This video tutorial from Prosperity Academics solves Cambridge Biology 9700 AS Level Paper 22 (Summer 2025), covering key topics including root hair cell structure and function, water transport pathways (symplast and apoplast), enzyme inhibition mechanisms, ELISA testing procedures, and carbohydrate chemistry. The session demonstrates how to analyze biological structures, explain transport mechanisms, and apply scientific reasoning to exam questions.
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Solving Biology 9700 AS Summer 25 Paper 22 (Theory) | M/J 25 QP22Added:
So we'll be solving question one, question two and question five today.
Root hair cells are specialized plant cells located in the outer layer of young roots of plant. Outer layer is basically your epidermis.
Root hair cells have an essential role in the uptake of water and dissolve mineral ions from the soy solution.
The transport of water across the root to reach the central xyllem tissue can occur by the simplass pathway or epoplast pathway.
Figure 1.1 is a diagram of a photo microraph of a root hair cell. On figure 1.1 name cell structures X, Y and Z. So X is basically our large permanent vacule.
Large permanent vacule.
Y is basically the membrane of the vacule which is known as the toal blast.
Z is basically the nucleus because it's like a spherical structure round structure shaped but because it's neither of the two. So this is basically your nucleus.
In the simplest pathway, water passes through the cells of the different tissues in the root before entering the xyllem vessels of xyllem tissue. Name the tissues of the root in the correct sequence through which water passes in the simplass pathway. So from the root hair cells which cells does which cell which cells which make up the tissues of the root does the water pass before it enters the xy. So root hair cell root hair cell is basically your epidermal cell.
So root hair cell the water moves towards the cortex cells.
Okay.
Cortex tissue from the cortex it moves towards the endodermis.
So endodmis endodmal tissue and endodmal tissue important if you want to write about percle you can write but your examiner is essentially expecting you to write at least about these three. So if you write these three you will get your marks over here.
tissue.
Let's move to the next part.
In the epoplast pathway which is the cell wall to cell wall pathway, water passes along the cell walls of adjacent plant cells and through the intercellular spaces. This is more efficient than the simplass pathway. Explain the structural features of plant cell walls. That mean the apoplast pathway an efficient pathway for the transport of water. So basically the plant cell walls are made up of cellulose fibers and they're arranged in a way in a crisscross pattern. So let's say this is one cellulose fiber and this is another cellulose fiber and this is another cellulose fiber. So between these fibers we have empty spaces or tiny gaps. Empty spaces or tiny gaps. Um these basically make the cell wall permeable fully permeable. Anything which can cross these gaps can move past the cell wall. So the fact that there are these gaps and water can freely move through these gaps make it possible for the water to move via the amoblast pathway.
So we'll write cell wall is made up of cellulose fibers arranged in a crisscross manner.
due to which there are small spaces within the cell wall.
Making it permeable to water.
Water can easily move through these space these spaces and also is moving through these spaces. It is forming adhesive bonds with the cellulose fibers.
Water can easily move through these spaces and forms adhesive hydrogen bonds with cellulose fibers.
Root hair cells also synthesize and secrete substances into the soil.
Electron microscopy of the structure of root hairs have identified endopplasmic reticulum, a number of small gi bodies and numerous vesicles. Rootar cells of the barley plant secrete enzymes known as acid phosphotasis which catalyze the release of inorganic phosphate ions from organic phosphates in the soil. Root hair cells of the sorum plant secrete a hydrophobic lipid compound known as surrogalon which slows down the growth of neighboring plants. Explain why the proportion of rough ER to smooth ER may be different in the root cells of barley plants compared with the root here cells of sorghum plants. Now the root hair cells of the barley plants are secretreting enzymes.
Enzymes are proteins.
So there would be more protein synthesis happening in the barley plants which means that there would be a greater proportion of reer compared to STR in the barley plants. On the other hand, when we're talking about the swing plant, it is basically secretreting a lipid compound, which means that lipid synthesis is taking place. And if lipid synthesis is taking place, that means that there would be more STR present compared to because STR is involved in lipid synthesis.
So barley plant would have a higher proportion of rough endopplasmic reticulum.
Since it is involved in protein or you can say enzyme synthesis.
So plant would have a higher proportion of STER since it is involved in lipid synthesis.
