The genome of a typical bacterium contains about 5 x 106 base pairs, and can be replicated in about 30 minutes. The human genome is 600 times larger (3 × 10° base pairs), and at the rate of a bacterium, it would require 300 hours (~12 days) to be replicated; yet the entire human genome can be replicated within several hours. How is this possible?
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- The chromosome of E. coli contains 4.6 million bp. How long will it take to replicate its DNA? Assuming that DNA polymerase III is the primary enzyme involved and that it can actively proofread during DNA synthesis, how many base pair mistakes will be made in one round of DNA replication in a bacterial population containing 1000 bacteria?In the very first round of PCR using genomic DNA, the DNA primers prime synthesis that terminates only when the cycle ends (or when a random end of DNA is encountered). Yet, by the end of 20 to 30 cycles - a typical amplification - the only visible product is defined precisely by the ends of the DNA primers. Explain how.To achieve a 10x depth coverage for a genome of length 10 Mbp, how many reads of 100 bp are required? a) 1 million b) 5,000 c) 100,000 d) 50,000
- Assume 2x108 reads of 75 bps long are obtained from a next-generation sequencing experiment to sequence a human genome. Suppose the length of the human genome is 3x109 bps. What is the depth (i.e., coverage) of the sequencing?How many Illumina clusters would you need to generate 15X coverage of the human genome, assuming your read lenght is 600 bp? (assume a human genome size of 3 billion bp) - 2.5 billion - 75 million - 7 million - 100 millionA 2.0kb bacterial plasmid ‘BS1030’ is digested with the restriction endonuclease Sau3A; the plasmid map is depicted in the diagram below and the Sau3A (S) restriction sites are indicated. Which of the following DNA fragments do you expect to see on an agarose gel when you run Sau3A-digested plasmid ‘BS1030’ DNA? a. 250 bp, 450 bp, 550 bp, 1.1 kb, 1.5 kb and 2.0 kb b. 2.0kb c. 250 bp, 400 bp, 450 bp, 500 bp and 550 bp d. 100 bp, 200 bp, 250 bp, 400 bp, 500 bp and 550 bp
- In a transformation experiment, donor DNA was obtained from aprototroph bacterial strain (a+b+c+), and the recipient was a tripleauxotroph (a-b-c-). What general conclusions can you draw aboutthe linkage relationships among the three genes from the followingtransformant classes that were recovered? a+ b- c- 180 a- b+ c- 150 a+ b+ c- 210 a- b- c+ 179 a+ b- c+ 2 a- b+ c+ 1 a+ b+ c+ 3Consider a genome whose length is 1000 bp. "Shotgun" sequencing techniques are applied to the genome, resulting in 20 reads, with an average length of 50 bp. A very important point is that, even though 20×50 = 1000, there is no guarantee that ALL 1000 bp of the genome are represented in the fragments. Calculate the coverage. What does this value mean? Why would it be a good idea to have a coverage greater than 1?The exponential nature of PCR allows spectacular increases in the abundanceof a DNA sequence being amplified. Consider a 10-kbp DNA sequence in agenome of 1010 base pairs. What fraction of the genome does this sequence represent? That is, what is the fractional abundance of this sequence in this genome?Calculate the fractional abundance of this target sequence after 10, 15, and 20 cycles of PCR, starting with DNA representing the whole genome and assuming that no other sequences in the genome undergo amplification in the process.
- During your experiment you analysed only a few of the recombinant clones for the presence of the highly repeated Aluelements. If you wanted to screen for a single-copy gene, you would need to screen a much larger genomic library. Assuming, that you already know the amino acid sequence of unicorn (a species with a similar physiology to humans) insulin, how would you construct a probe which would enable you to use nucleic acid hybridisation to screen a unicorn genomic DNA library for the insulin gene? Hint: you have access to any molecular biology reagents and equipment you might need, such as vectors, enzymes, and DNA sequencers.A linear piece of DNA that is 30 kb long is first cut with BamHI, then with HpaII, and, finally, with both BamHI and HpaII together. Fragments of the following sizes were obtained from this reaction: BamHI: 20-kb, 6-kb, and 4-kb fragments Hpall: 21-kb and 9-kb fragments BamHI and Hpall: 20-kb, 5-kb, 4-kb, and 1-kb fragments Draw a restriction map of the 30-kb piece of DNA, indicating the locations of the BamHI and HpaII restriction sitesFor the following short sequence of double stranded DNA and the given primers, there will be one major duplex DNA product after many cycles (imagine 10 cycles) of PCR. Provide the sequence of this one major duplex product and label the 5’ and 3’ ends of each strand. Sequence to be amplified: 5’- GGTATTGGCTACTTACTGGCATCG- 3’ 3’- CCATAACCGATGAATGACCGTAGC- 5’ Primers: 5’-TGGC-3’ and 5’-TGCC-3’