Comprehensive Guide to Tetrad Analysis, Sex Determination, and Pedigree Analysis

This technical reference manual provides an exhaustive, highly detailed examination of lower eukaryote tetrad analysis, animal and plant sex determination mechanisms, genic balance theories, sex-influenced/limited inheritance patterns, and human pedigree analysis.

1. Tetrad Analysis in Fungi and Algae

Certain species of unicellular algae and fungi spend the majority of their life cycles in a haploid state. These lower eukaryotes—particularly the sac fungi (Ascomycetes)—have become invaluable model systems in transmission genetics due to their unique sexual reproduction cycles.

1.1 Fungal Life Cycle and Spore Formation

Fungal cells are typically haploid (1n) and can reproduce asexually. When environmental conditions trigger sexual reproduction, two haploid cells of compatible mating types fuse to form a diploid zygote (2n).

• Karyogamy: The fusion of the two haploid nuclei to form the 2n zygote.
• Meiosis: The diploid zygote immediately undergoes meiosis to produce four haploid cells, termed meiospores.
• Mitosis: In many species, meiosis is immediately followed by a single mitotic division, doubling the four meiotic products to yield eight haploid cells, which are enclosed within a microscopic sac-like structure called an ascus (plural: asci). The individual haploid cells inside the ascus are called ascospores or an octad.

                   FUNGAL SEXUAL REPRODUCTION FLOW

       Haploid Hyphae (1n) + Haploid Hyphae (1n)
                          |
                          ▼
                  Plasmogamy (Cell Fusion)
                          |
                          ▼
                 Karyogamy (Nuclear Fusion)
                          |
                          ▼
                 Diploid Zygote Mother Cell (2n)
                          |
                          ▼  [ Meiosis I ]
                     2 Nuclei (1n)
                          |
                          ▼  [ Meiosis II ]
                Tetrad of 4 Nuclei (1n)
                          |
                          ▼  [ Mitosis ]
                Octad of 8 Nuclei (1n)
                          |
                          ▼  [ Differentiation ]
                 8 Ascospores in Ascus

1.2 Ordered vs. Unordered Tetrads and Octads

The physical arrangement of spores within an ascus varies by species:

• Unordered Tetrads/Octads: Found in species such as Saccharomyces cerevisiae (baker’s yeast). The ascus is spacious enough to allow the four meiotic products to mix randomly. The orientation of the spores does not reflect their developmental or meiotic lineage.
• Ordered Tetrads/Octads: Found in species such as Neurospora crassa (bread mould). N. crassa possesses a very narrow, tubular ascus that physically constrains the dividing cells. After karyogamy, the spindle fibers of meiotic division align strictly along the long axis of the ascus. As a result, the physical order of the mature ascospores inside the ascus directly reflects their meiotic lineage and centromeric segregation patterns.

2. Analysis of Ordered Tetrads

Because Neurospora crassa asci are narrow tubes, the position of each of the eight spores corresponds to its lineage. The first meiotic division yields two nuclei arranged linearly. The second meiotic division yields four nuclei (a tetrad) in a straight row. The subsequent mitotic division duplicates each nucleus to create eight spores (an octad) arranged in four distinct pairs.

                      MEIOTIC LINEAGE IN A LINEAR OCTAD

      [ Zygote 2n ] ──► [ Meiosis I ] ──► [ Meiosis II ] ──► [ Mitosis ]
                             (1-2)             (1) ────────► Spores 1 & 2
                               |               (2) ────────► Spores 3 & 4
                             (3-4) ────────►   (3) ────────► Spores 5 & 6
                                               (4) ────────► Spores 7 & 8

This structural constraint allows geneticists to map the recombination frequency between a target gene and its centromere. There are two primary segregation patterns observed:

2.1 First-Division Segregation (FDS)

If no crossing over occurs between the target gene locus and its centromere during Prophase I of meiosis, the alternative alleles (A and a) segregate during Meiosis I.

