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

2. What genetic principles account for the passing of traits from parents to offspring?
The “blending” hypothesis is the idea that genetic material from the two parents blends together (like blue and yellow paint blend to make green)
The “particulate” hypothesis is the idea that parents pass on discrete heritable units (genes)
Mendel documented a particulate mechanism through his experiments with garden peas.
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3. Fig. 14-1
Figure 14.1 What principles of inheritance did Gregor Mendel discover by breeding garden pea plants?

5. Concept 14.1: Mendel used the scientific approach to identify two laws of inheritance
Mendel discovered the basic principles of heredity by breeding garden peas in carefully planned experiments.
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6. Mendel’s Experimental, Quantitative Approach
Advantages of pea plants for genetic study:
There are many varieties with distinct heritable features
Mating of plants can be controlled
Cross-pollination (fertilization between different plants) can be achieved by dusting one plant with pollen from another
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7. TECHNIQUE Parental generation (P) Stamens Carpel RESULTS First filial
Fig. 14-2
TECHNIQUE 1 2
Parental
generation
(P)
Stamens
Figure14.2 Crossing pea plants
Carpel 3 4
RESULTS
First
filial
gener-
ation
offspring
(F1) 5

8. He also used varieties that were true-breeding (plants that produce offspring of the same variety when they self-pollinate)
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9. The true-breeding parents are the P generation
In a typical experiment, Mendel mated two contrasting, true-breeding varieties, a process called hybridization
The true-breeding parents are the P generation
The hybrid offspring of the P generation are called the F1 generation
When F1 individuals self-pollinate, the F2 generation is produced.
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10. The Law of Segregation
When Mendel crossed contrasting, true-breeding white and purple flowered pea plants, all of the F1 hybrids were purple
When Mendel crossed the F1 hybrids, many of the F2 plants had purple flowers, but some had white
Mendel discovered a ratio of about three to one, purple to white flowers, in the F2 generation.
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11. EXPERIMENT P Generation (true-breeding parents) Purple flowers White
Fig
EXPERIMENT
P Generation
(true-breeding
parents) 
Purple
flowers
White
flowers
Figure 14.3 When F1 hybrid pea plants are allowed to self-pollinate, which traits appear in the F2 generation?

12. F1 Generation (hybrids)
Fig
EXPERIMENT
P Generation
(true-breeding
parents) 
Purple
flowers
White
flowers
F1 Generation
(hybrids)
Figure 14.3 When F1 hybrid pea plants are allowed to self-pollinate, which traits appear in the F2 generation?
All plants had
purple flowers

13. EXPERIMENT P Generation (true-breeding parents) Purple flowers White
Fig
EXPERIMENT
P Generation
(true-breeding
parents) 
Purple
flowers
White
flowers
F1 Generation
(hybrids)
Figure 14.3 When F1 hybrid pea plants are allowed to self-pollinate, which traits appear in the F2 generation?
All plants had
purple flowers
F2 Generation
705 purple-flowered
plants
224 white-flowered
plants

14. What Mendel called a “heritable factor” is what we now call a gene
Mendel reasoned that only the purple flower factor was affecting flower color in the F1 hybrids
Mendel called the purple flower color a dominant trait and the white flower color a recessive trait
Mendel observed the same pattern of inheritance in six other pea plant characters, each represented by two traits
What Mendel called a “heritable factor” is what we now call a gene
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15. Table 14-1

16. Mendel’s Model
Mendel developed a hypothesis to explain the 3:1 inheritance pattern he observed in F2 offspring:
1. Alternative versions of a gene exist - are now called alleles. The position of the gene is called the locus.
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17. Allele for purple flowers
Fig. 14-4
Allele for purple flowers
Homologous
pair of
chromosomes
Figure 14.4 Alleles, alternative versions of a gene
Locus for flower-color gene
Allele for white flowers

18. Alternatively, the two alleles at a locus may differ – heterozygous.
2. For each character an organism inherits two alleles, one from each parent.
Mendel made this deduction without knowing about the role of chromosomes!
The two alleles at a locus on a chromosome may be identical – homozygous.
Alternatively, the two alleles at a locus may differ – heterozygous.
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19. 3.The third concept is that if the two alleles at a locus differ, then one (the dominant allele) determines the organism’s appearance, and the other (the recessive allele) has no noticeable effect on appearance
In the flower-color example, the F1 plants had purple flowers because the allele for that trait is dominant
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20. 4. The law of segregation, states that the two alleles for a heritable character separate (segregate) during gamete formation and end up in different gametes.
Thus, an egg or a sperm gets only one of the two alleles that are present in the somatic cells of an organism
This segregation of alleles corresponds to the distribution of homologous chromosomes to different gametes in meiosis.
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22. Punnett Squares P Generation Appearance: Purple flowers White flowers
Fig
Punnett Squares
P Generation
Appearance:
Purple flowers
White flowers
Genetic makeup: PP pp
Gametes: P p
Figure 14.5 Mendel’s law of segregation

