61Solved problemsfor the character of color, the cross must have been het-erozygote3homozygous recessive. Letting Y5yellowand y5green, we haveY/y3y/yspFor the character of shape, because all the progenyare round, the cross must have been homozygous domi-nant3homozygous recessive. Letting R5round and r5wrinkled, we haveR/R3r/rspR/r (round)Combining the two characters, we havespNow, cross B becomes crystal clear and must havebeenY/Y;R/R3y/y;r/rspY/y;R/rbecause any heterozygosity in pea B would have givenrise to several progeny phenotypes, not just one.What about C? Here, we see a ratio of 50 yellow:50green (1:1) and a ratio of 49 round:51 wrinkled (also1:1). So both genes in pea C must have been heterozy-gous, and cross C wasY/y;R/r3y/y;r/rwhich is a good demonstration of Mendel’s second law(independent behavior of different genes).How would a geneticist have analyzed these crosses?
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This note was uploaded on 01/10/2011 for the course BIOL BIOL taught by Professor Johnson during the Spring '08 term at Aberystwyth University.