• Molecular Process: The homologous chromosomes carrying the alleles A and a separate at Anaphase I. Consequently, the two nuclei produced by Meiosis I are already genotypically distinct (one is A, the other is a).
• Mitotic Outcome: The final octad exhibits a strict 4:4 parental arrangement (e.g. four A spores adjacent to four a spores).
• FDS Patterns: Can be AAAAaaaa or aaaaAAAA.

                              FIRST-DIVISION SEGREGATION (FDS)

  Homologues            Meiosis I             Meiosis II              Mitosis
 (No Crossover)        Segregation           Segregation             (Octad)

   ┌───[ A ]             ┌───[ A ]             ┌───[ A ]             ┌─── Spore 1 (A)
   ├───[ A ]             └───[ A ]             └───[ A ]             ├─── Spore 2 (A)
   |                                                                 ├─── Spore 3 (A)
   |                                           ┌───[ A ]             └─── Spore 4 (A)
   |                     ┌───[ a ]             └───[ A ]             ┌─── Spore 5 (a)
   ├───[ a ]             └───[ a ]                                   ├─── Spore 6 (a)
   └───[ a ]                                   ┌───[ a ]             ├─── Spore 7 (a)
                                               └───[ a ]             └─── Spore 8 (a)
                                                                         Pattern: 4:4

2.2 Second-Division Segregation (SDS)

If a single crossover occurs between the gene locus and its centromere during Prophase I, the alleles A and a remain bound together on sister chromatids and do not segregate during Meiosis I.

• Molecular Process: Meiosis I separates the homologous centromeres, but because of the crossover, each resulting nucleus contains one chromatid with the A allele and one chromatid with the a allele. The alleles do not separate until the sister chromatids pull apart during Anaphase II (Meiosis II).
• Mitotic Outcome: This delayed segregation produces a non-parental arrangement of spores in the octad, deviating from the 4:4 pattern.
• SDS Patterns: The resulting octads will show 2:2:2:2 or 2:4:2 arrangements of alleles (e.g., AAaaAAaa, aaAAaaAA, AAaaaaAA, aaAAAAaa).

                              SECOND-DIVISION SEGREGATION (SDS)

  Homologues            Meiosis I             Meiosis II              Mitosis
 (Crossover)           Segregation           Segregation             (Octad)

   ┌───[ A ] ─ ─ ┐       ┌───[ A ]             ┌───[ A ]             ┌─── Spore 1 (A)
   ├───[ A ]     |       └───[ a ]             └───[ A ]             ├─── Spore 2 (A)
   |         X   |                                                   ├─── Spore 3 (a)
   |             |                             ┌───[ a ]             └─── Spore 4 (a)
   |         X   |       ┌───[ a ]             └───[ a ]             ┌─── Spore 5 (A)
   ├───[ a ]     |       └───[ A ]                                   ├─── Spore 6 (A)
   └───[ a ] ─ ─ ┘                             ┌───[ A ]             ├─── Spore 7 (a)
                                               └───[ A ]             └─── Spore 8 (a)
                                                                       Pattern: 2:2:2:2

2.3 Calculation of Gene-Centromere Map Distance

Because a single crossover event involves only two of the four chromatids in a bivalent, only half of the meiotic products in an SDS ascus are recombinant. Therefore, to calculate the genetic distance between the gene and its centromere, we divide the percentage of SDS asci by two:

$$ \text{Map Distance} = \frac{\frac{1}{2} \times \text{Number of SDS asci}}{\text{Total number of asci}} \times 100 $$

Worked Problem:
In a genetic cross of Neurospora crassa, 300 asci were analyzed for spore pattern. 240 asci displayed First-Division Segregation (4:4) patterns, while 60 displayed Second-Division Segregation (2:2:2:2 or 2:4:2) patterns. Calculate the gene-to-centromere distance.

$$ \text{SDS Percentage} = \frac{60}{300} \times 100 = 20\% $$
$$ \text{Map Distance} = \frac{1}{2} \times 20\% = 10 \text{ cM} $$

3. Analysis of Unordered Tetrads

In species like Saccharomyces cerevisiae, the spores are arranged randomly in a spherical ascus. While this prevents gene-centromere mapping, unordered tetrads can be used to determine whether two genes are linked and to map the distance between them in dihybrid crosses (Ab × aB or AB × ab).

