Utilisateur
Radiation greatly increases mutation rates in all organisms.
two thymine bases covalently bind and block DNA replication.
Translesion Polymerases
The SOS system allows bacterial cells to bypass replication blocks through a mutation-prone pathway.
DNA polymerase η (eta) replaces damaged DNA with AA in eukaryotic cells to bypass replication blocks.
They remove electrons from atoms, altering the chemical structure of bases and causing double-stranded DNA breaks.
UV radiation causes pyrimidine dimer formation (usually TT, but also TC/CC), which stalls or stops DNA replication.
he SOS system bypasses pyrimidine dimers during DNA replication, but at a high error rate.
To prevent cell death and survive DNA damage, even though it increases the risk of mutations. Bypassing blocks allows replication to continue under stress, though it’s error-prone.
Double-stranded breaks (DSBs) are when both strands of the DNA helix are severed, which can lead to significant damage. They can be caused by high-energy radiation (X-rays, gamma-rays) or certain chemicals and pose a risk to the integrity of the genome.
Hiroshima was destroyed by an atomic bomb inn
1945.
Survivors of the initial blast suffered from radiation sickness and somatic mutations, most commonly leading to leukemia.
Estimates showed no increased germline mutations, likely because those with the highest exposure did not survive.
Base analogs are chemicals that resemble DNA bases in structure. When incorporated into DNA, they can lead to incorrect base pairing and cause mutations.
Alkylating agents donate alkyl groups (like a methyl or ethyl group) to DNA, causing mutations. An example is Ethylmethylsulfonate (EMS) and mustard gas, which was used in WWI chemical warfare (now banned).
Deamination, caused by chemicals like nitrous acid, removes an amine group from a base, altering its pairing ability and leading to mutations.
Hydroxylamine adds a hydroxyl group to bases, causing changes in base-pairing that can lead to mutations.
Oxidative reactions, like those caused by superoxide radicals or hydrogen peroxide, can damage DNA by altering base structures, leading to mutations.
Intercalating agents like proflavin, acridine orange, and ethidium bromide insert themselves between DNA bases, causing frameshift mutations during DNA replication.
5-Bromouracil resembles thymine, but it has a bromine atom in place of the methyl group on the 5-carbon atom. It can cause mutations by incorrectly pairing with adenine instead of guanine during DNA replication.
2-Aminopurine resembles adenine but has an amino group at the 2-position. It can pair with thymine or cytosine, causing mutations by mispairing during DNA replication.
if 5-bromouracil is incorporated into DNA in place of thymine, it may mispair with guanine in the next round of replication, the next replication this G pairs with C leading to permamnent muation, if it pairs with A then no replicated error occurs.
basically incorporation of bromouracil followed by mispairing leads to a TA--> CG transtion mutation
transition mutation during DNA replication
CG--> TA
TA--> CG
Nitrous acid (HNO₂) causes deamination of cytosine, converting it into uracil. During DNA replication, uracil pairs with adenine instead of guanine, leading to a transition mutation (cytosine to thymine) after several rounds of replication.
CG-->TA
TA-->CG
hydroxylamine (NH₂OH) adds a hydroxyl group to cytosine, converting it into hydroxycytosine. This altered base mispairs with adenine instead of guanine, leading to a transition mutation (cytosine to thymine) during DNA replication. only CG-->TA way not vice versa
C-->T
G-->A
EMS is an alkylating agent that adds an ethyl group to normal nucleotides, resulting in mis-pairing during DNA replication. This causes point mutations, such as G→A or C→T transitions.
Oxidative radicals convert guanine into 8-oxy-7,8-dihydrodeoxyguanine, which can mispair with adenine during DNA replication, leading to G→T transversions.
Transition: A mutation where a purine (A or G) is replaced by another purine, or a pyrimidine (C or T) is replaced by another pyrimidine (e.g., A→G or C→T).
Transversion: A mutation where a purine is replaced by a pyrimidine, or a pyrimidine is replaced by a purine (e.g., A→C, G→T).
Intercalating agents insert themselves between adjacent bases in the DNA, distorting the helix's three-dimensional structure. This leads to insertions and deletions during DNA replication, causing frameshift mutations.
The Ames test uses his⁻ strains of bacteria to test chemicals for their ability to cause his⁻ to his⁺ mutations. Since mutagenic activity is closely linked to carcinogenic potential, the Ames test is widely used to screen chemicals for their cancer-causing ability.
the test uses his⁻ strains of bacteria (bacteria that cannot grow without histidine) that have been genetically modified to be sensitive to mutations.
