1. Mendelian Genetics
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Mendelelian Genetics

2. Responsible for the Laws governing Inheritance of Traits
Mendelian Genetics
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Gregor Mendel ( )
Responsible for the Laws governing Inheritance of Traits

3. Gregor Johann Mendel Austrian monk
Mendelian Genetics
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Gregor Johann Mendel
Austrian monk
Studied the inheritance of traits in pea plants
Developed the laws of inheritance
Mendel's work was not recognized until the turn of the 20th century

4. Mendelian Genetics
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Gregor Johann Mendel
Between 1856 and 1863, Mendel cultivated and tested some 28,000 pea plants
He found that the plants' offspring retained traits of the parents
Called the “Father of Genetics"

5. Mendel’s Laws of Inheritance

6. 1. Genes In Pairs: Genetic characters are controlled by genes that exist in pairs of alleles in individual organisms and are passed from parents to their offspring. When two organisms produce offspring, each parent gives the offspring one of the alleles from each pair.

7. 2. Dominance and Recessiveness: When two unlike alleles responsible for a single character are present in a single individual, one allele can mask the expression of another allele. That is, one allele is dominant to the other. The latter is said to be recessive.

8. 3. The Law of Segregation: During the formation of gametes, the paired alleles separate (segregate) randomly so that each gamete receives one allele or the other.

9. 4. The Law of Independent Assortment: During gamete formation, segregating pairs of alleles assort independently of each other. Example: genes on different chromosomes will segregate independently. Linked genes (close together on one chromosome) do not follow this law.

10. Rule of Dominance
The trait that is observed in the offspring is the dominant trait (uppercase)
The trait that disappears in the offspring is the recessive trait (lowercase)

11. Law of Segregation
The two alleles for a trait must separate when gametes are formed
A parent randomly passes only one allele for each trait to each offspring

12. Law of Independent Assortment
The genes for different traits are inherited independently of each other.

13. Site of Gregor Mendel’s experimental garden in the Czech Republic
Mendelian Genetics
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Site of Gregor Mendel’s experimental garden in the Czech Republic
Fig. 5.co

14. Particulate Inheritance
Mendelian Genetics
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Particulate Inheritance
Mendel stated that physical traits are inherited as “particles”
Mendel did not know that the “particles” were actually Chromosomes & DNA

15. Mendelian Genetics
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Genetic Terminology
Trait - any characteristic that can be passed from parent to offspring
Heredity - passing of traits from parent to offspring
Genetics - study of heredity

16. Types of Genetic Crosses
Mendelian Genetics
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Types of Genetic Crosses
Monohybrid cross - cross involving a single trait e.g. flower color
Dihybrid cross - cross involving two traits e.g. flower color & plant height

17. Designer “Genes” Alleles - two forms of a gene (dominant & recessive)
Mendelian Genetics
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Designer “Genes”
Alleles - two forms of a gene (dominant & recessive)
Dominant - stronger of two genes expressed in the hybrid; represented by a capital letter (R)
Recessive - gene that shows up less often in a cross; represented by a lowercase letter (r)

18. Mendelian Genetics
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More Terminology
Genotype - gene combination for a trait (e.g. RR, Rr, rr)
Phenotype - the physical feature resulting from a genotype (e.g. red, white)

19. Here are the steps used to solve a Monohybrid Mendelian Genetic Problem:
Figure out the genotypes of the parents (Pure Dom, Hybrid, Pure Rec) Use the Capital letter of the DOMINANT trait and then its lower case form to represent the recessive trait.
Step 2:
Figure out what kinds of gametes the parents can produce.
Step 3:
Set up a Punnett Square for your mating.
Step 4:
Fill in the babies inside the table by filling each square by going Down and to the Left.
Step 5:
Figure out the genotypic ratio for your predicted babies.
How Many of the offspring are--Pure Dom: Hybrid: Pure Rec
Step 6:
Figure out the phenotypic ratio for your predicted babies.
How Many of the offspring are-- Dom : Rec
Step 7:
Answer the question you've been asked.

20. Solving Genetics Problems I: Monohybrid Crosses
Classical genetics is a science of logic and statistics. While many find the latter intimidating, the mathematical side of most classical genetics puzzles is relatively simple — and there are actually ways to get around most of the math. The logic part is inescapable. All genetics problems are solved using the same basic logic structure. If you learn the sense of the approach, you can solve virtually any genetics problem, provided you are given enough basic information.

