9700/12/F/M/2023/guided solution for as biology (2023)

Introduction

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Content

Hello, everyone and welcome in this video we'll be checking out.

The brewery March, 2023, 9700, variant 2, of course, McQ.

So if you have any specific questions that you would like to check out in this variant, I have timestamps in my comments.

And if you have any specific gears that you would like me to solve feel free to recommend those in the comments down below let's, get started number one.

The photo micrograph shows cells from a human blood smear, which calculation shows the correct method to calculate the actual diameter of the neutrophil shown in the photo micrograph in micrometers.

Now in order to calculate the actual diameter you have to remember the triangle, it's important for calculating among the diameter, the magnification also the actual, um, yeah and the image lens.

So first off, it's, it's, I am a I for image length a for actual lens or diameter and M for magnification.

So in this case, if we want to get the actual diameter, then a will equal to I over M.

So in other words, it's the length of PQ divided by the magnification, which they mentioned was basically, oh this didn't.

Mark it down, but it's here in the in the answer.

So of course, basically the length of PQ, but of course, we're going to measure it in millimeters never in centimeters.

So if one of the options uses centimeters, make sure to cancel that out because they don't accept centimeters in Cambridge, and of course, basically turn down to convert from millimeter to micrometers, which is what they want you multiply by 1000.

And if you want to convert from Micro Machines back to millimeters make sure to divide by thousands, therefore, the answer should be the length of the line PQ in millimeters times, 1000 divided by the magnification of the photomicrograph.

Okay.

So this is a tricky question, but it's also very intriguing as students calibrated an IPS gratically using a stage micrometer.

Each division of the stage micrometer was 0.01 millimeters.

Okay, using with the timestamp magnification objective lens, uh, 10, IPS, graphical units match 10 divisions on the stage micrometer.

Now the same microscope is used with a times 40, instead of a timestamp magnification objective lens to measure the diameter of the alveolus interesting.

So let's immediately match up the stuff.

So we have so far one of stage micrometer unit is 0.01 millimeters and 10 stage micrometer units match up with 10 IPS graduates.

Therefore we can assume that one eyepiece graphical units is the same length as you know where one stage micrometer units under the same magnification all right.

But now you change that to be from time to from times 10 to times 40.

So basically, if we were to imagine like what we would see, we would see something like this like it used to be one, two, three, four, five, six, seven, eight and blah, blah, blah, all right it's occupying like this at this same number right? But now you change the magnification.

So we literally zoomed in now we know that the IP scratch screen is only visual is only placed onto the eyepiece lens and it's, not placed on them.

You know, it's not placed on a slide.

And so it doesn't really get affected by changing the objective lenses.

So that means basically that it will not be zoomed in it's, only the stage my computer that's zoomed in.

So after zooming in that means basically that you're going to visualize much less of these divisions.

And instead let's say, you visualize an example, you visualize only one right so it's, four times larger because here from times 10 to times 14 that's, four times larger.

And so when this is four times larger, the length of the eyepiece critical unit will obviously like it used to be one to one now it's, one-fourth, right.

So now what's it called it's four times smaller in other words, it's one-fourth.

In other words, you have to divide the length of one eyepiece classical units by four to get the the lens of the hyperscratical unit under this magnification.

So it's, 0.01, divided by 4, which is was 0.0025, which is in millimeters.

And of course, we're going to have to convert that to micrometers so that's, 2.5, micrometers, okay, of course, I multiplied by 1000 in order to get 10 micrometers.

So basically millimeters times thousands, mom now 2.5 multiply that by 96.

And you should get the length or basically the diameter of the ambulance under this magnification, which is C so I'm going to introduce another idea, because some people are actually familiar with the way of using a formula.

And maybe you'd find formulas interesting too, because they actually help you out a lot when it comes to the eyepiece critical as we know that there's this formula, we're basically we're going to be using, yeah, the number of what's, it called stage micrometer units.

And we divide that by the number of I piece, graphical units.

And basically we multiply that by the length of one IP, scratch I'm, sorry, stage micrometer.

So length of stage macrometer units and multiply that by 1000 because it's usually given in millimeters.

And we want to convert that to micrometers so that's, just like a rapid formula.

And if we rely on what's given to us here, then we can immediately say, okay, it's 0.01 and then I'm sorry that basically we can't, wait to say, it's, 10, divided by 10.

times, 0.01 times, 1000, which should give us 10 just like the previous one all right.

But now we know that it's changed magnification from times 10 to times 14.

And so the length of the IP's critical will be 1 4 of that that we assume so we're going to divide the symmetically by 4 to get 2.5 and then 2.5 multiply that by the number 96 because basically multiply by the number of Ip, scratch screen is occupied by the alveolus.

And that is the other one I'm gonna give you an example, just in case, you're interested.