Student X stated that acid phosphotasis and serogulene could be transported out of root hair cells using the same process. Student Y stated that acid phosphotasis and cerogelene are transported out of root hair cells using different processes. Suggest the reasons given by student X and by student Y to support their statement. So this is a very interesting question.
Acid phosphotasis. Acid phosphotasis is an enzyme which means it is protein in nature and soreline is basically a lipid in nature. Same process. What do you think?
What could be a same process through which an enzyme or a lipid is moved out.
endopplasmic proportion. Do you guys see this point?
What does numerous vesicles suggest?
process form and we say transport exocytosis because transport we don't see what is the nature of the substance which was being transported outside of the cellarantityis it would move. So same process enzyme and lipid both if they're large in quantity or they're large in size they would cause the cell membrane to fold and they would actually cause the cell me they'll push the cell membrane push sorry they'll fuse with the cell membrane become the part of the cell membrane and release their content outside of the cell. So vesicles may ensided those vesicles would fuse with the cell surface membrane causing the release of the enzymes and lipid within the vesicles.
So student X would have suggested that both are transported via exo cytosis because both are large molecules.
Thus use bulk transport to exit the cell. Now student why acid phosphotas and would be transported by different processes. Now enzyme can an enzyme directly cross the phospholipid billayer.
Can a protein directly cross the phospho lipid billayer?
It could not. But can a lipid molecule directly cross the phospholipid billayer?
Yes, a lipid molecule can directly cross the phospholipid billayer. Which means that the two different processes through which acid phosphotasis and soreline could move is that the acid phosphotasis is moving via facilitated diffusion because it is hydrophilic in nature. So it would require a transport protein for its movement across the phosphoipid barrier. Whereas the surrogill lipid is moving with a simple diffusion as it is hydrophobic in nature and can cross the phosphoipid barrier. Does that make sense?
Soon moves via simple diffusion as it is hydrophobic in nature.
Whereas acid phosphotasis move via facilitated diffusion as it is hydrophilic in nature.
Okay. Next question. The transport of respiratory gases involves blood plasma and red blood cell. Red blood cells contain globular protein hemoglobin.
Describe features of a hemoglobin molecule that are typical of a globular protein other than having an approximately spherical shape. So basically a globular protein is a protein which has the amino acids with hydrophilic R groupoups pointing towards the outside aquis environment.
Whereas the amino acid with the hydrophobic R groupoups are pointing inwards away from the aquis external environment.
So a globular protein has the amino acids with hydrophilic R groups.
pointing towards the outside aquis environment.
Whereas the amino acids with hydrophobic R groups are pointing towards the center of the molecule. fuel away from the aquis or external environments.
A number of substances are involved in the transport of respiratory gases.
Complete table 2.1 by stating the name of the substances that matches the description of its role in the transport of respiratory gases. The first row has been completed for you. of the substance is carbon dioxide.
Its role in the transport of respiratory gases is that in the capillaries of respiring tissues, this combines with water to form carbonic acid. Okay, let's fill the remaining five boxes. In the capillaries of respiring tissues, this enters red blood cells through a membrane transport protein. So in the capillaries of respiring tissues, this enters red blood cells through a membrane transport protein.
Who do you think enters in the red blood cell through a membrane transport protein?
Let's recall red blood cells.
Red blood cells respiring tissue.
We have carbon dioxide entering.
It combines with water to form carbonic acid. Okay, let me make it a little bigger.
Yeah.
This is a red blood cell.
Carbon dioxide enters combines with water to form carbonic acid which dissociates into hydrogenion which eventually forms hemoglobinic acid causing the exit of oxygen to respiring tissues and the hemoglo the carbonic acid dissociates into the bicarbonate ion which exits hits and a chloride ion enters. Now a chloride ion is a charged molecule. A charged molecule would obviously need a membrane transport protein to enter because anything which is charged or polar in nature cannot directly cross the phospholipid biler membrane. The substance would be chloride ion. Very good Alia. In red blood cells this is formed when carbon dioxide binds to hemoglobin. If carbon dioxide is binding to hemoglobin, it is known as carb aminohemoglobin because the carbon dioxide binds to the amine group of the amino acids present in the hemoglobin. We don't call caroxyhemoglobin because caroxyhemoglobin is the substance which forms when carbon monoxide combines with the hem group of hemoglobin. carbon monoxidemoglobin.
So this would be carb amino hemoglobin in alvar capillaries.
