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Aqa a level biology > Topic 5: energy transfers between organisms > Flashcards

Topic 5: energy transfers between organisms Flashcards

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1
Q

describe the function of a chloroplast

A

where photosynthesis ocuurs
-thykaloid discs containing chlorophyll absorb light to produce carbohydrates and sugars

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2
Q

features of chloroplasts

A

-Features of chloroplasts –> starch grain, ribosomes (70S), outer and inner membrane, stroma, thykaloid, granum

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3
Q

thykaloid

A

Thykaloid –> contains chlorophyll and absorbs light

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4
Q

starch grain

A

Starch grain –> stores polymers of glucose

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5
Q

lamellae

A

Lamellae –> proteins that hold granum in optimum position to absorb light

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6
Q

absorbance and reflection by chlorophyll

A

Chlorophyll reflect green light and absorbs all other colours such as red

Accessory pigment –> maximize the amount of pigment that can be absorbed e.g carotenoids

-chlorophyll is a mixture of pigments each absorbing different wavelengths of light

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7
Q

Describe how you would present the data in the table as a graph

A

discrete data (distinct groups) = bar chart with standard deviation lines

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8
Q

In leaves at the top of trees in a forest. CO2 is often the limiting factor for photosynthesis. Explain why

A

-light = not limiting as there is now shading
-temperature = not limiting as no shading (fast reactions of enzymes in LDR)

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9
Q

structure of chloroplasts

A

Grana –> the stacked collection of thykaloid discs

-Thykaloid –> a small membrane bound sac or disc containing chloropyll

-Stroma –> a fluid filled space within the chloroplast containing an abnormally high concentration of protons (H+)

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10
Q

light dependent reaction

A

1) the first photosystem recieves light energy from the sun and uses this for photolysis (breaking w light) of water into H+ ions, O2 and electrons

2) The electrons are elevated to a high energy state allowing them to move through the photosystems giving energy to each membrane protein. It eventually recieves and is accepted by an electron accepting molecule called NADP forming NADPH (reduced NADP)

3) the energy given to each photosystem by the electron is used to actively transport protons (H+) from the stroma into the thykaloid to build a chemiosmotic gradient (conc gradient for protons)

4) The protons in the thykaloid are pumped through an enzyme called ATP synthase which provides energy to turn the head of the enzyme and force the reaction of ADP and Pi to form ATP

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11
Q

Calvin cycle (light independent reaction)

A

1) 5 C chain – ribulose bisphosphate (RuBP)

2) CO2 –> carbon fixation with an enzyme called rubisco (catalyses joining of CO2 and RuBP)

3) unstable 6 carbon intermediary molecule -> quickly breaks down to form 2 molecules of glycerite-3-phosphate (GP)

4) ATP from LDR to ADP

5) Reduced NADP from LDR to NADP return to LDR

6) triose phosphate x2 = glucose (some of triosephosphate is used to form sugars)

7) ATP from LDR to ADP and Pi

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12
Q

DCPIP practical

A

DCPIP (LDR):

-DCPIP competes with NADP as an additional electron acceptor. Reduces rate of LDR as there is less reduced NADP

-different concentrations of CDCPIP + plant cells (IV)

-DV = rate of photosynthesis measured by mass of sugar produced

-CV = light, CO2, temp, H2O

-higher conc of DCPIP the lower the mass of sugar produced

Limitations –> need balances w high resolution, less reliable - can measure glucose conc after using benedicts reagent and a colorimeter

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13
Q
A
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14
Q

chromatography practical

A

-used chlorophyll based substance

-use capillary tube to add a drop of solution to a pencil drawn baseline

-dip chromotography paper into a solvent below baseline

-remove before it reaches top and identify solvent front

-calculate Rf vaue = distance travelled by compound / distance travelled by solvent front

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15
Q

heat stress decreases the light-dependent reaction of photosynthesis

Explain why this leads to a decrease in the LDR

A

-there will be less ATP and less reduced NADP formed which are both needed in the LDR

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16
Q

why would a decrease in the activity of the enzyme rubisco would limit the rate of photosynthesis

A

-decreasing activity of RUBISCO reduces amount of carbon fixation of CO2 and RuBP to for intemediary 6 carbon unstable molecule therefore reducing amount of GP made

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17
Q

where is rubisco found in a cell

A

stroma

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18
Q

what part of the chloroplast does LDR occur

A

thykaloid membrane

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19
Q

define the term photolysis

A

splitting of water using light

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20
Q

why is photolysis also known as photoionisation

A

-loss of an electron in the prescence of light

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21
Q

equation for photolysis

A

H2O –> 2H+ + 2e- + 1/2O2

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22
Q

site of LDR

A

thykaloid mebrane

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23
Q

light energy in LDR is converted to

A

ATP and NADPH

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24
Q

photolysis of water is important form

A

generating an electrochemical gradient

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25
Q

describe how oxygen is produced during the light-dependent reactions of photosynthesis