3.1 Tetrad Types

• Parental Ditype (PD): Contains four haploid spores, all exhibiting the parental allele combinations.
  • If parents were AB × ab, the PD tetrad contains: 2 AB and 2 ab.
  • If parents were Ab × aB, the PD tetrad contains: 2 Ab and 2 aB.
• Non-Parental Ditype (NPD): Contains four haploid spores, all exhibiting non-parental (recombinant) allele combinations.
  • If parents were AB × ab, the NPD tetrad contains: 2 Ab and 2 aB.
  • If parents were Ab × aB, the NPD tetrad contains: 2 AB and 2 ab.
• Tetratype (TT): Contains all four possible genotypes: one of each parental type and one of each recombinant type. (1 AB, 1 ab, 1 Ab, 1 aB).

                       UNORDERED TETRAD SPORE GENOTYPES

         [ Parental Ditype (PD) ]        [ Non-Parental Ditype (NPD) ]
               ┌──────────┐                     ┌──────────┐
               |  AB  AB  |                     |  Ab  Ab  |
               |  ab  ab  |                     |  aB  aB  |
               └──────────┘                     └──────────┘

                              [ Tetratype (TT) ]
                                 ┌──────────┐
                                 |  AB  Ab  |
                                 |  aB  ab  |
                                 └──────────┘

3.2 Distinguishing Linkage from Independent Assortment

• Unlinked Genes: If two genes reside on different chromosomes, they assort independently. During Metaphase I, the alignment of the two chromosome pairs is random. This produces equal numbers of PD and NPD tetrads ($\text{PD} \approx \text{NPD}$). Tetratypes (TT) are produced only when there is a crossover between one of the genes and its centromere.
• Linked Genes: If the two genes are located on the same chromosome, parental combinations are preserved unless crossing over occurs. Therefore, PD tetrads will significantly outnumber NPD tetrads ($\text{PD} \gg \text{NPD}$).
  • A single crossover (SCO) between the two genes produces a Tetratype (TT).
  • A two-strand double crossover (DCO) produces a PD tetrad.
  • A three-strand double crossover (DCO) produces a TT tetrad.
  • A four-strand double crossover (DCO) produces an NPD tetrad.

3.3 Recombination Frequency Formula

Because each NPD tetrad contains 100% recombinant spores (all 4 spores are recombinants) and each TT tetrad contains 50% recombinant spores (2 out of 4 spores are recombinants), the recombination frequency (RF) is calculated as:

$$ \text{Recombination Frequency (RF)} = \frac{\text{NPD} + \frac{1}{2} \text{TT}}{\text{Total number of tetrads}} \times 100 $$

This equation accounts for single and double crossovers to estimate the map distance between two linked genes.

4. Sex Chromosomes and Sex Determination

In most animals, individuals show sexual dimorphism. Sex determination is governed by specialized sex chromosomes, while the remaining chromosomes are termed autosomes.

4.1 Structure of Human Sex Chromosomes

Humans possess 46 chromosomes: 22 pairs of autosomes and 1 pair of sex chromosomes.

• Females (XX): Homomorphic sex chromosomes. Females produce only X-bearing gametes and are therefore the homogametic sex.
• Males (XY): Heteromorphic sex chromosomes. Males produce both X- and Y-bearing gametes and are the heterogametic sex.

