Chapter 3 Notes: Mendelian Inheritance, Probability, Chi-square

Mendel and Monohybrid Crosses (Lecture 2.1)

• Learning objectives

  • Explain how Gregor Mendel discovered the principles of heredity.
  • Predict progeny produced in a simple genetic cross.
  • Identify factors that led to Mendel’s success.
  • Explain how the principle of segregation and the concept of dominance account for results in crosses involving one gene or trait.
  • Explain how chromosome separation in meiosis leads to allele inheritance.
  • Predict progeny in a genetic cross using a Punnett square.

• Chapter opener (p 47) note

  • Blond hair occurs in 5–10% of dark-skinned Solomon Islanders; blond hair in this group has a recessive trait and a different genetic basis from blond hair in Europeans.

• Key concepts interwoven in this chapter

  • Mendel’s principles of segregation and independent assortment.
  • Probability.
  • The behavior of chromosomes during meiosis.
  • These concepts are interrelated views of the same phenomenon; integrate them seamlessly.

• Mendel’s experimental subject and rationale (3.1)

  • Subject: Pisum sativum (pea plant).
  • Qualities for selection: easy cultivation, short generation time, many offspring, many pure-breeding varieties, seven characters each with two contrasting forms.
  • Emphasized good experimental methodology and accurate records.
  • Interpreted results with mathematics; formulated and tested hypotheses.
  • Mendel’s success summarized: systematic approach, quantitative analysis, hypothesis testing.

• The seven pea traits Mendel examined (Table/Figure on p 6)

  • Seed color: Yellow vs Green
  • Seed shape: Round vs Wrinkled
  • Seed coat color: Gray vs White
  • Flower position: Axial vs Terminal
  • Stem length: Short vs Tall
  • Pod color: Yellow vs Green
  • Pod shape: Inflated vs Constricted
  • Notes: Each trait had two contrasting forms.

• Genetic terminology (Table 3.1)

  • Gene: An inherited factor (encoded in DNA) that helps determine a characteristic.
  • Allele: One of two or more alternative forms of a gene.
  • Locus: Specific place on a chromosome occupied by an allele.
  • Genotype: Set of alleles possessed by an individual.
  • Heterozygote: An individual possessing two different alleles at a locus.
  • Homozygote: An individual possessing two of the same alleles at a locus.
  • Phenotype or trait: The appearance or manifestation of a characteristic.
  • Characteristic or character: An attribute or feature possessed by an organism.

• Example: Heterozygous at the R-locus

  • An individual may have genotype Rr; other individuals could be RR or rr at the same locus.

• Multiple loci on one homologous chromosome pair

  • Homozygous: two identical alleles at a locus (e.g., BB).
  • Heterozygous: two different alleles at a locus (e.g., Gg, Aa).
  • Example layout shows loci for characteristic 1 (G), 2 (A), 3 (B).

• Genotype vs phenotype; environmental influence

  • Inheritance: An individual inherits alleles of the genotype.
  • Phenotype results from genotype (allele interaction at a locus) plus environmental factors (P = G + E).
  • Mendel focused on phenotypes to deduce genotypes and inheritance rules.

• 3.1 Monohybrid Crosses: segregation and dominance

  • Monohybrid cross: cross between two parents differing in a single trait (e.g., ♂ round seeds × ♀ wrinkled seeds).
  • Pure-breeding lines used; reciprocal crosses involve opposite phenotypes.
  • Mendel’s scientific method applied (repeats, controls).

• Mendel’s conclusions from monohybrid crosses

  • Conclusion 1: One character is encoded by two genetic factors (two alleles for a gene).
  • Conclusion 2: The two genetic factors (alleles) separate when gametes are formed; one allele per gamete.
  • Conclusion 3: Existence of dominant and recessive traits.
  • Conclusion 4: Two alleles segregate with equal probability into gametes.

