Behaviour of Genes

Question 1

Haploinsufficiency

Answer 1

Haploinsufficiency is the phenomenon where a diploid organism has only a single functional copy of a gene (with the other copy inactivated by mutation) and the single functional copy of the gene does not produce enough gene product (typically a protein) to bring about a wild-type condition, leading to an abnormal or diseased state.

• incomplete dominance = intermediate phenotype

Question 2

Function loss of dominant phenotype

Answer 2

• most common
• haploinsufficiency

Question 3

Function gain as dominant trait

Answer 3

• dominant negative
• 50 WT: 50 mutant: protein >> displays mutant phenotype
• function gain almost always dominant

Question 4

Dominant negative

Answer 4

• antimorph

Haploinsufficiency means that you need the expression of both alleles for normal function; a 50% reduction (heterozygous loss-of-function) causes phenotypes by insufficiency.

Dominant negative (antimorph) means that the mutant protein is expressed, and somehow actively interferes with normal function, for example forming inactive dimers with the wild type protein, or binding to partners, but clogging up signaling (since it doesn’t work).

Question 5

Function gain as co-dominant

Answer 5

• 50 WT: 50 mutant
• both protein contribute to phenotype e.g. ABO blood group

Question 6

The distinction between loss-of-function and gain-of-function is not always super-clear

Answer 6

Loss-of-function usually means that less of a protein is made or that some function of the protein has been compromised.

Loss-of-function mutations are usually recessive, since in most cases, a single “good” copy of the gene will suffice

2 common types of exceptions:

“Haploinsufficiency”: One copy is not enough

“Dominant negative” or “antimorphic” mutations: The defective gene interferes with the function of the wild-type copy. This is common with proteins that form polymeric structures, such as filaments.

Question 7

Co-dominance

Answer 7

• all alleles of a gene co-dominant at the DNA sequence level

Question 8

Recessive lethal

Answer 8

• homozygosity = death

A^yA^y = death

Question 9

Pleiotropy

Answer 9

• more than one phenotype
• A pleiotropic gene is a single gene that controls more than one trait

Pleiotropy: where one gene affects multiple characteristics

Question 10

Polygenic traits

Answer 10

• multiple genes converge to result in a single phenotype
• mutation, in some cases, can produce the exact same phenotype

Question 11

Complementation

Answer 11

• allelic = mutation in the same gene
• mutant phenotype = mutation in the same gene
• WT phenotype = mutation in different genes, at least 1 functional copy at each gene

in ploygenic traits

Question 12

Complementation test

Answer 12

• determine whether mutants due to mutation in same or different gene

Question 13

Physical vs genetic linkage

Answer 13

Physical linkage and genetic linkage are not interchangeable terms.

This is because there are cold spots and hotspots for recombination so the relationship between physical linkage and genetic linkage is not linear.

Just because two genes are on different chromosomes it also doesn’t mean they are unlinked. This is because translocations create pseudo-linkage.

pseudolinkage = characteristic of a heterozygote for a reciprocal translocation, in which genes located near the translocation breakpoint behave as if they are linked even though they originated on nonhomologous chromosomes

Question 14

Chiasmatic vs achiasmatic gametogenesis

Answer 14

Drosophila = male drosophila do not have recombination in their gametes because they have achiastmatic spermatogenesis.

Question 15

Synthetic interaction

Answer 15

• parallel/independent function

Synthetic phenotypes = when genes affect the same phenotype but operate in different biochemical pathways.

Mutations in any of the genes will result in a non-wild type phenotype.

- genes occuring in different steps in same synthetic pathway explains 9:7 ratio in F2, dependent on where mutation is and where the pathway is blocked

Question 16

Genetic interaction

Answer 16

• linear, dependent function e.g. G1>>G4 linear, any mutation in G1>>G4 loss of phenotype
• interaction between alleles
• Multiple genes can work together to contribute to a phenotype
• in general, phenotype is determined by the properties of the proteins produced by the gene
• regulatory:
• 1 gene activates another gene, produces activation protein
• mutation in regulatory protein
• mutation in structural protein

A gene can have many different alleles in a certain population and this is called its allelic series.

