1. Classical Genetics

2. Why Peas?
Read Concept 2
Click Animation and View

3. Mendel’s Work Traits are passed on by factors (genes)
Factors (genes) have more than 1 form called alleles
There are at least 2 alleles for each trait

4. MENDELIAN INHERITANCE
Mendelian inheritance patterns
Involve genes directly influencing traits
Obey Mendel’s laws
Law of segregation
Law of independent assortment
Include
Dominant / recessive relationships
Gene interactions
Phenotype-influencing roles of sex and environment
Most genes of eukaryotes follow a Mendelian inheritance pattern 4 4

5. I. Principles of Genetics
Random segregation- dominant and recessive alleles separate randomly during meiosis II

6. I. Principles of Genetics
Independent assortment- parental chromosomes (homologs) separate randomly during meiosis I

7. II. Types of Crosses
Monohybrid Cross- looking at 1 trait in individuals that are mated (AA x Aa).
A 4-square Punnett
Dihybrid Cross- looking at 2 traits in individuals that are mated (AaBb x AaBB).
A 16 Square Punnett.
Looking at more than one gene at a time.

8. II. Breeding Experiments
B. The difference between genotype and phenotype 1. Genotype- pattern of alleles (think gene). This is the genetic appearance. AA, Bb, cc 2. Phenotype- physical appearance (phenotype=physical) resulting from the genotype. Ex: A=brown hair, b=no tail 3. Homozygous- 2 copies of the same allele. Either homozygous dominant (AA) or homozygous recessive (aa). 4. Heterozygous- 1 copy of each allele (Aa or Bb).

9. III. Play with Punnett Squares
A. The language of crosses
1. P= parent generation in a cross (what you start with). Example: P=Aa x AA
2. F1= First generation of offspring from the cross
3. F2 = Second generation of offspring, produced by crossing organisms in F1 with themselves.

10. III. Play with Punnet Squares
B. Simple dominance: one allele is completely dominant
P=Aa x Aa A=red, a=clear
A a A a
F1 Genotype=
AA: Aa: aa 1: 2: 1
F1 Phenotype=
A (red): a (clear) : 1 AA Aa aa

11. III. Play with Punnett Squares
C. Incomplete dominance- a recessive allele is not completely recessive and shows through the dominant allele.
Example: breeding red snapdragons and white snapdragons produces pink snapdragons.
P= RR x rr R =red, r =white
R R r Rr
F1 Genotype= Rr
Phenotype= pink

12. III. Play with Punnett Squares
D. Codominance- both alleles are equally strong and produce a heterozygous phenotype that is a mixture of both
Like in incomplete dominance, heterozygotes have a unique phenotype (appearance).
Example: breeding white beans and purple beans produces white beans with purple spots.
P=CWCP x CWCP CW=white, CP=purple
CW CP CW CP
F1 Genotype=
AA: Aa: aa
1: 2: 1
Phenotype=
A : Aa (spotted): a
CWCW
CWCP
CPCP

13. Mendel’s Work
Alleles are represented by the first letter of the dominant trait
Pea plant flower color (trait)
Purple or white
Purple is dominant over white
Alleles are represented as
P=purple p=white

14. Mendel’s Work You try a few…
Round seeds are dominant over wrinkled seeds
Yellow seeds are dominant over green seeds
Tall plants are dominant over short plants
1. R=round r=wrinkled
Y=yellow y=green
T=tall t= short

15. Mendel’s Work Describing Traits (Tall or short plants) Genotype
The actual genetic make-up of an organism
the “genes”
Phenotype
The physical appearance or form observed
the “physical”

16. Mendel’s Work Possible Genotypes (TT Tt tt) Homozygous
The two alleles for the trait are identical
TT homozygous dominant (purebred dominant)
tt homozygous recessive (purebred recessive)
Heterozygous
The two alleles for the trait are different
Tt heterozygote (hybrid)

17. Mendel’s Conclusions 1. Law of Dominance
Alleles for a trait are either dominant or recessive
The dominant form is expressed and the recessive form is hidden
The only way to express a recessive trait is if there are two copies of the recessive allele

18. Mendel’s Conclusions Law of Segregation
The two alleles for a trait separate during gametogenesis

19. II. Types of Crosses
C. Test Cross-an individual of unknown genotype (A?) is mated with a homozygous recessive individual (aa) to determine the unknown genotype.
-(P= A? x aa)

20. C. A Test Cross PP Pp pp Pp P ? P ?
If all of the offspring have a dominant phenotype, you know the unknown parent genotype was homozygous dominant (left Punnet, AA).
If some of the offspring have the recessive phenotype, you know the unknown parent genotype was heterozygous (right Punnett, Aa).

