Based on the reactivity with Tollen’s, Benedict’s or Fehling’s reagent, carbohydrates are classified as;
Reducing sugars
Carbohydrates that can reduce Tollen’s, Benedict’s or Fehling’s reagents are called reducing sugars (sugar with free aldehyde or ketone group). All monosaccharides and most of the disaccharides are reducing sugars. Some examples are Maltose and Lactose.
Non-reducing sugars
Carbohydrates that cannot reduce Tollen’s, Benedict’s or Fehling’s reagents are called non-reducing sugars. Sucrose is a non-reducing sugar.
Based on the reactivity with Tollen’s, Benedict’s or Fehling’s reagent, carbohydrates are classified as;
Reducing sugars
Carbohydrates that can reduce Tollen’s, Benedict’s or Fehling’s reagents are called reducing sugars (sugar with free aldehyde or ketone group). All monosaccharides and most of the disaccharides are reducing sugars. Some examples are Maltose and Lactose.
Non-reducing sugars
Carbohydrates that cannot reduce Tollen’s, Benedict’s or Fehling’s reagents are called non-reducing sugars. Sucrose is a non-reducing sugar.
This ppt explains the properties of monosaccharides, polysaccharides. the properties like mutarotation, reduction, optical activity, caramerlization, osazone is given in the ppt. Also the determination of ring size of the monosaccharide is explained/
Test for carbohydrates, qualitative test of carbohydrates, identification tes...RajkumarKumawat11
Test for carbohydrates, qualitative test of carbohydrates, identification test of carbohydrates, carbohydrates test , carbohydrates by Raj kumar, qualitative test of carbohydrates for students
This slide will help you to understand about chemical reactions of monosaccharides and Disaccharides. The carbohydrate can can undergo several reactions like oxidation, reduction, esterification, dehydration and tautomerization to give various products.
This ppt explains the properties of monosaccharides, polysaccharides. the properties like mutarotation, reduction, optical activity, caramerlization, osazone is given in the ppt. Also the determination of ring size of the monosaccharide is explained/
Test for carbohydrates, qualitative test of carbohydrates, identification tes...RajkumarKumawat11
Test for carbohydrates, qualitative test of carbohydrates, identification test of carbohydrates, carbohydrates test , carbohydrates by Raj kumar, qualitative test of carbohydrates for students
This slide will help you to understand about chemical reactions of monosaccharides and Disaccharides. The carbohydrate can can undergo several reactions like oxidation, reduction, esterification, dehydration and tautomerization to give various products.
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.
The increased availability of biomedical data, particularly in the public domain, offers the opportunity to better understand human health and to develop effective therapeutics for a wide range of unmet medical needs. However, data scientists remain stymied by the fact that data remain hard to find and to productively reuse because data and their metadata i) are wholly inaccessible, ii) are in non-standard or incompatible representations, iii) do not conform to community standards, and iv) have unclear or highly restricted terms and conditions that preclude legitimate reuse. These limitations require a rethink on data can be made machine and AI-ready - the key motivation behind the FAIR Guiding Principles. Concurrently, while recent efforts have explored the use of deep learning to fuse disparate data into predictive models for a wide range of biomedical applications, these models often fail even when the correct answer is already known, and fail to explain individual predictions in terms that data scientists can appreciate. These limitations suggest that new methods to produce practical artificial intelligence are still needed.
In this talk, I will discuss our work in (1) building an integrative knowledge infrastructure to prepare FAIR and "AI-ready" data and services along with (2) neurosymbolic AI methods to improve the quality of predictions and to generate plausible explanations. Attention is given to standards, platforms, and methods to wrangle knowledge into simple, but effective semantic and latent representations, and to make these available into standards-compliant and discoverable interfaces that can be used in model building, validation, and explanation. Our work, and those of others in the field, creates a baseline for building trustworthy and easy to deploy AI models in biomedicine.
Bio
Dr. Michel Dumontier is the Distinguished Professor of Data Science at Maastricht University, founder and executive director of the Institute of Data Science, and co-founder of the FAIR (Findable, Accessible, Interoperable and Reusable) data principles. His research explores socio-technological approaches for responsible discovery science, which includes collaborative multi-modal knowledge graphs, privacy-preserving distributed data mining, and AI methods for drug discovery and personalized medicine. His work is supported through the Dutch National Research Agenda, the Netherlands Organisation for Scientific Research, Horizon Europe, the European Open Science Cloud, the US National Institutes of Health, and a Marie-Curie Innovative Training Network. He is the editor-in-chief for the journal Data Science and is internationally recognized for his contributions in bioinformatics, biomedical informatics, and semantic technologies including ontologies and linked data.
