Comprehensive Genetics Study Guide: Linkage, Mapping, and Inheritance

Genetic Mapping, Linkage, and Test Crosses

• Homologous Chromosomes and Notation:

  • An individual possesses pairs of homologous chromosomes, inheriting one copy from the father and one copy from the mother.
  • For two loci ($A$ and $B$), an individual heterozygous at both loci can be represented using several equivalent notations:
  • Chromosomal line notation: $\frac{A \quad B}{a \quad b}$ or $\frac{A \quad b}{a \quad B}$
  • Slash notation: $AB / ab$ or $Ab / aB$
  • Standard genotype notation: $AaBb$
  • Horizontal lines explicitly represent physical chromosome strands.

• Standard Two-Point Test Cross Setup:

  • A test cross pairs a dihybrid (heterozygous at two loci) with a homozygous recessive individual:     $\frac{A \quad B}{a \quad b} \times \frac{a \quad b}{a \quad b}$
  • The homozygous recessive parent ($aabb$) produces only recessive gametes ($ab$). Therefore, the phenotypic outcome of the offspring directly reveals the gametic output of the heterozygous parent.

• Unlinked Genes (Independent Assortment):

  • When loci reside on separate chromosomes or are located very far apart on the same chromosome, gametes mix in equal proportions.
  • Expected gametic frequency from $\frac{A \quad B}{a \quad b}$:
  • $AB = \frac{1}{4}$ ($25\%$
  • $Ab = \frac{1}{4}$ ($25\%$
  • $aB = \frac{1}{4}$ ($25\%$
  • $ab = \frac{1}{4}$ ($25\%$
  • Expected offspring phenotypic ratio: $1 : 1 : 1 : 1$ (or $\frac{1}{4}$ for each combination).

• Linked Genes and Linkage Disequilibrium:

  • When genes are physically close to each other on the same chromosome, alleles tend to be inherited together rather than assorting independently.
  • Deviations from the $1 : 1 : 1 : 1$ expected test cross ratio indicate genetic linkage.
  • Example observed frequency under linkage:
  • Non-recombinant (Parental) types: $35\%$ $AB$, $35\%$ $ab$
  • Recombinant types: $15\%$ $Ab$, $15\%$ $aB$
  • Terminology:
  • Linkage Disequilibrium: Historical term describing allele combinations present at frequencies different from those expected if loci were assorting independently in equilibrium.
  • Linkage Groups: Genes located on the same chromosome form a linkage group. Genes separated by more than $50\,\text{centimorgans (cM)}$ behave as if they are on separate chromosomes due to obligatory recombination events.

• Chromosomal Configurations:

  • Cis Configuration (Coupling): Mutant or dominant alleles are present on the same homologue ($\frac{A \quad B}{a \quad b}$).
  • Trans Configuration (Repulsion): Dominant and recessive alleles are on opposite homologues ($\frac{A \quad b}{a \quad B}$).
  • Parental Phenotypes: Possess the highest numerical counts in test cross offspring because crossing over is a relatively rare event.
  • Recombinant Phenotypes: Possess the smallest numerical counts resulting from meiotic crossing over between non-sister chromatids.
  • Maximum possible recombination frequency between two loci is $50\%$.

• Worked Example of Map Distance Calculation:

  • Test cross offspring counts for loci $A$ and $B$ ($N = 650$):
  • $AB = 230$
  • $Ab = 99$
  • $aB = 97$
  • $ab = 224$
  • Step 1: Identify Parentals vs. Recombinants:
  • Parentals (highest counts): $AB = 230$, $ab = 224$
  • Recombinants (lowest counts): $Ab = 99$, $aB = 97$
  • Step 2: Calculate Recombination Frequency:     $\text{Recombination Frequency} = \frac{\text{Total Recombinants}}{\text{Total Offspring}} = \frac{99 + 97}{650} = \frac{196}{650} \approx 0.3015$
  • Step 3: Convert to Genetic Distance:     $\text{Map Distance} = 30.15\% = 30.15\,\text{cM (or Morgan units)}$

Statistical Tests for Inheritance Patterns

• Distinguishing Assortment from Segregation:

  • Mendel's Law of Independent Segregation: Two alleles of a single locus separate during gamete formation such that half the gametes carry one allele and half carry the other ($1 : 1$ ratio).
  • Mendel's Law of Independent Assortment: Alleles at different loci separate independently of one another during gamete formation ($1 : 1 : 1 : 1$ ratio for a dihybrid test cross).