21. Mendel’s 3rd Law of Independent Assortment
Alleles of different genes are assorted independently of one another during the formation of gametes
This means that calculating the probability of several traits appearing together is the product of the probability of each trait taken separately
The Rule of Multiplication

22. Studying the inheritance of two characters simultaneously
THE DIHYBRID CROSS
Studying the inheritance of two characters simultaneously
© 2007 Paul Billiet ODWS

23. Dihybrid Crosses
the natural progression for Mendel was to study 2 characteristics at the same time.
thus, the study of 2 pairs of contrasting traits at the same time = a dihybrid cross
ex.. round yellow seeds X wrinkled green seeds

24. Mendel’s Fourth Law The Law of Independent Assortment
During gamete formation, segregating pairs of unit factors assort independently of each other.
the two traits are inherited totally independently of each other.
i.e. color is inherited independently of seed shape.
Example:
Homozygous round green X wrinkled homozygous yellow

25. Mendel’s peas Character Trait Allele Seed shape Round R Wrinkled r
Cotyledon colour
Yellow Y
Green y
© 2007 Paul Billiet ODWS

26. Combinations Genotype Phenotype RRYY Round Yellow RRYy RrYY RrYy RRyy
Round Green
Rryy
rrYY
Wrinkled Yellow
rrYy
rryy
Wrinkled Green
© 2007 Paul Billiet ODWS

27. Is the inheritance of one character affected by the inheritance of another?
P Phenotypes
Round Yellow x
Wrinkled Green
(Pure Bred)
F1 Phenotypes
All Round Yellow
(Selfed)
F2 Phenotypes
Seed
colour
Yellow
Green
TOTAL
RATIO
Round
315
108
423
shape
Wrinkled
101 32
133
416
140
556
© 2007 Paul Billiet ODWS

28. Is the inheritance of one character affected by the inheritance of another?
P Phenotypes
Round Yellow x
Wrinkled Green
(Pure Bred)
F1 Phenotypes
All Round Yellow
(Selfed)
F2 Phenotypes
Seed
colour
Yellow
Green
TOTAL
RATIO
Round
315
108
423
3.18
shape
Wrinkled
101 32
133 1
416
140
556
2.97
© 2007 Paul Billiet ODWS

29. 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 =
Probability of an F2 seed being yellow =
Probability of an F2 seed being green =
© 2007 Paul Billiet ODWS

30. 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
© 2007 Paul Billiet ODWS

31. Therefore
Probability of an F2 seed being round and yellow = ¾ x ¾ = 9/16 = %
Probability of an F2 seed being round and green =
Probability of an F2 seed being wrinkled and yellow =
Probability of an F2 seed being wrinkled and green =
© 2007 Paul Billiet ODWS

32. Therefore
Probability of an F2 seed being round and yellow = ¾ x ¾ = 9/16 = %
Probability of an F2 seed being round and green = ¾ x ¼ = 3/16 = %
Probability of an F2 seed being wrinkled and yellow = ¼ x ¾ = 3/16 = %
Probability of an F2 seed being wrinkled and green = ¼ x ¼ = 1/16 = %
© 2007 Paul Billiet ODWS

33. Predicting how many seeds we could expect to get in a sample
What Mendel observed
556 x 9/16 round yellow
313
315
556 x 3/16 round green
108
556 x 3/16 wrinkled yellow
101
556 x 1/16 wrinkled green 32
© 2007 Paul Billiet ODWS

34. Predicting how many seeds we could expect to get in a sample
What Mendel observed
556 x 9/16 round yellow
313
315
556 x 3/16 round green
104
108
556 x 3/16 wrinkled yellow
101
556 x 1/16 wrinkled green 35 32
© 2007 Paul Billiet ODWS

35. THE LAW OF INDEPENDENT ASSORTMENT
It appears that the inheritance of seed shape has no influence over the inheritance of seed colour
The two characters are inherited INDEPENDENTLY
The pairs of alleles that control these two characters assort themselves independently
© 2007 Paul Billiet ODWS

36. Mendel & Meiosis
The pairs of chromosomes could orientate in different ways at Anaphase 1
© 2007 Paul Billiet ODWS

37. Dihybrid cross genetic diagramP
Phenotypes
Round Yellow seed X
Wrinkled Green seed
(Pure bred)
Genotypes
RRYY
rryy
meiosis
Gametes RY ry
fertilisation F1
RrYy
(Selfed)
Round Yellow
Proportions
100%
© 2007 Paul Billiet ODWS

38. Dihybrid cross genetic diagramF1
Phenotypes
RrYy
(Selfed)
Genotypes
Round Yellow
Proportions
100%
meiosis
Gametes Y y R RY Ry r rY ry
© 2007 Paul Billiet ODWS

