LECTURE 5: LINKAGE AND GENETIC MAPPINGReading: Ch. 5, p. 113-131Problems: 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; 5-23The dihybrid crosses that we’ve considered up to this point are those segregating for genes ondifferent chromosomes. What genotypes might we expect to see if the genes are located on thesame chromosome?Genes on the same chromosome assort together more often than notIn dihybrid crosses, departures from a 1:1:1:1 ratio of F1 gametes indicates that the two genesare on the same chromosome (linked). Among the F2 progeny, there are more parental typesthan recombinant types.Consider two Drosophila genes linked to the X-chromosome, the eye color gene white and abody color gene yellow. The wild-type white allele is w+ (red eyes); the recessive mutant alleleis 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 red, yellow male w y+ /w y+ x w+ y / YF1: red, brown female x white brown male w+ y / w y+ x w y+ / YF2 males:white, brown males w y+ / Y Parental type 4484red, yellow males w+ y / Y Parental type 4413red, brown males w+ y+ / Y Recombinant type 76white, yellow males w y / Y Recombinant type 53If the white and yellow genes sort independently of one another, than the F1 female wouldproduce 4 kinds of games with equal probability (w y+, w+ y, w+ y+, w y). We can reveal theratio of her gametes by observing only the male progeny, as sons receive their only X from theirmother. As you can see from the data, the four kinds of gametes do not occur with equalfrequency. Instead parental types vastly outnumber recombinant types (99% to 1%).Bottom line: When two genes assort independently, the numbers of parental and recombinant F2progeny are equal. Two genes are considered linked when the number of F2 progeny withparental 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 frequencyof less than 50%. The example we use above indicates tight linkage (the genes are closetogether), whereas other gene pairs give different percentages. A RF of 1% indicates tightlinkage, whereas a RF closer to 50% would indicate that the genes lie farther apart.Autosomal traits also exhibit linkage; using the testcrossTransmission of sex-linked traits is easier to follow, since the genotypes of the F1 femalechromosomes can be determined by examining her sons (they receive an X from their motherand a Y from their father). Autosomal traits are a bit harder to follow because all progeny receivetwo 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 x brown, curved male b c+ / b c+ x b+ c / b+ cF1: brown, straight progeny All b c+ / b+ cTest cross: brown, straight F1 female x black, curved male b c+ / b+ c x b c / b cTest cross progeny:black, straight b c+ / b c Parental type 2934brown, curved b+ c / b c Parental type 2768black, curved b c / b c Recombinant type 871brown, straight b+ c+ / b c Recombinant type 846The test cross shows that these autosomal genes exhibit linkage; they do not assortindependently. 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 aprobability 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 thishypothesis gives a precise prediction: unlinked genes will assort independently, giving halfparental and half recombinant progeny. If the chi square test shows that the observed data aredifferent from that expected from independent assortment, we can reject the null hypothesis andassume the two genes are linked. χ2 = Sum of: (# observed – # expected)2 # expectedFor the example above, the total number of progeny is 7419; if the two genes are not linked, wewould expect 25% progeny in each class, or 1855.χ2 = (2934 – 1855)2 /1855 + (2768 – 1855)2 /1855 + (871 – 1855)2 /1855 + (846 – 1855)2 /1855 = 628 + 494 + 522 + 549 = 2193p 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 ofindependently varying parameters in the experiment; with 4 (N) classes, there are 3 degrees offreedom) to determine a p value. A p value is the probability that a particular set of observedexperimental results represents a chance deviation from the values predicted by a particularhypothesis. A low p value suggests that the data showing deviation from the values predicted bythe hypothesis are significant enough to reject the null hypothesis. A p value of 0.001 means thatit is highly unlikely (1 chance in a 1000) that the observed distribution of phenotypes arose fromtwo unlinked genes.Recombination results when crossing over during meiosis separates linked genesReciprocal exchanges between homologous chromosomes are the physical basis ofrecombination. We will cover the actual mechanism of recombination in another lecture.Using chromosomes that had cytologically visible abnormalities, Creighton and McClintockworking with maize, and Stern, working with Drosophila, showed that recombination dependsupon the physical exchange of equal parts between maternal and paternal chromosomes duringmeiosis.Recombination frequencies for pairs of genes reflect the distances between
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