is imagining the DNA polymerase must sequence backwards (in the opposite direction of the helicase) before it can jump up forward.

In the image above, the heart shape is the DNA ligase which connected all necessary nucleotides together to complete the two DNA strands.

Something else that I found very interesting was how we have nucleotides on command ready to be sequenced into a new strand of DNA. It originally did not make sense to be because of the law of conservation of mass. I did not understand how we just had an “infinite” number of nucleotides waiting. It now makes more sense because everything we eat contains genetic information and those will contain the 4 different nucleotides that our body needs. So basically, we absorb nucleotides from the food we eat.

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4. Today’s modeling activity was needed to show the steps involved in DNA replication. What did you do to model the complementary base pairing and joining of adjacent nucleotides steps?

To show the complementary base pairing we had to show the DNA polymerase going in both directions from 3’ to 5’ on each strand (each strand 3’ to 5’ is a different direction).  We then showed nucleotides being put into position by the DNA polymerase right behind it.

To show the joining of adjacent pairs we had DNA Ligase (heart) connecting the nucleotide segments on the lagging strand together. It is important to show the DNA ligase only operates on the laggings strand and not the leading strand as it is not needed.

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For today’s activity I was at a soccer game and was not present. I had my group members provide me with images from the day, but all my information is coming from the OneNote files provided by Ms. York. Below are my answers to the questions about transcription and mRNA.

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5. How is mRNA different than DNA?

Messenger RNA (mRNA) is a polymer of nucleic acids just like DNA but has three main differences. First off, it is single stranded, and one strand of mRNA is much shorter than DNA. A single strand of mRNA only accounts for one gene on a strand of DNA and there are many genes on a strand of DNA. This makes mRNA a fraction of the size of a DNA molecule (difference in size varies on the type of gene, but mRNA will always be much smaller).

The second main difference is the sugar backbone. DNA contains the sugar backbone of deoxyribose which gives DNA the name Deoxyribonucleic Acid. With mRNA, it contains the 5-Carbon sugar Ribose which gives RNA the name Ribonucleic acid. The only difference between these two sugars is that Ribose contains a hydroxyl group at 2’ where deoxyribose will only contain a hydrogen.

The last main difference between DNA and mRNA is that mRNA contains the pyrimidine base Uracil in place of Thymine. This means that on mRNA, Cytosine will bond to Uracil instead of Thymine. Uracil is used because it uses less energy to produce in mRNA than Thymine.

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6. Describe the process of transcription

The general idea of transcription is that one strand of DNA will be translated into one strand of mRNA. On our super long strands of DNA there are individual sections of code that code for a protein. These are our genes. A specific gene on a strand of DNA will become exposed by the unwinding and unzipping of this certain section of DNA where the gene is present. Along the sense strand, RNA bases will bond. This will create the code for the protein. Then adjacent nucleotides will form covalent bonds and build the RNA backbone of Ribose and a phosphate. This is all done with the help of RNA Polymerase. The mRNA is then released from the DNA strand and the DNA will reform its double helix structure and we will be left with a strand of mRNA.

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

I was not present this day, but from looking at the images, it did a good job of modeling because we had a different colour for the ribose sugar (red) then the deoxyribose (green). This made it clear which strand was apart of DNA and which was the RNA. We also had a different enzyme for RNA Polymerase which looked different than DNA Polymerase. One thing that maybe wasn’t so accurate is that we separated the entire DNA molecule on our whiteboard which could infer that the entire DNA molecule gets unwind and unzipped. This would not be true because only a specific gene on DNA gets unwind and unzipped to be transcribed and the rest of the DNA molecule is still in its double helix structure. Overall, the images my group provided me with made a lot of sense than they did a good job.

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Today is October 8 and in our activity, we had to model the process of translation. We learned how the code in mRNA turns into a polypeptide of amino acids. Below are the questions answered from this activity.

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

Once transcription has occurred, we have an mRNA molecule which is ready to leave the nucleus and into the cytoplasm where it will be read by ribosomes in a process called translation. It is important to know that mRNA contains a code in three letter words called codons. These three letter words are all associated with a specific one amino acid. One more interesting thing to note before I explain the process, is that multiple ribosomes can be acting on the same strand of mRNA at once creating a polysome.

Translation can occur in three main steps: initiation, elongation and termination.

• Initiation: the mRNA will bond to the small ribosome subunit and then the large ribosome subunit will come and bind the two subunits together. The entire ribosome looks like squidwards head, where the mouth is the small subunit, and the eyes are the large subunit. The codon AUG on the mRNA tells the ribosome to start the process. The first amino acid is Methionine.

• Elongation: Elongation is the process that allows the complimentary tRNA to attach to the binding sites on the ribosome. The tRNA will bring the amino acid to the ribosome where it will become a polypeptide. This can be done from the anticodon on the tRNA. The anticodon has the complementary 3 bases to the codon on the mRNA strand and will allow the tRNA to attach itself to the ribosome to deliver the protein. With each of the 64 possible codons on the mRNA strand, each will code for a specific amino acid. When the tRNA delivers the amino acid, it will first connect to the P site on the ribosome and then the next tRNA will connect to the A site. This bonding will then allow the amino acid in the P site to leave the tRNA and then the empty tRNA will leave the ribosome. Because the ribosome now has an empty P site, it will move along the strand of mRNA and move the tRNA that was previously in the A site into the P site. This will allow a space in the A site for another tRNA to deliver its amino acid and attach to the next codon on the mRNA. With this process, the chain of amino acids in the P site will move on top of the new one in the A site, making a chain of the “oldest” amino acid on the top and the “newest” on the bottom of the polypeptide strand.

  

• Termination: The process of elongation will continue until the ribosome reads a STOP codon. There is no matching tRNA for this stop codon which will indicate to the ribosome that the process is finished. No new amino acid is added, and the ribosome separates into its two subunits and the finished polypeptide of amino acids is complete. The STOP codon on the mRNA is either UAG or UAA.

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

I thought our model was very accurate with the use of tRNA and showing its anticodons attaching to the certain ribosome sites to deliver amino acids. I also liked how it was clear which amino acids were which and how they moved through the ribosome to create the polypeptide. On the contrary, our model can become confusing through the unfortunate sequence of amino acids we were given. In our pictures, lysine is the third amino acid that was added. We modeled the process up to three amino acids very well. We then jumped to the end where lysine was also the last amino acid to appear and so it may become deceiving that the final lysine on the bottom of the chain was the same one as the one that was third on the chain. This can be confusing, but our group understands how to model the sequence correctly and it was just an unfortunate sequence of amino acids which can appear to look like a process that it is not.

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This is the end of my blog reflection post, I hope you learned something new.

The majority of the images I used in this reflection came from the internet, but I would still like to demonstrate the models we did in class. I could not find a good place to integrate them in the questions so I will insert the images of our model below.

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Day 1: DNA replication

In this process of DNA replication, you can clearly see that we started with one strand of DNA and ended the process with 2 identical DNA strands.

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Day 2: Transcription

I was not present this day. Looking at the pictures it is clear that we started with 1 double strand of DNA and ended with 1 double strand of DNA and 1 single strand of mRNA.

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Day 3: Translation

In this process of translation it is clear that we start with a single strand of mRNA and end with a primary structure of amino acids.