StudySheetMBG353
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Chemistry
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Name of
Media
Test For
Organism
Observation
Of Media
Results/Conclusion
Lactose
Broth
Lactose
1
Still Red, No Gas
formation
(-/-), fermentation -ve
2
Red to yellow
color, Bubble
trapped
(+/+), fermentation +ve
Glucose
Broth
Glucose
1
Still Red, no gas
formation
(-/-) fermentation -ve
2
Red to Yellow
Colour
Bubble trapped
(+/+) fermentation +ve
Starch Agar
Plate
Presence of
amylase
1
Clear Zone Around
Growth
(+,starch hydrolyzed +
ve amylase)
2
Blue Colour
Formed around
growth
-,starch not hydrolyzed
-ve amylase
Urea
Slant
Presence
Of urease
1
Bright Pinkish
purple on slant
+, urease +ve
2
No colour change
still yellow
-,urease -ve
Tryptophan
Broth
Test for
Indole
1
No change, still
work
-
2
Red Ring Formed
on top
+
MR
Broth
Mixed
Acid
Fermen
tation
1
No Red formation,
still Clear
-
2
Red to Light Red
colour
+
VP
Broth
Acetyl
Methyl
Carbinol
presence
1
Light Red colour
formed
+
2
No colour change,
still clear
-
Citrate
Citrate
1
Bright blue color
+
Slant
growth
on slant
2
No colour change,
Still Dark Blue
-
2. The IMVIC pattern for the bacterium E.coli as follows: ++--
In the same manner, write down the IMVIX patterns for organisms #1 and #2 from the
results in your table:
#1 - -++
#2 ++- -
1.
What is the function of Iodine in the starch hydrolysis test?
The function of Iodine is to react with starch to form a dark blue-colored complex.
If the area is clear, that means the organism broke down the starch that its due to
its production of amylase.
2.
What is the function of the Durham tube in the carbohydrate broth?
The function of the Durham tube is to detect the gas production.
Gas will be trapped in Durham tube and a bubble will form. The presence of a
bubble indicates fermentation and gas production of fermentation.
3.
a) Both the MR and VP tests determine whether a bacteria can use an organic
compound as a source of energy. What is this compound?
b) Even though the energy source is the same, the MR and VP tests determine
separate metabolic. What are the end products of these pathways for each test?
For the MR, it will be mixed stable acids (lactic, acetic, formic acid).
For the VP, it will be acetyl methyl carbinol.
4.
How is indole production detected in the indole test?
The presence of indole is present when a red ring is present at the surface of the
broth.
The presence of Indole when a red ring is formed at the top of the broth.
Starch Agar Plate:
Testing for Amylase
Typtophan Broth:
Test for Tryptophan broth transforms to Indole via the addition of kovac reagent
MRVP Broth
Mixed Acid Fermentation
Metabolizes acids to other acids, from glucose
Lactic acid, Formic acid, etc
VP
Acetyl methyl, carbinol from glucose
Metabolizes pyruvic acid to other acids (ie. lactic, formic, and acetic acid), from
glucose
Lactate, Formic, etc
VP
Neutral products like Acetyl methyl carbinol and ethyl alcohol from glucose
metabolism
Interpretation of Results
Record your results in the table. Follow the interpretation guidelines below.
Carbohydrate Fermentation
Positive test for fermentation = acid production (yellow colour). You must also
record
whether or not gas is produced. Do this by checking the inverted tube for the presence
of a “good size” air bubble.
Negative test for fermentation = original red/orange colour (no acid has been produced).
Indicate your results as acid/gas.
Choose between:
+ / + (acid & gas), + / -
(acid but no gas),
or
- / - (no acid or gas)
Starch Hydrolysis
After incubation, place several drops of Gram’s iodine onto the bacterial cultures. Try to
cover both the culture and the surrounding media – ie.
aim for an ‘edge’.
Examine the plate for a clear zone surrounding the growth --- a positive result.
Negative result? – the media immediately surrounding the bacterial growth appears
deep blue.
Urea Hydrolysis
Following incubation, observe the tubes for growth and any colour changes. Positive reaction:
a deep
pink-purple colour.
IMViC Tests
For the indole test, add about 5 drops of Kovac’s reagent directly to the tubes of
tryptone broth. Do not shake the tube. Observe the tube immediately for the formation
of a bright red ring at the top of the broth (positive). If the top portion remains yellowish
or brownish, this is a negative result.