                         HUMAN KARYOTYPE SCHEMATIC

                         ┌───────────────────────┐
                         |   Total Chromosomes   |
                         |         (46)          |
                         └───────────┬───────────┘
                                     |
                  ┌──────────────────┴──────────────────┐
                  ▼                                     ▼
        [ Autosomes (22 pairs) ]              [ Sex Chromosomes (1 pair) ]
                  |                                     |
                  |                        ┌────────────┴────────────┐
                  ▼                        ▼                         ▼
         Identical in both          [ Female: XX ]             [ Male: XY ]
          males & females            (Homomorphic)            (Heteromorphic)

The Y-chromosome is significantly smaller than the X-chromosome. It is structurally divided into three distinct regions:

1. Pseudoautosomal Regions (PAR1 and PAR2): Located at the tips of the short (p) and long (q) arms of both X and Y chromosomes.
   • PAR1: 2.6 Mb on the short arm, containing 16 active genes.
   • PAR2: 320 kb on the long arm, containing 4 active genes.
   • Function: These regions contain identical genes that do not undergo X-inactivation. They pair and recombine during male meiosis, ensuring correct chromosome segregation.
2. Male-Specific Region of the Y (MSY): The non-recombining portion of the Y-chromosome (comprising about 95% of its length). It contains 23 protein-coding genes and many pseudogenes.
3. Heterochromatin Region: Genetically inactive, condensed chromatin located primarily on the long arm of the Y.

                  STRUCTURAL MAP OF THE HUMAN Y CHROMOSOME

   PAR1 (2.6 Mb) ──────[=========]
                       [         ]
                       [         ]  Male-Specific Region (MSY)
                       [   SRY   ]  (Contains SRY gene)
                       [         ]
                       [=========]  Centromere
                       [         ]
                       [         ]  Heterochromatin Region
                       [         ]
   PAR2 (320 kb) ──────[=========]

5. Molecular Genetics of Mammalian Sex Determination

In mammals, the presence of the Y-chromosome determines male sex, specifically driven by the master regulatory gene SRY (Sex-determining Region Y).

5.1 The Bipotential Gonad

Early in embryonic development, the mammal develops an indifferent or bipotential gonad. It has the capacity to differentiate into either a testis or an ovary.

5.2 The SRY-Sox9 Regulatory Cascade

• The SRY Gene: Located on the short arm of the Y-chromosome, it encodes the SRY protein, also known as the Testis-Determining Factor (TDF).
• DNA Binding: SRY is a transcription factor containing a conserved 79-amino-acid high-mobility group (HMG) box domain that binds and bends DNA, activating downstream autosomal genes.
• Sox9 Activation: SRY upregulates the transcription of the autosomal gene Sox9 (SRY-related HMG box gene 9) in the bipotential gonad.
• Positive Feedback Loop: Once Sox9 expression is initiated, it self-regulates via a feedback loop, upregulating its own transcription and activating Fgf9 (fibroblast growth factor 9).
• Pathway Inhibition: Sox9 acts to directly block the β-catenin pathway, which is the master driver of ovarian development.

                  MALE SEX DETERMINATION MOLECULAR CASCADE

                             [ Y Chromosome ]
                                    |
                                    ▼
                               [ SRY Gene ]
                                    |
                                    ▼
                              [ SRY Protein ]
                                    |
                                    ▼
                         [ Autosomal Sox9 Gene ] ◄──┐ (Feedback
                                    |               |  Loop)
                 ┌──────────────────┴───────────────┴─┐
                 ▼                                    ▼
            [ Fgf9 Active ]                    [ Blocks β-catenin ]
                 |                                    |
                 ▼                                    ▼
         [ Testis Developed ]                    [ Inhibits Ovary
                 |                                Development ]
                 ▼
       ┌─────────┴─────────┐
       ▼                   ▼
[ Sertoli Cells ]   [ Leydig Cells ]
       |                   |
       ▼                   ▼
  [ anti-Mullerian     [ Testosterone ]
    hormone (AMH) ]        |
       |                   ▼
       ▼             Development of male
 Regression of       secondary characteristics
Mullerian ducts

5.3 Cellular Differentiation and Hormonal Control

Once Sox9 is expressed, cells in the indifferent gonad differentiate into male-specific lineages:

• Sertoli Cells: Differentiated cell type that secretes anti-Mullerian hormone (AMH) (also called Mullerian-inhibiting factor). AMH causes the regression of the embryonic Mullerian ducts (which would otherwise form the uterus, oviducts, and upper vagina).
• Leydig Cells: Differentiated cell type that secretes the androgen testosterone. Testosterone drives the development of the male reproductive tract and secondary sexual characteristics.

