(Lecture 6) Genetic Risk Assessment for Single Gene Disorders

Course Overview & Learning Objectives

• Course Identification: IBSSD 1520 & 1521, Module 3, Lecture 6
• Instructor: Dr. Susan Viselli
• Topic: Genetic Risk Assessment For Single Gene Disorders
• Terminal Objective:
  • Understand the central tenets of modern medical genetics.
• Enabling Objectives:
  • Distinguish phenotypic and genotypic ratios from each other and use Punnett squares to diagram monohybrid and dihybrid crosses.
  • Describe the principles of segregation and independent assortment, and determine when to apply the multiplication and addition rules.
  • Use the Hardy-Weinberg principle to calculate genotype and phenotype frequencies from allele frequencies.
  • Perform precise probability and genetic risk calculations to answer clinical study questions.

Historical Foundation & Principles of Mendelian Genetics

• Johan "Gregor" Mendel (1822–1884):

  • Born in 1822 as Johann; entered the Augustinian St. Thomas Abbey in Staré Brno in 1843, taking the name Gregor.
  • Conducted plant hybridization and meteorology research across a 42-year career; eventually served as Abbot.
  • Mendel Museum Exhibition at Masaryk University: Gregor Johann Mendel: The Story of a Humble Genius (Author: Ondřej Dostál; Architect: Zdeněk Fránek; Design: Marek Nedělka, Ondřej BÁcher; Graphic design: Hrdina Pavlík; Translations: Drew Barry, Adam Prentis; Editing/presentation: Dominik Krajíček; Realization: LIKO-S, a.s., MIVIS-CZ, s.r.o.).

• Mendelian Inheritance:

  • Defined as the pattern of inheritance where a phenotype is caused by the inheritance of a single mutant allele.

• Mendel's Seven Pea Traits in Garden Peas ($Pisum\ sativum$):

  • Seed shape: Round vs. Wrinkled
  • Seed color: Yellow vs. Green
  • Seed coat color: Gray vs. White
  • Pod shape: Inflated (Smooth) vs. Constricted
  • Pod color: Green vs. Yellow
  • Flower position: Axial vs. Terminal
  • Plant height: Tall vs. Short (Dwarf)      

• Core Mendelian Principles:

  • Principle of Segregation:

  • Sexually reproducing organisms possess paired genes (alleles) for each locus; during gamete development (meiosis), these two copies segregate so that each gamete receives only one factor.

  • Directly refutes the 19th-century "blending theory" of inheritance.          

  • Principle of Independent Assortment:

  • Genes located at different genetic loci are transmitted to offspring independently of one another.

  • Exception/Caveat: Does not hold true when genes are located close to each other on the same chromosome (genetic linkage).

  • Dominant and Recessive Alleles:

  • Alleles are alternative forms of a gene located on autosomal chromosomes (2 copies per gene).

  • Dominant Allele ($A$): Usual form that masks the phenotypic expression of a recessive allele.

  • Recessive Allele ($a$): Form whose expression is masked in the presence of a dominant allele.

Monohybrid Crosses and Genotype-Phenotype Relationships

• Genotype Definitions:
  • Homozygous Dominant ($AA$ or $HH$): Carrying two identical dominant alleles.
  • Homozygous Recessive ($aa$ or $hh$): Carrying two identical recessive alleles.
  • Heterozygous ($Aa$ or $Hh$): Carrying two different alleles.
• Expression Rules:
  • Dominant traits are observed in both homozygous dominant ($AA$) and heterozygous ($Aa$) states.
  • Recessive traits are observed exclusively in the homozygous recessive ($aa$) state.
• Monohybrid Cross Example ($Hh \times Hh$):
  • Parental Cross ($P$): Homozygous tall ($HH$) $\times$ Homozygous short ($hh$) produces an $F_1$ generation of all heterozygous tall ($Hh$) plants.
  • Self-Fertilization of $F_1$ ($Hh \times Hh$) yields the $F_2$ generation:
  • Gametes: $H$ ($50\%$) and $h$ ($50\%$) from each parent.
  • Punnett Square Outcomes:
    • $H \times H \rightarrow HH$ (Tall)
    • $H \times h \rightarrow Hh$ (Tall)
    • $h \times H \rightarrow Hh$ (Tall)
    • $h \times h \rightarrow hh$ (Short)
  • Genotypic Ratio: $1\text{ }HH : 2\text{ }Hh : 1\text{ }hh$ ($1:2:1$).
  • Phenotypic Ratio: $3\text{ Tall} : 1\text{ Short}$ ($3:1$).