Dominance shows the way that the alleles interact in a heterozygote.

Single sequence repeats can be indicative of different alleles in a gene so thus are used in fingerprinting.

When the frequency of each allele in the population is known then the probability of getting a certain result can be determined and it’s usually significant enough to conclude that the fingerprint belongs to a particular individual.

Question 17

Phenotypic Ratios

Answer 17

9:3:3:1 = no interaction, independent assortment

9:7 = gene is in same pathway. A mutation in any of the genes involved will break the enzymatic chain and result in the end product not being produced

9:3:4 = recessive epistasis.

12:3:1 = dominant epistasis.

13:3 = suppressor has no phenotype

10:6 = suppressor is like mutant

Question 18

Genetic Modifier

Answer 18

genetic modifier exists at another locus and changes degree of phenotypic expression of mutated gene at first locus

Modifier mutations = these mutations can result in either of 13:3 or 10:6 ratios. Modifier mutations are those that exist at another locus and reverse the effect of a mutation in another gene.

The ratio is 13:3 if the suppressor mutant has no detectable phenotype and 10:6 if they do (recessive suppressor has same phenotype as target mutation).

suppressor is a mutant allele, reverses the effect of mutation of a different gene, implies interaction between suppressor and target gene = WT or near WT phenotype

Question 19

Epistasis

Answer 19

recessive phenotype overrides a dominant phenotype, generally the double mutant overriding one of the single mutants.

dominant phenotype masks the rest causing one of the single mutant variations to appear as the dominant phenotype.

Divergent pathways can lead to epistasis

divergent pathways often lead to epistasis where the phenotype of one gene masks the expression of the phenotype of another gene.

The epistatic gene is the overriding mutation and the hypostatic gene is the overridden mutation.

Question 20

Reciprocal Cross

Answer 20

• male only contributes nucleus during fertilisation - chromosome in nucleus
• male = XO or XY
• female = XX
• results of reciprocal cross different = sex linked

X^wX^w white x X^w+Y red

X^w+X^w+ red x X^wY white

Sex linked:

father >> daughter

mother >> son

Matroclinous:

mother (X^WX^W)>> daughter

X^WX^WY

Patroclinous:

father (X^W+Y) >> son

X^W+O

Y-linked inheritance

both due to non-disjunction at meiosis I

Question 21

Sex determination - XY

Answer 21

presence of Y infers male

XO hemizygous

XX females only can be ‘testers’ for test cross

Question 22

Drosophila Sex Linkage

Answer 22

Female

X/Automsomal X = ratio of 1 = female

XXY/XX = female, fertile

XXX/XXX = female, fertile

threshold amount of sxl gene on X chromosome promote female development

Male

X/Autosomal X = ratio of 0.5 = male

XO = sterile (w/o Y = sterile in males or intersex)

XY/XX = male, fertile

XO/XX = male, sterile

Intersex

XX:3X (autosomes) = sterile, intersex 2:3

Question 23

Sex determination Z/W

Answer 23

ZZ = male homogametic

ZW = female heterogametic

• birds, butterflies, moth, some fish
• Z-linked genes

Plants

In plants separate sexes can exist (XX or XY) Or male and female sex organs in the same plant

Question 24

Sex determination - environmental

Answer 24

Incubation temperature

<28 celsius = male (cool guys)

28-32 celsius = mix

>32 celsius = female (hot chicks)

Question 25

Linkage

Answer 25

• the closer the two genes are, the less likely recombination will occur
• >50mU = unlinked
• recom. frequency = indirect measure of physical distance between two genes
• measure linkage in dihybrid cross:

2AB:1aB:1Ab = totally linked (phenotypic ration in F2 zygotes)

9AB:3aB:3Ab:1ab = independently segregating (phenotypic ratio in zygotes)

Question 26

Test cross

Answer 26

• cross with individual that is homozygous recessive

AA x aa

• phenotype of offspring represents gametes of unknown parent

Question 27

Recombination

Answer 27

• recom. distances are additive
• RF underestimates number of cross over events (DR look like parent)
• to increase map resolution, need more markers e.g.

