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Three hourly exams plus final exam (450 pts), You will have 1.5 hours to complete each exam,

…may seem really hard. …should be relatively easy. Grades. Three hourly exams plus final exam (450 pts), You will have 1.5 hours to complete each exam, You will be allowed one (1) 11” x 8.5” crib sheet, both sides, for each exam, Exams - 150 points each, Final Exam cumulative.

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Three hourly exams plus final exam (450 pts), You will have 1.5 hours to complete each exam,

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  1. …may seem really hard. …should be relatively easy. Grades • Three hourly exams plus final exam (450 pts), • You will have 1.5 hours to complete each exam, • You will be allowed one (1) 11” x 8.5” crib sheet, both sides, for each exam, • Exams - 150 points each, Final Exam cumulative. • Quizzes will be givenevery Wednesday (total 100 pts), • will cover the basics of the assigned reading (including that day's assignment), • quizzes 12.5 points each, ~15 minutes, • No Make-up Quizzes, absolutely no exceptions, • can drop two (2) lowest quiz scores (except 1&2). • Total course points - 550

  2. Know This… 1 1/2 1/2 1 1/4 1/2 1/4 1/2 1/2 1 Assignment: Correlate this with the observed phenotype.

  3. = 9 = 3 = 3 = 1 Mendel’s Results, F2 DihybridP generation cross: YYRR x yyrr F1 generation cross: YyRr x YyRr • Y_ R_ = 315 • yyR_ = 108 • Y_rr = 101 • yyrr = 32

  4. 1/4 x 1/4 = 1/16 YYRR 1/4 RR 1/2 Rr 1/4 x 1/2 = 1/8 YYRr 1/4 YY 1/4 rr 1/4 x 1/4 = 1/16 YYrr 1/4 RR 1/2 x 1/4 = 1/8 YyRR 1/2 Rr 1/2 x 1/2 = 1/4 YyRr 1/2 Yy 1/4 rr 1/2 x 1/4 = 1/8 Yyrr 1/4 x 1/4 = 1/16 yyRR 1/4 RR 1/2 Rr 1/4 x 1/2 = 1/8 yyRr 1/4 yy 1/4 rr 1/4 x 1/4 = 1/16 yyrr Forked-Line Method

  5. 1/4 x 1/4 = 1/16 YYRR 1/4 x 1/2 = 1/8 YYRr 1/2 x 1/4 = 1/8 YyRR 1/2 x 1/2 = 1/4 YyRr 1/16 + 2/16 + 2/16 + 4/16 Genotypes Y--R-- 1/4 RR 1/4 YY 1/2 Rr 1/4 RR 1/2 Yy 1/2 Rr = 9/16 yellow/round

  6. 1/4 x 1/4 = 1/16 YYRR 1/4 RR 1/2 Rr 1/4 x 1/2 = 1/8 YYRr 1/4 rr 1/4 x 1/4 = 1/16 YYrr 1/4 RR 1/2 x 1/4 = 1/8 YyRR 1/2 Rr 1/2 x 1/2 = 1/4 YyRr 1/4 rr 1/2 x 1/4 = 1/8 Yyrr 1/4 x 1/4 = 1/16 yyRR 1/4 RR 1/2 Rr 1/4 x 1/2 = 1/8 yyRr 1/4 rr 1/4 x 1/4 = 1/16 yyrr Genotypes Y--rr 1/4 YY 1/2 Yy 1/4 yy

  7. 1/4 x 1/4 = 1/16 YYrr 1/2 x 1/4 = 1/8 Yyrr 1/16 + 2/16 Genotypes Y--rr 1/4 YY 1/4 rr 1/2 Yy 1/4 rr = 3/16 yellow/wrinkled

  8. 1/4 x 1/4 = 1/16 YYRR 1/4 RR 1/2 Rr 1/4 x 1/2 = 1/8 YYRr 1/4 rr 1/4 x 1/4 = 1/16 YYrr 1/4 RR 1/2 x 1/4 = 1/8 YyRR 1/2 Rr 1/2 x 1/2 = 1/4 YyRr 1/4 rr 1/2 x 1/4 = 1/8 Yyrr 1/4 x 1/4 = 1/16 yyRR 1/4 RR 1/2 Rr 1/4 x 1/2 = 1/8 yyRr 1/4 rr 1/4 x 1/4 = 1/16 yyrr Genotypes yyR-- 1/4 YY 1/2 Yy 1/4 yy

  9. 1/4 x 1/2 = 1/8 yyRr 1/4 x 1/4 = 1/16 yyRR 2/16 + 1/16 Genotypes yyR-- 1/4 yy 1/2 Rr 1/4 yy 1/4 RR = 3/16 green/round

  10. 1/4 x 1/4 = 1/16 YYRR 1/4 RR 1/2 Rr 1/4 x 1/2 = 1/8 YYRr 1/4 rr 1/4 x 1/4 = 1/16 YYrr 1/4 RR 1/2 x 1/4 = 1/8 YyRR 1/2 Rr 1/2 x 1/2 = 1/4 YyRr 1/4 rr 1/2 x 1/4 = 1/8 Yyrr 1/4 x 1/4 = 1/16 yyRR 1/4 RR 1/2 Rr 1/4 x 1/2 = 1/8 yyRr 1/4 rr 1/4 x 1/4 = 1/16 yyrr Genotypes yyrr 1/4 YY 1/2 Yy 1/4 yy

  11. 1/4 x 1/4 = 1/16 yyrr 1/16 Genotypes yyrr 1/4 yy 1/4 rr green/wrinkled

  12. F2 via Forked Line • Y--R-- yellow/round 9/16 • Y--rr yellow/wrinkled 3/16 • yyR-- green/round 3/16 • yyrr green/wrinkled 1/16

  13. Why use Forked-Line Method? • Based on a classic dihybrid cross (YyRr x YyRr), what is the probability that an organism in the F2 generation will have round seeds and breed true for green cotyledons?

