Rupinder Kaur - DNA Structure Worksheet.pdfOpen with -E DNA StrName:MissingOmubo Jack JanActivity #20: DNA Structure100 pointsder Kaur - D..Directions: You may need your textbook to identify the nucleotides1. Color all of the phosphates Purple2. Color all of thedeoxyriboses REDComplete through3.Color all of the Thyminesd or createOrangeO Class comme4.Color all of the AdeninesBlue5. Color all of the CytosinesBrown6. Color all of the GuaninesGreenTurn inAdd a class comm7.Label the phosphates,deoxyriboses, adenines,thymines, guanines,cytosines, covalent bondsbetween molecules (notelements), and hydrogennmentsto Omubo JackbondsPage2 I 2F Q +

Name: Rupinder Kaur - DNA Structure Worksheet.pdfOpen with -E DNA StrName:M
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Description: Rupinder Kaur - DNA Structure Worksheet.pdfOpen with -E DNA StrName:MissingOmubo Jack JanActivity #20: DNA Structure100 pointsder Kaur - D..Directions: You may need your textbook to identify the nucleotides1. Color all of the phosphates Purple2. Col

DNA is the double helical structure that is found in the living organisms like humans, plants,…

Science

Biology

Name: Activity #20: DNA Structure Directions: You may need your textbook to identify the nucleotides 1. Color all of the phosphates Purple 2. Color all of the deoxyriboses RED 3. Color all of the Thymines Orange 4. Color all of the Adenines Blue 5. Color all of the Cytosines Brown 6. Color all of the Guanines Green 7. Label the phosphates, deoxyriboses, adenines, thymines, guanines, cytosines, covalent bonds between molecules (not elements), and hydrogen bonds Page 21 2 F Q +

Name: Activity #20: DNA Structure Directions: You may need your textbook to identify the nucleotides 1. Color all of the phosphates Purple 2. Color all of the deoxyriboses RED 3. Color all of the Thymines Orange 4. Color all of the Adenines Blue 5. Color all of the Cytosines Brown 6. Color all of the Guanines Green 7. Label the phosphates, deoxyriboses, adenines, thymines, guanines, cytosines, covalent bonds between molecules (not elements), and hydrogen bonds Page 21 2 F Q +

Biochemistry

6th Edition

ISBN:9781305577206

Author:Reginald H. Garrett, Charles M. Grisham

Publisher:Reginald H. Garrett, Charles M. Grisham

Chapter28: Dna Metabolism: Replication, Recombination, And Repair

Section: Chapter Questions

Problem 11P

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Similar questions

Which of the following does the enzyme primase synthesize? DNA primer RNA primer Okazaki fragments phosphodiester linkage
During proofreading, which of the following enzymes reads the DNA? primase topoisomerase DNA pol helicase
Homologous Recombination, Heteroduplex DNA, and Mismatch Repair Homologous recombination in E. coli leads to the formation of regions of heteroduplex DNA. By definition, such regions contain mismatched bases. Why doesnâ€™t the mismatch repair system of E. coli eliminate these mismatches?
Functional Consequences of Y-Family DNA Polymerase Structure The eukaryotic translesion DNA polymerases fall into the Y family of DNA polymerases. Structural studies reveal that their fingers and thumb domains are small and stubby (see Figure 28.10). In addition, Y-family polymerase active sites are more open and less constrained where base pairing leads to selection of a dNTP substrate for the polymerase reaction. Discuss the relevance of these structural differences. Would you expect Y-family polymerases to have 3-exonuclease activity? Explain your answer.
Semiconservative or Conservative DNA Replication If 15N-Iabeled E. coli DNA has a density of 1.724 g/mL, 14N-labeled DNA has a density of 1.710 g/mL, and E. coli cells grown for many generations on 14NH4+as a nitrogen source are transferred to media containing 15NH4+as the sole N-source, (a) What will be the density of the DNA after one generation, assuming replication is semiconservative? (b) Suppose replication took place by a conservative mechanism in which the parental strands remained together and the two progeny strands were paired. Design an experiment that could distinguish between semiconservative and conservative modes of replication.
DNA Profiles as Tools for Identification A PCR-based paternity test is conducted using STRs that consistently produce a unique DNA fragment pattern from a single chromosome. Examining the results of the following Southern blot, which male(s) can be excluded as the father of the child? Which male(s) could be the father of the child?
Using Figures 8.7 and 8.9 as a guide, draw a dinucleotide composed of C and A. Next to this, draw the complementary dinucleotide in an antiparallel fashion. Connect the dinucleotides with the appropriate hydrogen bonds. FIGURE 8.9 The two polynucleotide chains in DNA run in opposite directions. The left strand runs 5 to 3, and the right strand runs 3 to 5. The base sequences in each strand are complementary. An A in one strand pairs with a T in the other strand, and a C in one strand is paired with a G in the opposite strand. FIGURE 8.7 Nucleotides can be joined together to form chains caled polynucleotides. Polynucleotides are polar molecules with a 5 end (at the phosphate group) and a 3 end (at the sugar group). An RNA polynucleotide is shown at the left, and a DNA polynucleotide is shown at the right.
Which of the following phases is characterized by preparation for DNA synthesis? G0 G1 G2 S
Heteroduplex DNA Formation in Recombination From the information in Figures 28.17 and 28.18, diagram the recombinational event leading to the formation of a heteroduplex DNA region within a bacteriophage chromosome.
Review Figures 8.12 and 8.13. In cells, the primers for DNA synthesis are short strands of RNA, so each newly-synthesized strand of DNA has a segment of RNA al its 5 end. As replication proceeds, DNA polymerases remove these RNA segments and fill in the resulting gaps with DNA. However, the gaps at the very 5 ends of the new strands cannot be filled in with DNA. Why not? DNA replication leaves exposed about 100 nucleotides al the 5 end of each template strand, and these single-stranded ends are removed. What are the effects of this "end problem" on a cell's DNA as it continues to divide? FIGURE 8.12 DNA replication. Green arrows show the direction of synthesis for each strand. The Y-shaped structure where the DNA molecule is being unwound is called a replication fork. FIGURE 8.13 Discontinuous synthesis of DNA. This close-up of a replication fork shows that only one of the two new DNA strands is assembled continuously. The other is assembled in short segments.
The two complementary strands of the DNA double helix are held to each other by (a) ionic bonds between deoxyribose molecules (b) ionic bonds between phosphate groups (c) covalent bonds between nucleotide bases (d) covalent bonds between deoxyribose molecules (e) hydrogen bonds between nucleotide bases
Helicase Unwinding of the E. coli Chromosome Hexameric helicases, such as DnaB, the MCM proteins, and papilloma virus El helicase (illustrated in Figures 16.22 to 16.25), unwind DNA by passing one strand of the DNA duplex through the central pore, using a mechanism based on ATP-dependent binding interactions with the bases of that strand. The genome of E. coli K12 consists of 4,686,137 nucleotides. Assuming that DnaB functions like papilloma virus El helicase, from the information given in Chapter 16 on ATP-coupled DNA unwinding, calculate how many molecules of ATP would be needed to completely unwind the E. coli K 12 chromosome.