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Probability and Pedigrees in Human Inheritance

Probability and Pedigrees in Genetics

Introduction to Inheritance Patterns in Humans

  • Inheritance patterns in humans are analyzed by applying the principles of probability to pedigrees.

Pedigree Analysis

Pedigree Structure

  • Pedigrees use standardized symbols and lines to represent family relationships and trait inheritance across generations.
  • Roman numerals (I, II, III, ext{etc.}) designate generations.
  • Arabic numerals (1, 2, 3, ext{etc.}) identify individuals within each generation.

Common Pedigree Symbols

  • Sex:
    • Circle: Female
    • Square: Male
    • Diamond: Unspecified sex
  • Trait Expression:
    • Unfilled shape: Does not express trait (unaffected)
    • Filled shape: Expresses trait (affected)
  • Status:
    • Shape with a line through it: Deceased (often with date of death: d. 0000)
  • Family Relationships (Lines):
    • Horizontal line connecting a circle and a square: Parents (mating)
    • Double horizontal line connecting parents: Closely related by blood (consanguineous parents)
    • Vertical line from parents to a horizontal line: Connects parents to their children.
    • Vertical broken line: Adoption
    • Horizontal line connecting vertically descended lines: Siblings
    • Triangle from a sibling line: Identical (monozygotic) twins
    • Two separate lines from a sibling line: Fraternal (dizygotic) twins

Rules of Probability

General Principles

  • Probability helps determine the likelihood of certain outcomes, such as coin flips or genetic inheritance.

Sum Rule (Mutually Exclusive Events)

  • Definition: The probability of two or more mutually exclusive events occurring is the sum of their individual probabilities.
  • Formula: P( ext{event 1} ext{ or } ext{event 2}) = P( ext{event 1}) + P( ext{event 2})
  • Application: Most commonly used in genetics when considering alternative genotypes or phenotypes.
  • Example (coins): The probability of both coins being heads OR both being tails. P( ext{HH}) + P( ext{TT}) = (1/4) + (1/4) = 1/2 (Assuming fair coins, P( ext{H})=1/2, P( ext{T})=1/2).

Product Rule (Independent Events)

  • Definition: The probability of two or more independent events occurring together is the product of their individual probabilities.
  • Formula: P( ext{event 1} ext{ and } ext{event 2}) = P( ext{event 1}) imes P( ext{event 2})
  • Application: Common in genetics for combinations of unlinked genes.
  • Example (coins): The probability of both coins being heads: P( ext{H on coin 1}) imes P( ext{H on coin 2}) = (1/2) imes (1/2) = 1/4.

Applying Probability to Genetics Problems

Example 1: Multihybrid Cross

  • Problem: Two plants heterozygous for four unlinked genes (RrYyAaTt) are bred. What is the probability of an offspring being wrinkled, green, axial, and tall?
  • Solution:
    • Wrinkled (rr) from Rr x Rr: 1/4
    • Green (yy) from Yy x Yy: 1/4
    • Axial (A_) from Aa x Aa: 3/4
    • Tall (T_) from Tt x Tt: 3/4
    • Using the Product Rule: P( ext{rr yy A_ T_}) = (1/4) imes (1/4) imes (3/4) imes (3/4) = 9/256

Example 2: Dihybrid Cross (Mendel's F2 Generation)

  • Problem: What is the probability that one of Mendel's F2 generation pea plants (produced via a dihybrid cross of F1 RrYy) would have either yellow/round seeds or green/wrinkled seeds?
  • Solution:
    • Yellow/Round (RY): P(R) imes P(Y) = (3/4) imes (3/4) = 9/16
    • Green/Wrinkled (rryy): P(rr) imes P(yy) = (1/4) imes (1/4) = 1/16
    • Using the Sum Rule (mutually exclusive events): P( ext{yellow/round OR green/wrinkled}) = 9/16 + 1/16 = 10/16 = 5/8