• Single-Locus Chi-Square Test for Independent Segregation:

  • Using dataset ($AB = 230$, $Ab = 99$, $aB = 97$, $ab = 224$; Total $N = 650$):
  • Locus A Segregation:
  • Total $A$ allele count: $230 + 99 = 329$
  • Total $a$ allele count: $97 + 224 = 321$
  • Expected count for each allele ($\frac{1}{2}$ ratio): $\frac{650}{2} = 325$
  • Chi-Square ($\chi^2$) calculation:       $\chi^2 = \frac{(329 - 325)^2}{325} + \frac{(321 - 325)^2}{325} = \frac{16}{325} + \frac{16}{325} = 0.0985$
  • Degrees of Freedom ($df$): $\text{Categories} - 1 = 2 - 1 = 1$
  • Critical value for $df = 1$ at $\alpha = 0.05$ is $3.841$
  • Decision: Calculated $\chi^2 = 0.0985 < 3.841$ ($p > 0.5$). The failure to reject the null hypothesis confirms that Locus A obeys independent segregation.
  • Locus B Segregation:
  • Total $B$ allele count: $230 + 97 = 327$
  • Total $b$ allele count: $99 + 224 = 323$
  • Expected count: $325$ each
  • Calculated $\chi^2 = 0.0246$ ($p > 0.5$). Confirms locus B obeys independent segregation.

• Four-Class Chi-Square Test for Independent Assortment:

  • Null Hypothesis ($H_0$): Genes $A$ and $B$ assort independently ($1 : 1 : 1 : 1$ ratio).
  • Expected count per phenotype: $\frac{650}{4} = 162.5$
  • Chi-Square ($\chi^2$) calculation:     $\chi^2 = \frac{(230 - 162.5)^2}{162.5} + \frac{(99 - 162.5)^2}{162.5} + \frac{(97 - 162.5)^2}{162.5} + \frac{(224 - 162.5)^2}{162.5}$$\chi^2 = \frac{4556.25}{162.5} + \frac{4032.25}{162.5} + \frac{4290.25}{162.5} + \frac{3782.25}{162.5} = 28.038 + 24.814 + 26.402 + 23.275 = 102.53$
  • Degrees of Freedom ($df$): $4 - 1 = 3$
  • Critical value for $df = 3$ at $\alpha = 0.05$ is $7.815$
  • Decision: Calculated $\chi^2 = 102.53 \gg 7.815$ ($p < 0.05$). Null hypothesis rejected; the two genes are linked.

• Gene Symbol Conventions:

  • Wild-type alleles are typically designated with a plus superscript (e.g., $vg^+$, $b^+$, $bw^+$).
  • Mutant alleles are indicated by letter codes (e.g., $vg$, $b$, $bw$).
  • Multi-letter allele abbreviations (e.g., $vg^+$) refer to a single locus, not separate genes.

Drosophila Melanogaster Genetics & Three-Point Mapping Lab

• Drosophila Model Organism Biology:

  • Drosophila melanogaster are colloquially termed fruit flies, but feed on yeast growing on decaying fruit.
  • Recombination behavior: No meiotic crossing over occurs in male Drosophila melanogaster.
  • Anesthetization techniques: Carbon dioxide ($CO_2$), freezing, or commercial compounds like FlyNap (which immobilizes flies without immediate lethality).

• Loci and Phenotypic Descriptions:

  • Vestigial wings ($vg$ vs $vg^+$): Recessive mutation causing shriveled, stubby, non-functional wings versus long wild-type wings.
  • Black body ($b$ vs $b^+$): Recessive mutation causing dark thorax/body pigmentation versus wild-type gray/yellowish body color.
  • Brown eyes ($bw$ vs $bw^+$): Recessive mutation producing dull brown/off-red eyes versus bright red wild-type eyes.

• Class Dataset for Three-Point Test Cross ($N = 236$):

  • Offspring phenotypes scored across three loci:
  • Phenotype 1 ($+ \quad + \quad +$): $51$
  • Phenotype 2 ($vg \quad bw \quad b$): $51$
  • Phenotype 3 ($+ \quad bw \quad +$): $24$
  • Phenotype 4 ($vg \quad + \quad b$): $24$
  • Phenotype 5 ($+ \quad + \quad b$): $33$
  • Phenotype 6 ($bw \quad vg \quad +$): $34$
  • Phenotype 7 ($+ \quad vg \quad +$): $10$
  • Phenotype 8 ($bw \quad + \quad b$): $9$

• Step-by-Step Three-Point Mapping Procedure:

  • Step 1: Group Offspring by Frequency Class:

  • Parental Types (highest counts): $+ \quad + \quad +$ ($51$) and $vg \quad bw \quad b$ ($51$).

  • Double Crossovers / DCOs (lowest counts): $+ \quad vg \quad +$ ($10$) and $bw \quad + \quad b$ ($9$).

