1. Genetic Terminology
Gene - Segment of DNA that codes for formation of a protein
Locus – Position of gene on a chromosome
Trait - any characteristic that can be passed from parent to offspring
Heredity - passing of traits from parent to offspring
Genetics - study of heredity

2. Alleles
A gene that controls one function can exist in different forms. These different forms are called alleles.
Each different allele is identified by its specific phenotypic action.
Alleles are commonly represented by letters of the alphabet.
Eg. The gene LDLR controls blood cholesterol levels. Located on chromosome 19, it has two allelic forms:
B = abnormally high cholesterol levels
b = normal range

3. Autosomal Genotype
Remembering that non-sex chromosomes occur in homologous pairs in the diploid cell – there are two copies of each gene.
The double set of genetic instructions present makes up the genotype.
The number of possible genotypes depends on the number of allelic forms of the gene.

4. Genotype terminology
Homozygous (pure) genotype - gene combination involving 2 dominant or 2 recessive genes (e.g. RR or rr) 
Heterozygous (hybrid) genotype - gene combination of one dominant & one recessive allele    (e.g. Rr)

5. Phenotype
The visible expression of the genotype is called the phenotype. The expression may be a physical, biochemical or physiological trait.
Dominant trait: require only a single copy of the responsible allele for its phenotypic expression
Recessive trait: refers to a trait that is not expressed in a heterozygote
Co-dominant trait: both alleles in the heterozygote are expressed in the phenotype

6. e.g. 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

7. Genes and Environment Determine Characteristics

8. The relationship between genotype and phenotype is rarely simple!
Phenotype = Genotype + Environmental Factors
Hydrangeas: pink or blue?
Both plants have the pigment for colour called anthocyanin.
In acidic soils (low pH) the flowers are blue.
In alkaline soils (high pH) the flowers are pink).
Same genotype – different phenotype

9. Multiple Alleles or B AB O Allele Combination Phenotype or A
For some genes, three or more alleles can be present in the population.
You will only inherit two alleles (one on each chromosome)
The combination of any two alleles determines the final phenotype.
The ABO blood groupings in humans is an example.
Three alleles are involved in controlling blood group IA, IB and i. AA Ai BB Bi AB ii

10. Monogenic Traits
Monogenic traits are due to the action of a single gene with two or more allelic forms.
These traits show discontinuous variation - the members of the population can be grouped into a few discrete and non-overlapping classes.
E.g. blood types

11. Polygenic Traits
Polygenic traits are due to the actions of many genes (and their allelic forms). These traits show continuous variation (e.g. height).

12. Human Sex Chromosomes Traits (genes) located on the sex chromosomes
XX genotype for females
XY genotype for males
The X chromosome may carry up to 1500 genes. Genes located on the X chromosome are said to be X-linked.
Females have two alleles of a particular gene whereas males have only one (hemizygous genotype). This accounts for why many X-linked diseases show up more frequently in males than in females.
The Y chromosome has less than 300 genes. Genes located on the Y chromosome are said to be Y-linked. Males are also hemizygous for Y-linked genes.

13. X Inactivation in Female Mammals
Females who have two X chromosomes but one is subject to inactivation.
75% of alleles on one X chromosome are switched off in early embryonic development.
15% remain activated with another 10% altering their activation state in different females and in different cells within the same female.
See page 315 for case study

14. Mendel’s Pea Plant Experiments

15. 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
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"

16. Site of Gregor Mendel’s experimental garden in the Czech Republic
Mendelian Genetics
9/21/2018
Site of Gregor Mendel’s experimental garden in the Czech Republic
Fig. 5.co

17. Particulate Inheritance
Mendel stated that physical traits are inherited as “particles”
Mendel did not know that the “particles” were actually chromosomes & DNA

18. Reproduction in Flowering Plants
Mendelian Genetics
9/21/2018
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

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

22. Mendel’s Experimental Results

23. 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 the results approximate to the theoretical ratio

24. 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

25. 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

26. Monohybrid Cross
A trait determined by one gene with two or more allelic forms.

27. Punnett Square
Used to help solve genetic problems

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

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

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

32. 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

33. …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

34. F2 Monohybrid Cross (1st)
Trait: Seed Shape
Alleles: R – Round r – Wrinkled
Cross: Round seeds x Round seeds
RR x Rr
Genotype: RR, Rr
Phenotype: Round
Genotypic Ratio: 1:1
Phenotypic Ratio: All alike R r RR Rr R R RR Rr

35. F2 Monohybrid Cross (2nd)
Trait: Seed Shape
Alleles: R – Round r – Wrinkled
Cross: Wrinkled seeds x 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

36. F2 Monohybrid Cross Review
Homozygous recessive x heterozygous(hybrid)
Offspring: 50% Homozygous rr 50% Heterozygous Rr
Phenotypic Ratio is 1:1

37. Test crosses
Can be used to determine if an individual of dominant phenotype is homozygous or heterozygous
There are two possible testcrosses:
Homozygous dominant x Homozygous recessive = All heterozygous dominant Hybrid x Homozygous recessive = Mix of dominant and recessive phenotypes

38. Monohybrid cross Practice Problems

39. 1. Breed the P1 generation
tall (TT) x dwarf (tt) pea plants t T

40. 2. Breed the F1 generation
tall (Tt) vs. tall (Tt) pea plants T t T t

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

42. 2. Solution: tall (Tt) x tall (Tt) pea plants T t TT Tt tt
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

43. 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

44. 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)

45. Law of Dominance

46. 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.

47. Applying the Law of Segregation

48. 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.

49. 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)

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

51. 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

52. All possible gamete combinations
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

53. Dihybrid Cross Try filling in the punnet square and work out ratios RY

54. 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

55. Dihybrid Cross
Round/Yellow: 9 Round/green: 3 wrinkled/Yellow: 3 wrinkled/green: 1
9:3:3:1

56. 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

57. Test Cross Possible results: bC b___ bc bbCc C bC b___ bc bbCc bbcc or

58. Summary of Mendel’s laws
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

59. Incomplete Dominance
F1 hybrids have an appearance somewhat in between the phenotypes of the two parental varieties.
Example: snapdragons (flower)
red (RR) x white (rr)
RR = red flower
rr = white flower r R

60. Incomplete Dominance r produces the R Rr F1 generation All Rr = pink
(heterozygous pink)
produces the
F1 generation Rr

61. Incomplete Dominance

62. 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

63. 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

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

65. 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)

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