Chlorophyll was extracted from the leaf to test for the presence of products from photosynthesis. After detaching the leaf, all chemical reactions need to be stopped to accurately analyze what was produced during photosynthesis. Iodine is used to test for starch - a blue-black color indicates starch is present. While the leaf initially showed no color change with iodine, it later turned blue-black, suggesting starch was present and photosynthesis had occurred, though this test does not indicate starch is the first product formed. Other products besides starch may be revealed through different tests.
Distingish test for aldehyde and ketoneTanay Debnath
I have discussed all possible tests for distinguishing aldehydes and ketones. At first I have discussed the detection tests of carbonyl groups. Two important detection tests I have discussed, 1. 2,4-DNP test. 2. Sodium bisulfite test. After that I have discussed distinguish test for aldehydes..Here I have discussed 1. Fehling test 2. Benedict test. 3. Tollen's test or silver mirror test.4. Schiff test, 5. Chromic acid test. After that I have discussed two tests to distinguish aldehyde and ketone having alpha hydrogen..The tests are 1.nitropusside test 2. Janovsky test. Finally I have discussed about Iodoform test through which aldehyde and ketone having CH3C=O group can be detected.
Distingish test for aldehyde and ketoneTanay Debnath
I have discussed all possible tests for distinguishing aldehydes and ketones. At first I have discussed the detection tests of carbonyl groups. Two important detection tests I have discussed, 1. 2,4-DNP test. 2. Sodium bisulfite test. After that I have discussed distinguish test for aldehydes..Here I have discussed 1. Fehling test 2. Benedict test. 3. Tollen's test or silver mirror test.4. Schiff test, 5. Chromic acid test. After that I have discussed two tests to distinguish aldehyde and ketone having alpha hydrogen..The tests are 1.nitropusside test 2. Janovsky test. Finally I have discussed about Iodoform test through which aldehyde and ketone having CH3C=O group can be detected.
Determination of enthalpy of ionisation of acetic acidMithil Fal Desai
The acetic acid is a weak acid as it does not completely dissociate in dilute aqueous solutions into hydrogen (H+) and acetate (CH3COO-) ions. When acetic acid is neutralized with a strong base (NaOH), heat is evolved during the neutralization that is used in the process of dissociating the acetic acid further that will facilitate the completion of neutralization.
CH3COOH + NaOH = CH3COONa + H2O + heat
The enthalpy change associated with the neutralization of acetic acid with a strong base is lower than that of the enthalpy of neutralization of a strong acid with a strong base. The difference in the enthalpy of neutralization of a strong acid (HCI) with a strong base (NaOH) and enthalpy of neutralization of weak acid (CH3COOH) with a strong base (NaOH) is the enthalpy of ionization of the weak acid (CH3COOH).
MCQ on Chemical Reactions and Equations PDFAbdur Rohman
The PDF file contains MCQ (Multiple Choice Questions) related to Chemical Reactions and Equations. Chemical Reactions and Equations is the Chapter 1 of Class 10 Science (Chemistry).
Photosynthesis and the leaf - starch testSHANIQUECAREY
This document contains information on photosynthesis including the: raw materials needed, conditions and products. The steps needed by science students to carry out a starch test may also be found in this document. Lastly, you will find a critical thinking question involving a variegated leaf.
Determination of enthalpy of ionisation of acetic acidMithil Fal Desai
The acetic acid is a weak acid as it does not completely dissociate in dilute aqueous solutions into hydrogen (H+) and acetate (CH3COO-) ions. When acetic acid is neutralized with a strong base (NaOH), heat is evolved during the neutralization that is used in the process of dissociating the acetic acid further that will facilitate the completion of neutralization.
CH3COOH + NaOH = CH3COONa + H2O + heat
The enthalpy change associated with the neutralization of acetic acid with a strong base is lower than that of the enthalpy of neutralization of a strong acid with a strong base. The difference in the enthalpy of neutralization of a strong acid (HCI) with a strong base (NaOH) and enthalpy of neutralization of weak acid (CH3COOH) with a strong base (NaOH) is the enthalpy of ionization of the weak acid (CH3COOH).