So it's aluter capillaries. This combines with a hydrogen ion in red blood cells to form a carbonic acid.
So the bicarbonate which was in the plasma enters chloride moves outside the bicarbonate ion combines with hydrogen to form carbonic acid. This carbonic acid is then converted by car carbonic andhydrates into water and carbon dioxide and the carbon dioxide eventually leaves out into the lungs or lungs oxygen to form hemoglobin.
So this would be bicarbonate or hydrogen carbonate ion capillaries. This is formed in red blood cells as a result of oxygen binding to a hem group oxyhemoglobin.
In red blood cells, this combines with hemoglobin to form hemoglobinic acid. So um hydrogen binds with hemoglobin to hydrogen ion binds with hemoglobin to form hemoglobinic acid.
Okay. Over here you either have to write is it as H+ or write hydrogen ion. If you write hydrogen, hydrogen basically means H2 hydrogen gas. So that would be wrong. So you have to write hydrogen ion or write H+ to get the correct answer.
Okay. Then let's move towards question number five.
Okay, there is a global shortage of blood for transfusions.
Researchers can culture bone marrow stem cells in the laboratory to manufacture red blood cells for potential use as an artificial blood product.
The researchers collect bone marrow from stem cells that are present in small quantities in blood rather than extracting them from bone marrow. So stem cells present instead of extracting them directly from the bone marrow.
Antibodies specific to bone marrow stem cells are attached to tiny magnetic beads.
Third glycoids.
The glycoalics act as what? It acts as the cell recognition marker. Cell recognition marker antigen.
So each cell has an antigen.
response body. So antigens. So what happens is if we have to um get the stem cells from the blood what we do is we have antibodies which have complimentary binding sides complimentary antigen binding sides to the antigens present on the bone marrow stem cells introduce and they're attached to tiny magnetic beads. The beads are added to a sample of blood. So blood draw and blood sample what we do is that an electric field is applied to immobilize a bone marrow stem cells can be collected. So what happens so that um the tiny magnetic beads or the antibodies do not move and they stay where they are so that the bone marrows the stems themselves can be collected.
Are we clear so far? Any questions?
Okay. The antibodies are attached to the bead so that the sides used for binding to specific molecules on the bone marrow stem cells are left exposed.
Name the term given to the specific molecules on the bone marrow stem cells that attach to the antibbody binding sites.
Okay.
So although your concept is correct antibodies antibodies bind to antigens glyco is acting as a receptor.
Okay. Suggest and explain the advantages of using bone marrow stem cells from the blood sample to manufacture artificial red blood cells. Okay. So they're asking us question.
There's a global shortage of blood transfusion. Researchers can culture bone marrow stem cells in the laboratory to manufacture red blood cells for potential use as an artificial blood product.
RBCs. Can RBCs divide via mitosis?
Can RBC's divide 12 days 120 days is 4 months I think so 12 days is the lipan of an of an RBC because they don't have a nucleus mitochondria so where do the RBC's come from they come from the differentiation of stem cells so Stem cells use as a blood product in order to overcome blood transfusion. Blood transfusion continuously divide to form identical cells. And those identical cells can then further specialize into red blood cell. So whenever there is a need of transfusion blood available blood bands does that make sense.
So instead of taking red blood cells so they would be they'll continue to survive and not just survive they would continuously make their genetically identical cells and then those stem cells can specialize or differentiate into red blood cells.
This way we would have a large amount of red blood cells present in the present for blood transfusion whenever that is needed.
Simple.
process.
It just takes one prick to draw a blood sample. Compare that to actually undergoing a whole surgery to extract a bone marrow just so that you can get stem cells. Which is better? Uh drawing a blood sample or doing a bone marrow extraction? Rather prefer giving a blood sample or getting a bone marrow extraction.
Blood sample.
taking bone marrow stem cells from the blood sample is less invasive compared to extraction of the bone marrow. So that's why it is a better or advantageous methodology. So stem cells can divide via mitosis to produce genetically identical daughter cells.
Which can in turn differentiate into red blood cells.
this way large quantities of RBC's can be produced to overcome problems.
related to blood transfusion.
Collection of bone marrow stem cells from blood sample is less invasive than bone marrow extraction.
One desirable feature of artificial blood products such as artificial red blood cells is that they should be economical to produce. So just other desirable features of artificial blood products.
