Follow ups from last time

1. Follow ups from last time • Modified Mastering Genetics: • Pearsonmastering.com, click ‘student’ • Course ID = ray68015 • Code = USSGAT-CTTTT-INANE-TOGUE-METIS-RULES • Learning Catalytics • https://www.pearson.com/us/higher-education/products-services-teaching/learning-engagement-tools/learning-catalytics/training-support/students/buy-access.html • Or just search for ‘pearson learning catalytics purchase’

2. Follow ups from last time • Laboratories start this week • What to expect: • First lab exercise on drosophila genetics • Weekly quizzes start next week (1/28) • Based on lectures, end-of-chapter questions, and lab exercises • TAs are expecting you to discuss end-of-chapter problems • Information on timing of Current Events essay topics and the essays themselves

3. Follow ups from last time • Conflicts with evening exams • Will need confirmation of conflict with a class or lab • A copy of your schedule will do • Must arrange a separate time for exam within 48 hours of time on the syllabus (before or after)

4. Follow ups from last time • Second attempt at Learning Catalytics

5. The Mendelian View of the World • Gregor Mendel’s (1822-1884) discoveries • Goal: to determine the pattern by which inheritable characteristics were transmitted to the offspring • Generate testable hypotheses and experiment to determine the answer to innumerable questions. • Five keys to his success • 1. Controlled crosses • 2. Use of pure-breeding strains to begin • 3. Use of dichotomous traits • 4. Quantification of the results • 5. Use of replicate, reciprocal, and test crosses

6. The Mendelian View of the World • Five keys to his success • 1. Controlled crosses • Did not allow for random fertilization, only fertilizations that he determined • 2. Use of pure-breeding strains to begin • Strains consistently produce the same offspring, suggesting the same homogeneous genetic information • 3. Use of dichotomous traits • Reduces the number of possible outcomes, makes analyses easier • 4. Quantification of results • Numbers don’t lie • 5. Use of replicate, reciprocal, and test crosses • Replication is a must in science • Reciprocal and test crosses allow for all combinations of crosses

7. The Mendelian View of the World • Five keys to his success • 1. Controlled crosses • 2. Use of pure-breeding strains to begin • 3. Use of dichotomous traits • 4. Quantification of the results • 5. Use of replicate, reciprocal, and test crosses

8. Mendelian Inheritance • A classic experiment • A monohybrid cross • What did it tell Mendel? • What observations and conclusions can be made? • Pod shape was inherited ‘discretely’ • At least some organism characteristics are carried across generations as discrete ‘factors’

9. Mendelian Inheritance • A classic experiment • What did it tell Mendel? • What observations and conclusions can be made? • One of the alleles is able to block the other (is dominant vs. is recessive) • There was a definite mathematical pattern to the occurrence of the traits (3:1) in F2 ~75% ~25%

10. Mendelian Inheritance • A classic experiment • What did it tell Mendel? • What conclusions can be drawn? • To achieve these ratios, ‘factors’ are likely inherited in pairs and those pairs are randomly rearranged each generation • We now refer to factors as genes or loci • Homozygous vs. heterozygous • Phenotype vs. genotype

11. Monohybrid cross • Parents differ by a single trait. • Crossing two pea plants that differ in stem size, one tall one short T = allele for Tall t = allele for dwarf TT = homozygous tall plant t t = homozygous dwarf plant T T  t t

12. Monohybrid cross for stem length: P = parentals true breeding, homozygous plants: T T  t t (tall) (dwarf) T t (all tall plants) F1 generation is heterozygous:

13. Using a Punnett Square STEPS: 1. determine the genotypes of the parent organisms 2. write down your "cross" (mating) 3. draw a p-square Parent genotypes: TT and t t Cross T T  t t

14. Punnett square 4. "split" the letters of the genotype for each parent & put them "outside" the p-square 5. determine the possible genotypes of the offspring by filling in the p-square 6. summarize results (genotypes & phenotypes of offspring) T T T T  t t t t Genotypes: 100% T t Phenotypes: 100% Tall plants

15. Monohybrid cross: F2 generation • If you let the F1 generation self-fertilize, the next monohybrid cross would be: T t  T t (tall) (tall) Genotypes: 1 TT= Tall 2 Tt = Tall 1 tt = dwarf Genotypic ratio= 1:2:1 T t T t Phenotype: 3 Tall 1 dwarf Phenotypic ratio= 3:1

16. Mendelian Inheritance • A classic experiment • What did it tell Mendel? • What conclusions can be drawn? • To achieve these ratios, ‘factors’ are likely inherited in pairs and those pairs are randomly rearranged each generation • We now refer to factors as genes or loci • Homozygous vs. heterozygous • Phenotype vs. genotype ~75% ~25%

17. Mendelian Inheritance • A review of Gregor Mendel’s work • Goal: to determine the pattern by which inheritable characteristics were transmitted to the offspring • Major conclusions • Characteristics were governed by distinct units of inheritance (genes) • Each organism has/inherits two copies each gene, one from each parent • Thetwogenes may be identicalto one anotheror non-identical (alleles) • Gametes (reproductive cells) must therefore carry only one copy of the gene for each trait • Principle of Independent Segregation - an organism's alleles separate from one another during gamete formation and are independently transmitted to offspring • Principle of Independent Assortment - Each pair of alleles segregates independently from other pairs during gamete formation

18. Mendelian Inheritance • What happens if you look at 2+ traits simultaneously? • The alleles for each trait segregate independently AND • do not influence the inheritance (independent assortment) of the second trait

19. Mendelian Inheritance • The progeny phenotypes tend to occur in definite ratios based on the initial frequencies

20. Mendelian Inheritance • The same extends to trihybrid crosses and other combinations • You could draw a Punnett square for any number of traits but the squares get big and confusing fast.