• Worked example: monohybrid cross (seed shape)

  • Character: seed shape; forms Round (R) and Wrinkled (r).
  • P generation: RR × rr (homozygous dominant × homozygous recessive).
  • F1 generation: all Round seeds (Rr).
  • F2 generation (self-fertilization): genotypes RR, Rr, rr in a 1:2:1 ratio; phenotypes Round:Wrinkled in a 3:1 ratio.
  • Modern interpretation: P generation gametes are R and r; F1 genotype is Rr; self-fertilization yields RR, Rr, rr in 1:2:1; phenotypes 3 Round : 1 Wrinkled.

• What Monohybrid Crosses reveal (Fig. 3.3–3.4 style)

  • P generation: homozygous round (RR) × homozygous wrinkled (rr).
  • Gamete formation: each gamete gets one allele; fertilization forms F1 genotype Rr (Round).
  • F2 generation: genotypes RR, Rr, rr with ratios 1:2:1; phenotypes Round (dominant) : Wrinkled = 3:1.
  • Cross diagrams illustrate gamete combinations (R, r) and self-fertilization outcomes.

• Mendel’s principle of segregation and dominance (Lecture 2.1, p 17)

  • Principle of segregation (Mendel’s First Law):
  • Each diploid organism possesses two alleles for a trait; these alleles segregate during gamete formation; one allele per gamete; equal proportions in gametes.
  • Concept of dominance: when two different alleles are present, one (dominant) determines the phenotype; the other (recessive) is hidden in the presence of the dominant allele.

• Table 3.2: Segregation vs Independent assortment

  • Segregation (Mendel's First Law):
  • State of Meiosis: Anaphase I.
  • Summary: Alleles separate into gametes in equal proportions.
  • Independent assortment (Mendel's Second Law):
  • Alleles at different loci separate independently; applies when loci are on different chromosomes or far apart on the same chromosome.
  • Note: Crossing over can affect assortment, potentially involving Anaphase II if present.

• The Role of DNA in Mendelian Inheritance

  • A chromosome is a linear molecule of DNA wrapped around histone proteins.
  • A gene is a specific DNA sequence that encodes a product (RNA/protein) determining a trait.
  • Alleles are alternative forms of the same gene; variants of DNA sequence that encode variant forms of a trait.
  • Example: R/r alleles for seed shape; locus on chromosome 5 encodes the enzyme SBEI; R-allele is functional; r-allele is a mutation leading to non-functional enzyme.
  • Genotype RR and Rr yield sufficient enzyme for normal starch conversion and water balance; rr results in wrinkled seeds due to disrupted water balance.

• Relating Crosses to Meiosis

  • Chromosome theory of heredity: genes are carried on chromosomes; Mendel’s principles reflect chromosome behavior during meiosis.
  • Diagrammatic idea: DNA replication, Prophase I, crossing over (as shown in concept diagrams).
  • Without crossing over, allele segregation aligns with random alignment of homologs.
  • Crossing over can alter linkage and assorting behavior, depending on distance between loci.

• Predicting the outcome of genetic crosses: Punnett square

  • P generation with tall vs short (for a single locus) example yields genotypic and phenotypic ratios.
  • In foxes (coat color): recessive r with silver vs dominant R with red; cross outcomes yield expected genotypic and phenotypic ratios (to be computed via Punnett square).
  • In rabbits (hair length): dominant H for short hair vs recessive h for long hair; a cross of a short-haired female and a long-haired male can yield a 1:1 phenotypic ratio; if observed, explain the parental genotypes and the expected ratios; address why a single long-haired offspring may occur.

• Summary notes: Mendel’s framework links to modern genetics

  • Mendel’s monohybrid crosses establish segregation and dominance.
  • The genotypic and phenotypic ratios from monohybrid crosses underlie simpler and more complex crosses.
  • All crosses can be interpreted via gamete formation, fertilization, and subsequent offspring ratios.

• Additional notes and problems mentioned

  • The chapter includes worked problems on p65–p67 and p71–p73; review for practice.