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.
This pdf is about the Schizophrenia.
For more details visit on YouTube; @SELF-EXPLANATORY;
https://www.youtube.com/channel/UCAiarMZDNhe1A3Rnpr_WkzA/videos
Thanks...!
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 .
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.
Slide 1: Title Slide
Extrachromosomal Inheritance
Slide 2: Introduction to Extrachromosomal Inheritance
Definition: Extrachromosomal inheritance refers to the transmission of genetic material that is not found within the nucleus.
Key Components: Involves genes located in mitochondria, chloroplasts, and plasmids.
Slide 3: Mitochondrial Inheritance
Mitochondria: Organelles responsible for energy production.
Mitochondrial DNA (mtDNA): Circular DNA molecule found in mitochondria.
Inheritance Pattern: Maternally inherited, meaning it is passed from mothers to all their offspring.
Diseases: Examples include Leber’s hereditary optic neuropathy (LHON) and mitochondrial myopathy.
Slide 4: Chloroplast Inheritance
Chloroplasts: Organelles responsible for photosynthesis in plants.
Chloroplast DNA (cpDNA): Circular DNA molecule found in chloroplasts.
Inheritance Pattern: Often maternally inherited in most plants, but can vary in some species.
Examples: Variegation in plants, where leaf color patterns are determined by chloroplast DNA.
Slide 5: Plasmid Inheritance
Plasmids: Small, circular DNA molecules found in bacteria and some eukaryotes.
Features: Can carry antibiotic resistance genes and can be transferred between cells through processes like conjugation.
Significance: Important in biotechnology for gene cloning and genetic engineering.
Slide 6: Mechanisms of Extrachromosomal Inheritance
Non-Mendelian Patterns: Do not follow Mendel’s laws of inheritance.
Cytoplasmic Segregation: During cell division, organelles like mitochondria and chloroplasts are randomly distributed to daughter cells.
Heteroplasmy: Presence of more than one type of organellar genome within a cell, leading to variation in expression.
Slide 7: Examples of Extrachromosomal Inheritance
Four O’clock Plant (Mirabilis jalapa): Shows variegated leaves due to different cpDNA in leaf cells.
Petite Mutants in Yeast: Result from mutations in mitochondrial DNA affecting respiration.
Slide 8: Importance of Extrachromosomal Inheritance
Evolution: Provides insight into the evolution of eukaryotic cells.
Medicine: Understanding mitochondrial inheritance helps in diagnosing and treating mitochondrial diseases.
Agriculture: Chloroplast inheritance can be used in plant breeding and genetic modification.
Slide 9: Recent Research and Advances
Gene Editing: Techniques like CRISPR-Cas9 are being used to edit mitochondrial and chloroplast DNA.
Therapies: Development of mitochondrial replacement therapy (MRT) for preventing mitochondrial diseases.
Slide 10: Conclusion
Summary: Extrachromosomal inheritance involves the transmission of genetic material outside the nucleus and plays a crucial role in genetics, medicine, and biotechnology.
Future Directions: Continued research and technological advancements hold promise for new treatments and applications.
Slide 11: Questions and Discussion
Invite Audience: Open the floor for any questions or further discussion on the topic.
4. Table1.2 Benedict’s test for reducing sugars
Compound Color Reducing Sugar (yes/no)
water blue No
glucose red-orange Yes
fructose red-orange Yes
sucrose blue-green No
lactose red-orange Yes
maltose reddish brown Yes
starch blue No
glycogen orange Yes
Benedict’s test for reducing sugar is a test that determines the amount and presence of
reducing sugar in a particular solution.
Sugars are classified as either reducing or non-reducing sugar based on their ability to act
as a reducing agent during Benedict’s test. The reagent in Benedict’s test for reducing sugar is
the Benedict’s solution which contains copper (II), sodium carbonate, and sodium citrate. The
copper solution gives a blue coloration. It reacts with the electrons from the ketone or aldehyde
group of the free reactive carbonyl group on the carbohydrate to form cuprous oxide (a red-
brown precipitate). This precipitate is formed due to the reduction of the Cu2+ ions to Cu+ ions,
In the process, the carbohydrate is oxidized. The color of the precipitate is dependent on the
volume of reducing sugar present. This means that the more free carbonyl groups present, the
increase in the amount of precipitate formed.
