how to solve ibo probelms_2

8/10/2019 How to Solve IBO Probelms_2

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Lets continue with past paper of IBO 2012 Part A.

To answer this question you need to remember the difference between C3 and C4 plants. C3 plants are not

prosperous in hot dry conditions because their photosynthetic efficiency is affected by a process calledphotorespiration. This problem of photorespiration is overcome in C4 plants by a two-stage strategy which

keeps high CO2 concentrations and low oxygen concentrations in the bundle sheath cells where the Calvin

cycle takes place.

Therefore, we can conclude that a. is correct because of high CO2 assimilation at high temperatures. b. is

incorrect as we stated above that C3 plants are not adapted to high temperatures. c. must be correct because

if we chose the first graph for C4 plants, then the second graph is for C3 plants and we see that at

temperature 10onet assimilation of CO2 is higher as compared to the first graphs curve at 10o.

Answers:


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Another question concerned with photosynthesis:

You should remember the main steps of photosynthesis which consists of light reactions and dark reaction.

The processes that occur during the light reactions are depicted below:

The main products of light reaction are ATP and NADPH as well as oxygen.

During dark reactions Kalvin cycle takes place during which carbon dioxide is fixed, then reduced by

NADPH followed by regeneration of RuBP with G3P as a by-product.


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Having all this in mind, we can start tackling the problems:

a. is right because light reaction involves the generation of ATP.b. is incorrect because oxygen is just a by-product and the essential initial step is excitation of P680 electrons

which are transferred to the primary electron acceptor. You can come to the answer by remembering about

photorespiration in which oxygen interferes with photosynthesis and thus is unwanted product of

photosynthesis.

c. in photosynthesis ATP is synthesized in linear electron flow involving cytochrome complex. ATP synthase

complex resides in the inner membrane of mitochondrion, no chloroplast. Therefore, the answer is wrong.

d. the linear electron flow proceeds from photosystem II to photosystem I and cyclic electron flow involves

only photosystem I. Take a look at the diagram below. Therefore the statement is incorrect.

e. is correct because as you know light reactions takes place on thylakoid membrane but Kalvin cycle takes

place in the stroma.


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For such types of question you need to know some concepts because it is hard to make guesses if you do not

need the steps of photosynthesis.

The answers:

Lets take a look at another question:

18.1 The key point in this part is to realize that the smaller the surface area per unit volume ratio, the bigger

the organism (remember that cells are really small so that they have a big surface are to volume ratio to

maintain high levels of exchange between intracellular and extracellular environments). Bigger organisms

have smaller surface area for exchange and thereby their metabolic rate is quite sluggish, they do not lose the

heat that rapidly and therefore do not need to compensate for the loss by having high heart or breathing rates.On the contrary, smaller organisms have bigger surface area and thus lose heat quicker and in order to

maintain a constant temperature they have high metabolic rate reflected by higher breathing rate and heart

rate. Having this in mind, you can rank organisms as follows: A will have the highest ratio of surface area to

volume, followed by C, then B and finally D.

18.2 Total volume of blood will be highest in biggest organisms in order to provide nutrients and oxygen to

more cells and will be lowest in small organisms due to small size. This means that the answer will be the

reverse of 18.1: first would be D, followed by B, then C and finally A.


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Lets look at the human anatomy:

You should tick a. as a right one if you remember the renin-angiotensin-aldosterone system which involves a

change in permeability of some ion channels in the kidneys as well as constriction of vessel in the kidneys.

Not to mention reabsorption of water and sodium ions taking place in the nephron. b. is also right because of

the effects of ADH and aldosterone which both increase water reabsorption in the blood concurrently

increase blood volume. The c. is incorrect because the loop of Henle contains only ion channels in the

ascending part and aquaporins only in descending part. Try to remember the diagram showed below to

remind yourself of kidneys functional capabilities.


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In d. you should remember that hyperosmotic urine means that urine is rich in solutes and low in water. As

loop of Henle helps to concentrate the urine by regulating NaCl and water reabsorption, thus directly

affecting concentration of urine. The longer the loops, the more concentrated the urine could be because of

large area for reabsorption. Therefore the statement in question d. is incorrect. Some knowledge of chemistry

can help you with question e. If you remember, CO2 dissolves in aqueous intracellular fluid of erythrocytesand forms a weak H2CO3acid which almost immediately dissociates into hydrogen ion (which alters pH)

and bicarbonate ion which diffuses into the plasma. When bicarbonate reaches the lung alveoli, it combines

with hydrogen ions and produce carbon dioxide which we expire. This reaction helps regulate pH and

complements regulation of blood pH by kidneys which secrete hydrogen ions and reabsorb bicarbonate ions.

Thus, e. is correct.

I would choose f. as a right one, however, the answers below shows that it should be wrong. From the

diagram above we can clearly see that hydrogen ions are secreted and bicarbonate ions are reabsorbed.

Kidneys regulate pH by reducing excretion of hydrogen ions and thereby acidifying overly alkaline urine or

they may increase reabsorption of bicarbonate in order to neutralise overly acidic urine. As for question g.

you should remember that all acids produced in the body are non-volatile apart from carbonic acid, which is

the sole volatile acid because it can dissociate into carbon dioxide and water. If you look at kidneys diagram

above you will see that kidneys do not excrete carbon dioxide which is volatile, but instead reabsorbs

bicarbonate ions. Therefore, the answer is wrong.

