Comprehensive Notes on Linkage, Recombination, and Gene Mapping

• Linkage: The tendency of genes located close to each other on a chromosome to be inherited together, impacting the inheritance patterns observed in offspring.

• Recombination: The process during meiosis where segments of DNA are exchanged between homologous chromosomes, leading to genetic variation.

• Gene Mapping: Techniques used to determine the relative positions of genes on a chromosome, often utilizing data from linkage and recombination frequencies.

Independent Assortment

• Mendel's Principle of Independent Assortment: States that two alleles at a locus separate independently of alleles at other loci.

• Experiment (Sweet Peas):

  • Question: Do the genes for flower color (Purple/Red) and pollen shape (Long/Round) in sweet peas assort independently?

  • Methods:

    1. Cross two homozygous strains:

       • P generation: Purple flowers, long pollen ($PPL L$) $imes$ Red flowers, round pollen ($ppll$)

    2. F1 generation: All Purple flowers, long pollen ($P p L l$)

    3. Self-fertilize F1 generation.

  • Results (F2 generation):

    • Purple flowers, long pollen: $284$

    • Purple flowers, round pollen: $21$

    • Red flowers, long pollen: $21$

    • Red flowers, round pollen: $55$

  • Expected Independent Assortment Ratio for F2: $9:3:3:1$

  • Observed Ratio vs. Expected: The F2 progeny observed (approximating $12:1:1:2$) do not appear in the $9:3:3:1$ ratio expected with independent assortment.

  • Conclusion: The genes for flower color and pollen shape do not assort independently, suggesting they might be linked.

Linked Genes Do Not Assort Independently

• Linked Genes: Genes located close together on the same chromosome are inherited as a unit.

• Linkage Group: All genes on a single chromosome form a linkage group because they tend to travel together during meiosis and go to the same gamete.

• Unlinked Genes: Genes are considered unlinked if:

  • They are on different chromosomes (assort independently).

  • They are far apart on the same chromosome (leading to frequent crossing over, making them behave as if unlinked).

Crossing Over

• Mechanism: Genes occasionally switch from one homologous chromosome to the other through the process of crossing over.

  • This involves the physical exchange of genetic material between non-sister chromatids of homologous chromosomes during prophase I of meiosis.

• Effect: Crossing over breaks up the linkage of genes that are physically close together on a chromosome.

Recombination

• Definition: The sorting of alleles into new combinations.

• Recombinant: Refers to new combinations of alleles compared to the parental chromosomes.

• Nonrecombinant (Parental): Refers to the original combinations of alleles found on the parental chromosomes.

• Single Crossover Event: Affects $2$ out of $4$ chromatids in a homologous pair.

  • Results in half of the gametes being recombinants and half being nonrecombinants.

• Crossing Over vs. Linkage:

  • Linkage: Keeps genes together; they are inherited as a unit.

  • Crossing Over: Breaks up linked genes.

  • These two processes have opposite effects on allele combinations.

Linkage Notation

• For crosses involving linked genes, specific notation is used to represent the arrangement of alleles on homologous chromosomes.

• Examples:

  • $A B / a b$

  • $A B imes a b$ (representing two homologous chromosomes)

  • $A B / A B imes a b / a b$ (representing homozygous parents)

• Example Problem: Write $A<em>1 A</em>2 / B<em>1 B</em>2$ using linkage notation.

  • Given alleles: $A ext{ and } b$ on one chromosome, $a ext{ and } B$ on the homologous chromosome.

  • Correct notation: $A b / a B$ (Option B provided from a slide)

Recombinant Gamete Identification Practice

• Question: Which is an example of a recombinant gamete from a parent with alleles $A B / a b$?

  • Original parental combinations (nonrecombinant): $A B$ and $a b$

  • New combinations (recombinant) formed by crossing over: $A b$ and $a B$

• Options:

  1. $a B$

  2. $A b$

  3. $A B$

  4. $a b$

  5. $A a$

• Answer: A) 1 and 2 ($a B$ and $A b$ are recombinant gametes).

Types of Linkage

• Linked Genes: Genes located close to one another on a chromosome.

• Complete Linkage: Occurs when genes are so close together on a chromosome that no crossing over can occur between them.

  • Outcome: Produces only nonrecombinant (parental) gametes and, subsequently, only nonrecombinant progeny.

  • In a testcross ($MD/md imes md/md$): If completely linked, produces only $MD/md$ (normal leaves, tall) and $md/md$ (mottled leaves, dwarf) progeny in a $1:1$ ratio. No recombinant progeny ($Md/md$ or $mD/md$) are formed.

• Incomplete Linkage: Some crossing over occurs between genes.

  • Outcome: Produces both nonrecombinant and recombinant gametes, but with a higher proportion of nonrecombinant gametes.

  • Consequently, produces both nonrecombinant and recombinant progeny, with more nonrecombinant progeny.

