HEREDITY

HEREDITY

The Organisms that still exist until now have adaptive characteristic inherited from their parents.

Offspring can be produced by reproduction process both asexsual and sexual.

Asexual reproduction creates offspring  that are genetically identical to their parent.

Sexual reproduction creates offspring has combination traits gained from both parents. 

Offspring produced from combination traits because male’s traits are bequeathed through sperm, while female’s traits are bequeathed through ovum.

The traits of male or female are bequeathed I, that is heredity material.

THE DEFINITION

Heredity material is a material that contains traits or bequeathed information.

Heridity material in living organism is DNA (=deoxyribose nucleic acid)

DNA : - Containts of polynucleotide chain.

-         Double helix

Each nucleotide consists of:

                      DNA                                                                  RNA

a.     Phosphate                                                   a) Phosphate

b.    Ribosa                                                         b) Ribosa

c.     Nitrogen base, consist of :                           c) Nitrogen base consist of:

a)     Purin consist of : adenine and guanine         a) purin : A and G

b)    Pirimidin consist of : timin and cytosine      b)Pirimidin: Urasin and C

Chromosome of structure is sentromer and chromosome body

Base on the place of sentromer, chromosome devided 4 kinds, that is:

1. Metasentric is sentromer there are in central

2. Submetasentric: chromosome body has the different long and curved shaped

3. Akrosentric : sentromer there has the different long and straight line shaped

4. Telosentric, : sentromer is the tip of chromosome

Pea plants were used as experimental material because it has some advantages as follow:

a.     Easy to grow

b.     Producing a lot of seeds

c.      Mature at fast rate berumur pendek

d.     Self pollinated

e.      Easy to pollinate

f.       Has seven discrete characteristic : tall-short of plant, inflated -constricted of pod shape, round-wrinkled of seed shape, yellow-green of seed, flower position in axial-terminal, purple-white flower, green-yellow of pod color

Trait =sifat

Recessive Traits is - the traits suppressed  in the F₁ generation

                              - Symbolized with small letter

                              - The closing of the trait

Dominant traits is  - traits cover up the other traits

                              - Symbolized with capital letter

                              - The trait that appears.

Example:

Mendel’s try  to cross monohybrid, the Tall pea plant is more dominant from short

Tall use of symbolized with capital letter “ T ” and it’s must be a pair “ TT ”

Short use of symbolize  with small letter “ t ” and it’s must be a pair “tt”

Crossing tall pea plants with short pea plants. Allowing plants growing from F₁ seed to self-pollinate (F₁ X F₁) to produce the second generation (F₂)

Genotypic  is gene formation (allele) which determines traits

Phenotypic is trait that observable  by human senses

Based on the result of  the crossing above, it means the genotypic ratio and phenotypic ratio is ………….

Homozygote (=pure-breeding) is genotype be set either by similar gene a pair

Heterozygote is genotype be set either by different gene a pair.

Mendel’s first law say “ During gamete formation, the gene separated into two offspring cells”.

Homozygous Dominant means the genotype is set (composed) by dominant gene a pair. For example: TT------ is tall

Homozygous recessive means the genotype is set (composed) by recessive gene a pair. For example: tt

Heterozygous means the genotype is set (composed) by dominant gene and recessive gene. For example: Tt

Mendel conducted experiment of two discrete trait ( dihybrid ) and many discrete traits ( polyhybrid ) and he found that each trait does not depend on other trait.

For example: the height of pea plant is not influenced by seed form.

Mendel’s second law: “ in fertilization, segregating genes assort (will join) independently with other genes”.

P = parental, that is parents

F = filial is offspring produced by marriage.

Exercise:

Pea plant has tall of  homozygous and round homozygote of seed shape traits dominant crossed with short homozygote and wrinkled homozygote of seed shape traits recessive. Calculate F₂ (= F₁ x F₁) ratio phenotypic and genotypic of traits above!

Answer:

TT = tall of dominant homozygous on pea plant

RR = round of dominant homozygous on pea plant

tt = short of recessive homozygous on pea plant

rr = wrinkled of recessive homozygous on pea plant

Inheritance pattern (pola pewarisan sifat)

1) Monohybrid Cross = that is cross pollination with two different property

A) dominant monohybrid cross has characteristics as follow:

     1) Traits of dominant genes completely suppress recessive genes

     2) All F₁ generation have similar phenotype to parent with dominant genes

     3) The F₂ generation have 3 : 1 ratio of dominant phenotype and recessive phenotype

B) Intermediate Monohybrid Cross (codominance)

     1) Traits of dominant genes completely covers recessive genes

     2) Heterozygote offspring express intermediate traits, which are the mix of dominant and recessive traits.

     3) The F₂ generation have 1 : 2 : 1 rotio of dominant phenotype and recessive phenotype

        Conclusion of  Intermediate property is The pair of properties that did not show full dominance or full recessive property.

