② Allelic character: Characters that can be clearly contrasted with each other among traits (e.g., The shape is more clearly defined as round in contrast to wrinkled.)

③ Trait (phenotype): Type of character represented by an individual (e.g., Purple, Red)

○ Wild type (wt), mutant (mt), dominant trait, recessive trait

④ Genotype: Type of gene an individual has.

○ Homozygote: The case that both allele are the same (e.g., AA, BB, OO)

○ Heterozygote: The case that two alleles are different (e.g., AB, AO)

⑤ Recessive trait: Phenotypes of allele only in homozygotes (e.g., OO)

⑥ Dominant trait: Phenotypes that also appear in heterozygotes (e.g., AA, AO; BO, BB)

⑦ Carrier: Heterozygote for recessive disease

⑧ Mating

○ Unisexual mating (e.g., Aa), bisexual mating (e.g., AaBb)

○ Self-pollination (↔ Cross pollination): Breeding between individuals with the same genotype, often used for improving a breed.

○ Testcross: By mating an individual with a homozygous recessive individual, the genotype can be determined.

○ Reciprocal cross: Breeding a maternal line with the father’s genotype and a paternal line with the mother’s genotype. This can determine whether there is sex-linked inheritance.

○ p generation, F₁ generation (hybrid first generation), F₂ generation (hybrid second generation)

⑵ Mendel’s law

① Mendelian traits: Traits for which the pattern of inheritance can be easily determined, as in Mendel’s experiments.

② Plant hybrid research (1860s): Study on the inheritance of seven allelic traits in peas, with clear dominance and recessiveness.

○ Shell of peas: Round gray, Wrinkled white

○ An allele encodes an enzyme called SBE (starch branching enzyme), which is related to amylopectin.

○ Wrinkled peas have higher osmotic pressure: they become wrinkled when they absorb water and swell, then dry.

○ Color of peas: Yellow, Green

○ Color of flowers: White, Purple (Red)

○ Shape of pods: Smooth pods, Creased pods

○ Color of pods: Yellow, Green

○ The place where flowers bloom: Between the branches, or at the top.

○ Stem size: Big, Small

③ Advantages of peas

○ Peas have a short generation (3 months).

○ Pea leaves a lot of offspring.

○ Pea is easy to grow and easy to control reproduction.

○ Peas are capable of both self-fertilization and cross-fertilization.

○ Pea traits and dominance/recessiveness are very clear.

○ Although the alleles are linked and positioned far apart, they seem to adhere to the law of independence.

④ Highlights the design of experiments and the importance of mathematical thinking (statistics) of results.

○ Mendel conducts testcross to determine if an individual is true-breeding.

⑤ Mendel’s law

○ Several types of allele are used to determine traits.

○ The Law of Dominance: When pure lines with opposing traits are crossed, only one of the parent’s traits appears in the F₁ generation.

○ This means that in a heterozygous individual, only one allele’s trait is expressed.

○ The trait that appears in heterozygotes is dominant, while the other is recessive.

○ This contrasts with the blending inheritance hypothesis.

○ The Law of Segregation: Since reproductive cells contain half the genetic material, only one of the two alleles is passed on.

○ An individual inherits only two alleles, one from each parent.

○ The forked line method can be applied.

○ The Law of Independence: Genes that are not in an allelic relationship are assumed to assort independently during gamete formation. Doesn’t consider linkage along the same chromosome.

○ A 9:3:3:1 ratio in the case of A_B_ : aaB_ : A_bb : aabb suggests independence.

⑶ Genotyping via Punnett square

① Example: Round yellow and wrinkled green

Figure 2. How to draw a Punnett square in Mendel pea experiment

⑷ Pedigree Analysis

Figure 3. Example of pedigree

⑸ Complementary Test (Cis-Trans Test)

① Definition: It is a test to determine whether certain genes related to a specific mutant trait are the same gene or different genes.

② Example: In the example below, there are a total of three mutant genes: Gene 1 (= Gene 3 = Gene 5), Gene 2, and Gene 4.

