module 5

Administrative and Examination Information

• Course Identification: Genetics BIO2133_C00.

• Instructor: Hadi Khalil Ph.D.

• Midterm 1 Coordination:

  • Date for First Session: 10 Feb, 2026 @ 11:30.

  • Date for Second Session: 13 Mar, 2026 @ 01:00.

  • Type: Midterm.

  • Format: Paper.

  • Duration: 75 minutes.

  • Accommodation Requirements: Students with accommodations must submit a Notice of Examination (NOE) for each evaluation at least ten (10) days in advance.

Criteria for Homologous Chromosomes

• To classify two chromosomes as a homologous pair, they must meet the following criteria:

  • Both must be the same size.

  • They must exhibit identical centromere locations.

  • They must form pairs or synapses during the stages of meiosis.

  • They must contain an identical linear order of gene loci.

  • Origin: In each pair, there is a paternal homolog (derived from the sperm) and a maternal homolog (derived from an oocyte).

  • Exceptions: The X and Y chromosomes in mammals are excluded from these criteria.

Mendel’s Four Postulates

• Postulate 1: Unit Factors in Pairs: Genetic characters are controlled by unit factors that exist in pairs in individual organisms (e.g., $BB$, $Bb$, and $bb$).

• Postulate 2: Dominance/Recessive: When two unlike unit factors responsible for a single character are present in one individual, one unit factor is dominant to the other, which is recessive. Only the dominant factor is visible in the $F_1$ generation.

• Postulate 3: Segregation: During gamete formation, the paired unit factors separate (segregate) randomly so that each gamete receives one or the other with equal likelihood.

  • Example: An individual with genotype $Bb$ produces gametes with a $50\%$ probability of receiving either the $B$ or the $b$ unit factor.

• Postulate 4: Independent Assortment: During gamete formation, segregating pairs of unit factors assort independently of each other. This results in all possible combinations of gametes being formed with equal frequency.

• Definition: Mendel's "Unit factors" are now known as Alleles.

Allele Symbols and Gene Nomenclature

• General Symbols:

  • Dominant Alleles: Indicated by an italic uppercase letter ($D$) or letters ($W^r$).

  • Recessive Alleles: Indicated by an italic lowercase letter ($d$) or a group of italic letters ($w^r$).

  • Genetically Altered Alleles: Indicated by an italic letter ($e$).

  • Wild-type Alleles: Indicated by an italic letter plus a superscript ($e^+$).

• Gene and Protein Nomenclature Examples ($\beta$-actin):

  • Human Gene: Uppercase and italicized ($ACTB$).

  • Human Protein: Uppercase, NOT italicized (ACTB).

  • Mouse Gene: First letter uppercase, rest lowercase, italicized ($Actb$).

  • Mouse Protein: Uppercase, NOT italicized (ACTB).

• Locus Notations:

  • Separated Alleles: For a given locus, two alleles are written separated by a slash (e.g., $A/a$).

  • Different Chromosomes: If two loci are on different chromosomes, they are separated by a semi-colon (e.g., $A/a \, ; \, B/b$).

  • Same Chromosome (Linked): If two loci are on the same chromosome, the alleles are combined and the slash separates the homologs (e.g., $AB/ab$).

Laws of Probability: Product Law and Sum Law

• Product Law (AND Rule):

  • Definition: The probability of two or more independent genetic events occurring simultaneously is equal to the product of their individual probabilities.

  • Formula: $Pr(A \text{ and } B) = Pr(A) \times Pr(B)$.

  • Genetics Example: Probability of inheriting $aa$ from $Aa \times Aa$ parents:

    • $P(a) \text{ from parent 1} = 1/2$

    • $P(a) \text{ from parent 2} = 1/2$

    • $P(aa) = 1/2 \times 1/2 = 1/4$

  • Industrial Example: Drug Quality Control. Inspection A pass rate = $95\%$, Inspection B pass rate = $90\%$. Probability of passing both: $0.95 \times 0.90 = 0.855$ ($85.5\%$).

• Sum Law (OR Rule):

  • Definition: The probability of an outcome that can be produced by multiple, mutually exclusive genetic events equals the sum of their individual probabilities.

  • Formula: $P(A \text{ or } B) = P(A) + P(B)$.

  • Genetics Example: Probability of a dominant phenotype in $Rr \times Rr$ breeding:

    • $P(RR) = 1/4$

    • $P(Rr) = 1/2$

    • $P(\text{dominant}) = 1/4 + 1/2 = 3/4$

  • Inheriting at least one recessive allele r: $Pr(Rr) + Pr(rR) + Pr(rr) = 1/4 + 1/4 + 1/4 = 3/4$.

  • Medical Diagnosis: If a disease is caused by Mutation in gene A ($30\%$ cases) or Gene B ($20\%$ cases) and causes are mutually exclusive: $0.30 + 0.20 = 0.50$.

Probability Practice Problems

• Case Study: Arabidopsis plants:

  • Loci: $A/a$ (pigment), $B/b$ (leaf edge), $C/c$ (trichomes), $D/d$ (height), $E/e$ (seed shape).

  • Cross: $A/a \, ; \, b/b \, ; \, C/c \, ; \, D/d \, ; \, E/e$ and $A/a \, ; \, B/b \, ; \, C/c \, ; \, D/d \, ; \, e/e$.

  • Problem: Probability of $F_1$ showing fully recessive phenotype (green, smooth, no trichomes, dwarf, wrinkled):

    • $a/a$ probability: $1/4$

    • $b/b$ probability: $1/2$

    • $c/c$ probability: $1/4$

    • $d/d$ probability: $1/4$

    • $e/e$ probability: $1/2$

    • Calculation: $1/4 \times 1/2 \times 1/4 \times 1/4 \times 1/2 = 1/256$.

