BME 80H Lecture Notes - Lecture 13: Sex Linkage, Zygosity, Reciprocal Cross

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13th lecture
example mapping problem (test cross)
screen shot
10/100 x 100 = 10mu
c. What would be the map units if there were no recombinants
apparent in the gametes? (the answer is not 0)
-if you looked at 1,000 gametes the answer would be
<1/1000 x 100 = <0.1%
d. 50 map units represent no linkage between the genes.
Why?
-this would be the same result as if the alleles assorted
independently
IV. Violating Mendel's Law of Segregation
A. Sex-Linkage: Genes located on the X chromosome (figure 4.9, 4.10)
XX = female (homologous chromosomes)
XY = male (not homologous chromosomes)
-males with only one X chromosome cannot have
equally segregating alleles for x-linked genes
Males, having only one X chromosome, cannot be homozygous or
heterozygous for genes on the X. They are instead termed
"hemizygous" for X-linked alleles.
1. Expectations based on Mendel's work
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Document Summary

13th lecture example mapping problem (test cross) screen shot. 10/100 x 100 = 10mu: what would be the map units if there were no recombinants apparent in the gametes? (the answer is not 0) If you looked at 1,000 gametes the answer would be. <1/1000 x 100 = <0. 1: 50 map units represent no linkage between the genes. This would be the same result as if the alleles assorted independently. Violating mendel"s law of segregation: sex-linkage: genes located on the x chromosome (figure 4. 9, 4. 10) Males with only one x chromosome cannot have equally segregating alleles for x-linked genes. Males, having only one x chromosome, cannot be homozygous or heterozygous for genes on the x. hemizygous for x-linked alleles: expectations based on mendel"s work, for a cross of 2 pure breeding lines, one expressing the dominant trait and one expressing the recessive trait, the results are as follows:

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