The reducing agent for reducing sugars is the aldehyde functional group (COOH), the
formyl group. Reducing sugars have either an aldehyde functional group or a ketone group in an
open chain form which is then converted into an aldehyde.
Reducing sugars are simple sugars and they include all monosaccharides and most
disaccharides. Examples of monosaccharides are glucose, fructose and galactose and examples
5. for reducing disaccharides are the lactose and maltose. Disaccharides sucrose is not a reducing
sugar. In fact, sucrose is the most common non-reducing sugar.,
Based on the Benedict’s test that was conducted, water and starch had no change in color.
Therefore, water and starch have no non-reducing sugars present. Glucose and fructose had its
change color of red-orange while the food sample which is the cake turned orange; this means
that a moderate amount of reducing sugar is present on the said compounds. On the other hand,
maltose had its color change of reddish brown; this means that a large amount of non-reducing
sugar was present. We didn’t obtain the accurate result for lactose, but lactose is a reducing
sugar.
Table 2.3 Barfoed’s test for monosaccharides
Compound Color Monosaccharides (yes/no)
water blue No
glucose light orange Yes
fructose red orange Yes
sucrose blue No
lactose blue No
maltose blue No
starch blue No
food sample (cake) blue No
Barfoed’s test is used to distinguish monosaccharides from disaccharides.
Monosaccharides are reducing sugars and they react faster with the reagent in order to form such
color change. Whereas, reducing disaccharides reacts slower than monosaccharides.
Based on the results that were obtain from the experiment, water, sucrose, lactose,
maltose, starch, and the food sample had the same color of blue upon heating. Only glucose and
fructose had a change in its color. Glucose turned into light orange while the fructose turned into
red orange. This means that only glucose and fructose were monosaccharides. Barfoed’s test
6. used copper (II) ions in a slightly acidic medium, reducing monosaccharides were oxidized by
the copper ion in solution in order to form carboxylic acid and an orange precipitate of copper
(Ii) within five minutes. Reducing disaccharides goes the same reaction but do so at a slower
rate. The non-reducing sugar gave a negative result. Barfoed’s reagent comprises of cupric and
acetic acid in solution. Monosaccharides readily react with the reagent in order to cause a
reduction in the 𝐶𝑢2+
ions to 𝐶𝑢+
ions forming𝐶𝑢2 𝑂. The reason behind this is that
monosaccharides, they oxidized readily in weak acid solutions. Disaccharides can as well reduce
the cupric ions; however, its reaction is much slower.
Table2.4 Seliwanoff’s test for ketohexoses
Compound Color Ketohexose/Aldohexose
water light orange Aldohexose
glucose light pink Aldohexose
fructose red Ketohexose
sucrose red Ketohexose
lactose light pink Aldohexose
maltose light pink Aldohexose
starch light pink Aldohexose
food product (cake) flesh colored Aldohexose
Seliwanoff’s test is used to distinguish between aldohexose and ketohexose. Seliwanoff’s
reagent consists of resorcinol crystals dissolved in equal amounts of water and hydrochloric acid.
Upon heating the solutions, the formation of a red precipitate indicates a positive result.
Based on the experiment, only fructose and sucrose had the positive result. They both
formed a red precipitate which indicated that it is a ketohexose. Other compounds had light
pink/light orange color; this indicates that they are aldohexose.
The acid hydrolysis of polysaccharides and oligosaccharides yields simple sugars. The
dehydrated ketose then reacts with the resorcinol in order to produce a red color. Aldose may
7. react slightly to produce a light pink color. Fructose and sucrose are two common sugars that
yield to a positive result. Sucrose is a disaccharide that consist fructose and glucose, that is why
it had a positive result. The test reagent dehydrated ketohexose in order to form 5-
hydroxymethylfurfural. Aldohexose reacts slowly much slower that ketohexose to give 5-
hydroxymethylfurfural, that is why it had only a light red color. Once the 5-
hydroxymethylfurfural was produced, it then reacts with resorcinol that gave a dark red
condensation product. Sucrose hydrolyzes to give fructose, which eventually reacted, that is why
it produced a dark red color.
