65e806c5627be0001873d5e6_##_Principle of Inheritance and Variation in 1 shot_ Class Notes

Introduction

  • Genetics: study of heredity & variation.
  • Heredity: transfer of characters from parent to offspring.
  • Variation: offspring differ from parents.
  • Terms:
    • Genetics: Bateson
    • Gene: Johannsen
    • Factor: Mendel
  • Inheritance: transmission of characters from parent to offspring.

Gene

  • Definition: Segment of DNA which forms protein & this protein control particular character.
  • Character: Plant height
  • Traits: Tall, dwarf
  • Allele: slightly different form of same gene present on same locus/position on Homologous chromosance.
  • Homozygous: Both allele/trait are same (TT or tt).
  • Heterozygous: Both allele/trait are different (Tt).
  • Dominant allele: Which trait express in both homozygous (TT) & heterozygous (Tt) condition.
  • Recessive allele: Which trait express only in homozygous condition (tt).
  • 2n: chromosome number.
    • 2n = TT: Homozygous dominant allele, Tall.
    • 2n = Tt: Heterozygous dominant allele, Tall.
    • 2n = tt: Homozygous recessive allele, dwarf.

Conclusion from Table

  • Dominant allele can express
    • In both Homozygous & Hetrozygous condition
    • In presence of identical (TT) & non-identical allele (Tt)
  • Recessive allele can express
    • Only in Homozygous condition
    • Only in presence of its identical allele (tt)

Mendelism

  • Born: 22 July 1822
  • Place: Austria
  • Worked: Augustinian monastery
  • Name of Plant: Pea (Pisum sativum)
  • How many years: 7 years
  • He started experiment in: 1856
  • Experiment work till: 1863 (1856-1863)
  • Scientist of which century: 19th century.
  • Total plant variety: 14
  • Total character studied: 7
  • Total pair of contrasting trait: 7pairs = 14 Trait

Selection of pea plant

  • Annual plant: which complete their life cycle in one year, 3-4 months
  • Life cycle & offspring: short, maximum offspring can be produced in one year
  • Bisexual: Natural self pollination present
  • Cross pollination Can Be performed by Removal of anther (Emasculation) & Bagging (Female plant)
  • Cultivation : easy to cultivate.

Reason for Mendel success

  • First time: Mathmetical Tools & stastics.
  • Record: He kept Record of His work.
  • Sampling Size: Large, He Repeated same experiment on 100 to 1000 plant
  • Result proved By experiment Technique.
    • Emasculation
    • Bagging
  • Experiment: He studied one or Two Character at a Time.
  • He was lucky Because He didn't find LINKAGE.

Chromosome/allele/gene/Factor

  • Formation of Gamete: segregation or seperation.
  • Number and Type of gamete.
    • 2n=TT: No. of Gamete: 1, Type of gamete: T
    • 2n=Tt: No. of Gamete: 2, Type of gamete: T, t
    • 2n=tt: No. of Gamete: 1, Type of gamete: t

True Breeding

  • True Breeding show stable Trait for many generation.
    • TT × TT
    • tt × tt
  • Gamete
    • T (n)
    • t (n)

Inheritance of one gene

  • one gene = one character (Plant Height) = Two Trait.
  • MONOHYBRID CROSS
  • 2 parent /2 plant = 2 variety: Tall & dwarf.
  • True Breeding / Homozygous.

Monohybrid Cross

  • Parent: (TT) Tall × (tt) dwarf.
  • Gamete: T & t.
  • F_1 generation: Tt = Tall.
  • Selfing: Tt × Tt.
  • Punnett square: graphical Representation to Find out the offspring.
  • Scientist: RC Punnett, Butsch Geneticst.
  • Phenotypic ratio: 3(Tall): 1(dwarf)
  • No. of phenotype : 2 (Tall, dwarf)
  • Genotypic ratio: 1(TT): 2(Tt): 1(tt)
  • No. of genotype: 3
  • Phenotype: Physical appearence.
  • Genotype: genetic constitution of organism.