• Independent Assortment Comparison:

  • Results in nonrecombinant and recombinant gametes in equal proportions ($1/4$ of each type). For a heterozygous individual ($Mm Dd$), gametes would be $MD$, $md$, $Md$, $mD$ in equal frequencies.

  • Results in nonrecombinant and recombinant progeny in equal proportions ($1:1:1:1$ ratio for a testcross).

Practice Questions on Gamete Types

These questions assume a heterozygous parent formed from true-breeding $MD/md$ (meaning alleles M and D are on one chromosome, m and d on the homologous chromosome).

• Question 1: How many types of gametes are formed if the genes assort independently?

  • Answer: C) $4$ (MD, md, Md, mD)

• Question 2: How many types of gametes are formed if the genes are completely linked?

  • Answer: B) $2$ (MD, md - the parental combinations)

• Question 3: How many types of gametes are formed if the genes are incompletely linked?

  • Answer: C) $4$ (MD, md, Md, mD - but in unequal proportions, with parental types most frequent)

Testcross Results Summary (Recap from Table 5.1)

• Independent Assortment (Testcross $Aa Bb imes aa bb$):

  • Progeny: $25\%$ $Aa Bb$ (nonrecombinant), $25\%$ $aa bb$ (nonrecombinant), $25\%$ $Aa bb$ (recombinant), $25\%$ $aa Bb$ (recombinant).

• Complete Linkage (Genes in Coupling: $AB/ab imes ab/ab$):

  • Progeny: $50\%$ $Aa Bb$ (nonrecombinant), $50\%$ $aa bb$ (nonrecombinant). No recombinants.

• Linkage with Some Crossing Over (Genes in Coupling: $AB/ab imes ab/ab$):

  • Progeny: More than $50\%$ nonrecombinant ($Aa Bb$, $aa bb$), less than $50\%$ recombinant ($Aa bb$, $aa Bb$).

Calculating Recombination Frequency

• Formula:
  $\text{recombination frequency} = \frac{\text{number of recombinant progeny}}{\text{total number of progeny}} \times 100\%$

• Example Data (Cucumber Testcross):

  • Progeny: Normal leaves, tall ($MD/md$): $55$

  • Progeny: Mottled leaves, dwarf ($md/md$): $53$

  • Progeny: Normal leaves, dwarf ($Md/md$): $8$

  • Progeny: Mottled leaves, tall ($mD/md$): $7$

• Identification:

  • Nonrecombinant progeny: $55 + 53 = 108$

  • Recombinant progeny: $8 + 7 = 15$

  • Total progeny: $108 + 15 = 123$

• Calculation:
  $\text{recombination frequency} = \frac{15}{123} \times 100\% \approx 12.2\%$

• Conclusion: With linked genes and some crossing over, nonrecombinant progeny predominate.

• Important Rule: A recombination frequency of $\ge 50\%$ means that the genes are unlinked (either on different chromosomes or very far apart on the same chromosome).

Practice Problem: Silkmoths

• Genes: Red eyes (re) and white-banded wings (wb) are recessive mutants. Wild-type are $re^+$ and $wb^+$. The genes are on the same chromosome.

• Cross 1 (P generation): Homozygous red eyes, white-banded wings ($re wb / re wb$) $imes$ Homozygous wild-type ($re^+ wb^+ / re^+ wb^+$)

• F1 Generation: Heterozygous, normal eyes, normal wings ($re^+ wb^+ / re wb$)

• Cross 2 (Testcross): F1 ($re^+ wb^+ / re wb$) $imes$ red eyes, white-banded wings ($re wb / re wb$)

• Testcross Progeny:

  • Wild-type eyes, wild-type wings ($re^+ wb^+ / re wb$): $418$ (Nonrecombinant)

  • Red eyes, wild-type wings ($re wb^+ / re wb$): $19$ (Recombinant)

  • Wild-type eyes, white-banded wings ($re^+ wb / re wb$): $16$ (Recombinant)

  • Red eyes, white-banded wings ($re wb / re wb$): $426$ (Nonrecombinant)

• Calculation:

  • Number of recombinant progeny: $19 + 16 = 35$

  • Total number of progeny: $418 + 19 + 16 + 426 = 879$

  • Recombination frequency: $\frac{35}{879} \times 100\% \approx 3.98\%$

Coupling and Repulsion Configuration of Linked Genes

• Configuration: Refers to the arrangement of alleles on homologous chromosomes.

  • Crucial to Outcome: The arrangement affects which phenotypes appear as nonrecombinant and recombinant.

• Coupling (cis configuration): Wild-type alleles are on one homologous chromosome, and mutant alleles are on the other homologous chromosome.

  • Example: $p^+b^+ / pb$

  • Testcross progeny: $p^+b^+$ and $pb$ are nonrecombinant (more numerous), $p^+b$ and $pb^+$ are recombinant (less numerous).