21. Sample Problem using Steps:
This problem involves two gerbils named Honey and Ritz. The gene in question is a fur color gene which has two alleles — dominant brown (B) and recessive black (b). It's a very good idea to write down the information you are given in a problem so that it will be easy for you to refer to it when necessary. So begin by writing something like this at the top of your work page:

23. Step One: Figure out the genotypes of the parents.
Each of our parent gerbils is heterozygous for this gene. So here is our mating:
Step One: Figure out the genotypes of the parents.

24. Step One: Figure out what kinds of gametes the parents can produce..
Once you've got that settled, you need to address the question of all of the possible kinds of babies they could produce. Before any parent makes babies, of course, that parent makes gametes. So in order to find what kinds of babies they can have, you must first determine what kinds of gametes they can produce. Since Honey is a heterozygote (and paying attention to Rule #1), she can produce two kinds of eggs: B eggs and b eggs.

25. Ritz is also a heterozygote, so he can produce two kinds of sperm: B sperm and b sperm. Something like this:

26. Step Three: Set up a Punnett Square for your mating
Now you need to determine all the possible ways that his sperm can combine with her eggs. There are several different techniques used for this operation. The most popular among students is the Punnett Square. Punnett Squares are probability tables — a way to do statistics while avoiding as much math as possible.

27. Step 4: fill in the Punnett Square, down and to the left
Setting up a Punnett Square is easy. You need to create a chart with one column for each of the female's egg types, and one row for each of the male's sperm types. For Honey and Ritz, your table would look like this, then fill in the babies genotype by going down and to the left: Bb BB Bb bb

28. Step Five: Figure out the genotypic ratio for your predicted babies.
So we have now figured out that, if Honey and Ritz have a lot of babies, we can predict that 1/4 of them should be BB, 1/2 of them (2/4) should be Bb, and 1/4 should be bb.

29. This conclusion is often expressed as a genotypic ratio:1BB:2Bb:1bb
This conclusion is often expressed as a genotypic ratio:1BB:2Bb:1bb. This means that we are predicting that, for every BB baby, they should have 2 Bb babies (twice as many), and one bb baby.

30. Step Six: Figure out the phenotypic ratio for your predicted babies.
To do this, you need to ask yourself one question: do any of these different genotypes produce the same phenotype? In other words, do any of these babies look alike? This is where dominance enters the picture. If B is completely dominant to b, all gerbils with at least one B will look pretty much alike, no matter whether their second allele is B or b. So BB and Bb have the same phenotype, and we can add them together. Thus, our phenotypic ratio is 3 Brown:1 Black. Or, there should be three times as many brown babies as black babies.

31. So the answer to our question is, 3/4 of the babies should be brown.

32. Step Seven: Answer the question you've been asked.
The mating scheme we've just worked through is called a monohybrid cross. This means that we were paying attention to only one gene (mono=1), and both of our parents were heterozygous for that gene (hybrid=heterozygous).

33. Punnett Square Used to help solve genetics problems Mendelian Genetics
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Punnett Square
Used to help solve genetics problems

34. Mendelian Genetics
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35. Genotype & Phenotype in Flowers
Mendelian Genetics
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Genotype & Phenotype in Flowers
Genotype of alleles: R = red flower r = yellow flower
All genes occur in pairs, so 2 alleles affect a characteristic
Possible combinations are:
Genotypes RR Rr rr
Phenotypes RED RED YELLOW

36. Mendelian Genetics
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Genotypes
Homozygous genotype - gene combination involving 2 dominant or 2 recessive genes (e.g. RR or rr); also called pure 
Heterozygous genotype - gene combination of one dominant & one recessive allele    (e.g. Rr); also called hybrid

37. Genes and Environment Determine Characteristics
Mendelian Genetics
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Genes and Environment Determine Characteristics

38. Mendel’s Pea Plant Experiments
Mendelian Genetics
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Mendel’s Pea Plant Experiments

39. Why peas, Pisum sativum? Can be grown in a small area
Mendelian Genetics
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Why peas, Pisum sativum?
Can be grown in a small area
Produce lots of offspring
Produce pure plants when allowed to self-pollinate several generations
Can be artificially cross-pollinated

