Biology
5th Edition
ISBN: 9781260487947
Author: BROOKER
Publisher: MCG
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Chapter 19.4, Problem 2CC
Genetic Properties of Bacteria
Concept Check: Describe the similarities and differences between a bacterial chromosome and a plasmid.
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Chapter 19 Solutions
Biology
Ch. 19.1 - Prob. 1CSCh. 19.2 - Prob. 1CSCh. 19.2 - Viral Reproductive Cycles Concept Check: From the...Ch. 19.4 - Genetic Properties of Bacteria Core Skill:...Ch. 19.4 - Prob. 1CCCh. 19.4 - Genetic Properties of Bacteria Concept Check:...Ch. 19.4 - Prob. 3CCCh. 19.5 - Prob. 1CCCh. 19.5 - Prob. 1EQCh. 19.5 - Prob. 2EQ
Ch. 19.5 - Gene Transfer Between Bacteria CoreSKILL The gene...Ch. 19.5 - Prob. 2CCCh. 19.5 - Gene Transfer Between Bacteria Core Skill:...Ch. 19.5 - Gene Transfer Between Bacteria Concept Check: Is...Ch. 19 - Prob. 1TYCh. 19 - The characteristics of viral genomes show many...Ch. 19 - During viral infection, attachment is usually...Ch. 19 - Prob. 4TYCh. 19 - Prob. 5TYCh. 19 - Prob. 6TYCh. 19 - Prob. 7TYCh. 19 - Prob. 8TYCh. 19 - Prob. 9TYCh. 19 - Prob. 10TYCh. 19 - How are viruses similar to living cells, and how...Ch. 19 - Prob. 2CQCh. 19 - Prob. 3CQCh. 19 - Prob. 1COQCh. 19 - Prob. 2COQ
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- Conjugation: Diagram the process of conjugation in bacterial cells (using F plasmid transfer):a. In you diagram, label the parts of both the donor and recipient cell, draw an arrow, then show the resulting cells and their resulting phenotypes. Label any other special features needed for this process.arrow_forwardClassifiation of Llfe A new organism was discovered in a deep sea vent at the bottom of the ocean. The researchers collected the sample and made careful observations of it. The picture at right shows a microscopic image captured at 1000X its actual size. Below are some other observations made: • When DNA was isolated from the cell, it was mixed with an equal portion of histone proteins • There are no visible organelles present inside the cell • The specimen seem to die in the presence of oxygen • Able to live in conditions that mimic deep sea vents without any external food source provided Based on these observations, determine each of the following and provide a clear rationale. A) To which domain would you assign this organism? Why? B) Propose a hypothesis of how they obtain nutrients. Are these most likely autotrophs or heterotrophs? Provide a rationale.arrow_forwardPlease ASAP. Thankyouarrow_forward
- Please ASAP. Thankuarrow_forwarddraw structure of plasmidarrow_forwardGenome for C. diphtheriae have about 2,500,000 nucleotides, 87% of them are coding. This ingle circular chromosome contains 2,389 genes from which 2,272 proteins are coded. It does not contain any plasmids. The genome contains Pathogenicity Islands (PAIs), which C. diphtheriae has 13. What is a PAI and what are their characteristics?arrow_forward
- bioarrow_forwardart B-Processes occurring at a bacterial replication fork The diagram below shows a bacterial replication fork and its principal proteins. Drag the labels to their appropriate locations in the diagram to describe the name or function of each structure. Use pink labels for the pink targets and blue labels for the blue targ Replaces RNA primers with DNA nucleotides. Catalyzes phosphodiester bond formation, joining DNA fragments. Leading strand Breaks hydrogen bonds, unwinding DNA double helix. Synthesizes DNA 5' to 3 on leading and lagging strands. Lagging strand Coats single-stranded DNA, preventing duplex formation. Relaxes supercoiled DINA. Synthesizes RNA primers on leading and lagging strands. e Overall direction of synthesis reset ? helparrow_forwardAGUC 24. Use the table below to answer the following question. The template strand of a gene contains this sequence: 3'-TAC TAG GCT AGT TGA-5'. A mutation occurs due to exposure to a chemical mutagen that changes the gene sequence to 3'-TAC TAG ACT AGT TGA-5'. How does this mutation affect the resulting amino acid sequence? (LS3-2) * Second mRNA base A UCU UGU Cys UGC UUU UAU Tyr UAC Phe UUC U UCC Ser UCA Phe Gly (G) Leu Glu (F) (L) UUA UAA Stop UGA Stop A Ser Asp (E) (D) Leu (S) UCG UUG UAG Stop UGG Trp Tyr (Y) Ala CUU CCU CAU CGU U (A) His C G U A CỤC Leu CUA CC Pro CCA CAC CGC Arg CGA Cys (C) G A Val CAA (V) Gln G Trp (W) CUG CCG CAG CGG G Arg (R) G А С A Leu AUU ACU AAU AGU (L) Asn Ser Ser (S) AUC Ile ACC Thr ACA AAC AGC C/ Lys (K) UGA AUA Pro AAA Lys AAG AGA Arg AGG Asn (N) (P) AUG Met or start ACG His Thr (T) Gin (H) (Q) GUU GCU GAU GGU lle Asp Arg (R) (1) GUC Val GUA GCC GAC GGC Ala GCA Gly GGA GAA Glu GAG GUG GCG GGG The mutation changes a single amino acid to another amino…arrow_forward
- Yeast artificial chromosome how the synthesised?arrow_forwardQuestion: How does the interaction of covalent w/ hydrogen bonds and the structure of RNA encourage this molecule to replicate its own self? -Please include words: triphosphate vs. monophosphate RNA monomer, ribose, phosphate, base, carbon 5/carbon 3, polymerization, A-U, G-C, template-directed polymerization, complement, complement of complement, replicaarrow_forwardSmall extrachromosomal rings of DNA, often contain genes that give bacterial cells genetic advantages: * none is correct O chromatin Onucleoplasm nucleoid plasmidsarrow_forward
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Bacterial Genomics and Metagenomics; Author: Quadram Institute;https://www.youtube.com/watch?v=_6IdVTAFXoU;License: Standard youtube license