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.
An example of codominance is the ABO blood group system:
Blood group A has A antigens on red blood cells (RBC).
Blood group B has B antigens on RBC.
Blood group AB has both A and B antigens on RBC (codominance).
Blood group O has no antigens on RBC.
In the heterozygous genotype A/B (blood group AB), both A and B antigens are expressed on the red blood cells, showing codominance.
A and B alleles are dominant over O, meaning:
A/O = Blood group A (expresses A antigens).
B/O = Blood group B (expresses B antigens).
A/B = Blood group AB (expresses both A and B antigens).
Dominance and recessiveness describe the relationship between alleles but do not provide information about the type of defect that has occurred in the gene. Thus, these terms are less useful when considering the nature of the mutation itself.
A more useful system considers the type of defect in the gene. This system classifies mutant alleles based on their morph (form) or functional impact, rather than just their dominance or recessiveness.
H.J. Muller contributed significantly to radiation genetics and the classification of mutant alleles. He was awarded the Nobel Prize in 1946 for his work in this field.
H.J. Muller designated five classes of mutant alleles as "morphs," meaning "form." These classifications are still used today, with some modern equivalents also being employed.
: A Class 1: Amorph (also known as a null mutation) refers to the complete loss of function of a gene. It can result from mutations that entirely abolish the gene's function.
Examples of mutations that result in amorphs include:
Complete deletion of the entire gene.
A missense point mutation that abolishes all functions of the protein.
A nonsense point mutation that creates a truncated, non-functional protein.
No, the gene does not need to be deleted to be considered an amorph. The classification is based on the phenotype (i.e., complete loss of function), not the type of molecular lesion. A missense mutation that completely loses the function of the gene can still result in an amorph.
An example of an amorph allele is the fc- allele in flower color. Homozygous fc-/fc- plants have completely white flowers, indicating that the allele has lost the ability to make purple pigment, which is a complete loss of function.
A Class 2: Hypomorph is an allele that is partially functional but not as active as the wild-type allele. It represents a reduced function of the gene.
The modern equivalent of a hypomorph is a "partial loss of function" or "leaky" mutation, since the gene still functions to some degree but not fully like the wild-type allele.
Examples of mutations that can cause hypomorphs include:
A mutation in a regulatory element that reduces gene expression.
A point mutation that reduces the activity of the gene product but does not eliminate it entirely.
The key feature of a hypomorph is that it still retains some functional activity, whereas an amorph allele results in a complete loss of function.
Myostatin is a repressor of muscle growth, meaning it normally limits the production of muscle tissue.
A loss-of-function mutation in myostatin upregulates muscle production, leading to increased muscle growth.
In cattle, a stop mutation in the myostatin gene leads to a complete loss of function (amorph), resulting in increased muscle growth.
In cattle, a point mutation in the non-coding region of the myostatin mRNA causes a hypomorph allele, reducing the expression of myostatin and promoting more muscle growth.
In sheep, a mutation in the non-coding part of the myostatin mRNA creates a new function by attracting a repressor molecule, which adds a new property to the mRNA, making it a gain-of-function mutation.
In cattle, mutations in myostatin typically cause loss-of-function (amorph or hypomorph), leading to more muscle growth. In sheep, a mutation in the non-coding region of the myostatin mRNA results in a gain-of-function, giving the mRNA a new property that affects muscle growth regulation.
A neomorph allele is an allele that is active but has acquired a new or novel function that the wild-type gene does not have.
The modern equivalent of a neomorph is a "gain-of-function" mutation. However, this term can also refer to hypermorphs, making it imprecise in some contexts.
An example of a neomorph is a mutation in the active center of an enzyme that changes its substrate specificity, giving it a new or altered function compared to the wild-type enzyme.
A neomorph mutation in a regulatory region can lead to incorrect activation of a gene in the wrong tissues or at the wrong times, causing misexpression of the gene.
In Burkitt lymphoma, a neomorph mutation in the regulatory region activates the c-myc gene, leading to uncontrolled cell growth, a key feature of the cancer.