Table 2.5 Iodine test for polysaccharides
Compound Color Polysaccharides (yes/no)
water yellow No
glucose light orange No
fructose light orange No
sucrose yellow No
lactose red orange No
maltose red orange No
starch black Yes (amylase)
food product (cake) black Yes (amylase)
Iodine test is used for the detection of starch in the solution. The black color is due to the
formation of starch-iodine complex. Starch contain polymer of α-amylose and amylopectin
which forms a complex with iodine to give the black color.
Based on the experiment, only the starch and the food product cake had the positive
result. They both formed into a black precipitate. Therefore, both the food product and starch are
polysaccharides. Other compounds had a negative result.
8. Starch is composed of a straight chain subunit know as the amylase. Amylose is
responsible for the reaction with iodine. The amylase subunit of starch is made up of a straight
chain of alpha-glucose monomers that are connected by alpha 1á4 glycosidic bonds. In solution,
the amylase exists as a helically coiled structure. When the iodine was added, the molecules of
iodine became trapped within the helical structure and formed a complex one. Iodine has usually
a black color that causes the black coloration of starch which indicates that amylase is present.
The same with the food product cake, cake also result to a black precipitate because cake has an
ingredient of starch that had caused its black coloration.
Table 2.6 Hydrolysis of di- and polysaccharides
Compound Benedict’s Iodine
Sucrose(hydrolyzed) Yellow orange light yellow
Starch (hydrolyzed) blue black
This test is performed in order to breakdown to the monomer monosaccharide units. The
hydrolysis of the sucrose enabled the following results; in benedict’s test, the solution reduced
the 𝐶𝑢2+
ions to 𝐶𝑢+
that resulted to the formation of the yellow orange precipitate. This means
that there was the presence of a reducing sugar in the solution, and since sucrose is not a
reducing sugar; the hydrolysis had created a monosaccharide that contains a free reactive
carbonyl group. In the Iodine’s test, the presence of amylase facilitated the reaction with the
reagent to form a black compound.
HCl breakdown the 1á2 glycosidic bond that links the glucose and fructose monomer
together to form sucrose. This break causes the release of two monosaccharides. Sodium
hydroxide was then added in order to neutralize the pH of the solution since the presence of acid
in the solution might disrupt the alkalinity of the benedict’s and iodine reagent that were added
after. Once it were hydrolyzed, the two reducing sugar were formed. This means that when
benedict’s test was performed, the solution was able to reduce the reagent and formed a yellow
9. orange precipitate. In Iodine’s test, the presence of amylase gave a positive indication which
means that it is a polysaccharide.
On the case of starch, after adding a significant amount of hydrochloric acid to the starch
solution and heating the test tubes for a while, the iodine proved that a starch is a polysaccharide
and there is the presence of amylase because it formed into a black precipitate.
Once enough sodium hydroxide was added to the starch solution, it became slightly
basic. This experiment showed that starch can be rehydrated with HCl and heat. This breaks the
molecules up into single glucose molecules again causing the positive result in iodine’s test. H=
molecules break up the glucose molecules. However, because these have strong bonds, heat was
performed in order to create movement and weaken the bonds. This same process can also be
applied to sucrose in order to achieve a similar result, but this will happen much faster because
there are just enough bonds in sucrose compared to starch.
Compare the results of the Benedict’s test with sucrose before and after hydrolysis. Why are they
different?
Before hydrolysis, sucrose formed into a blue-green precipitate, whereas after hydrolysis it
formed into a yellow precipitate.
Sucrose is called a non-reducing sugar because it does not reduce copper sulphate, thus the
Benedict’s solution cannot break down the bonds of sucrose like other sugars. If sucrose is
hydrolyzed to its constituent monosaccharides (glucose and fructose), it will give a positive
Benedict's test. So sucrose is the only sugar that will give a negative Benedict's test before
hydrolysis and a positive test afterwards. Boiling the test solution with dilute hydrochloric acid
for a few minutes will hydrolyze the glycosidic bond forming the two monosaccharides or
(reducing sugars) that will now be capable of reacting with the Benedict's solution.
10. Compare the results of the Benedict’s test with starch before and after hydrolysis. Why are they
different?
Before hydrolysis, starch formed a blue precipitate, whereas after hydrolysis it had a different
color. Benedict’s reagent reacts with reducing sugars in order to produce a precipitate. Since
starch is not a reducing sugar, Benedict’s test had a negative result. However, the hydrolysis of
starch produces glucose which will react to Benedict’s reagent. As a result, benedict’s test after
hydrolysis had a positive result.
