General Genetics - Transmission Genetics II Notes

Transmission Genetics II: Key Concepts

• Today’s topics recap monohybrid crosses, then extend to dihybrid and beyond (tri- and tetrahybrids). Focus on (i) Mendel’s principles, (ii) probability rules (product and addition), (iii) Punnett squares for multiple genes, and (iv) hypothesis testing with the Chi-square test.

Test cross: determining genotype of a black moth

• Phenotype controlled by a single gene: dominant allele B encodes black; recessive allele b encodes red.
• Unknown genotype of a black moth could be BB or Bb.
• To reveal genotype, cross the black moth to a homozygous recessive (bb) tester:
  • Cross: black (BB or Bb) × red (bb).
• Expected outcomes:
  • If the black parent is BB: all offspring are Bb (black).
  • If the black parent is Bb: offspring are 1/2 Bb (black) and 1/2 bb (red).
• Correct approach: cross to a red moth (bb) to determine genotype of the black parent: D) A red one (bb).

Mendelian inheritance basics

• Monohybrid crosses recap:
  • Pure-breeding (homozygous) parental genotypes produce F1 all heterozygotes for one trait.
  • F1 cross (heterozygous × heterozygous) yields phenotypic ratio: 3:1 (dominant:recessive) and genotypic ratio: 1:2:1 (AA:Aa:aa).
• Di-hybrid crosses extend to two genes with independent assortment: phenotypic ratio 9:3:3:1 for two traits.
• Product rule: If two events are independent, the probability of both is the product: P(A ext{ and } B) = P(A) imes P(B).
• Addition rule (mutually exclusive events): The probability that either event occurs is the sum: P(A ext{ or }B) = P(A) + P(B). In general, P(A ext{ or } B) = P(A) + P(B) - P(A ext{ and }B).
• Chi-square test is used to test whether observed data fit Mendelian ratios: oxed{ \chi^2 = \sum \frac{(Oi - Ei)^2}{Ei} } where Oi are observed counts, E_i are expected counts, and degrees of freedom df = (number of classes − 1).
• Critical value for df=1 at p=0.05 is 3.841. If \chi^2 > 3.841, reject H0 (data do not fit the expected ratio). If \chi^2 < 3.841, fail to reject H0 (data fit the ratio).

Gene naming conventions and dominance

• Dominant alleles are uppercase; recessive alleles are lowercase.
• Genes are often named after the mutant phenotype (e.g., W = wrinkled, w = wrinkled’s recessive phenotype; W = round is dominant). Wild type is the phenotype commonly found in natural populations or in the lab.
• Wild-type phenotype is not always the dominant phenotype; dominance refers to phenotype expression within an individual, not population frequency.
• Example: B = brown eyes, b = blue eyes; in a population, brown (dominant) may be less frequent than blue (recessive) depending on allele frequencies.

Mendel’s single-trait analyses

• Seed color: yellow vs green; Seed shape: round vs wrinkled.
• Typical results (example from Mendel’s work):
  • Seed color (yellow dominant): 75% yellow, 25% green in F2 from a dihybrid cross? (Monohybrid example shows 3:1 for single trait in phenotypes.)
  • Seed shape (round dominant): 75% round, 25% wrinkled in F2.
• These results illustrate the 3:1 phenotypic ratio and the 1:2:1 genotypic ratio for a single gene with complete dominance.

Inheritance of multiple genes: dihybrid crosses

• When studying two traits (e.g., seed color and seed shape) with independent assortment:
  • F1 genotype: AaBb (two heterozygous loci).
  • F2 phenotypic ratio: 9:3:3:1 for two-trait phenotypes (e.g., yellow-round : yellow-wrinkled : green-round : green-wrinkled).
• Dihybrid cross Punnett square shows 16 genotype/phenotype combinations due to two loci, each with two alleles.
• Gametes produced by meiosis carry one allele from each gene; for AaBb, possible gametes: AB, Ab, aB, ab.

Independent assortment and gamete formation

• Alleles at different genes on different chromosomes segregate independently during gamete formation.
• Each gamete contains one copy of each chromosome (one allele per gene).
• Result: all combinations of alleles for the two genes appear in offspring with the 9:3:3:1 phenotypic ratio for the dihybrid cross.