40. Reproduction in Flowering Plants
Mendelian Genetics
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Reproduction in Flowering Plants
Pollen contains sperm
Produced by the stamen
Ovary contains eggs
Found inside the flower
Pollen carries sperm to the eggs for fertilization
Self-fertilization can occur in the same flower
Cross-fertilization can occur between flowers

41. Mendel’s Experimental Methods
Mendelian Genetics
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Mendel’s Experimental Methods
Mendel hand-pollinated flowers using a paintbrush
He could snip the stamens to prevent self-pollination
Covered each flower with a cloth bag
He traced traits through the several generations

42. Mendelian Genetics
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How Mendel Began
Mendel produced pure strains by allowing the plants to self-pollinate for several generations

43. Eight Pea Plant Traits Seed shape --- Round (R) or Wrinkled (r)
Mendelian Genetics
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Eight Pea Plant Traits
Seed shape --- Round (R) or Wrinkled (r)
Seed Color ---- Yellow (Y) or  Green (y)
Pod Shape --- Smooth (S) or wrinkled (s)
Pod Color ---  Green (G) or Yellow (g)
Seed Coat Color ---Gray (G) or White (g)
Flower position---Axial (A) or Terminal (a)
Plant Height --- Tall (T) or Short (t)
Flower color --- Purple (P) or white (p)

44. Mendelian Genetics
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45. Mendelian Genetics
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46. Mendel’s Experimental Results
Mendelian Genetics
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Mendel’s Experimental Results

47. Did the observed ratio match the theoretical ratio?
Mendelian Genetics
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Did the observed ratio match the theoretical ratio?
The theoretical or expected ratio of plants producing round or wrinkled seeds is 3 round :1 wrinkled
Mendel’s observed ratio was 2.96:1
The discrepancy is due to statistical error
The larger the sample the more nearly the results approximate to the theoretical ratio

48. Mendelian Genetics
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Generation “Gap”
Parental P1 Generation = the parental generation in a breeding experiment.
F1 generation = the first-generation offspring in a breeding experiment. (1st filial generation)
From breeding individuals from the P1 generation
F2 generation = the second-generation offspring in a breeding experiment. (2nd filial generation)
From breeding individuals from the F1 generation

49. Following the Generations
Mendelian Genetics
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Following the Generations
Cross 2 Pure Plants TT x tt
Results in all Hybrids Tt
Cross 2 Hybrids get 3 Tall & 1 Short TT, Tt, tt

50. Mendelian Genetics
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Monohybrid Crosses

51. P1 Monohybrid Cross r r Rr Rr R R Rr Rr Trait: Seed Shape
Mendelian Genetics
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P1 Monohybrid Cross
Trait: Seed Shape
Alleles: R – Round r – Wrinkled
Cross: Round seeds x Wrinkled seeds: P1 = RR x rr
Genotype: Rr
Phenotype: Round
Genotypic Ratio: All alike
Phenotypic Ratio: All alike r r Rr Rr R R Rr Rr

52. P1 Monohybrid Cross Review
Mendelian Genetics
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P1 Monohybrid Cross Review
Homozygous dominant x Homozygous recessive
Offspring all Heterozygous (hybrids)
Offspring called F1 generation
Genotypic & Phenotypic ratio is ALL ALIKE

53. F1 Monohybrid Cross R r RR Rr R r Rr rr Trait: Seed Shape
Mendelian Genetics
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F1 Monohybrid Cross
Trait: Seed Shape
Alleles: R – Round r – Wrinkled
Cross: Round seeds x Round seeds: P1 = Rr x Rr R r
Genotype: RR, Rr, rr
Phenotype: Round & wrinkled
G.Ratio: 1:2:1
P.Ratio: 3:1 RR Rr R r Rr rr

54. F1 Monohybrid Cross Review
Mendelian Genetics
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F1 Monohybrid Cross Review
Heterozygous x heterozygous
Offspring: 25% Homozygous dominant RR 50% Heterozygous Rr 25% Homozygous Recessive rr
Offspring called F2 generation
Genotypic ratio is 1:2:1
Phenotypic Ratio is 3:1

55. What Do the Peas Look Like?
Mendelian Genetics
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What Do the Peas Look Like?