Mendelian Traits in Humans & Pedigrees

• Human Mendelian Traits:
  • Principles derived from pea plants apply directly to human traits and single-gene inherited disorders.
  • Examples of Single-Gene Autosomal Human Traits:
  • Mid-digital hair (Dominant)
  • Tongue rolling ability (Dominant)
  • Widow's peak hairline (Dominant) vs. Straight hairline (Recessive)
  • Free earlobes (Dominant) vs. Attached earlobes (Recessive)

• Phenotype Concept & Modifiers:
  • Phenotype is the observed physical, biochemical, or clinical manifestation of an organism.
  • Two distinct genotypes can yield identical phenotypes (e.g., homozygous dominant $AA$ and heterozygous carrier $Aa$ for cystic fibrosis are both clinically unaffected).
  • Environmental Influence: The same genotype can produce different phenotypes depending on the external or internal/genetic environment (e.g., Phenylketonuria / PKU clinical severity is modified by dietary phenylalanine intake).
  • Formula: $\text{Genotype} + \text{Environment \& Lifestyle} = \text{Phenotype}$.

Dihybrid Crosses & Probability Analysis

• Definition: A cross between two individuals who are both heterozygous for two independent autosomal recessive loci located on different chromosomes ($AaDd \times AaDd$).
• Example Model: Dual carrier cross for Pigmentation ($A$ = normal, $a$ = albinism) and Hearing ($D$ = normal, $d$ = deafness).
• Gamete Generation: Each dual heterozygote produces 4 distinct gametes in equal proportions: $AD$, $Ad$, $aD$, $ad$.
• 16-Square Punnett Grid Breakdown:
  • 9 Unique Genotypes:
  • $1 \times AADD$
  • $2 \times AADd$
  • $2 \times AaDD$
  • $4 \times AaDd$
  • $1 \times AAdd$
  • $2 \times Aadd$
  • $1 \times aaDD$
  • $2 \times aaDd$
  • $1 \times aadd$
  • Genotypic Probabilities:
  • $P(AADD) = \frac{1}{16}$
  • $P(aadd) = \frac{1}{16}$
  • $P(AAdd) = \frac{1}{16}$
  • $P(aaDD) = \frac{1}{16}$
  • $P(AADd) = \frac{1}{8}$ (or $\frac{2}{16}$)
  • $P(aaDd) = \frac{1}{8}$ (or $\frac{2}{16}$)
  • $P(Aadd) = \frac{1}{8}$ (or $\frac{2}{16}$)
  • $P(AaDD) = \frac{1}{8}$ (or $\frac{2}{16}$)
  • $P(AaDd) = \frac{1}{4}$ (or $\frac{4}{16}$)
  • Phenotypic Ratio ($9:3:3:1$):
  • $\frac{9}{16}$ Normal pigment, normal hearing ($A\_D\_$)
  • $\frac{3}{16}$ Normal pigment, deaf ($A\_dd$)
  • $\frac{3}{16}$ Albinism, normal hearing ($aaD\_$)
  • $\frac{1}{16}$ Albinism, deaf ($aadd$)
• Dihybrid Risk Calculation Example (Study Question 4):
  • Scenario: Both members of a couple are dual carriers for Cystic Fibrosis (CF) and Sickle Cell Disease (SCD).
  • Question: What is the probability that they will have a child with EITHER Cystic Fibrosis OR Sickle Cell Disease, but NOT BOTH?
  • Solution: Based on the $9:3:3:1$ phenotypic ratio, the two "$3$" classes represent offspring affected by one disease but not the other.
  • Formula:     $\text{Probability} = \frac{3 + 3}{16} = \frac{6}{16} = 0.375 \text{ or } 37.5\%$
  • Answer: C. 37.5%.

Rules of Probability for Genetic Assessment

• Fundamental Definition: Probability is the proportion of times a specific outcome occurs in a series of independent events. Probabilities range from $0$ (impossible) to $1$ (certain).
• Summation Law: The probabilities of all mutually exclusive possible outcomes must add up to $1$ ($100\%$).
• Basic Formula:   $\text{Probability} = \frac{\text{Number of Favorable Outcomes}}{\text{Total Number of Possible Outcomes}}$
• Multiplication Rule ("AND" Rule):
  • Used to calculate the probability of two or more independent events occurring simultaneously or sequentially.
  • $P(A \text{ and } B) = P(A) \times P(B)$
  • Example: Probability of tossing two heads in a row with a fair coin:     $\frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$
• Addition Rule ("OR" Rule):
  • Used to calculate the probability of obtaining either one outcome OR another mutually exclusive outcome.
  • $P(A \text{ or } B) = P(A) + P(B)$
  • Example: Probability of tossing two heads in a row ($\frac{1}{4}$) OR two tails in a row ($\frac{1}{4}$):     $\frac{1}{4} + \frac{1}{4} = \frac{1}{2}$
• Clinical Risk Assessment Principles:
  • Always BEGIN with the MOST LIKELY situation.
  • Consider the most common inheritance mechanism and dominant population factors first before evaluating rare exceptions.

