Pierce Genetics A Conceptual Approach 3e：皮尔斯遗传学的概念：

1、GENETICS ESSENTIALSConcepts and Connections SECOND EDITIONBenjamin A. Pierce 2013 W. H. Freeman and CompanyCHAPTER 5Linkage, Recombination, and Eukaryotic Gene MappingChapter 5 Outline 5.1 Linked Genes Do Not Assort Independently, 114 5.2 Linked Genes Segregate Together and Crossing Over Produces Re

2、combination Between Them, 116 5.3 A Three-Point Testcross Can Be Used To Map Three Linked Genes, 128 5.4 Genes Can Be Located with Genomewide Association Studies, 137The case of male baldness X-linked or not? Study of linkage of a SNP and the association with male pattern baldness shows that they do

3、 not assort independently, they are transmitted together5.1 Linked Genes Do Not Assort Independently Linked genes: genes located close together on the same chromosome-Belong to the same linkage groupRecombination and alleles Flower color and pollen shapeF1 generation gives purple color and long poll

4、enAppears that the purple is dominant to red and long is dominant to short pollenWhat with the F2, it is expected to get 9:3:3:1BUTThere is no 9:3:3:1 ratio; what happened5.2 Linked Genes Segregate Together and Crossing Over Produces Recombination Between Them Notation for crosses with linkage Compl

5、ete linkage leads to nonrecombinant Gametes and nonrecombinant progeny Crossing over with linked genes leads to recombinant gametes and recombinant progenyLabeling alleles of two genes on a same chromosome Cross AA BB x aa bb can be represented byA B a bA B x a bThe F1 will be:A Ba bAlso it can be r

6、epresentedA Ba bThe two test cross cases Linked . not linkedLinked genes give only nonrecombinant gametesUnlinked genes give both nonrecombinant and recombinant Concept Check 1For single crossovers, the frequency of recombinant gametes is half the frequency of crossing over because:a. a test cross b

7、etween a homozygote and heterozygote produces heterozygous and homozygous progeny.b. the frequency of recombination is always 50%.c. each crossover takes place between only two of the four chromatids of a homologous pair.d. crossovers take place in about 50% of meiosis.What happens in recombination

8、and the gamete formation5.2 Linked Genes Segregate Together and Crossing Over Produces Recombination Between Them Calculating Recombination Frequency Recombination frequency = (No. recombinant progeny/Total No. of progeny) 100% Coupling and Repulsion Configuration of Linked Genes Coupling (cis confi

9、guration): wild-type alleles are found on one chromosome; mutant alleles are found on the other chromosome.Calculating frequenciesCalculating frequenciesRec. Frequency = x100% Rec. Frequency = x100% 8 + 7 55+53+8+7# of recombinant progenyTotal # of progenyRec. Frequency = x100% 15123 5.2 Linked Gene

10、s Segregate Together and Crossing Over Produces Recombination Between Them Coupling and Repulsion Configuration of Linked Genes Repulsion (trans configuration): wild-type allele and mutant allele are found on the same chromosome. Testing for Independent Assortment Coupling and repulsionp+ b+p bp+ bp

11、 b+Coupling and repulsion Gene Mapping with Recombination Frequencies: Genetic maps are determined by recombinant frequency. Map unit and centiMorgans Constructing a Genetic Map with the Use of Two-Point Testcrosses5.2 Linked Genes Segregate Together and Crossing Over Produces Recombination Between

12、ThemRecombination frequencies and distances of genesBased on recombination frequencies genetic maps are generatedActual distances are on physical mapsRecombination maps are approximations and are measured in map units or cM (centiMorgans)One map unit equals to 1% recombination frequencyGenetic dista

13、nces measured with recombination rates are approximately additive: if the distance from gene A to gene B is 5 ., the distance from gene B to gene C is 10 ., and the distance from gene A to gene C is 15 .Limitations of gene mapping If the genes are far apart and the frequency is 50% we cannot say if

14、the genes are on different chromosomes or just to far on same chromosome Two genes that are relatively far will give underestimate of recombination events due to possibility of multiple crosses between the genes.Example: two point cross to determine map of 4 genes5.3 A Three-Point Testcross Can Be U

15、sed to Map Three Linked Genes Constructing a Genetic Map with the Three-Point Testcross Determining the gene order Determining the location of crossoversDouble crossover :three genesA Three-Point Testcross Can Be Used to Map Three Linked Genes Calculating the recombination frequencies Interference a

16、nd coefficient of coincidence Effect of multiple crossovers5.3 A Three-Point Testcross Can Be Used to Map Three Linked Genes Calculating the recombination frequencies Based on the numbers in the fruit fly testcross for three loci calculate the distances between the loci. Recombinant progeny with a c

17、hromosome that underwent crossing over between the eye-color locus (st) and the bristle locus (ss) include the single crossovers (st+ / ss e and st / ss+ e+) and the two double crossovers (st+ / ss / e+ and st / ss+ / e); Total of 755 progeny; so the recombination frequency between ss and st is:Calc

18、ulating the recombination frequencies Based on the numbers in the fruit fly testcross for three loci calculate the distances between the loci. The map distance between the bristle locus (ss) and the body locus (e) is determined in the same manner. The recombinant progeny that possess a crossover bet

19、ween ss and e are the single crossovers st+ ss+ / e and st ss / e+ and the double crossovers st+ / ss / e+ and st / ss+ / e. The recombination frequency is:Calculating the recombination frequencies Based on the numbers in the fruit fly testcross for three loci calculate the distances between the loc

20、i. Finally, calculate the map distance between the outer two loci, st and e. This map distance can be obtained by summing the map distances between st and ss and between ss and e (14.6 . + 12.2 . = 26.8 .). We can now use the map distances to draw a map of the three genes on the chromosome:Interfere

21、nce and coefficient of coincidence Calculate the proportion of double-recombinant gametes by using the multiplication rule of probability Applying this rule, the proportion (probability) of gametes with double crossovers between st and e is equal to the probability of recombination between st and ss

22、 multiplied by the probability of recombination between ss and e, or 0.146 0.122 = 0.0178. Multiplying this probability by the total number of progeny gives us the expected number of double-crossover progeny from the cross: 0.0178 755 = 13.4. Only 8 double crossoversconsiderably fewer than the 13 ex

23、pectedwere observed in the progeny of the crossInterference and coefficient of coincidence Crossovers are frequently not independent events: the occurrence of one crossover tends to inhibit additional crossovers in the same region of the chromosome, and so double crossovers are less frequent than expected. The degree to which on