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Lecture notes

Gene amplification and rearrangement

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Detailed notes on gene amplification and rearrangement

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  • February 21, 2022
  • 5
  • 2018/2019
  • Lecture notes
  • Dr andrew cuming
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BLGY1232 Gene amplification and rearrangement

Gene amplification (make as many genes as necessary)
 Under certain circumstances, amplification of specific genes may occur, when the
organism has a specific need for high levels of production of a specific gene product.
 Some genes are already present within genomes in multiple copies, and it is
important to distinguish between these multiple gene copies, and specific, additional
amplification of the numbers of particular genes in certain cell types.
 Ribosomal RNA; Where there is only one “amplificatory step” in the path from DNA
to product, then having multiple copies of the rRNA genes enables rRNA and protein
to be synthesised in equivalent quantities  in some cases, there is an additional
requirement for rRNA synthesis (amphibian development)
 Xenopus laevis - the South African clawed frog; embryonic development
begins with the production (in females) of the Oocyte (pre-meiotic
unfertilised egg cell)  After fertilisation, the frog embryo develops without
any new transcription until it reaches the 4096-cell stage  everything
needed for early embryo formation must be pre-synthesised and stored in
the oocyte  Xenopus makes 1012 ribosomes and stores them in the mature
oocyte  This is a massive demand for rRNA synthesis which cannot be met
by transcription of the 500 rRNA genes within the Xenopus genome.
 To make this many ribosomes;
o 1 rRNA transcript (45S) = 12kb
o 1 rDNA transcription unit can bind an RNA polymerase every 100bp 
we can make 120 x 500 rRNA transcripts simultaneously
o 120 x 500 = 60,000
o To make 1012 transcripts, we have to do this (1012/60,000) times =
16.7 million times
o Polymerisation rate = 20 nucleotides per second To polymerize
12kb therefore takes 600 seconds (10 minutes)
o To do this 16.7 million times will therefore take 167 million minutes =
317.5 years
o (actually, this calculation is based on the number of genes per haploid
(N) genome: the premeiotic oocyte is 4N, so this figure should be
divided by 4 = 79 years)  FROGS DON’T LIVE THIS LONG
o Solution; During oogenesis, the rDNA repeat unit is amplified so that
the cell contains 2.106 copies of the rRNA genes  The rDNA is
amplified as extrachromosomal circular DNA  replicated by rolling-
circle replication to generate a population of these circular molecules,
each of which can contain between 1 and 12 rDNA transcription units
 These are all transcribed, so that the requisite number of rRNA
molecules can be accumulated within a few weeks (with 2.10 6 rDNA
repeats units we can make 2.106 x 120 transcripts simultaneously so
the process would only take 28.9 days)  Because these circular DNA
molecules do not contain origins of replication or centromeres, they
are rapidly lost when cell division is initiated, following fertilisation
Note: This applies only to the rDNA unit encoding 28S, 18S and 5.8S rRNA. (PolI
genes) The genes encoding the 5S rRNA component (PolIII genes) are “preamplified”

, as a permanent part of the Xenopus genome: there are over 10,000 (per haploid
genome) of these that are specifically transcribed only in oocytes
 Insect eggshells;
 Insect egg cells are protected by a shell, or chorion, made of proteins
synthesised by the surrounding nurse cells and follicle cells of the maternal
fly  the chorion is made of protein encoded by 4 genes that are amplified
during oogenesis
 In Drosophila, the single copies of the genes encoding the chorion proteins
are insufficient to produce enough mRNA to support the synthesis of the
chorion proteins by translation  in the follicle cells of the mother fly, the
chorion genes are selectively amplified by activating an origin of replication
adjacent to the chorion genes  32 to 64-fold amplification of the number
of these genes in this specialised cell type  these genes remain as an
integral part of the follicle cell genome




 The follicle cells are somatic cells, not germ-line cells, so this amplification is
not passed on in the next generation
 the silk moth possesses multiple copies of the chorion genes in its germ line
(and therefore in all other cells in every generation), and the expression of
the chorion protein is controlled by regulating the transcription of these “pre-
amplified” genes

Genome rearrangements (bring together genes to be coregulated for a particular
outcome): generation of antibody diversity
 In nitrogen-fixing Cyanobacteria, vegetative cells differentiate to form heterocysts in
which N-fixation takes place  this differentiation process involves DNA
rearrangements
 The immune system has the ability to generate a response to foreign substances
introduced to the body - A huge diversity of antibodies can be generated, including
antibodies capable of recognising completely novel, synthetic compounds, never
before seen in Nature
 Key features: Recognition – response – diversity
 An antigen is introduced to the body  it is recognized by cells of the
immune system; lymphocytes – T cells ( phagocytes) and B cells (antibody
secretion)  antibodies released by B cells circulate in the blood to
immobilize antigens
 Recognition and Response; when an antigen encounters and specifically
binds to an antibody molecule bound to the surface of a B-cell  stimulates
that B-cell to reproduce itself as a clone of cells and each cell in the clone

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