A

-oxygen is produced via the photolysis of water (splitting of water using light)

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26
Q

ATP is synthesised from ADP and Pi during the LDR. Describe the structures in a chloroplast that are involved in the reaction

A

-H+ ions pumped from the stroma and through thykaloid membrane
-thylakoid membrane contains protons that acts as electron carriers (electron transport chain)
-ATP synthase

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27
Q

explain how a herbicide which reduces the transfer of electrons reduces the rate of photosynthesis in weeds

A

-less energy re;eased as less electrons move down electron transport chain
-less H+ ions pumped out of thylakoid membrane and through ATP synthase
-fewer H+ ions = weaker electrochemical gradient
-less energy to phosphorylate the production of ATP by ADP and Pi
-less H+ ions to reduce NADP

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28
Q

electron transport chain establishes

A

chemiosmotic gradient
an electrochemical gradient for the active transport of H+ ions

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29
Q

describe what happens during the light dependent reaction (5)

A

-chlorophyll absorbs light energy
-electrons are exicted and exit the chloroplasts
-electron transport chain releases energy to pump H+ ions from stroma and through thylakoid membrane
-energy used to join ADP and Pi into ATP
-photolysis of water (write equation)
-NADP reduced by electrons into NADPH

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30
Q

stage 1 of light dependent reaction

A

a photon of light hits a molecule of chlorophyll in photosystem II. The chlorophyll molecule becomes excited and loses an electron, becoming oxidised. This is called photoionisation. The electron is transferred to the primary pigment molecule via a series of REDOX reactions

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31
Q

stage 2 of LDR

A

-Stage 2 = the organisation of the chlorophyll molecules means that energy becomes focused on the primary pigment reaction centre. This releases a high energy electron which then enters the electron transport chain

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32
Q

stage 3 of LDR

A

the lost electron must be replaced in order for the process to continue. To provide the electrons a molecule of water needs to be split.

H2O –> 2H+ + 2e- + ½O2

This only happens in the presence of light (photoionisation)

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33
Q

stage 4 of LDR

A

the electron lost from photosystem II passes down the electron transport chain via a series of REDOX reactions, with each reaction energy is lost. This energy is used to pump H+ ions from the stroma into the thylakoid space. This begins the process of chemiosmosis

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34
Q

stage 5 of LDR

A

at the same time that the chlorophyll molecules in photosystem II is being excited, the chlorophyll molecule in photosystem I is also excited. This means that it also loses an electron. This electron along with H+ ions from the photolysis of water, bind with NADP to form reduced NADP/ NADPH. The electron lost form te photosystem I is replaced by the electron lost from photosystem II

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35
Q

chemiosmotic theory

A

-the concentration of H+ ions in the thylakoid space increases

-the H+ ions then diffuse down a proton gradient via ATP synthase. The movement of H+ ions catalyses the formation of ATP

-ADP + Pi –> ATP

-this is known as photophosphorylation (addition of a phosphate molecule using light)

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36
Q

products of light dependent reaction

A

ATP
NADPH
oxygen

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37
Q

photosystem

A

Photosystem = protein containing lots of chlorophyll pigment molecules

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38
Q

electron transport chain

A

provides energy for chemiosmosis = series of proteins embedded into the membrane along which electrons can pass

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39
Q

reduced NADP

A

NADP gains H+ (from water) and e- (from chlorophyll) to be reduces

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40
Q

stage 1 of calvin cycle (light independent reaction)

A

1) CO2 is fixed by combining with a 5 carbon compound (RuBP) ribulose biphosphate. This is catalysed by the enzyme RuBISCO

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41
Q

stage 2 of calvin cycle

A

2) The 6-C compound produced is unstable so quickly breaks down into two 3-C compounds of glycerate-3-phosphate (GP)

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42
Q

stage 3 of calvin cycle

A

3) The GP is then reduced using energy from the breakdown of ATP and an electron from reduced NADP to form triose phosphates

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43
Q

stage 4 of calvin cycle

A

4) 80% of molecules of TP made are converted back into RuBP. The TP molecules re-arrange themselves to form 5-C compounds and are then phosphorylated to form RuBP

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44
Q

stage 5 of calvin cycle

A

5) 20% of the molecules of TP are used to create a 6-C compound known as glycose by binding with another molecule of TP. This can be isomerised into fructose and converted into another range of organic molecules

