Fft1033 3 genetics inheritance-2013
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1⁄2 Aa 1⁄2 Bb 1⁄4 bb 1⁄4 CC 1⁄2 Cc 1⁄4 cc 1⁄4 CC 1⁄2 Cc 1⁄4 cc Phenotypes: 9/16 ABC 3/16 ABc 3/16 aBC 1/16 abc Gregor Mendel conducted experiments on pea plants to study inheritance, observing traits such as plant height, seed color, and seed texture. Through his work, he discovered that traits are controlled by discrete factors (now known as genes) which are inherited based on his two principles: 1) dominance,
Fft1033 3 genetics inheritance-2013
- 2. Introduction Genetics is the branch of biology that studies heredity Genetics is the branch of biology that studies the storage, duplication, and transfer of information Organisms inherit characteristics from their parents The information for these characteristics is contained in an organism’s DNA.
- 3. The History of Modern Genetics Modern genetics began early in the 20th century with the pioneering work of Gregor Mendel Mendel worked on the garden pea (Pisum sativum), and he established that certain traits could be passed from generation to generation Gregor Mendel Prior to Mendel, heredity was regarded as a "blending" process and the offspring were essentially a "dilution" of the different parental characteristics.
- 5. Gregor Mendel’s Peas Mendel studied a number of characteristics in pea plants: Plant height: short or tall Seed color: green or yellow Seed shape: wrinkled or round Flower color: white or purple Pod shape: constricted or smooth Pod color: yellow or green Flower position: terminal or axial
- 6. Gregor Mendel’s Peas 7 True-breeding phenotypes in pea
- 8. Gregor Mendel’s Peas 1. Self-fertilization 2. Cross-pollination
- 9. Mendel’s Monohybrid Crosses Mendel crossed tall and dwarf pea plants to investigate how height was inherited He emasculated the male plant then applied pollen the stigma of other variety The hybrids were uniformly tall The same result for the reciprocal crosses The dwarf seemed to disappear in the progeny Mendel allowed them to undergo selffertilization among 1064 progeny: 787 tall and 277 dwarf ratio of approximately 3:1
- 14. Monohybrid Crosses Yielded Consistent Results
- 15. Mendel’s Monohybrid Crosses The hybrids had the ability to produce dwarf progeny The latent (masked) factor: recessive; the expressed factor: dominant The cross was called a monohybrid cross Each trait was controlled by a heritable factor (genes) The dominant and recessive forms are called alleles Each of the parental strains carried two identical copies of a gene homozygous the hybrid two different alleles heterozygous.
- 16. Mendel’s Monohybrid Crosses Mendel used symbols to represent the hereditary factors The allele for dwarfness, (recessive): d The allele for tallness (dominant): D The tall and dwarf pea strains: DD and dd The allelic constitution: genotype the physical appearance (tall or dwarf): phenotype As the parental strains, the tall and dwarf pea plants form the P generation of the experiment.
- 17. Mendel’s Monohybrid Crosses The hybrid progeny first filial generation (F1) The genotype F1: Dd (heterozygous) The F1 phenotype: tall because D is dominant over d During meiosis, these F1 produce two kinds of gametes, D and d, in equal proportions They separate, or segregate, from each other during gamete formation Upon self-fertilization, the two kinds of gametes produced by heterozygotes can unite in all possible ways.
- 18. Mendel’s Monohybrid Crosses Thus, they produce four kinds of zygotes: DD, Dd, dD, and dd Because of dominance, three of these genotypes have the same phenotype Thus, in the next generation, called the F2, the plants are either tall or dwarf, in a ratio of 3:1.
- 19. Genetics terms you need to know: Gene: a unit of heredity; a section of DNA sequence encoding a single protein Genome: the entire set of genes in an organism Alleles: two genes that occupy the same position on homologous chromosomes and that cover the same trait (Different forms of a specific gene) (A or a) Locus: a fixed location on a strand of DNA where a gene or one of its alleles is located.
- 20. Genetics terms you need to know: Homozygous: having identical alleles (one from each parent) for a particular (AA or aa) characteristic Heterozygous: having two different alleles (Aa) for a particular characteristic Dominant: the allele of a gene that masks or suppresses the expression of an alternate allele; the trait appears in the heterozygous (A) condition Recessive: an allele that is masked by a (a) dominant allele; does not appear in the heterozygous condition, only in homozygous.
- 21. Genetics terms you need to know: Genotype: the genetic makeup (AA, Aa or aa) of an organisms Phenotype: the physical appearance of an organism Monohybrid cross: a genetic cross involving a single pair of genes (one trait); parents differ by a single trait P: Parental generation F1: First filial generation; offspring from a genetic cross F2: Second filial generation of a genetic cross.