Monohybrid Cross Formula

  • n: no. of Hetrozygous.
  • No. of gamete: 2^n
  • No. of phenotype: 2^n
  • No. of genotype: 3^n
  • No. of zygote/offspring: (game)^2 = (2)^2

Gamete Formation

  • Calculate number of gametes, phenotype and zygote formed.
    • Gamete no. =2^n
    • Zygote no. =2n \times 2n

Question Based on Gamete formation

  • Different types of gametes from different genotypes (TT, Tt, tt, AABB, AaBB, AaBb).

Segregation of alleles

  • Segregation of alleles is a random process so the chances of a gamete containing either allele is 50%.

Sexual reproduction in F1 individual

  • If F_1 individual of genotype (Tt) go through sexual reproduction, then it's gamete with genotype (T) have 50% chances to pollinate eggs of the genotype (T).

Pure tall and pure dwarf plant cross

  • A pure tall and a pure dwarf plant were crossed to produce offsprings. Offsprings were self crossed, then the ratio between true breeding tall to true breeding dwarf is 1:1.

Homozygous tall plant cross with a dwarf plant

  • If a homozygous tall plant is crossed with a dwarf plant, the ratio of plants in offsprings is All heterozygous tall.

Gametes formed by F1 progeny

  • The number of different types of gametes that can be formed by F_1 progeny resulting from the cross: AA BB CC x aa bb cc is 8.

Monohybrid cross question

  • In F1-generation with genotype (AABbCC) on selfing of this plant what is the phenotypic ratio in F2-generation is 3:1.

Types of gametes from the organism

  • The number of types of gametes are expected from the organism with genotype AABBCC is One.

TTRr x ttrr cross

  • Due to the cross between TTRr x ttrr the resultant progenies showed 50% plants would be tall, red flowered.

Aa BB X aaBB cross

  • A cross between Aa BB X aaBB yields a genotypic ratio of 1 AaBB: 1 aaBB.

True breeding line

  • The odd one w.r.t. true breeding line is Shows expression for few generations only.

Test Cross

  • Find out genotype of Unknown parent in F_2 gen".
  • Unknown parent x Recessive Parent
  • Phenotypic ratio: 1:1 (Tall: Dwarf)
  • No. of phenotype : 2
  • Genotypic ratio: 1(Tt): 1(tt)
  • No. of genotype : 2
  • Phenotypic & genotypic Ratio are same.

Test Cross Example

  • Examples of Test Cross (NCERT)
  • Flower colour in pea:
    • Violet (D) white (R)
    • Find out genotype of Onknown parent.
    • all violet Flower
    • 1:1

Law of Inheritance

  • On Basis of monohybnd cross. Some Rules are given principle of inhentence
    • 1st Law: Law of dominance
    • 2nd law: Law of segregation

Law of Dominance

  • Characters are controlled by discrete units called factors/gene.
  • Factors occur in pairs (TT/Tt/tt).
  • In a dissimilar pair of factors one member of the pair dominates (dominant) the other (recessive).
  • The law of dominance is used to explain the expression of only one of the parental characters in a monohybrid cross in the F1, and the expression of both in the F2.
  • It also explains the proportion of 3:1 obtained at the F_2.
  • It is not Universal law, Some Exception is there.

Law of Segregation

  • This law is based on the fact that the alleles do not show any blending and that both the characters are recovered as such in the F2 generation though one of these is not seen at the F1 stage.
  • Though the parents contain allele do not show Blending
  • Universal law
  • Melosis: gamete formation:
  • Chromosome: seperate/segregate in such a way So that each gamete Receive only one Chromosome / one allele/ one factor
  • Homozygous & Hetrozygous produce different gametes.

Ratio of monohybrid test cross:

  • The genotypic and phenotypic ratio is 1:1.

Terminologies

  • Mendel's experiments, colour of seed, colour of flower, position of flower, colour of pod, height of stem, are called Phenotype.

