23. Replication and Recombination

Question 1

what are transposable elements?

Answer 1

• “Jumping Genes”
• chromosomes of bacteria, viruses, eukaryotic cells contain piences of DNA that move around genome
  – process called transposition
  – important role in generation and transfer of new gene combinations
• DNA segments carry genes required for this process
  – consequently, move baout chromosome are transposable elements (transposons)
• transposition does not require extensive homology between transposon and destination site

Question 2

what is the behaviour of transposable elements?

Answer 2

• similar to lysogenic phages
  – but originate in one position of chromosome, then move to different location in same chromosome
• transposon differ from phages
  – lacking virus cycle and plasmids
  – not able to reproduce autonomously
  – exist apart from chromosome

Question 3

what are the types of transposons?

Answer 3

• simple transposons
• complex transposons

Question 4

what are simple transposons considered as?

Answer 4

• contain palindromic sequences at each end
  – nucleotide sequences read same backwards and forwards
• bound by inverted repeat (IR) sequences on each side

Question 5

what are complex transposons considered as?

Answer 5

• include, not only, transposase gene
  – in insertion sequence
• other genes as well
  – bound by insertion sequences
• R-factors and cell marker genes
  – often contained in complex transposons

Question 6

what are insertion sequences (IS elements)?

Answer 6

• short sequence of DNA (750-1600 bp)
  – containing only genes for enzymes required for transposition
• bound at both ends by identical sequences of nucleotides
  – in reverse orientation
  – Inverted Repeats (15-25 bases long, vary between IS elements)
• terminal inverted repeats are inverted complements of each other
  – complement of ACGCTA (inverted repeat on right side of TE) is TGCGAT (reverse order of terminal inverted repeat on left side)
• one of roles of terminal inverted repeats are to be recognised by transposase
  – between inverted repeats is gene coding for enzyme transposase, required for transposition
• flanking direct repeats
  – no actually transposable element, rather play role in insertion of TE
• after TE excises, repeats left as ‘footprints’
  – can alter expression of gene in which they have been left, even after related TE has moved to another location on genome

Question 7

what is the mechanism of transposition?

Answer 7

• Tn5 transposase
• first step
  – individual molecules of transposase bind to specific sits at ends of transposon DNA
• next step
  – looping of transposon DNA forms synaptic complex, bringing two ends of transposable element close together
• once synaptic complex formed
  – Tn5 transposase cuts transposon DNA away from flanking donor DNA
• after cleavage
  – Tn5 transposase/DNA complex moves freely until encounters and binds to target DNA
• process called
  – strand transfer
  – transposase catalyses insertion of transposon DNA into traget DNA, completing transposition process

Question 8

Review Paper**

Answer 8

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Question 9

What is the replication of the bacterial genome?

Answer 9

• duplication of chromosome and separation of copies
• continues elongation of cell and movement of copies
• division into two daughter cells

Question 10

what is the bacterial genome?

Answer 10

• one circular DNA genome
  – single origin of replication
  – bi-directional DNA replication
• bacteria may contain plasmids
  – smaller circular DNA molecules
  – autonomously replicated
• bacteria divide by binary fission
  – asexual reproduction
• progeny are genetically identical
  – clonal replication

Question 11

what is the process of bacterial genome replication?

Answer 11

• enzymes unwind parental double helix
• proteins stabilise unwound parental DNA
• leading strand is synthesised continuously by DNA polymerase
• lagging strand is synthesised discontinuously
  – RNA polymerase synthesises short RNA primer which is extended by DNA polymerase
• DNA polymerase digests RNA primer and replaces with DNA
• DNA ligase joins discontinuous fragments of lagging strand

Question 12

what is bacterial DNA mutation?

Answer 12

• stability of nycleic acid sequence important for life
• sequence chages occur
  – resulting in altered phenotypes
• changes in base sequence of DNA
  – inheritable
• can be:
  – harmful / lethal
  – helpful
  – silent

Question 13

what are the causes of mutations?

Answer 13

• spontaneous mutations
  – in cells, result of errors in replication
  – develop in absence of any added agent
• physical and chemical agents
  – induced mutations as result of exposure to mutagen
  – UV radiation

Question 14

what type of mutations occur?

Answer 14

• silent
• missense
• nonsense
• frameshift
• large deletions/insertions

Question 15

what is a silent mutation?

Answer 15

• mutation occurs but no resultant phenotypic change
  – due to codon degeneracy
• mutation not detected
  – except at level of DNA / mRNA sequence

Question 16

what is a missense mutation?

Answer 16

• single base substitution in DNA
  – changes a codon for one amino acid into codon for another
• EG.
  – GGC specifies Glycine; changes to AGC specifying Serin
• expression / effect of mutation can vary
  – expressed at protein level, resulting effect on protein function may range
  — from complete loss of activity -> no change

Question 17

what are nonsense mutations?

Answer 17

• causes early termination of translation
  – results in truncated polypeptide
• results in conversion of sense codon to nonsense / stop codon
• phenotypic effect largely dependent on relative location of mutation

Question 18

what are frameshift mutations?

Answer 18

• arise from insertion/deletion of 1 or 2 base pairs within coding region of a gene
• usually very deleterious and yield mutant phenotypes
  – result from synthesis of nonfunctional proteins
• often produces stop codon and shorter peptide

Question 19

how do we calculate rate of incidence of mutations?

Answer 19

• if doubling time = 20 min
  – 2^3 cells/hr from single cell
• if spontaneous mutation rate = 1*10^-7 /gene
  – (12 hr)(day)(10^10)(10^-7)=10^3 mutations/gene/day
• if bacteria = 4000 genes
  – (410^3 genes) = 410^9 mutations/day