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Unformatted text preview: Lecture 5 Lecture 5 Lecture 5 Lecture 5 Lecture 5 Until now our analysis of genes has focused on gene function as determined by phenotype differences brought about by different alleles or by a direct test of function the complementation test. For the next six lectures our analysis will be concerned with the tests of gene position starting with the position of genes on chromosomes and finally mapping point mutations at the resolution of single nucleotide pairs. Weve taken it for granted that genes reside on chromosomes, but how do we know this? Lets review the properties of gene segregation. Consider two different traits. A / A , B / B x a / a , b / b The gametes from one parent will be A , B and from the other parent a , b These gametes will then give an F 1 generation of all A / a , B / b Crosses between F 1 individuals will give an F 2 generation with a 9 : 3 : 3 : 1 phenotypic ratio as shown before. A better way to look at segregation is by a test cross of the F 1 heteroxygote to a homozygous recessive individual. A / a , B / b x a / a , b / b The possible gamete genotypes from the F 1 will be: A , b a , b A , B a , B (recomb.) (parental) (parental) (recomb.) The corresponding genotypes of the offspring in the testcross will be: A / a , b / b a / a , b / b A / a , B / b a / a , B / b Each offspring receives either one or the other parental allele: gene segregation gene segregation gene segregation gene segregation gene segregation ....
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This note was uploaded on 09/03/2009 for the course BIOL 7.03 taught by Professor Chriskaiser during the Fall '04 term at MIT.
- Fall '04