Analytical Assignment #9_ Chargaff's rule and table ONLINE_locked

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Dec 6, 2023

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Scientific Skills Exercise: Working with Data in a Table Report Form Complete this assignment page after you have read the background material for the Chapter 16 Analytical Assignment #9. You only need to complete this form if you want feedback from me. This form will NOT be graded. The grade for this assignment will be assigned after you complete the Analytical Assignment #9 Quiz. Instructions: For all of the following questions you may write your answers however you chose. However, if I cannot understand your response I will not be able to give you any feedback. Your responses should answer the question that has been asked and not more than has been asked. Be sure the responses are written clearly, using proper rules of English grammar, punctuation, and spelling. DO NOT assume I know the answer you are intending to submit; I can only respond to the answer you actually submit. To type your answers on the page, download this page onto your computer. Type your answer in the [type here] space. When you have completed the page, save a copy to your computer. Return the page to me by Canvas email. If you are not able to type in the space, print a copy of this page and write you answers in by hand. When you have completed the page, save a copy to your computer. Return the page to me by Canvas email. 1. From the reading: a. Who is the main investigator of these studies? Chargaff b. Chargaff and his fellow investigators were studying the composition of the desoxypentose nucleotides, what is a more modern name for desoxypentose nucleic acid? (Hint: write out the name, do not use an abbreviation) Deoxyribonucleic acid c. In the units for genome size, what do the abbreviations ‘M’ and ‘G’ mean? (include the value for each, in your definition) M - Mega G - Giga
2. Define the variables: a. Reorganize the list of organisms from the largest genome to the smallest genome: Explain how you determined your answer. Wheat Human Salmon=Ox Sea Urchin E.coli Type here b. For Wheat, E. coli , Human, and Ox, list an estimated quantity for the nitrogenous bases not listed in the table. Show how you determined the concentration. Wheat: Concentration of Thymine = Type here E. coli : Concentration of Cytosine = Type here Concentration of Thymine = Type here Human: Concentration of Guanine = Type here Concentration of Cytosine = Type here
Ox: Concentration of Guanine = Type here Concentration of Cytosine = Type here Concentration of Thymine = Type here
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Use the answers to Q2b to complete the data table Source of DNA Size of complete genome % Adenine % Guanine % Cytosine % Thymine Sea urchin 814 Mb* 32.8 17.7 17.3 32.1 Salmon 3 Gb 29.7 20.8 20.4 29.1 Wheat 16 Gb 28.1 21.8 22.7 27.4 E. coli 4.6 Mb 24.7 26.0 24.6 24.7 Human 3.2 Gb 30.4 19.75 19.75 30.1 Ox 3 Gb 29.0 21 21 29.0 Averages 29.12 21.18 21.19 28.50 3. Describe the data: a. For each organism determine whether the G/C concentration is less than or greater than the A/T concentration. Source of DNA Greater than or Less than Sea urchin G/C < A/T Salmon G/C < A/T Wheat G/C > A/T E. coli G/C > A/T Human G/C < A/T Ox G/C < A/T
b. The data table in the introduction gives total size of each genome and the percentage of some of the nucleotides in the genome of each organism. From this data the total number (i.e. NOT percentage) of each nucleotide can be calculated. Below is an illustration of how to determine the number of each nucleotide in the Sea urchin. Data Given: Source of DNA Size of complete genome % Adenine % Guanine % Cytosine % Thymine Sea urchin 814 Mb* 32.8 17.7 17.3 32.1 Since Adenine if 32.8% of the total sea urchin genome the total number of Adenine nucleotides is 814 Mb x 0.328 = 266.992 Mb Adenine Since Guanine if 17.7% of the total sea urchin genome the total number of Adenine nucleotides is 814 Mb x 0.177 = 144.078 Mb Guanine Since Cytosine if 17.3% of the total sea urchin genome the total number of Adenine nucleotides is 814 Mb x 0.173 = 140.822 Mb Cytosine Since Thymine if 32.1% of the total sea urchin genome the total number of Adenine nucleotides is 814 Mb x 0.321 = 261.294 Mb Thymine Using this example calculations complete the table Total Number of each nucleotide Source of DNA Size of complete genome Adenine Guanine Cytosine Thymine Sea urchin 814 Mb* 266.992 Mb 144.078 Mb 140.822 Mb 261.294 Mb Salmon 3 Gb 0.891 Mb 0.624 Mb 0.612 0.873 Wheat 16 Gb 4.496 Mb 3.488 Mb 3.632 4.384 E. coli 4.6 Mb 1.1362 Mb 1.196 Mb 1.1316 1.1362 Human 3.2 Gb 0.9728Mb 0.632 Mb 0.632 0.9632 Ox 3 Gb 0.870 Mb 0.63Mb 0.63 0.87
4. Using data to make conclusions: a. In your own words write out the two Chargaff rules discussed in the introduction. Chargaff Rule #1 (within species rule) A equals T and G equals C Chargaff Rule #2 (between species rule) The percentage of A in one species does not determine the percentage of A in another species. b. Comparing the data listed for the sea urchin and salmon species, explain how the data demonstrate both of Chargaff’s rules. T 5. Summary analysis: If Chargaff’s rule #1 is valid, that within a species the amount of A equals the amount of T and the amount of C equals the amount of G, then a hypothesis has been proposed that we could extrapolate the rule to the combined DNA of all species on Earth (like one huge Earth genome). To see whether the data in the table supports this hypothesis, use the calculated average percentages from the table (bottom row) and answer the question “Does Chargaff’s equivalence rule hold true for one Earth genome?” Explain Type here
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