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Genetics Exam II Summary and Problem Sets – Key Concepts and Practice Opportunities

Prepare for Exam II in Genetics with this comprehensive summary and problem set overview. The content covers critical topics such as sex linkage, calculation of probabilities from pedigrees, and the extension of Mendelian genetics. Explore multiple alleles determining one trait and the role of multiple genes with additive effects. Dive into study questions that assess your understanding of trait inheritance patterns, including autosomal dominant and recessive traits. Use this guide to optimize your office hours and enhance your exam preparation.

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Genetics Exam II Summary and Problem Sets – Key Concepts and Practice Opportunities

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  1. GeneticsNews • Summary for Exam II- Do it NOW! • Office hours underutilization • Problem Set 7 available • Problem Set 8 available • Exam I … not available Our answers and comments on web. Click on Course at a Glance; then Click on Exams; Answer Keys

  2. Topics for Today • Sex linkage/Sex limitation • Calculation of probabilities from pedigree • Extension of Mendelian Genetics - Multiple alleles determining one trait - Multiple genes determining one trait - Multiple genes with additive effect

  3. Study Question 6 How is the trait inherited? x Dd Dd Autosomal dominant Dd

  4. Study Question 6 How is the trait inherited? Dd Dd Autosomal dominant x Autosomal recessive dd Dd Dd Dd Dd dd dd

  5. Study Question 6 How is the trait inherited? XDY XXD Autosomal dominant Autosomal recessive XDY x Sex-linked dominant

  6. Study Question 6 How is the trait inherited? XXd Autosomal dominant Autosomal recessive XXd XY XXd XdY XY Sex-linked dominant XdY XdY Sex-linked recessive

  7. Study Question 6 Could it be an autosomal dominant trait limited to males? Dd Dd Yes dd Dd Dd Dd dd Dd Dd

  8. Study Question 8 What is the probability that IV.1 will get the disease? Specify: P(III.2 x III.3 XdY) XXd Parse: Solve: XXd? XY P(III.2 is XXd) AND P(III.2 passes Xd) ? 1/2 1/2 XdY ? AND P(III.3 is XY) AND P(III.3 passes Y) 1 Probability = 1/8 1/2

  9. Multiple Genes/One Trait Conventional dihybrid cross Substrate1 Product1 Substrate2 Product2A B

  10. Multiple Genes/One Trait Dihybrid cross of interacting genes 9 : 3 : 3 : 1  =9 : 7 Substrate1 Substrate2 ProductA B

  11. Multiple Genes/One Trait Dihybrid cross of interacting genes 9 : 3 : 3 : 1  =9 : 7 chorismate compound X tryptophanM N

  12. Study Question 17 You cross a vermilion-eyed fly with a cinnibar-eyed fly to obtain F1 progeny, then you cross these progeny to get F2 progeny. What phenotypic ratio do you expect?

  13. Study Question 17 Dihybrid cross of interacting genes Substrate1Substrate2ProductV C

  14. Study Question 17 Dihybrid cross of interacting genes 9 : ? : ? : ? Substrate1Substrate2ProductV C

  15. Study Question 17 Dihybrid cross of interacting genes 9 : 3 : ? : 1  Substrate1Substrate2ProductV C

  16. Multiple Genes/One Trait Dihybrid cross of interacting genes 9 : 3 : 3 : 1 =9 : 3 : 4 9 : 3 : 3 : 1 Substrate1Substrate2ProductV C

  17. Study Question 19 Conventional dihybrid cross Pathway where dihybrid cross gives 15:1 ratio? Substrate1 Product1 Substrate2 Product2A B

  18. Study Question 19 What kind of genetic situation can you imagine that would lead to a 15:1 phenotypic ratio? 9 : 3 : 3 : 1  9 : 3 : 3 : 1  15 : 1 Substrate1 Product1 Substrate2 Product2A B

  19. Study Question 20 You find corn with kernels in the ratio: Yellow 247 Golden 154 Sickly green 23 Hypothesis? Hypothesis: One gene, multiple alleles AB x CD 1:(1:1):1= 1:2:1

  20. Study Question 20One gene hypothesis Exp Obs You find corn with kernels in the ratio: Yellow 247 Golden 154Sickly green 23 TOTAL 424 212106106424 Unit = 424/4 = 106

  21. Study Question 20Two gene hypothesis You find corn with kernels in the ratio: Yellow 247 Golden 154 Sickly green 23 Hypothesis? Hypothesis: Two genes for one trait AaBb x AaBb 9:3:3:1

  22. Study Question 20Two gene hypothesis Exp Obs You find corn with kernels in the ratio: Yellow 247 Golden 154Sickly green 23 TOTAL 424 238.5 = 26.5 x 9159 = 26.5 x 6 26.5 = 26.5 x 1424 Unit = 424/16 = 26.5

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