DNA markers (molecular variation) = SNP, indels

protein markers = change to protein function, different properties i.e. charge, size

biochemical marker = antigen/antibody interaction, alloenzyme

Alloenzyme

• phosphoglucoisomerase:
• subunits can have charge variations
• dimer
• homozygous: fast-fast, slow-slow
• heterozygous: fast-slow

Question 28

3 factor cross

Answer 28

• establishes genetic map
• allele arrangement
• order of loci
• how far apart the loci are

Question 29

Coefficient of coincidence (CC)

Answer 29

• interference (I) = occurence of 1st crossver interferes with 2nd cross over

I = 1 - CC

CC = ratio of actual observed double cross over expected double cross over events

• Double mutants/ No. of progeny * map distances (expressed as decimals)*
  e. g. 12:16 = 0.75

I = 1 - 0.75 = 0.25

Question 30

Copy number variants (CNV)

Answer 30

• also known as VNTR (variable number tandem repeat) or minisatellites
• STR (short tandem repeat) or microsatellites

We can analyse them using:

• restriction fragment length polymorphism analysis (RFLP), involves RE digest and hybridisation
• STR analysis, involves PCR with primers flanking CNV

Question 31

RFLP

Answer 31

The first DNA fingerprints

1. cut DNA with restriction enzyme
2. run sample on gel
3. southern hybridisation with probe to variable locus

4 detect allele

• RFLP detects changes in restriction site position

SNPs or indels can create/abolish restriction endonucleus restriction site therefore affecting quantities and legnth of DNA fragment resulting from RE digestion

Question 32

STR Analysis

Answer 32

• PCR based
• microsatellites
• primers flank repeated sequences
• more repeats = larger band
• primers isolate particular loci
• different combinations of DNA markers provide unique profiles
• frequency of alleles = probability of allele occurring
• correlation of microsatellite with disease/trait e.g. presence of M repeat correlates with presence of P (dominant disease allele)
• STR can occur in exons: cause mutations, repeated sequence in coding sequence could result in a non-functional protein

Question 33

SNPs

Answer 33

• used for identification purposes
• correlated SNPs outside and within gene = no effect
• causative SNPS within gene = non-coding SNPs effects how much protein is produced, coding SNPs changes aa sequence
• uncorrelated = far from gene, on same chromosome or different chromosome

Question 34

Southern Blot

Answer 34

detection of a specific DNA sequence in DNA samples.

Southern blotting combines transfer of electrophoresis-separated DNA fragments to a filter membrane and subsequent fragment detection by probe hybridization.

Question 35

Aneuploidy

Answer 35

• 2n + 1 = trisomy

trisomy 21 = down syndrome

XXY = klinefelter

• 2n - 1 = monosomy
• can occur spontaneously but rarely in gamete formation
• can be induced by drugs that disrupt meiotic spindle

Question 36

Polyploid

Answer 36

• extra set of chromosomes
• induced in plants by applying vinblastin to pre-meiotic tissue = large, favoured agriculturally
• triploids are sterile

Question 37

Triploids

Answer 37

• 3n
• in animals, only produced in drosophila = problematic, sterile
• homologous chromosomes do not evenly separate during anaphase = produces aneuploid gamete (extra chromosome in some cells)
• In plants, other polyploids are fertile and can reproduce normally.

Question 38

Allopolyploid

Answer 38

• fertilisation between two difference species
• chromosome different in numbers, size and shape, doesn’t have homologous pair = doesn’t align properly = missing chromosomes, aneuploid
• if two co-existing genomes are dissimilar, pairing of homologous chromosomes does not occur
• often sterile

Question 39

Allotetraploid

Answer 39

• spontaneous doubling:

2x(n1+n2) = 2n1 + 2n2 = 4n

• contains 2 different diploid genome
• pairing of homologous chromosome occurs
• if mate with original parent

(n1 + n2) + n1 = 3n = problem

(n1+n2) + n2 = 3n = problem

if allotetraploid mate with each other

(2n1 + 2n2) = new species?