Figure 4. Example of complementary test

⑹ Hereditary diseases

① Achondroplasia

3. Linkage and Cross

⑴ Drosophila research

① Theoretical formulation of the linkage is made by studying fruit flies, starting with Morgan.

② Benefits of Drosophila as a Genetic Study

○ Readily available.

○ Breeding is easy and does not harm person.

○ Breeds a lot and breeds every 2 weeks.

○ Small number of chromosomes: 4 pairs of chromosomes.

⑵ Linkage

① Two or more non-alleles exist on the same chromosome and behave identically.

② Man has 23 linkage groups.

③ Cis-configuration and Trans-configuration

○ Cis-configuration (coupling): Individuals with AB and ab chromosomes

○ Trans-configuration (repulsion): Individuals with Ab and aB Chromosomes

○ When doing cross experiments, you should first determine whether you are based on cis-configuration or trans-configuration.

○ Case 1. Cis-configuration AaBb × Cis-configuration AaBb

○ The probability that the offspring have the same genotype as the parents is 1/2: (AB | ab) → (AB | ab)

○ There are 2 types of phenotypes in the offspring: A_B_, aabb

○ Case 2. Trans-configuration AaBb × Trans-configuration AaBb

○ The probability that the offspring have the same genotype as the parents is 1/2: (Ab | aB) → (Ab | aB)

○ There are 3 types of phenotypes in the offspring: A_B_, A_bb, aaB_

○ Case 3. Cis-configuration AaBb × Trans-configuration AaBb

○ There are 3 types of phenotypes in the offspring: A_B_, A_bb, aaB_

○ In Cases 1 to 3, A is dominant over a, and B is dominant over b.

○ Example problem

④ Complete linkage and incomplete linkage

○ Complete linkage: Linkage that produces only parent type. If two genes are very close and have fewer offspring.

○ In other words, when there is 100% genetic linkage.

○ A_B_ : aaB_ : A_bb : aabb = 3:0:0:1 means cis-configuration complete linkage.

○ A_B_ : aaB_ : A_bb : aabb = 2:1:1:0 means trans-configuration complete linkage.

○ Incomplete linkage: Produces both parental and recombinant (crossover) traits.

⑶ Crossover

① Exchange between sister chromatid segments occurs when bivalent chromosomes (tetrad) are formed during the prophase of the first meiotic division.

○ First-Division Segregation (FDS): When there is no crossing over

○ Second-Division Segregation (SDS): When there is crossing over

○ Chiasma: The location where crossing over occurs. The number of chiasmata is proportional to chromosome length.

② The recombination rate is proportional to the distance between two linked genes.

③ Formula 1: Crossover rate = (Number of gametes where crossing over occurs ÷ Total number of gametes) × 100

④ Formula 2: Crossover rate = (Number of offspring resulting from crossing over in a test cross ÷ Total number of offspring obtained in a test cross) × 100

⑤ Multiple crossovers: The probability of multiple crossovers is theoretically equal to the product of the probabilities of each individual single crossover.

○ 1-2 crossover events typically occur for each chromosomal pair in each gamete.

⑥ The crossover rate is a maximum of 50%: When the distance between two genes is very large, they satisfy the law of independence, and in the case of independence, the crossover rate is 50%.

⑦ Issues with calculating crossover rate

○ If an even number of crossovers occur between two linked genes, they do not contribute to the crossover rate.

○ Interference: One crossover event can inhibit (negative interference, most commonly) or promote (positive interference) another crossover event.

○ As the distance between two genes increases, the crossover rate becomes less accurate.

○ In most cases, an equal and equivalent amount of DNA is exchanged. However, unequal crossover can result in copy number variants (CNVs) as well as inversions and translocations.

○ Crossover is non-random; over 25,000 small regions of elevated recombination have been characterized as recombination hotspots.

⑷ Gene map

① A plot of the location of genes on a chromosome based on crossover rates obtained by three-point test crosses.

○ Test cross: A cross between an individual that is heterozygous for all genes and a homozygous recessive individual.

○ 1^st. Group the 8 organisms into pairs.

