unit 2 -7 Flashcards

0
Q

Discuss bacterial mating

A

Called Conjugation
There is a long thin projection called the pilus used in mating- the pilus is coded for in a gene on a small double stranded, circular DNA molecule called the F plasmid. Bacteria that carry a particular F plasmid (+) can only mate with those that do not (-). The pilus forms a conjugation bridge which the F plasmid is transferred to a new host by DNA replication called rolling circle replication.
in certain instances part of the bacteria genome can also be transferred - frequent way in which antibiotic resistance is transferred

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1
Q

The basis of E. coli

A
haploid
genome is encoded on a single dsDNA, circular chromosome 
genome size is - 5 x10^6
doubles every 20 minutes 
mutation rate -1x10^9
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2
Q

Describe the process of transformation

A

the process by which some bacteria are able to pick up pieces of DNA from their environment -the exogenous DNA can originate from a lysed neighbor, it may contain anti-biotic resistant genes.

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3
Q

Describe the process of homologous recombination or the formation of “holiday junctions”

A

recombination is catalyzed by a set of enzymes referred to as the rec enzymes (rec A, B, C, and D)- specialized to mediate exchange between homologous sequences.

  • the homologous DNA sequences align -(bacterial DNA)
  • a nick is made in one strand (nick is made in both strands in humans) allowing it to invade and anneal to the complementary strand of the homologue
  • the ends are ligated to complete formation of the cross-strand exchange or Holiday junction (its like their joining arms)
  • at this point it they were cut and re-ligated no DNA exchange would have occurred
  • The strands have to do gymnastics to cross-
  • at the end of the flips and flops crossing strands are cut and ligated to one another
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4
Q

Describe the process of gene transfer via bacterial virus transduction and the two pathways

A

bacteriophage lambda as an example which has a liner dsDNA genome.
-viral capsid allows the virus to bind to new bacterial host and inject its DNA
-DNA becomes circularized
-two fates
LYTIC
1. -viral genome replicated
-new virus particles assembled
-the cell is lysed to release the new viral particles to environment
LATENT OR PROPHAGE
2. -the viral DNA is integrated into bacterial chromosome via a encoded integrase catalyzing a site specific recombination
-when the cell is signaled by environmental insult then the DNA is excised and the DNA enters lytic cycle.

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5
Q

Describe viruses briefly

A

background viruses are parasitic organisms that require a host machinery to be replicated

  • covered by a protein coat called a capsid
  • capsid serves to protect an transmit viral DNA to new host
  • viral genome typically carries gene encoding for capsid protein
  • enzymes required to integrate the viral genome into the host cell
  • bacterial viruses may carry DNA from previous hosts
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6
Q

Transduction

A

when bacteriophages transfer genes, such as antibiotic resistance - between bacteria by picking up flanking host DNA during sloppy excision events

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7
Q

Transposable elements

A

transposable elements - relatively short DNA sequences that have the ability to move from place to place within the genome of the organism by a process called transposition

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8
Q

What are the 5 most ways transposable elements contribute to genetic diversity?

A
  1. disrupting genes in which they land
  2. altering the expression of neighboring genes
  3. facilitating the rearrangement of the bacterial chromosome through homologous recombination btwn transposon copies
  4. by hopping into plasmids or bacteriophage and transferring their genes to other bacteria via these vectors
  5. amplifying copies of antibiotic resistance genes
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9
Q

What are the two ways transposons can move?

A

1- cut and paste (non-replicative)

2- copy and paste (replicative) - leads to amplification

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10
Q

what function does “junk” DNA play?

A

so called “junk” DNA provides greater plasticity/genetic diversity in mammalian cells

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11
Q

What is Mendel’s second law?

A

The law of independent assortment
- paternal chromosome pairs separate from one another independently during meiosis
2^n (where n= the number of homologous pairs)

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12
Q

Describe homologous recombination in eukaryotes

A

The general process is the same

  • double strand break (diff)
  • The Rad50 complex then resects the 5’ end via a 5’ to 3’ exonuclease activity
  • exposed 3’ end overhangs can find their homologous sequences and with help of RecA homologue Rad51 carry out strand invasion
  • gaps from resection filled in by DNA synthesis -> makes holliday junction
  • flip flop
  • cut and ligation
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13
Q

When does homologous recombination occur?

A

During meiosis after DNA replication

there are on average 3 cross overs per chromosome per meiosis

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14
Q

What are the three main types of retrotransposons of humans?

A
  1. LINE-1 elements (6-8 kbp in length) account for about 21% of human genome sequence- they encode for their own reverse transcriptase and are still capable of transposition
  2. SINEs (smaller 100-300 bp) found in more copies-account for abut 13% of genome- use the reverse transcriptase of LINEs to move
  3. Alu sequence (small around 300) - 5% of genome -no longer competent to transpose
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15
Q

How do retrotransposons contribute to human genetic diversity?

A
  1. they can disrupt gene function by inserting within the exon of a gene; an L1 insertion into the Factor VIII gene causes one form of hemophilia
  2. They can affect the expression of genes. For example, line-1 elements lower gene expression by decreasing rates of transcription and elongation
  3. create sites for illegitimate homologous recombination known as unequal cross- overs
16
Q

What did the population-scale genome sequencing tell us?

A
  1. Each individual differs from the reference human genome sequence at 10,000 to 11,000 site that change protein sequence
  2. On average, every person inherits 250-300 loss of function alleles
  3. On average every person is heterozygous for 50-100 variants known to cause inherited disorders
  4. The germ line mutation rate is 1.0-1.2 x 10 ^-8 per generation per base pair so about 60 new mutations are generated per generation