Punnett squares and probability in multi-gene crosses

• Punnett squares scale up as the number of genes increases; they become large quickly.
• Trihybrid cross (three genes): Forked-line (branching) method is used to simplify visualization.
• For AaBbCc × AaBbCc, the phenotypic ratio is 27:9:9:9:3:3:3:1 across eight phenotypic classes (each class corresponds to a pattern of dominant/recessive expression across the three traits).
• Key ratio for tri-hybrids (Forked-line rule): 27 : 9 : 9 : 9 : 3 : 3 : 3 : 1, totaling 64 possibilities.
• For three unlinked genes, the number of phenotype classes is 8 (2^3).

Trihybrid and Forked-Line rule

• Trihybrid forked-line rule breakdown (AaBbCc × AaBbCc):
  • The 64 equally likely offspring outcomes distribute into eight phenotype classes with the 27:9:9:9:3:3:3:1 ratio.
• This demonstrates Mendel’s principle of independent assortment extending to three genes.

Tetrahybrid cross (four genes)

• In a tetrahybrid cross, a plant with genotype Aa Bb Cc Dd is crossed to itself.
• Probability of obtaining genotype Aa Bb CC Dd in offspring:
  • For locus A: Aa from Aa × Aa has probability \tfrac{1}{2}.
  • For locus B: Bb from Bb × Bb has probability \tfrac{1}{2}.
  • For locus C: CC from Cc × Cc has probability \tfrac{1}{4}.
  • For locus D: Dd from Dd × Dd has probability \tfrac{1}{2}.
  • Multiply: \tfrac{1}{2} \times \tfrac{1}{2} \times \tfrac{1}{4} \times \tfrac{1}{2} = \tfrac{1}{32}.
• Correct answer: $\dfrac{1}{32}$.

Using probability to tackle multiple gene inheritance

• Product rule (independence): P(A \text{ and } B) = P(A) P(B).
• Addition rule (mutually exclusive): P(A \text{ or } B) = P(A) + P(B); in general, P(A \text{ or } B) = P(A) + P(B) - P(A \text{ and } B).
• Example: Probability of rolling at least one 4 with two dice:
  • P(at least one 4) = 1 − P(no 4 on either die) = 1 − (5/6)^2 = \frac{11}{36}.
• Another classic: Probability of two 5’s when rolling a pair of dice: \left(\tfrac{1}{6}\right)^2 = \tfrac{1}{36}.

Practice questions (selected from the lecture slides)

• Question 1 (test cross concept): In a test cross of a black moth (B_) where B is dominant to b, which tester should be used to reveal genotype of the black moth?
  • A) A black BB
  • B) A black bb
  • C) A red Bb
  • D) A red bb
  • E) Not possible to determine by cross
  • Answer: D) cross with bb (tester).
  • Rationale: Crossing a black phenotype to bb reveals whether the black parent is BB (produces all black offspring) or Bb (produces 1:1 black:red offspring).
• Question 2 (probability with dice): Probability of getting two 5’s when rolling two dice? Options include 1/36, etc.
  • Answer: $\dfrac{1}{36}$ (independent events, each die shows 5 with probability 1/6; product rule gives $(1/6)^2$).
• Question 3–7 (dihybrid context): Various dihybrid cross questions involving genotype vs phenotype, heterozygosity, and conditional probabilities (examples shown in the lecture). Answers depend on applying the product rule and Mendel’s 9:3:3:1 framework for phenotypes, or the 1:2:1/3:1 rules for genotypes as appropriate.
• Question 8 (additive rule example): Probability of at least one 4 when rolling two dice is 11/36; by the additive rule: P(A or B) = P(A) + P(B) − P(A and B). Here, A = first die is 4, B = second die is 4; P(A) = P(B) = 1/6; P(A and B) = 1/36; thus P(at least one 4) = 1/6 + 1/6 − 1/36 = 11/36.
• Question 9 (tetrahybrid): Probability that Aa Bb Cc Dd × Aa Bb Cc Dd yields Aa Bb CC Dd: \dfrac{1}{32} as shown above.

