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Dihybrid Crosses and Other Patterns of Inheritance

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Presentation on theme: "Dihybrid Crosses and Other Patterns of Inheritance"— Presentation transcript:

1 Dihybrid Crosses and Other Patterns of Inheritance

2 Dihybrid Cross a mating of two organisms that are different in two observable traits must use a two-gene Punnett Square to solve (16 total boxes)

3 Dihybrid Cross Two Trait Punnett Squares

4 One Trait Punnett Squares and the Law of Segregation
Punnett Squares show two alleles from one parent splitting up demonstrates Mendel’s Law of Segregation homologous chromosomes that carry the same genes split during Meiosis 1 allows offspring created to have one allele for each gene from each parent

5 Studying Heredity Holt Biology, pg. 170

6 Punnett Square

7 Two Traits and Mendel’s Laws
when creating possible gametes in dihybrid cross Punnett Squares, each allele may combine with dissimilar gene alleles but not with similar gene alleles

8 Dihybrid Cross Two Trait Punnett Squares

9 Two Traits and Mendel’s Laws

10 Two Traits and Mendel’s Laws
when creating possible gametes in dihybrid cross Punnett Squares, each allele may combine with dissimilar gene alleles but not with similar gene alleles “R” may combine with “Y” or “y” but not with “r”

11 Two Traits and Mendel’s Laws

12 Two Traits and Mendel’s Laws
when creating possible gametes in dihybrid cross Punnett Squares, each allele may combine with dissimilar gene alleles but not with similar gene alleles “R” may combine with “Y” or “y” but not with “r” demonstrates Mendel’s Laws of Segregation and Independent Assortment

13 Dihybrid Cross Example
A plant that is heterozygous for seed shape (Rr) and heterozygous for seed color (Yy) is self-pollinated. What will the genotypic and phenotypic ratios be of the offspring?

14 Allele Identification:
Dihybrid Cross Example A plant that is heterozygous for seed shape (Rr) and heterozygous for seed color (Yy) is self-pollinated. What will the genotypic and phenotypic ratios be of the offspring? Allele Identification: Seed Shape: R = round, r = wrinkled Seed Color: Y = yellow, y = green

15 Parent Genotype Identification:
Dihybrid Cross Example A plant that is heterozygous for seed shape (Rr) and heterozygous for seed color (Yy) is self-pollinated. What will the genotypic and phenotypic ratios be of the offspring? Parent Genotype Identification: Parent #1: RrYy Parent #2: RrYy

16 How many different gametes can a RrYy plant make?
4 RY Ry rY ry

17 What law explains why 2 Rs cannot be in the same gamete?
Law of Segregation

18 What law explains why each “R” allele could be in a gamete with each Y allele?
Law of Independent Assortment

19 Dihybrid Example Punnett Square

20 Dihybrid Example Punnett Square
Parent RrYy

21 Dihybrid Example Punnett Square
Parent RrYy RY Ry rY ry

22 Dihybrid Example Punnett Square
Parent RrYy RY Ry rY ry RRYY RRYy RrYY RRyy Rryy rrYY rrYy rryy

23 Dihybrid Cross Example
Phenotypic Ratio:

24 Begin by listing all of the possible phenotype combinations.

25 Dihybrid Cross Example
Phenotypic Ratio: Round & Yellow: Round & Green Wrinkled & Yellow Wrinkled & Green

26 Then count them in the Punnett Square.

27 Dihybrid Example Punnett Square
Parent RrYy RY Ry rY ry RRYY RRYy RrYY RRyy Rryy rrYY rrYy rryy

28 Dihybrid Cross Example
Phenotypic Ratio: Round & Yellow: 9 Round & Green: Wrinkled & Yellow: Wrinkled & Green:

29 Dihybrid Cross Example
Phenotypic Ratio: Round & Yellow: 9 Round & Green: 3 Wrinkled & Yellow: Wrinkled & Green:

30 Dihybrid Cross Example
Phenotypic Ratio: Round & Yellow: 9 Round & Green: 3 Wrinkled & Yellow: 3 Wrinkled & Green:

31 Dihybrid Cross Example
Phenotypic Ratio: Round & Yellow: 9 Round & Green: 3 Wrinkled & Yellow: 3 Wrinkled & Green: 1

32 Polygenic Inheritance
several genes influencing one trait must use separate letters for separate genes but both genes influence the same trait Example in humans: height, weight, hair color, and skin color

33 We won’t use a Punnett Square with this type of inheritance.

34 Polygenic Inheritance in Labs
two genes determine the fur coat color of a Labrador Retriever. B gene – determines skin color BB & Bb – Black skin bb – Brown skin

35 Polygenic Inheritance in Labs
two genes determine the fur coat color of a Labrador Retriever. E gene – determines expression of pigment in fur EE & Ee – Fur color will be identical to skin pigment color (black or brown) ee – Fur color will be yellow BB & Bb – Black skin bb – Brown skin