Question 40

Autopolyploidy

Answer 40

• when several sets of the same genome from the same species are found in an organism.

Question 41

Translocation

Answer 41

The movement of a chromosome segment to another location in the genome with no gain of loss of genetic material.

Translocations can be intrachromosomal (to the same chromosome) or interchromosomal (to another chromosome) and reciprocal and non- reciprocal.

Translocations between non-homologous chromosomes can result in unbalanced genes especially if the translocation is heterozygous.

The homologous regions still pair up at meiosis except the way they segregate determines whether the genes are balanced or not:

• Adjacent segregation = both chromosomes proceed directly away from each other resulting in unbalanced chromosomes
• Alternate segregation = the chromosome on one side swaps sides and this results in balanced chromosomes.

Question 42

Unbalanced chromosome

Answer 42

• not viable
• occurs 1/2 the time
• linked, not segregating independently
• from adjacent segregation

Question 43

Somatic cell translocation and cancer

Answer 43

• not associated with translocation itself
• translocation brings genes close together >> fuse >> cancer
• philadelphia gene
• Affect of transloca1on position and resulting in gene fusion, rather than chromosomal imbalance

Question 44

Robertsonian Translocation

Answer 44

• non-reciprocal
• fusion of 2 chromosomes
• translocation between acrocentric chromosomes
• seen in down syndrome
• extra chromosome material

Acrocentric chromosome = A chromosome in which the centromere is located quite near one end of the chromosome. Humans normally have five pairs of acrocentric chromosomes. Down syndrome is caused by an extra acrocentric chromosome (chromosome21)

Question 45

Compound chromosomes

Answer 45

• translocation-like event joins 2 identical chromosome arms to same chromosome
• can occur via robertsonian translocation
  e. g. 2 left arms (compound left)/2 right arms (compound right)

Question 46

Compound chromosome uses

Answer 46

Use to displace wild-type populations

Question 47

Inversion

Answer 47

To invert a segment of the chromosome is must be cut out, flipped, then replaced.

If the centromere is included in the inversion, then it is said to be pericentric however if it is not included then the inversion is paracentric.

Inversions do not themselves result in gene imbalance however they may produce a mutation if they cut a gene.

Meiosis

• homologous inversion okay, can still pair
• heterozygous inversion, problem, pair as loops

Heterozygous inversions

problems arise when homologous chromosomes are heterozygous for an inversion because an inversion loop has to form.

For both paracentric and pericentric inversion crossover within the inversion loop will result in duplication and deletion products and most often inviable gametes, non-sister chromatids participate in cross over

Dicentric = 2 centrioles, spindle pulls in 2 directions = random break

Acentric = no centrioles, acentric fragment lost, cannot attach to spindle

Question 48

Paracentric inversion

Answer 48

• can lead to deletion of products
• unbalanced gametes not viable
• 1/2 gametes affected
• only see parental arrangment of gene

Question 49

Pericentric inversion

Answer 49

• leads to unbalanced products
• non-viable
• only see parental arrangement

Question 50

Homozygous Inversion

Answer 50

• recombination can occur
• changes mU, different gene order = different distances
• no inversion loop

Question 51

Inversions “suppress” recombination

Answer 51

• only see the parental arrangement (recom. often results in unbalanced, unviable chromosomes, crossing over inside inverted region)
• reduce observable recom. freq. between markers
• recom. outside inverted region is normal
• recom. inside loop = zero (recover only parental genotype) therefore, recom. freq measures distance from flanking genes to inversion break-point

Question 52

Balancer Chromosomes

Answer 52

• dominant morphological mutation (morphological mutants are the direct result of a mutation in a biochemical pathway)
• contains multiple inversions
• used to maintain heterzygous strains
• used for genetically screening a population of organisms to select for heterozygotes i.e. homozygotes die, heterozygotes live

Question 53

Position effect

Answer 53

when a gene is placed next to heterochromatin due to the inversion it can cause inactivation because the heterochromatin can spontaneously spread (spread does not occur all the time). This phenomenon is known as positive effect variegation

Question 54

Genetic effects of inversions: summary

Answer 54

Depend on the position of the loci and where crossing over occurs.

