8 Linkage and Mapping

Recombination and Mapping : 

Recombination and Mapping Independent Assortment vs. Linkage Recombination and Linkage Linkage, Recombination Frequency and Map Distance Three Point Test Cross

Questions to think about……. : 

Questions to think about……. How is the outcome of a cross different when two genes are independently assorting as compared two genes that are linked on the same chromosome? What changes in the outcome of a cross when alleles of two genes are in cis rather than trans? How do we calculate recombination frequency? How are map distance and recombination frequency connected? Think about these topics as we go through the lecture and come back to them as you go over the lecture material and practice problem solving.

A Review of Mendel’s Laws : 

Mendel’s 1st Law: Law of Segregation: Diploid individuals possess 2 alleles that separate at Meiosis; 1 allele per gamete The Cell Biology: The 2 alleles are on homologous chromosomes and they separate at Meiosis; 1 homologous chromosome per gamete Mendel’s 2nd Law: The Law of Independent Assortment: 2 alleles at one locus separate independently of alleles at other loci, if they are on different chromosomes. The Cell Biology: Pairs of homologous chromosomes separate independently A Review of Mendel’s Laws

A Hypothetical Comparison of Di-hybrid Testcrosses: Independent Assortment vs. Complete Linkage : 

A Hypothetical Comparison of Di-hybrid Testcrosses: Independent Assortment vs. Complete Linkage M=normal leaves D= tall m=mottled leaves d = dwarf

Linked Genes and Chromosomal Recombination : 

Linked Genes and Chromosomal Recombination

Detecting Linkage with a Testcross : 

Detecting Linkage with a Testcross Parental

Detecting Linkage with a Testcross:A Chi-Squared Analysis : 

Detecting Linkage with a Testcross:A Chi-Squared Analysis Null Hypothesis: We hypothesize that the two genes are unlinked Expected Results: 1:1 ratio of Non-recombinants to Recombinants 61.5 Non-recombinants 61.5 Recombinants Observed Results: 108 Non-recombinants 15 Recombinants With 2= 70.2 and Df=1, the probability is <.005 that the genes are unlinked and differences are due to chance. Reject hypothesis.

A Problem to Try…… : 

A Problem to Try…… In tomatoes, tall (D) is dominant over dwarf (d) and smooth fruit (P) is over pubescent fruit (p), which is covered in fine hairs. A farmer has 2 tall and smooth tomato plants, which we will call A and B. The farmer crosses A and B with the same dwarf, pubescent plant and obtains the following progeny: What are the genotypes of plant A and B? Are the loci linked? If so calculate a map distance (RF). Why are the proportions of progeny different for plant A and B?

The Relationship Between Recombination and Map Distance:Building a Genetic Map : 

The Relationship Between Recombination and Map Distance:Building a Genetic Map From Essential Genetics by Peter J. Russell

Underestimation of Map Distance:Double Cross-overs : 

Underestimation of Map Distance:Double Cross-overs

Mapping With Three Genes:Three Point Test Cross : 

Mapping With Three Genes:Three Point Test Cross

Building a Genetic Map: An Example of a Three Point Test Cross : 

Building a Genetic Map: An Example of a Three Point Test Cross st+, wildtype eye color (dominant) st, scarlet eye color e+, gray body (dominant) e, ebony body ss+, normal bristles (dominant) ss, spineless bristles Determine: Gene Order Map Distances st+, e+, ss+ st , e , ss st , e , ss st , e , ss X Important: arbitrary gene order

The Results of the Cross : 

The Results of the Cross Genotype Phenotype Number Observed st+ e+ ss+ wildtype 283 st e ss all mutant 278 st+ e ss ebony, spineless 50 st e+ ss+ scarlet 52 st+ e+ ss spineless 5 st e ss+ scarlet, ebony 3 st+ e ss+ ebony 43 st e+ ss scarlet, spineless 41 Total 755

The Genetic Map of the Drosophila Genome : 

The Genetic Map of the Drosophila Genome