Concept explainers
Two NotI restriction enzymes cleave DNA on oppositesides of the Dbmgene in a species of yeast. A molecularprobe for Dbmdetects a DNA restriction fragment of
In haploid yeast with this dbmmutation, what is thelength of the restriction fragment detected by the probefollowing NotI digestion?
What DNA-fragment sizes are detected in a diploidyeast strain that is heterozygous for wild-type andmutant alleles at dbm?
Insertion of
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Genetic Analysis: An Integrated Approach (2nd Edition)
- After Drosophila DNA has been treated with a restrictionenzyme, the fragments are inserted into plasmids and selected as clones in E. coli. With the use of this “shotgun”technique, every DNA sequence of Drosophila in a librarycan be recovered.a. How would you identify a clone that contains DNAencoding the protein actin, whose amino acid sequenceis known?b. How would you identify a clone encoding a specifictRNAarrow_forwardn yeast cells, telomerase remains active and maintains telomeres of about 300 base pairs. Propose what would happen to the telomeres over time in a yeast lineage in which the following mutations were created. The template portion of the telomerase RNA is changed from 5’-ACACCCACA to 5’-ACAUCUACA.arrow_forwardAlthough DNA transposons are abundant in the genomes of multicellular eukaryotes, class 1 elements usually make up the largest fraction of very large genomessuch as those from humans (~2500 Mb), maize (~2500Mb), and barley (~5000 Mb). Given what you knowabout class 1 and class 2 elements, what is it about theirdistinct mechanisms of transposition that would accountfor this consistent difference in abundance?arrow_forward
- The sequence at one end of one strand of the Drosophilatransposon Mariner is shown below (dots indicatesequences within the transposon):5′ TTAGTTTGGCAAATATCTCCCTTCCGCCTTTTTGATCTTATGT... 3′You obtain a mutant bacterial strain tagged with anengineered Mariner transposon, cut the genomicDNA from this strain with the restriction enzymeMboI (whose recognition site is ^GATC), and circularize the resultant DNA fragments by diluting therestriction enzyme digest and adding DNA ligase.a. Design two 17 bp PCR primers that you could useto identify (by inverse PCR) the gene into whichthe transposon inserted.b. What DNA sequence will be amplified from thecircularized fragments of the mutant genome?Show the extent of this DNA sequence on a mapof the genome of the mutant strain, indicating thelocations of the transposon insertion and any relevant sites for the enzyme MboI.arrow_forwardHomologous recombination in E. coli forms heteroduplex regions of DNA containing mismatched bases. Why are these mismatches not eliminated bythe mismatch repair system?arrow_forwardMost laboratory strains of E. coli contain site-specific DNA methylases. The methylase encodedby the dam gene (Dam methylase) transfers a methyl group from S-adenosylmethionine (SAM) tothe N6 position of the adenine residues in the sequence GATC. The Dcm methylase (encoded bythe dcm gene) methylates the internal cytosine residues in the sequences CCAGG and CCTGG atthe C5 position.The E. coli strain used in this experiment is mutated so Dcm and Dam methylases are notfunctional. What is the advantage of using dcm or dam mutants when the E. coli will be a host torecombinant DNA? HINT – If the E. coli methylates the DNA in these cells, then how would thisaffect the activity of restriction endonucleasesarrow_forward
- The restriction endonuclease NotI recognizes the octanucleotide sequence GCGGCCGC. Calculate the expected number of NotI cleavage sites in the bacteriophage λgenome, a linear DNA duplex 48.5 kbp in length with a (G + C) content of 50%.arrow_forwardA group of overlapping clones, designated A through F, is isolated from one region of a chromosome. Each of the clones is separately cleaved by a restriction enzyme, and the pieces are resolved by agarose gel lectrophoresis,with the results shown below. There are nine different restriction fragments in this chromosomal region, with a subset appearing in each clone. Using this information, deduce the order of the restriction fragments in the chromosome.arrow_forwardTelomeres contain a 3′ overhang region, as shown. Does telomerase require a 3′ overhang to replicate the telomere region? Explain.arrow_forward
- A plasmid that is both ampicillin and tetracyclineresistant is cleaved with PstI, which cleaves within theampicillin resistance gene. The cut plasmid is ligated withPstI-digested Drosophila DNA to prepare a genomic library,and the mixture is used to transform E. coli K12. Question: How can you explain the presence of colonies thatare resistant to both antibiotics?arrow_forwardYou would like to add a nuclear localization sequence (NLS) of Lys-Lys-Lys-Arg-Lys to a protein that is usually found in the cytoplasm of a yeast cell. To accomplish this, you introduce the nucleotide sequence encoding the NLS into the gene that encodes the cytoplasmic protein of interest. a. What is the size of the nucleotide insert that will encode the NLS? Briefly explain. 5' 3' b. Below is a diagram of the gene encoding the cytoplasmic protein of interest in the yeast genome. If your goal is to put the NLS at the carboxyl (C) terminus of the protein, at which location (A-E) should the NLS be inserted? Briefly explain. A TATAA ATATT promoter +1 B ATG TAC D TAA ATT stop codon E 3' 5'arrow_forwardBelow is a portion of an exon from a gene that encodes protein X in the genome of the plant Arabidopsis. Wildtype DNA3’ TTC AAT GCT CCG AAT ACC 5’ template strand5’ AAG TTA CGA GGC TTA TGG 3’ non-template strand A new strain (Strain B) of Arabidopsis is identified with the same region of the gene coding for protein X: 3’ TTC AAT GCT CCC AAT ACC 5’ template strand5’ AAG TTA CGA GGG TTA TGG 3’ non-template strand Compare the two DNA sequences and look for any differences. Based on what you find a. There is no mutation in Strain B compared to Strain A. b. After the point of the mutation, all the amino acids encoded by the Strain B template will be different than the Strain A protein X. c. Protein X made from the Strain B template will be much shorter than protein X made from the Strain A template d. Protein X from Strain B will have one amino acid difference that would not affect protein function. e. There is a mutation but there will not be any difference in the…arrow_forward
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