- Concentration can be expressed in several ways including mass/volume percent, mass/mass percent, volume/volume percent, parts per million (ppm), and molar concentration (mol/L).
- Mass/volume percent is the mass of solute divided by the volume of solution. Volume/volume percent is the volume of solute divided by the volume of solution.
- Mass/mass percent is the mass of solute divided by the mass of solution. Parts per million (ppm) and parts per billion (ppb) express very small concentrations.
- Molar concentration expresses the number of moles of solute per liter of solution and is the standard unit for expressing concentration in chemistry
Organic compounds are almost 60% of all compounds. because of carbons tendency to form a compound as it has more than1 electron(4electrons) to form covallent compounds. SO a wide range of everything we eat is formed from carbon and hydrogen, which is the second important element to form organic compounds.
Heterogeneous and Homogeneous Mixtures are discussed in this presentation. High School chemistry, physical science, environmental science, earth systems, and material science students will benefit from this presentation. All essential introductory concepts are presented here.
CHEMICAL REACTION
CHEMICAL EQUATION
CHEMICAL FORMULA
BALANCING
TYPES OF CHEMICAL REACTION
COLLISION THEORY
FACTORS AFFECTING THE RATE OF CHEMICAL REACTION
It is a powerpoint presentation that discusses about the lesson or topic: Percentage Composition. It also talks about the definition, concepts and examples about the Percentage Composition.
Organic compounds are almost 60% of all compounds. because of carbons tendency to form a compound as it has more than1 electron(4electrons) to form covallent compounds. SO a wide range of everything we eat is formed from carbon and hydrogen, which is the second important element to form organic compounds.
Heterogeneous and Homogeneous Mixtures are discussed in this presentation. High School chemistry, physical science, environmental science, earth systems, and material science students will benefit from this presentation. All essential introductory concepts are presented here.
CHEMICAL REACTION
CHEMICAL EQUATION
CHEMICAL FORMULA
BALANCING
TYPES OF CHEMICAL REACTION
COLLISION THEORY
FACTORS AFFECTING THE RATE OF CHEMICAL REACTION
It is a powerpoint presentation that discusses about the lesson or topic: Percentage Composition. It also talks about the definition, concepts and examples about the Percentage Composition.
Volumetric analysis is the process of determining the concentration of solution of unknown concentration with the help of solution of known concentration.
This is the power point presentation for the students of class XII. This includes: Types of solutions, concentration of solutions, Solution of solid in liquid, solution of gas in liquid: Henry's law, vapour pressure of solutions, Raoult's law, Ideal & non ideal solutions, azeotropic mixtures, Colligative properties - (1) relative lowering of vapour pressure of solution of volatile solute, (2) elevation in boiling point of solution (3) depression in freezing point of solution (4) osmotic pressure, abnormal molar mass of solute, Van't Hoff's factor, numerical problems.
THE IMPORTANCE OF MARTIAN ATMOSPHERE SAMPLE RETURN.Sérgio Sacani
The return of a sample of near-surface atmosphere from Mars would facilitate answers to several first-order science questions surrounding the formation and evolution of the planet. One of the important aspects of terrestrial planet formation in general is the role that primary atmospheres played in influencing the chemistry and structure of the planets and their antecedents. Studies of the martian atmosphere can be used to investigate the role of a primary atmosphere in its history. Atmosphere samples would also inform our understanding of the near-surface chemistry of the planet, and ultimately the prospects for life. High-precision isotopic analyses of constituent gases are needed to address these questions, requiring that the analyses are made on returned samples rather than in situ.
Introduction:
RNA interference (RNAi) or Post-Transcriptional Gene Silencing (PTGS) is an important biological process for modulating eukaryotic gene expression.
It is highly conserved process of posttranscriptional gene silencing by which double stranded RNA (dsRNA) causes sequence-specific degradation of mRNA sequences.
dsRNA-induced gene silencing (RNAi) is reported in a wide range of eukaryotes ranging from worms, insects, mammals and plants.
This process mediates resistance to both endogenous parasitic and exogenous pathogenic nucleic acids, and regulates the expression of protein-coding genes.
What are small ncRNAs?
micro RNA (miRNA)
short interfering RNA (siRNA)
Properties of small non-coding RNA:
Involved in silencing mRNA transcripts.
Called “small” because they are usually only about 21-24 nucleotides long.
Synthesized by first cutting up longer precursor sequences (like the 61nt one that Lee discovered).
Silence an mRNA by base pairing with some sequence on the mRNA.
Discovery of siRNA?
