Introduction to Genetic Analysis 62

Introduction to Genetic Analysis 62 - 44200_02_p27-72...

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61 Solved problems for the character of color, the cross must have been het- erozygote 3 homozygous recessive. Letting Y 5 yellow and y 5 green, we have Y/y 3 y/y s p For the character of shape, because all the progeny are round, the cross must have been homozygous domi- nant 3 homozygous recessive. Letting R 5 round and r 5 wrinkled, we have R / R 3 r/r s p R/r (round) Combining the two characters, we have s p Now, cross B becomes crystal clear and must have been Y / Y ; R / R 3 y / y ; r / r s p Y / y ; R / r because any heterozygosity in pea B would have given rise to several progeny phenotypes, not just one. What about C? Here, we see a ratio of 50 yellow:50 green (1:1) and a ratio of 49 round:51 wrinkled (also 1:1). So both genes in pea C must have been heterozy- gous, and cross C was Y/y ; R/r 3 y/y ; r/r which 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.

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