Hardy-Weinberg Principle & Population Genetics

• Purpose: Relates gene (allele) frequencies to genotype frequencies in a population under equilibrium.

• Allele Frequencies:

  • $p$ = frequency of dominant allele ($A$ or $S$)
  • $q$ = frequency of recessive allele ($a$ or $s$)
  • Equation: $p + q = 1.0$ ($100\%$)

• Genotype Frequencies:

  • $p^2$ = frequency of homozygous dominant genotype ($AA$ or $SS$)
  • $2pq$ = frequency of heterozygous carrier genotype ($Aa$ or $Ss$)
  • $q^2$ = frequency of homozygous recessive affected genotype ($aa$ or $ss$)
  • Expansion Equation: $p^2 + 2pq + q^2 = 1.0$

• Five Population Requirements for Hardy-Weinberg Equilibrium:

  1. Large population size.
  2. Absence of natural selection for or against any genotype.
  3. Random mating (panmixia) with respect to the locus of interest.
  4. Negligible rate of new mutations.
  5. Absence of migration or gene flow.

• Worked Application 1: Sickle Cell Disease (SCD):

  • Prevalence: $\sim \frac{1}{600}$ births in African Americans; virtually non-existent in Northern European populations.
  • Inheritance: Autosomal recessive ($ss$).
  • Calculation Steps for Carrier Frequency ($2pq$):
  1. Set affected genotype frequency: $q^2 = \frac{1}{600} \approx 0.001667$
  2. Calculate recessive allele frequency $q$:        $q = \sqrt{\frac{1}{600}} = \frac{1}{24.5} \approx 0.04$
  3. Calculate dominant allele frequency $p$:        $p = 1.0 - 0.04 = 0.96$
  4. Calculate carrier frequency ($2pq$):        $2pq = 2 \times 0.96 \times 0.04 = 0.0768 \approx 0.08$
  • Clinical Interpretation: Approximately $1$ in every $12$ African Americans is a carrier ($Ss$) of Sickle Cell Disease. Universal newborn screening detects carriers and affected individuals.

• Decimal to Fraction Conversion Method:

  • Step 1: Write decimal over 1 (e.g., $\frac{0.75}{1}$).
  • Step 2: Multiply numerator and denominator by $10^n$ based on decimal places ($100$ for 2 digits: $\frac{0.75 \times 100}{1 \times 100} = \frac{75}{100}$).
  • Step 3: Simplify fraction:     $\frac{75}{100} \xrightarrow{\div 5} \frac{15}{20} \xrightarrow{\div 5} \frac{3}{4}$

• Worked Application 2: Cystic Fibrosis (CF):

  • Prevalence Variations:
  • European populations: $\sim \frac{1}{2500}$ births.
  • Asian populations: $\sim \frac{1}{90,000}$ births.
  • United States overall: $\sim 1$ in $3,300$ people.
  • Calculation Steps for European Carrier Frequency ($2pq$):
  1. Set affected genotype frequency: $q^2 = \frac{1}{2500}$
  2. Calculate recessive allele frequency $q$:        $q = \sqrt{\frac{1}{2500}} = \frac{1}{50} = 0.02$
  3. Calculate dominant allele frequency $p$:        $p = 1.0 - 0.02 = 0.98$
  4. Calculate carrier frequency ($2pq$):        $2pq = 2 \times 0.98 \times 0.02 = 0.0392 \approx 0.04$
  • Clinical Interpretation: Approximately $1$ in every $25$ individuals of European descent is a carrier ($Aa$) of Cystic Fibrosis.

Molecular Pathophysiology & Genetics of Cystic Fibrosis

• Gene Locus & Function:
  • Caused by mutations in the CFTR gene encoding an ATP-Binding Cassette (ABC) transporter.
  • Functions as a low-conductance $Cl^-$ selective channel gated by ATP binding and hydrolysis at nucleotide-binding domains (NBDs) and regulated by cyclic AMP (cAMP)-dependent phosphorylation of its regulatory domain.
• Pathophysiological Mechanics:
  • Defective chloride ion transport across mucosal surfaces leads to cell dehydration, hyper-thickening of mucus, and pathognomonic "salty" sweat.
• Multisystem Clinical Spectrum:
  • Respiratory & Pancreatic Systems (Primary targets):
  • Recurrent, life-threatening respiratory infections and pneumonia.
  • Pancreatic exocrine insufficiency, resulting in severe malabsorption, malnutrition, and failure to thrive.
  • Gastrointestinal System:
  • Severe constipation, intestinal obstruction, and meconium ileus in neonates.
  • Genitourinary System (Biological Males):
  • Congenital bilateral absence of the vas deferens leading to azoospermia and infertility.
• Prognosis & Survival Trends:
  • Historical median survival: $31$ years.
  • Modern median survival: Increased to $36.9$ or $47$ years (depending on reference registry data), with children born today expected to live significantly longer due to targeted therapies.

