Extranuclear Genes and Nuclear Genes 



417 



rate, producing the exconjugants of what we 

 shall consider to be the next generation. 

 You will recognize that both exconjugants 

 are chromosomally identical, since each con- 

 jugant contributes an identical haploid nu- 

 cleus to each fertilization micronucleus. The 

 chromosomal identity of exconjugants can be 

 proven by the use of various marker genes. 

 (Of course, if the conjugants were homo- 

 zygous for different alleles, the exconjugants 

 would be identical heterozygotes.) The dip- 

 loid micronucleus in each exconjugant divides 

 once mitotically, one product forming a new 

 macronucleus, the other remaining as the 

 micronucleus. 



What would be the normal consequence of 

 mating a killer with a sensitive individual? 

 (Mating can occur before a killer can kill a 

 sensitive individual; in fact, during conjuga- 

 tion, all conjugants are resistant to killing 

 action.) Normally, the cytoplasmic interiors 

 of conjugants are kept apart by a boundary 

 which is probably penetrated only by the 

 migrant haploid nuclei. As a result, little or 

 no cytoplasm is exchanged, so that, kappa- 

 wise, the exconjugants are the same as the 

 conjugants, namely, one is a killer and one is 

 a sensitive individual. However, using certain 

 experimental conditions, a wide bridge can 

 be seen to form between the conjugants 

 through which the cytoplasmic contents of 

 both mates can flow and mix (Figure 45-5). 

 Moreover, the extent of cytoplasmic mixing 

 can be controlled experimentally. If one of 

 the conjugants is killer and the other sensitive, 

 and if cytoplasmic mixing is extensive enough, 

 both exconjugants are found to be killers 

 because of the flow of kappa particles into 

 the sensitive conjugant. 



Consider now precisely how nuclear genes 

 are distributed in conjugation. Suppose each 

 conjugant is a micronuclear heterozygote, Aa. 

 It would be a matter of chance in each mate 

 which one of the four haploid nuclei produced 

 by meiosis, A, A, a, a, fails to disintegrate, 

 divides mitotically once, one of whose prod- 



ucts migrates to help form the fertilization 

 nucleus in the other mate. Accordingly, both 

 exconjugants will be AA 25% of the time, 

 both exconjugants will be Aa 50% of the time, 

 and both aa 25% of the time. This result 

 would be obtained whether or not the conju- 

 gants mix cytoplasms. Note again that, 

 regardless of the genotype of the conjugants, 

 the members of a pair of exconjugants are 

 identical with respect to micronuclear genes 

 and will give rise to clones phenotypically 

 identical with respect to the micronuclear 

 gene-determined trait under consideration. 

 When dealing with a trait determined by a 

 cytoplasmic particle hke kappa, on the other 

 hand, we have seen that the results may be 

 different. In this particular example, a cross 

 of sensitive with killer produces exconjugants 

 whose type depends upon the occurrence or 

 nonoccurrence of cytoplasmic mixing. 



Suppose, to generalize, two paramecia, 

 phenotypically different with respect to a 

 given trait, are mated together. It may be 

 found, when there has been no cytoplasmic 

 exchange, that the two exconjugant clones 

 remain different, each resembling its cyto- 

 plasmic parent in this trait. Suppose, more- 

 over, that only when there is cytoplasmic 

 mixing are the two exconjugant clones found 

 to be the same phenotypically. Such results 

 would prove that the trait under test is 

 determined at least in part by an extranuclear 



FIGURE 45-5. Silhouettes of conjugating Para- 

 mecium. A. Normal, no cytoplasmic mixing. 

 B. Wide bridge, permitting cytoplasmic mixing. 