So because sometimes it would give you let's say, you want to find the length of a nucleus right? So in order to find the length of the nucleus let's say, he occupied the nucleus occupied, um let's say, seven like this classical units.

Okay.

And one stage micrometer was in front of 50 ipractical units.

So that means basically that it would be one, which is divided by the number of ips card units that say, 15.

And then multiply that by the again, length of one stage micro material, which let me give you an example, 0.1 millimeters.

And then multiply the power 1000 to convert it to micrometers.

And you should get two micrometers, and we set it to the nucleus occupied, seven by seven IPS of critical units.

Well, two by seven equals to 14 micrometers that's, the length of the nucleus, usually it's smaller.

But this is the answer for this example, hope you understood I'm going to move on to the next question.

Now it is wrong.

They can transcribe the circular DNA through.

They can Transit the money through the enclosed button and brains true.

So two three and four just c number four, uh, which says structure is not usually found in cells from the root and I would suppose it's so said, one is actually usually found and chloroplast is not found because it doesn't really support undergo food sentences.

Okay, with poor size of approximately 2.5 nanometers extremely tiny because molecules have a diameter approximately 1.5 nanometers and can pass through the pores in the brain, what can unpass through Sports too so bacteria, obviously not because they have size between 0.5 to 5 micrometers, which is like way, too huge, five thousand nanometers.

You know, hemoglobin, absolutely not it's.

A polymer.

It cannot pass through 2.5 nanometer opening.

Ribosomes are 20 nanometers.

So even ribosomes cannot pass fructose is approximately the same size as glucose and maybe even smaller so d.

What is present in all viruses? Ribose, deoxyribose, adenine thymine.

Now we understand that ribos and deoxyribose, they form either parts of DNA or more nucleotides.

And so by screen, sometimes viruses could be retroviruses, meaning, they have RNA sometimes they're adenoviruses, maybe they have DNA.

So that is known and no thymine too because it's only from DNA.

If they also said, duracellular, that's only found in fdm in them, the RNA nucleotides.

So that's also wrong.

So the answer is C.

Okay.

So for this question, seven to estimate, the Constitution we're causing an unknown, uh solution, X L, equal volumes of a range of known concentrations of glucose where each Max is mixed with the same accessible.

Independent solution after mixing.

The solutions were placed in the themistically, controlled water bath at 9 degrees Celsius for three minutes.

Now the unknown solution was then treated in the same way.

And the colors of the known and unknown Solutions were compared what's.

The infinite variable lens procedure dependent variable.

This is dependent.

And so obviously it is concentration of glucose right? So that is a independent variable.

We have to focus if you said independent or dependent because dependent would be the final color of the solutions.

Okay, temperatures control variables.

One glucose solution.

Another, controlled variable, number eight, which it means are correct for amylose and also for amyl pectin.

So they're, both carbohydrate molecules.

It is true.

They are formed by condensation reactions through there are linear molecules.

Nominal pectin is Branched, and they contain I want I want for alphagocytic bonds.

Correct so that's, one, three and four, which is same I mean, one two and four excuse me, it is B.

Okay.

So let's move on.

Now, which statements up as autism is correct for comparison between saturated triglycerides molecules and unsaturated triglyceride molecules of approximately the same molecular mass so saturated and unsaturated, uh, the students correct for comparison.

So unsaturated, acrosides have multiple bonds and fewer hydrogen atoms than unsaturated unsaturated.

That is actually true.

Okay, unsaturated trigosides have fewer double bonds.

That is wrong.

Those tragicides have more loved ones and more hydrogen atoms.

Now if you have more double bonds, then you're going to look something like this.

So basically, this is a chain where the bonds each carbon atom should be surrounded by, you know, four four bonds.

Okay.

So if I said, there's, two here then there'll only be two like 100, hydrogen a bonds with hydrogen.

And another bond is carbon.

So let's have one less for each like one less hydrogen for each carbon atom.

Therefore, to always have less hydrogen atoms, unsaturated because I have a few other ones again wrong wrong.

So the answer the diagram represents a molecule from a cell surface membrane.

So what description of one of the labels is correct? So for one of the labels, we have what you call the fatty acids that at the heart.

And at the hydrophilic end of the molecule.

So let's see a is actually, yes, it is a hydrophilic and it's a hard for weekend hydropic end of the triggers the right molecule.

This is a phospholipid because you can see there's only two fatty acid chains and a phosphate group right here.

And the phosphate group at the hydrophilic end of the molecules, definitely see again, there's, no, there is glycerol, but it's, not that the hydrophobic and it's, not the hydrophilic ends because the against is hydrophilic.

Okay.

So which row correctly shows levels of protein structure that can be held together by each type of interaction.

We have hydrogen bonds.

Hydrogen bonds, hold the secondary, tertiary and quaternary structures, sometimes but never the primary structure, the primary structure, which is a sequence or this.

The specific sequence of amino acids in a polypeptide chain is basically only maintained by the peptide bonds, because if you have an amino acid.