The first small RNA:
In 1993 Rosalind Lee (Victor Ambros lab) was studying a non- coding gene in C. elegans, lin-4, that was involved in silencing of another gene, lin-14, at the appropriate time in the
development of the worm C. elegans.
Two small transcripts of lin-4 (22nt and 61nt) were found to be complementary to a sequence in the 3' UTR of lin-14.
Because lin-4 encoded no protein, she deduced that it must be these transcripts that are causing the silencing by RNA-RNA interactions.
Types of RNAi ( non coding RNA)
MiRNA
Length (23-25 nt)
Trans acting
Binds with target MRNA in mismatch
Translation inhibition
Si RNA
Length 21 nt.
Cis acting
Bind with target Mrna in perfect complementary sequence
Piwi-RNA
Length ; 25 to 36 nt.
Expressed in Germ Cells
Regulates trnasposomes activity
MECHANISM OF RNAI:
First the double-stranded RNA teams up with a protein complex named Dicer, which cuts the long RNA into short pieces.
Then another protein complex called RISC (RNA-induced silencing complex) discards one of the two RNA strands.
The RISC-docked, single-stranded RNA then pairs with the homologous mRNA and destroys it.
THE RISC COMPLEX:
RISC is large(>500kD) RNA multi- protein Binding complex which triggers MRNA degradation in response to MRNA
Unwinding of double stranded Si RNA by ATP independent Helicase
Active component of RISC is Ago proteins( ENDONUCLEASE) which cleave target MRNA.
DICER: endonuclease (RNase Family III)
Argonaute: Central Component of the RNA-Induced Silencing Complex (RISC)
One strand of the dsRNA produced by Dicer is retained in the RISC complex in association with Argonaute
ARGONAUTE PROTEIN :
1.PAZ(PIWI/Argonaute/ Zwille)- Recognition of target MRNA
2.PIWI (p-element induced wimpy Testis)- breaks Phosphodiester bond of mRNA.)RNAse H activity.
MiRNA:
The Double-stranded RNAs are naturally produced in eukaryotic cells during development, and they have a key role in regulating gene expression .
Cancer cell metabolism: special Reference to Lactate PathwayAADYARAJPANDEY1
Normal Cell Metabolism:
Cellular respiration describes the series of steps that cells use to break down sugar and other chemicals to get the energy we need to function.
Energy is stored in the bonds of glucose and when glucose is broken down, much of that energy is released.
Cell utilize energy in the form of ATP.
The first step of respiration is called glycolysis. In a series of steps, glycolysis breaks glucose into two smaller molecules - a chemical called pyruvate. A small amount of ATP is formed during this process.
Most healthy cells continue the breakdown in a second process, called the Kreb's cycle. The Kreb's cycle allows cells to “burn” the pyruvates made in glycolysis to get more ATP.
The last step in the breakdown of glucose is called oxidative phosphorylation (Ox-Phos).
It takes place in specialized cell structures called mitochondria. This process produces a large amount of ATP. Importantly, cells need oxygen to complete oxidative phosphorylation.
If a cell completes only glycolysis, only 2 molecules of ATP are made per glucose. However, if the cell completes the entire respiration process (glycolysis - Kreb's - oxidative phosphorylation), about 36 molecules of ATP are created, giving it much more energy to use.
IN CANCER CELL:
Unlike healthy cells that "burn" the entire molecule of sugar to capture a large amount of energy as ATP, cancer cells are wasteful.
Cancer cells only partially break down sugar molecules. They overuse the first step of respiration, glycolysis. They frequently do not complete the second step, oxidative phosphorylation.
This results in only 2 molecules of ATP per each glucose molecule instead of the 36 or so ATPs healthy cells gain. As a result, cancer cells need to use a lot more sugar molecules to get enough energy to survive.
Unlike healthy cells that "burn" the entire molecule of sugar to capture a large amount of energy as ATP, cancer cells are wasteful.
Cancer cells only partially break down sugar molecules. They overuse the first step of respiration, glycolysis. They frequently do not complete the second step, oxidative phosphorylation.
This results in only 2 molecules of ATP per each glucose molecule instead of the 36 or so ATPs healthy cells gain. As a result, cancer cells need to use a lot more sugar molecules to get enough energy to survive.
introduction to WARBERG PHENOMENA:
WARBURG EFFECT Usually, cancer cells are highly glycolytic (glucose addiction) and take up more glucose than do normal cells from outside.
Otto Heinrich Warburg (; 8 October 1883 – 1 August 1970) In 1931 was awarded the Nobel Prize in Physiology for his "discovery of the nature and mode of action of the respiratory enzyme.
