1. Mendel and the Gene Idea
Chp 14
Mendel and the Gene Idea

2. Blending Model vs. Particulate Model
genetic material mixes like paint
Parent’s traits inseparable
Discreet inheritable units
Traits retain separate identities

3. Web Lab: Mendel and his Peas
Figure 14-01
Gregor Mendel
Austrian Monk
Did not know about DNA, genes, or chromosomes!
Tried to prove particulate model of inheritance using pea plants
Web Lab: Mendel and his Peas

4. Advantages of working with pea plants:
Many different varieties
Character = inheritable feature (GENE)
Trait = variation of that character (ALLELE)
Can control plant matings (paintbrush)

5. LE 14-2 Removed stamens from purple flower Transferred sperm-
bearing pollen from
stamens of white
flower to egg-
bearing carpel of
purple flower
Parental
generation
(P)
Stamens
Carpel
Pollinated carpel
matured into pod
Planted seeds
from pod
Examined
offspring:
all purple
flowers
First
generation
offspring
(F1)

6. Started with true-breeding varieties = HOMOZYGOUS
If allowed to self-pollinate, all offspring had the same traits as their parent plant
Then, cross-pollinated two different true-breeding varieties = HYBRIDIZATION

7. Hybridization

9. Monohybrid Cross – tracks a single character/gene
P Generation
Appearance:
Purple
flowers PP
White
flowers pp
Genetic makeup:
Gametes: P p
F1 Generation
hybrid
Appearance:
Genetic makeup:
Purple flowers Pp
Monohybrid Cross – tracks a single character/gene

10. Mendel’s Law of Segregation: Parental alleles separate (segregate) during gamete formation
occurs in Metaphase I of Meiosis I
only one allele per gamete

12. Punnett Square shows possible offspring after fertilization gametes
P Generation
Appearance:
Purple
flowers PP
White
flowers pp
Genetic makeup:
Gametes P p
F1 Generation
Appearance:
Genetic makeup:
Purple flowers Pp
gametes
Gametes: 1 2 P 1 2 p
F1 sperm P p
F2 Generation
Punnett Square shows possible offspring after fertilization P PP Pp
gametes
F1 eggs p Pp pp 3
: 1

14. Phenotype Genotype PP (homozygous Purple 1 Pp (heterozygous 3 Purple 2
White 1
Ratio 3:1
Ratio 1:2:1

16. R r F1 Rr F2 R r Rr Rr rr
3 round: 1 wrinkled
1 RR : 2 Rr : 1 rr

17. Law of Dominance: The dominant allele (trait) is fully expressed and the recessive allele has no noticeable effect
Recessive ≠ Bad

18. Test Cross
Used to determine the genotype of an individual with a dominant phenotype
Cross it with a homozygous recessive and observe offspring ratios

20. 100% tall % Tt
1 tall : 1 dwarf Tt : 1 tt T T T t Tt Tt tt t t Tt Tt Tt Tt tt t t

21. Law of Independent Assortment: Each allele pair segregates independently of other allele pairs during Meiosis (Metaphase I).

22. Dihybrid Cross: tracks two genes (characters)

24. 9 tall purple 3 dwarf purple 3 tall white 1 dwarf white 12:4 or 3:1TP Tp tP tp TP
TTPP
TTPp
TtPP
TtPp
Tall purple Tp
TTPp
TtPp
TTpp
TtPp
Ttpp tP
TtPP
TtPp
ttPP
ttPp TP Tp tP tp tp
TtPp
Ttpp
ttPp
ttpp
9 tall purple
3 dwarf purple
3 tall white
1 dwarf white
12:4 or 3:1
12:4 or 3:1

25. Rules of Probability & Genetics
Probability ranges from 0 – 1
Rule of Multiplication: Used to determine the chance of two or more independent events occurring together
Determine probability of each independent event
Multiply probabilities together
Ex: What is the probability (chance) that when 2 coins are flipped they will both end up heads?
½ x ½ = ¼ (or .25)

28. Multiplication Rule Probability Problems
What is the probability (chance) that 2 dice will both show a 4 when rolled?
1/6 x 1/6 = 1/36

29. Ex (F1): T t P p x T t P p F2: t p t p ½ x ½ x ½ x ½ = 1/16 t t p p ½

30. Rule of Addition
What is the probability (chance) that the sum of the numbers shown on 2 dice will equal 5?
Rule of Addition: used to determine the probability of an event that can occur in two or more different ways/combinations
Calculate probability for each possible combination/way using Rule of Multiplication
Add the probabilities for each separate combination to get the total probability

31. Rule of Addition
What is the probability (chance) that the sum of the numbers shown on 2 dice will equal 5?
Possible combinations:
1/6 x 1/ /6 x 1/ /6 x 1/ /6 x 1/6
1/ / / / =
4/36 ( 1/9 )