Probability as a Tool in Genetics (Lecture 2.2)

• Learning objectives

  • Apply basic probability in predicting the outcome of genetic crosses.
  • Identify when to use multiplication rule, addition rule, and conditional probability.
  • Apply binomial expansion and probability to genetic scenarios.
  • Solve problems using Punnett squares or probability rules.

• Probability basics

  • Probability (P) expresses the likelihood of an event: P = \frac{\text{number of times an event occurs}}{\text{number of all possible outcomes}}
  • Examples: card deck, dice faces; probability can be expressed as a fraction or decimal.
  • How to determine probability: (i) how an event occurs (HOW) and (ii) how often it occurs (HOW OFTEN).

• The multiplication rule (and)

  • For two or more independent events, the probability of both events occurring is the product of their probabilities:
  • Example: probability of rolling a 4 and then a 6 with a fair die: P(4 \text{ on roll 1}) = \frac{1}{6}, \; P(6 \text{ on roll 2}) = \frac{1}{6} \; \Rightarrow \; P(4 \text{ then } 6) = \frac{1}{6} \times \frac{1}{6} = \frac{1}{36}.

• The addition rule (or)

  • For mutually exclusive events, the probability is the sum of their probabilities:
  • Example: probability of rolling a 3 or a 4 on one roll: P(3) = P(4) = \frac{1}{6} \Rightarrow P(3 \text{ or } 4) = \frac{1}{6} + \frac{1}{6} = \frac{2}{6} = \frac{1}{3}.

• Using probability in genetics

  • The multiplication and addition rules can replace Punnett squares for predicting outcomes, especially in multi-locus crosses.
  • Example: Aa × Aa cross: probability of AA progeny is P(AA) = \tfrac{1}{4}.
  • Probability of showing the dominant phenotype (A) in Aa × Aa is P(A) = P(AA) + P(Aa) = \tfrac{1}{4} + \tfrac{1}{2} = \tfrac{3}{4}.$n
  • Punnett squares are convenient for simple monohybrid crosses; probability methods are efficient for complex crosses.

• Conditional probability

  • When additional information is available (e.g., considering only tall plants in a Tt × Tt cross), the probability that a tall plant is heterozygous is: P(\text{heterozygous} \mid \text{tall}) = \frac{P(\text{Tt})}{P(\text{tall})} = \frac{\tfrac{1}{2}}{\tfrac{3}{4}} = \tfrac{2}{3}.

• The binomial expansion and probability

  • Use when there are multiple births or events with two outcomes (mutually exclusive), independent events, and order not specified.
  • Binomial expansion: (p+q)^n = \sum_{k=0}^{n} {n \choose k} p^{n-k} q^{k}
  • Coefficients follow Pascal’s triangle. Examples for small n:
  • n = 1: p+q
  • n = 2: p^2 + 2pq + q^2
  • n = 3: p^3 + 3p^2q + 3pq^2 + q^3
  • n = 4: p^4 + 4p^3q + 6p^2q^2 + 4pq^3 + q^4
  • n = 5: p^5 + 5p^4q + 10p^3q^2 + 10p^2q^3 + 5pq^4 + q^5
  • Coefficients in expansions are given by Pascal’s triangle; row n provides the coefficients for the expansion of (p+q)^n.

• Practical binomial calculation in genetics

  • Example: Aa × Aa with two outcomes (A or a) and n offspring.
  • Computation of the probability of exactly s occurrences of outcome X (with probability p) in n trials is: P = \frac{n!}{s!(n-s)!} p^{s} q^{n-s}, \quad q = 1-p.
  • Example: If p = \tfrac{1}{4} (aa) and q = \tfrac{3}{4} (A_ or AA), for n = 5 and s = 3 (three anemia cases), then P = \frac{5!}{3!2!} (\tfrac{1}{4})^{3} (\tfrac{3}{4})^{2} = 10 (\tfrac{1}{4})^{3} (\tfrac{3}{4})^{2} = 0.088.