For the MR-VP tests, aseptically transfer 1 ml of each of the cultures to two separate
sterile plastic tubes.
One plastic tube will be used for the MR test, the other plastic tube
will be used for the VP test. Add the following test reagents to the tubes as indicated:
MR test:
Add 3 drops of the methyl red reagent to one tube. Positive result
= red colour
(indicates acid production)
VP test:
Add 15 drops of Barritt’s reagent A, mix well. Then add 5 drops of Barritt’s
reagent B, mix well. Let tube sit for 10-15 minutes, shaking often. Positive result
= light
red colour
For the citrate test, observe the tube for growth and colour changes. Positive result
=
growth + deep blue colour, Negative result
= no growth, media remains green
True motility
True motility (self-propulsion) may be due to structural appendages called flagella
(singular, flagellum). These are thin protein tubes attached to the bacterial cell wall
which rotate to provide propulsion for the cell. Not all bacteria have flagella; hence,
some types of bacteria are non-motile. Flagella are generally too thin to be seen in live,
unstained bacteria even at high magnification.
While it is not possible to see flagella using the light microscope, true motility can be
detected in live bacteria by observing rapid changes in cell position. True motility will
result in the bacteria spinning around (rotational movement) or traveling long distances
across the field of view (translational movement), while changing directions from time to
time.
Brownian motion and streaming
All microbes suspended in liquid medium will appear to move quickly back and forth. This type of
movement is called Brownian motion and is due to the random vibrations produced when the microbes
collide with water molecules. Brownian motion should not be confused with true motility, as even non-
motile organisms will display this type of behavior.
In this exercise you will examine live bacteria for motility using two techniques: the wet mount and the
hanging drop mount.
Motility Media
Motility media is a semi-solid growth media that is used to determine bacterial motility. It
is an example of an indirect method of motility testing. In addition to basic nutrients
motility media also contains:
•
0.5 % agar
– Routine solid media usually contains 1.5 % agar. The lower concentration of agar in
the motility media allows the bacteria to migrate away from the line of inoculation. The
presence of agar prevents diffusion of the cells, so that any distribution of bacteria throughout
the media results only from true motility.
triphenyl tetrazolium (TPT)
– tetrazolium salts are used as an indicator of bacterial
growth. During growth, bacteria are able to take up and reduce the tetrazolium,
producing an insoluble, red compound called
formazan
. In this way, the site of bacterial
growth in the tube can be easily identified by the presence of a dark red color.
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Related Questions
Which control test tubes contained reducing sugars? Are these the results consistent with the sugars tested, explain?
Sucrose and lactose are both disaccharides, explain why the test results are the same or different?
Sucrose
8 minutes
Blue color like Benedict's solution, not reaction.
-
Lactose
8 minutes
orange-red color
+
Glucose
8 minutes
Orange-red color
+
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E-book - Elementary Bioorgani ×
My Course - Elementary Bioorg X
endocrine system - Google Do X
Homework 1 Matter and Materia X
Question 12 of 23 - Homework X
assessments.macmillanlearning.com/sac/5338340#/5338340/11/1
Update :
O Assignment Score:
Lx Give Up?
O Hint
Check Answer
52.2%
Resources
< Question 12 of 23
Attempt 3
Carry out the given conversions from one metric unit of length to another.
83.5 Mm =
km
4.75 nm =
mm
* TOOLS
х10
!!!
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Couple
E (M
-0.50
Co/glucose (-0.43) 24 e
· 2H*/H2 (-0.42) 2 e
- -0.40
CO/methanol (-0.38) 6 e
NAD+/NADH (-0.32) 2 e
--0.30
CO2/acetate (-0.28) 8 e
-0.20
S/H,S (-0.28) 2 e
-0.10
so,/H,S (-0.22) 8 e
-0.0
Pyruvate/lactate (-0.19) 2 e
S,0.-/S,0,2- (+0.024) 2 e
+0.10
· Fumarate/succinate (+0.03) 2 e
-+0.20
Cytochrome boxired (+0.035) 1 e
Fe3+/Fe2* (+0.2) 1 e, (pH 7)
+0.30
Ubiquinoneox/red (+0.11) 2 e
+0.40
Cytochrome cox/red (+0.25) 1 e
+0.50
Cytochrome aoxlred (+0.39) 1 e
· NO3/NO, (+0.42) 2 e
+0.60
+0.70
NO3½N2 (+0.74) 5 e
Fe3+/Fe2+ (+0.76) 1 e, (pH 2)
+0.80
k O/H,0 (+0.82) 2 e
+0.90
Using the redox tower above, rank the following equations from most energy
produced to least energy produced (Where 1 is the most energy produced and 4 is the
least energy produced).
