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Genetics
Genetics
Genetics
Genetics
Genetics
Genetics
Genetics
Genetics
Genetics
Genetics
Genetics
Genetics
Genetics
Genetics
Genetics
Genetics
Genetics
Genetics
Genetics
Genetics
Genetics
Genetics
Genetics
Genetics
Genetics
Genetics
Genetics
Genetics
Genetics
Genetics
Genetics
Genetics
Genetics
Genetics
Genetics
Genetics
Genetics
Genetics
Genetics
Genetics
Genetics
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Genetics

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2nd year

2nd year

Published in: Education, Lifestyle, Technology
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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%

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