BICD100 FINAL KEY

BICD100 FINAL KEY - BICD100 GENETICS Prof. Chisholm FINAL...

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Unformatted text preview: BICD100 GENETICS Prof. Chisholm FINAL EXAM KEY Thursday March 20, 2008 Class results: Distribution of total scores (both midterms + nal) (excluding Ws) Total scores 1 5 2 2 5 3 3 5 4 4 5 5 5 5 6 6 5 7 7 5 8 8 5 5 10 15 20 25 Bin Center number of students Number of students completing class: 153 Mean score = 517 SD 153 (Distribution is normal) Maximum score = 849 Grades: > 673 = A > 517 = B > 361 = C > 206 = D FINAL EXAM KEY 1 Question 1 The true-breeding wild-type strain of peas has yellow seeds. Fritz Mendel nds three independent true-breeding strains with green seeds (green 1,2 and 3). Fritz crosses each green-seeded strain to the true breeding yellow-seeded strain. From each cross he takes F 1 plants and allows them to self fertilize to yield the F 2 . Parental yellow x green 1 yellow x green 2 yellow x green 3 F 1 All yellow All yellow All green F 2 3:1 yellow:green 3:1 yellow:green 3:1 green:yellow (a) For each strain make up a symbol for the dominant and recessive alleles. State the symbols and the phenotype of each allele, making it clear which is dominant and which recessive. Use different symbols for the different strains. 5 points Uppercase alleles are dominant: Strain 1 G1 confers yellow g1 confers green Strain 2 G2 confers yellow g2 confers green Strain 3 g3 confers yellow G3 confers green (b) Using the symbols you dened in part (a), what are the genotypes of the green seeds in the F 2 of each strain? 5 points Cross 1 green F2: g1g1 Cross 2 green F2: g2g2 Cross 3 green F2: G3_ (G3G3 or G3g3) (c) If Fritz were to cross the F 1 from cross 2 (i.e. from yellow x green 2) with parental green strain 2, what phenotypic ratio should he expect in the progeny of that cross? 15 points The cross is G2g2 x g2g2. The progeny should be yellow (G2g2): green (g2g2) in 1:1 ratio. FINAL EXAM KEY 2 (d) Fritz crosses a plant from green strain 1 to a plant from green strain 2. He nds the F1 from this cross all have yellow seeds. Fritz allows the F 1 to self fertilize and counts the seeds of the F 2 . He nds that 71 plants have yellow seeds and 57 have green seeds. Devise an explanation for these results, stating the type of genetic interaction and drawing out the genotypes and phenotypes of the F 2 in a Punnett square. 25 points Green 1 and 2 must be recessive mutations in two different genes, as they complement (F1 are yellow). The segregation resembles a 9:7 ratio, suggesting recessive epistasis (10 points for stating this) in which G1_G2_ are yellow, g1g1__ and ___g2g2 are green. 15 points for a correct Punnett Square FINAL EXAM KEY (d) Fritz crosses a plant from green strain 1 to a plant from green strain 2. He nds the F1 from this cross all have yellow seeds. Fritz allows the F 1 to self fertilize and counts the seeds of the F 2 . He nds that 71 plants have yellow seeds and 57 have green seeds. Devise an explanation for these results, stating the type of genetic interaction and drawing out the genotypes and phenotypes of the F 2 in a Punnett square. in a Punnett square....
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BICD100 FINAL KEY - BICD100 GENETICS Prof. Chisholm FINAL...

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