23. P Generation Appearance: Purple flowers White flowers Genetic makeup:
Fig
P Generation
Appearance:
Purple flowers
White flowers
Genetic makeup: PP pp
Gametes: P p
F1 Generation
Appearance:
Purple flowers
Figure 14.5 Mendel’s law of segregation
Genetic makeup: Pp
Gametes:
1/2 P
1/2 p

24. P Generation Appearance: Purple flowers White flowers Genetic makeup:
Fig
P Generation
Appearance:
Purple flowers
White flowers
Genetic makeup: PP pp
Gametes: P p
F1 Generation
Appearance:
Purple flowers
Figure 14.5 Mendel’s law of segregation
Genetic makeup: Pp
Gametes:
1/2 P
1/2 p
Sperm
F2 Generation P p P PP Pp
Eggs p Pp pp 3 1

25. phenotype, or physical appearance
An organism’s phenotype also includes internal anatomy, physiology, molecular pathways, and behavior
genotype, or genetic makeup
In the example of flower color in pea plants, PP and Pp plants have the same phenotype (purple) but different genotypes
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26. Phenotype Genotype PP Purple 1 (homozygous) 3 Purple Pp (heterozygous)
Fig. 14-6
Phenotype
Genotype PP
Purple 1
(homozygous) 3
Purple Pp
(heterozygous)
Figure 14.6 Phenotype versus genotype 2
Purple Pp
(heterozygous) pp 1
White 1
(homozygous)
Ratio 3:1
Ratio 1:2:1

27. Important ratio: for monohybrid cross, Aa x Aa 3:1 dominant to recessives

28. The Law of Independent Assortment
A monohybrid cross follows one trait.
A dihybrid cross follows two traits.
Using a dihybrid cross, Mendel developed the law of independent assortment
The law of independent assortment states that each pair of alleles segregates independently of each other pair of alleles during gamete formation
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30. What does that mean in terms of meiosis?
Strictly speaking, this law applies only to genes on different, nonhomologous chromosomes.
So a AaBb parent can produce gametes:
AB, aB, Ab, ab

31. Predicted outcomes of heterozygous dihybrid cross: AaBb x AaBb 9 both dominant traits 3 one dominant, one recessive trait 3 other dominant, other recessive trait 1 both recessive traits

32. EXPERIMENT RESULTS Fig. 14-8 P Generation F1 Generation Hypothesis of
YYRR
yyrr
Gametes YR  yr
F1 Generation
YyRr
Hypothesis of
dependent
assortment
Hypothesis of
independent
assortment
Predictions
Sperm or
Predicted
offspring of
F2 generation
1/4 YR
1/4 Yr
1/4 yR
1/4 yr
Sperm
Figure 14.8 Do the alleles for one character assort into gametes dependently or independently of the alleles for a different character?
1/2 YR
1/2 yr
1/4 YR
YYRR
YYRr
YyRR
YyRr
1/2 YR
YYRR
YyRr
1/4 Yr
Eggs
YYRr
YYrr
YyRr
Yyrr
Eggs
1/2 yr
YyRr
yyrr
1/4 yR
YyRR
YyRr
yyRR
yyRr
3/4
1/4
1/4 yr
Phenotypic ratio 3:1
YyRr
Yyrr
yyRr
yyrr
9/16
3/16
3/16
1/16
Phenotypic ratio 9:3:3:1
RESULTS
315
108
101 32
Phenotypic ratio approximately 9:3:3:1

33. Fig. 14-UN11
REMEMBER!
Each gamete has to
have one of each allele!

35. Fig. 14-UN6

36. The laws of probability govern Mendelian inheritance
Probability of an event occurring = expected over possibilities
EX: Probability of throwing a tail with a coin is?
1/2
Segregation in a heterozygous plant is like flipping a coin: Each gamete has a 1/2 chance of carrying the dominant allele and a 1/2 chance of carrying the recessive allele

37. 1/2 1/2 1/2 1/4 1/4 1/2 1/4 1/4 Rr Rr  Segregation of Segregation of
Fig. 14-9 Rr Rr 
Segregation of
alleles into eggs
Segregation of
alleles into sperm
Sperm
1/2 R
1/2 r
Figure 14.9 Segregation of alleles and fertilization as chance events R R
1/2 R R r
1/4
1/4
Eggs r r
1/2 R r r
1/4
1/4

38. Mendel’s laws of segregation and independent assortment reflect the rules of probability
The multiplication rule states that the probability that two or more independent events will occur together is the product of their individual probabilities

