Comprehensive Notes: Mendelian Genetics, Deviations, and Genetic Mapping

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• 6.0 GENETIC INHERITANCE
  • Overview of genetic inheritance topics covered: Mendelian Genetics, Deviations from Mendelian Inheritance, and Genetic Mapping.
  • Emphasis on how traits are inherited and how they can deviate from classic Mendelian patterns.
• 6. Mendelian Genetics
  • Core framework for understanding how single genes with different alleles determine traits in offspring.
• 6.2 Deviations from Mendelian Inheritance
  • Introduction to deviations from simple Mendelian patterns (e.g., codominance, incomplete dominance, multiple alleles, linked genes).
• 6.2 Genetic Mapping
  • Introduction to genetic mapping: locating genes on chromosomes and estimating distances via recombination frequencies.

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• 6.1 MENDELIAN GENETICS
  • Terminologies to be reviewed: Monohybrid cross, Dihybrid cross.

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Terminologies

• Allele: one of the alternate forms of a gene that occupies the same location on homologous chromosomes and causes different development of the same trait. Example: flower colour gene has two alleles: purple and white.

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• 2. Alleles
  • Symbols of alleles: Genes -> Flower colour. Allelic pairs and genotype-phenotype mapping.
  • Genotype-Observed traits example (for a single gene with two alleles):
  • For two alleles a and A with dominance:
  • PP
    ightarrow ext{Purple}
  • Pp
    ightarrow ext{Purple}
  • pp
    ightarrow ext{White}
  • The table conveys how alleles map to phenotypes in a straightforward dominance relationship.

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• 2. Gene
  • A segment of DNA that acts as a unit of hereditary information.
  • Codes for a specific protein that determines a specific character (trait).
  • Examples: genes for flower colour, seed colour, seed shape.

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• 3. Locus
  • The location of a specific gene on a chromosome.
  • Alleles of a gene are located at the same locus on homologous chromosomes.

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• 4. Genotype
  • The genetic constitution of an individual with respect to the alleles under consideration.
  • The combination of alleles for a specific gene.
  • Example: $Pp$

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• 5. Phenotype
  • Observable characteristic of an individual.
  • Determined by genotype and the environment during development.
  • Example: Genotype $Pp$ → Phenotype: Purple flower.

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• 6. Homozygous
  • Diploid with two identical alleles for a character at a given locus.
  • Examples: $PP$ (homozygous dominant); $pp$ (homozygous recessive).

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• 7. Heterozygous
  • Diploid with two different alleles for a character at a given locus.
  • Example: $Pp$ (heterozygous).

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• 8. Dominant
  • Allele that is fully expressed when present, in both homozygous and heterozygous states.
  • Represented by a capital letter (e.g., $P$).
  • Example: $PP$ → Purple flower; $Pp$ → Purple flower.

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• 9. Recessive
  • Allele that expresses its trait only in the homozygous state (presence of another identical allele).
  • Represented by a small letter (e.g., $p$).
  • Example: $pp$ → White flower.

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• 10. Self cross
  • A cross between male and female from the same plant or same generation.
• 1. Test cross
  • A cross between a recessive homozygous (e.g., $tt$) with an organism of dominant phenotype but unknown genotype (e.g., TT or Tt).
  • If all offspring show the dominant characteristic, unknown is homozygous (TT).
  • If half show the recessive characteristic (tt), unknown is heterozygous (Tt).

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Character vs Trait

• Character: a heritable feature that varies among individuals (e.g., flower colour, height).
• Trait: each variant for a character (e.g., purple flower, white flower).

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• Gregor Mendel was an Austrian monk who began to study the inheritance of garden peas in 1856.

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• Mendel studied genetic inheritance in pea plants (Pisum sativum) for five reasons:
  • 1) Easy to grow in large numbers.
  • 2) Short life cycle.
  • 3) Many easily observable characteristics.
  • 4) Pollination could be easily controlled (has close reproductive structure inside petals).
  • 5) Have both male & female parts (can be self-pollinated or cross-pollinated and manipulated).