• Worked probability problems from the text

  • Five children, Aa × Aa (sickle cell anemia, recessive aa):
  • (1) Five affected: P(aa, aa, aa, aa, aa) = (\tfrac{1}{4})^{5}.
  • (2) First normal, next three affected, last normal: P = (\tfrac{3}{4})(\tfrac{1}{4})(\tfrac{1}{4})(\tfrac{1}{4})(\tfrac{3}{4}) = (\tfrac{3}{4})^2 (\tfrac{1}{4})^3.
  • (3) Any three affected and two normal: P = {5 \choose 3} (\tfrac{1}{4})^{3} (\tfrac{3}{4})^{2} = 10 (\tfrac{1}{4})^{3} (\tfrac{3}{4})^{2} = 0.088.
  • The note: formula vs binomial expansion yield the same result; the coefficient in the expansion corresponds to the number of ways to obtain that combination.

• Conditional probability and multi- locus problems

  • When order does not matter, use the binomial approach; when order or specific sequences matter, use product rule with permutations considered.

• Summary notes

  • Probability is a powerful tool to predict genetic outcomes, especially with multiple loci.
  • The multiplication rule, addition rule, conditional probability, and binomial expansion are core techniques.

Dihybrid crosses and Branch diagrams (Lecture 2.4)

• Learning objectives

  • Use branch diagrams and probability rules to work crosses involving two or more genes.
  • Predict genotypic and phenotypic offspring ratios.
  • Use probability to handle crosses with more than two loci.
  • Explain the value of a testcross in multi-locus contexts.

• Mendel’s dihybrid cross (independent assortment) overview

  • Classic dihybrid cross: Round, Yellow seeds (RRYY × rryy) → F1: Round, Yellow (RrYy); F2: phenotypic ratio ≈ 9:3:3:1 for round yellow : round green : wrinkled yellow : wrinkled green.
  • Principle: alleles at different loci separate independently of each other (Mendel’s Second Law).
  • The segregation principle still applies at each locus (two alleles segregate for each locus).

• Di-hybrid cross mechanics and outcomes

  • For two loci, consider all four loci combinations during gamete formation: RY, Ry, rY, ry.
  • When combining gametes, the progeny phenotypes follow the 9:3:3:1 ratio in the F2 generation due to independent assortment of two loci.
  • Example diagram shows parental gametes RY and ry forming F1, then self-fertilization producing the four phenotypes in the 9:3:3:1 ratio.

• Linkage caveat and crossing over

  • Genes on different chromosomes assort independently regardless of distance.
  • Genes on the same chromosome do not assort independently unless crossing over occurs between them (distance or recombination rate matters).

• Branch diagrams vs Punnett squares in dihybrids

  • Branch diagrams provide a faster route to probabilities for multi-locus crosses.
  • Example cross Aa Bb cc Dd Ee × Aa Bb Cc dd Ee can be analyzed by multiplying individual locus probabilities or by building a branch diagram.
  • If order of events is not specified, use the binomial/product approach; if order matters, a branch diagram helps to organize terms.

• Dihybrid testcross and testcross utility

  • A dihybrid testcross (RrYy × rryy) can be analyzed using Punnett squares or branch diagrams to predict proportions for each phenotype.
  • Testcross value: helps determine genotype of an individual with a dominant phenotype by comparing offspring with a homozygous recessive tester.
  • Example outcomes: using branch diagrams yields expected proportions for each phenotype (e.g., Round/Yellow, Round/Green, Wrinkled/Yellow, Wrinkled/Green).

• Dihybrid cross worked example with cucumber traits

  • Three characters (each encoded by distinct loci on separate chromosomes): dull vs glossy (D/d), orange vs cream (R/r), bitter vs nonbitter cotyledons (B/b).
  • If the two parental lines are homozygous for the extreme phenotypes and are crossed, the F1 are heterozygous for all three loci; intercrossing F1 yields the F2 with a 9:3:3:1:1:1:1:1:1 distribution across the eight possible phenotypes (conceptual).
  • The problem set asks to determine phenotypes/proportions and then to perform a cross with a tester line to predict progeny phenotypes.