56. Mendelian Genetics
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…And Now the Test Cross
Mendel then crossed a pure & a hybrid from his F2 generation
This is known as an F2 or test cross
There are two possible testcrosses: Homozygous dominant x Hybrid Homozygous recessive x Hybrid

57. F2 Monohybrid Cross (1st)
Mendelian Genetics
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F2 Monohybrid Cross (1st)
Trait: Seed Shape
Alleles: R – Round r – Wrinkled
Cross: Pure Round seeds x Hybrid Round seeds
P =RR x Rr
Genotype: RR, Rr
Phenotype: Round
Genotypic Ratio: 1:1
Phenotypic Ratio: All alike R r RR Rr R R RR Rr

58. F2 Monohybrid Cross (2nd)
Mendelian Genetics
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F2 Monohybrid Cross (2nd)
Trait: Seed Shape
Alleles: R – Round r – Wrinkled
Cross: Wrinkled seeds x Hybrid Round seeds = rr x Rr R r
Genotype: Rr, rr
Phenotype: Round & Wrinkled
G. Ratio: 1:1
P.Ratio: 1:1 Rr rr r r Rr rr

59. F2 Monohybrid Cross Review
Mendelian Genetics
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F2 Monohybrid Cross Review
Homozygous x heterozygous(hybrid)
Offspring: 50% Homozygous RR or rr 50% Heterozygous Rr
Phenotypic Ratio is 1:1
Called Test Cross because the offspring have SAME genotype as parents

60. Mendelian Genetics
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Mendel’s Laws

61. Results of Monohybrid Crosses
Mendelian Genetics
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Results of Monohybrid Crosses
Inheritable factors or genes are responsible for all heritable characteristics
Phenotype is based on Genotype
Each trait is based on two genes, one from the mother and the other from the father
True-breeding individuals are homozygous ( both alleles) are the same

62. Mendelian Genetics
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Law of Dominance
In a cross of parents that are pure for contrasting traits, only one form of the trait will appear in the next generation.
All the offspring will be heterozygous and express only the dominant trait.
RR x rr yields all Rr (round seeds)

63. Mendelian Genetics
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Law of Dominance

64. Mendelian Genetics
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Law of Segregation
During the formation of gametes (eggs or sperm), the two alleles responsible for a trait separate from each other.
Alleles for a trait are then "recombined" at fertilization, producing the genotype for the traits of the offspring.

65. Applying the Law of Segregation
Mendelian Genetics
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Applying the Law of Segregation

66. Law of Independent Assortment
Mendelian Genetics
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Law of Independent Assortment
Alleles for different traits are distributed to sex cells (& offspring) independently of one another.
This law can be illustrated using dihybrid crosses.

67. Sex-linked Traits Traits (genes) located on the sex chromosomes
Mendelian Genetics
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Sex-linked Traits
Traits (genes) located on the sex chromosomes
Sex chromosomes are X and Y
XX genotype for females
XY genotype for males
Many sex-linked traits carried on X chromosome

68. Sex-linked Traits Example: Eye color in fruit flies Sex Chromosomes
Mendelian Genetics
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Sex-linked Traits
Example: Eye color in fruit flies
Sex Chromosomes
XX chromosome - female
Xy chromosome - male
fruit fly
eye color

69. Sex-linked Trait Problem
Mendelian Genetics
Sex-linked Trait Problem
4/15/2017
Example: Eye color in fruit flies
(red-eyed male) x (white-eyed female) XRY x XrXr
Remember: the Y chromosome in males does not carry traits.
RR = red eyed
Rr = red eyed
rr = white eyed
XY = male
XX = female XR Xr Y

70. Sex-linked Trait Solution:
Mendelian Genetics
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Sex-linked Trait Solution: XR Xr Y
50% red eyed female
50% white eyed male
XR Xr
Xr Y

71. Mendelian Genetics
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Female Carriers

72. Incomplete Dominance and Codominance
Mendelian Genetics
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Incomplete Dominance and Codominance

73. Mendelian Genetics
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Incomplete Dominance
F1 hybrids have an appearance somewhat in between the phenotypes of the two parental varieties.
Example: snapdragons (flower)
red (RR) x white (WW)
RR = red flower
WW = white flower W R

74. Incomplete Dominance W R W All RW = pink (heterozygous pink)
Mendelian Genetics
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Incomplete Dominance W R W
All RW = pink
(heterozygous pink)
produces the
F1 generation RW

75. Incomplete Dominance Problem:
In cattle when a red bull(RR) is mated with white(WW) cow the offspring are roan(RW) a blending of red and white. Mate a red bull with a roan cow. Show the P1, the Punnett Square, and give the genotypic and phenotypic ratios for this cross.