A. H2 + NO3 → NO2 + H20
B. C6H1206 + 602 → 6CO2 + 6H2O
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An enzyme catalyzed the reaction:
S -> P has a AG" value of +4.8 kJ/mol.
If you start with a solution containing 500 mM substrate [S] prior to the addition of enzyme and initiating the
catalyzed reaction, then what is the substrate concentration [S] after reaching equilibrium?
[R = 8.315 J/ mol* K; T = 25 C]
NOTE - Provide your numeric answer in mM units as a TWO digit significant figure, and do NOT include units in your answer (report the numeric
answer ONLY).
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|||
Select a School
Module Knowledge Check
Question 3
Make the following changes on the molecule in the drawing area below:
1. First, make sure this is the a anomer. Edit the molecule if necessary.
2. Then replace the hydroxyl group on carbon #4 with a chlorine atom.
I Don't Know
Type here to search
CH₂OH
2
X
OH
OH
H
OH
H
H
H
Submit
H
OH
II
مع
DELL
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optimise, produce and market a commercial assay kit for the enzyme-linked determination of glucose in a range of samples
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20
15
0 mM fructose-2,6-bisphosphate
I mM fructose-2,6-bisphosphate
5 mM fructose-2,6-bisphosphate
25 mM frictove-2,6-bisphosphate
Vaited
50
100 200
Fructose-1,6-bisphosphate, jM
Use the data in the graph above to calculate the percent inhibition of fructose-1,6-bisphosphatase by 25 M fructose-2,6-bisphosphate under the following
conditions:
When fructose-1,6-bisphosphate is 20 ut.
Percent inhibition - 93.33
Submit
F ose-1.6-bisphosphatase activity,
units/mg protein
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20
20
21
-000
Calculate the DGO' for the following reaction at 37 °C:
Glucose + Pi + H+ → Glucose 3-phosphate + H2O
[Glucose 3-phosphate]eq = 0.0114 μM; [Glucose] eq = 4.32 mM; [Phosphateleq =
8.83 mM;
[H2O] eq = 55.6 M; [H+] eq = 1X 10-7 M;
Type your answer...
Use the following information (T = 37 °C)
ADP+ H2O
kJ/mol
I
What is the AGO' for the following reaction?
AMP + ATP
Type your answer...
--->
-DO
AMP+Pi + H+
AG°' = -35.7
ADP+ Pi + H+
AGO' = -30.5
2ADP
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Your Learning Points 3
Osiagwu, Chiemeka K
A reaction is carried out by mixing together a solution of K,CO, and a solution of Fe(NO,),.
K,Co, (aq) + Fe(NO,), (aq) - FeCo, (s) + 2KNO, (aq)
Assume the purple particles below each represent 1 mole of particles. Construct the balanced net
ionic equation by dragging the appropriote components into the gray boxes. Drag each unreacted
spectator ion along with its balancing coefficient into the blue box that represents the aqueous
solution.
Spectator ions
Balanced Net lonic Equation
+
FeCo, (s)…
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12
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(1) Is the molecule a mixed or a simple triglyceride?
(2) What reaction took place with the TAG above if one of the products is the structure below?
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2. Go to NutritionCalc Plus and search for (USDA) beef, ground, pan browned, 30% fat. How many grams of total
fat əre provided in a 3 oz serving?
O 41 grams
O 6 grams
O 22 grams
O 15 grams
3. Go to NutritionCalc Plus and search for (USDA) beef, ground, pan browned, 30% fat and (USDA) beef, ground,
pan browned, 10% fat. Evaluate the nutrition information for a 3 oz serving. How does the saturated fat content of
the ground beef, 30% fat, compare to the saturated fat content of the ground beef, 10% fat?