○ 2^nd. The majority of pairs will display the parental type.

○ 3^rd. At least one pair of individuals contains the gene located in the middle on the map: Using the double crossover rate

○ 4^th. Calculate the crossover rate between the intermediate gene and each of the other genes to estimate map distance of genes.

② 1 cM (centi-Morgan): A map distance where the crossover rate is 1%.

③ Double crossovers

○ For values below 25 cM, there is an almost proportional relationship between recombination frequency and map distance.

○ For values above 25 cM, the crossover rate appears lower compared to the map distance.

○ Reason: This is due to the occurrence of double crossovers. In each gamete, typically 1–2 crossovers occur per chromosome.

○ Application: The crossover rate between the two most distant genes is smaller than the sum of the distances between the respective intermediate genes.

4. Non-mendel Genetics

⑴ Co-dominance: Both alleles are expressed as phenotypes.

① MN blood type

② ABO blood type: I^A and I^B are co-dominant. i is recessive.

Figure 5. ABO blood type inheritance

○ The gene encoding the glycoprotein on the surface of red blood cells located on chromosome 19.

○ AB type: I^AI^B

○ A type: I^AI^A or I^Ai

○ B type: I^BI^B or I^Bi

○ O type: ii

⑵ Incomplete dominance (intermediate inheritance)

① A complete phenotype is expressed only in homozygotes, while the phenotype of heterozygotes is intermediate.

② Quantitative inheritance: Usually pigments, receptors (e.g., Inheritance of hypercholesterolemia and LDL receptors)

③ Flower color of snapdragon: Rr × Rr

○ RR : Rr : rr = Red : Pink : White = 1:2:1

○ Tip: The inheritance of color is not a common dominant but an incomplete dominant.

⑶ Multiple allele inheritance: A case where more than three alleles exist at a single gene locus, often originating from mutations.

① The ABO blood type gene is determined by three alleles (multiple alleles).

⑷ Pleiotropy: The effect of a single allele influencing two or more phenotypes.

① Cystic fibrosis (CF): Recessive genetic disease

○ 1^st. A defect in the CFTR gene on chromosome 7.

○ 2^nd. Mutations occur in channels that transport chlorine ions out of epithelial cells.

○ 3^rd. Excessive chloride ions are secreted, causing a significant amount of water to be released outside the epithelial cells.

○ Symptom: Accumulation of thick mucus, repeated lung infections, digestive problems, impaired liver function, shortened lifespan

○ Among white Americans, it occurs in approximately 1 in 2,500 individuals.

○ Pseudomonas aeruginosa lung infections are a major cause of death in CF patients.

② Hemophilia(factor Ⅶ): Excessive bleeding, bruising, joint pain and swelling, loss of vision, etc.

③ Sickle cell anemia

○ Definition: A disease caused by the substitution of the 6th amino acid in the β-chain of hemoglobin from glutamic acid (Glu, hydrophilic) to valine (Val, hydrophobic).

○ N terminal - Val - His - Leu - Thr - Pro - Glu/Val - Glu - ···

○ Types of hemoglobin variants: Approximately 500 Types

○ Most are caused by the substitution of a single amino acid in the globin polypeptide chain.

○ Example 1. HbS (sickle-cell Hb): Substitution of a single Glu with Val on the surface.

○ Example 2. Hb Cowtown: Removal of an ion pair involved in T-state stabilization.

○ Example 3. Hb Memphis: Substitution of a similar-sized residue on the surface with a polar but uncharged residue. Nearly a neutral mutation.

○ Example 4. Hb Bibba: Substitution of one Leu with Pro, affecting the α-helix.

○ Example 5. Hb Milwaukee: Substitution of one Val with Glu.

○ Example 6. Hb Providence: One Lysine (Lys) that normally sticks out at the center of the tetramer is replaced with Asparagine (Asn). This decreases BPG binding and increases oxygen saturation.

○ Example 7. Hb Philly: One Tyrosine (Tyr) is replaced with Phenylalanine (Phe), breaking a hydrogen bond at the α₁β₁ junction site. This increases BPG binding and decreases oxygen saturation.