76. P1 = __RR__ x __RW__ Genotypic ratio: ____ : _____ : _____
Phenotypic ratio: ____ : _____ : _____

77. Mendelian Genetics
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Incomplete Dominance

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

79. Mendelian Genetics
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Codominance Problem
Example: homozygous male Type B (IBIB) x heterozygous female Type A (IAi) IB IA i
IAIB
IBi
1/2 = IAIB
1/2 = IBi

80. Another Codominance Problem
Mendelian Genetics
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Another Codominance Problem
Example: male Type O (ii) x female type AB (IAIB) i IA IB
IAi
IBi
1/2 = IAi
1/2 = IBi

81. Mendelian Genetics
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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)

82. Codominance Answer: IB IA i IAIB ii Parents: genotypes = IAi and IBi
Mendelian Genetics
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Codominance
Answer: IB IA i
IAIB ii
Parents:
genotypes = IAi and IBi
phenotypes = A and B

83. Mendelian Genetics
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Dihybrid Cross
A breeding experiment that tracks the inheritance of two traits.
Mendel’s “Law of Independent Assortment”
a. Each pair of alleles segregates independently during gamete formation
b. Formula: 2n (n = # of heterozygotes)

84. Mendelian Genetics
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Question: How many gametes will be produced for the following allele arrangements?
Remember: 2n (n = # of heterozygotes)
1. RrYy
2. AaBbCCDd
3. MmNnOoPPQQRrssTtQq

85. Answer: 1. RrYy: 2n = 22 = 4 gametes RY Ry rY ry
Mendelian Genetics
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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

86. All possible gamete combinations
Mendelian Genetics
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Dihybrid Cross
Traits: Seed shape & Seed color
Alleles: R round r wrinkled Y yellow y green
RrYy x RrYy
RY Ry rY ry
RY Ry rY ry
All possible gamete combinations

87. Mendelian Genetics
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Dihybrid Cross RY Ry rY ry RY Ry rY ry

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

89. Mendelian Genetics
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Dihybrid Cross
Round/Yellow: 9 Round/green: 3 wrinkled/Yellow: 3 wrinkled/green: 1
9:3:3:1

90. Mendelian Genetics
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Test Cross
A mating between an individual of unknown genotype and 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 bC
b___ bc

91. Test Cross Possible results: bC b___ bc bbCc C bC b___ bc bbCc bbcc or
Mendelian Genetics
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Test Cross
Possible results: bC
b___ bc
bbCc C bC
b___ bc
bbCc
bbcc or c

92. Summary of Mendel’s laws
Mendelian Genetics
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Summary of Mendel’s laws
LAW
PARENT CROSS
OFFSPRING
DOMINANCE
TT x tt tall x short
100% Tt tall
SEGREGATION
Tt x Tt tall x tall
75% tall 25% short
INDEPENDENT ASSORTMENT
RrGg x RrGg round & green x round & green
9/16 round seeds & green pods 3/16 round seeds & yellow pods 3/16 wrinkled seeds & green pods 1/16 wrinkled seeds & yellow pods

93. Genetic Practice Problems
Mendelian Genetics
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Genetic Practice Problems

94. Breed the P1 generation tall (TT) x dwarf (tt) pea plants t T
Mendelian Genetics
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Breed the P1 generation
tall (TT) x dwarf (tt) pea plants t T

95. tall (TT) vs. dwarf (tt) pea plants
Mendelian Genetics
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Solution:
tall (TT) vs. dwarf (tt) pea plants T t
All Tt = tall
(heterozygous tall)
produces the
F1 generation Tt

96. Breed the F1 generation tall (Tt) vs. tall (Tt) pea plants T t T t
Mendelian Genetics
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Breed the F1 generation
tall (Tt) vs. tall (Tt) pea plants T t T t

97. tall (Tt) x tall (Tt) pea plants
Mendelian Genetics
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Solution:
tall (Tt) x tall (Tt) pea plants T t
produces the
F2 generation
1/4 (25%) = TT
1/2 (50%) = Tt
1/4 (25%) = tt
1:2:1 genotype
3:1 phenotype TT Tt tt

98. Mendelian Genetics
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