Lecture+8+End+Topic+V+Correction+to+Slide+39+typo

Lecture+8+End+Topic+V+Correction+to+Slide+39+typo - Topic V...

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Unformatted text preview: Topic V. Slides 38-40 February 8, 2010 Note: Correction of typographical error on slide 39. Next part Mapping genes in human chromosomes Mapping • X chromosome linkage • Handout #9 • Somatic cell hybridization technique. Tetrad analysis Tetrad • Handout #8 (will not be used in lecture; not on exam) February 8, 2010 BIS101-001, Winter 2010—Genes and Gene Expression BIS1012010— 2 Mapping genes in human chromosomes X chromosome linkage To get accurate results, the data collected form human chromosome recombination analysis are from many pedigrees of the same trait. Handout #9 Recombination analysis in humans: X-linked chromosome linkage February 8, 2010 BIS101-001, Winter 2010—Genes and Gene Expression BIS1012010— 3 Mapping genes in human chromosomes Summary X chromosome linkage Parents: male is affected (has two rare X-linked recessive alleles: a, b) female is normal (homozygous dominant for A, B alleles on the X chromosome). F1: all offspring are unaffected; however, the females are carriers for both traits (a and b). This is because she gets one X chromosome from the affected father. Look at 4 progeny: the genes are distributed equally between the male offspring (we are only interested in the male offspring). -thus without recombination, we expect ½ AB male and ½ ab male. -however, recombination occurred so we get two more possibilities: Ab male aB male. BIS101-001, Winter 2010—Genes and Gene Expression BIS1012010— 4 F2: February 8, 2010 ...
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