And then if this is, if you ever see such a structure like a bead, this is one amino acid.

Another another.

What is this Bond it's, a peptide bond? What is this structure it's, a primary protein structure? You don't, see any hydrogen bonds? You don't, see any covalent bonds besides from the peptide ones you don't, see any hydrophobic interactions or disulfide bonds or whatever now so hydrogen bonds we can say, secondary and tertiary.

And then for hydrophobic interactions, tertiary structure for covalent bonds, primary and tertiary that is true because for primary the covalent bond is peptide bond for tertiary.

The covalent bond is disulfide bonds.

These are also covalent bonds.

However, hydrogen bonds and ionic bonds.

They're, not, of course, which molecules contain at least three double bonds saturated triglycerides, hemoglobin or collagen.

Now at least three here, we only have one actually because if you draw a saturated triglyceride, then remember it is going to contain like glycerol, B and O.

This is the Ester Bond, and it will look something like this.

The only double bond here.

So this zigzag shape represents the this exact shape represents the fatty acid chain.

And here because it's saturated that means there are no double bonds within the fatty acid chain.

So it's, literally only a carbon to carbon to carbon and then hydrogens all over the place.

No double bonds at all.

Okay.

If I wanted to draw it unsaturated, one I would say something like this.

Anyway, it is essential.

So the only double one here is this one.

So actually, yes, we have at least three dry drugs.

Right? Three double bonds.

So that is true.

It contains at least 300 Trend.

So it is true.

The trigger starts contain at least three double bonds.

So we can do a check here for collagen.

Collagen has lots of amino acids and amino acid has a structure, which is something like this hydrogen hydrogen, and then our group hydrogen.

And then this is the carboxyl group.

So you can see there's one double ones in each one of them, because if you wanted to draw the peptide bond, it should be something like this.

And we know that collagen is a popular type.

So it has more than three amino acids.

So obviously it has more than three the double bonds.

So we can do a check here same thing for hemoglobin.

Therefore, the answer is d.

Okay, which diagrams correctly shows hydrogen bonding between two water molecules.

Uh, obviously, the first one is wrong.

Because of the signs it's meant to be something like this.

The Delta sign should be present, because this represents a strong positive and negative strong, positive and negative charges.

However, the hydrogen bond is totally a weak Bond.

Therefore, it has Delta positive.

Delta negative.

And obviously we have to make sure of the distribution.

Hydrogen ion atoms that are never Negative.

They don't carry it an active charge and that's, because they only have one proton in their nucleus.

If I were to draw a hydrogen atom.

I would just like, for example, hydrogen.

And then for example, if I want to draw the nucleus it's only going to be one proton, one Neutron and one electron that is going to be a hydrogen atom.

Right? However, if you draw an oxygen atom, it's going to have 16 protons.

So if you place a water molecule, which looks something like this, one oxygen, two hydrogen, right? Basically, if you place it somewhere in a tank, for example, where there are charges, obviously, the negative charges will all be attached to the oxygen atom's nucleus.

However, the positive charges will all be accumulating at the hydrogen and atom's nucleus because everything left all that like all the negative charges left and went here, therefore it's, obviously going to be positive around here.

And that is why it's going to be carrying a partially positive charge.

However, the oxygen will be carrying the partial negative charge.

And therefore, the answer is B for, um, which technique describes an example of an extracellular enzyme, amylase and okay.

Extracellular enzyme is an enzyme that is produced inside the cell, but it's going to be used outside the cell so amylase and saliva is an enzyme that catalyzes the breakdown of starch in the mouth.

That is true amylase is produced in your salivary glands and the cells of your salivary glands and then secretes it into the mouth into the mouth cavity and therefore it's an extracellular enzyme.

So we can go ahead and check that let's.

Look at the next enzymes.

Carbonic anhydrous is an enzyme that helps in transport of carbon dioxide in the blood.

The function is correct, but it's, not an extracellular enzyme because Carbonic anhydrates is found as it is produced in red blood cells.

And it remains inside the red blood cells, DNA, polymerase, again, it's, not exocellular, because it remains inside the nucleus where it's going to be used during DNA replication same thing for RNA polymerase.

It remains inside the cellular to be used during transcription.

Therefore the answer is a excuse me about that a okay, control is correct for the enzymes that causalize your actions using local hypothesis.

Okay, remember locking Q hypothesis, it's, just like its name.

You have the enzyme, which acts as the lock.

And you have the substrate to access the key.

And obviously, basically they are complementary in shape.

You can see this fits perfectly with this.

And the enzyme with specific for the substrate.

And obviously once they're bind, they form the enzyme substrate complex.

And then this is maintained by the temporary hydrogen bonds between the r groups of the amino acids, forming the enzyme and basically the and the substrate atoms.

And of course, basically, if you compare this to the induced foot mechanism, this is the second mode of action that could that enzyme could and have.

