3. The speed of DNA replication at a replication fork is about 100 nucleotides per second (on each strand) in human cells. What is the minimum number of origins of replication that a human cell must have in order to replicate its DNA once every 20 hours? Recall that a human cell contains two copies of the human genome, one inherited from the mother and one from the father each consisting of 3 x 10° base pairs

3. The speed of DNA replication at a replication fork is about 100 nucleotides per second (on each strand) in human cells. What is the minimum number of origins of replication that a human cell must have in order to replicate its DNA once every 20 hours? Recall that a human cell contains two copies of the human genome, one inherited from the mother and one from the father each consisting of 3 x 10° base pairs

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Description: 3. The speed of DNA replication at a replication fork is about 100 nucleotides per second (on each strand) in human cells. What is the minimum number of origins of replicationthat a human cell must have in order to replicate its DNA once every 20 ho

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 8P: Human Genome Replication Rate Assume DNA replication proceeds at a rate of 100 base pairs per second...

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Replication 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?
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.
4. (2 points) Pseudomonas aeruginosa's circular chromosome is 7.5x106 bp long and the replication fork in Pseudomonas aeruginosa moves at a constant 1000 nt/s using bidirectional replication. Approximately how long will it take a P. aeruginosa to replicate its own genome?
1a. What do DNA polymerases need to be able to synthesize a new strand of DNA? In 1970, Fred Sanger and colleagues published a DNA sequencing procedure based on the principles of DNA replication. This procedure uses in vitro DNA synthesis in the presence of radioactive nucleotides and specific chain-terminators. These specific chain terminators lack a 3' hydroxyl group and are called 2', 3' – dideoxyribonucleoside triphosphates. This means that once a chain terminator is built into the newly synthesized strand, no further synthesis can occur on that particular strand. They are most usually labelled with radioactive 355 (isotope 35 of sulphur). Four reactions are assembled, each containing one each of 2', 3' - dideoxythymidine triphosphate (ddTTP), ddCTP, ddATP or ddGTP. In each reaction tube, all the fragments will end with same base (T, C, A or G). These reactions are each loaded in their own lane and separated by gel electrophoresis, exposed to autoradiograph film (X-ray film) and…
Which statements are true? Explain why or why not.1 The different cells in your body rarely havegenomes with the identical nucleotide sequence.2 In E. coli, where the replication fork travels at 500nucleotide pairs per second, the DNA ahead of the fork—in the absence of topoisomerase—would have to rotate atnearly 3000 revolutions per minute.3 In a replication bubble, the same parental DNAstrand serves as the template strand for leading-strandsynthesis in one replication fork and as the template forlagging-strand synthesis in the other fork.4 When bidirectional replication forks from adja-cent origins meet, a leading strand always runs into a lag-ging strand.5 DNA repair mechanisms all depend on the exis-tence of two copies of the genetic information, one in eachof the two homologous chromosomes
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5' 3' For numbers 6 to 10, refer to the image above and answer the questions. 6. Which among the two strands will have a continuous replication? 7. Which is the lagging strand? 8. Along which strand will Okazaki fragments appear? 9. How are Okazaki fragments joined together? 10. Where should the 3' end of the lagging strand be located? On the right or left side? in
4A. In the human gene for the beta chain of haemoglobin (the oxygen-carrying protein in the red blood cells), the first 30 nucleotides in the amino-acid-coding region is represented by the sequence: 3'-TACCACGTGGACTGAGGACTCCTCTTCAGA-5'. What is the sequence of the partner strand? 4B. If the DNA duplex for the beta chain of haemoglobin above were transcribed from left to right, deduce the base sequence of the RNA in this coding region.
5’ TAAGCGTAACCCGCTAA CGTATGCGAAC GGGTCCTATTAACGCAC 3’ 3’ ATTCGCATTGGGCGATT GCATACGCTTG CCCAGGATAATTGCGTG 5’ Imagine that the double-stranded DNA molecule shown above was broken at the sites indicated by spaces in the sequence and that before the breaks were repaired the DNA fragment between the breaks was reversed. What would be the base sequence of the repaired molecule? Show the sequence, label the 5’ and 3’ ends and briefly explain the reasoning supporting your answer
2. Dr. Kim at Research Center performed shotgun Sanger sequencing on an unknown DNA sample, and obtained the following reads (12 reads). Since the length of each read is quite short, Dr. Kim ran the original sample in a gel electrophoresis, and realized that the original DNA is just 50 base pairs long. (Please note that the resolution of gel electrophoresis is not so good. Thus, we cannot figure out the exact length of DNA using gel electrophoresis in the real world.) 2) AGCCG AGATG GCTG 4) CTGCA AGCCG A 1) САСТС ССAGT GTACC T 3) GGAGT CAATC GC 5) GGCTG TGCTT GG 7) GATGG CTGTG 9) CAGTG TACCT GCA 11) TGCAA GCCGA G 6) CTTGG AGTCA ATCGC 8) САСТС ССAG 10) GCTGT GСТTG G 12) TGCTT GGAGT (a) Find the sequence of the original DNA (reconstruction), and align each read with the reconstructed DNA sequence. (hint: put all reads on a ppt slide, and move them around to find overlaps.) (b) Calculate coverage at each nucleotide position of the reconstructed DNA, i.e., how many reads cover that…
(a) How fast does template DNA spin (expressed in revolutions per second) at an E. coli replication fork? (b) What is the velocity of movement (in micrometers per second) of DNA polymerase III holoenzyme relative tothe template?
The human genome contains about 3 billion base pairs. During the first cell division after fertilization of a human embryo, S phase is approximately three hours long. Assuming an average DNA polymerase rate of 50 nucleotides/second over the entire S phase, what is the minimum number oforigins of replication you would expect to find in the human genome? Show your solution.