WARNBURG EFFECT : cancer cells under aerobic (well-oxygenated) conditions to metabolize glucose to lactate (aerobic glycolysis) is known as the Warburg effect. Warburg made the observation that tumor slices consume glucose and secrete lactate at a higher rate than normal tissues.
Observation of Io’s Resurfacing via Plume Deposition Using Ground-based Adapt...Sérgio Sacani
Since volcanic activity was first discovered on Io from Voyager images in 1979, changes
on Io’s surface have been monitored from both spacecraft and ground-based telescopes.
Here, we present the highest spatial resolution images of Io ever obtained from a groundbased telescope. These images, acquired by the SHARK-VIS instrument on the Large
Binocular Telescope, show evidence of a major resurfacing event on Io’s trailing hemisphere. When compared to the most recent spacecraft images, the SHARK-VIS images
show that a plume deposit from a powerful eruption at Pillan Patera has covered part
of the long-lived Pele plume deposit. Although this type of resurfacing event may be common on Io, few have been detected due to the rarity of spacecraft visits and the previously low spatial resolution available from Earth-based telescopes. The SHARK-VIS instrument ushers in a new era of high resolution imaging of Io’s surface using adaptive
optics at visible wavelengths.
A brief information about the SCOP protein database used in bioinformatics.
The Structural Classification of Proteins (SCOP) database is a comprehensive and authoritative resource for the structural and evolutionary relationships of proteins. It provides a detailed and curated classification of protein structures, grouping them into families, superfamilies, and folds based on their structural and sequence similarities.
2. Concentration = amount of solute per
quantity of solvent
Mass/volume % = Mass of solute (g) x 100%
Volume of solution (mL)
CONCENTRATION AS A MASS/VOLUME PERCENT
Usually for solids dissolved in liquids
3. SAMPLE PROBLEM:
2.00mL of distilled water is added to 4.00g of a
powdered drug. The final volume is 3.00mL. What is the
concentration of the drug in g/100mL of solution? What
is the percent (m/v) of the solution?
CONCENTRATION AS A MASS/VOLUME PERCENT
4. = (4.00g) x 100%
(3.00mL)
CONCENTRATION AS A MASS/VOLUME PERCENT
SAMPLE PROBLEM:
2.00mL of distilled water is added to 4.00g of a powdered drug. The final volume
is 3.00mL. What is the concentration of the drug in g/100mL of solution? What is
the percent (m/v) of the solution?
Mass/volume % = Mass of solute x 100%
Volume of solution
Therefore the mass/volume percent is 133%
= 133%
this also means that the concentration is 133g/100mL
5. 1.7% = Mass of solute x 100%
(2000mL)
SAMPLE PROBLEM:
Many people use a solution of Na3PO4 to clean walls before
putting up wallpaper. The recommended concentration is 1.7%
(m/v) . What mass ofNa3PO4 is needed to make 2.0L of solution?
CONCENTRATION AS A MASS/VOLUME PERCENT
Mass/volume % = Mass of solute x 100%
Volume of solution
Mass of solute = 34g
Therefore the mass required is 34g
6. CONCENTRATION AS A MASS/MASS PERCENT
Concentration = amount of solute per
quantity of solvent
Mass/Mass % = Mass of solute (g) x 100%
Mass of solution (g)
Usually for solids dissolved in liquids
7. Mass/Mass % = Mass of solute x 100%
Mass of solution
SAMPLE PROBLEM:
Calcium chloride, CaCl2, can be used instead of road salt to melt the
ice on roads during the winter. To determine how much calcium
chloride had been used on a nearby road, a student took a sample of
slush to analyze. The sample had a mass of 23.47g. When the
solution was evapourated, the residue had a mass of 4.58g (assume
that no other solutes were present).
a) What was the mass/mass percent of calcium chloride in the slush?
b) How many grams of calcium chloride were present in 100g solution?
CONCENTRATION AS A MASS/MASS PERCENT
8. Mass/Mass % = Mass of solute x 100%
Mass of solution
= (4.58g) x 100%
(23.47g)
= 19.5%
Therefore the mass/mass percent of calcium chloride is
19.5%
SAMPLE PROBLEM:
Calcium chloride, CaCl2, can be used instead of road salt to melt the ice on roads during the winter. To
determine how much calcium chloride had been used on a nearby road, a student took a sample of
slush to analyze. The sample had a mass of 23.47g. When the solution was evapourated, the residue had
a mass of 4.58g (assume that no other solutes were present).
a) What was the mass/mass percent of calcium chloride in the slush?