These are both theories by the way it's just suggestions of how the enzyme can work.

So basically, um, it could be something like this.

Maybe the enzyme has them weird and not fixed like flexible shape at the beginning.

And then it meets the substrate once it sees the substrate.

It remembers its Origins.

And it decides to change back to its original shape.

And so sometimes also the substrate changes a little bit to allow the perfect fit between the enzyme and the substrate.

And so this happens and obviously this, uh, this happens when the substrate enters the enzyme active site.

And then basically the Reformation of the enzymes are complex.

And then same thing, the temporary enzyme product complex.

And then the products will be released.

The enzyme returns back to its original shape and it's ready to be reused.

So let's, look at the effect of the enzyme on activation energy of the reaction being catalyzed.

However, the mechanism of the enzyme it is always going to lower the activation energy, okay, the shape of the active site in comparison to the substrate as we mentioned it's complementary.

Okay before during.

And after the formation of this officer complex that is correct let's move on.

So a scientist character, investigated the rate of breakdown of hydrogen peroxide for experiments require that during different using different mixtures.

The results are sketched in the diagram.

So obviously substrate only it's not going to be breaking down unless basically like in temperature is high or something like that.

And in the case of hydrogen peroxide, I mean, I can see why it's going to be breaking down substrate and enzyme obviously it's going to have the highest rate of reaction if there's an inhibitor that's going to lower down negative reaction, but for competitive inhibitor, we have to understand its effects versus the non-competitive inhibitor.

Now we know what's.

The V Max is the V Max is the maximum rate of the reaction right now.

The competitive inhibitor is not going to lower the Vmax you're going to eventually go back to the Vmax.

If you increase the substrate concentration, because we understand that the end the competitive inhibitor from its name, it's competing with the substrate.

If you increase the amount of the substrate, obviously the substrate is going to win right.

And so basically, the maximum rate of reaction does not change.

However, the Affinity of the enzyme to the substrate is going to be decreased.

Uh, what is the Affinity? You say, it's, basically, the ability of the enzyme to bond to the substrate as soon as it sees it.

Of course, when you see that there's a distraction like the competitive inhibitor, the enzyme is going to be distracted and it's, not going to immediately bind to the substrate.

Unlike when the enzyme is alone with the substrate right? And so the K we say the km is what represent what allows us to understand the Affinity of the enzyme to the substrates.

The km is the substrate concentration at which half the V Max is reached.

So this is, this is the V Max here we have half the VMAX.

Okay, basically, if we know that half the V Max is go is this then of course we can draw some type of line over here.

And we can see the V Max for this enzyme is lower compared to, uh, this enzyme.

And we understand that for one one does not have any inhibitor involved.

It doesn't have any distractions.

So it has a higher affinity for the substrate.

So obviously, and it has a lower km.

So that shows that the lower the km, the higher definitive for the enzyme of the enzyme for its substrate.

However, the higher the km, the lower the Affinity of the enzyme Force substance.

Let me write this down again, the higher the km.

The lower the affinity, the lower the km, the higher the Affinity, which you can see in the graph.

However, the effect of a non-competitive inhibitor is less like the shape of graph.

Number three, okay.

So for graph number three, you can see actually that the V Max is never reached.

Okay.

So here, um, let's, look at the Vmax and get the half the BMX because as I mentioned again, at the substitute concentration at half the V Max, this is the km value.

And based on our results, we were able to assume that as you increase the km value, the Infinity for the enzyme of the enzyme for substrate decreases.

So let's see effect of the non-competitive inhibitor, which we now know the graph looks something like this.

The V Max is lowered on the km value.

So this is the V Max for them enzyme with non-competitive inhibitor.

Now we want to see the effect of the non-competitive inhibitor on the km because that's where it's different, the non-competitive inhibitor has no effect on the Vmax.

However, it has an effect on the km value.

Let's see if the non-competitive inhibitor has an effect on the km.

The km again is absolute concentration at half, the V Max or half the maximum effectivity of the enzyme.

And so if we know this is the V Max, we can assume that maybe this is, um, half the Vmax.

We could say, okay, however, if we know this is a V Max for this one let's say, this is, um, half the V Max.

So this is half and I suppose, basically, then this is half the V Max I'm just trying to approximate.

It it's, not exactly true.

But if this is to have the Remax a line from here and a line from here, we can see that the substitute concentration for both is the same.

Okay.

So that means that the km value for both is equal.

So that means it has no effect on the km value.

So let's just summarize this.

If you see a graph looking like this, the most normal, this is the enzyme alone and a graph where if you increase the substrate concentration, you'll, eventually reach back to the V Max.

This is the non-composite the competitive inhibitor.

And if you see a graph where you where it doesn't really reach the V Max, it reaches a plateau, but the km value.

It Remains.

The Same.

This is the effect of a non-competitive inhibitor.