39. Dihybrid cross genetic diagram
Gametes Y y R RY Ry r rY ry
fertilisation F2
Genotypes
RRYY
RRYy
RrYY
RrYy
RRyy
Rryy
rrYY
rrYy
rryy
© 2007 Paul Billiet ODWS

40. Dihybrid cross proportions
Phenotypes
Proportions
Round Yellow
9/16 or
56.25%
Round Green
Wrinkled Yellow
Wrinkled Green
© 2007 Paul Billiet ODWS

41. Dihybrid cross proportions
Phenotypes
Proportions
Round Yellow
9/16 or
56.25%
Round Green
3/16 or
18.75%
Wrinkled Yellow
Wrinkled Green
1/16 or
6.25%
© 2007 Paul Billiet ODWS

42. Dihybrid test cross
In monohybrid crosses, to know if a dominant trait is homozygous (RR) or heterozygous (Rr) it is necessary to carry out a test cross
This is done with a homozygous recessive (rr) 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 (rryy)
© 2007 Paul Billiet ODWS

43. Dihybrid test cross Phenotypes Round Yellow X Wrinkled Green Genotypes
RrYy
rryy
Gametes
RY, Ry, rY, ry ry
Genotypes RY Ry rY ry
RrYy
Rryy
rrYy
rryy
Phenotypes
Round Yellow
Round Green
Wrinkled Yellow
Wrinkled Green
Proportions
1/4 or 25%
© 2007 Paul Billiet ODWS

44. Solving Dihybrid Crosses
Follow the steps
If you were to cross a Homozygous round green X wrinkled homozygous yellow plant, what would your phenotypic and genotypic ratios be?

45. Step 1: Make a Key….
G= yellow
g = green
W = round
w = wrinkled

46. Step 2: Give the genotypes of the parents. . .
Homozygous round green X wrinkled homozygous yellow plant
P1 = WWgg X wwGG

47. Step 3: Determine the Gametes . . .
WWgg X wwGG
WWgg = Wg, Wg, Wg, Wg
wwGG = wG, wG, wG, wG
HINT: Foil to get the gametes

48. Step 4: Fill in the Punnett square . . .Wg wG
WwGg

49. Step 5: Answer any questions . . .
Genotypic ratio
= all WwGg
Phenotypic ratio
= all round yellow
G= yellow
g = green
W = round
w = wrinkled

50. F1 X F1 = WwGg X WwGg WwGg X WwGg
Remember to determine the gametes for your punnett square.
WwGg X WwGg

51. Fill in the Punnett SquareWG Wg wG wg
WWGG
WWGg
WwGG
WwGg
WWgg
Wwgg
wwGG
wwGg
wwgg
Do you see a pattern?

52. The PATTERN . . . WG Wg wG wg WWGG WWGg WwGG WwGg WWgg Wwgg wwGG wwGg

53. What are the ratios? Genotypic Ratio: 1 WWGG 2 WWGg 1 WWgg 2 WwGG

54. Phenotypic Ratio: Phenotypic Ratio: Genotypic Ratio: 9 Yellow Round
G= yellow
g = green
W = round
w = wrinkled
Phenotypic Ratio:
Genotypic Ratio:
1 WWGG
2 WWGg
1 WWgg
2 WwGG
4 WwGg
2 Wwgg
1 wwGG
2 wwGg
1 wwgg
Phenotypic Ratio:
9 Yellow Round
Dominant Dominant
3 Yellow Wrinkled
Dominant Recessive
3 Green Round
Recessive Dominant
1 Green Wrinkled
Recessive Recessive
** Explain the patterns

55. The Short Cut: can only be used if you have
(Big little Big little X Big little Big little)
BbTt x BbTt
Do these fit?
BBTt X BbTt
BbTt x Bb tt

56. Try this: C= Inflated Pod c = constricted pod D = Tall d = dwarf
The cross: a plant heterozygous for inflated pod and heterozygous tall is crossed with a another heterozygous inflated pod and heterozygous tall plant. What are the genotypic and phenotypic ratios?
The parents are:
CcDd x CcDd

57. Example: CcDd x CcDd Does this fit the short cut? Yes C= Inflated Pod
c = constricted pod
D = Tall
d = dwarf
Example:
Genotypic Ratio:
1 CCDD
2 CCDd
1 CCdd
2 CcDD
4 CcDd
2 Ccdd
1 ccDD
2 ccDd
1 ccdd
Genotypic Ratio: 1 2 4
CcDd x CcDd
Does this fit the short cut?
Yes
Phenotypic Ratio:
(D) (D)
9 Inflated Tall
(D) ®
3 Inflated Dwarf
® (D)
3 Constricted Tall
® ®
1 Constricted Dwarf
Phenotypic Ratio:
(D) (D) 9
(D) ® 3
® (D)
® ® 1