Dihybrid Cross and Mendelian Assumptions

• Dihybrid Cross: Observations involving two pairs of contrasting traits (e.g., pea color and shape).

• Ratios: Monohybrid cross yields a $3:1$ ratio; Dihybrid cross yields a $9:3:3:1$ ratio.

• Implicit Assumptions for Predicted Ratios:

  1. Each allele is strictly dominant or recessive.

  2. Segregation is random.

  3. Independent assortment occurs.

  4. Fertilization is random.

• If observed ratios deviate significantly from the predicted ratios, at least one of these assumptions is violated (e.g., non-independent assortment or non-random fertilization).

Chi-Square ($\chi^2$) Analysis

• Purpose: A statistical method to determine if observed offspring ratios differ significantly from expected Mendelian ratios due to chance.

• Variables Affecting Deviations:

  1. Independent Assortment: Subject to random fluctuations.

  2. Sample Size: Average deviation decreases as the total sample size increases.

• Formula:

  • $\chi^2 = \sum \frac{(O-E)^2}{E}$

  • $O$ = observed number of offspring; $E$ = expected number of offspring.

• Degrees of Freedom ($df$):

  • Calculated as $(n-1)$, where $n$ is the number of possible outcomes/categories.

  • Monohybrid cross (2 phenotypes): $df = 1$.

  • Dihybrid cross (4 phenotypes): $df = 3$.

• Distribution Characteristics:

  • Asymmetric and right-skewed.

  • Defined for non-negative values ($\chi^2 \geq 0$).

  • Shape depends on the $df$.

  • Different from the symmetric, bell-shaped normal distribution.

The Null Hypothesis ($H_0$) and P-values

• Null Hypothesis ($H_0$):

  • Assumes the data fits a given ratio (e.g., $3:1$).

  • Assumes no real difference between measured and predicted values; differences are attributed to chance.

• Decision Rules (for $\alpha = 0.05$, and critical value $3.84$ at $df=1$):

  • If \chi^2 < 3.84: Do not reject the null hypothesis; deviation is consistent with chance.

  • If \chi^2 > 3.84: Reject the null hypothesis; deviation is too large to be attributed to chance.

• Probability ($p$-value): The probability that the observed deviation occurred strictly as a result of chance.

  • High $p$-value means the observed ratio is likely due to chance (cannot reject $H_0$).

Chi-Square Exercise: Mendel’s F2 Data

• Data: $315$ round yellow (RY), $108$ round green (Rg), $101$ wrinkled yellow (rY), $32$ wrinkled green (rg). Total $N=556$.

• Testing 9:3:3:1 Ratio ($df=3$):

  • Expected values: $RY = 312.75$, $Rg = 104.25$, $rY = 104.25$, $rg = 34.75$.

  • $\chi^2 = 0.016 + 0.135 + 0.101 + 0.218 = 0.47$.

  • $p \approx 0.93$. Conclusion: Accept the ratio (no significant deviation).

• Testing 3:1 Ratio (Round vs Wrinkled, $df=1$):

  • Observed: Round ($315+108 = 423$); Wrinkled ($101+32 = 133$).

  • Expected: Round ($417$); Wrinkled ($139$).

  • $\chi^2 = \frac{(423-417)^2}{417} + \frac{(133-139)^2}{139} = 0.086 + 0.259 = 0.35$.

  • $p \approx 0.56$. Conclusion: Accept the observed ratio.

Extensions of Mendelian Genetics

• Types of Dominance:

  1. Complete Dominance: One allele masks the expression of the other.

  2. Incomplete Dominance: Neither allele is dominant; the heterozygote shows a blend of phenotypes (intermediate).

     • Example: Familial hypercholesterolemia (FH) involving the $LDLR$ gene.

       • Normal: $LDLR^+/LDLR^+$.

       • Heterozygote: $LDLR^+/LDLR^-$ (moderately high cholesterol).

       • Homozygote mutant: $LDLR^-/LDLR^-$ (very high cholesterol, early heart disease).

  3. Codominance: Both alleles are fully and simultaneously expressed (e.g., ABO blood type).

• Polygenic Traits: Traits affected by multiple genes (gene interaction), resulting in continuous variation.

  • Skin Color: Influenced by over $378$ genes; $7$ discrete phenotypes are produced from a 3-loci cross ($2^3 = 8$ gametes).

• Multiple Alleles: Presence of three or more alleles for one gene increases diversity (e.g., ABO system).

• Lethal Alleles: Cause nonviable offspring by disrupting essential life functions, altering expected ratios.

ABO Blood Groups and the Bombay Phenotype

• ABO System: Determined by three alleles ($I^A$, $I^B$, and $i$).

  • $I^A$ and $I^B$ are codominant to each other and dominant to $i$.

  • Antigens: Carbohydrates bound to lipids on red blood cells (Isoagglutinogens).

  • H Substance: A precursor molecule. The wild-type $FUT1$ allele adds fucose to a molecule to create the H substance. Antigens A and B are then added to this H substance.

• Bombay Phenotype ($O_h$):

  • Mechanism: Individuals have the genotype $hh$ (homozygous recessive at the H gene on chromosome 19).

  • Result: No H antigen is produced. Therefore, A and B antigens cannot be formed, even if the person carries $I^A$ or $I^B$ alleles. The blood type phenotypically appears as O.

  • Genes involved: ABO gene (chromosome 9) and H gene ($FUT1$, chromosome 19).