1 / 99

EVOLUTION: Molecular Genetics

Biology 11 Enriched. EVOLUTION: Molecular Genetics. Mendelian Genetics. Tuesday, December 4 th ,2 012. Introduction. Human have observed changes in population for many centuries.

tim
Télécharger la présentation

EVOLUTION: Molecular 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. Biology 11 Enriched EVOLUTION:Molecular Genetics

  2. Mendelian Genetics Tuesday, December 4th,2 012

  3. Introduction • Human have observed changes in population for many centuries. • Humans have domesticated animals and plants and have artificially selected for various traits in our livestock and crops. • Character – an observable physical feature • Trait – a particular form of a character

  4. Introduction • In the 19th Century, 2 theories on how traits are passed on from one generation to the next: • THEORY OF BLENDING INHERITANCE • THEORY OF PARTICULATE INHERITANCE

  5. 19th Century theories on inheritance BLENDING INHERITANCE PARTICULATE INHERITANCE That both hereditary determinants remain in fertilized zygote. Both parental traits will be exhibited in the offspring. • That “hereditary determinants” (genes) were found in gametes • Physical traits from both parents are blendedand the offspring will exhibit an intermediate trait between the two.

  6. Gregor Mendel • He studied biology, physics, and mathematics   • Developed the fundamental laws of heredity • Took some ideas from blended model and particulate model • Mendel chose garden peas (Pisumsativum) as his subjects as they are easily grown and their pollinationis easily controlled. • He controlled pollination by manually moving pollen between plants  Funfact: Mendel originally wanted to breed mice, but wasn't allowed to because it was considered scandalous

  7. Gregor Mendel • Pea flowers have both male and female sex organs so Mendel was able to developed “true-breeding” plants by self-pollination.

  8. Mendel examined varieties of peas for inheritable characters and traits for his study. (stem length, pod shape, seed shape, seed color, etc.)

  9. Gregor Mendel • In 1865, Mendel published his findings in a paper called Experiments on Plant Hybridization, which was mostly ignored at the time due to a number of reasons. • First, Mendel was not well known in scientific community. • Second, his theory ran against the popular model of blended inheritance.

  10. Mendel’s Three Laws of Inheritance Mendel’s Three Laws Law of Dominance Law of Segregation Law of Independent Assortment

  11. Monohybrid Cross • A monohybrid cross involves one (mono) character and different (hybrid) traits. • Mendel crossed a parental generation with 2 different traits of the same character (in this example, flower color) • The F1 seeds were all purple • The whiteflower trait failed to appear at all. • There was also no “blending” of colors

  12. Monohybrid Cross • Mendel then took purple flowers from the F1 generation and allowed them to self-fertilize. • The flowers in the F2 generation showed a 3:1 ratio of purple:white flowers

  13. Monohybrid Cross • The purple trait completely masks the white trait when true-breeding plants are crossed • the purpleflower trait is called dominant • the whiteflower trait is called recessive. • The Law of Dominance

  14. Monohybrid Cross • When Mendel repeated the procedure but for other characters and traits, he would observe a F1 generation with only 1 trait and the 3:1 ratio in the F2 generation.

  15. Mendel’s Conclusion / Law of Segregation The Law of Segregation • Mendel concluded that each gametemust hold 1 copy of a gene and the zygotewill hold 2 copies (1 from each parent) • Character = genes • traits = alleles of the gene. But why that 3:1 ratio?

  16. Why 3:1 Ratio? • An organism has 2 alleles of a gene (1 from each parent). • If both alleles are the same  homozygous • If alleles are different  heterozygous • Let’s assign the each allele a letter/symbol.P = purple p = white

  17. Why 3:1 Ratio? BY CONVENTION: • The dominant trait is given a capital letter, the lowercase of that same letter is the recessive trait.  DO NOT MIX LETTERS.  Pick one and stick to it. • Also, some letters are better than others.  Capital S looks a lot like a lowercase (s).  Pick a different letter... Okay                                     Better (use H for hair) Short  hair  = SS                                 HH Short hair = SsHh Long hair = sshh

  18. Why 3:1 Ratio? • Both parents are homozygous.

  19. Why 3:1 Ratio? • We can use a Punnett Square to show us the allele combinations All offspring in the F1 generation are heterozygous dominant

  20. Why 3:1 Ratio? • In his experiment, he then crossed the F1 generation to produce F2 generation • 3 out of 4 are purple • 1 is homozygous • 2 are heterozygous • 1 out 4 is white • Homozygous recessive

  21. Why 3:1 Ratio?

  22. Practice picking letters.... the following traits are found in the common Shirtuscanadianus. Polka dots are dominant to stripes. Long sleeves are dominant to short sleeves. Collared shirts are recessive. Buttons are dominant over snaps. Pockets are recessive.   

  23. Practice with Punnett Squares A  round seeded plant (RR) is crossed with a wrinkle seeded plant (rr).  What are the phenotypes of the offspring? Two heterozygous purple flowered pea plants are crossed.  What are the phenotypes of their offspring and in what proportion? A plant with green seeds (yy) is crossed with a heterozygous plant.  What percentage of their offspring have yellow seeds?

  24. In dragons...wings are a dominant trait, but some dragons are born wingless. What are the chances that two heterozygous dragons have a whelp that is wingless? If a wingless dragon is crossed with one that is heterozygous, how many of its offspring will also be wingless?