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• Seven characteristics of pea plant (with clear opposite traits):
  1) Flower Color: Purple vs White
  2) Flower Position: Axial vs Terminal
  3) Seed Color: Yellow vs Green
  4) Seed Shape: Round vs Wrinkled
  5) Pod Shape: Inflated vs Constricted
  6) Pod Color: Green vs Yellow
  7) Stem Length: Tall vs Dwarf

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• Continued listing of the seven characteristics with their two variants as shown above.

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• Mendel studied genetic inheritance in pea plants because: (additional reasons)
  • 4) Pollination could be easily controlled (has close reproductive structure inside petals)
  • 5) Have both male & female parts (can be manipulated; can self pollinate, cross pollinate)

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• Types of Pollination
  • 1. Self pollination: Transfer of pollen from anther to stigma of the same flower or another flower of the same plant. Self fertilization is the fusion of male and female gametes from the same individual.

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• Diagram illustration of stigma and anther to depict self-pollination.

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• 2. Cross Pollination: Transfer of pollen from anther to stigma of a different plant (same species). Fusion of male & female gametes from different individuals.

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• Monohybrid Cross
  • Only one characteristic is observed (e.g., flower colour).
  • Involves crossing two true-breeding individuals with different phenotypes:
  • Purple flower (homozygous dominant, $PP$)
  • White flower (homozygous recessive, $pp$)

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• Mendel’s three-step experiment for a single trait:
  1) Produce the P generation (true breeding)
  2) Produce the F1 generation
  3) Produce the F2 generation
• Mendel studied traits across three generations of peas.

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• True Breed or Pure Breed
  • Organisms that are homozygous for the trait under consideration.
  • When self-fertilized, they produce offspring with the same phenotype generation after generation.
  • Example: Pure breed for white flower color; $pp$.

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• Step 1: Produce P generation (true-breeding plants)
  • Mendel selected two varieties with different traits (e.g., purple and white flowers).
  • These plants were allowed to self-pollinate for several generations.

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• Step 1 (example continued)
  • Seeds from purple-flowered plants always produce purple-flowered offspring; seeds from white-flowered plants always produce white-flowered offspring.
  • The two varieties are the parental or P generation.

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• Step 2: Produce F1 generation
  • Stamen removed from purple flower plant; pollen from white flower transferred to stigma of purple flower (cross-fertilized).
  • Seeds obtained are planted.

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• Results
  • All offspring in F1 are purple flowered.
  • Offspring population represents the first filial generation, FI.

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• Genetic Diagram for Monohybrid Crossing
  • P generation: PP × pp → F1: Pp
  • F1 × F1 → F2: genotype and phenotype ratios
  • F2: Genotype ratio: $1:2:1$ (PP : Pp : pp)
  • F2: Phenotype ratio: $3:1$ (Purple : White)

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• Step 3: Produce F2 generation
  • F1 generation self-fertilizes to produce F2 generation.

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• Summary diagram for Mendel’s monohybrid cross (P, F1, F2)
  • All F1 plants have purple flowers in the given cross.
  • F2 generation displays the Mendelian segregation in a predicted 3:1 phenotypic ratio.

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• Mendel conducted experiments with other characteristics as well:
  • Flower color (Purple vs White), Flower position (Axial vs Terminal), Seed color (Yellow vs Green), Seed shape (Round vs Wrinkled), Pod shape (Inflated vs Constricted), Pod color (Green vs Yellow), Stem length (Tall vs Dwarf).

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• Conclusions from Mendel’s Experiment
  1) Principle of dominance: Purple flower allele is dominant over white flower allele.
  2) Principle of Segregation: Each characteristic is determined by a factor (gene) existing in pairs. During gamete formation, the two alleles separate; a gamete receives only one allele from each pair.

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• Mendel’s 1st Law: Law of Segregation
  • Each characteristic is controlled by 2 factors (alleles) which separate and pass into different gametes.
  • Gametes receive only one allele from each pair.