It is very important that of blood product that is not mistakenly um assumed as a non-self antigen by the patient's immune system and it launches an immune response. So we have to make sure that the artificial blood product does not trigger an immune response in the person's body who's receiving the artificial blood product. Second thing as Ala mentioned So that when we transferring or transfusing these artificial blood products into the patient, we're not transfusing pathogen in the patient's blood that should be long because products then obviously it is not serving a good purpose. They're not as useful compared to an artificial blood products which can be used or which can be saved for up to a year or maybe two years.
Does that make sense?
So, artificial blood products should not trigger an immune response in the person's body.
who receives them.
They should not contain any pathogens.
They should have a long shelf life.
Okay. Carrots are root vegetables of the carrot plant. The carrot plant is an important food crop that is grown throughout the world. Carrots have a sweet taste because sugars from a proportion of the total carbohydrate present. Plant breeding has produced many different varieties of carrot with different levels of sweetness. One of the sugars found in carrots is galactose. Galactose has the same molecular formula C6 H2O6 as alpha glucose. Figure 3.1 shows the molecular structure of galactose found in carrots.
This is similar but not identical to the molecular structure of alpha glucose. In figure 3.1, the six carbon atoms are numbered 1 to six.
State the differences between the molecular structure of galactose shown in figure 3.1 and the molecular structure of alpha glucose. So in alpha glucose A for alpha B for below the O of the first carbon is pointing downward. So alpha glucose or hydrogen above. So that's one difference. Similarly, carbon number four that points downwards and hydrogen points upwards.
Carbon one or carbon 4 to arrangement placement of O group that is opposite in alpha glucose.
The positions of O and H are other way around in alpha glucose compared to galactose.
An investigation was carried out to determine the sugar content of mature carrots produced by different local varieties of carrot plants that are grown in Tunisia, North Africa. Carrot plants have grown. Carrot plants were grown from seed under standardized conditions. Different varieties of carrot plants.
All the environmental conditions necessary for growth of a plant were kept same for all the different varieties which were grown. When mature, the carrots were harvested. The different sugars present in the carrots were identified. Measurements of sugar content for each of the sugars were made. Table 3.1 shows the results for five different local varieties of carrot plant A to E. So five different varieties of carrot plants different sugar content for each of the sugars was seen. So the sugars and sugar content carrot types, galactose, glucose and sucrose. After studying the results, student table 3.1, a student concluded that the carrots from the local varieties contain the same four sugars. So all the five uh types of carrots have fructose, galactose, glucose and sucrose. The student made three other conclusions from the data in table 3.1. Conclusion one was that there are non-reducing and reducing sugars in the carrot which is correct.
Reducing sugars and sucrose is a non-reducing sugar. So this is a correct statement. Conclusion two, there are monossaccharide and disaccharide sugars in the carrots. So over here fructose, galactose and glucose monossaccharides all three of them are monossaccharides whereas tucose is a disaccharide.
So this is also a correct statement. The carrots have the same pattern of results.
of results.
So for all of them you can see highest proportion highest sugar content in all of the five varieties fructose for example 183.04 04.
So fructose is highest glucose.
Then we have sucrose or say content is galactose.
So this is the highest proportion. This is the lowest proportion to descending order even in variety we have the highest proportion of fructose then fructose glucose second third sucrose or fourth be galactose in variety C we are seeing the same trend subcru then we have glucose then sucrose and subaras galacto then For variety D the same thing is observed. Highest content is of fructose then we have glucose then sucrose and then we have galactose. And the same thing is observed in variety E as well that the highest content is of fructose then we have glucose then we have sucrose and the least content is of galactose that is the same in all five varieties.
Explain the evidence to table 3.1 that supports these three other conclusions.
So discuss conclusion one conclusion sugar sorry carrots may non-reducing and reducing sugars are present. So we'd say that evidence is that sucrose is a non-reducing sugar whereas fructose glucose and galactose are reducing sugars.
which are present in all varieties.
Conclusion two, the conclusion two stated that there are monossaccharides and disaccharide sugars in the carrot.
So we would state that fructose, galactose, glucose they are monossaccharides and sucrose the disaccharide and they are found in all varieties.
Soup.
Sucrose is a disaccharide whereas fructose, glucose and galactose are monossaccharides which are all found in five varieties and the third conclusion was that the same pattern of results is observed in all five varieties. So we would say yes all varieties >> have the highest content of fructose.
Lowest content of galactose.