And so let's see what they want for this question, which draw shows the correct lines for the two experimental mixtures, uh, substrate, only it's, obviously, four because it doesn't look like there's, actually any much breakdown going on so I'd, say, it's, four, but substrate and enzyme and competitive inhibitor.

We've established that it's two.

So that is B.

Okay.

So which group of these substances can pass directly through the soil surface membranes without using a calorie protein or channel protein.

So things that can pass directly without transportation carry proteins have to either be lipid soluble or like an organic solvent.

Something that is hydrophobic or nonpolar or sometimes it's extremely extremely small that even if it's spoiler that doesn't really matter like in the case of water, oxygen carbon dioxide, anuria so let's, see what we have here mineral ions, obviously not because, um, their requirement instance, they are hydrophilic, and they are large in size same thing for glucose or fructose.

Whatever this molecule is, however, carbon dioxide is as I mentioned one of the molecules.

Okay, what happens to the surface area to volume ratio of a cube when the length of each side is doubled, actually the ratio halves.

Now you don't need to know this you don't need to memorize this.

You can actually just experiment you understand that the surface area to volume ratio is literally what's the name.

Okay, you have to get the surface area in order to calculate it for a cube it's just going to be 6s squared.

And then basically divide that by the volume, which is literally side by side by side, or in other words, it's not cubed.

Okay.

So we can, for example, let me give you the example of the cube with the surface area or let's say, the lens of one width of one and height of one.

Okay.

So basically, of course, it's a cube so it's, literally just one squared and two one to the power of three that is six divided by one, which is six a really high surface area.

Bonding ratio.

Now let's say, I doubled the side length.

So let's say, it's down two so it's, six multiplied by two all of it squared and then divide that by two to the power of three right three.

So that halves in the exam, if you do not know this piece of information, you can just calculate it.

Okay, we're moving on.

Now we have which events are part of the mitotic cell cycle.

Okay, again, matotic cell cycle.

It's, actually, it includes everything.

But when it comes to what's, it called, um, the mitosis itself.

Actually it only includes prophase, metaphase, anaphase and telophase.

So remember the bar are the pie charts.

So here we have what's it called.

Okay.

I hope.

This is okay, something like this.

So this is G1.

This is a cell cycle.

Okay, G1.

And then we've got S phase right? And then here we have G2 second growth period.

And then here we have mitosis.

Okay.

So mitosis is part of the mitotic cell cycle.

Okay, but my.

But the cell cycle is not part of mitosis.

So if you were to ever ask you, what is part of mitosis, you would only say p, matte, which is the phases that make up mitosis.

There are four phases called prophase, metaphase, anaphase and telophase.

Okay, this is also cytokinesis it's part of the cell cycle, but not part of mitosis, therefore, all of them are parts of it it's a.

Now, the estimated total number of red blood cells in human body is 2.5 by 10 to the power of 13.

And it is estimated that each day 2.5 button to the power of 11 red blood cells are removed from the circulation and are replaced by stem cells in the bone marrow.

What is the percentage of total number of weapon cells is replaced each day.

So what percentage is replaced each day? We just divide 2.5 by 10 to the power of 11 by 2.5 by 10 to the power of 13., multiply that by 100 to get the percentage.

So that equals to one percent of the red blood cells are placed each day.

Now, DNA polymerase catalyzes, the condensation reaction between molecules during semiconservative DNA replication.

What is joined by DNA polymerase.

This is a repeated question, actually, it's, nucleotide and nucleotides.

Okay.

So that is C.

And basically two polynucleotides turns make up DNA molecule, which description is correct.

The percentage of Citizen is 50 that of guanine and one molecule, um, actually, no representative.

Cytosine is correct.

Again.

It's one to one the percentage of cyan is the same as that we've gone in eastern no not in eastern because they are complementary.

And hence, they're going they're going to line up against each other according to complementary whispering, but they're, not going to be the same on each strand it's going to be different.

Okay, this the percentage of cytosine in each it's, the same you need to start with the molecule that is wrong.

Because obviously the reason for variation between us is basically the fact that we have different sequences and different numbers of each of the bases right.

And so basically, you cannot say that the percentage of Citizen is always going to be the same.

Okay.

So the answer is V.

Okay.

Now, for this question, the ptx1 protein is found in a wide range of animal species where it is regarded as a tension for normal metabolism and sandras.

The gene coding for PDX1 protein has a much lower proportion of a and t, nucleotides A and C nucleotides.

Lower proportion interesting.

Okay.

And basically then be the exponent from other animals, it's, not finding the sequel and the complete DNA sequence of ptx1g and sandrats has been difficult and been difficult sequencing involves splitting off nucleotides one at a time from signature and DNA what could account for difficulty and finding the DNA sequence of pre-dx1 Gene and sandrats.

So I suppose be the X1 genius presence and only a small portion of nucleola and nuclei actually it's present in all nuclear nuclei.