39. What is the probability of getting an F2 (Aa x Aa) offspring that is homozygous (aa) for both traits?
½ x ½ = 1/4

40. The rule of addition states that the probability that any one of two or more exclusive events will occur is calculated by adding together their individual probabilities
The rule of addition can be used to figure out the probability that an F2 plant from a monohybrid cross will be heterozygous rather than homozygous
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41. Using the addition rule:
In the cross of Rr x Rr, what is the
probability of Rr in offspring?
If sperm is R and egg is r = ½ x ½ = ¼
If sperm is r and egg is R = ½ x ½ = ¼
So adding those together is ¼ + ¼ = 1/2

42. Solving Complex Genetics Problems with the Rules of Probability
We can apply the multiplication and addition rules to predict the outcome of crosses involving multiple characters
A dihybrid or other multicharacter cross is equivalent to two or more independent monohybrid crosses occurring simultaneously
In calculating the chances for various genotypes, each character is considered separately, and then the individual probabilities are multiplied together

43. Examples
AaBb X Aabb
What is probability of obtaining a Aabb offspring?
Aa = ½
bb = ½
Answer = 1/4

44. Probability of obtaining AabbCc?
AaBbCc x aabbCc
Probability of obtaining AabbCc?
For Aa from Aa x aa = ½
For bb from Bb x bb = ½
For Cc from Cc x Cc = ½
The answer is ½ x ½ x ½ = 1/8

45. AaBbcc x AabbCC a) Probability of offspring that are A_B_C_?
A_ = ¾, B = ½, C = 1/1 answer 3/8
b) A_bbC_
aaB_C_
A_B_C_
c) aabbC_

46. Used to tell the genotype of an individual with the dominant phenotype
The Testcross
Used to tell the genotype of an individual with the dominant phenotype
Such an individual must have one dominant allele, but the individual could be either homozygous dominant or heterozygous
A testcross: breeding the mystery individual with a homozygous recessive individual
If any offspring display the recessive phenotype, the mystery parent must be heterozygous.
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47. Fig. 14-7 TECHNIQUE RESULTS Dominant phenotype, unknown genotype:
Dominant phenotype,
unknown genotype:
PP or Pp?
Recessive phenotype,
known genotype: pp
Predictions
If PP
If Pp or
Sperm
Sperm
Figure 14.7 The testcross p p p p P P Pp Pp Pp Pp
Eggs
Eggs P p Pp Pp pp pp
RESULTS or
All offspring purple
1/2 offspring purple and
1/2 offspring white

48. Extending Mendelian Genetics for a Single Gene
Inheritance of characters by a single gene may deviate from simple Mendelian patterns in the following situations:
When alleles are not completely dominant or recessive
When a gene has more than two alleles
When a gene produces multiple phenotypes
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49. In codominance, both alleles are expressed fully
Degrees of Dominance
Complete dominance occurs when phenotypes of the heterozygote and dominant homozygote are identical
In incomplete dominance, the phenotype of F1 hybrids is somewhere between the phenotypes of the two parental varieties
In codominance, both alleles are expressed fully
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51. Important ratio Aa x Aa IF INCOMPLETE DOMINANCE 1:2:1 1 dominant 2 mix 1 recessive

52. The Relation Between Dominance and Phenotype
A dominant allele does not subdue a recessive allele; alleles don’t interact
Alleles are simply variations in a gene’s nucleotide sequence
For any character, dominance/recessiveness relationships of alleles depend on the level at which we examine the phenotype
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53. Tay-Sachs disease is fatal; a dysfunctional enzyme causes an accumulation of lipids in the brain
At the organismal level, the allele is recessive
At the biochemical level, the phenotype (i.e., the enzyme activity level) is incompletely dominant
At the molecular level, the alleles are codominant
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54. Inheritance of Tay-Sachs
Organism level
Imperfect enzymes
The disease is caused from mutations on chromosome 15 in the HEXA gene. This gene defect would be codominant with a normal gene.

55. Frequency of Dominant Alleles
Dominant alleles are not necessarily more common in populations than recessive alleles
For example, one baby out of 400 in the United States is born with extra fingers or toes, the result of a dominant allele
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56. Multiple Alleles
Most genes exist in populations in more than two allelic forms
For example, the four phenotypes of the ABO blood group in humans are determined by three alleles for the enzyme (I) that attaches A or B carbohydrates to red blood cells: A, B, AB, O. O is recessive.
The enzyme encoded by the A allele adds the A carbohydrate, whereas the enzyme encoded by the B allele adds the B carbohydrate; the enzyme encoded by the O allele adds neither