The first step is to expose the his⁻ bacteria to the chemical being tested, along with a plate containing minimal histidine, which allows only bacteria with mutations that restore their ability to make histidine to grow.
If the chemical causes mutations, some his⁻ bacteria will mutate to his⁺, meaning they can now make histidine and grow. The number of his⁺ colonies indicates the mutagenic potential of the chemical.
A higher number of his⁺ colonies compared to a control (untreated) sample indicates that the chemical is mutagenic and may have carcinogenic potential.
Since mutagenic chemicals are often carcinogenic, the Ames test serves as a fast, preliminary screening method to identify chemicals that could potentially cause cancer.
EMS is an alkylating agent that induces point mutations in the genome. In forward genetic screens, it is used to create mutations in a population of organisms. Researchers then observe the phenotypes and use these to identify genes involved in specific biological processes or traits.
Forward genetic screens aim to identify genes that contribute to a specific phenotype or trait by inducing mutations and observing the resulting changes in the organism's traits.
After a genetic screen, researchers have a collection of new mutants, but they don’t know if the same genes have been mutated multiple times in different individuals.
Complementation tests are used to determine if mutations in different mutants affect the same gene. If two mutants with the same phenotype complement each other, they are in different genes. If they don’t complement, they are likely mutations in the same gene.
1. Forward Genetic Screen: Use mutagenesis (like EMS) to create mutations in a population of organisms, then observe how these mutations affect learning and memory behaviors.
2. Behavioral Assays: Test the organism for learning and memory tasks (e.g., maze learning, fear conditioning) to identify mutants with impaired learning.
3.Gene Identification: Once mutants are found, use genetic mapping or next-generation sequencing to identify the specific genes involved.
4.Complementation Tests: If multiple mutants share similar learning defects, perform complementation tests to determine whether mutations are in the same or different genes.
5.Gene Expression Analysis: Analyze the expression of candidate genes in the brain during learning-related tasks to confirm their involvement in memory processes.
1.Feed EMS to male flies and allow them to mate.
2.Take the progeny and establish heterozygous +/m stocks (~200 progeny).
3.Test homozygous m/m flies for learning ability.
4.Keep stocks where homozygotes show poor learning (e.g., 10% of stocks = ~20 lines).
5.Two extreme possibilities:
Extreme I: 20 mutations in 20 different genes.
Extreme II: 20 mutations in the same gene.
purpose: Determine if two mutations (e.g., m1, m2, ...) are alleles of the same gene.
Procedure:
1.Cross m1/+, m2/+, and m3/+ heterozygotes with wild-type alleles.
2.All heterozygotes (e.g., m1/+, m2/+, m3/+) show wild-type phenotypes (normal learners).
3.Homozygotes (m1/m1, m2/m2, m3/m3) show similar poor learning phenotypes.
If two mutations produce normal phenotypes when crossed, they are likely in different genes (complementation). If they don’t, they are likely mutations in the same gene.
1.Cross the parental m1/m1 and m2/m2 mutants.
2.F1 progeny: Test for learning ability.
3.If F1 progeny are still poor learners, this indicates that m1/m2 do not complement, meaning they are likely mutations in the same gene.
Conclusion: m1/m2 mutations fail to complement, suggesting they are alleles of one gene.
1.Cross the parental m1/m1 and m3/m3 mutants.
2.F1 progeny: Test for learning ability.
3.If F1 progeny are normal learners, this indicates that m1/+ and +/m3 complement each other.
Conclusion: Since the mutant phenotypes complement, m1 and m3 are mutations in two different genes, not alleles of the same gene.
1.Test crosses are done between multiple mutant lines (e.g., m1, m2, m3, ... m6).
2.The results show whether the mutations complement (normal learners) or fail to complement (poor learners).
3.Based on the cross results, you group mutants into complementation groups:
-Complementation group I: m1, m2, m5 (mutations in the same gene).
-Complementation group II: m3 (mutation in a different gene).
-Complementation group III: m4, m6 (mutations in another gene).
Conclusion: There are 3 complementation groups, indicating mutations in 3 different genes.
-Solomon Island blonde has TYRP1- and KITLG mutations.
-wedish blonde has TYRP1- and KITLG+.
The F1 would likely NOT have blonde hair because of the genetic interactions between TYRP1 and KITLG.
F1 offspring could have brown or black hair, depending on the influence of other genes involved in hair color.
