Linkage, Recombination, and Gene Mapping Vocabulary

Principles of Independent Assortment and Linkage

  • Mendel’s Law of Independent Assortment: This law assumes that the four possible allele combinations (e.g., TB,Tb,tB,tbTB, Tb, tB, tb) are all equally likely to occur in gametes.

  • Mechanism of Independent Assortment:   - Occurs for genes located on different chromosomes.   - During Metaphase I of meiosis, two chromosome pairs can align in two different orientations (A/A,B/BA/A, B/B vs a/a,b/ba/a, b/b OR A/A,b/bA/A, b/b vs a/a,B/Ba/a, B/B).   - Resulting gametes will contain either ABAB and abab alleles or aBaB and AbAb alleles.

  • The Concept of Recombination:  

  •  - Nonrecombinant (Parental) Chromosomes: These are identical to the chromosomes inherited from the parents (e.g., ABCABC from Dad or abcabc from Mom).

  •   - Recombinant Chromosomes: These result from crossing over during meiosis, creating new allele combinations different from what was inherited (e.g., ABcABc or abCabC).

  • Distance and Assortment:   

  • - Independent Assortment: Only occurs for genes that lie at a significant distance from each other.  

  •  - Genetically Linked Genes: Genes that are located close together on the same chromosome (e.g., Gene AA and Gene BB) and do not assort independently.   - Example: If A/aA/a and C/cC/c assort independently, the probability of combinations AC,Ac,aC,acAC, Ac, aC, ac is all equally likely at 25%25\%.   - Example: If A/aA/a and B/bB/b are linked, knowing if a gamete contains AA or aa provides specific information about whether it contains BB or bb.

Recombination Frequency (Rf) Mechanics and Limitations

  • Quantitative Definition of Rf:   - Rf=number of recombinant progenytotal number of progeny×100Rf = \frac{\text{number of recombinant progeny}}{\text{total number of progeny}} \times 100

  • Calculations for Linked vs. Unlinked Genes:   - For unlinked genes (AA and CC), all four combinations occur at 25%25\% each. Precentage of recombinants is Ac+aCAc + aC, which is 25%+25%=50%25\% + 25\% = 50\%. Therefore, Rf=50%Rf = 50\%.   - For completely linked genes (AA and BB), recombinant allele combinations (Ab,aBAb, aB) never appear. Therefore, Rf=0%Rf = 0\%.

  • Maximum Value of Rf: The maximum RfRf is 50%50\%. Even if a recombination breakpoint always falls between two genes, only two of the four chromatids in a tetrad participate in recombination, leaving the other two nonrecombinant.

  • Intermediate Distances: Genes that lie a small-to-medium distance apart have RfsRfs between 0%0\% and 50%50\%.   - In some meiosis, the breakpoint falls between the genes (yielding 50%50\% recombinant and 50%50\% nonrecombinant gametes).   - In other meiosis, the breakpoint falls elsewhere (yielding no recombinant gametes).   - Over many meioses, the overall RfRf reflects the average.

  • Influences on Recombination:   - Distance: The closer the two genes, the smaller the RfRf.   - Chromosomal Position: In humans, recombination is more frequent near telomeres than near the centromere.   - Chromosome Type: Different chromosomes have different RfsRfs for the same physical distance.   - Group Differences: RfRf can vary between men and women or between different ethnic groups.

Impact of Linkage on Phenotypic Ratios

  • Mendelian Dihybrid Test Cross Expectations:   - If parents are AaBb×aabbAaBb \times aabb: In independent assortment, gametes are AB,Ab,aB,abAB, Ab, aB, ab at 25%25\% each.   - Progeny results: 25%25\% Dominant/Dominant, 25%25\% Dominant/Recessive, 25%25\% Recessive/Dominant, and 25%25\% Recessive/Recessive.

  • Linkage Distortions:   - Complete Linkage: If genes are completely linked (no crossing over), only nonrecombinant progeny are produced.   - Independent Assortment (Unlinked): Half the progeny are recombinant and half are not.

  • Violation of the 9:3:3:1 Ratio:   - The standard 9:3:3:19:3:3:1 ratio for a dihybrid cross (AaBb×AaBbAaBb \times AaBb) relies on two assumptions: allele combinations appear equally often in gametes, and all combinations are equally likely to create viable offspring.   - If Rf < 50\%, the first assumption is violated. Some meioses will not produce recombinant chromosomes, limiting the possible offspring genotypes.

Genotype Configuration: Coupling and Repulsion

  • Notation: Linkage is often noted as p+b+pb\frac{p^+ b^+}{p b} or p+b+/pbp^+ b^+ / p b.

  • Coupling (Cis Orientation): Two wild-type alleles (or dominant alleles) are on the same chromosome, and two mutant alleles are on the other (p+b+pb\frac{p^+ b^+}{p b}).

  • Repulsion (Trans Orientation): One wild-type allele and one mutant allele are on each chromosome (p+bpb+\frac{p^+ b}{p b^+}).

  • Significance: The orientation (coupling vs. repulsion) determines which phenotypes will be the most common (nonrecombinants) and which will be rare (recombinants) in a test cross.   - Example: In coupling (Rf=20%Rf = 20\%), nonrecombinants might be Green thorax/brown puparium and Purple thorax/black puparium. In repulsion, these same phenotypes would be the recombinants (low frequency).