33. = 3/8 x 1 x A a AA Aa aa b B Bb bb C c Cc

34. A__ B__ C__ = A__ B__ c c = A__ b b C__ = a a B__ C__ = A a AA Aa aa b

35. A__ B__ C__ = 3/8 A__ B__ c c = 0 A__ b b C__ = 3/8 a a B__ C__ = 1/8
Sum = 7/8 A a AA Aa aa b B Bb bb C c Cc

36. Text book: pages 272 - 273 All problems (#1-17) due on
Tonight, work on problems:
# 2, 3, 4, 7, 8, & 10

37. Types of Dominance
Complete Dominance – one allele complete masks the other; the heterozygous (Rr) and homozygous dominant (RR) have the same phenotype
The dominant allele usually codes for some protein/enzyme and the recessive allele codes for a defective protein/enzyme. Both are “expressed”, but only the dominant allele is functional and observable

38. Types of Dominance
Codominance = two alleles expressed separately & both affect phenotype
Ex: Red & White rhododendron
“Roan” color in cattle CR CW
CR CW

39. Types of Dominance
Incomplete Dominance = F1 hybrids (heterozygotes) have an intermediate phenotype
Results in a THIRD phenotype
NOT the same as blending (F2 show all phenotypes)
1:2:1 phenotypic and genotypic ratios in F2

40. Fill in Interactive Question 14.6
Multiple Alleles = more than two alleles/varieties
IA & IB
are codominant over i
Fill in Interactive Question 14.6

43. Blood Type practice…
Suppose a father of blood type B and a mother of blood type A have a child of type O. What are the chances that their next child will be:
Type O?
Type B?
Type A?
Type AB?

44. Blood Type practice…
Suppose a father of blood type B and a mother of blood type A have a child of type O. What are the chances that their next child will be:
Type O? ¼
Type B? ¼
Type A? ¼
Type AB? ¼ IB i IA
IAIB
IAi
IBi ii

45. Pleiotropy Pleiotropy = one gene has multiple phenotypic effects
Ex: cystic fibrosis & sickle-cell disease

47. Epistasis = (“force upon”) one gene affects the expression of another gene
Ex: coat color in mice
BbCc
BbCc
Sperm 1 BC 1 bC 1 Bc 1 4 4 4 4 bc 1 BC 4
BBCC
BbCC
BBCc
BbCc 1 bC
BbCC 4
bbCC
BbCc
bbCc 1
BBCc
BbCc
BBcc
Bbcc 4 Bc 1
bbcc bc
BbCc
bbCc
Bbcc 4 9 3 4 16 16 16

48. M__B__
¾ * ¾ = 9/16
brown
¾ * ¼ = 3/16
white
mm __ __
¼ * 1 = 4/ (¼)

49. # alleles + 1 = # phenotypes
Polygenic = additive effect of 2 or more genes on one phenotype
Ex: skin color height
AaBbCc
AaBbCc
aabbcc
Aabbcc
AaBbcc
AaBbCc
AABbCc
AABBCc
AABBCC
20/64
15/64
Fraction of progeny
6/64
# alleles + 1 = # phenotypes
1/64

51. Seven (6 alleles + 1 = 7 phenotypes)
AaBbCc = 25 cm
Seven (6 alleles + 1 = 7 phenotypes)

52. Nature vs. Nurture….
Acidic soil
Basic soil

53. Pedigree
Family tree showing relationships between family members and the pattern of inheritance across the generations

55. Figure 14-16

58. No – either he or his wife is AA, but can’t be certain
Aa Aa
AA aa aa Aa A_ A_ A_
Aa Aa Aa Aa aa Aa Aa A_ A_ A_ A_ A_ Aa Aa AA Aa Aa
No – either he or his wife is AA, but can’t be certain

59. Recessive Disorders
Carriers = heterozygotes (Aa) have the defective allele, but are not affected by disorder
“carry” allele, but do not express it
Can pass it on to offspring (50% chance) A a AA Aa aa
2/3

60. Recessive Disorders Cystic Fibrosis Tay-Sachs Disease
Sickle-cell Disease

61. ? 2/3 Aa x 2/3 Aa x ¼ aa = 4/36 = 1/9 A a AA Aa aa Aa Aa Aa Aa aa A_

62. Dominant Disorders RARE – nature will weed out (no carriers!)
Achondroplasia Dwarfism
Huntington’s Disease

63. LE 14-17a Amniocentesis Amniotic fluid withdrawn A sample of
amniotic fluid can
be taken starting at
the 14th to 16th
week of pregnancy.
Fetus
Centrifugation
Placenta
Uterus
Cervix
Fluid
Fetal
cells
Biochemical tests can be
performed immediately on
the amniotic fluid or later
on the cultured cells.
Biochemical
tests
Several
weeks
Fetal cells must be cultured
for several weeks to obtain
sufficient numbers for
karyotyping.
Karyotyping

64. LE 14-17b Chorionic villus sampling (CVS) A sample of chorionic villus
tissue can be taken as early
as the 8th to 10th week of
pregnancy.
Fetus
Suction tube
inserted through
cervix
Placenta
Chorionic villi
Fetal
cells
Biochemical
tests
Karyotyping and biochemical
tests can be performed on
the fetal cells immediately,
providing results within a day
or so.
Several
hours
Karyotyping