5.4 The Female Default Pathway

In the absence of the Y-chromosome (and thus the SRY gene):

• Sox9 remains inactive, allowing the activation of the β-catenin pathway.
• The cortex of the bipotential gonad develops into an ovary.
• Germ cells differentiate into oocytes, and somatic support cells become follicle cells (rather than Sertoli cells) and theca cells (rather than Leydig cells).
• Estrogen is secreted instead of testosterone.
• In the absence of AMH, the Mullerian ducts develop into the oviducts, uterus, cervix, and upper vagina. This is referred to as the “default” pathway.

6. Sex Determination Mechanisms in Animals and Plants

Different organisms have evolved diverse genetic and environmental systems to establish sex.

6.1 Animal Sex-Determining Systems

Honey Bee Haplodiploidy: Diploid females develop into sterile worker bees or fertile queen bees depending on larval diet (royal jelly). Haploid males develop parthenogenetically from unfertilised eggs into drones, producing sperm via mitosis.

6.2 Environmental Sex Determination (ESD) in Reptiles

In many reptiles (such as crocodilians, turtles, and lizards), sex is determined by the ambient incubation temperature of the egg during a critical thermosensitive period (TSP). This is known as Temperature-Dependent Sex Determination (TSD). Reptilian species exhibit three main patterns of TSD:

• MF Pattern: High proportions of males are produced at low temperatures; females are produced at high temperatures (e.g. some marine turtles).
• FM Pattern: Females are produced at low temperatures; males are produced at high temperatures (e.g. some lizards).
• FMF Pattern: Females are produced at both extreme high and extreme low temperatures; males are produced at intermediate temperatures (e.g. crocodiles, tuatara).

                      REPTILIAN TSD CURVES (PROPORTION MALE)

     Prop. Male                     Prop. Male                     Prop. Male
    1.0 ┌─\                        1.0 ┌─         /               1.0 ┌─   /        |  \                           |         /                    |   /
    0.5 |   \                      0.5 |        /                 0.5 |  /            |    \                         |       /                      | /
    0.0 └───┴───► Temp             0.0 └──────┴───► Temp          0.0 └┴──────┴─► Temp
         [ MF Pattern ]                 [ FM Pattern ]                 [ FMF Pattern ]

The Molecular Mechanism of TSD:
The physiological control of TSD relies on the activity of the enzyme aromatase, which converts male androgens (testosterone) into female estrogens (17β-estradiol).

• High Temperature (MF/FMF): Upregulates aromatase gene transcription and enzyme activity, driving estrogen synthesis, which directs the bipotential gonad to form ovaries.
• Low/Intermediate Temperature: Aromatase activity remains low, allowing androgen levels to remain high, directing the gonad to form testes.

                          AROMATASE TEMPERATURE REGULATION

                               [ Incubation Temp ]
                                       |
                ┌──────────────────────┴──────────────────────┐
                ▼ (High Temperature)                          ▼ (Low/Int. Temp)
      [ High Aromatase Activity ]                    [ Low Aromatase Activity ]
                |                                             |
                ▼                                             ▼
       Androgen ──► Estrogen                         Androgens Predominate
                |                                             |
                ▼                                             ▼
            [ Ovary ]                                     [ Testis ]

6.3 Sex Determination in Plants

• Monoecious Plants: Individual plants bear both male and female reproductive organs (e.g. corn, where the tassel is male and the ear is female).
• Dioecious Plants: Individual plants are unisexual, bearing either male or female flowers (e.g. Melandrium album / white campion). M. album utilizes an XY sex chromosome system: females are XX (22 autosomes + XX) and males are XY (22 autosomes + XY).

7. Genic Balance Theory in Drosophila

In Drosophila melanogaster, sex is not determined by the presence or absence of a Y-chromosome. Instead, sex is determined by the ratio of X-chromosomes to the number of haploid autosomal sets (A). This concept was discovered by Calvin Bridges in 1921.