To come to the answer of h. you need to remember the diagram of the nephron and that acidosis means a

decrease in pH of blood. Ammonia (NH3) is secreted by the cells in the proximal tubule which acts as a

buffer to trap hydrogen ions producing ammonium ions (NH4+). The more acidic the filtrate is in the

proximal tubule, the more ammonia is produced. So the statement is correct.

The statement i. is correct if you remember that glomerular filtration is influenced by high blood pressure

resulting from differences between diameter of afferent and efferent arterioles. Also try to remember theeffect of angiotensin II which involves constriction of kidney vessels and thus decreasing filtration rate

thereby conserving water.

j. is a hard nut if you are not acquainted with hormones but you should remember that antidiuretic hormone

and oxytocin are two hormones secreted by posterior pituitary. Thus, statement j. is incorrect.


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The answers are:

The following question is interesting because it does not require previous knowledge:

Before starting answering the question you should realize that chewing time is directly related to the type of

food, that is vegetarian food is hard to chew because it contains insoluble cellulose whereas meat is quite

easy to chew. Also you should take into account the size of animal and its energy requirements. Lets begin

by classifying those four animals into two groups: horse and cattle are herbivores whereas wolf and human

are carnivores (but humans are more omnivores nowadays). Thus, as plant food is harder to chew, herbivores

need more saliva to degrade the material then carnivores. And as cattle are bigger than horse, they need more

food for energy and therefore you have one part of the chain which is b c. I would place humans in the

category of 0.75 to 1.5, that is, they should secrete less then herbivores but more that carnivores due to somecellulose-rich food in the human diet. The least amount of saliva, to my mind, should be secreted by wolf

which takes relatively easily digestible diet. In my opinion, the answers should be a


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A brief introduction tells us that the eel was removed from water, therefore respiration by gills should reach

the lowest levels and this suggests that II. corresponds to B. Total blood saturation should decrease a little bit

so that the eel could survive the transition to air from water and this indicates that I. corresponds to A. Then

you should remember that air bladder contains a limited supply of oxygen and therefore it initially provides

quite a lot of oxygen but then gradually decreases is the oxygen is consumed. This means that IV

corresponds to D. Lastly, III must correspond to C as the eel uses its skin both in water and in the air and this

gives a relatively flat line.

The answers are shown below:


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This one is a really interesting one, again requiring no previous knowledge, just logic:

Lets first find two fastest and two slowest swimmers. The key to the answer is by considering fish surface

area and diameter which all have impact on friction. The sharper the front of a fish, the smaller the surface

area is in contact with water thereby reducing friction. Thus, you can see that D and H would be the fastest

swimmers. The slowest swimmers will have the highest area of body in contact with water and you can see

that A and G will have to detail with the greatest levels of friction.Lets then consider the habitats. First of all notice the flat shapes of C and E which provide an evolutionary

advantage of being invisible in the bottom of the sea. You should also notice that seahorse in G has a shape

that reduces mobility and speed and therefore they use camouflage for defense by rapidly changing colors,

allowing them to blend into their surroundings (sea grass beds). Rock crevices are suitable only for slim and

flexible fish, therefore, B (I would say D is a bit bulky and could not maneuver in the crevices) could be

found there. In the surface you would find small non-aggressive fish such as F. I cannot come up with the

rationale of putting fish A at the bottom as I would suggest it should be somewhere in the middle due to its

aggressive appearance which makes this fish a formidable predator. As there are not such many preys at the

bottom in the sea, this predator would not find what to eat. However, the answers suggest it should reside inthe bottom. Lastly, fish D could be found in the middle layer as they are quite big (and thus would be easily

spotted by predators if it swam in the surface), not flat enough for bottom habitat, not that flexible for rock

crevices, and too big for sea grass beds. This leaves you with the answer that D is found in the middle.


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Another task involving graphs:

29.1 part asks to fill the table:

L/D < 1

> 45

L/D < 1 means that drag is higher that lift and thus from the introduction above you can highlightthat animals that parachute maximize D as they often have no significant surface area to produceenough L. If you compare surface area of flying squirrel and flying frog, you can see that frog hassmaller surface area and therefore produces L/D < 1. Consequently, you should choose answer B.

The way I approach the second part when > 45, is by drawing a small picture with vectors andapplying law of tangents.

The number 1.1 means that D is higher than L, the same as in the first part, thus we can concludethat organism b. will have a higher angle than 45 degrees.


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Now lets look at the graphs to make correct conclusions.

a. statement is true because if you look at the first graph you should notice that there are no points ofDraco

volanslog weight beyond point 1.25, whereas log weight ofPtychozoon kuhli extends to 1.5. If you

evaluate the lines in the first graph of two organisms you can see that they quite coincide.

b. statement is incorrect because when you look at the second graph and compare y values forPtychozoon

kuhli andDraco Volans when x value is the same for both species, you see that the line of Draco volans

is significantly higher than that ofPtychozoon kuhli.

c. Statement is incorrect because when you look at graph 3, you can see that values forPtychozoon kuhli

are higher as compared toDraco volansvalues.

d. is correct because you have to look at graph 4 and the values quite neatly coincides.

e. is correct because you compare graphs 2 and 3 and see thatDraco volans; has bigger patagium area

whereasPtychozoon kuhli has bigger accessoryarea.

The answers are provided below:

how to solve ibo probelms_2

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