1 / 41

GENETICS

GENETICS. Genetics. The study of heredity . Gregor Mendel (1860’s) discovered the fundamental principles of genetics by breeding garden peas. Genetics. Alleles 1. Alternative forms of genes. 2. Units that determine heritable traits.

olesia
Download Presentation

GENETICS

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. GENETICS

  2. Genetics • The study of heredity. • Gregor Mendel (1860’s) discovered the fundamental principles of genetics by breedinggarden peas.

  3. Genetics • Alleles 1. Alternative forms of genes. 2. Units that determine heritable traits. 3. Dominant alleles (TT - tall pea plants) a. homozygous dominant 4. Recessive alleles (tt - dwarf pea plants) a. homozygous recessive 5. Heterozygous (Tt - tall pea 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.

  7. T T t t Breed the P generation • tall (TT) vs. dwarf (tt) pea plants

  8. T T produces the F1 generation Tt Tt t Tt Tt t All Tt = tall (heterozygous tall) tall (TT) vs. dwarf (tt) pea plants

  9. T t T t Breed the F1 generation • tall (Tt) vs. tall (Tt) pea plants

  10. T t produces the F2 generation Tt TT T 1/4 (25%) = TT 1/2 (50%) = Tt 1/4 (25%) = tt Tt tt t 1:2:1 genotype 3:1 phenotype tall (Tt) vs. tall (Tt) pea plants

  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. eye color locus B = brown eyes eye color locus b = blue eyes Paternal Maternal Homologous Chromosomes This person would have brown eyes (Bb)

  13. B sperm B B Bb haploid (n) b b diploid (2n) b meiosis II meiosis I Meiosis - eye color

  14. B b male gametes B Bb x Bb b female gametes Monohybrid Cross • Example: Cross between two heterozygotesfor brown eyes (Bb) BB = brown eyes Bb = brown eyes bb = blue eyes

  15. B b 1/4 = BB - brown eyed 1/2 = Bb - brown eyed 1/4 = bb - blue eyed BB Bb B Bb x Bb b Bb bb 1:2:1 genotype 3:1 phenotype Monohybrid Cross

  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 ry 2. AaBbCCDd: 2n = 23= 8 gametes ABCD ABCd AbCD AbCd aBCD aBCd abCD abCD 3. MmNnOoPPQQRrssTtQq: 2n = 26= 64 gametes

  19. RY Ry rY ry x RY Ry rY ry possible gametes produced Dihybrid Cross • Example: cross between round and yellow heterozygous pea seeds. R = round r = wrinkled Y = yellow y = green RrYy x RrYy

  20. RY Ry rY ry RY Ry rY ry Dihybrid Cross

  21. RY Ry rY ry Round/Yellow: 9 Round/green: 3 wrinkled/Yellow: 3 wrinkled/green: 1 RY RRYY RRYy RrYY RrYy Ry RRYy RRyy RrYy Rryy rY RrYY RrYy rrYY rrYy ry RrYy Rryy rrYy rryy 9:3:3:1 phenotypic ratio Dihybrid Cross

  22. bC b___ bc Test Cross • A mating between an individual ofunknown genotypeand a homozygous recessive individual. • Example:bbC__ x bbcc BB = brown eyes Bb = brown eyes bb = blue eyes CC = curly hair Cc = curly hair cc = straight hair

  23. bC b___ C bC b___ c bc bbCc bbCc or bc bbCc bbcc Test Cross • Possible results:

  24. R R r r 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) RR = red flower rr = white flower

  25. R R produces the F1 generation Rr Rr r r Rr Rr All Rr = pink (heterozygous pink) Incomplete Dominance

  26. Codominance • Two alleles are expressed (multiple alleles) in heterozygous individuals. • Example: blood 1. type A = IAIA or IAi 2. type B = IBIB or IBi 3. type AB = IAIB 4. type O = ii

  27. IB IB IAIB IAIB IA 1/2 = IAIB 1/2 = IBi i IBi IBi Codominance • Example: homozygous male B (IBIB) x heterozygous female A (IAi)

  28. IA IB IAi IBi i 1/2 = IAi 1/2 = IBi i IAi IBi Codominance • Example:male O (ii) x female AB (IAIB)

  29. Codominance • 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. IA i IAIB IB i ii Codominance • Answer: Parents: genotypes = IAi and IBi phenotypes = A and B

  31. Sex-linked Traits • Traits (genes) located on the sex chromosomes • Example:fruit flies (red-eyed male) X (white-eyed female)

  32. fruit fly eye color XX chromosome - female Xy chromosome - male Sex-linked Traits Sex Chromosomes

  33. XR y Xr Xr 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 Rr = red eyed rr = white eyed Xy = male XX = female

  34. XR y XR Xr Xr y Xr Xr XR Xr Xr y Sex-linked Traits 1/2 red eyed and female 1/2 white eyed and male

  35. Population Genetics • The study of genetic changes in populations. • The science of microevolutionary changes in 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 homozygousdominant.

  40. q2 = .04 2. then use 1 = p + q 3. for heterozygous use 2pq 4. For homozygous dominant use p2 Answer: 1. q2 = 4% or .04 q = .2 1 = p + .2 1 - .2 = p .8 = p 2(.8)(.2) = .32 or 32% .82 = .64 or 64%

  41. Hardy-Wienberg equation 1 = p2 + 2pq + q2 • 64% = p2 = homozygous dominant • 32% = 2pq = heterozygous • 04% = q2 = homozygous recessive • 100%

More Related