7.1 The X:A Ratio Rule

The balance of genes promoting female development on the X-chromosome versus genes promoting male development on the autosomes determines the sexual phenotype:

$$ \text{Sex Index} = \frac{\text{Number of X chromosomes (X)}}{\text{Sets of Autosomes (A)}} $$

Role of the Y-Chromosome: The Y-chromosome is entirely irrelevant for sex determination in Drosophila. An XO fly has an X:A ratio of 0.50 and is anatomically male. However, the Y-chromosome carries genes essential for spermiogenesis; thus, XO males are completely sterile.

7.2 The Sxl Gene Molecular Switch

The master genetic switch in Drosophila is the Sex-lethal (Sxl) gene located on the X-chromosome.

• In XX Embryos (Ratio ≥ 1.0): High concentrations of sisterless proteins (encoded on the X-chromosome) activate the transcription of Sxl. The Sxl protein acts as a splicing regulator that ensures the production of functional Sxl protein and directs downstream female-specific splicing pathways.
• In XY or XO Embryos (Ratio ≤ 0.5): Low concentrations of X-encoded regulators fail to activate early Sxl expression. In the “off” state, the default splicing pathway is used, which introduces a premature stop codon, resulting in non-functional Sxl protein, directing male development.

8. Gynandromorphs and Mosaicism

8.1 Gynandromorphs in Drosophila

A gynandromorph is an individual in which one part of the body is male and the other part is female. This phenomenon is a striking demonstration of the cell-autonomous nature of insect sex determination.

• Bilateral Gynandromorph: A classic form where the left side of the fly is female (XX) and the right side is male (XO).
• Mechanism: This occurs due to mitotic non-disjunction or chromosome lagging during the first cleavage division of a female (XX) zygote. One of the daughter cells fails to receive an X-chromosome, becoming XO (male), while the other receives both, remaining XX (female). As development proceeds, the XX line forms the female side of the body and the XO line forms the male side.

                      BILATERAL GYNANDROMORPH ORIGIN

                              [ Zygote XX ]
                                    |
                        [ First Mitotic Division ]
                         (Lagging X Chromosome)
                                    |
                  ┌─────────────────┴─────────────────┐
                  ▼                                   ▼
          [ Cell Line XX ]                    [ Cell Line XO ]
                  |                                   |
                  ▼                                   ▼
            Female Half                           Male Half
        (Red eye, long wing)                 (White eye, short wing)

8.2 Genetic Mosaicism

• Mosaic: An individual with two or more genetically distinct cell lines that originated from a single zygote. Mosaicism can be caused by somatic mutations, mitotic chromosome loss, or epigenetic changes.
• Chimera: Distinct from a mosaic; a chimera arises from the fusion of two or more independent zygotes.
• Somatic vs. Germline Mosaicism: Somatic mosaics contain mutant cells only in body tissues, which are not inherited. Germline mosaics have mutations in the gamete-producing germ cells, allowing them to pass the mutation to offspring.

9. Sex-Linked, Sex-Limited, and Sex-Influenced Inheritance

Genetic traits located on the sex chromosomes exhibit unique patterns of inheritance that deviate from autosomal Mendelian ratios.

9.1 Sex-Linked Inheritance

• X-Linked Inheritance: Driven by genes located on the X-chromosome.
• Hemizygosity: Males have only a single X-chromosome (XY) and are hemizygous for all X-linked genes. They express any recessive allele carried on their X-chromosome, as there is no corresponding locus on the Y-chromosome to mask it (pseudodominance).
• Criss-Cross Inheritance: X-linked recessive traits are passed from affected fathers to carrier daughters, and then to affected grandsons (e.g. Morgan’s white-eye Drosophila experiments).
• Y-Linked (Holandric) Inheritance: Governed by genes on the male-specific region of the Y-chromosome. These traits are transmitted exclusively from father to son (100% penetrance in males, completely absent in females).