O The ground beef, 30% fat, contains 4 more grams of saturated fat than the ground beef, 10% fat
O The ground beef, 30% fat, contains 8 more grams of saturated fat than the ground beef, 10% fat
O The ground beef, 30% fat, contains 6 more grams of saturated fat than the ground beef, 10% fat
O The ground beef, 30% fat, contains 2 more grams of saturated fat than the ground beef, 10% fat
4. Go to NutritionCalc Plus and search for (USDA) cheese, parmesan, grated. How many grams of total fat…
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X
The body only breaks
down protein when
energy sources are
low.
True
false
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Read the page 1,2 and 3 carefully
Lab Report
Page:-2
If a yellow/green precipitate appears, this indicates that a reducing sugar is present. If the solution
You have to make two parts one is discussion/ analysis and another one is conclusion based on these remains blue, this means an absence of a reducing sugar. Some disaccharides (e.g. sucrose) and all
polysaccharides (e.g. starch) are not reducing sugars because they do not have free aldehyde
experiment procedure and results.
groups.
1. Discussion: In discussion, you have to explain the results that are given in page:-3 based on the
experiment procedure.
2. Conclusion: Conclusion must be in 2 or 3 lines.
EXPERIMENTI
Page:-1
DETECTING COMPOUNDS MADE BY LIVING THINGS
INTRODUCTION
Specific compounds are necessary for life processes such as growth, development, reproduction
maintenance and repair. Scientists have developed a series of chemical tests to determine the
presence and absence in food samples of the building blocks of living…
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BMR (kcal) = (4.54 × weight in lb) + (15.9 × height in inches) - (5 × age in years) - 161
Determine the basil metabolic rate of a 136 lb, 5' - 4" 20 year old woman in watts. You will need to convert from kcal to joules.
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be precise on the answers
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You are supplied with the following: / Jy word voorsien van die volgende:
NaCl(Mr= 58,443 g /mol)
2.5MTris-Cl, pH 8 solution /oplossing (1 Litre)
EDTA,natriumsout(Mr= 380,2g/mol)
10% sodium dodecyl sulphate solution / natriumdodecyl sulfaat oplossing
Proteïnase K solution / oplossing (50 mg dissolved / opgelos in 1 ml ddH2O)
You need a digestion buffer consisting of the following: / Jy moet 'n verteringsbuffer op maak wat uit die volgende bestaan:
15m M NaCl
75 mM Tris-Cl,pH 8
16 mM EDTA,pH 8
0.8% sodium sulphate / natrium dodecyl sulfaat
0,75 mg/ml proteïnase K
How will you prepare 500 ml of the digestion buffer? Show all your steps and calculations. Remember to explain exactly how you will make it up.
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PROCEDURES
1. Search at least five food products at home.
2. Determine the major fat component (Saturated or Unsaturated)
RESULTS
FOOD PRODUCTS
French Fries fried using
vegetable oil
French Fries fried using
Types of Fats
Unsaturated
Advantages/Disadvantages
It has health benefits/It is costlier
Saturated
It is much cheaper/ Not healthy – fats
will be deposited in veins
derived from animal oil
--
CONCLUSIONS
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Based on the graph below what is the carrying capacity for the species represented by the BLUE line with squares.
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As some people age their skin becomes lighter in color due to reduced activity of the pigment producing cells in the skin, melanocytes.
True or False
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A Life Science-1803035-377210-1
A Life Science-1803035-377210-1
1803035.agilixbuzz.com/student/142171010/activity/b13bdb49b1d84a84a3aff52cc918e067
Todigy Jackson-Wonder E.
E Edulastic: Formativ.
RL Home
Zearn Math: Top-rat.
A Who was the capta.
Netflix M Wes
mical Reactions - Show it
> Life Science-1803035-377210-15 / Unit 7: Chemical Reactions/ Week 23 - Chemical Reactions/ Monday, Lesson 107 - 2/8/21
Identifying the Parts of a Chemical Reaction. Identify the reactants and
products in the chemical reaction below.