○ Mechanism: When a hydrophilic amino acid is replaced with a hydrophobic amino acid, huge structural changes occur.

○ Clumping (fibrosis) of mutant hemoglobin: Due to hydrophobic interactions, hemoglobin molecules stick together and form fibers.

○ The formation of fibers is the reason why hemoglobin takes on a sickle shape.

○ The same genetic mutation is not observed in myoglobin: This is because myoglobin functions as a single monomer.

○ Trait based on genetic type

○ Dominant homozygous: Normal red blood cells.

○ Recessive homozygotes: Sickle-shaped red blood cells, significantly increased resistance to malaria (↑↑), but severely reduced oxygen transport efficiency (↓↓).

○ Heterozygotes: Some red blood cells are sickle-shaped, increased resistance to malaria (↑), and slightly reduced oxygen transport efficiency (↓).

○ Symptoms: Rheumatoid arthritis, atherosclerosis, dementia, weakened immunity, stroke, and heart weakness.

○ Treatment: Currently, there is no cure for sickle cell anemia, but bone marrow transplantation can help improve symptoms.

④ Tay-Sachs disease: Autosomal recessive disease

○ Lipid metabolism disease caused by hexoseaminidase A deficiency.

○ It manifests in newborns and typically leads to death around the age of 3.

⑤ Phenylketonuria (PKU): Autosomal recessive diseases

○ Definition: A disorder where PAH (Phenylalanine Hydroxylase) cannot be synthesized.

○ PAH (Phenylalanine Hydroxylase): An enzyme that hydroxylates phenylalanine into tyrosine.

○ Tyrosine: Essential for the production of thyroxine, epinephrine, norepinephrine, and melanin.

○ Phenylalanine accumulates in the blood and is eventually excreted in the urine, producing dark-colored urine.

○ Converts to phenylpyruvic acid, causing acidosis, neural tube developmental disorders, intellectual disability, and pale skin.

○ Around 3 to 5 months of age, infants show indifference and do not respond to their environment.

○ By the age of 1, significant intellectual decline becomes noticeable: Even in adulthood, the IQ often remains below 50.

○ Diagnosis: Phenylpyruvic acid, a kind of phenylketone, in the urine are detected using ferric chloride, turning green (normal individuals show a brown color).

○ Treatment: If a newborn consumes a low-phenylalanine diet from early infancy, normal nervous system development can be achieved.

⑸ Polygenic inheritance: A single phenotype is influenced by two or more genes (loci), leading to quantitative traits.

① It is distinguished from multiple alleles inheritance in that it pertains to a single phenotype.

② Skin color, eye color: Multiple melanin-synthesizing enzymes (e.g., tyrosinase) + transport proteins → Melanin content

○ Melanin is a dark brown pigment made from tyrosine.

○ Tyrosinase deficiency leads to albinism.

③ At least three genes determine height.

○ When F1 offspring (AaBbCc) resulting from a cross between AABBCC × aabbcc are self-crossed, a normal distribution is observed.

○ Normal distribution: ₆C₀ : ₆C₁ : ₆C₂ : ₆C₃ : ₆C₄ : ₆C₅ : ₆C₆

④ cis AB type

○ When a mutation causes the A gene and the B gene to exist at different loci on the same chromosome.

○ O children can be born from AB parents.

⑹ Epistasis: When one gene influences or affects the phenotype of another gene.

① Fur color 9:3:4

○ Pigment gene and pigment precipitation gene are involved.

○ In the 9:3:3:1 ratio, the pigment deposition gene combines the 3 and 1 categories (i.e., recessive homozygous for that gene) into 4.

○ Tip. If 12:3:1, 9:6:1, 9:3:4, or 9:7 is observed in any experiment, it is the case of epistasis.

② Complementary genes: Non-allelic genes that work together to express a single trait.

③ Bombay O phenotype

○ H gene: A gene that attaches a fucose sugar to the surface of red blood cells to determine blood types.

○ When the H gene is recessive homozygous, an O blood type child can be born even in combinations where an O blood type child is not normally possible. In other words, the H gene locus is epistatic to the I gene locus.