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45
Q

chemicals needed for LDR

A

NADP, ADP, Pi and water

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46
Q

describe what happens during photoionisation in the LDR

A

-chlorophyll absorbs light energy which causes electrons to be lost as they are exicted and move out the chlorophyll

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47
Q

explain why the studen marked the origin using a pencil rather than ink

A

ink and leaf pigments would mix

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48
Q

describe the method the student used to separate the pigments are the solution

A

-level of solvent below origin line
-remove before reaches the top end

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49
Q

suggest and explain the advantage of having different coloured pigments in leaves

A

absorbs more wavelengths for photosynthesis

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50
Q

explain how atrazine reduces the rate of photosynthesis in weeds

A

-reduced chemiosmotic gradient
-less ATP and NADP produced
-LDR slows

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51
Q

explain what would happen to the pH of the solution during this investigation

A

-pH would increase
-CO2 removed for photosynthesis

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52
Q

suggest why the rate of photosynthesis was low between these wavelengths of light

A

-less absropton
-light required for LDR

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53
Q

why does iron deficency result in a decrease in uptake of CO2

A

-less TP converted to RuBP
-CO2 combines with RuBP

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54
Q

how does temperature affect enzyme activity

A

low temperatures = low kinetic energy = less chance of collisions = less E-S complexes

high temp = enzyme denature = ionic bonds break in tertiary structure = active site changes shape = no longer complimentary

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55
Q

glucose + nitrates =

A

amino acids

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56
Q

photosynthesis limiting factors

A

CO2
H2O
light intensity
temperature

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57
Q

how does disruption in an electron transport chain reduce growth

A

less energy for chemiosmosis = less ATP and NADPH produced = less GP reduced to TP = less glucose synthesised = less other organic compounds such as amino acids

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58
Q

why do low CO2 levels affect the LDR

A

Less CO2 is fixed to RuBP = less GP reduced to TP = less NADPH oxidised = less NADP to accept electrons in electron transport chain of LDR = LDR slows

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59
Q

reduced TP production in iron-deficient plants

A

if electron transport is reduced then there is a lower chemiosmotic gradient to pump H+ ions through ATP synthase to produce ATP from Pi = less ATP and less NADP by electrons = less energy and electrons to reduced GP to TP

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60
Q

what is the site of photosynthesis in a plant

A

leaf

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61
Q

ATP production

A

ATP production in both photosynthesis and respiration is formed when protons travel down an electrochemical gradient through molecules of ATP synthase

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62
Q

adaptations of the leaf

A

-large surface area
-thin
-transparent cuticle and epidermis
-long narrow packed mesophyll cells
-stomata that respond to changes in light intensity

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63
Q

stroma + thylakoid

A

-H+ ions are pumped from the stroma using protein carriers in the thykaloid membrane

-photolysis of water increases H+ ion concentration in the thykaloid space and low concentration in the stroma

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64
Q

adaptations of chlorophyll in the light dependent reaction

A

-thylakoid membrane provides large surface area for attachment of chlorophyll/electrons
-proteins in the granumn hold in place for maximum absorption
-selectivley permeable = establish protein gradient
-contain DNA + ribosomes

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65
Q

dehydrogenase

A

Dehydrogenase​​ is an ​enzyme ​​found in plant chloroplasts that is crucial to the ​light dependent stage​​ of photosynthesis. In the light dependent stage, ​electrons ​​are accepted by ​NADP. ​​ Dehydrogenase ​catalyses​​ this reaction.

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66
Q

carbon numbers in light independent reaction

A

RuBP = 5 C
GP = 3
TP = 3
Glucose = 6

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67
Q

explain why the light independent reaction slows down at low temperatures

A

-light indepednent reaction involves enzymes
-enzymes denature = less kinetic energy = less collision

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68
Q

how much triose phosphate is converted to RuBP

A

83%

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69
Q

which process is the source of ATP used in the conversion of GP to triose phosphate

A

light dependent reaction / photophosphorylsation

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70
Q

why were all tubes placed at the same distance from the lamp

A

so all tubes recieve the same amount of heat

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71
Q

describe the part played by chlorophyll in photosynthesis

A

-absorbs light energy
-excites electrons
-form ATP

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72
Q

explain how a lack of light caused the amount of radioactively labelled glycerate 3 phosphate to rise

A

-ATP and reduced NADP not formed
-less GP to form RuBP

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73
Q

true or false -> ATP synthase is a transport protein not an enzyme

A

false -> ATP synthase is an enzyme and catalyses the formation of ATP from ADP + Pi

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74
Q

active transport in LDR

A

There is a higher concentration of protons in the thylakoid space compared to the stroma. To maintain this proton gradient, there is active transport of protons from the stroma into the thylakoid space.