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• Punnett Square
  • A diagram to show all possible combinations of alleles in gametes produced by the parental generation to predict possible offspring genotypes.
  • Male gametes are written along the top, female gametes along the side.
  • Grid contains all possible combinations.

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• Test cross
  • Cross between a homozygous recessive individual and an organism of dominant phenotype but unknown genotype.
  • Purpose: determine the exact genotype of the unknown parent.

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• Monohybrid FI Test Cross
  • If the dominant phenotype is from a homozygous dominant (PP), all offspring will be purple; if from a heterozygous (Pp), half will be purple and half white.

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• Monohybrid FI Test Cross (Genotype and Phenotype Ratios)
  • F1 cross: Pp × pp
  • F2 genotype ratio: $1:1$ for Pp:pp
  • F2 phenotype ratio: $1:1$ purple : white

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• Summary – Monohybrid Crossing
  • F1 self cross (Pp × Pp): F2 genotype ratio $1 ext{ PP} : 2 ext{ Pp} : 1 ext{ pp}$
  • F2 phenotype ratio: $3 ext{ Purple} : 1 ext{ White}$
  • F1 test cross (Pp × pp): F2 genotype ratio $1 ext{ Pp} : 1 ext{ pp}$
  • F2 phenotype ratio: $1 ext{ Purple} : 1 ext{ White}$

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Dihybrid Cross

• Two characteristics are observed (e.g., seed colour and seed shape).
• Involves crossing between two true-breeding individuals with different phenotypes:
  • Example parental: Yellow round seed (homozygous dominant, $YYRR$) × Green wrinkled seed (homozygous recessive, $yyrr$).

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• Mendel’s Experiment (three steps) [reiterated structure]
  • Produce P generation, F1, and F2 generations for dihybrid traits.

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• Step 1: Produce P generation (true-breeding parents)
  • Example: Yellow round seeds $YYRR$ × Green wrinkled $yyrr$.

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• Step 2: Produce F1 generation
  • Cross fertilization of the two true-breeding varieties.
  • All F1 offspring have yellow and round seeds (YyRr).

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• Step 3: Produce F2 generation
  • F1 × F1 to produce F2; or FI is self-fertilized to produce F2.

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• Y and R alleles
  • Y: dominant allele for yellow seed; y: recessive allele for green seed.
  • R: dominant allele for round seed; r: recessive allele for wrinkled seed.
  • Parental: $YYRR$ × $yyrr$.
  • F1 genotype: $YyRr$ (Yellow round all).

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• F1 × F1 cross (Gametogenesis)
  • Gametes from both parents: $YR, Yr, yR, yr$.
  • F2 genotype combinations include: $YYRR, YYRr, YyRR, YyRr, YYrr, Yyrr, …$
  • F2 phenotype: all yellow, all round in the basic example, with ratios consistent with independent assortment.

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• F2 genotype and phenotype from the dihybrid cross
  • Genotype examples: $YYRR, YYRr, YyRR, YyRr, YYrr, Yyrr, …$
  • Phenotype examples: Yellow round, Yellow wrinkle, Green round, Green wrinkle.
  • F2 phenotype ratio: $9:3:3:1$ (Yellow round : Yellow wrinkled : Green round : Green wrinkled).

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• Dihybrid Test Cross
  • F1: Yellow round (YyRr) × recessive homozygous (yyrr).
  • F2 phenotype ratio (without considering crossing over): typically the 1:1:1:1 when all combinations are equally likely in a test cross with independently assorting genes.

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• Summary – Dihybrid Crossing
  • F1 self cross (YyRr × YyRr): F2 phenotype ratio → $9:3:3:1$.
  • F1 test cross (YyRr × yyrr): F2 genotype ratio → $1 ext{ YyRr} : 1 ext{ Yyrr} : 1 ext{ yyRr} : 1 ext{ yyrr}$.
  • F2 phenotype ratio → $1:1:1:1$.

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6.2 Deviations from the Mendelian Inheritance

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• Codominant Alleles
  • Both alleles of a pair are fully expressed in heterozygous condition.
  • Heterozygous and homozygous individuals have different phenotypes.
  • Example: MN blood group in humans.