MCQ on Chemical Reactions and Equations PDFAbdur Rohman
The PDF file contains MCQ (Multiple Choice Questions) related to Chemical Reactions and Equations. Chemical Reactions and Equations is the Chapter 1 of Class 10 Science (Chemistry).
Photosynthesis and the leaf - starch testSHANIQUECAREY
This document contains information on photosynthesis including the: raw materials needed, conditions and products. The steps needed by science students to carry out a starch test may also be found in this document. Lastly, you will find a critical thinking question involving a variegated leaf.
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.
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.
Nutraceutical market, scope and growth: Herbal drug technologyLokesh Patil
As consumer awareness of health and wellness rises, the nutraceutical market—which includes goods like functional meals, drinks, and dietary supplements that provide health advantages beyond basic nutrition—is growing significantly. As healthcare expenses rise, the population ages, and people want natural and preventative health solutions more and more, this industry is increasing quickly. Further driving market expansion are product formulation innovations and the use of cutting-edge technology for customized nutrition. With its worldwide reach, the nutraceutical industry is expected to keep growing and provide significant chances for research and investment in a number of categories, including vitamins, minerals, probiotics, and herbal supplements.
Professional air quality monitoring systems provide immediate, on-site data for analysis, compliance, and decision-making.
Monitor common gases, weather parameters, particulates.
Earliest Galaxies in the JADES Origins Field: Luminosity Function and Cosmic ...Sérgio Sacani
We characterize the earliest galaxy population in the JADES Origins Field (JOF), the deepest
imaging field observed with JWST. We make use of the ancillary Hubble optical images (5 filters
spanning 0.4−0.9µm) and novel JWST images with 14 filters spanning 0.8−5µm, including 7 mediumband filters, and reaching total exposure times of up to 46 hours per filter. We combine all our data
at > 2.3µm to construct an ultradeep image, reaching as deep as ≈ 31.4 AB mag in the stack and
30.3-31.0 AB mag (5σ, r = 0.1” circular aperture) in individual filters. We measure photometric
redshifts and use robust selection criteria to identify a sample of eight galaxy candidates at redshifts
z = 11.5 − 15. These objects show compact half-light radii of R1/2 ∼ 50 − 200pc, stellar masses of
M⋆ ∼ 107−108M⊙, and star-formation rates of SFR ∼ 0.1−1 M⊙ yr−1
. Our search finds no candidates
at 15 < z < 20, placing upper limits at these redshifts. We develop a forward modeling approach to
infer the properties of the evolving luminosity function without binning in redshift or luminosity that
marginalizes over the photometric redshift uncertainty of our candidate galaxies and incorporates the
impact of non-detections. We find a z = 12 luminosity function in good agreement with prior results,
and that the luminosity function normalization and UV luminosity density decline by a factor of ∼ 2.5
from z = 12 to z = 14. We discuss the possible implications of our results in the context of theoretical
models for evolution of the dark matter halo mass function.
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.
1. Photosynthesis 2.02
Experiment 2. Discussion
I What is the reason for extracting the chlorophyll from the leaf?
2 If one is testing a leaf for the products of photosynthesis, why is it necessary to arrest
all chemical reactions after detaching it ?
3 For what substance is iodine a test ? What result do you see if this substance is
present ?
4 What was the colour of the leaf (a) immediately before adding iodine, (b) after adding
iodine?
5 How do you interpret this change ?
6 In subsequent experiments we are going to claim that the presence of starch in a leaf
is strong evidence that photosynthesis has taken place. Why is the result in the present
experiment not good evidence of this ?
7 What products of photosynthesis might be present which are not revealed by this test?
8 Assuming that the leaf has, in fact, been photosynthesizing and has produced starch,
do your results indicate whether starch is the first product of photosynthesis to be
formed?
inflammable
vapour
BUNSEN OUT
boiling
alcohol
leaf hot water
Extracting chlorophyll from a leaf