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75
Q

oxidation vs reduction

A

oxidation = loss of H+ ions

reduction = gain of H+ ions

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76
Q

no electron transport chain =

A

protons cant be actively transported into thylakoid space = no proton gradient in stroma = no ATP

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77
Q

differences between glycerate-3 phosphate and TP

A

glycerate = negative O while TP has a H so needs to be reduced by NADPH to gain the hydrogen

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78
Q

adaptions of chloroplast

A

-large surface area = many thylakoid
-chlorophyll pigment allows maximum absoroption of light energy
-grana surrounds stroma which allows products to move to the stroma from LDR
-chloroplast has ribosomes to make enzymes for photosynthesis
-contains accessory pigments to absorb different wavelengths of light
-thylakoid has small internal volume which maximises H+ conc

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79
Q

what happens when GP is converted to TP

A

GP is reduced
-NADPH is oxidised into NADP
-ATP is hydrolysed into ADP + Pi

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80
Q

how many cycles of the calvin cycle are needed to form 1 glucose molecule

A

6

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81
Q

true or false -> ATP is needed to convert RuBP to GP

A

true

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82
Q

dark =

A

stomata close
CO2 not used/no uptake
not used in photosynthesis

some CO2 uptake through upper surface

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83
Q

explain why GP remained constant

A

-GP being used and reformed at same rate
-GP reduced to TP

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84
Q

factors affecting photosynthesis

A

-CO2
-temperature
-pH
-water supply
-light intensity / wavelength / duration

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85
Q

CO2

A

Low CO2 affects the Calvin Cycle. If CO2 levels are low, rubisco cannot convert RuBP to GP in step one of the Calvin Cycle. This leads to accumulation of RuBP and an overall slowing of the Calvin Cycle, which results in a fall in the production of TP/GALP.
High CO2 can cause stomata to close. However, if CO2 rises too high, then some stomata begin to close, which can lead to less CO2 uptake by the plant.
The ideal CO2 concentration is 0.4%. The atmospheric CO2 concentration is 0.04%. Increasing this by ten times to 0.4% will increase the rate of photosynthesis

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86
Q

water

A

Water is needed for photolysis in the light-dependent stage. Low levels of water prevent efficient photolysis occurring during the light dependent reactions. This will disrupt production of ATP and reduced NADP, both of which are needed for the Calvin Cycle.
Water vapour and soil are sources of water. Plants can obtain water through their stomata from water vapour in the air, and also by absorbing water from soil via their roots.

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87
Q

light intensity

A

Light is needed for the light-dependent stage. Light energy is needed to excite the electrons and for photolysis of water in the light-dependent stage. Without light, there would be little ATP and reduced NADP produced for the Calvin Cycle.
Certain wavelengths of light are absorbed. Light is absorbed by the photosynthetic pigments (e.g. chlorophyll a), as we learnt before. Green light is not absorbed but is instead reflected.

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88
Q

role of dehydrogenase

A

catalyses the reduction of NADP to NADPH at end of the electron transport chain

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89
Q

role of DCPIP

A

An electron acceptor that can accept electrons at the end of the electron transport chain to prevent NADP from being reduced to NADPH.

blue to colourless when reduced

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90
Q

purpose of control experiments

A

To ensure we can see if the independent variable is the only factor affecting the investigation. In order to make comparisons.

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91
Q

purpose of ice

A

Cold reduced the enzyme activity which prevents to hydrolysis of organelles.

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92
Q

why did the student set up tube 1

A

to show light doesnt affect DCPIP
show reaction cant occur without chloroplasts

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93
Q

explain the advantage of using IC50 in the investigation

A

to compare different chemicals

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94
Q

explain how chemicals which inhibiti the decolourisation of DCPIP could slow the growth of weeds

A

-reduces energy in electron transport chain
-DCPIP accepts less electrons
-less ATP produced
-less NADPH produced
-less GP reduced to TP

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95
Q

the teacher said we could not draw definite conclusions = why

A

no error bars to show if overlap occurs
-cannot determine significance

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96
Q

weeds have been shown to give off small amounts of heat - why

A

energy released from electrons excited from chlorophyll molecules

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97
Q

why did the relative amounts of GP and RuBP remain the same

A

-temperature = limitng factor below optimum
-light intensity = limiting factor

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98
Q

purpose of the 3 leaf treatments

A

1) compare = see if open stomata reduces CO2 uptake
2) stops CO2 uptake
3) CO2 uptake cannot occur at all