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• Codominant Alleles – MN blood group
  • Determined by two alleles: LM and LN.
  • LM expresses antigen M; LN expresses antigen N.
  • P: LMLM (M) and LNLN (N)
  • F1: LMLN (MN)
  • F1 × F1: LMLN × LMLN → F2 genotypes: LMLM, LMLN, LNLN
  • F2 phenotype: 1 M : 2 MN : 1 N

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• Incomplete dominance alleles
  • Dominant allele does not fully inhibit the recessive allele in the heterozygous condition; phenotype of heterozygote is intermediate.
  • Example: Antirrhinum (snapdragon) flower color.
  • Alleles: CR (red) and CW (white).

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• Incomplete dominance – Snapdragon example
  • P generation: CRCR (Red) × CW CW (White)
  • F1: CRCW (Pink)
  • F1 × F1: CRCW intercross -> F2: CRCR (Red), CRCW (Pink), CW CW (White)
  • F2 genotype ratio: $1:2:1$
  • F2 phenotype ratio: $1 ext{ Red} : 2 ext{ Pink} : 1 ext{ White}$

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• Multiple Alleles
  • A gene controlling a character can have more than two alleles, though an individual has at most two alleles for that gene.
  • Phenotype is determined by the combination of two alleles.
  • Example: ABO blood group in humans.

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• Multiple Alleles – ABO system (summary)
  • Three alleles: IA, IB, i (often written as O).
  • Genotypes and phenotypes:
  • IAIA or IAi → Type A
  • IBIB or IBi → Type B
  • IAIB → Type AB (codominance of IA and IB)
  • ii → Type O
  • Note: i is recessive; IA and IB are codominant.

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• Multiple Alleles (continued) – Genotype/Antigen mapping
  • IAIA or IAi: Antigen A present
  • IBIB or IBi: Antigen B present
  • IAIB: Both antigen A and B present
  • ii: No antigen

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• P generation – ABO example
  • Type A parent (IAIA or IAi) × Type B parent (IBIB or IBi)
  • F1: AB (IAIB) or A or B phenotypes depending on combinations; examples shown demonstrate how heterozygotes express multiple phenotypes.

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• More on the ABO cross examples
  • F1 genotype and phenotype results: mixtures of IA, IB, i alleles yield various phenotypes (A, B, AB, O).

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• Continued ABO examples with different parental genotypes
  • Demonstrates segregation of IA, IB, and i across offspring with phenotype frequencies reflecting codominance and recessivity.

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• Linked Genes
  • Genes located on the same chromosome.
  • They are always inherited together and do not undergo independent assortment.
  • They do not obey the Second Mendel’s Law (Law of Independent Assortment).

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• Linked Genes (example setup)
  • In Drosophila, two genes: Body color (G vs g) and Wing size (N vs n).
  • Both genes are located on the same chromosome, so they tend to be inherited together unless crossing over occurs.

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• Two-characters example continued
  • Body color: grey (G) vs black (g); Wing size: normal (N) vs vestigial (n).

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• Differences between dihybrid vs Linked genes
  • Dihybrid crosses involve two genes on different chromosomes; they undergo independent assortment.
  • Linked genes are on the same chromosome and do not independently assort unless crossing over occurs.

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• Genotype example for linked genes
  • Genotype: GgNn
  • Parental arrangement for two genes on the same chromosome: GN/gn
  • Meiosis I and II diagrams show how crossing over can generate recombinant chromatids.

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• Possible gametes in a linked gene scenario without crossing over
  • GN, Gn, gN, gn

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• F1 generation (linked genes)
  • Self-cross of F1: GN/gn × GN/gn yields four gamete types: GN, Gn, gN, gn
  • With no crossing over, parental phenotypes are more frequent; recombinants are rare without crossing over.