• Repulsion (trans configuration): A wild-type allele and a mutant allele are found on the same homologous chromosome.

  • Example: $p^+b / pb^+<br>$

  • Testcross progeny: $p^+b$ and $pb^+$ are nonrecombinant (more numerous), $p^+b^+$ and $pb$ are recombinant (less numerous).

• Conclusion: The phenotypes of the offspring are the same (e.g., green thorax, brown puparium), but their numbers differ depending on whether the alleles in the heterozygous parent were in coupling or repulsion.

Practice Problem: Identifying Coupling/Repulsion

• Testcross Progeny:

  • $10$ $AB/ab$

  • $40$ $Ab/ab$

  • $40$ $aB/ab$

  • $10$ $ab/ab$

• Analysis:

  • Nonrecombinant progeny are typically more numerous.

  • In this data, $Ab/ab$ (40) and $aB/ab$ (40) are the most frequent.

  • These represent the parental combinations of alleles if the parent was in repulsion configuration: $Ab/aB$

• Answer: The A and B alleles in the parent were in repulsion configuration ($Ab/aB$).

Recombination Frequencies for Predicting Cross Outcomes

• Usefulness: Geneticists use recombination frequencies to predict the results of crosses involving linked genes.

• Example: Cucumbers (Warty/Smooth, Dull/Glossy Fruit)

  • Assume recombination frequency between gene T (warty/smooth) and D (dull/glossy) is $16\%$. Parent is $TD/td$

  • This means $16\%$ of gametes will be recombinant ($Td$ or $tD$). Since there are two types of recombinant gametes, each will be $16\%/2 = 8\%$ ($0.08$).

  • The remaining $100\% - 16\% = 84\%$ will be nonrecombinant ($TD$ or $td$). Each nonrecombinant type will be $84\%/2 = 42\%$ ($0.42$).

  • Predicted Frequencies of Gametes from heterozygous parent ($TD/td$):

    • Nonrecombinant: $TD = 0.42$, $td = 0.42$

    • Recombinant: $Td = 0.08$, $tD = 0.08$

• Testcross with $td/td$ parent (produces only $td$ gametes):

  • Predicted Progeny Frequencies:

    • $TD/td$ (Warty, dull fruit): $0.42 imes 1.00 = 0.42$

    • $td/td$ (Smooth, glossy fruit): $0.42 imes 1.00 = 0.42$

    • $Td/td$ (Warty, glossy fruit): $0.08 imes 1.00 = 0.08$

    • $tD/td$ (Smooth, dull fruit): $0.08 imes 1.00 = 0.08$

• Conclusion: The total proportion of recombinant gametes directly corresponds to the recombination frequency. These frequencies can then be used to predict progeny ratios in a testcross.

Gene Mapping

• Physical Maps: Represent the actual physical distances between genes along a chromosome, typically measured in numbers of base pairs.

• Genetic Maps: Illustrate the order of genes and their relative distances from each other, based on recombination frequencies.

• Relationship: Physical distances between genes are generally related to recombination frequencies: genes physically farther apart are more likely to undergo recombination.

• Units of Genetic Maps:

  • Map Units (m.u.): One map unit is defined as $1\%$ recombination frequency.

  • centiMorgans (cM): Map units are also called centiMorgans ($1 ext{ cM} = 1 ext{ m.u.} = 1\% \text{ recombination}$). Named after Thomas Hunt Morgan, a pioneer in genetic mapping.

More Crossing Over Events Scenario

• Question: Which scenario (A or B) would show more crossing over events?

  • Scenario A: Genes A and B are $10$ units apart.

  • Scenario B: Genes A and D are $200$ units apart.

• Answer: Scenario B. Genes that are farther apart on a chromosome ($200$ units vs. $10$ units) have a greater chance for crossing over to occur between them.

Using Genetic Maps to Estimate Recombination Frequency

• Example: Predict the frequency of recombination between the genes for spineless bristles ($58.5 ext{ m.u.}$) and pink eyes ($48.0 ext{ m.u.}$), assuming they are on the same chromosome (based on the Drosophila map). Given these are map positions, not distances from a common point.

• Calculation: The recombination frequency is the difference in map units.
  $|58.5 ext{ m.u.} - 48.0 ext{ m.u.}| = 10.5 ext{ m.u.}$

• Answer: c. $10.5$ (This implies a $10.5\%$ recombination frequency).

Genetic Map vs. Physical Map (Think-Pair-Share)

• Genetic maps: Based on rates (frequencies) of recombination between genes. The farther apart genes are, the higher the recombination frequency.

• Physical maps: Based on actual physical distances (e.g., base pairs) along the DNA molecule of a chromosome.

• Key Difference: While generally correlated, discrepancies can arise due to