○ H gene is on chromosome 19.

○ ABO allele is present on chromosome 9.

④ Double recessive gene: 9:3:3:1 becomes 15:1.

⑺ Recessive lethal: The condition often manifests at a young age.

① In the case of AB × AB, the offspring are AA, AB, BA, and BB, but if BB is lethal, only AA, AB, and BA are observed.

○ If A is dominant over B, the ratio of A trait to B trait is 1:0.

○ If A is recessive to B, the ratio of A trait to B trait is 1:2.

② Example: Cystic fibrosis, sickle cell anemia, albinism, Tay-Sachs disease, etc.

⑻ Dominant lethal: The condition often manifests later in life.

① Genotypes and traits

② Example: Huntingtons’ disease

○ It occurs at 30-40 years old, so it can be easily inherited.

○ A progressive incurable condition characterized by the death of brain nerve cells, cognitive decline, central nervous system disorders, and symptoms of chorea.

○ Died 10-20 years after onset of a physiological disorder.

○ HD gene on chromosome 4: Encodes a 3114 amino acid protein → Accumulates in the nuclei of brain cells, forming toxic protein aggregates, leading to neuronal death.

○ 5’-CAG repeat sequence: Promotes the detachment of RNA polymerase.

○ 5’-CAG repeated 9-36 times: Normal.

○ 5’-CAG repeated 37-66 times: Encodes polyQ (polyglutamine). Onset occurs at age 40 with 40 repeats.

○ 5’-CAG repeated 33-36 times: Premutation, with a high likelihood of expansion in offspring.

○ 5’-CAG repeated over 100 times: Onset at age 2.

○ Genetic anticipation: As generations progress, the number of repeats increases, leading to a higher likelihood of disease onset.

③ Incomplete dominant lethal: When heterozygotes do not result in death but exhibit severe or fatal symptoms, such as deformities.

○ Example: Achondroplasia

⑼ Genomic imprinting and extranuclear inheritance

① Genomic imprinting

○ Phenomenon where the expression of a trait differs depending on whether it is inherited from the mother or the father.

○ Occurs due to DNA methylation (methylation of C nucleotides): Influenced primarily by the methylation of CpG islands.

○ In all genes during gametogenesis, methylation is erased and re-established according to whether the gene comes from a paternal or maternal origin.

○ A male individual, who is heterozygous and carries a recessive trait due to maternal imprinting, may exhibit a dominant trait because of paternal imprinting.

○ Example: Igf-1 gene in mice

○ Example 1. Igf-1 gene in mice

○ Paternal imprinting: Methylation of the insulator → Activates transcription.

○ Maternal imprinting: Methylation of the promoter → Suppresses transcription.

○ Example 2. Prader-Willi syndrome and Angelman syndrome

○ Common feature: A recessive trait caused by the suppression of specific gene expression due to chromosomal deletion and genomic imprinting.

○ Prader-Willi syndrome (PWS)

○ Deletion of paternal chromosome 15

○ The non-deleted region of the maternal chromosome is imprinted through methylation, resulting in the suppression of the PWS gene.

○ Angelman syndrome (AS)

○ Deletion of maternal chromosome 15

○ Known as the “happy puppet syndrome” due to characteristic laughter and smiling.

○ The non-deleted region of the paternal chromosome is imprinted through methylation, resulting in the suppression of the AS gene.

② Maternal effect: Cytoplasmic determinants

○ Proteins required in the early stages of development are delivered as mRNA from the mother, as there is not enough time for transcription.

○ Regardless of the genotype of the germ cells, if the mother has at least one allele, the corresponding cytoplasmic determinant is delivered to the germ cells.

○ Genotype and phenotype must be considered separately.

○ Example: There are cases where individuals homozygous for a recessive lethal gene can survive but are unable to reproduce.

○ The mother’s genotype must be inferred from the offspring’s phenotype.

○ If all offspring exhibit the dominant trait, the mother’s genotype is dominant.

○ If all offspring exhibit the recessive trait, the mother’s genotype is recessive.

