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Black Lab Breeding: Predicting Offspring Colors

Discover the possible coat colors of puppies from a mating between a black lab female and a black lab male. Learn about the inheritance of traits and the principles of genetics.

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Black Lab Breeding: Predicting Offspring Colors

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  1. A Black Lab Female mates with a black lab Male: • All Black puppies • 5 Black and 3 Yellow puppies • All Yellow Puppies • 5 Yellow and 3 Black • What about Gender of the offspring?

  2. Your are worried that you might become bald (the ring in the top of your head). Which of the following would be the best indicators? • Your Mom • Your Dad • Your Maternal Uncle • Your Paternal Uncle • Materal Grandpa • Paternal Grandpa

  3. Genetics • Gregor Mendell -- Founder (1850’s) • Austrian Monk • Pea Plants • Pure strains (true breeding plants) • Developed hybrids

  4. Mendel’s Conclusions • Multiple alleles for the same trait • Organism inherits two alleles for each trait (one from mom / one dad) • Dominant allele is fully expressed, recessive allele has no noticeable effect • Two alleles of a trait segregate independently of each other

  5. Chromosomes Allele for white flower Allele for Purple flower Specific gene locus Homologous Chromosomes

  6. Mendel’s Laws • Law of segregation (alleles): Alleles separate with equal probability of being passed on to the offspring- the alleles separate so each gamete produced has one randomly • Law of independent assortment (Homologous chromosomes): Each chromosome in a pair has equal probability of being found in a gamete (passed on to offspring)-meiosis

  7. Terms • Dominant:allele that expresses itself at the expense of an alternate allele • Recessive:allele not expressed (covered up) by dominant allele • Homozygous/Purebred: Both alleles are the SAME • Heterozygous/Hybrid: Allele pair is Different • Genotype: Actual allele combination, written as letters • Phenotype: Outward characteristics, written as words • Linkage groups: Two different traits (genes) that are found (and usually inherited) together

  8. Two Round parent pea plants (P1) • Sexually reproduce and create seeds that can be grown the next year (F1 generation). • These offspring (F1 generation) can sexually reproduce and produce seeds which will become the F2 generation.

  9. Genotype Rules: • Must use the SAME letter for a specific trait. Flower color: Use letter P and p Seed Color: Y and y • Dominant Trait always has the Capital Letter Purple Dominant = P White Recessive = p Shape Round dominant = R Wrinkled is recessive = r

  10. Determine the genotypes of the following: Homozygous Tall: _______ Heterozygous Tall: _______ Hybrid Green: _______ Yellow: ________ Purebred Purple:

  11. Tt Tt Tt Tt Punnett Squares • Determine probability of offspring having a trait • Mendel’s cross: Pure Tall * Pure short Genotype of Tall = _____ Short = _____ T T All Tt Genotype: Phenotype: t All Tall t

  12. R r r r • Practice: Round is dominant to wrinkled. A heterozygous Round is crossed with a wrinkled plant. What is the possible genotypes and phenotypes? Rr and rr Parents: ______________ Rr rr Genotype: Phenotype: 1 Rr : 1 rr Rr 1 Round : 1 Wrinkled rr

  13. G g G g • Practice: Green is dominant to yellow. Two green parents have a yellow offspring. What is the genotype of the parents and what is the chance of having another yellow offspring? Gg and Gg Parents: GG Gg Genotype: Phenotype: 1GG: 2Gg: 1gg Gg 3 Green : 1 Yellow 25% chance of another yellow gg

  14. Having eye color is dominant to Blue. A blue eyed mom has a blue eyed child with a father that has brown eyes. PhenotypeGenotype Dad = Mom = Punnet Square =

  15. Practice: A breeder of Black Labrador retrievers wants to know if a potential sire has any possibility of passing on an allele for yellow, which is a single gene recessive trait. The Sire is black, how could the breeder determine the genotype of his male sire? What would be a positive result of the male being homozygous? Perform a test cross by breeding the dog to a yellow female. If one puppy was yellow, dad is guaranteed to be heterozygous. The size of the litter would play a role in being sure if all puppies are black. If only two puppies are born and both are black, this would not guarantee dad is homozygous. If 8 puppies are whelped and all are black, this would be a higher degree of certainty. Show the Punnett Square of this cross

  16. B ____ b Bb * b b Bb * b

  17. Test Cross • Determine the genotype of an individual by crossing with the recessive • If ANY offspring is recessive, the individual in question is heterozygous • Used in purebred breeding (dogs), Now do DNA analysis • Multiple offspring must be produced (Why?) Ex. Dog with normal hip is crossed with female who has a bad hip (recessive test) and one puppy has bad hips. Show the cross. Would you breed this Male to other females? h H Genotype: Phenotype: 1 Hh: 1 hh 1 normal : 1 bad Hips NO h hh Hh h

  18. Dihybrid Cross • Two traits / assume independent assortment (two traits are on different chromosomes) • Must FOIL genotypes • Ex: TtPp has four different gametes: TP, Tp, tP, tp • Determine the gametes for the following: (Tall, Round, and Purple are dominant) Heterozygous tall and wrinkled Short and heterozygous Round Purebred Round and white Ttrr = Tr, Tr, tr, tr so Tr, tr ttRr = tR, tr, tR, tr so tR , tr RRpp = Rp, Rp, Rp, Rp so Rp