So that's wrong pdx12 is transcribed in only some cells that is true.

But this does not make it difficult in finding the basically the sequence it actually has nothing to do with final sequence, it's, totally fine to find some cells.

So I think this is also wrong.

And the strength of the hydrogen bonding between the two strands of pre-dx-1 gene is unusually high.

Now this makes sense considering that he has a lower proportion of adenine and thymine and between adenine and time.

You have only two hydrogen bonds.

If you substitute those by saturated and guanine, which have 300 bonds.

And obviously increasing the proportion of the number of the increase in number of hundreds and bonds, which makes it more difficult to split the strands and form single standard DNA, I think that's.

The reason why it's so difficult so 23 that is d, different tissues and plants.

And the plants were supplied with radioactively enabled substances to identify which tissues were actively sanitizing mRNA.

So obviously which radioactively enabled substance would be suitable for these experiments anything that is specific to RNA and not DNA, because there are differences between ammonia nucleotides and DNA nucleotides right? So mRNA, DNA nucleotides have only like both of them are going to have mental sugars, but DNA has a deoxyribose while rnas can have ribose that's.

Hence, the name it's called the oxidium nucleic acid.

Right? Another difference could also be the fact that bases right so there's, adenine, assamine, cytosine and guanine in DNA.

But however, in in RNA there's, only adenine, cytosine, uracil and well, guanine there's, no thymine.

So what can we do in order to basically identify tissues that were synthesizing mRNA? We can have a security actively labeled duracin right? And we can also have radioactively enabled ribose or an organic sponsor and in our present in both DNA and RNA.

So that would be stupid because all cells have DNA.

Okay, two and four only C now, 25 which row correctly matches the structure and function of the fluency tube elements, a structure preference, cytoplasm, okay, the fluency of tube elements, let's, go ahead and draw them.

Okay.

So basically let's say, this is the one that shares with the companion cell.

Okay.

So basically here we have nucleus and mitochondria and lots of what's called cell structures.

However, here we only have preference cytoplasm, not many cell structures, you can see maybe some smooth or rough reticulum, but no nucleus.

So basically, mononucleus the correct function right acetate resistance as possible to the flow that is also right and was a modified form, safe place.

There's.

Another transport of soil is actually no in order to reduce resistance into another continuous flow of the solutes and elongated cells joint end to end sure thing why not to form a tube to transport, soft mineral ions and water.

The function here is wrong.

This is the function of design investment element to rent loss of water.

Absolutely not actually like is what prevents the leakage of water.

If you do not have leg, then you're, not going to prevent the loss of water.

And water is not the main thing that's being transported here.

The important part is avoiding loss of The Flume, except it's really precious.

And hence I'm going to be choosing a now, which feature of Zone investor elements, help adhesion during transpiration.

Okay.

So Legend forms a complete secondary wall.

Okay, this does not help adhesion because adhesion is supposed to like be between the water and the water right? So basically it's the bonding between the water and sidewalls of the design investor element or any tube.

So like the information was not the right thing because Legend is impermeable to water.

Okay.

So we don't like Legend does not like water.

Okay, new vessels carry extra water.

As the plant grows.

There are no cross walls between the Zone investor elements, I think this more helps with cohesion, because if there's no cross walls and basically water will not suddenly, you know, stop flowing as one continuous column, elements form, a narrow tube after they form a narrow tube, then it's going to increase the number of water molecules that are exposed to the tube ones.

And that means basically that it's going to help with adhesion I think, it's, D.

Okay.

Mass flow is the bulk movement of materials from one place to another.

How many vessels is that carry fluids by mass flow, the artery from system and vein and Xanax all of them.

Of course, it's D.

Okay, uh.

Moving on submarine are components of a variance of various plant cell walls, descriptions of this thermal components are listed.

A component of the experienced it's gonna be submarine redirects water into the simplest pathway.

Another submarine and another one about the brain and hardcore components as I said, was you could say so you can say legend or submarine.

And basically you want to interacts with a small kid in a fast pathway definitely cellulose.

So which row correctly matches the three descriptions of with the cell wall components cellulose as I mentioned is four lignin it's, a natural component could be three.

It could be.

Yeah, it's only three.

And then one could be Supreme could be one or two.

Now, the arrow shows the direction of water movement across the plant's root from water and soil to the stem.

Okay, which arrow shows water movement only in a plus pathway.

Okay.

So a plus pathway consists of two things, it's, the intercellular spaces and cellulose cell one.

So literally from its name it's called like the dead pathway.

Okay, anything dead anything.

Not alive is going to be forming a plus boundaries.

And plus pathway is going to be a live pathway, such plasm cell surface membrane plasma dismantown, like here, it is cytoplasmic extensions.

Same plus pathway.

There's also the vascular pathway, but related to be considering it's more as simple as pathway.