21. LC questions

22. In chickens, the dominant allele Cr produces the creeper phenotype (having extremely short legs). However, the creeper allele is lethal in the homozygous condition. If two creepers are mated, what proportion of the living progeny will be creepers? • Crcr – short legs • crcr – normal legs • CrCr – no legs, it’s lethal • What must be the genotypes if the parents are creepers? --- Crcr Cr cr Cr cr

23. A dihybrid cross is made between a pea plant that has constricted pods, which is a recessive trait (smooth is dominant), and is heterozygous for seed color (yellow is dominant to green) and a plant that is heterozygous for both pod texture and seed color.  Construct a Punnett Square for this cross.  What is the probability that the offspring will be heterozygous for both traits? • S=smooth, s = constricted • Y=yellow, y =green • ssYy x SsYy SY SysYsy sY sy

24. A dihybrid cross is made between a pea plant that has constricted pods, which is a recessive trait (smooth is dominant), and is heterozygous for seed color (yellow is dominant to green) and a plant that is heterozygous for both pod texture and seed color.  Construct a Punnett Square for this cross.  What is the probability that the offspring will be heterozygous for both traits? • S=smooth, s = constricted • Y=yellow, y =green • ssYy x SsYy SY SysYsy sY sy

25. The genotype of parent one is A/a; B/b; C/c and parent two is A/a; B/b; c/c. Normal dominance relationships are implied by the capitalization of each allele. Assume the three genes are independently assorting and control three different phenotypes. What fraction of progeny will share the phenotype of the first parent? • The easiest way to figure out the answer is to use the fact that for independently assorted traits, the probability of an organism sharing two traits is the product of each individual trait • The probability of an offspring sharing the first trait with the first parent is 3/4 (A/A, A/a, or a/A, but not a/a). • The probability of an offspring sharing the second trait with the first parent is 3/4 (B/B, B/b, or b/B, but not b/b). • The probability of an offspring sharing the third trait with the first parent is 1/2 (C/c). • ¾ x ¾ x ½ = 9/32

26. Mendelian Inheritance • Keep in mind that all of these inheritance patterns are probabilistic • The ratios are what is predicted, not what is guaranteed • That means you need a basic understanding of probability

27. Mendelian Inheritance • Rules of probability theory apply • Product rule • If two or more events are independent of one another, their joint probability is the product of the individual probabilities • “What are the chances of this AND that happening?” • P(g) = ½, P(g) = ½ • P(gg) = ½ x ½ = ¼

28. The genotype of parent one is A/a; B/b; C/c and parent two is A/a; B/b; c/c. Normal dominance relationships are implied by the capitalization of each allele. Assume the three genes are independently assorting and control three different phenotypes. What fraction of progeny will share the phenotype of the first parent? • The easiest way to figure out the answer is to use the fact that for independently assorted traits, the probability of an organism sharing two traits is the product of each individual trait • The probability of an offspring sharing the first trait with the first parent is 3/4 (A/A, A/a, or a/A, but not a/a). • The probability of an offspring sharing the second trait with the first parent is 3/4 (B/B, B/b, or b/B, but not b/b). • The probability of an offspring sharing the third trait with the first parent is 1/2 (C/c). • ¾ x ¾ x ½ = 9/32

29. Mendelian Inheritance • Sum rule • The joint probability of any of two or more mutually exclusive (not independent) events is the sum of the probabilities of each event • The probability of obtaining any heterozygote is equal to the sum of the probabilities of each possible heterozygote • “What are the chances of this OR that happening?” • P(one Gg) = ¼, P(other Gg) = ¼ • P(any Gg) = ¼ + ¼ = ½

30. Mendelian Inheritance • Conditional probability • The probability of an event (A), given that another (B) has already occurred • Ex. What is the likelihood that a yellow-seeded F2 is heterozygous, like the parents? • Yellow could be GG or Gg • 2/3 • Ex. If the yellow-seeded F2s are allowed to self-fertilize, what proportion would be expected to breed true (produce only yellow offspring)?

