LECTURE 5 LINKAGE AND GENETIC MAPPING Reading Ch 5 p 113 131 Problems Ch 5 solved problems I II 5 2 5 4 5 5 5 7 5 9 5 12 5 16a 5 17 5 19 5 21 5 22a e 5 23 The dihybrid crosses that we ve considered up to this point are those segregating for genes on different chromosomes What genotypes might we expect to see if the genes are located on the same chromosome Genes on the same chromosome assort together more often than not In dihybrid crosses departures from a 1 1 1 1 ratio of F1 gametes indicates that the two genes are on the same chromosome linked Among the F2 progeny there are more parental types than recombinant types Consider two Drosophila genes linked to the X chromosome the eye color gene white and a body color gene yellow The wild type white allele is w red eyes the recessive mutant allele is w white eyes The wild type allele is y brown bodies the recessive mutant allele is y yellow bodies Cross P white brown female x w y w y x F1 red brown female w y w y F2 males white brown males red yellow males red brown males white yellow males red yellow male w y Y x white brown male x w y Y w y Y w y Y w y Y wy Y Parental type 4484 Parental type 4413 Recombinant type 76 Recombinant type 53 If the white and yellow genes sort independently of one another than the F1 female would produce 4 kinds of games with equal probability w y w y w y w y We can reveal the ratio of her gametes by observing only the male progeny as sons receive their only X from their mother As you can see from the data the four kinds of gametes do not occur with equal frequency Instead parental types vastly outnumber recombinant types 99 to 1 Bottom line When two genes assort independently the numbers of parental and recombinant F2 progeny are equal Two genes are considered linked when the number of F2 progeny with parental genotypes exceeds the number of F2 progeny with recombinant genotypes The recombination frequency RF the percentage of total progeny that are recombinant depends upon the gene pair under consideration Linked genes have a recombination frequency of less than 50 The example we use above indicates tight linkage the genes are close together whereas other gene pairs give different percentages A RF of 1 indicates tight linkage whereas a RF closer to 50 would indicate that the genes lie farther apart Autosomal traits also exhibit linkage using the testcross Transmission of sex linked traits is easier to follow since the genotypes of the F1 female chromosomes can be determined by examining her sons they receive an X from their mother and a Y from their father Autosomal traits are a bit harder to follow because all progeny receive two copies of autosomal genes This is where the test cross comes in handy In this cross we look at an autosomal gene for body color b is brown and recessive b is black and a gene for wing shape c is straight and recessive c is curved P black straight female b c b c F1 brown straight progeny All b c b c x x brown curved male b c b c Test cross brown straight F1 female x black curved male b c b c x bc bc Test cross progeny black straight b c b c Parental type 2934 brown curved b c b c Parental type 2768 black curved bc bc Recombinant type 871 brown straight b c b c Recombinant type 846 The test cross shows that these autosomal genes exhibit linkage they do not assort independently Instead the parental types are transmitted together 50 of the time Chi Square Test p 117 120 Read about this test and do the problems What if linkage is not very tight and the percentage of recombinant classes approaches 50 How can we determine statistically if the two genes are truly linked The Chi Square Test is a probability test that measures the goodness of fit between experimental and predicted results We test whether the data is consistent with the hypothesis that the genes are not linked as this hypothesis gives a precise prediction unlinked genes will assort independently giving half parental and half recombinant progeny If the chi square test shows that the observed data are different from that expected from independent assortment we can reject the null hypothesis and assume the two genes are linked 2 Sum of observed expected 2 expected For the example above the total number of progeny is 7419 if the two genes are not linked we would expect 25 progeny in each class or 1855 2 2934 1855 2 1855 2768 1855 2 1855 871 1855 2 1855 846 1855 2 1855 628 494 522 549 2193 p value 0 001 use Table 5 1 on page 120 of your book Use the chi square value together with the degrees of freedom measure of the number of independently varying parameters in the experiment with 4 N classes there are 3 degrees of freedom to determine a p value A p value is the probability that a particular set of observed experimental results represents a chance deviation from the values predicted by a particular hypothesis A low p value suggests that the data showing deviation from the values predicted by the hypothesis are significant enough to reject the null hypothesis A p value of 0 001 means that it is highly unlikely 1 chance in a 1000 that the observed distribution of phenotypes arose from two unlinked genes Recombination results when crossing over during meiosis separates linked genes Reciprocal exchanges between homologous chromosomes are the physical basis of recombination We will cover the actual mechanism of recombination in another lecture Using chromosomes that had cytologically visible abnormalities Creighton and McClintock working with maize and Stern working with Drosophila showed that recombination depends upon the physical exchange of equal parts between maternal and paternal chromosomes during meiosis Recombination frequencies for pairs of genes reflect the distances between them along a chromosome Since genes are arranged linearly on chromosomes Morgan proposed that different gene pairs exhibited different linkage rates because the closer together two genes are the less likely they will be separated by a recombination event That is the probability of a crossover occuring between two genes increases with the distance separating them Sturtevant an undergraduate in Morgan s lab suggested that recombination frequency could be used to gauge the physical distance between two genes 1 RF 1 cM 1 map unit Recombination frequency recombinants total progeny x 100 Experimental recombination frequencies between two genes are never greater than 50 Recombinants among the F2 progeny are never in the majority Genes on different
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