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99
Q

advantage of light being off

A

prevents water loss by transpiration
maintains water potential gradient in cells

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100
Q

why does CO2 uptake close to zero when light is off

A

stomata close in dark
no diffusion gradient for CO2 into the leaf

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101
Q

order of electron carriers

A

Y X W Z

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102
Q

explain why low rate of photosynthesis between 525 and 575nm of light

A

-pigments reflected - > less absorbed
-light required for photolysis
-represents green pigment

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103
Q

why did the conc of radioactive RuBP increases

A

no CO2 to combined with RuBP

104
Q

Describe how ATP is resynthesised in cells

A

-ATP synthase catalyses the condensation reaction between ADP + Pi to form ATP
-during respiration
-elimination of water molecule

105
Q

2 ways in which the hydrolysis of ATP is used in cells

A

1) phosphorylates other reactions making them more reactive
2) active transport

106
Q

why is ATP used over glucose as an energy store

A

-ATP is more manageable as it releases less energy

-ATP cannot leave the cell it was made in whilst glucose can

-ATP requires less energy to be broken down

-ATP doesnt take up storage space

ATP –> ADP + Pi (hydrolysis) (release 30.6 KJmol-1)

107
Q

uses of ATP

A

-Phosphorlyation = make more reactive

-biosynthesis (protein synthesis)

-cell division (mitosis)

-cell signalling (hormones)

-thermoregulation

-cell mobility (flagellum)

-active transport of substances across a membrane

108
Q

process of respiration

A

glycolysis –> link reaction –> krebs cycle –> oxidative phosphorylation

109
Q

where does glycolysis occur

A

-glycolysis takes place in the cytoplasm while the rest takes places in the mitochondria

-glucose = too large to fit into mitochondria

-glycolysis (anaerobic process)

-oxidative phosphorylation (Aerobic = more energy)

110
Q

products of glycolysis

A

Products –> 2 ATP molecules, x2 NADH, x2 pyruvate

111
Q

energy investment + energy harvesting phase (glycolysis)

A

Energy investment phase –> stable 6 Carbon glucose molecule is phosphorylated by 2 molecules of ATP form glucose phosphate.

Energy harvesting phase –> 6 carbon glucose phosphate is broken down into 2 molecules of TP (3 carbon molecules with Pi). Oxidation occurs where NAD is reduced to NADH involving the loss of x2 ATP molecules. This forms 2 molecules of pyruvate (stable 3 carbon molecule)

112
Q

steps of glycolysis

A

1) Glucose is phosphorylated using ATP

2) The phosphorylated glucose molecule quickly breaks down into 2 molecules of TP

3) TP is oxidised into pyruvate releasing ATP and reducing NAD into NADH

3) Net products –> x2 pyruvate, x2 ATP, x2 NADH

113
Q

The link reaction

A

-to enter the krebs cycle pyruvate needs to be converted into Acetyl CoA (a coenzyme)

-1) Decarboxylation of pyruvate to form CO2 biproduct (loss of CO2)

2) Pyruvate is oxidised resulting in the dehydrogenation of NAD to NADH + H+ (loss of hydrogen)

3) This forms acetate (2 carbon molecule)

4) This coenzyme A attaches to acetate to form acetyl CoA

114
Q

acetate

A

Acetate = x2 carbon molecule added to a co-enzyme to form acetyl CoA

115
Q

function of coenzyme A

A

Function of Co-enzyme A = transport remaining acetyl to the krebs cycle

116
Q

products of link reaction

A

Products of link reaction –> x2 acetyl CoA, x2 CO2, x2 NADH

117
Q

other than carbon compounds what other products are produced during the krebs cycle

A

ATP
NADH
FADH2

118
Q

aerobic vs anaerobic respiration

A

aerobic = requires oxygen –> produces ATP, CO2 and water

anaerobic –> takes place in the absence of oxygen + produces lactate in animals or ethanol + CO2 in yeast cells.

Both produce a little ATP

119
Q

summary of glycolysis

A

-activation of glucose by phosphorylation
-splitting of phosphorylated glucose into 2 TP
-TP is oxidised
-production of ATP

120
Q

true or false –> enzyme for glycolysis are only found in the cytoplasm

A

true

121
Q

coenzymes in photosynthesis and respiration

A

NAD + FAD = respiration
NADP = photosynthesis

NAD works with the dehydrogenase enzyme to catalyse the removal of hydrogen

122
Q

importance of the krebs cycle

A

-breaks down macromolecules into smaller ones
-produces hydrogen atoms
-regenerates 4c molecule to bind with acetylCoA
-involved in the manufacture of other compounds

123
Q

mitochondria in muscle cells

A
124
Q

importance of oxygen in respiration

A

-act as final acceptor for hydrogen atoms

125
Q

purpose of Krebs cycle

A

-Purpose of the krebs cycle is to reduce electron carriers to carry electrons and hydrogen in the electron transport chain to make ATP

-takes place in the mitochondrial matrix

126
Q

Net products of krebs cycle for 1 molecule of glucose

A

-Net products –> x2 ATP, x2 CoA, x2 FADH2, x6 NADH, x4 CO2 (must occurs twice for 1 glucose molecule bc 2 pyruvate)

127
Q

ATP + substrate level phosphorylation

A

-substrate level phosphorylation –> direct formation of ATP by transferring a phosphate group from a high energy compound (4C) to an ADP molecule

-In both respiration and photosynthesis, ATP production occurs when protons diffuse down an electrochemical gradient through molecules of the enzyme ATP synthase, embedded in the membranes of cellular organelles.