  25. What is a Test cross? Help, help!  I don't know what my genotype is!! Am I Dd or DD? I can help you!  Let's have offspring! D = winged d = wingless

  26. What is a Test Cross? Because we know wingless dragons are homozygous recessive, we can breed wingless dragons with a winged dragon. • By looking at the ratios, we can tell if the winged dragon is homozygous or heterozygous • If female dragon is PP then 100% heterozygous winged dragons

  27. What is a Test Cross? If female dragon is Pp then a 1:1 ratio is observed.

  28. Dihybrid Cross • Mendel's Law of Independent Assortment is illustrated by the dihybrid cross • The second law describes the outcome of dihybrid (two character) crosses, or hybrid crosses involving additional characters. • A dihybrid is an individual that is a double heterozygote (e.g., with the genotype RrYy - round seed, yellow seed).R = round/r = bumpy, Y = yellow/y = green • What are the gametes that can be produced by an individual that is RrYy? • RY, Ry, rY, ry

  29. Dihybrid Cross (RrYy x RrYy)

  30. Dihybrid Cross • The ratio that is seen is a 9:3:3:1 ratio • a total of 4 phenotypes are observed: • 9 round, yellow • 3 round, green • 3 bumpy, yellow • 1 bumpy, green (double recessive)

  31. Dihybrid Cross – Practice • You have 30 minutes to complete as much of this package as you can. • At 2:20, we are moving to notes.

  32. Probability and Inheritance • A Punnet Square for a dihybrid cross is pretty epic! • a Punnett Square is helpful for 1 or 2 genes but a little troublesome for more characters • A trihybridcross needs 64 boxes • A tetrahybridcross needs 256boxes • TOO MUCH EPIC!

  33. Probability and Inheritance • Mendel’s laws of segregation and independent assortment reflect the rules of probability

  34. Probability and Inheritance • Remember that the alleles of different (and unlinked) traits ending up in a gamete is independent of the chances of other alleles. • To find the probability of an series of INDEPENDENTevents happening together, the individual probabilities of the events are multiplied together: P(A and B) = P(A) x P(B)

  35. Probability and Inheritance • If an event is absolutely certain to happen, its probability is 1. • If it cannot possibly happen, its probability is 0. • All other events have a probability between 0 and 1.

  36. Probability and Inheritance An example of independent events is the flipping of a coin. Each flip of a coin is independent from the previous or next flips. They don’t influence each other.

  37. Probability and Inheritance • P(A and B) = P(A) x P(B) What is the probability of getting 5 “tails” in a row? If P(tails) = 0.5 (or ½) P(5 tails) = ½ x ½ x ½ x ½ x ½ = 1/32 or 0.03125 (never use %)

  38. Rr Rr  Segregation of alleles into sperm Segregation of alleles into eggs Sperm 1/2 1/2 R r R R R Heads ¼ + ¼ + ¼ = ¾ Tails1/4 1/2 r R 1/4 1/4 Eggs r r r R 1/2 r 1/4 1/4

  39. Probability and Inheritance • We used the rule of multiplication to find the probability of a certain genotype. • We used the rule of addition to find the probability of a certain phenotype. • In calculating the chances for various genotypes, each character is considered separately, and then the individual probabilities are multiplied together

  40. Calculate the probability that an F2 seed will be spherical and yellow. Remember, calculate each trait separately. (create 2 monohybrid squares rather than 1 dihybrid square)

  41. P(yellow,round) = ¾ yellow x ¾ round = 9/16 P(yellow, wrinkled) = ¾ yellow x ¼ wrinkled = 3/16 P(green, round) = ¼ green x ¾ round = 3/16 P(green,wrinkled) = ¼ green x ¼ wrinkled = 1/16 A 9:3:3:1 ratio!!

  42. For any gene with a dominant allele A and recessive allele a, what proportions of the offspring from an AA x Aacross are expected to be homozygous dominant, homozygous recessive, and heterozygous?

  43. Two organisms, with genotypes BbDDand BBDd are mated. Assuming independent assortment of the B/b and D/d genes, write the genotypes of all possible offspring from this cross and use the rules of probability to calculate the chance of each genotype occurring.

  44. Three characters (flower color, seed color, and pod shape) are considered in a cross between two pea plants (PpYyIi x ppYyii). What fraction of offspring are predicted to be homozygous recessive for at least two of the three characters?

  45. Three characters (flower color, seed color, and pod shape) are considered in a cross between two pea plants (PpYyIix ppYyii). What fraction of offspring are predicted to be homozygous recessive for at least two of the three characters? Pp = ¼ + ¼ = ½ pp = ¼ + ¼ = ½ YY = ¼Yy = ¼ + ¼ = ½ yy = ¼ Ii = ¼ + ¼ = ½ ii = ¼ + ¼ = ½

  46. Inheritance patterns are often more complex than predicted by simple Mendelian genetics

  47. Extension on Mendel • The relationship between genotype and phenotype is rarely as simple as in the pea plant characters Mendel studied • Many heritable characters are not determined by only 1 gene with 2 alleles. • However, the basic principles of segregationand independent assortment apply even to more complex patterns of inheritance

  48. Extending Mendelian Genetics for a Single Gene • Inheritance of characters by a single gene may deviatefrom simple Mendelian patterns in the following situations: • When alleles are not completely dominant or recessive • When a gene has more than two alleles • When a gene produces multiple phenotypes

  49. Degrees of Dominance • Complete dominanceoccurs when phenotypes of the heterozygote and dominant homozygote are identical • In incomplete dominance, the phenotype of F1 hybrids is somewhere betweenthe phenotypes of the two parental varieties • In codominance, two dominant alleles affect the phenotype in separate, distinguishable ways

More Related