23intetion of Lormhda.ne att siucated betwee deneatt sitesand gonsonete nvacteragedot utilizedootractose atola tdon sgal*ucIntegration toform prophageomSOwardcor coct some croon!ni imple ex23bio*gal*14. The figure provided portrays the integration of Lambda phage into a host chromosome at the att site, located between the gal+and bio+ genes. This prophage may disintegrate from the host carrying with it host genes, such as gal+ and/or bio+ and go on totransduce another host bacterium. How would one determine if a gal- host bacterium's phenotype was changed from gal- to gal+?To clarify, the minus version cannot utilize galactose as a carbon source for growth because it does not produce galactase, theenzyme that hydrolyzes galactose into monomeric sugars. There is a straight forward answer/solution to this – do not concoctsome crazy solution! Think simple experiment.

In order to determine the change of the host bacterium status from gal- to gal +, the galactose…

The figure provided portrays the integration of Lambda phage into a host chromosome at the att site, located between the gal+ and bio+ genes. This prophage may disintegrate from the host carrying with it host genes, such as gal+ and/or bio+ and go on to ransduce another host bacterium. How would one determine if a gal- host bacterium's phenotype was changed from gal- to gal+? To clarify, the minus version cannot utilize galactose as a carbon source for growth because it does not produce galactase, the

The figure provided portrays the integration of Lambda phage into a host chromosome at the att site, located between the gal+ and bio+ genes. This prophage may disintegrate from the host carrying with it host genes, such as gal+ and/or bio+ and go on to ransduce another host bacterium. How would one determine if a gal- host bacterium's phenotype was changed from gal- to gal+? To clarify, the minus version cannot utilize galactose as a carbon source for growth because it does not produce galactase, the

Biology: The Unity and Diversity of Life (MindTap Course List)

15th Edition

ISBN:9781337408332

Author:Cecie Starr, Ralph Taggart, Christine Evers, Lisa Starr

Publisher:Cecie Starr, Ralph Taggart, Christine Evers, Lisa Starr

Chapter20: Viruses, Bacteria, And Archaea

Section: Chapter Questions

Problem 1DAA: Bacteriophage-Inspired Antibiotics Although bacteriophages have been infecting bacteria for billions...

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Similar questions

As described in Figure, host DNA is hydrolyzed into smallpieces, which are occasionally assembled with phage proteins, creatinga phage with bacterial chromosomal DNA. If the breakage ofthe chromosomal DNA is not random (i.e., it is more likely tobreak at certain spots as opposed to other spots), how might nonrandombreakage affect cotransduction frequency?
Host DNA is digested into small pieces, which are occasionally assembled with phage proteins, creating a phage with bacterial chromosomal DNA. If the breakage of the chromosomal DNA is not random (i.e., it is more likely tobreak at certain spots as opposed to other spots), how might suchnonrandom breakage affect cotransduction frequency?
If you extract the DNA of the coliphage φ X174, you willfind that its composition is 25 percent A, 33 percent T,24 percent G, and 18 percent C. Does this compositionmake sense in regard to Chargaff’s rules? How wouldyou interpret this result? How might such a phage replicate its DNA
In the lysogenic cycle of bacteriophages, which of the fol-lowing is true?(a) Lysogeny is the term applied to a temperate phage thatdoes not replicate itself independently and does not lysethe bacterial host cell.(b) Phage DNA is incorporated into the host bacterium’sDNA, at which time it is called a prophage.(c) Temperate phages can be replicated either as a pro-phage along with bacterial chromosomal replicationor can suddenly revert to the lytic cycle by replicatingthemselves and assembling into new phages.(d) All the above are true.(e) a and c only.
Briefly describe the role of the Lambda repressor (cI) and cII activator to regulate the switch betweenthe lytic and lysogenic cycles of phage Lambda.
Before the integration of a transposon, its transposasemakes a staggered cut in the host target DNA. If thestaggered cut is at the sites of the arrows below, drawwhat the sequence of the host DNA will be after thetransposon has been inserted. Represent the transposonas a rectangle.TAATTTGGCCTAGTACTAATTGGTTGGTTAAACCGGATCATGATTAACCAACCc
Conjugation depends on direct cell-to-cell contactbetween a donor F+ carrying either a conjugative________ (the F plasmid) or an integrated conjugativeelement (as in Hfr strains), and a recipient lacking suchan element (F−).
If a single bacteriophage infects one E. coli cell present in a cultureof bacteria and, upon lysis, yields 200 viable viruses, howmany phages will exist in a single plaque if three more lytic cyclesoccur?
#4 BamI --- 5’ CCTAG ↓G 3’ 5’ ACGCCTAGGACGTATTATCCTAGGTAT CCGCCGCCGT CATCA 3’ 3’ TGCGGATCCTGCATAATAGGATCCATAGGCGGCGGCAGTAGT 5’ Restriction enzyme: Recognition sequence: Number of pieces of DNA: Type of cut:
Many viruses that infect eukaryotic cells express genes that alterthe regulation of host gene expression to promote viral replication.For example, herpes simplex virus-1 (HSV-1) expresses aprotein called ICP0, which is necessary for successful viral infectionand replication within the host. Lutz et al. (2017. Viruses9: 210) showed that ICP0 can act as a ubiquitin ligase and targetthe redundant transcriptional repressors ZEB1 and ZEB2,which leads to upregulation of the miR-183 cluster (a set of threemiRNAs transcribed from the same locus). Speculate on how miR-183 cluster upregulation may benefitthe virus.
Hi, can you tell me which gene(s) are being mutated in humans due to hepatic exposure to high levels of PCB toxin? Thanks!!
The following diagram describes the mRNA sequenceof part of the A gene and the beginning of the B geneof phage ϕX174. In this phage, some genes are read inoverlapping reading frames. For example, the code forthe A gene is used for part of the B gene, but the readingframe is displaced by one base. Shown here is the singlemRNA with the codons for proteins A and B indicated.aa# 5 6 7 8 9 10 11 12 13 14 15 16A AlaLysGluTrpAsnAsnSerLeuLysThrLysLeumRNA GCUAAAGAAUGGAACAACUCACUAAAAACCAAGCUGB MetGluGlnLeuThrLysAsnGlnAlaaa# 1 2 3 4 5 6 7 8 9Given the following amino acid (aa) changes, indicatethe base change that occurred in the mRNA and theconsequences for the other protein sequence.a. Asn at position 10 in protein A is changed to Tyr.b. Leu at position 12 in protein A is changed to Pro.c. Gln at position 8 in protein B is changed to Leu.d. The occurrence of overlapping reading frames isvery rare in nature. When it does occur, the extentof the overlap is…