# Lecture 5 - Lecture 5 Working with independent assortment...

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Lecture 5 Working with independent assortment

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As the number of loci increase, the Punnett  approach becomes cumbersome Branch diagrams? (argh!) Statistical rules can be applied to predict  genotypes… For example the ever useful PRODUCT RULE

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What proportion of progeny will be of a specific genotype? A/a ; b/b ; C/c ; D/d ; E/e and A/a ; B/b ; C/c ; d/d ; E/e 5 loci assorting independently A/a X A/a one fourth will be a/a b/b X B/b half will be b/b C/c X C/c one fourth will be c/c D/d X d/d half will be d/d E/e X E/e one fourth will be e/e ¼  X  ½  X  ¼  X  ½  X  ¼ = 1/256 Predict proportion of homozygous recessive for all loci:

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How many distinct genotypes will a given cross  produce? Self a tetrahybrid  (four loci assorting  independently) F1 genotypes: A/a ; B/b ; C/c ; D/d 3 genotypes for each pair: (A/A, A/a and a/a) For 4 gene pairs: 3 4 = 81 different genotypes
Determining validity of observed outcome Chi-square test. Useful for comparing an observation/result with an expected outcome. Class Observe d (O) Expected  (E) (O-E) 2 (O-E) 2 /E Red 55 60 25 25/60=0.4 2 White 65 60 25 25/60=0.4 2 Total = X 2 = 0.84 Testcross result Hypothesis: Plant is heterozygous for petal color trait A = red a = white (A/a) Testcross to homozygous recessive (a/a ; white); 120 progeny = Σ (O-E) 2 / E for all classes

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df = degrees of freedom = classes Classes = red or white df = 2-1=1 = 0.84 P>0.05 result is compatible with hypothesis
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Lecture 5 - Lecture 5 Working with independent assortment...

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