So the r plus pathway is now this is wrong because the past two cells of membrane and cytoplasm.

This is also noted known and I think it's going to be one and two because we only see them passing through.

But no one doesn't.

One two is wrong.

I think four is right two and four, which is C now moving on.

Okay, number 13, which are the current identify sources instincts of sugars, Roots cells, absorbing mineral ions, uh, I, don't think it's gonna be considered like a sorcerer sink like this is most so mineral ions are transported by design of this element, there's, no Source in sync and xylem like a like a transport only occurs in One Direction difference though in the film is that there's a source and zinc, and that can change at different times of the year when the cloud is growing right.

So and the sink is basically What receives the sugars and sources, what provides them when root cells are absorbing mineral ions, doesn't mean that they are supplying sugars, are they that's only time when something under the soil is going to be this source is when there's an endosperm.

And when there's a store a starch storage, right like in the case of a storage cells of seed that starting to grow these are sources.

So basically they are absorbinary ions and that's by active uptick, right and active uptake requires energy.

And how do they get this energy? It spiers firing glucose, where are they supplied to the glucose? Well that is where they are the sink.

Okay.

So that's, why I'm going to choose a moving on? We have 31 photo micrograph shows a section of threat to shoot and with an artery.

So which row correctly shows the type of artery.

And whether it's about inside the artery is oxidation or deoxygenated.

So an artery.

And this is more so clearly lung tissue, because of the alveoli that we can see over here.

So obviously the artery is going to be pulmonary artery.

And what we know about the mineral arteries is it carries the oxygen to the blood.

It is a muscular artery and let's see what's what so basically which rock actually shows the type of artery and whether the blood inside the cell.

Yeah.

So type of artery as I mentioned, a muscular and deoxygenated.

So that's, basically, a heart surgery may cause a disease and the transmission of impulse in particular tissue to the right side of the heart.

So what is a possible effect of this disease? So the impulse will be delayed by the atroventricular node I think because like the decrease in transmission to the right side that result, basically in like a weaker contraction by the muscles in the right ventricle.

So let's see they must have argument code contracts, slightly more slowly than the muscle of the left ventricle, possibly, yes.

They must have the right atrium would not contract for as fully as the muscle of the left atrium, really the procurrent tissue.

If you know the structure of the heart it's, um within the septum right? So it has no effect on the atrium, I'm, sorry for the turbo, drawing, but I.

Hope you get the point that I'm trying to give this is the the right atrium, right ventricle here we have this in arterial node, the artery ventricular node and then basically non-conducting band of fibers.

So here we have an insulation and the only way for blood, the ways of excitation to pass from The Atrium to the arterial ventricular to the basically ventricles is through the anterventricular old, then through the perkind tissue, which is right over here to the apex of the heart.

And then like the way up all the way up to towards the ventricle right? So it has no effect on the right atrium.

It has no effect on like what's.

It called the impulse will be made by that like this is already something that happens.

Okay, this is like something natural it's, not part of the disease, there's, no matter when you're transmitting fewer impulses.

It has nothing to do with that.

Okay, the transmission here it's the problem with the proton tissue.

So obviously let's see, okay, excuse me, I by mistake.

So see it's actually B.

Now the diagram shows a part network of advices and Supply blood to the muscles and human body.

So what is correct comparison between the blood at X? And about that y when the muscle tissue is at rest.

So when the muscle tissue is at rest blood pressure, compared at x with enough pressure at y.

So obviously the blood pressure here.

The artery region is higher the blood pressure at y because it's like a capillary and it's, even closer to the vein, better potential of that at X, compared to the water potential of 1.

So also water comes in with a higher water potential.

Because here water will already be going out and into the tissue fluid.

So this unit also has higher and higher I think, that's, C.

Okay.

Now for 34, uh, scientists have shown that basically the oxygen transmission curves for hemoglobin are smaller of smaller moments are to the right of those of larger mamons.

What does it suggest about hemoglobin of smaller mammals? So, uh, homogeneous is going to have a lower Affinity to oxygen, or like in other words, that's it's able to release oxygen more easily exactly.

So this is letter C, but let's read the other options at low partial pressures of oxygen.

It minus oxygen more strongly than hemoglobin of large numbers, that's wrong to the right means it's shifted like this.

So this is the graph.

And then this is the smaller moment.

So it binds to less oxygen, even at like them, lower partial pressures, which yeah is not suggested by this answer.

It saturates with oxygen at lower partial pressures of oxygen than hemoglobin over larger mammal.

Again, no Etc, it's, larger pressure, oxygen.

And as I mentioned C is the answer.

But when the partial pressure of oxygen is high, it carries more oxygen than hemoglobin of larger mammals.

No they carry the same amount if it's high enough, but usually also it will carry more the larger amount will carry more than a smaller mammal.

Okay, 35 what is the effect of an increased concentration of carbon dioxide in the blood? So carbon dioxide is going to decrease the ph and decrease the Affinity of microphone to oxygen.