                      RECIPROCAL X-LINKED RECESSIBLE CROSSES

    Cross A: Affected Male x Wild-Type Female   Cross B: Wild-Type Male x Affected Female

             (X^a Y)    x    (X^A X^A)                    (X^A Y)    x    (X^a X^a)
                        |                                             |
                        ▼                                             ▼
          F1: 100% Wild-Type Phenotype                  F1: Females Wild-Type (X^A X^a)
              (Females X^A X^a, Males X^A Y)                Males Affected (X^a Y)

9.2 Sex-Limited Traits

Sex-limited traits are determined by autosomal genes and are present in both sexes, but are expressed in only one sex. The penetrance of these genes in the opposite sex is zero, typically controlled by sex hormones.

• Example in Humans: Breast development is limited to females; beard growth is limited to males.
• Key Properties: Autosomal inheritance, expression restricted to one sex, influenced by systemic sex hormone profiles.

9.3 Sex-Influenced Traits

Sex-influenced (or sex-conditioned) traits are determined by autosomal genes and are expressed in both sexes, but the dominance relationship of the alleles differs between the sexes.

Example in Humans (Pattern Baldness): The allele B is dominant in males but recessive in females.

10. Human Pedigree Analysis

Pedigrees are structured genetic diagrams mapping family relationships and traits across generations using standardized symbols:

                            PEDIGREE SYMBOLS MAP

         [ Unaffected ]                     [ Affected ]
       Male      Female                   Male      Female
        ┌─┐        ▲                       ■          ●
        └─┘        ●  

         [ Mating ]                      [ Consanguineous Mating ]
       ┌─┐      ┌───┐                     ┌─┐      ╔═══╗
       └─┘──────│   │                     └─┘══════║   ║
                └───┘                              ╚═══╝

10.1 Key Patterns of Inheritance in Pedigrees

• Autosomal Recessive (e.g. Albinism, Cystic Fibrosis, Sickle-Cell Anemia)
  • Generational Skipping: The trait often skips generations, appearing in the offspring of unaffected carrier parents ($Aa \times Aa$).
  • Sex Ratio: Affects males and females with equal frequency.
  • Consanguinity: More common in marriages between close relatives.
  • Carrier Offspring: If both parents are affected ($aa \times aa$), 100% of their children must be affected.
• Autosomal Dominant (e.g. Huntington’s disease, Hypercholesterolemia)
  • No Skipping: The trait appears in every generation; affected individuals must have at least one affected parent.
  • Sex Ratio: Affects males and females with equal frequency.
  • Transmission: An affected heterozygote (Aa) crossed with an unaffected individual (aa) passes the trait to approximately 50% of offspring.
• X-Linked Recessive (e.g. Color blindness, Hemophilia A/B, G6PD deficiency)
  • Male Bias: Significantly more males are affected than females.
  • No Male-to-Male Transmission: An affected father ($X^a Y$) cannot pass the trait to his sons, but passes it to 100% of his daughters, who become carrier heterozygotes.
  • Transmission via Carriers: Unaffected carrier mothers ($X^A X^a$) pass the trait to 50% of their sons.
• X-Linked Dominant (e.g. Vitamin-D-resistant rickets)
  • No Skipping: The trait does not skip generations.
  • Transmission: Affected males ($X^A Y$) pass the trait to 100% of their daughters, but 0% of their sons. Affected heterozygous females ($X^A X^a$) pass the trait to 50% of their sons and 50% of their daughters.
• Y-Linked (Holandric) Traits
  • Exclusive Male Inheritance: Affects only males.
  • Transmission: All sons of an affected male are affected.

11. Solved Genetics Problems

Problem 1: Probability in Albinism

Two individuals who are heterozygous for albinism (Aa, a recessive autosomal condition) have four children. Calculate:

1. The probability that all four children will be phenotypically normal.
2. The probability that the first three will be normal and the fourth will be albino.
3. The probability that exactly three will be normal and one will be albino.

Solution:
For each child, the probability of being normal (A_) is $p = \frac{3}{4}$, and the probability of being albino (aa) is $q = \frac{1}{4}$.