31. Mendelian Inheritance • Some questions involve predicting the likelihood of a series of events • We use binomial probabilities

32. Mendelian Inheritance • Binomial Expansion • Two variables, each representing the frequency of one of two alternative outcomes • P(outcome 1) = p; P(outcome 2) = q • If p and q are the only possible outcomes, p+q = 1 • In examining probable outcomes (e.g. coin flips) we expand the expression by n events; (p+q)n=1

33. Mendelian Inheritance • Binomial probability • the probability that a binomial experiment results in exactly X successes

34. Mendelian Inheritance • Coin flips = 2 possibilities • p=heads, q=tails • Assuming a fair coin and one event, p = q = ½ = 0.5 • If you flip a coin three times, what are the possible outcomes? • HHH, HHT, HTT, HTH, THH, TTH, THT, TTT • There are three different ways to get two heads and one tails • 3 possible ways out of 8 total ways, 3/8 = 0.375 • Only one way to get three tails • 1 possible way out of 8 total ways, 1/8 = 0.125

35. Mendelian Inheritance • HHH, HHT, HTT, HTH, THH, TTH, THT, TTT • P(three tails) = 1/8 = 0.125 • P(exactly two tails) = 1/8 + 1/8 + 1/8 = 3/8 = 0.375 • P(at least two tails) = 1/8 + 1/8 + 1/8 + 1/8 = 1/2 = 0.5 • How does this equate to the binomial expansion? • Three flips  (p + q)3 = p3 + 3p2q + 3pq2 + q3 = 1 •  (1/2)3 + 3(1/2)2(1/2) + 3(1/2)(1/2)2 + (1/2)3 •  1/8 + 3/8 + 3/8 + 1/8 = 1

36. Mendelian Inheritance • How does this apply to genotypes/phenotypes? • Two possible phenotypes – yellow seeds, green seeds • Assuming Gg x Gg • What’s the probability of getting yellow seed? • Yellow = GG, Gg • In Gg x Gg, P(yellow) = P(GG) + 2*P(Gg) = ¾ • Or… p2 + 2pq = 3/4

37. Mendelian Inheritance • How does this apply to genotypes/phenotypes? • What if we repeated this 6 times? As some plants do when they make seed pods • If six seeds/pod, 26=64 possible combinations

38. Mendelian Inheritance • Application to genotypes/phenotypes • (p+q)6 = • p6 + 6p5q + 15p4q2 + 20p3q3 + 15p2q4 + 6pq5 + q6 = 1 • But who wants to do that? • Pascal’s triangle lets you cheat

39. Mendelian Inheritance • Ex. Cross Gg x Gg. If the result is always a pod with 8 seeds/pod, how often would you expect to find pods with 1 green seed? …

40. Mendelian Inheritance • Ex. Cross Gg x Gg. If the result is always a pod with 8 seeds/pod, how often would you expect to find pods with 1 green seed? 8/256 = 0.03125 …

41. Mendelian Inheritance • Ex. Cross Gg x Gg. If the result is always a pod with 8 seeds/pod, how often would you expect to find pods with exactly 4 green seeds? • p8 + 8p7q + 28p6q2 + 56p5q3 + 70p4q4 +56p3q5 + 28p2q6 + 8pq7 + q8 = 1

42. Mendelian Inheritance • Ex. Cross Gg x Gg. How often would you expect to find pods with at least 7 green seeds? • p8 + 8p7q + 28p6q2 + 56p5q3 + 70p4q4 +56p3q5 + 28p2q6 + 8pq7 + q8 = 1

43. LC question

44. Mendelian Inheritance • Normal Distributions and Chi-square tests • Remember, all of this is probabilistic. You will likely never get exactly what you expect. • How far away from the expected value can we get before suspecting that something funky is going on? • Remind you of anything?

45. Mendelian Inheritance • The normal distribution • A binomial distribution depicting all of the experimental outcomes • µ = the average outcome • σ = one standard deviation • A result is considered statistically significant if it falls more than ~2σ away from µ • It has a probability (p-value) of <0.05 • The p-value is the probability of a result deviating by at least that much by chance

46. Mendelian Inheritance • Chi-square (X2) analysis • A method to statistically evaluate the relationship between observed and expected values; are the results statistically significant? • Two steps to calculate X2 value • Calculate the difference between the observed (O) and expected (E), square it and then divide by E. • Sum for each outcome class

47. Mendelian Inheritance • Chi-square (X2) analysis • Dependent on three factors – sample size, number of outcome classes, number of observations in each class • Experiments with larger numbers of outcome classes or more observations will have larger X2 values. • Comparing experiments that have different numbers of outcome classes or different numbers of observations are therefore not comparable

48. Mendelian Inheritance • Chi-square (X2) analysis • Now, evaluate X2 to determine the p-value • In other words, find out where your X2 falls on a normal distribution for data like yours • You will need to know the degrees of freedom (df) for your experiment • each of a number of independently variable factors affecting the range of states in which a system may exist • df = number of outcome classes (n) – 1 • When flipping a coin, df = 1. There are two possible values (H, T) but they are not independent of one another • Use a handy table (Table 2.4 in Genetic Analysis)

50. Mendelian Inheritance • Testing Mendel’s data • Round vs. wrinkled seeds • F1 cross – RR x rr all Rr, no test needed • F2 cross – Rr x Rr • Predictions • ¼ RR, ½ Rr, ¼ rr • 0.75 round, 0.25 wrinkled • If 7324 seeds, predict 5493 round and 1831 wrinkled • Observations • 7324 seeds counted • 5474 round (0.747), 1850 wrinkled (0.253)