Welcome to IB 201!

1. Welcome to IB 201! Genetics and Evolution

2. A phenotype ratio of 9:3:3:1 in the offspring of a mating between two individuals that are heterozygous for two traits occurs when: A. the genes reside on the same chromosome B. each gene contains two mutations C. the gene pairs assort independently during meiosis D. only recessive traits are scored E. none of the above IB 201: Review Question

3. Course Outline • Genetic Data Analysis: • Probability & Statistics • Deviations from Mendelism: • Epistasis; Unusual Modes of Inheritance • Chromosomal Inheritance: • Chromosomal Abnormalities; Sex Determination • Mapping: • Gene and Genome Mapping • Traits Affected by Genes & Environment • Quantitative Traits • Genes in Populations • Genetic Mechanisms of Evolution; Population Genetics of Disease and Disease resistance

4. Genetic Data Analysis II Some simple rules of probability

5. Sum Rule • The combined probability of two events that are mutually exclusive is the sum of the individual probabilities. Clue: look for “or” Q: What’s the probability of rolling a ‘five’ or a ‘six’ on one six-sided die? A: 1/6 + 1/6 = 1/3

6. Genetic Example: Monohybrid Cross • P: GG x gg • F1: Gg • Gg x Gg ==> • F2: 1/4 GG: 1/2 Gg: 1/4 gg What is the probability that the F2 offspring has the dominant phenotype (is either GG or Gg)? 1/4 GG + 1/2 Gg = 3/4 G-

7. Genetic Example 2: Dihybrid Cross • P: GG ww x gg WW • F1: Gg Ww • Gg Ww x Gg Ww ==> • F2: 9/16 G-W- 3/16 G-ww 3/16 ggW- 1/16 ggww What is the probability that an F2 offspring will have the dominant phenotype (G-ww or ggW-) for only one of the two traits? 3/16 G-ww + 3/16 ggW- = 6/16=3/8

8. Product Rule • The probability of several independent events is the product of the individual probabilities. • Two events are independent if the occurrence of the first event has no effect on the probability of the second event. Clue: look for “and”. • Q: You roll two dice. What’s the probability of getting a ‘two’ on the first one and a ‘five’ on the second one. • A: 1/6 * 1/6 = 1/36

9. Genetic example of product rule • P: AA bb CC DD ee ff x aa BB cc dd EE FF • F1: Aa Bb Cc Dd Ee Ff x Aa Bb Cc Dd Ee Ff • Q: What proportion of F2 progeny will be • AA bb Cc DD ee Ff ? A: 1/4 * 1/4 * 1/2 * 1/4 * 1/4 * 1/2 = 1/1024

10. Deviations from Mendelism • Lethal Alleles • Epistasis • Unusual sex linkage • Sex influenced inheritance • Genetic Anticipation

11. Manx Cats

12. Lethal alleles F1: Mm x Mm F1: 1 MM 2 Mm 1 mm F2: 1 Lethal: 2 Manx: 1 Normal F2 phenotypic ratio: 2:1 instead of 3:1

13. Other lethal mutations • Achondroplasia (humans) • Yellow body color (domestic mice) • Curly wings (Drosophila)

14. Epistasis • Genetic interaction between two (or more) loci. • One gene modifies the phenotypic effects of another gene.

15. Agouti: wild type

16. BB CC x bb cc P: agoutialbino Bb Cc F1: agouti Simple dominant phenotype? F2: 9/16 B- C- 3/16 bb C- 3/16 B- cc 1/16 bb cc albino agouti albino black F2 Phen. ratio: 9 agouti : 3 black : 4 albino novel phenotype

17. Epistasis Normal dihybrid ratio is altered from 9:3:3:1 to 9:3:4 C and B gene have an epistatic interaction

18. Epistasis Locus 1 Locus 2 BB Bb bb CC Cc cc agouti agouti blackno effect no effectalbino

19. Biochemical model • CC or Cc: tyrosinase is produced (involved in production of melanin) • BB or Bb: controls distribution of the pigment

20. Figure 10.18b Crosses between pure lines produce novel colors. Parental generation rrYY RRyy X Yellow Brown Codominance? F1 generation R-Y- Red Self-fertilization F2 generation R-Y- rrY- R-yy rryy Red Yellow Brown Green 9/16 3/16 3/16 1/16

21. Figure 10.18c Model to explain 9 : 3 : 3 : 1 pattern observed above: Two genes interact to produce pepper color. yellow ----------------> green -----------------> brown pepper yy? chlorophyll R? red pigment yellow -------------------> yellow------------------> orange Y? No chlorophyll R? red pigment Genotype Color Explanation of color R-Y- Red Red pigment + no chlorophyll rrY- Yellow Yellow pigment + no chlorophyll R-yy Brown Red pigment + chlorophyll rryy Green Yellow pigment + chlorophyll Gene 1 Gene 2 Y = Absence of green (no chlorophyll) R = Red y = Presence of green (+ chlorophyll) r = Yellow (-) = Y or y (-) = R or r

22. Practice Problem • In Labrador retrievers, coat color is controlled by two loci each with two alleles B,b and E,e respectively. When pure breeding Black labs with genotype BB EE are crossed with pure breeding yellow labs of genotype bb ee the resulting F1 offspring are black. F1 offspring are crossed (Bb Ee x Bb Ee). Puppies appear in the ratio: • 9/16 black; 3/16 chocolate;4/16=1/4 yellow. • What genotypes correspond to these three phenotypes? 9/16 B- E-3/16 B- ee3/16 bb E- 1/16 eebb B- E- B- ee bb E- and bb ee

