Genetics380_lecture5_2009v1_Sakai

Genetics380_lecture5_2009v1_Sakai - Sept 17, 2009 Complete...

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Sept 17, 2009 • Complete Topic 4 (Dihybrid crosses and Chi square goodness-of-fit) • Topic 5 (Sex determination and Sex- linked characteristics)
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Dihybrid Ratios AaBb X AaBb -> 9 A_B_ : 3 aaB_ : 3 A_bb : 1 aabb AaBb X aabb -> 1 A_B_ : 1A_bb : 1 aaB_ : 1 aabb
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Applying Probability and the Branch Diagram to Dihybrid Crosses
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The Dihybrid Test Cross (wrong figure inserted in Lecture 4)
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Problem Solving Strategies (p.61 of text) 1. Determine the questions to be answered. 2. Write down the basic information given in the problem. 3. Write down any genetic information that can be determined by the phenotypes alone. 4. Break the problem down into smaller parts. 5. Work the different parts of the problem. 6. Check all work
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BREAK: In class problem solving In cats, curled ears result from an allele (Cu) that is dominant over an allele (cu) for normal ears. Black color results from an independently assorting allele (G) that is dominant over an allele for gray (g). A gray cat homozygous for curled ears is mated with a homozygous black cat with normal ears. All the F 1 cats are black and have curled ears. a. If two of the F 1 cats mate, what PHENOTYPES and proportions are expected in the F 2 ? b. An F 1 cat mates with a stray cat that is gray and possesses normal ears. What phenotypes and proportions of progeny are expected from this cross?
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How does real data compare to expected results? Let us take the data of Mendel from the F 2 generation of a WWGG x wwgg cross- 315 W-G- 108 W-gg 101 wwG- 32 wwgg
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Observed ratios of progeny may deviate from expected ratios by chance To evaluate the role of chance in producing deviations between observed and expected values, a statistical test called the goodness-of-fit chi-square test is used. Applied to numbers of progeny, not to proportions or percentages. Degrees of freedom (df) = n-1 where n is the number of phenotypes.
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Statistical Analysis of Genetic Data- the X 2 test X 2 = Sum of [d 2 /e] where d= Deviation = [Observed-Expected] and e= Expected value. Let us take the data of Mendel from the F 2 generation of a WWGG x wwgg cross- 315 W-G-: 108 W-gg : 101 wwG- : 32 wwgg A 9 : 3 : 3 : 1 ratio is expected, if the genes are assorting independently. In our case, Phenotypes = W-G- W-gg wwG- wwgg Observed = 315 108 101 32 Total = 556 Expected = 313 104 104 35 (9/16 x 556) (3/16 x 556) (3/16 x 556) (1/16 x 556) Deviation = +2 +4 -3 -3 [d= (obs-exp)] d 2 = 4 16 9 9 d 2/ e = 4/313 16/104 9/104 9/35 X 2 =sum of [d 2 /e] = 0.01 + 0.15 + 0.09 + 0.26 = 0.51 Degrees of freedom = 3 (n-1). P > 0.9 [see next slide]
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Most scientist accept the 0.05 probability level as cut-off.
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Goodness-of-Fit Chi Square Test • Indicates the probability that the difference between the observed and expected values are due to chance. • If the probability (p value) is high, then
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This note was uploaded on 10/03/2010 for the course RUT 146:356 taught by Professor Gol during the Spring '10 term at Rutgers.

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Genetics380_lecture5_2009v1_Sakai - Sept 17, 2009 Complete...

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