Gene Mapping Methodologies

  • Definitions:   - Physical Maps: Distance measured in basepairs (bpbp).   - Genetic Maps: Distance measured in RfRf.   - Units: Map units (m.u.m.u.), also known as centiMorgans (cMcM). 1m.u.=1cM=1%1 \, m.u. = 1 \, cM = 1\% recombination.

  • Additive Property: Distances can be summed. If AB=5m.u.A-B = 5 \, m.u. and BC=20m.u.B-C = 20 \, m.u., then AC=25m.u.A-C = 25 \, m.u..

  • Two-Point Test Crosses: Conducting dihybrid crosses for pairs of genes to determine relative order.   - Example: AB=35cMA-B = 35 \, cM, AC=7cMA-C = 7 \, cM, BC=42cMB-C = 42 \, cM implies the order is CABC-A-B.

  • Limitations of Genetic Mapping:   - Rf=50%Rf = 50\% does not indicate distance or even if genes are on the same chromosome.   - Observed RfRf often underestimates actual physical distance because double crossovers between two genes go undetected.

Three-Point Test Crosses and Interference

  • Process: Crossing a triple heterozygote (a+ab+bc+ca^+a b^+b c^+c) with a homozygous recessive (aabbccaa bb cc).

  • Key Principles:   - Nonrecombinant progeny: The most common class (> 50\%   - Double recombinant progeny: The least common class.   - Gene Order: A double recombination event moves only the center gene's alleles. By comparing nonrecombinants to double recombinants, the middle gene can be identified.

  • Calculating Recombination Frequencies:   - To find RfRf between Gene 1 and Gene 2: Single Crossovers (1-2)+Double CrossoversTotal Progeny×100\frac{\text{Single Crossovers (1-2)} + \text{Double Crossovers}}{\text{Total Progeny}} \times 100

  • Interference:   - One crossover may reduce the likelihood of another crossover occurring nearby.   - Coefficient of Coincidence: observed double crossesexpected double crosses\frac{\text{observed double crosses}}{\text{expected double crosses}}.   - Expected Double Crosses: Calculated by multiplying the probabilities of the two single crossovers (e.g., .146×.122×755=13.4.146 \times .122 \times 755 = 13.4).   - Interference Calculation: Interference=1coefficient of coincidenceInterference = 1 - \text{coefficient of coincidence}.   - If more double recombinations occur than expected, the coefficient is > 1.0 and interference is negative.

Human Linkage Analysis and LOD Scores

  • Polymorphic Markers: Specific DNA sequences with known locations (markers) used to trace chromosome inheritance.

  • Linkage Analysis: Finding a marker allele that is tightly linked to a disease-causing mutation across a pedigree. Affected family members should ideally share a specific marker allele.

  • LOD Scores (Logarithm of Odds):   - Used to handle the small sample sizes of human families.   - LODRf=log(P(data if linked at specific Rf)P(data if unlinked))LOD_{Rf} = \log\left(\frac{P(\text{data if linked at specific } Rf)}{P(\text{data if unlinked})}\right).   - Interpretation: A positive LOD score favors linkage; a negative score suggests linkage is unlikely.   - Significance: A score of 3.03.0 indicates it is 1,0001,000 times more likely the genes are linked than unlinked.

  • Calculation Example Overview:   - If finding 22 recombinants out of 1313 offspring:     - Unlinked probability: (0.5)13=0.000122(0.5)^{13} = 0.000122.     - Linked (Rf=15%Rf=15\%) probability: (0.15)2×(0.85)11=0.00377(0.15)^2 \times (0.85)^{11} = 0.00377.     - Ratio 30.8630.86; LOD=1.49LOD = 1.49.

Association Studies and GWAS

  • Association Study (Case-Control Study): Compares allele frequencies of a polymorphism between a group with the disease (cases) and a group without it (controls).

  • Genome-Wide Association Study (GWAS): Subjects are genotyped for a large number of markers across the entire genome in one analysis.

  • Risk Alleles: Alleles found more frequently in cases are "risk-increasing," while those found more in controls are "risk-decreasing" (protective).

  • Gene Identification:   - If a polymorphism is within a gene, it becomes a candidate gene.   - If the marker is near genes, the function of those nearby proteins is investigated.   - Confirmation studies include sequencing mRNA, measuring protein levels, and using animal models.

Questions & Discussion

  • Quiz 1: If a person is AaBbAaBb in trans configuration and undergoes recombination, what are the four outcomes? Which are recombinant/nonrecombinant?   - Response: Nonrecombinants are AbAb and aBaB; Recombinants are ABAB and abab.

  • Quiz 2 (Mapping Strategy): Draw a map for Genes A,B,C,D,EA, B, C, D, E given the following distances:   - AB=10,AC=50,AD=29,AE=28,BC=41,BD=39,BE=18,CD=50,CE=23,DE=50AB = 10, AC = 50, AD = 29, AE = 28, BC = 41, BD = 39, BE = 18, CD = 50, CE = 23, DE = 50.   - Reasoning: Start with closest pairs (AB=10AB=10). Incorporate BE=18BE=18 and AE=28AE=28 to see that the order is EABE-A-B (18+10=2818+10=28). Add CE=23CE = 23 to find CC. Finally, place DD using measurements like AD=29AD=29 and BD=39BD=39.