○ Examples: Early development in fruit flies, right-handed or left-handed coiling of snail shells, and inheritance of moth larva eye color.

③ Cytoplasmic inheritance

○ Plastid inheritance

○ Mitochondrial inheritance**: Maternal inheritance

○ Electron transport chain and ATP synthase are encoded by mitochondrial DNA.

○ Heteroplasmy: Due to the cooperation of mitochondria with numerous cellular organelles (mainly the nucleus), not all children of an affected mother inherit the disease.

○ Problem Type 1.: If mating with a mutant follows Mendelian inheritance → Abnormalities in nuclear DNA required for mitochondrial function.

○ Problem Type 2. If mating with a mutant follows maternal inheritance → Abnormalities in mitochondrial DNA.

○ Problem Type 3 Yeast: Mitochondria are inherited from both parents.

④ Sex-controlled inheritance: A gene is located on an autosome, but its expression (e.g., dominance relationship) varies depending on the sex of the individual.

○ Example 1: The gene for deer antlers exhibits dominant expression in males and recessive expression in females.

○ Example 2: The baldness gene encodes 5α-reductase, which converts testosterone into DHT (dihydrotestosterone). It shows dominant expression in males and recessive expression in females.

○ DHT is a substance that enhances male characteristics.

○ In females, instead of baldness, symptoms such as thinning of the hair occur.

⑤ Xenia: Paternal inheritance of endosperm traits.

⑴ Determination of sex

① Sex determination by chromosomes

○ XY type: ♂ (XY), ♀ (XO) (e.g., human, fruit fly)

○ XO type: ♂ (XO), ♀ (XX)

○ ZW type: ♂ (ZZ), ♀ (ZW) (e.g., poultry)

○ ZO type: ♂ (ZZ), ♀ (ZO)

○ Haploid-diploid type (e.g., honeybees)

② Sex determination by environment

○ Example: The reptile’s sex is determined by the incubation temperature of the eggs.

③ Sex determination by fertilization

○ Example: Unfertilized honeybee eggs develop into males, while fertilized eggs develop into females.

④ Gender conversion

○ Example: Some fish change into males based on social signals, but only certain individuals undergo this transformation.

⑤ Hermaphrodite

○ Example: C. elegans has both male and female reproductive organs.

⑵ Sex-linked inheritance: Refers to genes located on sex chromosomes. If significant differences are observed in the results of reciprocal crosses, it indicates sex-linked inheritance.

① There are 18 genes common to both the X and Y chromosomes.

② Y chromosome-linked inheritance (sex-limited inheritance)

○ SRY (Sex-determining Region of the Y Chromosome)

○ The SRY gene determines sex.

○ Contains very few genes.

○ In fruit flies, the sex-determining factor is located on the X chromosome.

○ Ear hair in men

○ Striped patterns in female silkworms

W chromosome-linked inheritance

③ X chromosome-linked inheritance (sex-linked inheritance): Males express X-linked traits more frequently than females.

○ Color blindness: Green and Red Opsin Genes (X Chromosome)

○ Blue opsin gene (chromosome 7)

○ Duchenne muscular dystrophy (DMD)

○ X-chromosome recessive inheritance, incidence of 1 / 3,500 males

○ The most severe of the 10 muscular dystrophy: Induces muscle cell degeneration.

○ Uses a wheelchair during puberty.

○ Dies before reaching adulthood due to respiratory and heart complications.

○ DMD gene: 2.4 million bp, consisting of more than 70 exons, encoding dystrophin protein.

○ Dystrophin

○ A protein that links six proteins regulating calcium ion channels in the muscle cell membrane with the cytoskeleton.

○ Stabilizes the structure of muscle cells by connecting actin filaments to glycoproteins in the cell membrane.

○ The DMD gene is unable to regulate channels due to exon deletion.

○ Hemophilia

○ White eyes in fruit flies

○ Z chromosome-linked inheritance

④ Sex-controlled inheritance: Actually not sex-related.

⑤ X inactivation: Observed only in mammals

○ Defition: Females have two X chromosomes, so inactivating one of them for quantitative compensation.