23. Other kinds of epistasis 9/16 A-B- 3/16 A-bb 3/16 aaB- 1/16 aabb Hint: usually given numbers, not fractions 27 agouti; 12 albino; 9 black 28 agouti; 11 albino; 4 black

24. Practice Problem • In the summer squash (Cucurbita pepo) fruit shape is determined by two genes. Two different true-breeding spherical types were crossed. The F1's were all disk, and the F2's segregated 35 disk, 25 spherical and 4 long. Explain these results. What’s the first step? Notice novel phenotype: disk, long. What’s the next step? Notice there are three F2 phenotypes. What kind of inheritance will give three F2 phenotypes? Genetic Model? Incomplete, codominance Epistasis Expected F2 ratio? 1:2:1 Variation on 9:3:3:1

25. Practice Problem, cont. • In the summer squash (Cucurbita pepo) spherical fruit is recessive to disk, True-breeding spherical types from different geographic regions were crossed. The F1's were disk, and the F2's segregated 35 disk, 25 spherical and 4 long. Explain these results. Are the phenotypic ratios closer 1:2:1 or to a variant of 9:3:3:1 ? If phenotypic ratios closer to a variant of 9:3:3:1, then what variant is it? Total # of individuals = 35 + 25 + 4 = 64 64/16 = 4 9*4 = 36 6*4 = 24 1*4 = 4 Phenotypic ratio close to 9:6:1

26. Practice Problem, cont. • In the summer squash (Cucurbita pepo) spherical fruit is recessive to disk, True-breeding spherical types from different geographic regions were crossed. The F1's were disk, and the F2's segregated 35 disk, 25 spherical and 4 long. Explain these results. If phenotypic ratios are close to 9:6:1, then what are the genotypes associated with each phenotype? 35 disk 25 spherical 4 long 9/16 A- B- 3/16 A- bb + 3/16 aa B- 1/16 aa bb What were the genotypes of the original spherical parents? AA bb aaBB

27. Xa Xa XA XA XA XA Y Y Sex Linkage: mammals, flies Heterogametic Sex Diploid XAXa XAY Adults Gametes Male XAXA XAY Female XAXa XaY

29. W W ZB ZB ZB ZB Zb Zb Sex Linkage: birds, butterflies Homogametic Sex Heterogametic Sex Diploid ZBW ZBZb Adults Female Male Gametes Male ZBZb ZBZb Female ZBW ZbW

30. Y-linked inheritance

31. Hairy ears

32. Male pattern baldness: what kind of inheritance?

33. Sex influenced phenotype • Genotype Female Male • bb Bald Bald • bb’ Not bald Bald • b’b’ Not bald Not bald

34. Environment-dependent dependent expression of a genotype

35. Siamese or “Himalayan” Different allele of the C locus that causes albinism. Temperature sensitive.

36. Phenotypes are not always a direct reflection of genotypes • Temperature-sensitive alleles: Siamese color pattern • Nutritional effects: phenylketonuria • Genetic anticipation: several genetic diseases

37. Phenylketonuria • Nutritional defect: can’t metabolize phenylalanine. • Can lead to severe physical and mental disorders in children, but only if they consume phenylalanine. • Disease phenotype can be avoided by eliminating phenylalanine from the diet

38. Genetic Anticipation • Huntington disease • Fragile-X syndrome • Kennedy disease • Myotonic muscular dystrophy

39. Fragile X syndrome • Symptoms: delayed development & mental retardation. More severe in males than females • Caused by expansion of triplet repeat (CGG) in a gene on the long arm of the X chromosome • Named for breakage of X chromosome in cell preparations.

40. Normal range: 7-52 (average=30) “Pre-mutation”: 60-200 repeats Full Mutation: > 230-1000s. DNA becomes abnormally methylated, promoter is inactivated, and gene silenced. Pre-mutation is unstable: maternally-inherited premutation with >100 repeats almost always expands to a full mutation Fragile X

41. Most common kind of inherited mental retardation. Named for “fragile site” Due to expansion of 3-base pair repeat (CGG) in a gene near the tip of the long arm of X chromosome. Genetic Anticipation: Fragile X

42. Pre-mutation is unstable: maternally-inherited premutation with >100 repeats almost always expands to a full mutation Fragile X

43. Genetic Anticipation causes subsequent generations in a family to be more severely affected by a disease. It does this by increasing the number of triplet repeats in the fragile area of the X chromosome through the generations.

44. Autosomal dominant lethal (chromosome 4) Progressive neurological deterioration First symptoms appear after reproductive age One of 8 known neurodegenerative diseases caused by expansion of (CAG) repeats All show inverse correlation with age of onset and number of repeats. Huntington Disease

45. Autosomal dominant lethal (chromosome 4) Progressive neurological deterioration First symptoms appear after reproductive age One of 8 known neurodegenerative diseases caused by expansion of (CAG) repeats All show inverse correlation with age of onset and number of repeats. Huntington Disease

46. Autosomal dominant lethal (chromosome 4) Progressive neurological deterioration First symptoms appear after reproductive age One of 8 known neurodegenerative diseases caused by expansion of (CAG) repeats All show inverse correlation with age of onset and number of repeats. Huntington Disease

47. Which is the pedigree of autosomal dominant (like HD)