- 24. Mendel’s Monohybrid Crosses Mendel’s two principles of heredity: 1. The Principle of Dominance: In a heterozygote, one allele may conceal the presence of another 2. The Principle of Segregation: In a heterozygote, two different alleles segregate from each other during the formation of gametes.
- 25. Example 1 In rabbits the allele for black coat color (B) is dominant over the allele for brown coat color (b). What is the genotypic ratio and phenotypic ratio be for a cross between an animal homozygous for black coat color and one homozygous for brown coat color?
- 26. In rabbits the allele for black coat color (B) is dominant over the allele for brown coat color (b). What is the genotypic ratio and phenotypic ratio be for a cross between an animal homozygous for black coat color and one homozygous for brown coat color? Let B: black coat color; and b: brown coat P: homozygous for black X homozygous brown coat BB bb Gamete: B b F1 : All Bb Genotypic ratio: 100% Bb Phenotypic ration: 100% Black coat
- 27. Example 2 In humans, polydactyly (an extra finger on each hand or toe on each foot) is due to a dominant gene. When one parent is polydactylous, but heterozygous, and the other parent is normal, what are the genotypic and phenotypic ratios of their children?
- 28. In humans, polydactyly (an extra finger on each hand or toe on each foot) is due to a dominant gene. When one parent is polydactylous, but heterozygous, and the other parent is normal, what are the genotypic and phenotypic ratios of their children? D: allele for polydactyly; d: alelle for normal finger D is dominant over d P: heterozygous polydactyly Dd Gamete: D, d F1 : d D d X normal dd d Dd (polydactyly) dd (normal) Genotypic ratio: ½ Dd : ½ dd Phenotypic ratio: ½ polydactyly : ½ normal
- 29. Dihybrid Cross Mendel investigated the inheritance of seed shape (smooth vs wrinkled) and seed color (green vs yellow) at the same time From his monohybrid crosses he knew that smooth seeds were dominant to wrinkled seeds and yellow seeds were dominant to green seeds He chose to cross plants that were pure breeding for both dominant features (smooth and yellow seeds) with plants that were pure breeding for both recessive features (wrinkled and green seeds).
- 30. Dihybrid Cross
- 31. Dihybrid Cross Seed color: 2 alleles g (green) and G (yellow), Seed texture: w (wrinkled) and W (round) The parental strains, doubly homozygous yellow, round plants: GG WW green, wrinkled plants: gg ww Gametes from GG WW plants: GW gametes from gg ww plants: gw Cross-fertilization produces F1 hybrids (doubly heterozygous) GgWw Phenotype yellow, round indicates that the G and W alleles are dominant.
- 32. Dihybrid Cross P: round, yellow x wrinkled, green F1: 100% round, yellow seeds F1 intercross: round, yellow x round, yellow F2 offspring: Total 556 seeds from F2 plants Ratio 315 108 101 32 9 3 3 1 round and yellow seeds round and green seeds wrinkled and yellow seeds wrinkled and green seeds
- 33. Dihybrid Cross
- 34. Dihybrid Cross
- 35. Dihybrid Cross
- 36. Dihybrid Cross A dihybrid cross can be treated as two separate monohybrid crosses The expected probability of each type of seed can be calculated: Probability of an F2 seed being round = 75% or ¾ Probability of an F2 seed being wrinkled =25% or ¼ Probability of an F2 seed being yellow = 75% or ¾ Probability of an F2 seed being green = 25% or ¼
- 37. Dihybrid Cross Therefore: Probability of an F2 seed being round and yellow = ¾ x ¾ = 9/16 = 56.25% Probability of an F2 seed being round and green = ¾ x ¼ = 3/16 = 18.75% Probability of an F2 seed being wrinkled and yellow = ¼ x ¾ = 3/16 = 18.75% Probability of an F2 seed being wrinkled and green = ¼ x ¼ = 1/16 = 6.25%.