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• F2 generation and phenotypic ratios for linked genes without crossing over
  • Phenotype ratio: 3 grey normal : 1 black vestigial (example given for a specific cross)

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• Linked Genes (F2 generation) – with crossing over
  • F2 from self-cross of F1 shows parental phenotypes in larger groups and recombinant phenotypes in smaller groups when crossing over is rare.
  • F2 phenotype ratios without crossing over reflect linkage; with crossing over, classic 9:3:3:1 is no longer guaranteed.

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• Linked Genes Test Cross
  • F1 test cross: G N I gn (grey normal) × gn/gn (black vestigial)
  • F2 generation yields genotype and phenotype proportions depending on recombination.
  • If no crossing over occurs, the offspring would show a 1:1 ratio for parental types; recombinant types would appear in different proportions depending on recombination.

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• Summary – Linked Genes
  • Linked genes are on the same chromosome and are inherited together most of the time.
  • Crossing over during meiosis can separate them, producing recombinant chromatids.
  • They do not obey Mendel's Law of Independent Assortment when assessed without considering recombination.

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• Chromosome mechanics and recombination visuals
  • Diagrams illustrating chromatid exchange (crossing over) during Prophase I and how recombinant chromosome types arise.

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• Four types of gametes in linked genes context
  • G g and N n combinations can produce four gametes: GN, Gn, gN, gn, with varying frequencies depending on crossing over.

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• Genotype GN/gn (linked) – Metaphase I and Meiosis II visuals
  • Demonstrates how parental and recombinant gametes are formed depending on whether crossing over occurs.

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• Linked Genes With Crossing Over
  • If self-cross is done on F1 offspring and no crossing over occurs, F2 phenotype ratio is 3:1 (for dihybrid where traits are still linked and no recombination is observed).
  • If test cross and no crossing over occurs, F2 phenotype ratio is 1:1.
  • If crossing over occurs, there is no definite phenotype ratio in F2.

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• F2 generation expectations in linkage scenarios
  • Generally, two large parental phenotype groups and two smaller recombinant phenotype groups when crossing over is limited.

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• Differences in crosses involving linkage
  • Distinguish between: (a) dihybrid crosses vs (b) crosses involving linked genes without crossing over and (c) those with crossing over.

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• Sex Chromosome basics
  • Sex is determined by sex chromosomes in many species.
  • In humans: X chromosome (larger) and Y chromosome (smaller).
  • Humans have 2 sex chromosomes and 22 autosome pairs.
  • Female: XX; Male: XY.
  • The X chromosome contains many vital genes.

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• Sex-linked Genes
  • Located on sex chromosomes.
  • Females can be homozygous or heterozygous for sex-linked alleles; males are hemizygous (only one X allele).
  • Males more frequently exhibit X-linked diseases due to having only one X.
  • Examples: haemophilia and colour blindness.

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• Haemophilia & colour blindness (sex-linked traits)
  • Haemophilia: a disease where blood does not clot properly; lack of clotting factor VIII or IX.
  • Symptoms: bleeding, increased internal bleeding risk.

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• Sex-linked inheritance: haemophilia and colour blindness
  • They are determined by 2 alleles: XH (normal) and Xh (haemophilia allele).
  • Males: XHY (normal) or XhY (haemophiliac).
  • Females: XHXH (normal), XHXh (carrier), XhXh (haemophiliac) depending on parental genotypes.

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• Haemophilia details
  • Sex-linked, X-linked disease. Factor VIII or IX deficiency.

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• Mechanism of inheritance for haemophilia (example cross)
  • Standard parental cross P: XHXH × XhY (normal female × haemophiliac male) or similar.
  • F1: XHXh (carrier female) and XHY (normal male).
  • F1 phenotypes: 50% females normal & 50% carriers; 50% males normal & 50% haemophiliac (illustrative).

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• Additional cross scenarios
  • If P: XhY (haemophiliac male) × XHXh (carrier female), F1 phenotypes include 50% carrier females and 50% haemophiliac females; 50% normal males and 50% haemophiliac males.

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• P, F1, and phenotype outcomes for another sex-linked cross
  • F1 phenotypes: various combinations of normal, carrier, and haemophiliac depending on parental alleles.

