RESPIRATORY (BREARTH) SYSTEM

Breathing is done by all organisms to produce energy.

Respiration is the exchange of Oxygen (O₂) and carbon dioxide (CO₂) gaseous that is done by all of living organisms to produce energy and oxidation waste of carbon dioxide and water vapour.

Human respiratory organs consists of nose, pharynx, larynx, trachea and lung.

1. Nose (Nasal Cavity)

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

                                                                                                                

 

pressure

PRESSURE in solid

What happen if you sticking the sharp nail compared with if you sticking to dull nail through the same force?

Off course easier  you sticking the sharp nail compared with if you sticking to dull nail, because the force acted an the first nail is smaller compared to the force acted on the second nail, so the pressure produced by the first nail is smaller compared to the pressure produced by the second nail.

Based on the explanation above:

      Pressure is magnitude of the force acting on the body for each unit surface area of pressure field.

CONCLUSION:  The magnitude of pressure on a body depends on the surface field and magnitude of force acted.

Mathematically, pressure is formulated as follow

      F

P=-----

      A

In which : P = pressure (N/m²)

                 F = force (N)

                 A = plane area/luas bidang (m²)

In SI system pressure un it is force unit per area unit, namely:

So : 1 Pa = 1 N/m²

In cgs (=centimeter gram second) system pressure unit is dyne/cm²

	1 atm = 76 cm Hg  1 N/m2 = 100.000 dyne/cm2

 

Exercise:

1. A body of pressure area 6 m² is given force of 54N. what is the pressure produced?

2. A cube has edge of 10 cm with pressure of 40 N/m². Calculate what is the magnitude of force?

3. A child of mass of 30 kg stays on the floor with base area of his feet each of  75 cm². If earth gravitational accelereation is 10 m/s². What is the magnitude of pressure of the child to the floor?

PRESSURE IN LIQUID

What happen, if dam is not strong to hold water pressure?

So, liquid also has pressure and the pressure in water to every direction.

Make hole at the plastic bottle by using nail vertically from the bottom to up word in parallel for 3 hole. In which hole does the water spout the most distance?

Of course, the most distance water spout come from the most bottom hole.

So the pressure in liquid is larger if its position is deeper.(semakin dalam)

Liquid pressure in motionless is called hydrostatics pressure.

The magnitude of hydrostatics pressure can be measured by using a HARTL DEVICE

Hydrostatics pressure depends on density, height or depth of liquid, also Earth gravitational acceleration, so it can be formulated as follow:

 

In which :   P  = Hydrostatics pressure (N/m² or Pa)

                   ρ  = density of liquid (kg/m³)

                   g  = earth gravitational acceleration (m/s²)

                   h = height or deep of the liquid (m)

exercise:

			Calculate hydrostatic pressure of water at point t (see figure besides). With depth of  h = 20 cm from water surface, if  given density of water = 500 kg/3 and earth gravitational acceleration = 10 m/s2.

 

 

1. Pascal Law:

Pascal law

Expressed that:

Pressure given to liquid in closed chamber will be continued by the liquid to every direction with uniform and the same magnitude

The pascal’s law a lot of used to make hydrolytic device which is used removing and altering the magnitude of force, exampleby giving a little force to obtaining a large force.

Such as :   - hydraulic  jack (dongkarak hidrolik)

                 - hydrolic lifter (alat pengangkat hidrolik)

                 - hydrolic brake (rem hidrolik)

Piston I has section area of A_1,                Piston II has section area of A₂

                                    A₁           <           A₂

Since A₂ > A₁ then F₂ > F_1. This causes the force working on section A₂ is large A₁

Exercise:

A car of weight 6000 N is lifted by using car lifter device. Section area of the big piston is 300 cm², while section area of small piston is 60 cm². What is the force that must be done in the small piston?

Connected vessel

If the connected vessel is filled by the same liquid and in balance condition, liquid surface in the vessel is located on one flat plane.

Connected vessel does not hold if:

1 . The vessel is filled with more than one kind of liquid

2.  The vessel is in closed condition

3. there is capillary pipe

                           P₁ = P₂

                   ρ₁ g.h₁= ρ₂ g.h₂

                       ρ₁ h₁= ρ₂ h₂

Exercise:

A connected vessel contains two liquids which are not mixed, namely and water (as in the figure). Given density of water is ρ_a = 2 g/cm³ and density ρ_m = 0,50 g/cm³. If the height h_a= 5 cm, find the height of oil surface h_m.