CONCENTRATION AS A MASS/MASS PERCENT
9. If your mass/mass % of CaCl2 is 19.5% and you
have 100g, then you have 19.5g of CaCl2
Ex: 19.5% x 100g = 19.5g
SAMPLE PROBLEM:
Calcium chloride, CaCl2, can be used instead of road salt to melt the ice on roads during the winter. To
determine how much calcium chloride had been used on a nearby road, a student took a sample of
slush to analyze. The sample had a mass of 23.47g. When the solution was evapourated, the residue had
a mass of 4.58g (assume that no other solutes were present).
b) How many grams of calcium chloride were present in 100g solution?
CONCENTRATION AS A MASS/MASS PERCENT
10. Concentration = amount of solute per
quantity of solvent
Volume/Volume % = Volume of solute (mL) x 100%
Volume of solution (mL)
CONCENTRATION AS A VOLUME/VOLUME PERCENT
Usually for liquids dissolved in liquids
11. CONCENTRATION AS A VOLUME/VOLUME PERCENT
SAMPLE PROBLEM:
Rubbing alcohol is sold as a 70% (v/v) solution of isopropyl alcohol in
water. What volume of isopropyl alcohol is used to make 500mL of
rubbing alcohol?
Volume/Volume % = volume of solute x 100%
volume of solution
(70%) = volume of solute x 100%
(500mL)
= 350mL
Therefore the volume is 350mL
12. CONCENTRATION AS PARTS PER MILLION (ppm)
ppm = Mass of solute (g) x 106
Mass of solution (g)
Usually mass/mass relationships BUT does not refer to
the number of particles
Mass of solute (g) = x
Mass of solution (g) 106
g of solution
OR
13. CONCENTRATION AS PARTS PER BILLION (ppB)
ppb = Mass of solute (g) x 109
Mass of solution (g)
To the power of 9 instead…
Mass of solute (g) = x
Mass of solution (g) 109
g of solution
OR
14. CONCENTRATION AS PARTS PER BILLION (ppB)
SAMPLE PROBLEM:
A fungus that grows on peanuts produces a deadly toxin.
When ingested in large amounts, this toxin destroys the liver
and can cause cancer. Any shipment of peanuts that contains
more than 25ppb of this fungus is rejected. A company
receives 20 tonnes of peanuts to make peanut butter. What is
the maximum mass (in g) of fungus that is allowed?
15. CONCENTRATION AS PARTS PER BILLION (ppB)
SAMPLE PROBLEM:
A fungus that grows on peanuts produces a deadly toxin. When ingested in large amounts,
this toxin destroys the liver and can cause cancer. Any shipment of peanuts that contains
more than 25ppb of this fungus is rejected. A company receives 20 t of peanuts to make
peanut butter. What is the maximum mass (in g) of fungus that is allowed?
ppb = Mass of fungus (g) x 109
Mass of peanuts (g)
Convert 20 t to grams:
20t x 1000kg/t x 1000g/kg = 20 x 106
g
25ppb = Mass of fungus x 109
20 x 106
g
25ppb = Mass of fungus x 103
20
500 = Mass of fungus
103
0.5g = Therefore the maximum mass of the
fungus allowed is 0.5g
17. MOLAR CONCENTRATION
The number of moles of solute in 1L of solution
(this is the standard measure of concentration in chemistry!!!)
Molar concentration (mol/L) = moles of solute
Volume of solution (L)
C = n
V
Simplified…
18. MOLAR CONCENTRATION
SAMPLE PROBLEM:
A saline solution contains 0.90g of NaCl dissolved in 100mL
of solution. What is the molar concentration of the solution?
C = n
V
Given:
m = 0.90g
n = m/M = 0.90g / 58.44g/mol = 0.0154mol
V = 100mL = 0.1L
C = (0.0154mol)
(0.1L)
= 0.154mol/L
Therefore the molar concentration of the saline solution is 0.15 mol/L
19. MOLAR CONCENTRATION
SAMPLE PROBLEM:
At 20°C, a saturated solution of CaSO4 has a concentration of
0.0153mol/L. A student takes 65mL of this solution and
evaporates it. What mass (in g) is left in the evapourating
dish?
C = n
V
Given:
C = 0.0153mol/L
V = 65mL = 0.065L
n = ? 0.0153mol/L = n
(0.065L)
= 0.000994mol
NOT DONE YET! Convert to grams!
MCaSO4
= 136.15g/mol
mCaSO4
= n x M
= 0.000994mol x 136.15g/mol
= 0.135g
Therefore 0.14g of
CaSO4 are left in the
evapourating dish