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2nd year

2nd year

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  • 1. GENETICS
  • 2. Genetics• The study of heredity. heredity• Gregor Mendel (1860’s) discovered the fundamental principles of genetics by breeding garden peas. peas
  • 3. Genetics• Alleles 1. Alternative forms of genes. 2. Units that determine heritable traits. 3. Dominant alleles (TT - tall pea plants) plants a. homozygous dominant 4. Recessive alleles (tt - dwarf pea plants) plants a. homozygous recessive 5. Heterozygous (Tt - tall pea plants) plants
  • 4. Phenotype• Outward appearance• Physical characteristics• Examples: 1. tall pea plant 2. dwarf pea plant
  • 5. Genotype• Arrangement of genes that produces the phenotype• Example: 1. tall pea plant TT = tall (homozygous dominant) 2. dwarf pea plant tt = dwarf (homozygous recessive) 3. tall pea plant Tt = tall (heterozygous)
  • 6. Punnett square• A Punnett square is used to show the possible combinations of gametes. gametes
  • 7. Breed the P generation• tall (TT) vs. dwarf (tt) pea plants T T t t
  • 8. tall (TT) vs. dwarf (tt) pea plants T T Tt produces thet Tt F1 generationt Tt Tt All Tt = tall (heterozygous tall)
  • 9. Breed the F1 generation• tall (Tt) vs. tall (Tt) pea plants T t T t
  • 10. tall (Tt) vs. tall (Tt) pea plants T t produces the TT Tt F2 generation T 1/4 (25%) = TT Tt tt 1/2 (50%) = Tt t 1/4 (25%) = tt 1:2:1 genotype 3:1 phenotype
  • 11. Monohybrid Cross• A breeding experiment that tracks the inheritance of a single trait.• Mendel’s “principle of segregation” a. pairs of genes separate during gamete formation (meiosis). b. the fusion of gametes at fertilization pairs genes once again.
  • 12. Homologous Chromosomeseye color locus eye color locusB = brown eyes b = blue eyes This person would have brown eyes (Bb) Paternal Maternal
  • 13. Meiosis - eye color B B sperm B Bb haploid (n) bdiploid (2n) b b meiosis I meiosis II
  • 14. Monohybrid Cross• Example: Example Cross between two heterozygotes for brown eyes (Bb)BB = brown eyes B b maleBb = brown eyes gametesbb = blue eyes B Bb x Bb b female gametes
  • 15. Monohybrid Cross B b 1/4 = BB - brown eyed B BB Bb 1/2 = Bb - brown eyedBb x Bb 1/4 = bb - blue eyed b Bb bb 1:2:1 genotype 3:1 phenotype
  • 16. Dihybrid Cross• A breeding experiment that tracks the inheritance of two traits.• Mendel’s “principle of independent assortment” a. each pair of alleles segregates independently during gamete formation (metaphase I) b. formula: 2n (n = # of heterozygotes)
  • 17. Independent Assortment• Question: How many gametes will be produced for the following allele arrangements?• Remember: 2n (n = # of heterozygotes) 1. RrYy 2. AaBbCCDd 3. MmNnOoPPQQRrssTtQq
  • 18. Answer:1. RrYy: 2n = 22 = 4 gametes RY Ry rY ry2. AaBbCCDd: 2n = 23 = 8 gametes ABCD ABCd AbCD AbCd aBCD aBCd abCD abCD3. MmNnOoPPQQRrssTtQq: 2n = 26 = 64 gametes
  • 19. Dihybrid Cross• Example: cross between round and yellow heterozygous pea seeds. R = round RrYy x RrYy r = wrinkled Y = yellow RY Ry rY ry x RY Ry rY ry y = green possible gametes produced
  • 20. Dihybrid Cross RY Ry rY ryRYRyrYry
  • 21. Dihybrid Cross RY Ry rY ry Round/Yellow: 9RY RRYY RRYy RrYY RrYy Round/green: 3Ry RRYy RRyy RrYy Rryy wrinkled/Yellow: 3rY RrYY RrYy rrYY rrYy wrinkled/green: 1ry RrYy Rryy rrYy rryy 9:3:3:1 phenotypic ratio