Example 3: Proportion of Specific Genotype

  • Problem: If a plant heterozygous for four unlinked genes (AaBbDdEe) is selfed, what proportion of the progeny will be genotypically AaBBddEe?
  • Solution:
    • P(Aa) = 1/2 (from Aa x Aa)
    • P(BB) = 1/4 (from Bb x Bb)
    • P(dd) = 1/4 (from Dd x Dd)
    • P(Ee) = 1/2 (from Ee x Ee)
    • Using the Product Rule: P(AaBBddEe) = (1/2) imes (1/4) imes (1/4) imes (1/2) = 1/64

Example 4: Proportion Exhibiting at Least One Dominant Phenotype

  • Problem: What proportion of the progeny produced by selfing a trihybrid (AaBbDd) will display the dominant phenotype for at least one gene?
  • Solution: It's easier to calculate the complement – the probability of all recessive phenotypes, then subtract from 1.
    • P( ext{aa}) = 1/4
    • P( ext{bb}) = 1/4
    • P( ext{dd}) = 1/4
    • Probability of all recessive (aabbdd): P( ext{aabbdd}) = (1/4) imes (1/4) imes (1/4) = 1/64
    • Probability of at least one dominant phenotype: 1 - P( ext{aabbdd}) = 1 - 1/64 = 63/64

Types of Inheritance Patterns

There are four main types of Mendelian inheritance patterns evaluated in human genetics:

  1. Autosomal Dominant (AD)
  2. Autosomal Recessive (AR)
  3. X-Linked Dominant (XD)
  4. X-Linked Recessive (XR)

Autosomal Dominant (AD) Inheritance

  • Key Characteristics:
    1. Males and females are affected in approximately equal frequency.
    2. Each individual who has the trait has at least one parent with the trait (the trait does not skip generations).
    3. Either gender can transmit the trait to a child.
    4. If neither parent has the trait, none of their children will have it.
    5. If the trait is rare (less than about 1% incidence), affected individuals are likely heterozygous. In such cases, if one parent has the trait and the other does not, approximately half the offspring will have the trait.
    6. If both parents have the trait, they may produce children who do not have it (indicating that the parents are likely heterozygous).
  • Incomplete Penetrance: An individual may carry the dominant allele but not express the associated phenotype, complicating pedigree analysis.

Autosomal Recessive (AR) Inheritance

  • Key Characteristics:
    1. Males and females are affected in approximately equal frequency.
    2. Individuals who have the trait are often born to parents who do not (the trait can skip generations, and parents are heterozygous carriers).
    3. If both parents have the trait, all children will have it, as both parents can only pass on the recessive allele.
    4. Normal parents with affected offspring must both be heterozygous carriers.
    5. Affected offspring can have normal parents (though this is not a requirement, just a telling hallmark).
    6. When both parents express the trait, ALL of their offspring must also express the trait.

Calculating Risks in Pedigree Analysis: Independent Events

  • Steps to Calculate Probability for the Next Child:
    1. Determine the most likely mode of inheritance from the pedigree.
    2. Determine the genotype of individuals for whom it is known with certainty.
    3. Determine the probability that individuals with unknown genotypes are carriers.
    4. If the trait is rare, assume individuals marrying into the family from outside are not carriers (homozygous dominant).
  • Probability Calculation for Specific Offspring: This involves multiplying the probabilities of three independent events:
    • P( ext{male is heterozygote for the disease allele})
    • P( ext{female is heterozygote for the disease allele})
    • P( ext{if both are heterozygote, they will both pass allele on to offspring}) (which is 1/4 for a recessive trait)
  • Pedigree/Probability Key Tip:
    • Every generation with an uncertain genotype in the direct line of the proband (the affected individual for whom the pedigree is drawn) represents another independent event.
    • To calculate the probability of a proband having a specific genotype, start with the proband and work your way up through the direct ancestral line until you know a genotype with certainty.
    • Work your way back down to the proband, multiplying the probability that each direct ancestor is a carrier.
    • Use Punnett Squares to determine these probabilities if you are unsure.