128
Q

aerobic vs anaerobic

A

-If respiration is only anaerobic, pyruvate can be converted to ethanol or lactate using reduced NAD. The oxidised NAD produced in this way can be used in further glycolysis.

If respiration is aerobic, pyruvate from glycolysis enters the mitochondrial matrix by active transport

129
Q

steps for krebs cycle

A

1) Acetyl CoA (2C) binds to 4C molecule to form Citrate (6C)

2) 6 carbon molecule is decarboxylated and oxidsed to form 5C molecule ketogluterate. NAD is reduced to NADH

3) 5 carbon molecule is oxidsed back into 4 carbon molecule (oxaloacetate). 2 molecules of NAD are reduced to NADH. FAD is reduced to FADH2. ADP + Pi participate in substrate level phosphorylation to form ATP, CO2 is produced during decarboxylation of 5 carbon molecule

4) This cycle must occur twice for 1 glucose molecule as 2 pyruvate are formed at the end of glycolysis and the purpose of the krebs cycle is to produce reduced electron carriers for the electron transport chain

5) Net products

130
Q

oxidative phosohpurlation

A

Reduced NAD and FAD release hydrogen atoms. Reduced NAD and reduced FAD from the previous steps release hydrogen atoms in the mitochondrial matrix, and in the process they become NAD and FAD again

Hydrogen atoms break up into protons and electrons. The hydrogen atoms split up into both H+ ions and electrons. The H+ stay in the matrix.

The electrons enter the electron transport chain (ETC). The electrons are taken up by the electron carriers in the ETC. The electrons move along the ETC, from carrier to carrier, and at each carrier the electrons release energy.

The energy from electrons is used to pump protons. The released energy at each carrier is used by the electron carriers to pump H+ from the mitochondrial matrix into the intermembrane space. This forms an electrochemical gradient – there is a higher H+ concentration in the inter-membrane space than in the matrix.

Protons diffuse down the gradient. Protons diffuse from the inter-membrane space to the matrix, down the electrochemical gradient. This movement of protons is chemiosmosis.

ATP synthase transports the protons. Protons are unable to diffuse through the phospholipid bilayer, so instead go through the enzyme ATP synthase. Proton movement provides potential energy, which rotates a section of ATP synthase, and causes phosphorylation of ADP (ADP + Pi → ATP).

Water is formed. The electrons leave the last electron carrier and pass into the matrix, where and are accepted by oxygen. H+ also joins, forming water – a product of respiration.

131
Q

chemiosmosis + oxidative phosphorylation

A

Chemiosmosis = movement of ions across a membrane bound structure down an electrochemical gradient. Movement of H+ ions to produce ATP.

-oxidative phosphorylation is the last stage of respiration and takes place in the inner mitochondrial membrane (cristae)

132
Q

purpose of glycolysis

A

-the purpose of glycolysis is to produce pyruvate to diffuse through the mitochondrial membrane for the link reaction as glucose is too large a molecule

133
Q

purpose of link reaction

A

-The purpose of the link reaction is to convert pyruvate by binding to a 2 carbon molecule to form acetyl CoA to enter the krebs cycle

134
Q

purpose of krebs cycle

A

-the purpose of the krebs cycle is to reduce electron carriers in preparation for oxidative phosphorylation

135
Q

respiration equation

A

C6H12O6 + 6O2 –> 6CO2 + 6H2O

-glucose = broken down into pyruvate during glycolysis

-CO2 is made in link/krebs cycle due to carboxylation

136
Q

absence of oxygen

A

Process of respiration would stop in the absence of oxygen as it wouldn’t be able to act as the final electron acceptor + H+ wouldnt be produced due to less pumping. NADH/FADH2 wouldn’t be oixdised so electron transport will stop as less energy is released. Weaker proton gradient to siddiuse back down ATP synthase, less ATP produced.

