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Linked Genes On The Same Chromosome Exhibit Distorted Mendelian Ratios It was not long from the time that Mendel's work was rediscovered that new anomalous ratio began appearing. One such experiment was performed by Bateson and Punnett with sweet peas. They performed a typical dihybrid cross between one pure line with purple flowers and long pollen grains and a second pure line with red flowers and round pollen grains. Because they new that purple flowers and long pollen grains were both dominant, they expected a typical 9:3:3:1 ratio when the F1 plants were crossed. The table below shows the ratios that they observed. Specifically, the two parental classes, purple, long and red, round, were over represented in the progeny. Observed Expected Purple, long (P_L_) 284 215 Purple, round (P_ll) 21 71 Red, long (ppL_) 21 71 Red, round (ppll) 55 24 Total 381 381 At the time of these experiments, Bateson and Punnett were not able to develop an acceptable hypothesis. The best explanation they posed was that in some manner the phenotypic classes (alleles) in the parents were coupled, and they did not sort independently into gametes as predicted by Mendel's second law. Proof that genes on the same chromosome can at times be inherited as blocks awaited the results of Thomas Hunt Morgan with Drosophila. Morgan crossed red eye, normal wing flies (pr+pr+ vg+vg+) with purple eye, vestigal wing (prpr vgvg) flies. The figure below shows the cross and the F1 genotypes. (The bars are used to shows that the genes reside on the same chromosome.)During meiosis, four different F1 gametes are produced. The parental gametes are developed without any processing. The recombinant gametes though occur by a process called called crossing over. (The X between the two F1 chromosomes represents the crossing over event.) Morgan performed a testcross by crossing prpr vgvg flies to F1. The testcross is powerful because it allows you to follow the meiotic events in one parent because all of the gametes from the test cross parent are homozygous recessive. For this example, the testcross genotype is pr vg. Therefore the testcross progeny will represent the distribution of the gametes in the F1. Remember that a testcross to F1 derived from a dihybrid cross gave a 1:1:1:1 ratio. But this is not what Morgan observed. The following table shows the result of this test cross. F1 Gamete Testcross Distribution Gamete Type pr+ vg+ 1339 Parental pr+ vg 151 Recombinant pr vg+ 154 Recombinant pr vg 1195 Parental These results confirm the Bateson and Punnett hypothesis that two genes do not always assort independently. A further confirmation experiment was performed by Morgan when he crossed red eye, vestigal wing flies and purple eye, normal wing flies. Whereas in the first cross, the two dominant alleles and two recessive alleles were on the same chromosome the F1, in the is cross a dominant allele was on the same chromosome as a recessive allele. The term for the first chromosomal arrangement of the F1 is called coupling, whereas the second arrangement is called repulsion. Another set of terms to describe these arrangements are cis and trans, respectively. The following shows the chromosomal arrangement for the cross of two parents in repulsion. As with the first cross, Morgan testcrossed these F1 flies. The following table shows the distribution of these F1 gametes. F1 Gamete Testcross Distribution Gamete Type pr+ vg+ 157 Recombinant pr+ vg 965 Parental pr vg+ 1067 Parental pr vg 146 Recombinant It was expected that both the coupling and repulsion crosses would yield 1:1:1:1 ratios. How can we determine if the results deviate from this ratio. As with any ratio, we can use the chi-square test to determine if the observed results fit or deviate from the expected ratio. The two tables below show the results for the chi-square for the two crosses. Coupling Cross Chi-Square Test F1 Gamete Observed Expected (O-E)2/E pr+ vg+ 1339 709.75 557.9 pr+ vg 151 709.75 439.9 pr vg+ 154 709.75 435.2 pr vg 1195 709.75 331.8 Total 2839 2839 X2=1764.8 Repulsion Cross Chi-Square Test F1 Gamete Observed Expected (O-E)2/E pr+ vg+ 157 583.75 312.0 pr+ vg 965 583.75 249.0 pr vg+ 1067 583.75 483.3 pr vg 146 583.75 328.3 Total 2335 2335 X2=1372.6 It is quite clear that both of these large chi-square values indicate that neither of these ratios fit the 1:1:1:1 Copyright © 1997. Phillip McClean