Identify the reactants: CH4 + 202 --> CO2 + 2H2O
Your answer
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Identify the products: CH4 - 202 --> CO2 + 2H2O
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Identify the reactants: NaCi Na CI
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Chemistry
Hi, just wondering how to calculate the glucose amount? thank you
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I need the correct answer fast
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C6H12O6 + a*O2 + b*NH3 --> c*CH1.64O0.52N0.16 + d*CO2 + e*H2O + f*C2H5OH
In the aerobic yeast growth process, overflow metabolism can be observed when the sugar feed rate exceeds the oxidizing capacity of the yeast. In this case, part of the substrate is treated anaerobically and the production of biomass is reduced and ethanol is formed. This can also be observed when the respiratory rate RQ goes above one. The chemical composition of this yeast is CH1.64O0.52N0.16 . The yield of ethanol from glucose is 0.14 g g-1 and the yield of biomass from glucose is 0.362 g g-1. Calculate the oxygen demand of the reaction.
In this calculation, use the following rounded even values as atomic weights for all calculations: C=12 g/mol, H=1 g/mol, O=16 g/mol ja N=14 g/mol.
So how many moles of oxygen are needed per mole of glucose?
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1. The fertilizer recommendation for a 1 hectare of rice farm during dry season is 120-60-60kg
N-P2O5-K20. Compute the following:
A. how many bags of Urea (46%) and Complete (14%) will be needed to satisfy the
recommendation @50kgs/bag?
B. how many kgs of Urea (46%) and Complete (14%) will be needed to satisfy the
recommendation?
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"best" to perform the following task.
Removing K+ ions from a protein sample
.
Oion exchange
salting out
differential centrifugation
O gel filtration
O dialysis
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In replicate analyses, the carbohydrate content of a glycoprotein (a protein with sugars attached to it) is found to be 12.6, 11.9, 13.0, 12.7, and 12.5 g of carbohydrate per 100 g of protein. Find the 50% and 90% confidence intervals for the carbohydrate content.
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H
H-C-OH
HO-C-H
H²-OH
H-C-OH
6
CH₂-OH
D-(+)-glucose
rotate
H
6
5 C-CH₂-OH
HI
OH
OH
I
OH C
H
21
OH
H
H
CH₂-OH
-OH
OH
H
H
OH
=
H
OH
CH₂-OH
5
H
OH
H
H
ß (a)
OH (H)
H (OH)
OH anomeric
carbon
D-glucopyranose
1. Redraw wrapping from C1-C6 clockwise. Keep all things on the right of the Fisher projection
pointing down, all things on the left pointing up.
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Chemistry
Hello! Can you please explain why these answers are wrong? Thank you.
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Related Questions
- Which control test tubes contained reducing sugars? Are these the results consistent with the sugars tested, explain? Sucrose and lactose are both disaccharides, explain why the test results are the same or different? Sucrose 8 minutes Blue color like Benedict's solution, not reaction. - Lactose 8 minutes orange-red color + Glucose 8 minutes Orange-red color +arrow_forwardPlease answer fast i give you upvote.arrow_forwardE-book - Elementary Bioorgani × My Course - Elementary Bioorg X endocrine system - Google Do X Homework 1 Matter and Materia X Question 12 of 23 - Homework X assessments.macmillanlearning.com/sac/5338340#/5338340/11/1 Update : O Assignment Score: Lx Give Up? O Hint Check Answer 52.2% Resources < Question 12 of 23 Attempt 3 Carry out the given conversions from one metric unit of length to another. 83.5 Mm = km 4.75 nm = mm * TOOLS х10 !!!arrow_forward