  19. Determine the Gametes • Hybrid Tall hybrid Round seed? Tt Rr TR Tr tR tr bJ and bj bbJj

  20. TTPP TTPp TTPp TTpp TtPP TtPp TtPp Ttpp • Practice: A purebred Tall heterozygous purple plant is crossed with a heterozygous tall heterozygous purple. • Gametes plant #1: • Gametes plant #2: TTPp = TP and Tp TtPp = TP , Tp, tP, tp Tp TP Genotype: TP 1 TTPP : 2 TTPp :1 TTpp: 1 TtPP : 2 TtPp : 1 Ttpp Tp Phenotype: tP 3 Tall Purple : 1 Tall White tp

  21. XXHh XXhh XyHh Xyhh • Practice: A Male with hitchiker’s thumb (Dominant) marries a woman with a straight thumb. They have two children, one with a straight thumb, one with hitchiker’s. What is the chance of having a girl with hitchiker’s for their third child? Parents = XyHh and XXhh gametes XH, Xh, yH, yh Xh XH Xh yH yh Genotype: Xh 1XXHh: 1XXhh: 1XyHh: 1Xyhh Phenotype: 1 H.H male: 1 HHfemale: 1 straight male: 1 straight Female 25%

  22. Pedigrees • Trace the family history of a specific condition • Males shown as squares, females circles • “Affected” individuals shaded in

  23. Determine the Genotypes of Each Individual Is the condition Dominant, Recessive, or Sex Linked (recessive)?

  24. What type of inheritance is Hemophilia? • Draw a punnett square to show why Leopold’s son’s were not affected.

  25. Sex-linkage • Trait found on X chromosome • Autosomal refers to anything not on X or y chromosome • Humans: XX = female Xy = male • y chromosome small / few genes (TDF) • Birds: XX = male Xy = female • Bees: Fertilized eggs = female Unfertilized = male Humans: Males afflicted with recessive condition more often (look at grandfather on mom’s side) since there is only ONE “X” Chromosome Color blindness, male pattern baldness, hemophelia

  26. Karyotype: 23 Pairs of Homologous Chromosomes. 24 chromosomes are shown since the sex chromosomes “X” and “y” are NOT identical in their gene sequences. http://k-12.pisd.edu/currInst/science/Genetic/genetic-home.htm

  27. Practice: A woman with a colorblind father marries a man who sees color normally. What is the % chance of having a colorblind child? Colorblind Daughter? A son being colorblind? (Colorblindness is recessive) X y Colorblind Child = 25% X Colorblind Daughter = 0% Xc Colorblind Son = 50%

  28. Male pattern baldness is found on the X chromosome. On a Jerry Springer show, a bald female talked about life without hair. Which of the following would be possible genetic reasons for her baldness? • Her Dad was bald and her mom was bald • Her Dad was normal, but her mom was bald • Both her dad and mom were normal, but mom was a carrier • Her Dad was a carrier and mom was normal • Her Dad was bald and her maternal grandfather was also bald *** Draw a Punnet square for one of the above cases that you circled as being possible

  29. Incomplete Dominance • “Blending” of two alleles • Heterozygous condition is intermediate (in-between) the two possible conditions • Purebred Blue winged butterfly * Purebred white winged butterfly. F1 = all light blue Genotypes: Phenotypes:

  30. Skin Color: Incomplete Dominance

  31. Codominance • Both possible alleles are Dominant • Both traits are expressed in some form of “spotted” pattern • Blood types in humans: Three possible alleles: IA, IB, Ii (IA, IB are codominant, Ii is recessive) • genes both code for glycoproteins on the surface of a blood cell (ID markers) • Ii gene codes for no glycoprotein (O is lack of ID) • Antigen / antibody: Antigen is anything foreign introduced to the body that will trigger the production of antibodies (IA would be an antigen for Type B blood)

  32. Addition of incompatible blood type will result in blood “clumping” which can clot vessels, cause death

  33. Blood Types Cont. Bill has type AB blood and Maria has type O blood. What are the potential blood types of their offspring? If Maria has a child with type O blood, could it be Bill’s? What is Bill’s genotype: What is Maria’s genotype: Genotypes: Phenotypes:

  34. Practical Applications of Genetics • Answering questions about the past • Solving Crimes • DNA Fingerprinting

  35. Protecting crops • Spider Toxins • Bananas

  36. Genes and Probability • Determine the genotypes by looking at offspring ratio • Larger the population, the more accurate the results Ex: A group of seeds all from the same parents were planted with the following results: 98 were short and yellow, 103 were short and green, 297 were tall and green, 303 were tall and yellow Assume each trait follows law of independent assortment (found on different chromosomes), look at each trait separately Parents genotypes:

  37. DNA Fingerprinting • DNA is “cut” by restriction enzymes • Produces unique band pattern • Different restriction enzymes produce different banding patterns Gel Electrophoresis http://gslc.genetics.utah.edu/units/biotech/gel/

  38. Bami1 G G A T C C C C T A G G ECORI C T T A AG G A A T T C ATACCAATGGATCCACATGGGATATTA TATGGTTACCTAGGTGTACCCTATAAT CTTAAGATTGGATCCATTACTTAAGGAT GAATTCTAACCTAGGTAATGAATTCCTA

  39. How Many Fragments

  40. Use of DNA fingerprint to determine Suspects Which Suspect is guilty?

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