Mendel's Principle of segregation

  • Mendel's Principle of segregation means that the gamete cells always receive one of the paired alleles.

Result of test cross:

  • More than one option is correct, Equal number of genotypes and phenotypes in progeny, Phen = 2 =Tall, dwarf and Geno= 2 = Tt, tt

Select the incorrect match:

  • Alleles = Slightly different forms of two genes (incorrect), Alleles = slightly different forms of same gene.

law of dominance Exception of Mendelism

  • Incomplete Dominance
  • Example Antirrhinum majus/dog Flower/snapdragon
    • Mirabilis Jalapa/4 o'clock plant
  • None of Two allele are domment
  • F_1 do not show Resemblance to any parent
  • Mixing of allele/colour/Blending Intermediate
  • Both phenotypic & genotypic Ratio are Same (1:2:1).

Multiple Allele

  • More than Two allele for a gene control one character.
  • Out of three allele: only two present on Homologous chromasome.
  • Human Blood group (phenotype) = 4
  • Human Blood group (genotype) = 6
  • Character : ABO Blood group

Types of Blood Group

  • Types of blood which can express in both homozygous and heterozygous condition.
  • Type of blood group which express in homozygous condition
  • Type of blood group which can express only in homozygous condition

Codominance

  • Both allele equally express, independently,
  • \F_1 Resemble to Both parents equally,
  • No mixing of allele / No Blending
  • Co-Dominance Examples:
    • Coat Colour in cattle
    • Sickle cell anemia

Mother and Father blood group-related Questions

  • If mother is B: Hetrozygous and Father is A: Heterozygous, possible blood groups in offsprings?
  • If Mother is O and Father is A: Homozygous, possible blood groups in offsprings?

ABO Blood group:

  • ABO Blood group controlled by one gene I, controlled by three allele I^A, I^B, I^O, Follow codominance I^AI^B, Follow multiple allele

Concept of Dominance

  • Dominant allele or Wild allele or Normal allele produce Normal enzyme that control particular phenotype (Character).
  • Modified allele/mutant allele either produce Non-Functional enzyme or No enzyme, so phenotypic character is change/altered. This is Recessive allele.

Pleiotropy

  • One gene control more than one character.
  • Ex: Starch synthesis in pea plant
  • SEEDS Size Control.
  • Shape of Seed.
  • Dominance is not autonomais feature of gene.
  • A Same gene control: more than one character.
  • size of seed (phenotype): Study: incomplete donnance
  • shape of seed (phenotype): study: Complete dominance

Inheritance of Two gene

  • Study of 2 gene = 2 character = 4 Trait.
  • DICOTYLEDONOUS CROSS
  • Male & female parent
  • Phenotypic Ratio 9:3:3:1
  • Genotypic Ratio 1:2:1:2:4:2:1:2:1

Dihybrid Cross

  • Round, yellow Seed Colour (RRYY) x Wrinkled green seed Colour (rryy).
  • F_1 = RrYy (Round yellow).

Ratio of trait

  • Two dominant trait controlled by same gene in DC: False.
  • Two recessive trait controlled by same gene in DC: False.
  • Ratio of two dominant trait controlled by two different gene is 12:4 = 3:1.
  • Ratio of two recessive trait controlled by two different gene is 12:4 = 3:1.
  • The number of plant which are heterozygous for both trait is 4.

Law of Independent assortment

  • Law of Independent assortment Can't Be explained by Monohybind cross ,given on Basis of Dihybind cross.
  • Exception: LINKAGE

Dihybind cross basis

  • On Basis of Dihybnd cross Law of Dom", Law of segregation and law of indep ass. Can be exp.

Monohybrid condition Questions:

  • Which is monohybrid condition (One: Hetrozygous)

Dihybrid condition questions:

  • Which represent dihybrid condition 2 Hetrozy

Fgeneration question

  • Total number of plants in Fgeneration where all alleles are in identical condition (Dihybrid cross).
  • Total number of plants in F2 generation which resemble with F1 generation phenotypically (Dibybrid cross).
  • Total number of