Template strand 5' 3' ТАСG GTAC ТA GTA GT A G тс G A T с G А А DNA Replication ТCАТ СА ТСА GCTA A GCTТ Direction 5' of synthesis DNA polymerase New strand FIGURE 7.6 The Proces of DNA Synthesis DNA polymerase synthesizes a new strand by adding T A one nucleotide at a time to the 3' end of the elongating strand. The base- pairing rules determine the specific nucleotides that are added. OH OH Considering that DNA is synthesized in the 5' to 3' direction, which direction P must DNA polymerase travel along the template strand: 5' to 3' or 3' to 5'?

Template strand 5' 3' ТАСG GTAC ТA GTA GT A G тс G A T с G А А DNA Replication ТCАТ СА ТСА GCTA A GCTТ Direction 5' of synthesis DNA polymerase New strand FIGURE 7.6 The Proces of DNA Synthesis DNA polymerase synthesizes a new strand by adding T A one nucleotide at a time to the 3' end of the elongating strand. The base- pairing rules determine the specific nucleotides that are added. OH OH Considering that DNA is synthesized in the 5' to 3' direction, which direction P must DNA polymerase travel along the template strand: 5' to 3' or 3' to 5'?

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 17P

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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.
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.
What does it mean to say thhat extension by DNA polymerase III proceed 5'---3'? A. The 5' end of a DNA polymerase molecule attaches to the 3' end of primase. B. DNA polymerase adds nucleotides to a growing strand, moving in the 5' to 3' direction. C. DNA polymerase seals nicks as it moves along a DNA strand toward the 3' end. D. DNA polymerase can only synthesize DNA at the 5' end of an existing strand of DNA.
Writing a Full Strand:1. A. Original DNA: CCT ATA TCT CTC TAT ATC TCT CAT ACT GTG TGT CTC TAT Complementary DNA:_______________________________________________________________________ B. Make identical strands of DNA CCT ATA TCT CTC TAT ATC TCT CAT ACT GTG TGT CTC TAT (original) ______________________________________________________________________ (new) _____________________________________________________________________ (new) ______________________________________________________________________ (complementary from 1A)2. A. Original DNA: CCG GAT TTT AAT TAG CTA CTA TCG TAC TAC GTT GGT GCT Complementary DNA: _______________________________________________________________________ B. Make identical strands of DNA CCG GAT TTT AAT TAG CTA CTA TCG TAC TAC GTT GGT GCT (original) _______________________________________________________________________ (new) ________________________________________________________________________ (new)…
DNA polymerase requires both a template, to be copied, and a primer, which provides a 3′ hydroxyl from which polymerase can extend. Yet, this molecule supports DNA polymerase activity. Explain. pTGACACAGGTTTAGCCCATCGATGGG -OH
a. Replicate this sense strand to create a double-stranded DNA helix. Write your answers in CAPS LOCK with NO SPACES between the nucleotides - e.g. ATGCCGAG..... TGAGGATGAAACTCACACCGGGGCGCAGTTTGGCACTTAGATTCTTGTACACGACCTAGTATAACACAGTT complementary strand: b. Using this DNA double helix, express the gene – i.e. determine the resulting polypeptide sequence by using the correct reading frame. Write your answers using the three letter abbreviation for each amino acid. polypeptide sequence: does the sense strand DNA sequence have 5’ and 3’ UTR sequences? 5'UTR = 3'UTR =
You are given a segment of DNA : 5’ - CATGTCAAC – 3’ What is the complimentary strand?
3’atgtaccatgcgcaaatttaaaggccc5’. a) Using this single template strand of a DNA as a template, write the base sequence of the complementary strand. b) List the molecules must be present for DNA to be replicated and briefly describe their function.
A. Create the complimentary strand for the DNA strand below. Make sure to label the parts and direction of the strand. 5’-AACGGTCCAGTCCAAGTTACG-3’ 2. Below is a segment of DNA that is ready to be replicated. Outline the processes that the segment will go through during replication. Make sure to include the names of the enzymes that are involved. AATTGCCTGCTAGTCTCAG TTAACGGACGATCAGAGTC B. DNA: G T A C G C G T A T A C C G A C A T T C RNA: C A U G C G CAU A U G G C U G U A G Codons: AUG - CGC - AUA -UGG - CUG - UAA Anti-codons: UAC - GCG -UAU - ACC - GAC - AUU Amino acids: Met- Arg - Ile - Try - Leu Using the example above transcribe the following DNA strands into m-RNA and translate that strand into a polypeptide chain identifying the codons, anti-codons and amino acid sequence. 3. DNA: A T A C G A A A T C G C G A T C G C G G C G A T T C G G 4. DNA: T T T A C G G C C A T C A G G C A A T…
Write down the double stranding sequence of the resulting DNA fragment(s) if the following DNA molecule were digested with XhoI: 5'-GGTATCCGCGGAGCTCAAATA-3' 3'-CCATAGGCGCCTCGAGTTTAT-5'
DNA strand 1 (template) TACATGCTCGTGACTTTT Mutation in strand 1 TACATGTCGTGACTTTT DNA Strand 5 (template) TACAGGTGTTCCCAGATCGGG Mutation in DNA strand 5 TACAGGTGTTCCCATATCGGG 1.use the original (template) strand of DNA to find the correct amino acid sequence in the DNA. 2.find and highlight the mutation in the DNA. 3.Find the new sequence of the amino acids for the protein in the new mutated strand of DNA. 4.Determine if this mutation is a point mutation of a Frame shift Mutation. 5.Determine if it is a silent mutations, Missense mutation, Nonsense mutation, Insertion or Deletion. 6.will this new protein function the same as the old protein? Why or why not?
DNA polymerase requires both a template, to be copied, and a primer, which provides a 3′ hydroxyl from which polymerase can extend. Yet this molecule supports DNA polymerase activity. Explain. pTGACACAGGTTTAGCCCATCGATGGG−OH Match the items in the left column to the appropriate blanks in the sentence on the right.