Tracking time of marker entry generates a chromosome map (a) 100 azi ton lac" gal 10 20 30 40 50 60 Time (minutes) (b) F factor 10 min Hir st Origin Origin Fstr 17 min 25 min FIGURE 5-12 in this intemupted mating conjugation experiment, F streptomycin-resistant cels with mutations in azi, ton, lac, and gal are incubated for varying times with Ht cels that are sensitive to streptomycin and cay wild-type aleles for these genes. (a) A plot of the frequency of donor aleles in exconjugants as a function of time after mating. (b) A schematic view of the transfer of markers (shown in diferent colors) with the passage of time. a Data trom EL Wolman, Facob, and W. Hayes, Cokd Sping Hartor Symp. Quant. Bol. 21, 1956, 141) charaders among str' excorjugants

Tracking time of marker entry generates a chromosome map (a) 100 azi ton lac" gal 10 20 30 40 50 60 Time (minutes) (b) F factor 10 min Hir st Origin Origin Fstr 17 min 25 min FIGURE 5-12 in this intemupted mating conjugation experiment, F streptomycin-resistant cels with mutations in azi, ton, lac, and gal are incubated for varying times with Ht cels that are sensitive to streptomycin and cay wild-type aleles for these genes. (a) A plot of the frequency of donor aleles in exconjugants as a function of time after mating. (b) A schematic view of the transfer of markers (shown in diferent colors) with the passage of time. a Data trom EL Wolman, Facob, and W. Hayes, Cokd Sping Hartor Symp. Quant. Bol. 21, 1956, 141) charaders among str' excorjugants

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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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?
Analyzing Karyotypes 1. Originally, karyotypic analysis relied only on size and centromere placement to identify chromosomes. Because many chromosomes are similar in size and centromere placement, the identification of individual chromosomes was difficult, and chromosomes were placed into eight groups, identified by the letters A to G. Today, each human chromosome can be readily identified. a. What technical advances led to this improvement in chromosome identification? b. List two ways this improvement can be implemented. c. What clinical information does a karyotype provide?
Let’s suppose you have isolated chromatin from some bizarreeukaryote with a linker region that is usually 300–350 bp inlength. The nucleosome structure is the same as in other eukaryotes.If you digested this eukaryotic organism’s chromatin with ahigh concentration of DNase I, what would be your expectedresults?
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.
Equalizing the Expression of X Chromosome Genes in Males and Females Individuals with an XXY genotype are sterile males. If one X is inactivated early in embryogenesis, the genotype of the individual effectively becomes XY. Why will this individual not develop as a normal male?
Homologous Recombination, Heteroduplex DNA, and Mismatch Repair Homologous recombination in E. coli leads to the formation of regions of heteroduplex DNA. By definition, such regions contain mismatched bases. Why doesnâ€™t the mismatch repair system of E. coli eliminate these mismatches?
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.
Functional Consequences of Y-Family DNA Polymerase Structure The eukaryotic translesion DNA polymerases fall into the Y family of DNA polymerases. Structural studies reveal that their fingers and thumb domains are small and stubby (see Figure 28.10). In addition, Y-family polymerase active sites are more open and less constrained where base pairing leads to selection of a dNTP substrate for the polymerase reaction. Discuss the relevance of these structural differences. Would you expect Y-family polymerases to have 3-exonuclease activity? Explain your answer.
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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?
The completely synthetic yeast chromosome Syn IIIcontains a loxP site in the 3′ UTR of every gene thatis potentially nonessential to yeast survival. As youwill recall from Chapter 6, loxP sites are targets ofsite-specific recombination. The researchers who constructed Syn III included these loxP sites as a way to“scramble” the chromosome, meaning that parts ofthe chromosome could easily be deleted or rearranged.The goal of these investigations is to drive the evolution of Syn III so as to define a minimal genome thatcan support the life of this organism. Outline the experiment the researchers would do to scramble Syn IIIin order to define a minimal genome.
. Genomes A, B, and C all have basic chromosomenumbers (x) of nine. These genomes were derivedoriginally from plant species that had diverged from each other sufficiently far back in the evolutionarypast that the chromosomes from one genome can nolonger pair with the chromosomes from any other genome. For plants with the following kinds of euploidchromosome complements, (i) state the number ofchromosomes in the organism; (ii) provide terms thatdescribe the individual’s genetic makeup as accurately as possible; (iii) state whether or not it is likelythat this plant will be fertile, and if so, give the number of chromosomes (n) in the gametes.a. AABBCb. BBBBc. CCCd. BBCCe. ABCf. AABBCC