• Branch diagram practice

  • Use branch diagrams to track independent assortment across multiple loci.
  • For each locus, apply the monohybrid ratio and multiply across loci to get the multi-locus genotype/phenotype ratios.

• Summary notes

  • Dihybrid crosses illustrate how independent assortment creates a 9:3:3:1 phenotypic ratio when loci assort independently.
  • Branch diagrams provide an efficient alternative to Punnett squares for multi-locus crosses.
  • Testcrosses remain valuable for determining genotype when multiple loci are involved.

Chi-Square Test (Lecture 2.5)

• Learning objectives

  • Use the chi-square goodness of fit test to determine if deviations between observed and expected progeny numbers can be due to chance.
  • Calculate the chi-square value and the associated probability.
  • Interpret the probability from the chi-square test.

• Why observed ratios may deviate from expected

  • Ratios are predictions based on Mendelian principles; experimental results often differ due to random sampling fluctuations.
  • Small sample sizes increase error.

• The Goodness-of-Fit Chi-Square Test

  • Formula: \chi^2 = \sum \frac{(O - E)^2}{E} where O = observed, E = expected.
  • Degrees of freedom: df =\text{(number of classes)} - 1.
  • Compare the calculated chi-square to a chi-square distribution table to obtain a probability (P).
  • If P is large (typically P > 0.05), differences are attributed to chance (do not reject H0). If P is small (P < 0.05), differences are considered significant (reject H0).

• Worked examples

  • Cockroaches: brown (B) vs yellow (b) with a Yy × yy cross; expected ratio is 1:1 (brown:yellow). In a sample of 40 offspring, observed could be 21 brown and 19 yellow, 27 brown and 13 yellow, or 7 brown and 33 yellow; apply chi-square to evaluate.
  • Mendel’s dihybrid cross data: 315 round yellow, 108 round green, 101 wrinkled yellow, 32 wrinkled green; test against the 9:3:3:1 expectation using \chi^2 = \sum \frac{(O - E)^2}{E} with E values derived from the 9:3:3:1 proportions.

• Critical values and interpretation

  • Table 3.7 lists critical values for the chi-square distribution at various degrees of freedom and probabilities.
  • Most scientists use the 0.05 significance level as a cutoff for significance.
  • Practical example: If the calculated chi-square yields P > 0.05, accept that deviations are due to chance.

• Worked calculation examples from the notes

  • Example: Purple vs White flowers with a P generation cross gives observed 105 purple and 45 white; expected based on 3/4 and 1/4 of 150: 112.5 and 37.5; chi-square value calculated as 2.0 with df = 1; P-value between 0.1 and 0.5; conclusion: no significant difference.
  • Example: Dihybrid cross (Mendel’s 9:3:3:1) with observed counts 315 round yellow, 108 round green, 101 wrinkled yellow, 32 wrinkled green; compute chi-square to test fit to 9:3:3:1 (procedure outlined above).

• Summary notes

  • Chi-square tests quantify whether deviations from expected Mendelian ratios are due to chance.
  • Key steps: (i) determine expected numbers, (ii) compute chi-square, (iii) determine df, (iv) consult table for probability, (v) interpret results in context of H0 (no deviance beyond chance).

• Homework and practice prompts mentioned

  • Achieve practice exercises: Chi-square tutorials, interactive assignments, and problem-solving videos.

• End-of-study-unit guidance

  • Review textbook and lecture material.
  • Use Achieve resources for practice.
  • Complete study guide questions and assignments.
  • Prepare for tutorials; use contact time effectively; seek help if needed.