A neomorph can result from translocations where breakpoints in genes create new hybrid genes, such as the ABL-BCR fusion gene, which causes chronic myelogenous leukemia (CML).
Oncogenes are neomorphic alleles of cancer genes that cause uncontrolled cell growth and contribute to the development of cancer.
Neomorphic alleles of oncogenes often result from mutations in the regulatory regions or structural genes, leading to new functions or misexpression that disrupt normal cell regulation and promote cancer development.
The MYC gene codes for a transcription factor that regulates cell growth and proliferation. Mutations or misregulation of MYC can lead to uncontrolled cell division, a hallmark of cancer.
: Immunoglobulin (IG) genes, typically produced by B lymphocytes, can be involved in oncogene formation, such as in Burkitt lymphoma, where translocations cause the MYC gene to be regulated by the IG gene promoter, leading to inappropriate expression and uncontrolled growth.
Translocations can bring the MYC gene under the control of the immunoglobulin (IG) gene's regulatory region, resulting in the overexpression of MYC, leading to cancer such as Burkitt lymphoma.
An antimorph is an allele that not only acts as a dominant negative but also overrides the function of the wild-type allele in a heterozygous setting. It’s different from normal dominance because it disrupts the normal gene function rather than promoting it
The modern equivalent of an antimorph is a "dominant negative" allele, which interferes with the normal function of the wild-type allele, causing a loss of function in a heterozygous state.
In a heterozygous setting, antimorphs can override the function of the wild-type allele. This happens because the antimorphic allele produces a non-functional protein that disrupts the function of the wild-type protein.
A common example of antimorphs is in protein dimers, where a mutated allele causes the protein to form non-functional dimers with the wild-type protein, rendering the entire complex dysfunctional. For example, in the case of a helicase, a mutant dnaB allele may prevent the helicase hexamer from functioning properly by forming defective complexes.
When an antimorphic allele is present, it can interfere with the function of the entire protein complex by creating non-functional dimers, which disrupt the activity of the wild-type proteins, even in heterozygotes.
The capital "M" represents a dominant mutant allele, where the mutated version of the allele can override or affect the function of the wild-type allele.
When the wild-type protein and mutant protein bind together, they form a non-functional dimer, which prevents the normal protein complex from functioning properly. This dominance is due to the mutant allele's ability to disrupt the wild-type protein's function.
The dominant mutant allele can cause a loss of function in the wild-type allele by forming a non-functional dimer. As a result, the normal function of the wild-type protein is disrupted, even in a heterozygous individual.
In heterozygotes carrying a mutant allele (M) and wild-type allele (+), the mutant allele can disrupt the normal protein function by interacting with the wild-type protein, leading to a non-functional dimer and affecting the phenotype.
The mutant allele is considered dominant because it interferes with the normal function of the wild-type allele, resulting in a dominant negative effect. Even a single copy of the mutant allele can cause a loss of function in the entire protein complex.
Heredity is the process by which traits or characteristics are passed from parents to offspring through genetic inheritance.
Gregor Mendel was an Augustinian priest, teacher, and the founder of the principles of heredity. He conducted groundbreaking genetic experiments with pea plants.
Mendel had a broad science education in mathematics, botany, and plant physiology, which helped him apply the scientific method in his genetic experiments.
Mendel performed breeding experiments with pea plants, studying how traits were inherited across generations.
Mendel discovered the fundamental principles of heredity—the laws of inheritance—without knowledge of meiosis, chromosomes, or the concept of genes.
Mendel’s work remained unrecognized until the 1900s, long after his death.
Pea plants are a good model organism because they have a relatively fast generation time (~1 year), produce many offspring, and have many varieties with diverse characteristics, which allows for mathematical calculations in genetic studies.
A characteristic is an attribute or feature possessed by an organism, while a trait (or phenotype) is the appearance or manifestation of that characteristic.
Mendel focused on 7 key characteristics that had distinct and easily discernible traits/phenotypes.
Mendel focused on characteristics that had binary traits (traits with only two possible outcomes), rather than those with a large range of variation.
Mendel's success was partly due to his use of the scientific method. He made new hypotheses and tested them, rather than just describing results as others had done.