○ Drosophila and C. Elegans double the expression of male X chromosomes for quantitative compensation.

○ Barr body: Inactivated X chromosome, cause of mosaic pattern (e.g., Cat fur color pattern)

○ Only in mammals.

○ Lyon hypothesis: Female early embryos randomly inactivate one X chromosome in each cell.

○ In animals, Barr bodies form either before or during cleavage, depending on the species.

○ Inactivation is irreversible, but it is reversed only during germ cell formation.

○ Once a Barr body is formed, all descendant cells produced through somatic cell division have the same region condensed (e.g., cancer cells).

○ Mechanism

○ 1^st. Generation of Xist (X-inactive specific transcript) from Xist RNA on X chromosome with inactivation determined.

○ 2^nd. X chromosome, determined to be activated, methylates Xist RNA to prevent inactivation mechanisms.

○ 3^rd. RNA transcribed from the Xist binds to the corresponding X chromosome.

○ 4^th. RNA interference blocks transcription entirely on the X chromosome that has been determined for inactivation.

○ 5^th. Xist additionally attaches methyl groups to cytosine (C) and deacetylates histones to condense the chromosome → inhibit transcription.

⑥ Fragile X syndrome

○ A mutation occurs if the CGG repeat sequence exceeds 200.

○ Prior to that, it is sequentially classified as permutation, intermediate, and normal (less than 40 repeats).

6. Quantitative Genetics-Genetics and the Environment

⑴ Quantitative traits (e.g., height, weight) are continuous variation traits, influenced by both genetic differences and environmental factors.

① Genetic factors: Polygenic inheritance

○ Traits that have high inheritance indices are consistently expressed regardless of the environment.

② Environmental factors: Environmental dependence of genes, environmental dependence of traits

○ Example: Study of identical twins grown in different environments

⑵ Penetrance: The extent to which the expected traits of a particular genotype actually appear.

① Complete penetrance: 100% trait is expressed as predicted by genotype.

② Incomplete penetrance

③ Example: The dominant disorder, polydactyly, has an 80% penetrance.

⑶ Expressivity: The severity of the phenotype expressed in an individual.

⑷ Heritability (Inheritance index): The proportion of variation within a genetic population attributable to genetic factors.

① Calculation through correlation analysis between groups

○ Example: Parents and offspring have the same degree of immunity.

Figure 5. Heritability example

② 0.2 ~ 0.4: Genetic

③ 0.4 ~: Very genetic

④ Heredity is based on populations, not individual differences.

⑤ Even traits with high heritability can be influenced by the environment.

7. Genetic testing

⑴ Pedigree analysis

① Tip 1: Determining Dominance

○ If non-affected children are born from two affected parents: Dominant inheritance.

○ If affected children are born from two non-affected parents: Recessive inheritance.

② Tip 2: Determining Sex-Linkage

○ Tip 2-1: If the daughter of an affected father is always affected: Dominant X-linked inheritance.

○ Tip 2-2: If the son of an affected mother is always affected: Recessive X-linked inheritance.

○ Y chromosome-related disorders are rare due to the small number of genes on the Y chromosome.

○ If dominance is determined in ①, and Tip 2-1 and Tip 2-2 confirm that it is not X-linked, then it is autosomal inheritance.

③ Pedigree problems that deviate from these cases are unlikely to appear, as only solvable problems are presented.

⑵ Prenatal testing

① Amniocentesis: Performed around 14–16 weeks of pregnancy. Fetal cells are collected from the amniotic fluid for biochemical tests and karyotype analysis.

② Chorionic villus sampling: Performed around 6–8 weeks of pregnancy. Chorionic villi from the fetus are collected for biochemical tests and karyotype analysis.

⑶ Genetic fingerprinting test

Input: 2015.6.29 22:24

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Chapter 7. Heredity and Genetics

1. Genes and Chromosomes

2. Mendel Genetics

3. Linkage and Cross

4. Non-mendel Genetics

6. Quantitative Genetics-Genetics and the Environment

7. Genetic testing

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