Classify each of the characteristics according to whether they describe Spo11, Cas9, or both enzymes. Functions in defense against bacteriophages Found in bacterial genomes Homologs found in yeast and many eukaryotic genomes Endonuclease activity Expressed during meiosis Functions in homologous recombination Guided to specific DNA sequence by RNA molecule Spo11 Cas9 Both Spo11 and Cas9 Reset

Classify each of the characteristics according to whether they describe Spo11, Cas9, or both enzymes. Functions in defense against bacteriophages Found in bacterial genomes Homologs found in yeast and many eukaryotic genomes Endonuclease activity Expressed during meiosis Functions in homologous recombination Guided to specific DNA sequence by RNA molecule Spo11 Cas9 Both Spo11 and Cas9 Reset

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

Chapter9: From Dna To Protein

Section: Chapter Questions

Problem 1SQ: A chromosome contains many different genes that are transcribed into different ___ . a. proteins b....

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Each species of bacteria has its own distinctive cell surface. Thecharacteristics of the cell surface play an important role in processessuch as conjugation and transduction. For example, certainstrains of E. coli have pili on their cell surface. These pili enableE. coli to conjugate with other E. coli and also enable certain bacteriophages (such as M13) to bind to the surface of the E. coli andgain entry into the cytoplasm. With these ideas in mind, explainwhich forms of genetic transfer (i.e., conjugation, transduction,and transformation) are more likely to occur between different speciesof bacteria. Discuss some of the potential consequences ofinterspecies genetic transfer.
The gene encoding the beta subunit of RNA polymerase fromEscherichia coli is said to be orthologous to the rpoB gene ofBacillus subtilis. What does that mean about the relationshipbetween the two genes? What protein do you suppose therpoB gene of B. subtilis encodes? The genes for the differentsigma factors of E. coli are paralogous. What does that sayabout the relationship among these genes?
2 3 inte tion of Lormhda. ne att si ucated betwee de ne att sites and go ns one te n vacter aged ot utilize doot ractose ato la t don s gal* uc Integration to form prophage om SOward cor coct some cro on! ni imple ex 2 3 bio* 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 to transduce 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 enzyme that hydrolyzes galactose into monomeric sugars. There is a straight forward answer/solution to this – do not concoct some crazy solution! Think simple experiment.
Please select all cellular processes that do NOT rely on RecA protein function as far as we considered them. bacteriophage lambda switch to lytic stage Lacl repression O LexA repression O natural transformation ODNA damage response O homologous recombination insertion of lambda phage into chromosome
pcc300ATAAADATATAOOTTAA 1. Use the genetic code table and the information in the diagram below to determine the amino acids that would make up the portion of the polypeptide shown. Include information for a key as well. DNA template 3' G CATA ACAGAGGATT-5' al bnsua AMAm pniwollot erfT E transcription s yd bnsita ebitgeqylog s sidmeaze of beae RNA strandUU UAOUOUU A-emoaodin 5'-CGUA AUUGUC UCCUUA- 3' J J JL erit o elinW (s) translation bluow terdt aspnso sigootiwsone polypeptide viemetis ns ebivo19 (d) ent ot etslanT Key:
Figure 1 is a bacterial gene (1-180). The first base to be transcribed is the base located at position 77. 45 5' TTGGT CTTGG TCGGA TTCCA GAGGA TGAAG TGTTG ACAGC GCATT 3' 3 AACCA GAACC AGCCT AAGGT CTCCT ACTTC ACAAC TGTCG CGTAA 5' 46 5 AATTG ACCTT GCTGT ATTAT AGCCA AGGAC AGATC TACGA GCATG 3' 3 TTAAC TGGAA CGACA TAATA TCGGT TCCTG TCTAG ATGCT CGTAC 5' 91 5 TGCGA ACCGC AAGCA TTCGT TCTCC TAGGC TACTC GATCC CGTAA 3' 3 ACGCT TGGCG TTCGT AACCA AGAGG ATCCG ATGAG CTAGG GCATT 5 77 90 110 135 136 5 TGATG TAGCT GATTC TGTTG AAAGG CTCCT TTTGG AGCCT TTTTT 3 3' ACTAC ATCGA CTAAG ACAAC TTTCC GAGGA AAACC TCGGA AAAAA 5 156 180 Figure 1. Illustrate how termination of transcription occurs in the gene above. (Hint: position from 156 to 180)
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
Actinomycin D inhibits DNA-dependent RNA synthesis. Thisantibiotic is added to a bacterial culture in which a specific proteinis being monitored. Compared to a control culture, into which noantibiotic is added, translation of the protein declines over a periodof 20 minutes, until no further protein is made. Explain theseresults.
About the technique of phage display: MOLECULAR BIOLOGY_advanced The Escherichia coli cell infected by the phage codifies for the optimized ligand when the phage DNA integrates in the DNA of the bacteria. Phages are selected if they express on their surface the optimized ligand. One selects Escherichia coli cells that are resistant to the phage infection. More than one optimized ligand can be selected during the panning procedure. The ligand to be selected on the surface of the phage is non-covalently linked to one of the surface proteins.
Understanding the life cycle of Plasmodium is important and we know that it is spread via mosquitos. The mosquito serves as O a mechanical vector in which the pathogen does not need the vector for replication O a biological vector in which the pathogen does not need the vector for replication O a mechanical vector in which the mosquito serves as a host for the multiplication of the pathogen during some stage of the pathogen's life cycle O a biological vector in which the mosquito serves as a host for the multiplication of the pathogen during some stage of the pathogen's life cycle Question 34 Patients are becoming inherently resistant to malaria resulting from O Sporogenic cells traveling a shorter distance O Climate changes O Genetic deficiency of glucose-6-phosphate dehydrogenase O Decrease in mosquito numbers 12 ....
Figure 1 is a bacterial gene (1-180). The first base to be transcribed is the base located at position 77. 45 5 TTGGT CTTGG TCGGA TTCCA GAGGA TGAAG TGTTG ACAGC GCATT 3' 3 AACCA GAACC AGCCT AAGGT CTCCT ACTTC ACAAC TGTCG CGTAA 5 46 77 90 5' AATTG ACCTT GCTGT ATTAT AGCCA AGGAC AGATC TACGA GCATG 3' 3 TTAAC TGGAA CGACA TAATA TCGGT TCCTG TCTAG ATGCT CGTAC 5' 91 110 5 TGCGA ACCGC AAGCA TTCGT TCTCC TAGGC TACTC GATCC CGTAA 3' 3 ACGCT TGGCG TTCGT AACCA AGAGG ATCCG ATGAG CTAGG GCATT 5' 135 136 5 TGATG TAGCT GATTC TGTTG AAAGG CTCCT TTTGG AGCCT TTTTT 3 3 ACTAC ATCGA CTAAG ACAAC TTTCC GAGGA AAACC TCGGA AAAAA 5 156 180 Figure 1. (i) Identify both the hexameric sequences of the promoter region in the coding strand above.
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