57. You may see IA , IB or i for the alleles.
Fig
Allele
Carbohydrate
You may see IA , IB or i for the alleles. A A B B O
none
(a) The three alleles for the ABO blood groups
and their associated carbohydrates
Red blood cell
appearance
Phenotype
(blood group)
Genotype
Figure Multiple alleles for the ABO blood groups
AA or AO A
BB or BO B AB AB OO O
(b) Blood group genotypes and phenotypes

58. If Mary has A blood and Ben has B blood, can they have a child with O blood?

59. Pleiotropy
Most genes have multiple phenotypic effects, a property called pleiotropy
For example, pleiotropic alleles are responsible for the multiple symptoms of certain hereditary diseases, such as cystic fibrosis and sickle-cell disease
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60. Epistasis
In epistasis, a gene at one locus alters the phenotypic expression of a gene at a second locus
For example, in mice and many other mammals, coat color depends on two genes
One gene determines the pigment color (with alleles B for black and b for brown)
The other gene (with alleles C for color and c for no color) determines whether the pigment will be deposited in the hair
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61. 1/4 1/4 1/4 1/4 1/4 1/4 1/4 1/4  BbCc BbCc Sperm Eggs BBCC BbCC BBCc
Fig 
BbCc
BbCc
Sperm
1/4
1/4
1/4
1/4 BC bC Bc bc
Eggs
1/4 BC
BBCC
BbCC
BBCc
BbCc
Figure An example of epistasis
1/4 bC
BbCC
bbCC
BbCc
bbCc
1/4 Bc
BBCc
BbCc
BBcc
Bbcc
1/4 bc
BbCc
bbCc
Bbcc
bbcc 9
: 3
: 4

62. Interactive Question 14.7 MmBb x MmBb Phenotype Genotype Ratio Black

63. Polygenic Inheritance
Quantitative characters are those that vary in the population along a continuum
Polygenic inheritance results in an additive effect of two or more genes on a single phenotype which results in quantitative variation.
Skin color in humans is an example of polygenic inheritance
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64. Fig 
AaBbCc
AaBbCc
Sperm
1/8
1/8
1/8
1/8
1/8
1/8
1/8
1/8
1/8
1/8
1/8
Figure A simplified model for polygenic inheritance of skin color
1/8
Eggs
1/8
1/8
1/8
1/8
Phenotypes:
1/64
6/64
15/64
20/64
15/64
6/64
1/64
Number of
dark-skin alleles: 1 2 3 4 5 6

65. Nature and Nurture: The Environmental Impact on Phenotype
Another departure from Mendelian genetics arises when the phenotype for a character depends on environment as well as genotype
The norm of reaction is the phenotypic range of a genotype influenced by the environment
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66. Fig
For example, hydrangea flowers of the same genotype range from blue-violet to pink, depending on soil acidity
Figure The effect of environment on phenotype

67. Norms of reaction are generally broadest for polygenic characters
Such characters are called multifactorial because genetic and environmental factors collectively influence phenotype
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68. The risk of heart disease follows a continuum
The risk of heart disease follows a continuum. It is multifactorial due to both genetic and environmental influences.

69. Role of environment?
An organism’s phenotype reflects its overall genotype and unique environmental history
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70. What is the dominant trait? ____________________________
EXPERIMENTAL DATA
In a cross of true-breeding green seed coat peas with true-breeding yellow seed coat peas, all of the F1 peas had green-seed coats. In the F2, you counted 122 green ones and 33 yellow seeded peas. The total number of individuals you counted, N, is 155.
What is the dominant trait? ____________________________
What is your hypothesis for the F2 population? Show a Punnett square to show the proposed results.
 Fill in the tables below with your observed data, calculate the expected result according to your hypothesis. Determine the Χ2 value for each experiment, and use the table of probabilities to accept or reject the null hypotheses.
Monohybrid cross
Phenotypes Observed Expected d = o –e d d2/e
  Total (X2 )_____
 X2 = _____     degrees of freedom: _____
Range of probability: ______________
Accept or reject null hypothesis? __________

71. Experimental Data for Chi Square Determination
In a cross of true-breeding rough seed coat peas with true-breeding smooth seed coat peas, all of the F1 peas had rough-seed coats. In the F2, you counted 79 rough coats and 33 smooth coats. The total number of individuals you counted is 112.
What is the dominant trait? ____________________________
What is your hypothesis for the F2 population? Show a Punnett square to show the proposed results.

72. Fill in the tables below with your observed data, calculate the expected result according to your hypothesis. Determine the Χ2 value for each experiment, and use the table of probabilities to accept or reject the null hypotheses.
Monohybrid cross
Phenotypes Observed (o) Expected (e) d = o – e d d2/e
Rough
Smooth
Total _____
Χ2 = _____     degrees of freedom: _____
Range of probability: ______________
Accept or reject null hypothesis? __________