Known: ρ_a = 2 g/cm³

             ρ_m = 0,50 g/cm³

             h_a= 5 cm

Ask : h_m = ?

Aswer:

Archimedes Law

Stated that:

“ a body which is immersed partly or entirely into liquid, will get buoyancy force which the magnitude is equal to the weight of liquid moved the object”.

Mathematically Archimedes law is

                    F_A = ρ.V.g

In which:

F_A = buoyancy force / gaya ke atas (N)

ρ = density of liquid(kg/m³)

V= volume of liquid moved (m³)

g = Earth gravitational acceleration (m/s²)

Exercise:

A body of volume 6000 cm³ and density 1200 kg/m³ is immersed in the water at all. Density of water is 1000 kg/m³, Earth gravitational acceleration is g = 10 m/s². What is the magnitude of the buoyancy experienced by the body all and the weight of the body in the water?

Answer:

Glass beaker filled by water, When stone inserted to glass beaker, do volume of water changed?

Actually, Abody if into liquid some are

a. sinking (tenggelam)= the body is lies on the base of a liquid

b. suspending (melayang)= the body is immersed at all but does not reach base of the liquid,      or

c. floating (terapung) = if the body is immersed only a part in the liquid, so volume of liquid moved by the body (V_a) is smaller than volume of the body

sinking,

                                            

                                        w > F_a

                                  m_b . g > m_a . g

                       (V_b . ρ_b) . g > (V_a . ρ_a) . g

                         Because V_b = V_a

                                      ρ_b >  ρ_a

in which:

ρ_a =  density of liquid (N/kg)

ρ_b = density of the body

V_a =volume part of the body immer (m³)

V_b = volume of the body

suspending,

                                        w = F_a

                                  m_b . g = m_a . g

                       (V_b . ρ_b) . g = (V_a . ρ_a) . g

                         Because V_b = V_a

                                      ρ_b =  ρ_a

Floating

 

                                        w < F_a

                                  m_b . g < m_a . g

                       (V_b . ρ_b) . g < (V_a . ρ_a) . g

                         Because V_b = V_a

                                      ρ_b <  ρ_a

       

 

Exercise

A body is floating in the water. Part of the body that appear on water surface is 75% of its volume. If the density of water is 1 g/cm³, what is the density of the body?

From explanation above can be concluded several things as follow:

a. a body can sink, if the density of body is large than the density of liquid

b. a body can suspend, if the density of the body is equal to density of the liquid

c. a body can float, if the density of the body is smaller than the density of the liquid.

PRESSURE IN GAS

What is The meaning of:

a. Troposphere: 

b. Stratosphere

c. Mesosphere

d. Thermosphere

e. Exosphere

known: density of mercury = 13.600 kg/m³

                   g = 9,8 N / Kg

             height of mercury coloum = 76 cm = 0,76 m

ask Pressure of mercury= ?

answer : P = ρ. g . h       

              P = 100.000 Pa

Relationship between height of place and air pressure

The lower position of a place from see level, the large air pressure

In which : h = height of place above sea level

                 x = pressure of place above sea level

Exercise:

Someone is in a place wuth certain height. After measured by using a barometer , the mercury height in the tube shows the number of 66 cm. What is the height of the place?

Air Pressure Measuring Device

The device used to measure air pressure is barometer.

Kinds of barometer:

a. mercury barometer : is the device to measure air pressure in open air.

b. water barometer: working principle is the same with mercury barometer

                    P = ρ_air . g . h_air

In which:              

ρ = density of liquid(kg/m³)

g = Earth gravitational acceleration (m/s²)

h = height of mercury coloumn

c. Aneroid barometer (metal) for  to measure air pressure by using metal

The Use of atmospheric pressure

a. The time when sucking beverage by using straw

     That caused atmosphere pressure outside of the straw push the beverage so it can go up through the straw to the mount.

b. When releasing condensed from the can

c. Rubber Sucker

d. Injector

The device to measure air pressure in closed chamber is called manometer

Kinds of manometer:

a. Liquid Manometer

b. metal manometer

BOYLE’S LAW

“At constant temperature, gas pressure in closed chamber is inversely with its volume”.

The statement above is known with boyle law.

                          P . V = C

In which:

                        P = pressure

                        V = volume

                       C = Constant

Or can be written as follows:

                       P₁  . V₁ = P₂ . V₂

Several example : water pump, air pump, and bicycle pump

Exercise:

A closed chamber of volume 5 m³ has gas pressure of  2.5 atm. If the pressure is made 10 atm, what is its volume now?