In genetics, "cross" refers to the mating of two organisms to produce progeny, typically denoted by an "X".
Unlike earlier investigators who merely described experimental results, Mendel used the scientific method, making hypotheses and testing them through controlled experiments.
How are traits passed on to offspring?
"Cross" refers to the mating of two pea plants to produce offspring (progeny), used to study inheritance patterns.
When yellow and green pea plants are crossed, all offspring are yellow because yellow is the dominant trait.
When pea plants with blended traits are crossed, all offspring show yellow-green as a blended phenotype.
Pea plants are hermaphroditic, meaning they have both male (anthers) and female (stigma) sex organs.
The anthers release pollen (male gametes), and the stigma is where the female gametes (eggs) are located. These organs work together for fertilization.
: Pea plants self-fertilize when pollen released from the anther enters the stigma and fertilizes the female gametes, resulting in the formation of a diploid zygote.
To avoid self-fertilization, remove the anthers from one plant before they are fully developed, ensuring no pollen is produced. This plant becomes the female in the cross.
To perform the cross, move pollen from the second plant's anthers (the male in the cross) to the stigma of the first plant (the female in the cross).
The female plant produces seed pods, and the seeds within these pods are the progeny from the cross.
Pure breeding means that all progeny from multiple generations have the same trait. For example, Mendel’s pure breeding strains were homozygous for a trait, like yellow seed color, and would produce offspring with the same trait.
Mendel used self-fertilization for multiple generations to ensure the traits were homozygous, thus creating pure breeding strains for his experiments.
The F1 generation will have all yellow seeds, and when the F1 generation self-fertilizes to create the F2 generation, both yellow and green seeds will appear in a 3:1 ratio.
A monohybrid cross is a cross between parents that differ in only one trait. For example, Mendel’s cross between round and wrinkled peas.
In the F1 generation, all the progeny had round seeds. However, in the F2 generation, the wrinkled trait re-emerged. This suggests that the F1 generation carried genetic information for both round and wrinkled seeds.
In the F3 generation, we see a 3:1 ratio of round to wrinkled peas, showing that the round trait is dominant and the wrinkled trait is recessive, even though both traits are carried by the F1 generation.
They show that traits are encoded by alleles, and some alleles can be dominant over others. Mendel observed that dominant traits, like round seeds, mask the expression of recessive traits, like wrinkled seeds.
F2 round parents produce 1/3 plants with only round peas (like the pure-breeding round parent) and 2/3 plants showing both round and wrinkled peas in a 3:1 ratio (like F1 generation). This shows that the round trait is dominant.
: F2 wrinkled parents produce only wrinkled peas, which is the same as the pure-breeding wrinkled parent, indicating the wrinkled trait is recessive and requires two copies to be expressed.
Each plant must possess two genetic factors (alleles) that encode a characteristic, and the re-emergence of the wrinkled trait in the F2 generation is explained by the presence of two copies of the recessive allele in the F2 generation.
Monohybrid crosses show that traits are encoded by two alleles. One allele may mask the other, as the dominant allele (e.g., round) hides the presence of the recessive allele (e.g., wrinkled). During reproduction, alleles separate, with one allele going into each gamete, and fertilization brings two alleles together, explaining why traits like wrinkled can re-emerge in later generations (F2/F3).
R (capital) = Dominant round trait
r (lowercase) = Recessive wrinkled trait
RR = Homozygous round (pure-breeding)
Rr = Heterozygous round (carrying both round and wrinkled alleles)
rr = Homozygous wrinkled (pure-breeding)
Mendel observed a 3:1 ratio of dominant to recessive phenotypes in the F2 generation.
From this, he deduced that each individual plant carries two alleles for a given trait, and these alleles separate with equal probability into the gametes during reproduction.
During gamete formation, two alleles of a gene segregate with equal probability, meaning each gamete receives one allele of the gene.
When two gametes join during fertilization, the resulting diploid organism inherits one allele from each parent, forming a pair of alleles for a characteristic.
