Genetics Chapter 3

Blending theory of inheritance

• Factors that control hereditary traits are malleable

• They can blend together from generation to generation

What theory did Mendel’s experiments with garden peas refute?

Blending theory of inheritance

Hybridization

The mating or crossing between 2 individuals that have different characteristics

• Purple-flowered plant X white-flowered plant

Hybrids

The offspring that result from such mating (hybridization)

Why did Mendel choose to experiment on garden peas?

• It existed in several varieties with distinct characteristics

• Its structure allowed for easy crosses where the choice of parental plants could be controlled

Self-fertilization

• Pollen and egg are derived from the same plant

• Naturally occurs in peas because a modified petal isolates reproductive structures

Cross-fertilization

• Pollen and egg are derived from different plants

• Required removing and manipulating anthers

Trait vs Character

• Trait- describes the specific properties of a character

• eye color is a CHARACTER, blue eyes is a TRAIT

True breeder

A variety that produces the same trait over several generations

7 characters Mendel studied

• Height

• Flower color

• Flower position

• Seed color

• Seed shape

• Pod color

• Pod shape

Variants Mendel studied

• Height (Tall vs Dwarf)

• Flower color (Purple vs White)

• Flower position (Axial vs Terminal)

• Seed color (Yellow vs Green)

• Seed shape (Round vs Wrinkled)

• Pod color (Green vs Yellow)

• Pod shape (Smooth vs Constricted)

Empirical approach

Mendel believed that a quantitative analysis of crosses may provide mathematical relationships that govern hereditary traits

• Used to deduce empirical laws

Monohybrid cross

Crossing 2 variants of the same characteristic

• a single characteristic is being observed

• Tall X dwarf

Dihybrid cross

Crossing individual plants that differ in 2 characters

• Character 1- Seed texture (round vs wrinkled)

• Character 2- Seed color (yellow vs green)

Single factor crosses

1. For each of seven characters, Mendel cross-fertilized two different true breeding strains. Keep in mind that each cross involved in two plants that differed in regard to only one of the seven characters studied. The illustration at the right shows one cross between a tall and dwarf plant. This is called a P (parental) cross.

2. Collect the F1 generation seeds. The following spring, plant the seeds and allow the plants to grow. These are the plants of the F1 generation.

3. Allow the F1 generation plants to self-fertilize. This produces seeds that are part of the F2 generation.

4. Collect the F2 generation seeds and plant them the following spring to obtain the F2 generation plants

5. Analyze the traits found in each generation

Particulate Theory of Inheritance

The genetic determinants that govern traits are inherited as discrete units that remain unchanged as they are passed from parent to offspring

Mendel’s Law of Segregation

During gamete formation, the paired factors for a given character segregate randomly so that half of the gametes receive one factor and half of the gametes receive the other

Genes

Mendelian factors

Alleles

Different versions of the same gene

Homozygous

An individual with 2 identical alleles

Heterozygous

An individual with 2 different alleles

Genotype

Specific allelic composition of an individual

Phenotype

Outward appearance of an individual

Principle of Segregation

• 2 alleles for a gene segregate during gamete formation and are rejoined at random, one from each parent, during fertilization

• Physical basis for allele segregation is the behavior of chromosomes during meiosis

• Mendel had no knowledge of chromosomes or meiosis

Mendel’s 2nd Law of Independent Assortment

• 2 genes on different chromosomes segregate their alleles independently

• The inheritance of an allele of 1 gene does not influence which allele is inherited at a 2nd gene

Principle of Independent Assortment

1. In a dihybrid cross, the alleles of each gene assort independently

2. The segregation of different allele pairs is independent

3. Independent alignment of different homologous chromosome pairs during metaphase I leads to the independent segregation of the different allele pairs

Pedigree Analysis

Pedigree analysis is commonly used to determine the inheritance pattern of human genetic diseases

• 2 normal heterozygous individuals will have, on average, 25% of their offspring affected

• 2 affected individuals will produce 100% affected offspring

Cystic Fibrosis (CF)

• A recessive disorder of humans

• Gene encodes a protein called the cystic fibrosis transmembrane conductance regulator (CFTR)

  • CFTR protein regulates ion transport across cell membrane

• Mutant allele creates altered CFTR protein that causes ion imbalance

  • Abnormalities in pancreas, intestine, sweat glands, and lungs

Probability

Probability = # of times an event occurs / Total # of events

• P(heads) = 1 heads/(1 heads + 1 tails) = ½ = 50%

Random sampling error

Accuracy of the probability prediction depends largely on the size of the sample

• Large for small samples and small for large samples

Product Rule

The probability that 2 or more independent events will occur is equal to the product of their respective probabilites

Two heterozygous individuals plan to start a family

• What is the probability that the couple’s first three children will all have congenital analgesia?

Applying the product rule

1. Calculate the individual probabilities

• Obtained via Punnett square

  • P(congenital analgesia) = ¼

2. Multiply the individual probabilites

• ¼ x ¼ x ¼ x 1/64

• 1/16 can be converted to 0/016

Therefore 1.6% of the time, the first 3 offspring of a heterozygous couple, will all have congenital analgesia

Binomial Expansion

Represents all of the possibilities for a given set of unordered events

Two heterozygous brown-eyed (Bb) individuals have 5 children

• What is the probability that 2 of the couple’s 5 children will have blue eyes?

1. Calculate the individual probabilities

• P(blue) = p = ¼

• P(brown) = q = ¼

2. Determine the # of events

• n = total # of children = 5

• x = # of blue-eyed children = 2

3. Substitute the values for p, q, x, and n in the binomial expansion equation