Exercise

 To cross flower of Mirabilis jalapa plant between red color dominant homozygote with white color recessive intermediately to result F₁ (pink flower). Calculate ratio of  Phenotypic and Genotypic of  F₂ from F₁ x F₁!

c)     Test cross = that is cross to done between individual that wanted to be known of their genotype with individual that had recessive homozygote genotype.

EXERCISE:

. If organism of genotype curl hair (CC) is crossed with recessive normal hair.

a)     determine F₁

b)    determine to comparation F₁ that crossed with in test cross method.

2)  Dihybrid

                   Is Crossing with two different properties.

      For example :  a yellow pea of round seed with wrinkled green seed. Round seed is dominant over wrinkled seed and yellow color is dominant over green color.

The number of gametes, genotype of  F₂ and phenotype of  F₂ are determined by the following formula:

a)     Formula to determine the number of gamete = 2ⁿ.

n = The number of different properties

Example : The genotype for round seed BB, Then its gamete is B and B or Bb genotype, then its gametes are B and b.

b)    Formula to determine the number of  kinds of  F₂ Genotype (3ⁿ)

F₂ from the number of different properties is 1 the number of kind F₂ genotype is 3¹ = 3

c)     Formula to determine the number of kinds of  F₂ phenotype = (2ⁿ) 

2) Dihybride cross

Exercise:

Pea plant has tall of  homozygous and round homozygote of seed shape traits dominant crossed with short homozygote and wrinkled homozygote of seed shape traits recessive. Calculate ratio phenotypic and genotypic of traits above!

3) Backcross and Testcross

    Backcross is a crossing of an offspring with one of  its parents.

    Testcross is the crossing occurs between the offspring with recessive pure breeding

Mendel’s Laws:

1)    Mendel’s law 1 = law of gene segregation that has allele (segregation)

2)    Mendel’s law 2 = law of gene classification.

The application of properties inheritance

1.     Discovery of superior germ (penemuan bibit unggul)

2.     The determination of sex (penentuan jenis kelamin)

3.     Inheritance of sex-related properties (pewarisan sifat terpaut sex)

1) Superior germ is germ that has good properties according to human needs, where this good properties can be collected in one individual through cross marriage.

2) The determination of sex.

- Sex kinds of  its heredity is determined by a pair of sex chromosomes.

    - Sex chromosome in human is separated into X chromosome and Y chromosome.

    - If women is XX, and in man is XY.

    - Y = that determines the formation of male, if Y is not found so a female is born.

    - The number of human chromosome is 46 (=23 pairs), consisting of 22 pairs of autosomes and 1 pair of sex chromosomes.

3) Inheritance of sex-related properties  consisting of:

    a) Color blindness (buta warna) = is recessive properties.

        X^CY = Normal man

        X^cY = color blindness man

        X^CX^C = normal woman

        X^CX^c = carier woman

        X^c X^c = color blindness woman

        Exercise : in a marriage between woman of color blindness properties carrier with normal man, calculate ratio phenotypic and ghenotypic on F₁!

    B) Hemophilia (hemofili)

        - The characteristic , if there is injury then the blood is hard to coagulate.

        - the properties carrier, ---its genotypes as follows:

             - X^HY = Normal man

             - X^hY = hemophilic man

             - X^HX^H = normaly woman

             - X^HX^h = carier woman

             - X^h X^h = hemophilic woman

        Exercive : Determine to produce  the crossing a woman of hemophilic properties carrier with a man of hemophilic sufferer:

    C) Albinism (albinisme / albino)

        - in Albinis is a heredity disease brought by recessive gen

        - in albino, that recessive gene is not related to sex chromosome, but in  autosom chromosome.

        - the properties carrier, ---its genotypes as follows:

         AA = normaly human

         Aa = Carier human

         aa = suffers albino (penderita albino)

        Exercise:

        The crossing  of  husband and wife of albino property carrier and the other is carrier of albino property, calculate ratio phenotypic and genotypic on F₁!

d)    Blood group / blood type:

Several possibilities of genotype found in each blood group are as follow:

-         Blood group A : I^AI^A = Blood group A homozygote

                         I^AI^o = Blood group A heterozygote

-         Blood group B : I^BI^B = Blood group B homozygote

                         I^BI^o = = Blood group B heterozygote

-         Blood  group AB : I^AI^B

-         Blood group O : I^oI^o

Exercise:

1.     Determine the blood group of children if the father has blood group A of homozygote and mother has blood group B heterozygote.

2.     Determine the blood group of children if the father has blood group A of heterozygote and mother has blood group B heterozygote.

Exercise:

Four grandchildren were born in the family. Two of male grandchildren have hemophilia. Determine the genotype of the other members of the family! Copy this following chart and write down the genotype in the box or circle!

                                                                                                             = male

 

                                                                                                             = female