Name one carbohydrate. Carbohydrates that are not sugars are also present in carrots. Name one carbohydrate that is present in carrots and that is not a sugar. Polysaccharides.
Polysaccharides are basically not all sugars. So yes, they're made up of sugars but they're not all sugars. So that would be a correct answer. So starch or cellulose starch correct answer. So any of the four would have been a correct answer. So for conclusion three, the carrots have the same pattern of results. They're not saying same results. Same results.
content fructose is the highest amount and about the glucose per sucrose per galactose. So they're not saying the carrots have the same results. They're saying that the carrots have the same pattern of results.
A okay correct. So obviously similar pattern.
Yeah, that obviously would come with practice definitely.
Now each carrot plant produces a carrot in the first year of growth. So we know Carrot plant first year of growth.
If the carrot is not removed from the plant after it matures, the plant passes through a dormant period and shoots develop from the carrot in the second year of growth.
move it would under the plant would undergo a dominant period and eventually during that period new shoot.
This allows flowers to be produced and seed formation to occur before the plant dies. So if carrot is present and it hasn't been harvested and once the new plant has developed the old plant would eventually die. Explain with reference to the life cycle of the carrot plant when the carrot acts as a source and when the carrot acts as a sink. So before dormant period and after dormant period um reference source carrot or sink carrot in which before or after a period of the dormant would we see carrot acting as a source and which period would we see it acting as a sink.
So before the dormant period or in the first year of growth am manipas the carrot is acting as a sink and in that it is basically um taking up nutrients um or assimilates for its growth and development and source after new produce and it would be providing for flowers and seed formation that time duration is when it is acting as a source. So carrot and store towards starch that would break down into disaccharides or simpler sugars and those sugars would move from the carrot to the growing to the new growing parts of the immature plant. This is summer 2025 paper 2.
So during Sorry. However, is not during after the dormant period.
The carrot transports it's stored starch or you can say assimilates.
It can transports sugars by breaking down the stored starch would be a better explanation by converting the stored starch to simpler sugars or you can say sucrose.
First of first year of growth the carrot.
Okay. Carrot virus Y is a pathogen of carrot plants. The virus which belongs to a group known as spotty virus replicates its viral nucleic acid and proteins within host carrot cells. The general structure of proteuses is shown in figure 3.2. It's an RNA virus. The synthesis of viral proteins and host carrot cells only involves the process of translation. Translation RNA is converted into poly peptide chain.
The process of transcription does not occur. Suggest why translation occurs in host carat cells during synthesis of viral proteins. your transcription does not occur. That is because it is an RNA virus transcription DNA is converted into RNA virus and virus DNA. So the DNA does not have to be transcribed into an RNA called protein synthesis. Does that make sense?
So the virus is RNA virus and thus there is no need to convert a DNA to RNA.
The viral RNA can be directly translated to form viral proteins.
Next question. Tuberculosis is an infectious disease that is caused by a bacterial pathogen. The pathogen has mechanisms to avoid digestion by fagosytes. Macrofasages may encases the bacteria remain alive within the cells instead of being killed. So even when uh macrofasages have ingested them via fagocytosis, still some of the bacteria can remain alive and not be digested. Although there are two main species of bacterium causing TB, it is rare for the species that causes bouine TB in cattle to infect humans. Name of the species of bacterium that is the main cause of TB in humans. So the main cause of TB is mcoacterium tuberculosis in humans.
Explain how the pathogen named in A is transmitted from a person with a disease to a person who is uninfected. So infected person, sneezes or cuffs And >> the pathogen is transferred to the uninfected person.
No.
as they inhale the droplet containing pathogen. State the type of therapeutic drug used to treat TB. So antibiotics are used to treat any bacterial infection. Figure 4.1 is a photo microraph of a section of lung tissue taken from a person who has not been infected with the bacterial pathogen and who does not have TB.
No type ofutic name of therapeutic drugarial antibiotic antivirals parisynamite is so type whether it's antimalarial drug antibiotic antivirus anti- viral blood vessel then some structures of the gas exchange system are visible in figure 4.1. On figure 4.1 use a label line and label a bronchus a bronchule a blood vessel. Okay. So bronchules they lead into alvular duct.
This is an alvular duct alv and they're obviously opening the air spaces are your alvuli. So structure which is opening into an alvular duct.
This is basically your bronchule.
So through the presence of alvular duct we're able to identify the bronchule.