- 38. Dihybrid Cross Phenotype We could What Mendel expect observed 556 x 9/16 round yellow 313 315 556 x 3/16 round green 104 108 556 x 3/16 wrinkled yellow 104 101 556 x 1/16 wrinkled green 35 32
- 39. Mendel’s Principles 3. Principle of Independent Assortment: The alleles of different genes segregate or assort independently of each other Genes get shuffled – these many combinations are one of the advantages of sexual reproduction
- 40. Trihybrid crosses P: Gamete: F1: AABBCC x aabbcc ABC abc AaBbCc Gametes: ABC, ABc, AbC, Abc, aBC, aBc, abC, abc
- 41. Trihybrid Crosses Types of gametes produced by AaBbCc B A b B a b C c C c C c C c ABC ABc AbC Abc aBC aBc abC abc
- 42. Punnett Square for Trihybrid crosses ♀ ♂ ABC ABc AbC Abc aBC aBc abC abc ABC AABBCC AABBCc AABbCC AABbCc AaBBCC AaBBCc AaBbCC AaBbCc ABc AABBCc AABBcc AAbBCc AABbcc AaBBCc AaBBcc AbC AABbCC AABbCc AAbbCC AAbbCc AaBbCC AaBbCc AabbCC AabbCc Abc AABbCc AABbcc AAbbCc AAbbcc AaBbCc AaBbcc AabbCc Aabbcc aBC AaBBCC AaBBCc AaBbCC AaBbCc aaBBCC aaBBCc aaBbCC aaBbCc aBc AaBBCc AaBBcc AaBbCc AaBbcc aaBBCc aaBBcc aaBbCc aaBbcc abC AaBbCC AaBbCc AABBCC AabbCc aaBbCC aaBbCc aabbCC aabbCc abc AaBbCc aabbCc aabbcc AaBbcc AabbCc Aabbcc aaBbCc aaBbcc AaBbCc AaBbcc
- 43. Forked-line Method for Trihybrid Crosses Split the individual gene Make crosses Combine them back Genotypes: Aa x Aa ¼ AA : ½ Aa : ¼ aa Bb x Bb ¼ BB : ½ Bb : ¼ bb Cc x Cc ¼ CC : ½ Cc : ¼ cc Phenotypes: Aa x Aa ¾ A- : ¼ aa Bb x Bb ¾ B- : ¼ bb Cc x Cc ¾ C- : ¼ cc
- 44. Trihybrid Cross: AaBbCc X AaBbCc ¼ BB ¼ AA ½ Bb Genotypes: ¼ bb ¼ BB ½ Aa ½ Bb ¼ bb ¼ BB ¼ aa ½ Bb ¼ bb ¼ CC ½ Cc ¼ cc ¼ CC ½ Cc ¼ cc ¼ CC ½ Cc ¼ cc ¼ CC ½ Cc ¼ cc ¼ CC ½ Cc ¼ cc ¼ CC ½ Cc ¼ cc ¼ CC ½ Cc ¼ cc ¼ CC ½ Cc ¼ cc ¼ CC ½ Cc ¼ cc ¼ x ¼ x ¼ = 1/64 AABBCC ¼ x ¼ x ½ = 2/64 AABBCc ¼ x ¼ x ¼ = 1/64 AABBcc ¼ x ½ x ¼ = 2/64 AABbCC ¼ x ½ x ½ = 4/64 AABbCc ¼ x ½ x ¼ = 2/64 AABbcc ¼ x ¼ x ¼ = 1/64 AAbbCC ¼ x ¼ x ½ = 2/64 AAbbCc ¼ x ¼ x ¼ = 1/64 AAbbcc ¼ x ¼ x ¼ = 1/64 AABBCC ¼ x ¼ x ½ = 2/64 AABBCc ¼ x ¼ x ¼ = 1/64 AABBcc ¼ x ½ x ¼ = 2/64 AABbCC ¼ x ½ x ½ = 4/64 AABbCc ¼ x ½ x ¼ = 2/64 AABbcc ¼ x ¼ x ¼ = 1/64 AAbbCC ¼ x ¼ x ½ = 2/64 AAbbCc ¼ x ¼ x ¼ = 1/64 AAbbcc ¼ x ¼ x ¼ = 1/64 aaBBCC ¼ x ¼ x ½ = 2/64 aaBBCc ¼ x ¼ x ¼ = 1/64 aaBBcc ¼ x ½ x ¼ = 2/64 aaBbCC ¼ x ½ x ½ = 4/64 aaBbCc ¼ x ½ x ¼ = 2/64 aaBbcc ¼ x ¼ x ¼ = 1/64 aabbCC ¼ x ¼ x ½ = 2/64 aabbCc ¼ x ¼ x ¼ = 1/64 aabbcc
- 45. Trihybrid Cross: AaBbCc X AaBbCc Phenotype ¾ B¾ A¼ bb ¾ B¼ aa ¼ bb (¾)(¾)(¾)A-B-C- 27/64 A-B-C- ¼ cc (¾)(¾)(¼)A-B-cc 9/64 A-B-cc ¾C- (¾)(¼)(¾)A-bbC- 9/64 A-bbC- ¼ cc (¾)(¼)(¼) A-bbcc ¾C- (¼)(¾)(¾) aaB-C- 9/64 A-bbcc ¾C- 3/64 aaB-C3/64 aaB-cc ¾C- (¼)(¾)(¼) aaB-cc (¼)(¼)(¾) aabbC- 3/64 aabbC- ¼ cc (¼)(¼)(¼) aabbcc 1/64 aabbcc ¼ cc Phenotypic ratio: 27:9:9:9:3:3:3:1
- 46. Monohybrid Test Cross How can you determine genotype from individual expressing dominant phenotype? DD or Dd? Cross individual with dominant phenotype to a homozygous recessive individual.