Problem: m3 has mutations in both Gene A and Gene B, and both cause non-functional products.
Solution: Perform complementation tests to rescue one mutation at a time.
-Cross m3 with a strain that has a functional Gene A or Gene B.
-If the m3 mutation is rescued by a functional copy of Gene A, the poor learning phenotype is likely due to the Gene B mutation.
-If m3 is rescued by a functional copy of Gene B, the poor learning phenotype is likely due to the Gene A mutation.
Conclusion: This allows you to identify which gene mutation is responsible for the poor learning phenotype.
1.Introduce a wild-type copy of Gene A into the m3 fly strain.
2.Test transgenic flies for learning ability:
-If m3 flies become normal learners, the mutation in Gene A is responsible for poor learning.
-If m3 flies are still poor learners, the mutation in Gene A is NOT responsible.
3.Next, perform a rescue complementation experiment with a wild-type copy of Gene B to test if Gene B is responsible for the poor learning phenotype.
4.The gene responsible for poor learning is identified based on which rescue experiment restores normal learning.
if there are staggered cuts in the target dna, transposable element can insert itself into the dna, the staggered cuts leave short single stranded pieces of DNA, the replication of this single stranded DNA creates the flanking direct repeats
Transposable elements are parasitic DNA sequences that can move within the genome, often causing mutations.
Transposition is the movement of transposable elements from one location to another within the genome.
Flanking direct repeats: Short sequences that are repeated on both sides of the transposon.
Terminal inverted repeats: Sequences that are reversed and complementary at the ends of the transposon.
Transposition can occur through DNA or an RNA intermediate.
In replicative transposition, a new copy of the transposable element inserts into a new location, while the old copy remains at the original site.
In nonreplicative transposition, the old copy excises from its original site and moves to a new site, leaving no copy behind.
RNA intermediate transposition requires reverse transcription to convert the RNA back into DNA before integrating into the target site.
DNA or an RNA intermediate.
A new copy of the transposable element inserts into a new location, and the old copy remains at the original site.
The old copy excises from its original site and moves to a new site, leaving no copy behind.
RNA intermediate transposition requires reverse transcription to convert the RNA into DNA before it integrates into the target site.
About 45% of the human genome comprises sequences related to transposable elements, mostly retrotransposons.
Transposons cause mutations by:
Inserting into another gene, disrupting its function.
Promoting chromosomal rearrangements (e.g., inversions, deletions).
Moving genes (deleting and inserting) as they move within the genome.
pairing by looping and crossing over between two transposable elements oriented in the same direction leads to deletion
pairing by bending and crossing over betwen two transposable elements oreinted opposite direction leads to inversion
Transposons can cause chromosomal rearrangements by:
Inserting at new locations, disrupting the normal structure of chromosomes.
Promoting inversions, deletions, or duplications in the genome as they move.
misalignment and uneqal exchnage between transposable elements located on sister chromatids
Insertion sequences (e.g., IS1) are simple transposable elements found in bacteria. They consist of:
Short DNA sequences that can move around within the genome.
transposase gene, then Typically include 23bp terminal inverted repeats and 9bp flanking direct repeats.
They do not carry additional genes, only the genes required for transposition.
Tn10 has the following structure:
Two IS elements (IS10) at the ends: These are insertion sequences that mediate transposition.
Central region: Contains genes, such as antibiotic resistance genes (e.g., tetracycline resistance).
The IS10 elements are identical or nearly identical, enabling the movement of the central region between different genomic locations.
flanking direct repeat seuqences on the outside
Barbara McClintock was the first to discover transposable elements in the 1940s while studying maize (corn). She identified that certain genes could move around within the genome, causing changes in the expression of other genes.
Ac (Activator) and Ds (Dissociation) are transposable elements in maize discovered by Barbara McClintock.
Ac is an autonomous element, meaning it can move on its own because it encodes the necessary transposase enzyme.
Ds is a non-autonomous element, meaning it cannot move on its own but relies on Ac for transposition.
Transposition causes variegated maize kernels by disrupting the C (color) gene in maize.
Ac element produces transposase which stimulates transposition of a Ds element in the C allele. this disrupts its pigment producing function, resulting cells have genotype Ctc and are colorless
Variegation occurs because some kernels have the C gene active (producing color), while others have the gene disrupted by the transposon, leading to a lighter color or no color at all.
an Ac element produces transposase which stimulates further transpositon of the DS element in some cells. As DS transposases, it leaves the C alle restoring alleles function. A cell in which Ds has transposed out of the C allele will produce pigment generating spots of color in an otherwise colorless kernel
Long terminal direct repeats and short flanking direct repeats at the target site.