This is your bronus because that is not opening into alessle blood.
So this is your blood vessel.
In some people with TB, areas known as granulomaomas may form in lung tissue as part of an immune response to the pathogen.
Okay.
airspace.
Okay. So in some people with DBN areas known as granulomaomas may form a lung tissue as part of an immune response to the pathogen. Pop 4.2 is a proto microrab of a granuloma in lung tissue.
Okay. So we have multi-ucleated giant cells formed by the fusion of many macrofasages. So Fusion of many multi-ucleated joint cells macrofasages fused together.
This region is showing us dead cells and infected cells mainly macrofus. So dead cells and cells infected cells will be mainly macrofasages.
Collectively multi-ucleated giant cells formed by the fusion of many macrofasages and the dead and infected cells which are mainly macrofasages.
They're collectively known as granuloma.
Darkly stained white blood cells in outer area of the granuloma are also shown.
With reference to figure 4.1 and figure 4.2, describe and explain how the changes that occur as a result of granuloma formation can affect gas exchange and harm the health of an infected person may help to prevent TB developing in other parts of the body.
What happens?
Would that affect our gas exchange? If a lot of granuloma form within our gas exchange system, the formation of granuloma in our alular duct or in our bronchules would cause the lumen of our airways to decrease.
Less would enter the less oxygen would be diffused or less oxygen less blood sorry less oxygenated blood would be formed less oxygen is reaching the body tissues less respiration less respiration to less ATP form does that make sense any confusions about So part one may that granoluma formation would result in a decrease in the lumen of airways.
Due to which less air would reach the alvuli leading to decreased oxygen apply to the respiring tissues.
Now may help to prevent TV developing in other parts of the body.
granuloma that is basically present in the lung tissue.
The presence of macrofasages within the granuloma would ensure that the pathogen does not leave the lungs and enter into another body tissue.
The presence of so many macrofasages multi-ucleated cells are formed from the fusion of macrofasages.
They would basically try to eradicate the pathogen and stop the pathogen from reaching to other parts of the body.
exit lymphocytes for adaptive immune response. So within the lungs adaptive immune response trigger which would basically result in the formation of antibodies and those antibodies would then further eradicate the bacterial pathogen.
Okay, any questions?
Let us know the distinctive five points of this five mark questions. Um, are these points clear? Um, Aisha.
Okay.
Okay. Question five. Let's do question six. Horseer radish per oxidase HRP is an enzyme that can be extracted from the roots of the horseer radish plant. So this is an enzyme which is found in the horse horseer radish plant or say extractor HRB is used extensively in industry and technology in the reaction catalyzed by HRB. Hydrogen peroxide is used to oxidize an organic substrate.
This is summarized by figure 6.1.
organic substrate hydrogen peroxide react in the presence of the hydrogen peroxidase enzyme to form oxidized substrate and water inhibitors can have an effect on Vmax the maximum rate of reaction on KM the micus mental constant of HRP figure 6.2 shows the effect of substrate concentration on the rate of reaction of HRP complete figure 6.2 do by drawing a curve to show how the presence of a non-comparative inhibitor will affect the rate of reaction of HRP.
Use the curve you have drawn to obtain an estimate of the KM.
First and foremost at the same rate unit is the Wax lower or is it the same?
Okay, as you guys mentioned at a lower um arbitrary unit max okay the curve you have drawn predict overlapping.
The substrate concentration at Vax upon 2 substrate non-competitive inhibitor 4 AU is the Vax. So two would be our Vax upon two. So now Scientists can design synthetic DNA nucleotide sequences to produce a synthetic HRP gene. These sequences will include a start codon and a stop codon so that translation of messenger RNA can occur. Explain what is meant by a start codon and a stop codon. So start codon is basically the codon which initiates the process of translation.
The codon which initiates the process of translation and we know start for the amino acid methine. I pass the start code on is a triplet which initiates translation and codes for methunine.
Yeah, you should know the name methionine. And stopcodon is a triplet which basically um stops the process of translation or terminates the process of translation.
Stop codon. It does not hold for any amino acid.
Stop codon stops or terminates.
translation.
However, it does not code for an amino acid.
HRP is used in an imological test known as a sandwich elya. One use of the test is to diagnose disease. Figure 6.3 outlines the main steps in a sandwich alysa in which a toxin released by a pathogen is detected in a sample of a body fluid taken from a person who is ill. The test involves two types of monoconal antibbody that can bind to the toxin. A capture antibbody and a detection antibody. So there is a capture antibody and there is a detection antibody.
