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BclI cleaves after the first T:
5’ T GATCA 3’
3’ ACTAG T 5’
Does cleavage by BclI result in a 5’ or 3’ overhang? What is the sequence of this overhang?
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- BamHI, cleaves after the first G: 5’ G GATCC 3’ 3’ CCTAG G 5’ Does cleavage by BamHI result in a 5’ or 3’ overhang? What is the sequence of this overhang?Given: BamHI, cleaves after the first G: 5’ G GATCC 3’ 3’ CCTAG G 5’ AND BclI cleaves after the first T: 5’ T GATCA 3’ 3’ ACTAG T 5’ THEN -- Given the DNA shown below: 5’ATTGAGGATCCGTAATGTGTCCTGATCACGCTCCACG3’ 3’TAACTCCTAGGCATTACACAGGACTAGTGCGAGGTGC5’ i) If this DNA was cut with BamHI, how many DNA fragments would you expect? ii) If the DNA shown above was cut with the enzyme BclI, how many DNA fragment would you expect?Show the expected labeled cleavage products if the following DNA segment were subjected to each of the four cleavage reactions: 32P-A-A-C-A-T-G-G-C-G-C-T-T-A-T-G-A-C-G-A
- Which statements are true? Explain why or why not.1 In terms of the way it interacts with DNA, thehelix–loop–helix motif is more closely related to the leu-cine zipper motif than it is to the helix–turn–helix motif.2 Once cells have differentiated to their final spe-cialized forms, they never again alter expression of theirgenes.3 CG islands are thought to have arisen during evo-lution because they were associated with portions of thegenome that remained unmethylated in the germ line.4 In most differentiated tissues, daughter cells retaina memory of gene expression patterns that were presentin the parent cell through mechanisms that do not involvechanges in the sequence of their genomic DNA.Assume the following DNA template strand: 3'-ATA GCG AGG AGT ATC-5' A) What would be the protein associated with this DNA template strand? Give the sequence of amino acids encoded by this fragment. Leave traces of your steps. B) In the synthesis of this protein, what are the codon and the anticodon for? Explain in one sentence for each. C) We find, in another cell, a mutation of this DNA template strand: 3' ATA GCG TGG AGT ATC-5’ 1. What type of point mutation is it? 2. Did this mutation arise during transcription, translation or DNA replication? D) If this mutation is found in a spermatozoon, will it have an effect on the individual, its offspring or both? Briefly explainFour cosmid clones, which we will call cosmids A, B, C, and D, werehybridized to each other in pairwise combinations. The insert size ofeach cosmid was also analyzed. The following results were obtained:
- A diploid human cell contains approximately 6.4 billion base pairs of DNA. Q.How many histone proteins are complexed with this DNA?Here is a DNA coding strand’s sequence and direction: 5’-ATGCCGATATAG-3’ . What would be the amino acid sequence in the polypeptide encoded by this DNA?IPSCs are nearly identical to human embryonic stem cells in terms of gene expression, but there may be other ways in which they are not equivalent. For example, the telomeres of IPSCs often vary in length, with many IPSCs cells having telomeres shorter than those of embryonic. How might shortened telomeres affect the life-span of IPSCs or of differentiated cells derived from them?
- 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.Multiple 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?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.