Suppress recombination in heterozygotes “heterozygous” = standard + inverted.

Inversion homozygous >> altered genetic distances between some loci.

Position effects - inversion moves genes to a diferent chromosomal environment >> might altered gene expression.

Mutations – genes might be disrupted/modified at inversion breakpoints.

Question 55

Deletion and Duplication

Answer 55

• changes amount of genetic material
• may be a result of unequal crossing over between homologous chromosomes

Question 56

Deficiency (deletion) mapping

Answer 56

• map against chromosome where we know deletion has occured
• pseudodominance

Question 57

Chromosomal pairing in duplication heterozygote

Answer 57

• chromosomal duplication important for evolution of multi-gene family

Question 58

Break point mutation

Answer 58

genes might be disrupted/ modified at translocation/inversion breakpoints Eg. Somatic cell cancers due to gene fusion events.

Question 59

The bar mutation - Drosophila

Answer 59

• bar mutation = reduces facets produced in eye
• more mutation copies = more severe = copy number effect
• phenotype more sever when bar alleles adjacent to each other = position effect

Question 60

Non-mendelian inheritance

Answer 60

Extra-chromosomal genomes - organelles and their genomes

• DNA that is not found in the nucleus (extra nuclear DNA) does not follow mendelian laws i.e. chloroplast and mitochondrial DNA.

These organelles contain lots of enzymes and proteins vital to the function of the cell.

The region within the organelles that contains the genome is called the nucleoid.

There are often multiple copies of chromosomes per nucleoid and multiple nucleoids per organelle. Despite this nowadays most of the proteins in the organelles are encoded in the nucleus and transported there.

Chloroplasts = chloroplasts use light energy and the fixation of carbon to create sugar molecules. All of their genomes are circular and are relatively similar in size and sequence between land plants.

Mitochondria = responsible for cellular respiration in eukaryotes. Most mitochondrial genomes are circular however some are linear. There is significant variation in genome size between the mitochondria of different organisms whilst the gene content remains fairly similar.

Question 61

Mitochondrial Genetic Code

Answer 61

The eukaryotic nuclear genome codes for 32 different tRNAs whilst the mitochondrial genome only codes for 22. This genetic code has changed over time and has resulted from many years of evolution.

Question 62

Endosymbiosis

Answer 62

• Ancestral eukaryote engulfed a prokaryote
• The oxidative phosphorylation activities in eukaryote gradually transferred from the nucleus to the mitochondrion.
• Oxidative phosphorylation was lost in the eukaryotic cell.
• The genome of the mitochondrion was reduced drastically over time. This occurred in 3 steps; firstly, the genes moved to the nucleus then they took over the function of the mitochondrial genes. Finally, the now redundant mitochondrial genes were lost through mutation.

Whilst eukaryotic organisms have varying mitochondrial genes the arrangement and consistency is very similar among them. This suggests that the organelle arose from a single progenitor i.e. it evolved once.

Question 63

Reasons for organelle genome reduction

Answer 63

Intracellular competition is a phenomena whereby multiple organelles or parts of a cell compete.

With respect to mitochondria they compete with how fast they replicate. A smaller genome means faster replication so they adapt to have smaller genomes.

The reason that any genes remain in the mitochondria at all is unclear.

Question 64

Transmission of extrachromosomal genes: non-mendelian

Heteroplasmy

Answer 64

Extrachromosomal genes do not abide by Mendel’s laws.

Heteroplasmy:

• Where there are two types of organelle in a cell due to mutation, the wild type and the mutant. When heteroplasmy is present the cells are somatically unstable which means that they can change phenotype in the process of division.
• Cells can become homoplasmic again through cytoplasmic segregation where the cytoplasm of the cells separate and two different daughter cells are created.

- Variegation results from heteroplasmy

Question 65

Transmission of extrachromosomal genes: non-mendelian

Uniparental inheritance

Answer 65

The inheritance of cytoplasmic genes in plants is usually uniparental and maternal.

The female egg determines the phenotype of the offspring because the male pollen cells have little to no cytoplasm and no chloroplasts.

