Linkage, Recombination and Genetic Mapping
Linkage, Recombination and Genetic Mapping
甲工 2 Male Female 巴日 Hemophilia a Hemophilia B Colorblind Hemophilic and 34 田田 colorblind
Linkage: Alleles of two genes segregate together during meiosis. W y >1/4 1144 1/2 114 +y 114 y W十 1/2 w y W 114 >114 //hWy Complete Linkage Partial Linkage No Linkage
Linkage: Alleles of two genes segregate together during meiosis. + + w y w w y y + + + + w w y y + + + + + + w y Complete Linkage Partial Linkage No Linkage 1/2 1/2 1/4 1/4 1/4 1/4 >1/4 >1/4 <1/4 <1/4
During meiosis crossing over between homologous chromosomes produces recombinant chromosomes
• During meiosis crossing over between homologous chromosomes produces recombinant chromosomes A a b B A a B b
The frequency of recombinant gametes is 12 the frequency of crossing over a b b b a B A B A b A B A B Each crossover event gives rise to two recombinant and two parental chromosomes. A crossover frequency of 100% gives a recombinant chromosome frequency of 50%
The frequency of recombinant gametes is 1/2 the frequency of crossing over • Each crossover event gives rise to two recombinant and two parental chromosomes. • A crossover frequency of 100% gives a recombinant chromosome frequency of 50%. a b a b A B A B a b a B A b A B
Tests for Linkage: Chi Square The chi square test is used to test the goodness of fit of observed data to expectations. This test can be used to test whether two genes are linked. Example. Flies heterozygous for black(b) and vestiial(vg) were testcrossed. Are these two genes linked? progeny phenotypes number grey normal 283 grey vestigial 1294 black normal 1418 black vestigial 241
Tests for Linkage: Chi Square • The chi square test is used to test the goodness of fit of observed data to expectations. This test can be used to test whether two genes are linked. • Example. Flies heterozygous for black (b) and vestiial (vg) were testcrossed. Are these two genes linked? progeny phenotypes number grey normal 283 grey vestigial 1294 black normal 1418 black vestigial 241
Chi Square Test Hypothesis: the genes are not linked Prediction: the progeny should be in a 1: 1: 1: 1 ratio x test: obs exp (obs-exp)/exp 283 809 342.0 1294 809 290.8 1418 809 458.4 241 809 398.8 Total 3236 3236 1490.0 Degrees of freedom=4-1=3 one less than the number of progeny classes)
Chi Square Test • Hypothesis: the genes are not linked. • Prediction: the progeny should be in a 1:1:1:1 ratio. • 2 test: obs exp (obs-exp)2 /exp 283 809 342.0 1294 809 290.8 1418 809 458.4 241 809 398.8 Total 3236 3236 1490.0 • Degrees of freedom = 4 -1 = 3 (one less than the number of progeny classes)
Chi Square Test From the calculated x2 value and the known degrees of freedom, the x2 table reveals a probability value that the deviation from predicted would occur randomly For this example: probability <0.001 This means that the deviation from predicted is expected to occur randomly less than 1 in 1000 times Conclusion? The hypothesis is not correct. The genes do not show independent assortment
Chi Square Test • From the calculated 2 value and the known degrees of freedom, the 2 table reveals a probability value that the deviation from predicted would occur randomly • For this example: probability <0.001 • This means that the deviation from predicted is expected to occur randomly less than 1 in 1000 times • Conclusion? The hypothesis is not correct. The genes do not show independent assortment
A.H. Sturtevant and linkage mapping Sturtevant: strength of linkage is relative to physical distance (1)Recombination is randomly distributed in the chromosome. (2) Each gene has a unique location (locus)on the chromosome (3) The frequency of recombination is proportional to the distance between two loci on the chromosome ie, a-B is smaller than b-c A B
A.H. Sturtevant and linkage mapping • Sturtevant: strength of linkage is relative to physical distance. • (1) Recombination is randomly distributed in the chromosome. • (2) Each gene has a unique location (locus) on the chromosome. • (3) The frequency of recombination is proportional to the distance between two loci on the chromosome. ie. A - B is smaller than B - C A B C
inkage mapping Linkage mapping is done using test crosses a b/a B x a b/a b 00 Parental a B 25 Recombinant Ab 25 Recombinant A B 100 Parental R=# recombinant progeny/fTotal progeny X 100=CM R=50250X100=20cM
Linkage mapping • Linkage mapping is done using test crosses. a b / A B X a b / a b a b 100 Parental a B 25 Recombinant A b 25 Recombinant A B 100 Parental R = # recombinant progeny/ #Total progeny X 100 = cM R = 50/250 X 100 = 20 cM