- Couple E (M -0.50 Co/glucose (-0.43) 24 e · 2H*/H2 (-0.42) 2 e - -0.40 CO/methanol (-0.38) 6 e NAD+/NADH (-0.32) 2 e --0.30 CO2/acetate (-0.28) 8 e -0.20 S/H,S (-0.28) 2 e -0.10 so,/H,S (-0.22) 8 e -0.0 Pyruvate/lactate (-0.19) 2 e S,0.-/S,0,2- (+0.024) 2 e +0.10 · Fumarate/succinate (+0.03) 2 e -+0.20 Cytochrome boxired (+0.035) 1 e Fe3+/Fe2* (+0.2) 1 e, (pH 7) +0.30 Ubiquinoneox/red (+0.11) 2 e +0.40 Cytochrome cox/red (+0.25) 1 e +0.50 Cytochrome aoxlred (+0.39) 1 e · NO3/NO, (+0.42) 2 e +0.60 +0.70 NO3½N2 (+0.74) 5 e Fe3+/Fe2+ (+0.76) 1 e, (pH 2) +0.80 k O/H,0 (+0.82) 2 e +0.90 Using the redox tower above, rank the following equations from most energy produced to least energy produced (Where 1 is the most energy produced and 4 is the least energy produced). A. H2 + NO3 → NO2 + H20 B. C6H1206 + 602 → 6CO2 + 6H2Oarrow_forwardAn enzyme catalyzed the reaction: S -> P has a AG" value of +4.8 kJ/mol. If you start with a solution containing 500 mM substrate [S] prior to the addition of enzyme and initiating the catalyzed reaction, then what is the substrate concentration [S] after reaching equilibrium? [R = 8.315 J/ mol* K; T = 25 C] NOTE - Provide your numeric answer in mM units as a TWO digit significant figure, and do NOT include units in your answer (report the numeric answer ONLY).arrow_forward× A ALEKS - Eniola Salvador - Know x ▸ Antibiotic Sensitivity - YouTube x A ALEKS-T www-awu.aleks.com/alekscgi/x/Isl.exe/10_u-IgNslkr7j8P3jH-IQUHIQg6bJxmeSyVpHOEB1plef9xyC5Ca9QIIX9h BC ||| Select a School Module Knowledge Check Question 3 Make the following changes on the molecule in the drawing area below: 1. First, make sure this is the a anomer. Edit the molecule if necessary. 2. Then replace the hydroxyl group on carbon #4 with a chlorine atom. I Don't Know Type here to search CH₂OH 2 X OH OH H OH H H H Submit H OH II مع DELLarrow_forward
- 6arrow_forward5 optimise, produce and market a commercial assay kit for the enzyme-linked determination of glucose in a range of samplesarrow_forwardThis question has multiple parts. Work all the parts 20 15 0 mM fructose-2,6-bisphosphate I mM fructose-2,6-bisphosphate 5 mM fructose-2,6-bisphosphate 25 mM frictove-2,6-bisphosphate Vaited 50 100 200 Fructose-1,6-bisphosphate, jM Use the data in the graph above to calculate the percent inhibition of fructose-1,6-bisphosphatase by 25 M fructose-2,6-bisphosphate under the following conditions: When fructose-1,6-bisphosphate is 20 ut. Percent inhibition - 93.33 Submit F ose-1.6-bisphosphatase activity, units/mg proteinarrow_forward
- 20 20 21 -000 Calculate the DGO' for the following reaction at 37 °C: Glucose + Pi + H+ → Glucose 3-phosphate + H2O [Glucose 3-phosphate]eq = 0.0114 μM; [Glucose] eq = 4.32 mM; [Phosphateleq = 8.83 mM; [H2O] eq = 55.6 M; [H+] eq = 1X 10-7 M; Type your answer... Use the following information (T = 37 °C) ADP+ H2O kJ/mol I What is the AGO' for the following reaction? AMP + ATP Type your answer... ---> -DO AMP+Pi + H+ AG°' = -35.7 ADP+ Pi + H+ AGO' = -30.5 2ADParrow_forwardPartial Enzyme Lab Report Submi x + Chapter 4 Competency 3 X A cloud.scorm.com/ScormEnginelnterface/defaultui/player/legacy.html?configuration=uJ3nrCSmzNPWgXiwoJ4YwRLimO8D6cRHCEe163Lap90fEcUCaNJhdd-I.. I Apps Google I MyMasteryPath o Welcome to LSU M. P MasteringBiology E FRENCH ALEKS - Adaptive L. Pt Periodic Table - Pta. Other bookmarks E Reading list Chapter 4 Competency 3 POWERED BY THE RUSTICI EN PREVIOUS NEXT CLOSE ITEM RETURN TO LMS Page 7 of 26 Your Learning Points 3 Osiagwu, Chiemeka K A reaction is carried out by mixing together a solution of K,CO, and a solution of Fe(NO,),. K,Co, (aq) + Fe(NO,), (aq) - FeCo, (s) + 2KNO, (aq) Assume the purple particles below each represent 1 mole of particles. Construct the balanced net ionic equation by dragging the appropriote components into the gray boxes. Drag each unreacted spectator ion along with its balancing coefficient into the blue box that represents the aqueous solution. Spectator ions Balanced Net lonic Equation + FeCo, (s)…arrow_forward12arrow_forward
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