So increase the movement of chloride ions out of the red blood cells grow the red blood cells.

So here we have CO2.

And then we have water a reversible reaction and then Carbonic acids formed this enzyme here is going to be Carbonic anhydrase.

This dissociates to hydrogen ions and hydrogen carbonate.

Ions hydrogen carbonate.

Ions are going hydrogen ions are going to be bonding with hemoglobinic acid, I'm, sorry, homogeneous hemoglobin to perform homogenic acid.

The important parts for this question here, hard recovery times are going to move out of this cell by the positive diffusion through channel proteins.

And in order to maintain the maintain a balancing balanced polarization state of the plasma membrane, we have chloride ions coming in through the cells.

Therefore, this is wrong.

Okay, it's supposed to move and two, not out of increased concentration of hemoglobinic acid.

That is true because karma as I mentioned here, hydrogen is a bind with hydrogen oxyhemoglobin to formulaic acid, exactly and decreased concentration of hydrogen carbonates ions in the blood plasma.

No as I mentioned it's going to increase decrease concentration of carbamino hemoglobin, actually, no, it will also increase because another way carbon dioxide is going to be transported is by bonding to the terminal a mind group of hemoglobin in order to form.

Carbon homogene answer is B.

Okay, two of the requirements of an efficient gas exchange system are a large surface area.

And a short diffusion distance, which row correctly describes how alveolar adapted to meet these requirements, a large surface area.

Okay.

So in order to increase in our surface area, it has, of course, the alveoliator that's, how we know them.

So, yes, there are photos that interconnectedness room short division distance.

And we can say, they're, one cent thick or let's, see an extracellular layer inside the alveolus contains what capillaries is.

It does not decrease diffusion systems.

Alveolar was our next capillaries.

Okay.

Okay.

I can see how let's look at the sensor guys it's up in the layer.

This does not increase surface area.

Okay walls of the alveoli are one side is thick.

So totally red blood cells are very close to the capillary ones.

Whoever was the alveoli are formed of screaming substitutes as doesn't increase the surface area, see what defines an infectious disease infectious is basically their deviations from health and they're caused by pathogen that could be transmitted from a person to another.

So let's see symptoms are caused by bacteria.

Not necessarily let's, see the other the other answers sometimes are caused by pathogens, totally symptoms are caused by microorganisms that is carried by a vector, not necessarily symptoms are caused by a virus.

It mutates the infect a new species again, not necessarily it could be caused by a virus by by prototists by yeah.

So anything really just basically any type of infectious disease, any type of like a pathogen that comes your body as long as it can be transmitted from one person to another and it's caused by pathogen that is an infectious disease.

And not infectious disease example, is cancer sex anemia.

It's.

A hurricane destroys a large town on an island.

People move away from the towns of that tense or sanitation is poor which this is the most likely to spread within a week of the change of living conditions.

Because sanitation is poor I'm going to assume cholera cholera depends heavily on Sanitation people who live in areas where there's no sanitation.

People have terrible hygienic, Habits, Like, someone doesn't wash his hands before preparing his food, right, he's going to end up transmitting cholera right that's.

Basically, the main thing for TB I would say like if they moved into tents, and it was overcrowded obviously in TB, if they basically moved into a really humid place, I would say, malaria, HIV I'd say, if they're not educated enough, that's 39 or statements correctly explain why viruses are infected by penicillin so here's, another action of penicillin for you to understand there's.

This enzyme known as auto license auto license is an enzyme that breaks down a cell wall and to basically blocks and what that's basically in like newly produced or like, you know and young bacterial cells.

Okay.

So just for growth to happen.

Okay.

So basically, it breaks it down into these subunits and then there's.

Another enzyme, I don't.

We do not really call it names because they really change the names a lot.

So we just refer to it as the enzyme.

It allows the formation of hydrogen bonds between effect and subunits right.

Now, what happens penicillin comes in it's going to accident inhibitor to this enzyme and hence, reducing or stopping the rate of like the formation of the hydrogen bonds between subunited against cell walls, all right.

So basically, the enzyme, the viruses do not have cell walls, they're, not half cells again and do not have the enzyme exercise against subunits, hopefully that made sense to you.

So the answer here is clearly Dean.

Penicillin only blocks come together again, synthesis.

So which process characterizes the mode of action of phagocytes.

Phagocytes We have antibody production, no that's the role of B lymphocytes and specifically plasma cells that are produced by the differentiation of B lymphocytes receptor binding receptor binding.

Okay, sure, because we have adherence that's one of the things that have, in fact, cytosis, okay, we have endocytosis.

True.

Exocytosis is true hydrolysis, uh, through also the action of lysosome, two, three, four and five let's see.

This was the end of this variant.

I hope.

This was useful and good luck.

Everyone on your tests.

Goodbye.

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