Bacteria 2 (MJC)
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State the differences between the chromosome of a bacterium and a human cell.
- 1) Bacterium has circular DNA while human chromosome is linear.
- 2) Bacterial chromosome possess few non-coding regions while a large portion of human chromosomes are non-coding
- 3) Introns are not present in bacterium chromosome but do in human chromosomes.
- 4) Bacterial chromosome is associated with nucleoid-associate proteins while human chromosomes are associated with histones.
- 5) Bacterium possess only 1 chromosome while humans have 46.
State the distinguishing features which show that the organism above is a bacterium.
- 1) Presence of numerous flagella on cell surface
- 2) Contains a dense nucleoid region which is not bound by a membrane
State the differences between mitosis in an eukaryotic cell and binary fissionin a prokaryotic cell.
- 1) Mitosis is separated into 4 different phases while binary fission do not have diferent phases.
- 2) Spindle fibers are necessary in mitosis but no spindle fibers are needed in binary fission.
- 3) Mitosis results in 2 genetically identical nucleus while binary fission results in 2 bacteria cells.
F+ E.Coli cells are grown in a culture medium without antibiotics. It was found after a few generations that some of the progenies do not have any sex pili on their cell urface membrane.
Account for why this has occurred.
- - The E.Coli cells divide via binary fission
- -In the course of binary fission, unequal distribution of cytoplasmic constituents results in some progenies not inheriting the F plasmid from the parental cells.
- - The F plasmid contains genes necessary for sex plius formation, the lack of the F plasmid would result in some progeny to be unable to form the sex phili.
State the similarities between a chloroplast and a bacterium cell.
- 1) Both contains 70s ribosomes.
- 2) Both the choloroplast and the bacterium contains circular DNA.
What are the similarities in the principles of control of both prokaryotic andeukaryotic gene expression?
- 1)Transcriptional processing is the most important level of control in both prokaryotes and eukaryotes.
- 2) Both involves regulatory proteins which either inhibits or enhance transcription
- 3) Both differentiates genes that are expressed continuously and genes that have to be turned on or off depending on the environment.
Describe how the presence of allolactose turns the expression of the lac operon on.
- - Allolactose will bind to the active repressor hence inactivating it.
- -This results in the inactive repressor to be unable to bind to the operator region.
- - RNA polymerase is hence able to bind to the promoter region to initiate transcription.
State the function of the following in the regulation of the lac operon expression.
i) cAMP receptor protein binding site
ii) LacI operon
i) The CRP binding site is where the activated CRP binds to to enhance the binding of the RNA polymerase to the promoter region and hence enhances the rate of transcription.
ii) The LacI operon contains genes that code for the active repressor protein of the lac operon and hence inhibits the transcription of the genes in the lac operon if the LacI operon is turned on.
Explain the existence of multiple origins of replication in eukaryotes while prokaryotes have only one.
- -Eukaryotic genome contains more non-coding DNA sequences as well as genes compared to the prokaryotic genome.
- -This results in eukaryotic genome to be much longer as compared to prokaryotic genome.
- -Thus eukaryotic genome have multiple origins of replication to speed up rate of DNA replication whereas prokaryotic genome just require only one origin.
Describe the differences between specialised transduction and conjugation.
- 1) The transfer of genetic material in specialised transduction occurs via a temperate phage while in conjugation, this transfer occurs through a cytoplasmic bridge formed by the sex pili.
- 2) Specialised transduction only transfers bacterial genes near the prophage site to another bacterium while conjugation involves the transfer of the entire F plasmid.
- 3) Specialised transduction involves the integration of viral DNA into the chromosomes of the host bacterium while in conjugation, there is no integration.
Define the following terms.
i) Homologous recombination
- i) It is where a portion of donor DNA sequence of a bacterium is cut and rejoined with a homologous recipient DNA sequence of another bacterium, thus resulting in genetic variation.
- ii) Conjugation is is the unidirectional transfer of genetic material in the form of a Fertility or F plasmid from a donor bacterial cell to a recipient bacterial cell via a temporary sex pilus.
Define and explain how natural transformation occurs.
- -Transformation is the incorporation of naked DNA (fragments of DNA) from the environment into the recipient cells to alter the bacteria genotype.
- -The DNA of the donor cell is fragmented and released into the environment.
- -One of the fragments is taken up by a recipient cell. (at random)
- -Homologous recombination occurs and the donor DNA
- is incorporated into the DNA genome of recipient cell.
Define the terms
i) Enzyme induction
ii) Enzyme repression
i) This involves the synthesis of an enzyme only when its substrate is present.
The substrate is known as the inducer.
- ii)If a product is present in the medium, enzymes catalyzing the synthesis of product are not synthesized.
- Hence this product functions as a corepressor
Explain the control of gene expression in the trp operon.
- -When tryptophan is present in the cell, the cell
- needs to stop synthesizing these enzymes to save energy
- and resources.
- -Tryptophan act as a corepressor and binds to inactive
- repressor protein to activate it.
- -The active repressor now binds to the operator and blocks transcription.
- -mRNA and the proteins it codes for are not made.
What happens to the lac operon in the cellin the presence of both lactose and glucose?
- -Lactose taken up by the cell is converted to its isomer, allolactose, which binds to the active repressor.
- -This inactivates the repressor and prevents it from binding to the operator.
- -This allows RNA polymerase to bind to the promoter
- to initiate low level of transcription.
- -Presence of glucose in the cell inhibits the synthesis of cAMP and stimulates its (the cAMP) transport out of the cell.
- -A low level of cAMP results in a low level of activated cAMP Receptor Protein (CRP), as CRP can only be
- activated when complexed to a small molecule of cAMP)
- -Low level of transcription results
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