X-Linked Inheritance

  • X-linked inheritance describes traits determined by genes located on the X chromosome. It is distinct from sex determination.

Sex Determination in Mammals

  • Step 1: In an XY zygote, the SRY gene on the Y chromosome produces Testis-Determining Factor (TDF). In an XX zygote, no Y chromosome means no SRY gene and no TDF.
  • Step 2: TDF induces the medulla of the embryonic gonads to develop into testes. The lack of TDF allows the cortex of the embryonic gonads to develop into ovaries.
  • Step 3: The testes produce testosterone, a hormone that initiates the development of male sexual characteristics. In the absence of testosterone, the embryo develops female sexual characteristics.

Thomas Hunt Morgan's Experiments: Genes Located on Chromosomes

  • Background: Morgan's work with Drosophila melanogaster (fruit flies) and eye color provided the first evidence that genes are located on chromosomes, specifically the X chromosome.
  • Reciprocal Crosses (Red vs. White Eye Color):
    • Cross A: P: Red-eyed female (X^WX^W) x White-eyed male (X^+Y). F1: All red-eyed offspring (females X^WX^+ and males X^WY).
    • Reciprocal Cross (e.g., Cross C - from slide text): P: White-eyed female (X^WX^W) x Red-eyed male (X^+Y).
      • F1 Prediction: Females (X^WX^+) would have red eyes, and males (X^WY) would have white eyes.
      • Difference from Cross A: This result differs significantly from Cross A, where all F1 offspring had red eyes. The differing outcomes of reciprocal crosses are a hallmark of X-linked inheritance.
  • Conclusions:
    • Genes are located on chromosomes.
    • X-linked genes show unique and distinct patterns of inheritance compared to autosomal genes.

X-Linked Recessive (XR) Traits

  • Key Characteristics:
    • Males are predominately affected because they are hemizygous for X-linked genes (they have only one X chromosome).
    • Only the mother has to be a carrier (X^AX^a) to have a son with the trait (X^aY).
    • A daughter with the trait (X^aX^a) must have an affected father (X^aY) and a mother who is at least a carrier (X^AX^a or X^aX^a).
    • The trait can skip generations through carrier females.
  • Examples of X-Linked Recessive Disorders:
    • Hereditary Colorblindness: Mutations in opsin 1 genes (red/green photoreceptors). The molecular basis often involves unequal crossover between homologous X chromosomes during meiosis, leading to duplication or deletion of opsin genes.
    • Hemophilia: A mutation in the factor VIII clotting gene, leading to impaired blood clotting.
    • Duchenne Muscular Dystrophy: Caused by a mutated dystrophin gene, which is crucial for muscular integrity, leading to progressive muscle degeneration.

X-Linked Dominant (XD) Traits

  • Key Characteristics:
    • Males and females may be equally affected, though sometimes females present with milder symptoms due to a second X chromosome.
    • All affected offspring must have at least one affected parent.
    • Affected fathers transmit the trait to 100% of their daughters, but never to their sons (as fathers pass their Y chromosome to sons).
    • Affected mothers transmit the trait to approximately half of their daughters and half of their sons (assuming heterozygosity).
  • Examples of X-Linked Dominant Disorders:
    • Rett Syndrome: A severe neurodevelopmental disorder almost exclusively affecting females (males with the mutation are often inviable or severely affected). Caused by a MeCP2 mutation/deletion.
    • Fragile X Syndrome: Characterized by a CGG repeat expansion in the FMR1 gene, leading to autism-like behaviors and mental retardation.
    • Alport Syndrome: A kidney disease caused by a mutation of collagen genes, leading to progressive kidney damage and hearing loss. There are also autosomal forms, but the X-linked dominant form is common. This is not strictly X-linked dominant (more often X-linked recessive with some dominant patterns depending on the mutation).

Conclusion

  • Understanding pedigree analysis and the rules of probability is fundamental to deciphering human inheritance patterns and predicting disease risk. Distinguishing between autosomal and X-linked, dominant and recessive modes is critical for genetic counseling and research.