Chi-square test in practice

• Use when comparing observed vs expected phenotypic counts to test if data fit a Mendelian ratio.
• Example setup:
  • Observed A-class: 85, B-class: 35; Total = 120.
  • Expected under H0 (3:1 ratio): A-class = \frac{3}{4} \times 120 = 90\,, B-class = \frac{1}{4} \times 120 = 30\,.
  • Compute: \chi^2 = \frac{(85-90)^2}{90} + \frac{(35-30)^2}{30} = \frac{25}{90} + \frac{25}{30} = 1.11.
• Decision rule: If \chi^2 < 3.841 (df = 1, p = 0.05), fail to reject H0; data are consistent with a 3:1 Mendelian ratio. If greater, reject.
• How to interpret: A small chi-square value means observed data fit expected Mendelian segregation; this supports single-gene control with segregation according to Mendel’s Law.

Key formulas to memorize

• Product rule (independence): P(A \text{ and } B) = P(A)P(B)
• Addition rule (mutually exclusive events): P(A \text{ or } B) = P(A) + P(B) (general: P(A \text{ or } B) = P(A) + P(B) - P(A \text{ and } B))
• Mendel’s dihybrid ratio: 9:3:3:1
• Mendel’s tri-hybrid (three genes) phenotype ratio: 27:9:9:9:3:3:3:1
• Mendel’s tetra-hybrid cross probability (example): For Aa Bb Cc Dd × Aa Bb Cc Dd, probability of Aa Bb CC Dd is \dfrac{1}{32}
• Chi-square test statistic: \chi^2 = \sumi \frac{(Oi - Ei)^2}{Ei}
• Critical value for df = 1 at p = 0.05: 3.841

Connections to foundational principles and real-world relevance

• Independent assortment underpins how different traits segregate and combine in offspring; explains why dihybrid crosses yield the 9:3:3:1 ratio when genes are unlinked.
• The product and addition rules are fundamental probabilistic tools across biology and other disciplines, enabling calculation of complex trait probabilities beyond single-gene inheritance.
• Chi-square testing provides a quantitative framework to evaluate whether observed data match expected Mendelian ratios, allowing inference about genetic architecture (single-gene vs polygenic, linkage, or other deviations).
• Understanding dominance, wild-type concept, and gene naming helps interpret genetic data in research and clinical contexts, where phenotypes may not always align with simple dominant-recessive expectations.

Quick reference: examples from the slides

• Test cross to reveal genotype of a black moth: cross to bb; if offspring are all black, parent is BB; if 1:1 black:red, parent is Bb.
• Monohybrid cross: P generation true-breeding → F1 all Aa → F2 phenotype ratio 3:1; genotype ratio 1:2:1.
• Dihybrid cross: AaBb × AaBb → F2 phenotype ratio 9:3:3:1; genotype combinations form a 16-panel Punnett square.
• Trihybrid cross: AaBbCc × AaBbCc → phenotypic ratio 27:9:9:9:3:3:3:1; forked-line method simplifies tracking.
• Tetrahybrid cross: Aa Bb Cc Dd × Aa Bb Cc Dd → probability of Aa Bb CC Dd is \dfrac{1}{32}.
• Chi-square example: observed 85 A-class, 35 B-class with total 120; expected 90 and 30; compute \chi^2 = \frac{(85-90)^2}{90} + \frac{(35-30)^2}{30} = 1.11; since 1.11 < 3.841, fail to reject H0 at p = 0.05.

Glossary of key terms

• Dominant allele: masks the expression of the other allele in a heterozygote; uppercase letter.
• Recessive allele: secreted in homozygotes in presence of a dominant allele; lowercase letter.
• Wild-type: most common phenotype in a natural population or lab setting.
• Homozygous: same allele on both chromosomes (AA or aa).
• Heterozygous: two different alleles (Aa).
• Phenotype: observable traits.
• Genotype: genetic makeup (alleles) of an organism.
• Independent assortment: alleles of genes on different chromosomes separate independently during gamete formation.
• P-value (in Chi-square): probability of observing data as extreme as the current data, assuming H0 is true. A small p-value (< 0.05) leads to rejecting H0.

End of notes