Because of this it can be concluded that heteroplasmy must arise through a new mutation in the mother’s germ line cells. In very rare cases it can be inherited paternally but this is unlikely.

Human mitochondrial diseases = these diseases are almost all inherited maternally so the mother will pass it on to all of her children.

An example is Lebers Hereditary Optic Neuropathy which leads to partial blindness due to malfunctioning electron transport chain in mitochondria (appears dominant).

Mitochondrial diseases/phenotypes are very prone to variable expression and penetrance because various environmental factors as well as somatic variation (heteroplasmic parent cells becoming homoplasmic through cytoplasmic segregation), and nuclear genes can affect them.

Neurospora the fungus are another example of maternal inheritance. The cytoplasm is always inherited from the mother

Heterokaryon = contains nuclei of both parents

• partition of nuclei, isolate nuclei, able to determine whether nuclear or cytoplasmic DNA

Question 66

Variable penetrance and expressivity

Answer 66

Not all individuals display the disease phenotype or to the same degree

• environmental influences
• genetic differences
• cytoplasmic segregation

Question 67

Main rules of non-mendelian inheritance

Answer 67

• The mendelian ratios are not observed
• Reciprocal crosses involving extra-nuclear genes differ from those with nuclear genes.
• Genes cannot be mapped to the nuclear chromosome
• Substituting the nucleus for another nucleus will not affect the progeny.

Question 68

Maternal inheritance vs. the maternal effect

Answer 68

The maternal effect does NOT involve extra nuclear genes.

Sometimes the products of nuclear genes (mRNA or proteins) from the mother are left in the egg and determine the phenotype of the zygote regardless of the genotype of the zygote.

You can distinguish between maternal inheritance and effect by conducting a cross of the progeny to determine their genotype.

Question 69

Plasmids

Answer 69

Plasmids are autonomously replicating DNA molecules separate to the chromosomal DNA of bacteria.

They are generally double stranded, circular, and between 2-200kb long.

The degree of copies found in a single cell ranges from 1-2 to >500.

Question 70

Integrative Plasmid

Answer 70

some plasmids are capable to becoming part of the chromosome of the host cell and then replicating with it.

Occasionally the plasmid comes out of the chromosome again and brings some of the chromosome with it.

Most plasmids are non- integrative and they replicate independently of the chromosome. However, they still frequently use the host’s replication machinery.

Question 71

Non-integrative plasmid

Answer 71

• replicate independent of host chromosome
• uses host replication machinery
• must be compatible with host replication machinery
  1. high copy number = relaxed plasmid

those that have 10-100 copies per cell and initiate cell replication until the copy number is reached

2. low copy number = stringent plasmid

those that only occur in 1-5 copies per cell and initiate replication twice per cell division.

Question 72

Control of copy number in plasmids

Answer 72

• The ColE1 gene is one of the most important in plasmids with respect to replication.

Most vectors have a ColE1 origin of replication.

• The gene encodes an inhibitor of replication and this concentration increases with plasmid number until a threshold is reached and replication stops.

Process:

if RNA 2 is expressed and nothing else occurs then replication will occur.

However, if RNA 1 and the repressor of primer (rop) protein are produced RNA 1 and RNA 2 will form duplexes and replication will cease.

Some plasmids rely on random partitioning during cell division because they don’t have the partitioning (par) function. This means they can be lost in some cell lines.

Question 73

Random Partitioning

Answer 73

• partioning ensures low copy number plasmids segregate during cell division

Question 74

Plasmid Incompatibility

Answer 74

The inability to two different types of plasmids to exist in the same cell in absence of selection pressure due to having the same mechanism of partition or copy number control. Plasmids in different compatibility groups can exist happily in the same cell.

Question 75

Function of plasmids

Answer 75

Cells lacking plasmid outcompetes cells that retain them - so why retain them?