Terminology and naming conventions for alleles (Notes from Lecture 2.3, 2.4)

• Allele naming rules

  • One gene is carried at a single locus on homologous chromosomes, but different alleles can exist.
  • For different alleles of the same gene, use the same letter(s) of the alphabet and distinguish alleles by:
  • Upper- / lowercase letters (A, a),
  • Superscripts or subscripts (e.g., Lfs1, Lfs2),
  • A plus superscript to denote wild-type allele (ye+) vs mutant allele (ye).
  • Slash to distinguish two alleles in a genotype (El+/ElR or +/ElR).
  • Spaces separate genotypes for multiple loci (E1+/E1R G/g).

• Acceptable genetic symbols for alleles

  • Dominant allele uses uppercase, recessive uses lowercase (A, a).
  • Multiple letters per allele (Hl, hl) or (Azh, azh).
  • Wild-type allele with plus superscript (ye+), mutant allele without plus (ye).
  • Superscripts and subscripts can be used (Lfr1, Lfr2).
  • Slashes distinguish two alleles in a genotype (El+/ElR or +/ElR).

• The complexity of genetic traits (multifactorial inheritance)

  • Not all traits follow simple Mendelian patterns.
  • Variations can arise from: genetic variation at multiple loci that yield the same phenotype, interactions between alleles at multiple loci, and environmental factors that influence phenotypes.
  • Multifactorial inheritance is common; genetic-environmental interaction is often the rule, not the exception.

Conceptual notes on inheritance and meiosis (General wrap-up)

• The core of Mendel’s success lies in observing how alleles segregate and how different loci assort independently under meiosis.

• The chromosome theory links genes to chromosomes, explaining why Mendel’s rules hold across generations.

• When analyzing crosses, consider:

  • The genotype at each locus (homozygous vs heterozygous).
  • How gametes are formed and what alleles they carry.
  • How many loci are involved and whether they assort independently.
  • The potential environmental influences on phenotype.

• Practical exam-ready points:

  • Know genotype-to-phenotype relationships for dominant/recessive traits.
  • Be able to construct and interpret Punnett squares for monohybrid and dihybrid crosses and to translate results into genotypic and phenotypic ratios.
  • Be able to apply probability rules (multiplication, addition, conditional) to predict cross outcomes and to use the binomial expansion for larger-family scenarios.
  • Be able to perform and interpret a chi-square goodness-of-fit test for observed vs. expected progeny counts.

Quick reference formulas and key ratios

• Monohybrid cross results (typical):
  • Genotypic ratio: 1:2:1
  • Phenotypic ratio (dominant vs recessive): 3:1
• Dihybrid cross results (independent assortment): 9:3:3:1
• Probability rules
  • Multiplication rule: if events are independent, P(A\text{ and } B) = P(A) \times P(B)
  • Addition rule: for mutually exclusive events, P(A\text{ or } B) = P(A) + P(B)
• Conditional probability: e.g., tall plant heterozygosity given tall phenotype in a Tt × Tt cross is P(\text{Tt} \mid \text{tall}) = \frac{P(\text{Tt})}{P(\text{tall})} = \frac{\tfrac{1}{2}}{\tfrac{3}{4}} = \tfrac{2}{3}
• Binomial expansion: (p+q)^n = \sum_{k=0}^{n} {n \choose k} p^{n-k} q^{k}
  • Examples: for n = 2: p^2 + 2pq + q^2; n = 3: p^3 + 3p^2q + 3pq^2 + q^3; n = 5: p^5 + 5p^4q + 10p^3q^2 + 10p^2q^3 + 5pq^4 + q^5
• Chi-square goodness-of-fit: \chi^2 = \sum \frac{(O-E)^2}{E}; df = \text{(number of classes)} - 1$$; compare to critical values to determine P.
• Testcross genetics
  • A testcross with a homozygous recessive tester (tt) reveals the genotype of a dominant-phenotype individual, e.g., TT × tt → all Tt; Tt × tt → 1:1 Tt:tt.