Independent Segregation 47 



garden peas. From the breeding behavior tion shows that the garden pea possesses a 



of hybrids it is possible to estabhsh the exist- diploid number of seven pairs of chromo- 



ence of seven different pairs of genes (each somes, so that the number of chromosome 



happening to show dominance in the hybrid pairs is large enough for each pair of genes 



condition), each pair segregating independ- to be located on a different pair of chromo- 



ently of all the others. Cytological observa- somes. 



SUMMARY AND CONCLUSIONS 



When two different traits were studied separately, in each case the phenotypic alternatives 

 were found to be due to the presence of a single pair of genes. Studies were then made 

 of the distribution of phenotypes in successive generations when these two pairs of traits 

 were followed simultaneously in the same individuals. 



The data obtained showed that each trait is due to the presence of a different pair of genes, 

 proving that the genetic material is made not of a single indivisible pair of genes but of a 

 number of pairs. The results, moreover, are best explained on the principle that the segre- 

 gation of one pair of genes is at random with respect to the segregation of all other pairs 

 tested. The simplest hypothesis for the physical basis of independent segregation is that 

 different pairs of genes reside in different pairs of chromosomes. 



REFERENCES 



Mendel, G. See references at the end of Chapter 2. 



Supplement I (at the end of this book). 



QUESTIONS FOR DISCUSSION 



6.1. Make genetic diagrams for the crosses and progeny discussed in the second paragraph 

 on p. 39. Be sure to define your symbols. 



6.2. Is a test cross or backcross used to determine genotypes from phenotypes in cases 

 of no dominance? Explain. 



6.3. What types and frequencies of gametes are formed by the following genotypes, all 

 gene pairs segregating independently? 



a. Aa Bb CC 



b. AA BB Cc DD 



c. Aa Bb Cc 



d. Mm Nn Oo Pp 



6.4. How many different genotypes are possible in offspring from crosses in which both 

 parents are undergoing independent segregation for the following numbers of pairs 

 of heterozygous genes — 1, 2, 3, 4, n? 



6.5. What conclusions could you reach about the parents if the offspring had phenotypes 

 in the following proportions? 



a. 3:1 



b. 1 :1 



c. 9:3:3:1 



d. 1:1:1:1 



6.6. Would you be justified in concluding that a pair of chromosomes can contain but a 

 single pair of genes? Explain. 