36 Polygenic Inheritance in Labs

37 Polygenic Inheritance
Twins!

38 Polygenic Inheritance in Horses
“B” gene for hair color brown coat color (B) is dominant over tan (b) “C” gene for pigment production in hair pigment production (C) is dominant gene (C) over the absence of pigment (c) a homozygous recessive for the second gene (cc), it will have a white coat regardless of the genetically programmed coat color (B gene) because pigment is not deposited in the hair

39 Polygenic Inheritance in Horses

40 Environmental influences
conditions around organisms also influence their phenotype

41 Environmental influences

42 Environmental influences

43 Environmental influences
conditions around organisms also influence their phenotype Examples in humans: height and weight are influenced by nutrition, skin and hair color are influenced by exposure to UV radiation

44 Allele Identification:
In rabbits, a black coat is due to a dominant gene (B), and a white coat to its recessive allele (b). Short hair length (H) is due to a dominant gene, and long hair to its recessive allele (h). In a cross between a heterozygous black, homozygous longhaired rabbit and a homozygous white, heterozygous shorthaired rabbit, what will the phenotypes be? Allele Identification: B = black fur b = white fur H = short hair h = long hair

45 Parent Genotype Identification:
In rabbits, a black coat is due to a dominant gene (B), and a white coat to its recessive allele (b). Short hair length (H) is due to a dominant gene, and long hair to its recessive allele (h). In a cross between a heterozygous black, homozygous longhaired rabbit and a homozygous white, heterozygous shorthaired rabbit, what will the phenotypes be? Parent Genotype Identification: Black, long haired parent Bbhh White, short haired parent bbHh

46 Dihybrid Practice Problem #1
Parents Bbhh & bbHh Bh bh bH BbHh bbHh Bbhh bbhh

47 Offspring Phenotypic Ration:
In rabbits, a black coat is due to a dominant gene (B), and a white coat to its recessive allele (b). Short hair length (H) is due to a dominant gene, and long hair to its recessive allele (h). In a cross between a heterozygous black, homozygous longhaired rabbit and a homozygous white, heterozygous shorthaired rabbit, what will the phenotypes be? Offspring Phenotypic Ration: Black short hair: 4 Black long hair: 4 White short hair: 4 White long hair: 4

48 Offspring Phenotypic Ration:
In rabbits, a black coat is due to a dominant gene (B), and a white coat to its recessive allele (b). Short hair length (H) is due to a dominant gene, and long hair to its recessive allele (h). In a cross between a heterozygous black, homozygous longhaired rabbit and a homozygous white, heterozygous shorthaired rabbit, what will the phenotypes be? Offspring Phenotypic Ration: Black short hair: 1 Black long hair: 1 White short hair: 1 White long hair: 1

49 Allele Identification:
In tomatoes, red fruit color (R) is dominant to yellow (r). Round shaped fruit (F) is dominant to pear shaped fruit (f). A gardener crosses a heterozygous red fruit, homozygous round fruit plant with a yellow fruit, heterozygous round fruit plant. Record the phenotypic ratios of the offspring. Allele Identification: R = red fruit r = yellow fruit F = round fruit f = pear shaped fruit

50 Parent Genotype Identification:
In tomatoes, red fruit color (R) is dominant to yellow (r). Round shaped fruit (F) is dominant to pear shaped fruit (f). A gardener crosses a heterozygous red fruit, homozygous round fruit plant with a yellow fruit, heterozygous round fruit plant. Record the phenotypic ratios of the offspring. Parent Genotype Identification: Red Round Parent RrFF Yellow Round Parent rrFf

51 Dihybrid Practice Problem #2
Parents RrFF & rrFf RF rF RrFF rrFF rf RrFf rrFf

52 Offspring Phenotypic Ration:
In tomatoes, red fruit color (R) is dominant to yellow (r). Round shaped fruit (F) is dominant to pear shaped fruit (f). A gardener crosses a heterozygous red fruit, homozygous round fruit plant with a yellow fruit, heterozygous round fruit plant. Record the phenotypic ratios of the offspring. Offspring Phenotypic Ration: Red round: Red pear: Yellow round: Yellow pear:

53 Offspring Phenotypic Ration:
In tomatoes, red fruit color (R) is dominant to yellow (r). Round shaped fruit (F) is dominant to pear shaped fruit (f). A gardener crosses a heterozygous red fruit, homozygous round fruit plant with a yellow fruit, heterozygous round fruit plant. Record the phenotypic ratios of the offspring. Offspring Phenotypic Ration: Red round: 8 Red pear: Yellow round: 8 Yellow pear:

54 Offspring Phenotypic Ration:
In tomatoes, red fruit color (R) is dominant to yellow (r). Round shaped fruit (F) is dominant to pear shaped fruit (f). A gardener crosses a heterozygous red fruit, homozygous round fruit plant with a yellow fruit, heterozygous round fruit plant. Record the phenotypic ratios of the offspring. Offspring Phenotypic Ration: Red round: 1 Yellow round: 1


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