Compare the results of the Iodine test with starch before and after hydrolysis. Why is there a
difference?
Before and after hydrolysis on iodine’s test, both starch turned into a black precipitate which
indicates a positive result because of the presence of amylase.
Table 2.7 Fermentation test
Carbohydrates Height of the bubbles Fermented (yes/no)
Glucose 3.3 cm Yes
Fructose 2.1 cm Yes
Sucrose 2.0 cm Yes
Lactose 1.8 cm Yes
Maltose 1.4 cm Yes
Starch 6.5 cm Yes
Fermentation test using yeast is used to determine which carbohydrates substrates
positively influence yeast fermentation since yeast seems to have greater ability to utilize certain
carbohydrates.
Fermentation is a process that is important in anaerobic conditions when there is no
oxidative phosphorylation to maintain the production of ATP (Adenosine triphosphate) by
11. glycolysis. During fermentation pyruvate is metabolised to various different compounds.
Homolactic fermentation is the production of lactic acid from pyruvate; alcoholic fermentation is
the conversion of pyruvate into ethanol and carbon dioxide; and heterolactic fermentation is the
production of lactic acid as well as other acids and alcohols. Typical examples of fermentation
products are ethanol, lactic acid, and hydrogen. However, more exotic compounds can be
produced by fermentation, such as butyric acid and acetone. Although the final step of
fermentation (conversion of pyruvate to fermentation end-products) does not produce energy, it
is critical for an anaerobic cell since it regenerates nicotinamide adenine dinucleotide (NAD),
which is required for glycolysis. This is important for normal cellular function, as glycolysis is
the only source of ATP in anaerobic conditions.
Based on the results that were obtained from the experiment, all compounds were
fermented. On the height of bubbles of each of the compounds, glucose was 3.3 cm, fructose was
2.1 cm, sucrose was 2.0 cm, lactose was 1.8 cm, maltose was 1.4 cm and starch was 6.5 cm.
Table 2.8 Analysis of food products
Food sample Benedict’s Test Barfoed’s Test Seliwanoff’s
Test
Iodine Test
biscuit + + - +
bread + + - +
rice + - - +
junk food
(piattos)
+ + - +
pork chop - - - +
cake + + - +
For the food samples that were brought by each group, the results on each tests had a
positive result and a negative result. The food samples biscuit, bread, and cake had a positive
12. result in Benedict’s, Barfoed’s, and Iodine test whereas on Seliwanoff’s test, it had a negative
result. On the other hand, the food sample rice had a positive result on both the Benedict’s test
and Iodine test but it had a negative result in both Barfoed’s and Seliwanoff’s test. The junk food
(Piattos) had a positive result on the three tests; the Benedict’s, Barfoed’s, and Iodine test but had
a negative result on Seliwanoff’s test. Pork chop had a negative effect in Barfoed’s, Benedict’s,
and Seliwanoff’s test but had a positive result on Iodine test.
Conclusion
Carbohydrates are the most abundant class of bioorganic molecules on planet earth.
Although their abundance in the human body is relatively low, carbohydrates constitute about
75% by mass of dry plant materials. Green (chlorophyll-containing) plants produce
carbohydrates via photosynthesis. In this process, carbon dioxide from the air and water from the
soil are the reactants, and sunlight absorbed by chlorophyll is the energy source.
There are several tests to use in order to determine a sample whether it is a carbohydrate
or not, if it is a monosaccharide or a disaccharide etc.
Some of the several tests use in determining carbohydrates includes, Benedict’s test for reducing
sugars, Barfoed’s test for monosaccharides, Seliwanoff’s test for ketohexose, Iodine test for
polysaccharides and hydrolysis of di- and polysaccharides.
Benedict’s test for reducing sugar is a test that determines the amount and presence of
reducing sugar in a particular solution. Barfoed’s test is used to distinguish monosaccharides
from disaccharides. Seliwanoff’s test is used to distinguish between aldohexose and ketohexose.
Iodine test is used for the detection of starch in the solution. Lastly, the hydrolysis of di- and
polysaccharides used to breakdown to the monomer monosaccharides.
Fermentation is also a test that can be used in carbohydrates. Fermentation test using
yeast is used to determine which carbohydrates substrates positively influence yeast
fermentation. Those following test that were stated above can help you determine the most
efficient methods for utilizing these test when dealing with carbohydrates.