• CFTR Mutation Classes (Over 2,000 specific mutations identified):
  • Class I: Protein Production Mutations (No functional CFTR created).
  • Mechanism: Nonsense, splice site, or frameshift deletions leading to unstable, shortened RNA.
  • Examples: $G542X$, $W1282X$, $R553X$.
  • Prevalence: $22\%$ of CF patients.
  • Therapies: Read-through compounds allowing full-length translation.
  • Class II: Protein Processing Mutations (CFTR misfolds).
  • Mechanism: Protein misfolding prevents trafficking to the apical plasma membrane.
  • Examples: $F508del$ (most prevalent globally), $N1303K$, $I507del$.
  • Prevalence: $88\%$ of CF patients.
  • Therapies: Correctors (e.g., lumacaftor, tezacaftor) to assist proper folding.
  • Class III: Gating Mutations (Channel gate fails to open).
  • Mechanism: Protein reaches cell surface, but channel gate opening is severely impaired.
  • Examples: $G551D$, $S549N$.
  • Prevalence: $6\%$ of CF patients.
  • Therapies: Potentiators (e.g., ivacaftor) to keep the channel gate open.
  • Class IV: Conduction Mutations (Faulty channel conductance).
  • Mechanism: Protein reaches surface with functional gate, but ion transit rate is reduced.
  • Examples: $D1152H$, $R347P$, $R117H$.
  • Prevalence: $6\%$ of CF patients.
  • Therapies: Potentiators (e.g., ivacaftor) to increase ion conductance.
  • Class V: Insufficient Protein Mutations (Reduced quantity of normal protein).
  • Mechanism: Splicing or promoter defects cause reduced synthesis of functional CFTR.
  • Examples: $3849+10\text{kb }C \rightarrow T$, $2789+5\text{G} \rightarrow A$, $A455E$.
  • Prevalence: $5\%$ of CF patients.
  • Refined Classification System: Class 1A (no mRNA produced) and Class 1B (no protein produced).

• Screening Guidelines & Diagnostic Evolution:
  • Historical Guidelines (ACOG, October 2001, ~25 years ago): Recommended carrier screening for individuals with a family history of CF, partners of individuals with CF, and Caucasian couples of European or Ashkenazi Jewish descent.
  • Modern Practice: Mandatory universal newborn screening in all 50 US states. Direct gene sequencing is used to eliminate low detection rates associated with targeted mutation panels.

Practice Genetic Risk Calculations

• Practice Problem 1:

  • Question: What is the probability that a European couple of unknown genotypes will have a child with cystic fibrosis?
  • Step 1: Probability father is a carrier ($Aa$) = $\frac{1}{25}$.
  • Step 2: Probability mother is a carrier ($Aa$) = $\frac{1}{25}$.
  • Step 3: Probability two carriers pass on recessive alleles ($aa$) = $\frac{1}{4}$.
  • Calculation:     $P = \frac{1}{25} \times \frac{1}{25} \times \frac{1}{4} = \frac{1}{2500} = 0.0004 \text{ or } 0.04\%$

• Practice Problem 2:

  • Question: What is the probability that a man known to be a carrier of cystic fibrosis, and his wife of European descent with unknown genotype, will have a child with cystic fibrosis?
  • Step 1: Probability father is a carrier = $1$ ($100\%$ confirmed).
  • Step 2: Probability mother is a carrier = $\frac{1}{25}$.
  • Step 3: Probability two carriers pass on recessive alleles ($aa$) = $\frac{1}{4}$.
  • Calculation:     $P = 1 \times \frac{1}{25} \times \frac{1}{4} = \frac{1}{100} = 0.01 \text{ or } 1\%$

• Practice Problem 3:

  • Question: What is the probability that a healthy sibling of a person with severe cystic fibrosis is a carrier of cystic fibrosis? (Both parents are healthy).
  • Step 1: Since healthy parents produced an affected offspring ($aa$), both parents are obligate carriers ($Aa \times Aa$).
  • Step 2: Standard offspring probabilities for $Aa \times Aa$ are 1\text{ }AA : 2\text{ }Aa : 1\text{ }aa$.\n * *Step 3*: Since the sibling is explicitly stated to be **healthy**, the homozygous recessive (aa) genotype is completely eliminated.\n * *Step 4*: The remaining sample space consists of 3 equal possibilities (1\text{ }AA$and$2\text{ }Aa).\n * *Calculation*:\n    P(\text{Carrier} \mid \text{Healthy}) = \frac{2}{3} \approx 66.7\%$$