137
Q

steps for oxidative phosphorylation

A

1) NADH and FADH2 drop of e-/H+ ions in mitochondrial matrix returning to krebs cycle as oxidised molecules

2) electrons travel down the electron transport chain via a series of redox reactions. Energy is lost from electron transport chain to pump H+ ions across the inner mitochondrial membrane

3) H+ ions diffuse back down a proton gradient to mitochondrial matrix via ATP synthase which allows ATP to be formed by the phosphorylation of ADP + Pi

4) oxygen acts the final electron acceptor

138
Q

products of oxidative phosphorylation

A

Net products = H2O, ATP, FAD, NAD

139
Q

the respiratory solution and ADP added after 5 minutes was part of a buffered isotonic solution .what other substance would the solution contain

A

inorganic phosphate

140
Q

explain the line between P and Q

A

-no aerobic respiration
-no substrate

141
Q

line between Q and R

A

-oxygen concentration falls
-aerobic respiration uses oxygen during oxidative phosphorylation
-oxygen is an electron acceptor to combine with H+ and electrons to form water

142
Q

describe and explain the difference between line R and S to R and T (plant vs human mitochondria)

A

-O2 conc falls in animals but stays constant in humans
-aerobic respiration continues in plants
-electron transfert still continues

143
Q

hydrogen atoms

A

split into electrons and protons (H+) for electron transport chain -> chemiosmotic theory

144
Q

in natural ecosystems most light is not used for photosynthesis. Why?

A

light is reflected
light is the wrong wavelength

145
Q

role of pyruvate in anaerobic respiration in animals

A

-hydrogen acceptor
-reduced to lactate so NAD can be regenerated
-so glycolsysi can continue and ATP is produced

146
Q

why can aerobic respiration not take place in the abscence of oxygen

A

-oxygen = final electron acceptor
-reduced chemiosmotic gradient + proton gradient
-no water formed leading to abundance of H+ and e-
-causes electron transport chain to stop as all e- are saturated
-NADH and FADH2 wont be oxidised
-less energy released to pump H+ ions through ATP synthase

147
Q

give one piece of evidence from the diagram which suggest that the conversion of pyruvate to ethanol involves reduction

A

reduced NAD is oxidised into NAD (Redox reaction)

148
Q

why converting pyruvate to ethanol is important in allowing the continued production of ATP in anaerobic respiration

A

-convert pyruvate into ethanol = produces ATP
-NAD is regenerated for glycolysis

149
Q

electrons in ETC move down energy gradient

A
150
Q

alternative respitory substances

A

-lipids + proteins

glycerol from lipds are phosphorylated and converted into TP and fatty acid component is converted to acetyl-CoA

proteins = hydrolysed into amino acids = either converted into pyruvate or intermediate for krebs cycle

151
Q

equation = fermenation in plants/yeast cells

A

pyruvate + NADH –> ethanol + CO2 + NAD+

152
Q

equation = fermentation in animals

A

pyruvate + NADH –> lactate + NAD+

153
Q

anaerobic respiration in animals

A

-glycolysis is the only process from aerobic respiration that can occur without oxygen

-purpose of anaerobic respiration = regeneration of NAD

-Glycolysis –> fermentation = anaerobic respiration

154
Q

glucose –> lactate

A

Glucose –> 2 pyruvate –> 2 Lactate

155
Q

lactic acid

A

Lactate dissolves into lactic acid which causes pH to lower thus denaturing proteins/enzymes

Lactate is eventually taken into rive to be regenerated into pyruvate then into glucose so can re-enter glycolysis. This requires ATP (produced during aerobic respiration). Regeneration leads to oxygen debt where athlete continues at high level of O2 consumption (oxygen debt).

156
Q

steps for fermentation in animals

A

1) Glycolysis products pyruvate

2) Pyruvate is reduced into lactate

3) NADH is oxidised into NAD

157
Q

gluconeogensis

A

Gluconeogenesis –> creation of new glucose using lactate (2 lactate –> glucose (requires 6 ATP))

158
Q

rapid ATP production in animals

A

-anaerobic respiration

-creatine phosphate

159
Q

phosphocreatine

A

1) in prescence of ADP, phosphorcreatine breaks down to release energy and phosphate group

2) no oxygen required + energy for 6-10 seconds = phosphocreatine regenerated at rest

Creatine (work) ⇌ phosphocreatine (generate energy ATP) (rest)

160
Q

anaerobic respiration in yeast cells

A

1) glucose is converted to pyruvate via glycolysis

2) ATP is formed and 2H converted into NADH and then NAD in a cycle. 2H is returned to ethanol = reduced

3) pryuvate is decarboxylated into ethanal

4) alcohol dehydrogenase (enzyme) catalyses the reduction of ethanal into ethanol

Glucose –> ethanol + CO2 + (ATP)