The segregation of alleles occurs during Meiosis I when homologous chromosomes, each carrying different alleles, are separated into different gametes.
Walter Sutton proposed the Chromosome Theory of Heredity, which states that genes are located on chromosomes, and Mendel's laws of inheritance are explained by the behavior of chromosomes during meiosis.
Homologous chromosomes are chromosome pairs, one from each parent, that carry the same genes but possibly different alleles. They are important because they segregate into different gametes during meiosis, explaining the inheritance of alleles.
A backcross involves crossing the F1 progeny back to one of the parental genotypes to test the inheritance pattern of traits.
Mendel hypothesized that tall (T) was dominant to short (t), so F1 progeny from a cross between pure-breeding tall and short plants would all be heterozygous (Tt).
Mendel tested his hypothesis by performing a backcross: crossing F1 progeny (Tt) with a short (tt) pure-breeding plant.
Genotypic ratio: 2 Tt : 2 tt (1:1)
Phenotypic ratio: 2 Tall : 2 Short (1:1)
If tall is dominant to short, the progeny should show a 1:1 phenotypic ratio, with half of the offspring being tall (Tt) and half being short (tt), supporting Mendel's hypothesis of tall dominance
A testcross is a genetic cross between an individual with an unknown genotype and a homozygous recessive individual to determine the genotype of the unknown individual.
The purpose of a testcross is to determine whether an individual with a dominant phenotype is homozygous (TT) or heterozygous (Tt) for the trait by analyzing the offspring's phenotypes.
To perform a testcross, you cross the individual with the dominant phenotype (unknown genotype) with a homozygous recessive individual (tt).
The phenotypic ratio for the offspring would be 1:1:
50% Tall (Tt)
50% Short (tt)
The phenotypic ratio would be 100% Tall (Tt) because all the offspring would inherit a T allele from the dominant parent and a t allele from the recessive parent.
A characteristic refers to a feature or attribute of an organism that can vary, such as seed shape, color, or size. (gene)
A trait (or phenotype) is the visible expression or manifestation of a characteristic. For example, the trait for seed shape could be round or wrinkled.
Alleles segregate independently, with each gamete receiving one allele with equal probability.
The probability of a gamete receiving a specific allele is 1/2.
Probability = # of times a specific outcome occurs / # of possible outcomes
Understanding probability allows you to predict more complicated genetic scenarios that a Punnett square cannot easily handle.
Use the multiplication rule when the question involves “and” (e.g., What is the probability of rolling a 4 and another 4?).
Formula: Probability = P(event 1) * P(event 2)
For rolling two dice and getting a 4 and another 4:
Probability = 1/2 * 1/2 = 1/32
Use the addition rule when the question involves “or” (e.g., What is the probability of rolling a 3 or a 4?).
Formula: Probability = P(event 1) + P(event 2)
For rolling a dice and getting a 3 or a 4:
Probability = 1/2 + 1/2 = 1/4
P(RR) = Probability progeny will be homozygous RR (R "and" R).
Using the multiplication rule:
P(RR) = P(gamete R) x P(gamete R) = 1/2 x 1/2 = 1/4
Phenotype: 3/4 round, 1/4 wrinkled (3:1 ratio)
Genotype: 1/4 RR, 1/4 rr, 1/2 Rr (1:1:2 ratio)
Use Punnett square to confirm these results.
P(rr) = Probability progeny will be homozygous rr.
Using the multiplication rule:
P(rr) = P(gamete r) x P(gamete r) = 1/2 x 1/2 = 1/4
There are two possible ways to achieve heterozygote genotype: Rr or rR.
Using the addition rule:
P(Rr) = P(R "and" r) + P(r "and" R)
P(Rr) = (1/2 x 1/2) + (1/2 x 1/2) = 1/4 + 1/4 = 1/2
Round phenotype can come from combinations of R and r alleles.
Probability calculation:
P(round) = P(R "and" r) + P(r "and" R) + P(R "and" R)
P(round) = (1/2 x 1/2) + (1/2 x 1/2) + (1/2 x 1/2) = 3/4
Wrinkled phenotype comes from the homozygous rr genotype.