  14. OK? YR Yr yR yr YYRR YYRr YyRR YyRr YR YYRr YYrr YyRr Yyrr Yr YyRR YyRr yyRR yyRr yR YyRr Yyrr yyRr yyrr yr 3/16 p = 0.1875

  15. 1/4 x 1/4 = 1/16 yyRR 1/4 RR 1/4 yy 1/2 Rr 1/4 x 1/2 = 1/8 yyRr 1/4 rr 1/4 x 1/4 = 1/16 yyrr Better 3/16 p = 0.1875

  16. 1/4 yy 3/4 R_ 1/4 x 3/4 = 3/16 yyR_ Sum Law: 1/4 RR + 1/2 Rr Best (?)

  17. 9/16 yellow round 3/4 round 3/4 yellow 1/4 wrinkled 3/16 yellow wrinkled 3/16 green round 3/4 round 1/4 green 1/16 green wrinkled 1/4 wrinkled Forked-Line Method(phenotypes)

  18. 1/4 RrYY 1/2 YY 1/2 Yy Example P Rr YY x rrYy Probability Rr YY in offspring; 1/2 Rr 1/2 rr

  19. 1/2 Rr 1/4 RrYy 1/2 RR 1/4 YY 1/4 yy Example P Rr Yy x RRYy Probability of Rr Yy in offspring; 1/2 Yy

  20. Independent Assortment gametes Random Segregation .5 .5 .25 .5 probability .25 Using ProbabilityLecture 3 Example YYSs x YySs YY x Yy Ss x Ss YY or Yy SS Ss ss (p) Y_ = 1 (p) S_ = .75 Product Rule: (p) Y_S_ = .75 (p) ss = .25 Product Rule: (p) Y_ss = .25

  21. Humans ? • Is it possible to ascertain the mode of inheritance of genes in organisms where designed crosses and the production of large numbers of offspring are not practical? Pedegree: an orderly diagram of a families relevant genetic features.

  22. Albinism is a recessive trait in humans. Assignment: figure out this pedigree.

  23. From Previous Page Assignment: figure out this pedigree.

  24. Symbols

  25. More Symbols

  26. And more… 2

  27. aa aa aa Where Do you Start? Aa Aa or AA Aa Aa Recessive Trait? or Dominant Trait?

  28. aa aa aa What More Can You Say? Aa Aa or AA Aa Aa Recessive Trait

  29. Predictions What if you were a genetic counselor? What are the odds that this individual carries the trait?

  30. Predictions What if you were a genetic counselor? What are the odds that this individual carries the trait?

  31. Monohybrid Cross Conditional Probability Aa Aa ? 1/4 AA 1/4 Aa 1/4 Aa 1/4 aa 1 : 1 : 1 (p)Aa = 2/3 = .66

  32. Conditional Probability …is the probability of an event occurring given that another event also occurs... P(event) without the condition p(condition)

  33. Conditional Probability Example: With a 6-sided die, what is the probability of rolling a 2, given that an even number is rolled on the die: p(2 roll | even #) = p(2 roll) p(even#) p(2 roll | even #) = 1/6 1/2 = 1/3

  34. Aa Aa ? p( heterozygous | A_ ) = 1/2 3/4 probability without the conditionprobability of the condition p(probability of being a heterozygote) = 1/2 p(probability of A_) = 3/4 = 2/3

  35. A a AA Aa A Aa aa a 1/4 AA 1/4 Aa 1/4 Aa 1/4 aa 1 : 1 : 1 Conditional Probability p(event) without the condition p(condition) • Use the formula, • Or use a Punnett Square, • Or... p(A|B)=

  36. Kidney Disease If 1 and 2 had an offspring, what is the probability that their first kid would show the phenotype?

  37. A Simplification • Unless otherwise specified (or the pedigree suggests otherwise), the traits that we will track will be rare, • We will assume a p = 0 that a non-familial mate carries the trait.

  38. Kidney Disease • non-familial mates: from outside of the family, • if k is the recessive trait, then these individuals are KK.

  39. p(P1) heterozygous x p(P2) heterozygous x p(FF) homozygous recessive 1/2 x 1 x 1/4 Kidney Disease If 1 and 2 had an offspring, what is the probability that their first kid would show the phenotype? = 1/8 = .125

  40. Kidney Disease If 1 and 2 had an offspring, what is the probability that the first kid would be a boy, and show the phenotype?

  41. And, what is the probability that this boy is a carrier? p(P1) heterozygous p (boy) x p(P2) heterozygous x p(FF) homozygous recessive 1/2 x 1 x 1/4 Kidney Disease If 1 and 2 had an offspring, what is the probability that the first kid would be a boy, and show the phenotype? x 1/2 = 1/16

  42. Practice #1 Round (R) and Yellow (Y) are dominant.

  43. Practice #2

  44. Practice #3

  45. Questions • Don’t rely on the answers in the back of the book to solve your problems… • Don’t just solve them, but understand the principles needed to solve them.

  46. a b f c e d

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