Contains a reverse-transcriptase gene (and sometimes other genes).
Transposes through an RNA intermediate.
Examples: Ty (yeast), copia (Drosophila), Alu (human).
Short terminal inverted repeats and short flanking direct repeats at the target site.
Contains a transposase gene (and sometimes other genes).
Transposes through DNA (either replicative or nonreplicative).
Examples: IS1 (E. coli), Tn3 (E. coli), Ac and Ds (corn), P elements (Drosophila).
Alleles are different forms of a gene that exist at a specific locus (location) on a chromosome.
A wild type allele is the most commonly found allele in a population and typically represents the normal or standard version of a gene.
A variant or mutant allele is different from the wild type allele. It may:
Not adversely affect the gene product.
Not result in a detectable phenotype (physical expression).
In certain populations, the most common allele for color (such as in flowers) is purple, which is considered the wild type because it is the most frequently observed color.
An organism is homozygous for an allele if it has identical alleles for that gene on both homologous chromosomes.
An organism is heterozygous for an allele if it has one wild type allele and one mutant allele for a given gene.
An allelic series or multiple alleles refers to the set of known mutant alleles for a gene plus its wild type allele.
A wild type allele is the most common allele in a population and is considered the functionally normal version of the gene.
Homozygotes are cells/organisms that have identical alleles for a gene of interest.
Example: FC/FC or fc/fc, where both alleles are the same on homologous chromosomes.
Heterozygotes are cells/organisms that have one wild type allele and one mutant allele for a gene of interest.
Example: FC/fc, where one allele is wild type (FC) and the other is a mutant (fc).
Hemizygous refers to a situation where a cell/organism has only one copy of a gene, locus, or chromosomal region.
In a deletion scenario, the corresponding gene/locus/region is deleted on the homologous chromosome, leaving only one copy of the gene in the organism.
In XY individuals (males), most genes on the X or Y chromosomes are hemizygous, meaning they only have one copy of the gene because the other chromosome is missing a second copy (i.e., males have one X chromosome and one Y chromosome).
A dominant allele is one that, in heterozygous individuals (with one copy of the allele), masks the effect of the other allele. The phenotype of these individuals is the same as those that are homozygous for the dominant allele.
A recessive allele only shows its phenotype when an individual is homozygous for that allele. In heterozygotes, the recessive allele's effect is masked by the dominant allele.
In complete dominance, one allele completely masks the effect of the other allele in heterozygotes. The phenotype of a heterozygote is indistinguishable from a homozygous dominant individual.
Dominant alleles are typically represented by uppercase letters (e.g., FC), and recessive alleles are represented by lowercase letters (e.g., fc).
The FC allele (dominant) causes a purple flower color, and the fc allele (recessive) causes a different color. Because FC/fc and FC/FC both result in purple flowers, the FC allele is dominant over the fc allele.
Heteroallelic refers to an individual that has two different mutant alleles for the same gene, one on each chromosome.
In a heteroallelic organism, the alleles are different mutant alleles for a given gene. For example, fc1 and fc2 are two different mutant alleles of the Flower Colour gene, resulting in a heteroallelic combination of fc1/fc2.
An example of a heteroallelic individual would be one with fc1/fc2 alleles for the Flower Colour gene, where fc1 and fc2 are both mutant alleles different from the wild type FC allele.
Incomplete dominance occurs when the alleles of a gene do not exhibit a simple dominance/recessive relationship. Instead, the heterozygote shows an intermediate (or blended) phenotype between the two homozygous parents.
An example of incomplete dominance is the flower color in Four-o'clock plants:
CRed/CRed = Red petals (homozygous)
CWhite/CWhite = White petals (homozygous)
CRed/CWhite = Pink petals (heterozygous, intermediate phenotype).
In incomplete dominance, heterozygotes show an intermediate or blended phenotype between the two homozygous parents. For example, in the Four-o'clock plants, the heterozygote CRed/CWhite has pink petals, which is a blend of red and white.
In incomplete dominance, heterozygotes have an intermediate phenotype, while homozygotes express the full trait of one allele. For example, CRed/CRed produces red petals, while CWhite/CWhite produces white petals, but CRed/CWhite produces pink petals.
Codominance occurs when heterozygotes express the phenotypes of both alleles simultaneously, rather than showing an intermediate phenotype. Both traits are fully expressed.