MT1 Key - BICD 100 GENETICS Winter 2008 MIDTERM 1 KEY...

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BICD100 GENETICS Prof. Chisholm MIDTERM 1 Approximate grade scheme: >222 = A 168.5-222 = B 114-168 = C 59-113 = D 0-58 = F (the other midterms and final will follow similar grade curves) Descriptive statistics (as of 2/1/08) Minimum 25.0 25% Percentile 132.875 Median 167.75 75% Percentile 210.0 Maximum 278.0 Mean of 5 best scores: 264.4 Mean 168.17 Std. Deviation 54.206 Midterm 1 distribution 2 0 4 6 8 1 0 10 20 30 40 Bin center number of exams BICD 100 GENETICS Winter 2008 MIDTERM 1 KEY 1
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Question 1 A highly prized breed of chickens has a mixture of light and dark feathers, a plumage known as ‘Erminette’. In 1964 Frederick Hutt decided to investigate the inheritance of the Erminette phenotype. When male and female Erminette chickens were crossed, their F1 progeny consisted of 130 with Erminette plumage, 62 pure white plumage, and 64 pure black plumage. (a) Assuming that this trait is determined by a single gene, formulate a hypothesis for the genetic basis of the Erminette phenotype. Calculate the chi-squared statistic for the observed data to test your hypothesis. (20 points) The simplest model is that the Erminette phenotype results from incomplete dominance. Denote the Dark allele as A and the light as a. If Erminette is Aa, then the cross Aa x Aa will give AA:Aa:aa in 1:2:1 ratio. The Expected numbers based on this model are 64:128:64. ! 2 = (4/128)+ (4/64) = 0.093. (Credit for saying codominance, which is a viable interpretation) 10 points for the ratio and expected numbers. 10 points for ! 2 . (b) What is the approximate P value of the chi-squared? Should you accept or reject your hypothesis, using a significance level or critical value of 5%? Give the degrees of freedom in your calculation. (10 points) There are three classes so two degrees of freedom, so 0.975 < ! 2 < 0.9. The critical value of ! 2 is 5.99. The hypothesis cannot be rejected (‘accept hypothesis’ fine). (c) A geneticist objected that Hutt’s results might also be explained by a two-gene model, in which the white phenotype is epistatic to black or Erminette. What phenotypic ratio does this hypothesis predict? Calculate a chi-squared value for the observations given the two-gene model and comment on the relative merits of the one gene vs two gene models. (20 points). An epistasis model would predict a 9:3:4 ratio of Erminette: black: white. The Erminette phenotype would be due to dominant alleles at two loci (A_B_). Black would be A_bb. aa__ would be white. The expected values would be 144:48:64 and the chi squared would be (196/144) +(196/48) = 5.44. Given a critical value of 5.99 we cannot reject the
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MT1 Key - BICD 100 GENETICS Winter 2008 MIDTERM 1 KEY...

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