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Honors Biology Assignment

A short paper I wrote in the 9th grade.
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  Honors Biology  –   DNA By Matt Wedekind Honor’s Challenge Activity #1:  Essay Question Set #1: 1.) The four nitrogenous bases found in DNA include guanine (G), cytosine (C), adenine (A), and thymine (T). In DNA, a purine nitrogenous base (guanine and adenine) always binds with a pyrimidine nitrogenous base (cytosine and thymine). Therefore, the base pairs found in DNA are guanine-cytosine and thymine-adenine, or the reverse of those two. (Cytosine-guanine and adenine-thymine). 2.) If a double-stranded DNA molecule has 15% thymine, this means that adenine, its reverse, must also consist of 15% of the molecule. This is because thymine and adenine are always bound together in a base pair. Knowing that thymine and adenine consist of 30% of the DNA molecule, we can subtract 30% from the 100% total of the DNA molecule to figure that cytosine and guanine consist of the other 70% of the molecule. Because cytosine and guanine always pair together, half of 70% is 35%. Therefore, to answer the question, adenine consists of 15% of the DNA molecule, cytosine consists of 35%, and guanine consists of 35%. 3.) If there were no base pair rules, DNA’s ability to serve as a source of genetic information would be rendered completely inoperable. We can know this by possessing a basic understanding of the processes of transcription and translation. Starting with transcription, when the messenger RNA reads the unzipped gene in the DNA, the RNA polymerase has to call for the opposite nitrogenous base, with uracil replacing thymine. However, if there was no rule for base  pairs, this entire process would be impossible and there would be no “code” read, as the opposite  nitrogenous bases spell out the code of the respective gene in the DNA. There would be no way for transcription to occur, and because of this, there would be no way for translation to occur, and because of this, no proteins would be created. The process wholly relies on base pair rules. 4.) First, let’s look at Virus I. We can tell right away that its genetic material consists of RNA. This is because it contains uracil, but no thymine, and we know this to be a property of RNA. We can also tell that it is double-stranded, because there is an equal amount of uracils to adenines, and an equal amount of cytosines to guanines. This is because double-stranded molecules have base  pairs, and because of this, there can never be more of one nitrogenous base than its respective base pair.  Next, let’s look at Virus II. Its genetic material consists of DNA, not RNA. We can kn ow this because it contains thymine, but no uracil, and thymine is only found in DNA. Additionally, this virus’s genetic material is double -stranded, because there is an equal amount of uracils to adenines, and an equal amount of cytosines to guanines, and this means that the virus’s genetic material consists of a molecule with base pairs. What about Virus III? Its genetic material consists of DNA, like Virus II. Just like Virus II, we know this because the molecule contains thymine, but no uracil, and thymine is found only in DNA. However, Virus III is different than Virus II. Virus III’ s genetic material is single-stranded. This is because there is not an equal number of nitrogenous bases in each base pair. There are more adenines than thymines, and more cytosines than guanines. This would be impossible in a double-stranded molecule, as the nitrogenous bases come in pairs. Finally, Virus IV’s genetic material consists of RNA. This is because the molecule contains uracil, but no thymine, and uracil is only found in RNA. Virus IV’s genetic material is  single-stranded. This is because there is not an equal number of nitrogenous bases in each base  pair. There are more uracils than adenines, and more guanines than cytosines. This would be impossible in a double-stranded molecule, as the nitrogenous bases come in pairs. Essay Question #2: 1.) See attached loose-leaf paper. 2.) The direction of replication is shown on the paper. All replication moves from 5’ to 3’. This causes problems, as the helicase unzips i n one direction only, thus creating lagging and leading strands. The leading strands replicates with the helicase, the 3’ lying at the very end of the strand and the 5’ lying where the helicase begins to unzip. The DNA polymerase adds respective nucleotides as the helicase unzips, making it very simple. However, in the case of the lagging strand, it is much more difficult. The helicase begins to unzip at the 3’, and replication does not begin at 3’. A complicated process begins in which DNA polymerase creat es new DNA in “pieces” known as Okazaki fragments, named for one of their discoverers, Reiji Okazaki. These fragments are linked together with DNA ligase, and new, replicated DNA is created discontinuously. 3.) Semi-conservative replication is one of three postulates related to how DNA could  possibly replicate. It can be described as two new, identical DNA molecules being created, each containing one of the srcinal strands and one new strand, a replica of the opposite strand. This is how DNA replicates, as the helicase separates the DNA into two separate strands, and the DNA  polymerase adds new DNA to the single strands. 4.) DNA is anti-  parallel, with the 5’ and 3’ being on opposite sides of each othee (On one strand, the 5’ is on one end, while the opposite strand has 5’ on the opposite end, rather than the  must be copied in a unique way, as described before. This is because while helicase unzips the DNA molecule in one direction, the DNA is not replicated in the same direction because of its anti-parallel nature, thus leading to the creation of Okazaki fragments. Option #1: Chargaff’s data   1.)  Mycobacterium tuberculosis  has a higher concentration of guanine and cytosine than adenine and thymine. However, the eukaryotes (the mammals, fish, invertebrates, plants, fungi) all have a higher concentration of thymine and adenine in their base composition. 2.) Humans: 31 + 19.1 = 50.1, 31.5 + 18.4 = 49.9 50.1 / 49.9 = 1.004, approx. 1 Mycobacterium tuberculosis: 15.1 + 34.9 = 50, 14.6 + 35.4 = 50 50 / 50 = 1 3.) This claim is true, and I have proven it using Chargaff’s data. In question 2, I demonstrated that the ratio of purines over pyrimidines in humans and in mycobacterium tuberculosis were exactly or almost exactly 1. This means that there is always a C to go with a G, or a T to go with an A. It means that the DNA is double-stranded. Because this ratio is 1, the amount of adenine to thymine is equal, and the amount of guanine to cytosine is equal. 4.) In terms of structure, the DNA is double-stranded, and this means that there is always a base pair of nitrogenous bases. If there is a cytosine, a guanine accompanies it, and if there is an adenine, a thymine accompanies it, and vice-versa. 5.) This would be because the viruses are single-stranded. They are not prokaryotes or eukaryotes, as they are not living things, therefore their genetic material can be single-stranded.


Jul 30, 2017
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