  • 22. Test Cross• A mating between an individual of unknown genotype and a homozygous recessive individual.• Example: bbC__ x bbcc BB = brown eyes Bb = brown eyes bC b___ bb = blue eyes bc CC = curly hair Cc = curly hair cc = straight hair
  • 23. Test Cross • Possible results: bC b___ C bC b___ cbc bbCc bbCc or bc bbCc bbcc
  • 24. Incomplete Dominance• F1 hybrids have an appearance somewhat in between the phenotypes of the two parental varieties.• Example: snapdragons (flower)• red (RR) x white (rr) R R RR = red flower r rr = white flower r
  • 25. Incomplete Dominance R R produces ther Rr Rr F1 generationr Rr Rr All Rr = pink (heterozygous pink)
  • 26. Codominance• Two alleles are expressed (multiple alleles) alleles in heterozygous individuals. individuals• Example: blood 1. type A = IAIA or IAi 2. type B = IBIB or IBi 3. type AB = IAIB 4. type O = ii
  • 27. Codominance• Example: homozygous male B (IBIB) x heterozygous female A (IAi) IB IB IA I AI B I AI B 1/2 = IAIB 1/2 = IBi i I Bi IBi
  • 28. Codominance• Example: male O (ii) x female AB (IAIB) IA IB i I Ai IBi 1/2 = IAi 1/2 = IBi i I Ai IBi
  • 29. Codominance• Question: Question If a boy has a blood type O and his sister has blood type AB, what are the genotypes and phenotypes of their parents.• boy - type O (ii) X girl - type AB (IAIB)
  • 30. Codominance• Answer: IA iI B I AI B Parents: genotypes = IAi and IBi phenotypes = A and Bi ii
  • 31. Sex-linked Traits• Traits (genes) located on the sex chromosomes• Example: fruit flies (red-eyed male) X (white-eyed female) red white
  • 32. Sex-linked Traits Sex Chromosomes fruit fly eye colorXX chromosome - female Xy chromosome - male
  • 33. Sex-linked Traits• Example: fruit flies (red-eyed male) X (white-eyed female)• Remember: the Y chromosome in males does not carry traits.RR = red eyed XR yRr = red eyedrr = white eyed XrXy = male XrXX = female
  • 34. Sex-linked Traits XR yXr XR Xr Xr y 1/2 red eyed and female 1/2 white eyed and maleXr XR Xr Xr y
  • 35. Population Genetics• The study of genetic changes in populations. populations• The science of microevolutionary changes in populations. populations• Hardy-Weinberg equilibrium: the principle that shuffling of genes that occurs during sexual reproduction, by itself, cannot change the overall genetic makeup of a population.• Hardy-Wienberg equation: 1 = p2 + 2pq + q2
  • 36. Question:• How do we get this equation?Answer: “Square” 1 = p + q ↓ 12 = (p + q)2 ↓ 1 = p2 + 2pq + q2
  • 37. Hardy-Wienberg equation• Five conditions are required for Hardy-Wienberg equilibrium. 1. large population 2. isolated population 3. no net mutations 4. random mating 5. no natural selection
  • 38. Important• Need to remember the following: p2 = homozygous dominant 2pq = heterozygous q2 = homozygous recessive
  • 39. Question:• Iguanas with webbed feet (recessive trait) make up 4% of the population. What in the population is heterozygous and homozygous dominant. dominant
  • 40. Answer:1. q2 = 4% or .04 q2 = .04 q = .22. then use 1 = p + q 1 = p + .2 1 - .2 = p .8 = p3. for heterozygous use 2pq 2(.8)(.2) = .32 or 32%4. For homozygous dominant use p2 .82 = .64 or 64%
  • 41. Hardy-Wienberg equation 1 = p2 + 2pq + q2• 64% = p2 = homozygous dominant• 32% = 2pq = heterozygous• 04% = q2 = homozygous recessive• 100%