• They encode toxins that kill other bacteria in Bacteriocins.
• Some encode functions required for infection of other organisms.
• Some encode enzymes for the breaking down of organic molecules.
• Some encode fertility or antibiotic resistance related genes, more than one resistance genes often located in tranposons

Question 76

The mobility of some plasmid genes

Answer 76

Some plasmid genes can jump onto the chromosome of the cell or other genomes through non-homologous recombination.

Insertion sequence elements and transposons are the main types of these in bacteria

Question 77

Transposons

Answer 77

transposons are more complex than insertion sequence elements because they encode for unrelated functions as well as mobility.

Simple transposons lack insertion sequence elements (SINE element?) and only have small flanking repeats.

However, composite transposons (LINE element?) have flanking repeats that ARE insertion sequence elements.

movement requires transposase

Transposition events can be conservative where the original transposon moves to a new location without replicating, or replicative where the gene replicates and the original remains.

autonomous?

replicative? conservative?

retrotransposon? does not contain introns, no reverse transcriptase

“active” transposable element requires terminal repeats (TR) and movement requirese transposase (encoded in mobile elements)

insert in genome not based on homology, can cause mutation, gene disruption, change regulation of genes, induce chromosomal mutations

Question 78

Conservative tranposition

Answer 78

• Transposase cleaves the target site.
• Transposition sequence inserts into the target site
• The short single stranded DNA gaps are repaired.
• 5b repeats flank the transposition event.
• no DNA replication
• ‘cut and paste’ transposition

Question 79

Replicative Transposition

Answer 79

• Transposase cuts DNA at transposition site and target site.
• Transposase fuses the transposon to the target site and the cointegrate is formed.
• Single stranded DNA is used to fill in the gaps.
• Recombination between the original and new molecules is catalysed by resolvase.
• copy of Tn inserts into new chromosome location, original Tn remains in place
• DNA replication, net gain of DNA

Question 80

Insertion Sequence elements (IS)

Answer 80

these parts of the sequence only encode for proteins necessary for the mobilisation of the sequence.

IS replication uses host replication enzymes and causes inactivation when it disrupts a gene.

Question 81

Dissociation Element (DS)

Answer 81

• causes gene disruption
• DS gene unstable in presence of Ac element
• Ac element (autonomous) activates DS to transpose to new site/break chromosome where it is located, increases tendency for chromosome breakage >> loss of dominant markers, loss of chromsome arms
• requires presence of unlinked Ac for breakage/movement of Ds
• DS is non-autonomous, needs Ac to move
• TR’s but does not encode transposase
• activated by trans-acting tranposase

Question 82

Activator element (Ac)

Answer 82

• autonomous
• TR’s encodes transposase
• also a mobile element

Question 83

Autonomous vs non-autonomous

Answer 83

• capacity of transposons to excise and re-insert itself
• presence/absence of tranposons
• transposase is tran-acting

Question 84

Transposase

Answer 84

• regulated
• activity generally low due to different regulatory mechanisms
• prevents spurious movement and mutations

Question 85

Retrotransposon

Answer 85

• use reverse transcriptase to tranpose through RNA intermediate
• viral retrotransposon, properties similar to retrovirus
• non-viral retrotransposon, are the most frequently encountered transposable elements in mammals

Question 86

R plasmids

Answer 86

Plasmids that encode resistance to antibiotics, heavy metals, antiseptics etc.

R plasmids are transmissible which means they are capable of being transmitted between organisms = superbug

Question 87

Hybrid dysgenesis

Answer 87

• appearance of defects when hybrids formed by crossing different strains
• developed as molecular genetic tool

Question 88

Transposable elements in humans

Answer 88

LINES (long interspersed elements)

1000-5000 bp

20,000-100,000 copies

~ 35,000 bp separation

SINES (short interspersed elements)

< 500 bp

1/2 - 1 million copies

1000-2000 bp separation

Alu (200bp, 10% genome)

Question 89

How do we function with so much mobile junk DNA in our genomes?

Answer 89

1. Locations of LINEs and SINEs, clustered in introns, heterochromatin region
2. Activity of LINEs and SINEs, regulatory mechanisms restrict movement
3. Domestication, gives selective advantage, serves purpose for host
   - genome of most higher eukaryotes = many ‘dead’ transposable elements