  • Single Crossovers / SCOs (intermediate counts): All remaining classes ($24$, $24$, $33$, $34$).

  • Step 2: Determine Gene Order:

  • Compare parental configurations ($+ \quad + \quad +$) with DCO configurations ($+ \quad vg \quad +$).

  • Identify the single locus that flips relative to the other two between Parental and DCO classes.

  • In the Parental, all three alleles are wild-type ($+ \quad + \quad +$). In the DCO class ($+ \quad vg \quad +$), the wing locus ($vg$) is mutant while the flanking loci remain wild-type ($+$).

  • Conclusion: Vestigial ($vg$) is the middle gene.

  • Gene Order: $bw - vg - b$ (or $b - vg - bw$).

  • Step 3: Calculate Inter-Gene Map Distances:

  • Distance between $bw$ and $vg$:

    • Single Crossovers: $+ \quad bw \quad +$ ($24$) and $vg \quad + \quad b$ ($24$) $\rightarrow 24 + 24 = 48$
    • Double Crossovers: $10 + 9 = 19$
    • Total recombinants = $48 + 19 = 67$
    • Map Distance = $\frac{67}{236} \approx 0.2839 = 28.4\,\text{cM}$

  • Distance between $vg$ and $b$:

    • Single Crossovers: $+ \quad + \quad b$ ($33$) and $bw \quad vg \quad +$ ($34$) $\rightarrow 33 + 34 = 67$
    • Double Crossovers: $10 + 9 = 19$
    • Total recombinants = $67 + 19 = 86$
    • Map Distance = $\frac{86}{236} \approx 0.3644 = 36.4\,\text{cM}$ (using rounded laboratory value: $0.35 = 35\,\text{cM}$)

  • Total Map Length ($bw$ to $b$):

    • Sum of short map intervals: $28.4\,\text{cM} + 35.0\,\text{cM} = 63.4\,\text{cM}$.
    • Map distances are additively calculated from shorter sub-intervals to correct for undetected double crossovers over long physical distances.

  • Step 4: Calculate Interference and Coincidence:

  • Observed DCO Frequency ($FL_{obs}$):       $FL_{obs} = \frac{9 + 10}{236} = \frac{19}{236} \approx 0.0805 \quad (8.05\%)$

  • Expected DCO Frequency ($FL_{exp}$):       $FL_{exp} = \text{Distance}_1 \times \text{Distance}_2 = 0.28 \times 0.35 = 0.098 \quad (9.8\%)$

  • Coefficient of Coincidence ($C$):       $C = \frac{\text{Observed DCO}}{\text{Expected DCO}} = \frac{0.0805}{0.098} \approx 0.82 \quad (82\%)$

  • Coefficient of Interference ($I$):       $I = 1 - C = 1 - 0.82 = 0.18 \quad (18\%)$

  • Biological Meaning: A coefficient of interference of $18\%$ indicates that the presence of one crossover event in the interval reduces the probability of a second crossover occurring in the adjacent interval by $18\%$.

Two-by-Two ($2 \times 2$) Contingency Table Analysis

• Purpose:

  • Evaluates independent assortment and independent segregation concurrently without requiring separate chi-square calculations for each allele pair.

• Example Test Cross Dataset ($N = 100$):

  • $AB = 35$
  • $Ab = 15$
  • $aB = 17$
  • $ab = 33$

• Constructing the Contingency Table:

  | Locus A / Locus B | $B$ | $b$ | Row Total |   | :--- | :--- | :--- | :--- |   | $A$ | $35$ | $15$ | $52$ |   | $a$ | $17$ | $33$ | $48$ |   | Column Total | $52$ | $48$ | $100$ |

• Computing Expected Values:

  • Formula for expected cell count:     $\text{Expected} = \frac{\text{Row Total} \times \text{Column Total}}{\text{Grand Total}}$
  • Expected count for $AB$ cell: $\frac{52 \times 52}{100} = 27.04$ (or $\frac{52 \times 50}{100} = 26$ using marginalized half-proportions)
  • Expected count for $Ab$ cell: $\frac{48 \times 50}{100} = 24$
  • Expected count for $aB$ cell: $\frac{52 \times 50}{100} = 26$
  • Expected count for $ab$ cell: $\frac{48 \times 50}{100} = 24$

• Calculating Chi-Square ($\chi^2$):   $\chi^2 = \frac{(35 - 26)^2}{26} + \frac{(15 - 24)^2}{24} + \frac{(17 - 26)^2}{26} + \frac{(33 - 24)^2}{24}$$\chi^2 = \frac{81}{26} + \frac{81}{24} + \frac{81}{26} + \frac{81}{24} = 3.115 + 3.375 + 3.115 + 3.375 = 12.98$

• Degrees of Freedom for Contingency Table:   $df = (r - 1) \times (c - 1) = (2 - 1) \times (2 - 1) = 1$

  • Critical value for $df = 1$ at $\alpha = 0.05$ is $3.841$.
  • Decision: $\chi^2 = 12.98 > 3.841$ ($p < 0.05$). Rejects independence; demonstrates significant genetic linkage.

