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Biochemistry
6th Edition
ISBN: 9781305577206
Author: Reginald H. Garrett, Charles M. Grisham
Publisher: Cengage Learning
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Slide 14
Please help me explain this on my report on cell dogma

Transcribed Image Text:SEMI-DISCONTINUOUS REPLICATION
New strand synthesis always in the 5'-3’ direction
a
Leading strand
Lagging strand
with Okazaki
OH
fragments
15
13'
05
S.3
OH
HOoo
3.
Sugar
Most recently
synthesized
DNA
15'
5h
3
3' HO
3%
13'
Base
5' triphosphate
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- Multiple Replication Forks in E. coli II On the basis of Figure 28.2, draw a simple diagram illustrating replication of the circular E. coli chromosome (a) at an early stage, (b) when one-third completed, (c) when two-thirds completed, and (d) when almost finished, assuming the initiation of replication at oriC has occurred only once. Then, draw a diagram showing the E. coli chromosome in problem 3 where the E. coli cell is dividing every 20 minutes.arrow_forwardReplication involves a period of time during which DNA is particularly susceptible to the introduction of mutations. If nucleotides can be incorporated into DNA at a rate of 20 nucleotides/second and the human genome contains 3 billion nucleotides, how long will replication take? How is this time reduced so that replication can take place in a few hours?arrow_forwardHeteroduplex 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.arrow_forward
- Which of the following components is not involved during the formation of the replication fork? single-strand binding proteins helicase origin of replication ligasearrow_forwardMultiple Replication Forks in E. coli I Assuming DNA replication proceeds at a rate of 750 base pairs per second, calculate how long it will take to replicate the entire E. coli genome. Under optimal conditions, E. coli cells divide every 20 minutes. What is the minimal number of replication forks per E. coli chromosome in order to sustain such a rate of cell division?arrow_forwardThe ends of the linear chromosomes are maintained by helicase primase DNA pol telomerasearrow_forward
- Human Genome Replication Rate Assume DNA replication proceeds at a rate of 100 base pairs per second in human cells and origins of replication occur every 300 kbp. Assume also that human DNA polymerases are highly processive and only two molecules of DNA polymerase arc needed per replication fork. How long would it take to replicate the entire diploid human genome? How many molecules of DNA polymerase does each cell need to carry out this task?arrow_forwardTranscribe and translate the following DNA sequence (nontemplate strand): 5'- ATGGCCGGTTATTAAGCA-3'arrow_forwardWhich of the following is not a true statement comparing prokaryotic and eukaryotic DNA replication? Both eukaryotic and prokaryotic DNA polymerases build off RNA primers made by primase Eukaryotic DNA replication requires multiple replication forks, while prokaryotic replication uses a single origin to rapidly replicate the entire genome DNA replication always occurs in the nucleus Eukaryotic DNA replication involves more polymerases than prokaryotic replication.arrow_forward
- Figure 9.10 You isolate a cell strain in which the joining together of Okazaki fragments is impaired and suspect that a mutation has occurred in an enzyme found at the replication fork. Which enzyme is most likely to be mutated?arrow_forwardHelicase 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.arrow_forwardReview 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.arrow_forward
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