• Probability of 4 normal children:
  $$ P = \left(\frac{3}{4}\right)^4 = \frac{81}{256} \approx 31.6\% $$
• Probability of specific order (Normal, Normal, Normal, Albino):
  $$ P = \frac{3}{4} \times \frac{3}{4} \times \frac{3}{4} \times \frac{1}{4} = \frac{27}{256} \approx 10.5\% $$
• Probability of exactly 3 normal and 1 albino (unordered). Using binomial expansion with n = 4, x = 3:
  $$ P = \frac{4!}{3! 1!} \left(\frac{3}{4}\right)^3 \left(\frac{1}{4}\right)^1 = 4 \times \left(\frac{27}{64}\right) \times \left(\frac{1}{4}\right) = \frac{108}{256} \approx 42.2\% $$

Problem 2: Probability in X-Linked Hemophilia

Hemophilia is a recessive X-linked trait in humans. If a heterozygous carrier woman ($X^H X^h$) has children with a normal man ($X^H Y$), calculate:

1. The probability of having an affected child.
2. The probability of having four unaffected children in a row.
3. The probability that exactly two out of five children will be affected.

Solution:

• Maternal gametes: 1/2 $X^H$ and 1/2 $X^h$.
• Paternal gametes: 1/2 $X^H$ and 1/2 $Y$.
• Offspring genotypes: 1/4 $X^H X^H$ (normal female), 1/4 $X^H X^h$ (carrier female), 1/4 $X^H Y$ (normal male), 1/4 $X^h Y$ (affected male).

Therefore, the probability of any child being affected is $q = \frac{1}{4}$ (specifically, 50% of the sons are affected), and the probability of a child being unaffected is $p = \frac{3}{4}$.

• Probability of having an affected child:
  $$ P = \frac{1}{4} = 25\% $$
• Probability of 4 unaffected children in a row:
  $$ P = \left(\frac{3}{4}\right)^4 = \frac{81}{256} \approx 31.6\% $$
• Probability that exactly 2 out of 5 children will be affected. Using binomial expansion with n = 5, x = 2:
  $$ P = \frac{5!}{2! 3!} \left(\frac{3}{4}\right)^3 \left(\frac{1}{4}\right)^2 = 10 \times \left(\frac{27}{64}\right) \times \left(\frac{1}{16}\right) = \frac{270}{1024} \approx 26.3\% $$

Problem 3: PTU Tasting Genetics

The ability to taste the chemical phenylthiocarbamide (PTU) is an autosomal dominant trait (T). Non-tasters are tt. A female taster (T_) whose father was a non-taster (tt) marries a male taster (T_) who had a non-taster daughter in a previous marriage. Calculate the probability that their first child will be:

1. A non-taster girl.
2. A taster girl.
3. A taster boy.

Solution:

• Maternal Genotype: The female is a taster (T_), but her father was a non-taster (tt). She must have inherited a t allele from her father. Therefore, her genotype is Tt.
• Paternal Genotype: The male is a taster (T_), but he has a non-taster daughter (tt) from a previous marriage. He must have passed a t allele to this daughter. Therefore, his genotype is Tt.
• The Cross: $Tt \times Tt \implies \frac{3}{4}$ Tasters (T_) and $\frac{1}{4}$ Non-tasters (tt).
• For any birth, the probability of a girl is 1/2 and the probability of a boy is 1/2.

• Probability of a non-taster girl:
  $$ P = P(\text{Non-taster}) \times P(\text{Girl}) = \frac{1}{4} \times \frac{1}{2} = \frac{1}{8} = 12.5\% $$
• Probability of a taster girl:
  $$ P = P(\text{Taster}) \times P(\text{Girl}) = \frac{3}{4} \times \frac{1}{2} = \frac{3}{8} = 37.5\% $$
• Probability of a taster boy:
  $$ P = P(\text{Taster}) \times P(\text{Boy}) = \frac{3}{4} \times \frac{1}{2} = \frac{3}{8} = 37.5\% $$