Systematic Pedigree Analysis

• Elimination Strategy for Modes of Inheritance:

  • Rather than asserting a mode of inheritance directly, systematically eliminate impossible modes using diagnostic rules:

• Diagnostic Rules for Elimination:

  • Y-Linked Inheritance:
  • Affected individuals are exclusively male.
  • Affected fathers MUST pass the trait to ALL sons.
  • If an unaffected father has an affected son, or an affected father has an unaffected son, rule out Y-linkage.
  • Sex-Linked vs. Autosomal:
  • Sex-linked traits show a strong male bias in affected proportion.
  • Autosomal traits show approximately equal ratios of affected males and females.
  • Dominant vs. Recessive:
  • Dominant traits do not skip generations; every affected individual must have at least one affected parent.
  • Recessive traits can skip generations; unaffected carrier parents can produce affected offspring.
  • X-Linked Recessive:
  • Affected females must have an affected father and an affected or carrier mother.
  • Affected mothers pass the trait to all of their sons.

• Evaluated Pedigree Case Studies:

  • Case 1: Unaffected parents produce two affected male offspring; no females affected $\rightarrow$ Rule out Y-linked (father unaffected), rule out dominant $\rightarrow$ X-linked Recessive.
  • Case 2: Trait passes strictly from affected father to all male offspring across generations, females never express or transmit the trait $ ightarrow$ Y-linked.
  • Case 3 (Royal Hemophilia): Affected males, unaffected obligate carrier females, skipped generations $ ightarrow$ X-linked Recessive.
  • Case 4: Affected parents pass trait to both male and female offspring in roughly equal numbers; trait present in every generation without skipping $ ightarrow$ Autosomal Dominant.

Advanced Inheritance Scenarios & Problem Solving

• Recessive Epistasis in Labrador Retrievers:

  • Coat color determined by two independent loci ($B/b$ and $E/e$):
  • Black coat: $B\_ \quad E\_$
  • Brown coat: $bb \quad E\_$
  • Yellow coat: $\_\_ \quad ee$ ($e$ allele is epistatic to the $B/b$ locus)
  • Cross Analysis 1:
  • Yellow female ($bb \quad ee$) crossed with Brown male 1 ($bb \quad E\_$) produces $\frac{1}{2}$ Brown ($bb \quad Ee$) and $\frac{1}{2}$ Yellow ($bb \quad ee$).
  • Deducing Male 1 Genotype: Must be heterozygous $bb \quad Ee$ to supply the recessive $e$ allele.
  • Cross Analysis 2:
  • Same Yellow female ($bb \quad ee$) crossed with Brown male 2 produces $100\%$ Brown offspring ($bb \quad Ee$).
  • Deducing Male 2 Genotype: Must be homozygous dominant at the E-locus ($bb \quad EE$).

• Lethal Alleles in Chinese Hamsters:

  • Spotting Locus ($A_1$ vs $A_2$):
  • Cross 1: Spotted hamster $\times$ Non-spotted hamster $\rightarrow \frac{1}{2}$ Spotted, $\frac{1}{2}$ Non-spotted.
    • Indicates a test cross structure ($A_1 A_2 \times A_2 A_2$).
  • Cross 2: Spotted hamster $\times$ Spotted hamster $\rightarrow \frac{2}{3}$ Spotted, $\frac{1}{3}$ Non-spotted.
    • The characteristic $2 : 1$ modified monohybrid ratio (in thirds rather than quarters) identifies a recessive lethal allele.
    • Homozygous embryonic lethality ($A_1 A_1$) alters the surviving adult ratio from $1 : 2 : 1$ to $2 : 1$.

• Sex-Influenced Traits (Bearded Goats):

  • Locus is autosomal ($B^b$ for bearded, $B^+$ for beardless), but phenotypic expression depends on sex hormones:

    | Genotype | Male Phenotype | Female Phenotype |     | :--- | :--- | :--- |     | $B^b B^b$ | Bearded | Bearded |     | $B^b B^+$ | Bearded | Beardless |     | $B^+ B^+$ | Beardless | Beardless |

• Heterozygotes ($B^b B^+$) present as bearded in males (dominant expression) and beardless in females (recessive expression).