So complete 6.4 for 6.3 6.3 steps involved.
Step one, prepare a test well by immobilizing the capture monoconal antibody to the support.
Remove a sample of the body fluid from the person who is ill and add the sample to the test well.
compliment to the toxin released by a pathogen present only. Wash out the test well to remove any substances not bound to the capture antibodies. So substances remove other than the antibodies and the substance the toxin that is being produced. So toxin detection antibodies detection antibodies basically bind toxin present. So add detection antibody to the test. So detection antibody attached. Wash out the test well to remove any detection antibbody that is not bound to the toxin. So antibodies binding wash out substrates for the catalysis by HRP and add a set time to stop the reaction.
inological test with the detection antibbody. HRP substrate substrate.
HRP was basically producing the oxidized substrate.
any questions about the steps complete 6.4 to show the situation in the test well after step five wash out the test well to remove any detection antibody that is not bound to the step basically figure toxins.
So antibbody that's how we're showing the antibodies being attached to toxins with reference 6.3 describe one difference between the capture antibbody and the detection antibody. So my capture antibody that basically Let's look at it. Difference between the capture antibody and the detection antibody.
This is the capture antibody and this is the detection antibody.
B capture antibody toxin or bind detection antin.
Now if you see toxin lower pointed triangular surface detection antipbody and the detection antibody are binding to different sides of the toxin and also as Abdullah mentioned detection antibody it consists HRP whereas the capture antibody does not consist HR. RB. So basically difference captur antibbody and detection antibbody bind to different locations or different sides of the oxide. Figure 6.5 outlines the reaction catalyzed by HRP. In step six, the organic substrate TMV is in access and changes color when it is oxidized.
This indicates a positive test result. A student suggested that a low concentration of toxin may be diagnosed as a negative result instead of a positive result. The student said it is possible.
Is that a valid conclusion or a suggestion that the student has made?
present.
So in so the student suggested to use a coloring meter after step six in the sandwich alisa test that it would provide a quantitative result and help to avoid this error.
With reference to figure 6.5, explain why using a coloring meter would provide a quantitative measurement for detection of a low concentration of the toxin and help to confirm a positive result.
Previously I think this is a meter.
I'm assuming and look at my arrow. Can you see tube glass?
Just a second.
This is a test tube present.
tube.
Change of color because of the reaction being proceeded.
You are aware.
So for exampid tell Transparently window surfaces transparent surfaces solution to absorbance of light zero transmittance% absorbance of light by the solution would increase and the transmitt of light across the test tube would decrease. Do we all agree with this?
Okay, repeat.
Um second question torch in front of the glass which contains water and in front of the glass which contains Coca-Cola.
Let's say tube. Let me draw the test tube. This is a test tube which is transparent.
And let's say this is the test tube which is colored.
Now solution.
Do you agree with this?
or light colored solution. Some amount of light would be absorbed by the solution and to come amount of light would be transmitted. Is that correct?
Before starting our experimentter reading, it is known as the zeroing of the calorie meter or calibration of the calorie meter.
Zero absorbance or 100% zero light was being absorbed and all the light was being transmitted.
Does that now make sense? It measures absorbance by the solution. Exactly.
Absorb measure.
Absence of lightance values.
Does that make sense?
Okay.
exact question with reference to figure 6.5 explain why using a meter would provide a quantitative measurement for detection of a low concentration of the toxin and help to confirm a positive result. The Maripas colorimeter basically measures the absorption of light by a solution.
Um even if there is a slight or a low change in the color that degree or that measure of absorbance would be recorded by the color meter.
If the absorbance values are lower absorbance values of the toxin present instead of saying toxin present instead of saying negative result does that make sense?
So we'll say calorie meter gives values of the absorption answer.
That was the last question of this paper.
Let me read your questions and then we'll call off the class.
Shall we shall know the significance of coloring meter and its uses. You should know the uses of coloring aquas reaction solution.
absorbance absorbance absorbance absorbance absorbance.
rate of reaction.
Does that um answer your question Aisha?
The MS accept the MS accepted range for CM value question was 0.55 to 6. Oh, let me see.
55 to 6 like this magic. Thank you so much Abdullah for correcting.
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