Homework Week 1-katecoffman-dr.pineda

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Santa Ana College *

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CLS315

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Biology

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Feb 20, 2024

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Name ________ Kate Coffman _________________ Date_____ 02/06/2024 _________ CLS 315 Molecular Diagnostics – Dr. Pineda Week 1 Homework 15pts 1. Cellular Location and Activity of RNA Polymerase in Eukaryotes. List the types and subcellular location of each. (1.5pts) RNA Polymerase I Located in the nucleolus RNA Polymerase II Located in the nucleus RNA Polymerase III Located also in the nucleus 2. Write the complimentary sequence to the following: 3’GGCATGCAG5’ (1.5pts) A (adenine) pairs with T (thymine) T (thymine) pairs with A (adenine) C (cytosine) pairs with G (guanine) G (guanine) pairs with C (cytosine) Given: ‘3GGCATGCAG5’ Complimentary: ‘5GGCATGCAG3’ 3. A plasmid was digested with the enzyme NotI . On agarose gel electrophoresis you observe three bands 500bp, 1000bp, and 1500bp. (4pts) a. How many NotI sites are present in this plasmid? Observed bands on agarose gel electrophoresis: 500bp, 1000bp and 1500bp Present: 500 Solution: We take 1000bp – 500bp = 500 b. What are the distances between each site? The distance between the first and second bands is 500bp. Solution: We take 1000 – 500 = 500bp The distance between the second and third bands is also 500bp. Solution: We take 1500 – 1000 = 500bp Since the distances between the bands are all the same, so I think there is only one Notl site in the plasmid. c. What is the size of the plasmid? The size of the plasmid is 1000bp The Notl side is located 500bp away from each end of the plasmid.
d. Draw a picture of the plasmid with the NotI sites. 5’- - - - - - - - - - Notl - - - - - - - - - - 3’ ‘3 - - - - - - - - - - Notl - - - - - - - - - 5’ The distance between the Notl site and the end of the plasmid is 500bp on each side. 4. Calculate the nucleic acid concentration in μg/mL from the following information: (4pts) a. DNA: A reading at 260 nm = 0.5 Concentration of DNA = 0.5 x 1 x 50 = 25 ug / ml b. RNA: A reading at 260 nm = 0.25 Concentration of RNA = 0.25 X 1 X 40 = 10 ug / ml c. DNA: A reading at 260 nm from a 1:100 dilution = 0.12 Concentration of DNA (diluted) = 0.12 x 100 x 50 = 600 ug ml d. RNA: A reading at 260 nm from a 1:200 dilution = 0.33 Concentration of RNA (diluted) = 0.33 x 200 x 40 = 2640 ug / ml 5. Calculate the yield for (a-d in question 4), if the volume of the solutions was 0.25ml. (4pt) a. Yield of DNA ¿ 25 X 0.25 = 6.25 ug b. Yield of RNA ¿ 10 X 0.25 = 2.5 ug c. Yield of DNA (diluted) ¿ 600 x 0.25 = 150 ug d. Yield of RNA (diluted) ¿ 2640 x 0.25 = 660 ug
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