- 47. Dihybrid test cross In monohybrid crosses, to know if a dominant trait is homozygous (DD) or heterozygous (Dd) it is necessary to carry out a test cross This is done with a homozygous recessive (dd) individual The same is true for a dihybrid cross where the test cross is made with an individual which is homozygous recessive for both characters (wwgg)
- 48. Dihybrid testcross Testcross results of four smooth round individuals WwGG WWGG WWGG xsx wwgg wg WG WwGg WwGG x wwgg wg WG WwGg wG wwGg WWGg WWGg x wwgg wg WG WwGg Wg Wwgg WwGg WwGg x wwgg wg WG WwGg Wg Wwgg wG wwGg wg wwgg Phenotypic ratio All smooth yellow All smooth yellow Phenotypic ratio ½ smooth yellow ½ wrinkle yellow All yellow Phenotypic ratio ½ smooth yellow ½ smooth green All smooth Phenotypic ratio ¼ smooth yellow ¼ smooth green ¼ wrinkle yellow ¼ wrinkle green Mixed
- 49. Monohybrid and Dihybrid Problems 1. White (W) hair in sheep is caused by the 2. dominant gene while black (w) hair is recessive. A heterozygous white male and a black female are parents of a black lamb. What is the probability that their next lamb will be white? What are the genotypic and phenotypic ratios? Albinism is recessive in humans. An albino man marries a woman who is not albino, but had an albino father. What is the probability of this couple having a child that is not an albino? What are the genotypic and phenotypic ratios?
- 50. Monohybrid and Dihybrid Problems Todd and Melissa are college students who are planning to get married. They are currently taking a genetics course and decided to determine the eye color of any possible children they might have. Blue eyes are recessive to brown eyes. Todd has brown eyes, like his three brothers. His mother and grandmother have blue eyes, but his father and all other grandparents have brown eyes. Brown eyed Melissa has one blue eyed sister and one brown eyed sister and a mother with blue eyes. Her father and all of her grandparents have brown eyes. Construct an accurate punnett square to determine the possible eye colors of their yet to be born children. What are the genotypic and phenotypic ratios? In Teenage Mutant Ninja Turtles, green shells are dominant over brown shells. Leonardo, who is heterozygous for a green shell, marries the lovely Mona Lisa, who has a brown shell. What are the genotypic and phenotypic ratios?
- 51. 5. Monohybrid and Dihybrid Problems In guinea pigs, rough coat (R) is dominant over smooth coat (r). A rough coated guinea pig is bred to a smooth one, giving eight rough and seven smooth progeny in the F1. a) What are the genotypes of the parents and their offspring? b) If one of the rough F1 animals is mated to its rough parent, what progeny would you expect? 6. In summer squash, white fruit (W) is dominant over yellow (w), and disk-shaped fruit (D) is dominant over sphere-shaped fruit (d). The following problems give the phenotype of the parents and their offspring. Determine the genotypes of the parents in each case: a) White, disk x yellow, sphere gives 1/2 white, disk and 1/2 white, sphere. b) White, sphere x white, sphere gives 3/4 white, sphere and 1/4 yellow, c) sphere. Yellow, disk x white, sphere gives all white, disk progeny.
- 52. Monohybrid and Dihybrid Problems 7. In human, aniridia, (a type of blindness resulting from absence of an iris) is due to a dominant gene. Migraine (a sickening headache) is due to a different dominant gene. A man with aniridia, whose mother was not blind, marries a woman who suffers from migraine. The woman’s father did not suffer from migraine. In what proportion of their children would both aniridia and migraine be expected to occur? 8. In watermelons, solid green color (G) is dominant over striped pattern (g), and short shape (S) is dominant over long shape (s). What is the probability of each phenotype of possible offspring if a heterozygous solid, long watermelon cross pollinates with a heterozygous solid, heterozygous short watermelon?
- 53. 9. Having two eyebrows is dominant (E) over having one large eyebrow (e). Also having six fingers (F) is dominant over having five fingers (f). What is the probability of each phenotype if a man that has one eyebrow and twelve fingers total (heterozygous), has children with a woman that is heterozygous for both traits?