Probability calculation:
P(wrinkled) = P(r "and" r) = (1/2 x 1/2) = 1/4
Principle of Segregation (Mendel’s First Law)
Concept of Dominance
A dihybrid cross is a genetic cross that involves the inheritance of two separate characteristics (e.g., seed shape and seed color).
Seed Shape: Round (R) is dominant, and wrinkled (r) is recessive.
Seed Color: Yellow (Y) is dominant, and green (y) is recessive.
The binomial probability formula is:
𝑃( k successes) = (𝑛/𝑘)⋅𝑝^𝑘⋅(1−𝑝)^𝑛−𝑘
Where:
𝑛 is the total number of trials,
𝑘 is the number of successes,
𝑝 is the probability of success,
(𝑛/k) is the binomial coefficient.
We are considering 3 trials (3 progeny).
binomial coefficent is 3
P92 rr out of 3 porgeny) = 3 x (1/4)^2 X (3/4), the final [roabbility of wrinkled seed is 9/64
A dihybrid cross involves the inheritance of two traits, each controlled by different genes located on different chromosomes.
Answer: Each parent will produce 4 different gametes (since there are 2 genes with 2 alleles each: 2 x 2 = 4).
he expected phenotypic ratio is:
9 Round, Yellow (double dominant)
3 Round, Green
3 Wrinkled, Yellow
1 Wrinkled, Green (double recessive)
Mendel’s Second Law states that genes at different loci (on different chromosomes) separate independently of one another during meiosis.
The Principle of Independent Assortment is true for genes located on different chromosomes or genes that are far apart on the same chromosome.
The Principle of Segregation states that alleles of a gene (e.g., R, r) separate independently during meiosis, specifically during the separation of homologous chromosomes.
The Principle of Independent Assortment explains the independent separation of non-homologous chromosomes (and the genes they carry, like R vs Y) during meiosis.
The Principle of Segregation is related to the separation of homologous chromosomes during meiosis, where alleles (e.g., R, r) for a gene segregate into separate gametes.
The probability of obtaining a round and yellow progeny is the product of the probabilities of inheriting round and yellow alleles independently.
Formula:
P(round, yellow) = P(round) × P(yellow)
Because of Mendel's Second Law (Principle of Independent Assortment), these events are independent.
The Binomial Expansion is used to calculate the probability of specific outcomes in situations with multiple progeny or events (e.g., a certain number of round vs. wrinkled progeny in a large cross). It helps solve probability questions with multiple outcomes by expanding the binomial expression.
p = probability of a specific outcome (e.g., progeny being wrinkled = 1/4)
q = probability of the opposite outcome (e.g., progeny being round = 3/4)
n = number of progeny (or number of trials)
s = number of times p occurs (e.g., number of wrinkled progeny)
t = number of times q occurs (e.g., number of round progeny)
Binomial expansion:
Probability= (n!/s!⋅t!) ⋅p ^s ⋅q ^t
Where:
n! = factorial of n
s! and t! = factorials of s and t
p = probability of one event
q = probability of the other event
The factorial n! represents the product of all positive integers up to n. It is used to account for all possible arrangements of events or outcomes.
Example: 5! = 5 × 4 × 3 × 2 × 1 = 120.
The Chi-square goodness of fit test is used to determine whether the observed ratios of progeny differ significantly from the expected ratios, and whether the differences are due to random chance or some other biological mechanism.
The Chi-square test provides a way to evaluate the likelihood that the observed differences between the expected and actual data are due to chance. It helps determine if there is a significant deviation between the observed and expected results.
The null hypothesis assumes that there is no significant difference between the observed and expected data, meaning that any difference is due to random chance.
χ 2=∑ (O−E) ^2/E
o determine statistical significance, compare the calculated Chi-square value to a critical value from the Chi-square distribution table, based on the degrees of freedom and the desired significance level (e.g., 0.05).
If the calculated χ² is greater than the critical value, the difference is significant, and the null hypothesis is rejected.
A low Chi-square value suggests that the observed and expected values are similar, and the differences could likely be due to chance, supporting the null hypothesis.