Question textA true breeding line of lizards with stumpy tails and grey eyes (P1) was crossed with a true-breeding line with long tails and black eyes (P2). The F1 progeny were all long tailed with grey eyes. The Geneticist in charge of the experiment was busy teaching and the F1 offspring were left to inter-breed and generate an F2 generation. The F2 consisted of 102 lizards with long tails and grey eyes, 36 lizards with stumpy tails and grey eyes, 33 lizards with long tails and black eyes and 12 lizards with stumpy tails and black eyes. a) From the F1 phenotype, deduce which phenotype is dominant and which is recessive for tail length and eye colour. The Answer 1 Question 4[select: , long, stumpy] tails is dominant to Answer 2 Question 4[select: , stumpy, long] tails. The Answer 3 Question 4[select: , black, grey] eyes is dominant to the Answer 4 Question 4[select: , black, grey] eyes. (1 mark) In this question, the alleles controlling tail length are the A and a alleles while the alleles controlling eye colour are the B and b alleles b) Using the A and B gene notation, give the following alleles: Stumpy tail allele is Answer 5 Question 4[select: , a, A, B, b] Long tail allele is Answer 6 Question 4[select: , b, B, A, a] Grey eye allele is Answer 7 Question 4[select: , b, A, B, a] Black eye allele is Answer 8 Question 4[select: , b, a, A, B] (1 mark) c) Using the A and B gene notation, give the following genotypes: P1 genotype Answer 9 Question 4[input] P2 genotype Answer 10 Question 4[input] F1 genotype Answer 11 Question 4[input] In entering your answers: 1) case is important; 2) don't leave spaces; 3) separate A and B genes with a semi-colon i.e. ; (2 marks) At this point we should really consider each of the traits separately, and perform chi square tests to show that the two alleles of each gene are segregating from each other in Mendelian proportions and that there are no viability differences. However due to time constraints we will assume that this is the case, and proceed to test if the alleles of one gene assort into gametes independently of the alleles of the other gene. d) What is the expected genotypic proportions in the F2 generation, assuming the genes are unlinked and are independently assorted? Give a whole number numerical answer for each space. Answer 12 Question 4[input] A_;bb :  Answer 13 Question 4[input] aa;bb :  Answer 14 Question 4[input] A_;B_ :  Answer 15 Question 4[input] aa;B_ (2 mark) A reminder before attempting the next question: The F2 consisted of 102 lizards with long tails and grey eyes, 36 lizards with stumpy tails and grey eyes, 33 lizards with long tails and black eyes and 12 lizards with stumpy tails and black eyes.e) Complete the chi-square table to test the Genetic Hypothesis that the genes for skin colour and eye shape were unlinked and assorting independently. Fill in any blank spaces: Give your answers to 2 decimal places. [table] Phenotypic class | Observed (O) | Expected (E) | (O-E) | (O-E)2 | (O-E)2/E long, grey | Answer 16 Question 4 | Answer 17 Question 4 | N/A | N/A | 0.009 stumpy, grey | Answer 18 Question 4 | Answer 19 Question 4 | N/A | N/A | 0.083 long, black | Answer 20 Question 4 | Answer 21 Question 4 | N/A | N/A | 0.05 stumpy, black | Answer 22 Question 4 | Answer 23 Question 4 | N/A | N/A | 0.028 χ2 value: | Answer 24 Question 4 [/table] Degrees of Freedom (Df) = Answer 25 Question 4[input] (3 marks)f) When the experiment was repeated, similar data were obtained, with a X2 value of 2.1. Using the Chi-sq table provided below, what would be the approximate P value associated with this X2 value? Multiple choice 1 Question 4>0.90.8 to 0.90.7 to 0.80.5 to 0.70.3 to 0.50.2 to 0.30.1 to 0.20.05 to 0.10.02 to 0.050.01 to 0.020.001 to 0.01<0.001 Would the null hypothesis be accepted? Answer 26 Question 4[select: , no, yes] Would the genetic hypothesis be accepted? Answer 27 Question 4[select: , no, yes] (3 marks)多项填空题