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• Final note on sex-linked crosses
  • Demonstrates how sex-linked inheritance produces different proportions of phenotypes in males and females due to X-linked alleles.

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• Polygenes
  • A characteristic controlled by many genes.
  • Each gene exists in multiple alleles; the phenotype is determined by the combination of two alleles.
  • A quantitative trait showing continuous variation and influenced by environmental factors (e.g., height, body weight, skin/eye color, intelligence).

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• 6.3 GENETIC MAPPING

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• Genetic Mapping
  • A technique to study the relative position and sequence of genes on a chromosome.
  • Can be done only for linked genes.
  • The distance between two genes equals the average number of crossovers between them.
  • Genes very close are inherited together (rare recombination); genes far apart have higher recombination probability.
  • Crossovers produce new gene combinations; recombinants are offspring with new gene combinations.
  • The proportion of recombinant offspring from a test cross is used to calculate recombination frequency (a.k.a. recombination frequency).
  • Recombination frequency is used to estimate the distance between two genes; lower value = closer genes.

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• Genetic Mapping (continued) – key concepts
  • Offspring with new gene combinations are recombinants.
  • Recombination frequency is used to calculate distance between genes (map units).
  • A map unit (mu) or centiMorgan (cM) equals a 1% recombination frequency.

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• Example: F2 offspring from a test cross with linked genes
  • Parental phenotype examples: Grey body, Normal wings and Black body, Vestigial wings correspond to parental chromatids.
  • Recombinant phenotypes: Grey body, Vestigial wings; Black body, Normal wings.
  • Data example: Counts for parental vs recombinant phenotypes: 237, 241 (parental); 44, 48 (recombinant).
  • Crossing over frequency = (recombinant offspring) / (total offspring) = (44 + 48) / (237 + 241 + 44 + 48).
  • In this dataset: total = 237 + 241 + 44 + 48 = 570; recombinants = 44 + 48 = 92; recombination frequency ≈ 92/570 ≈ 0.161 ≈ 16.1% ≈ 16 map units.

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• Relation between map unit and crossing over frequency
  • Distance between genes is measured in map units (mu) or centiMorgans (cM).
  • 1 map unit corresponds to 1% recombination frequency.

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• Example continued – calculation of map distance
  • Recombination frequency = $\frac{44 + 48}{237 + 241 + 44 + 48} imes 100 ext{ } = ext{approx } 16.1 ext{ map units}$
  • Equivalent to about $16 ext{ cM}$ between the genes in this dataset.

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• Genetic Mapping (another data set)
  • Example with genes CS/CS and cs/cs.
  • F1 test cross yields: parental phenotypes CS/CS and cs/cs occur more frequently; recombinant phenotypes CS/cs and cs/CS occur less frequently depending on recombination.
  • Data: 480 parental (CS/CS and cs/cs) and 20 recombinants (CS/cs and cs/CS) for a total of 1000 offspring.

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• Genetic Mapping (another data set – calculation)
  • Parental: 480 + 480 = 960; Recombinant: 20 + 20 = 40.
  • Crossing over frequency = 40/1000 = 0.04 = 4 map units.
  • Conclusion: Distance between CS and cs is 4 cM.

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• Genetic Mapping (four-gene example – P, Q, R, S)
  • Given crossing over frequencies between pairs:
  • P and Q: 35%
  • P and R: 5%
  • R and Q: 40%
  • Q and S: 10%
  • R and S: 30%
  • Task: Find the relative position of these genes along the chromosome.
  • Tip: Start with the highest crossing over frequencies (the farthest genes).

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• Genetic Mapping (solution sketch for four genes)
  • Distances derived from given frequencies lead to an arrangement around 40 map units total between the farthest genes.
  • A proposed arrangement (based on maximizing distance information):
  • Order: R — P — S — Q (with distances roughly: R–P ≈ 5, P–S ≈ 25, S–Q ≈ 10; total R–Q ≈ 40)
  • The total distance across the gene cluster is about 40 map units; the individual intervals sum to this total.

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• Thank you
  • End of the lecture/notes.