161
Q

compare fermentation in animals + yeast (3)

A

-both work to regenerate NAD for glycolysis

-both produce similar amounts of ATP

-Fermentation in animals produce lactate whilst in yeast ethanol and CO2 is produced

-pyruvate is decarboxylated into ethanol in yeast cells whilst in anmals pyruvate is reduced into lactate

162
Q

suggest an explanation for the effect of temperature on the rate of CO2 release

A

-enzymes = faster
-higher rate of respiration and CO2 replease
-spiracles open more often

163
Q

explain why the scientist did not used glucose as the respitory substrate for these isolated mitochondria

A

-glucose is broken down during glycolysis
-glucose cannot cross mitochondrial membrane but pyruvate can

164
Q

explain how malonate inhibitis the formation of fumarate from succinate

A

-it is a competitive inhibitor
-reduces enzyme substrate complexes from forming

165
Q

why did O2 uptake decrease when malonate was added

A

-krebs cycle inhibited
-hydrogens not passed to ETC therefore oyxgne is not used as final electron acceptor

166
Q

what measurements should the student have taken to calculate the rate of aerobic respiration

A

-mass of woodlice
-diamteter
-time and distance

167
Q

describe the part played by the inner membrane of a mitochondrion in producing ATP

A

-electrons transferred down electron transport chain
-provide energy to take protons into space between membranes
-protons pass into matrix
energy used to produce ATP

168
Q

explain why oxygen is needed for the production of ATP on the cristae of the mitochondrion

A

-ATP formed as electrons pass electron transport chain
-oxygen is terminal electron acceptor
-accepts H+ from reduced NAD

169
Q

describe how pyruvate is converted into lactate and NAD during anaerobic respiration

A

-pyruvate acts as a hydrogen acceptor
-forming NAD by accepting the H+ from NADH

170
Q

reject null hypothesis

A

calculate value is greater than critical value = statistically significant

171
Q

radioactive carbon converted to sugars during

A

photosynthesis (LIR) = can be moved during mass transport in phloem

172
Q

A = energy lost from gut
C = energy consumed in good
F = energy lost in faces
P = energy used in production of new tissue
R = energy lost by respiration
U = energy lost in urine

Complete the equation using the letters for the production of new tissue

A

P = C - (F + R) - U

173
Q

calculate the value of P

A

91.34 - (57.06 + 30.51) - 0.03
= 3.74

174
Q

it has been estimated that an area of 8100m2 of grassland is needed to keep one cow. The productivity is 21,135 m2 per year. What is the energy percentage used in the production of new tissue in one cow.

A

(3.74 x 10^6) x 100 / (21135 x 8100)

x 100 = percentage
denominator = total energy x productivity

= 2.18

175
Q

keeping cattle indoors in barns leads to a higher efficiency of energy transfer. Explain why

A

less energy lost as heat

176
Q

A population of deer consume 40 000 kJ m2 yr −1 of energy.
27 000 kJ m2 yr −1 of this energy is lost as excretion (urine and faeces).
5000 kJ m2 yr −1 energy is used in respiration.
Net production of consumers is calculated as

A

N = 40 000 − (27 000 + 5000)
N = 40 000 − 32 000
N = 8000 kJ m2 yr −1

177
Q

productivity

A

-productivity –> the rate at which this new biomass is generated in an ecosystem
Productivity is the rate of production of new biomass. There are two categories of productivity: primary and secondary. .

178
Q

biomass

A

-biomass –> mass of carbon/dry mass of tissue pre given area per given time

179
Q

what does the productivity of an ecosystem depend on

A

-sunlight

-CO2 levels

-pH of soil

-space/location of ecosystem

-how much energy’s captured by producers and the efficacy producers can convert the energy into new biomass

-how much energy is transferred to higher trophic levels (Excretion)

180
Q

how to obtain dry tissue sample

A

How would you attain dry tissue/sample to measure –> place in a drying room and heat to remove water. Check the mass regularly and stop drying once mass remains the same

181
Q

gross primary production

A

Gross primary productivity (GPP) –> the rate at which energy is incorporated into organic molecules by an ecosystem. Expressed as KJ m^-2yr^-1 or MJm^-2yr^-1

182
Q

net primary productivity

A

Net primary productivity (NPP) –> the chemical energy store in plant biomass of respiritory losses in the environment

183
Q

NPP =

A

GPP - respiration

184
Q

The net annual primary productivity of a particular wetland ecosystem is found to be 8000kcal/m1. If respiration by the aquatic producers is 12,000 per year, what is the gross annual primary productivity for this ecosystem.

A

1) 8000 = GPP – 12000 so GPP = 20,000

185
Q

If you measure the available biomass for a patch of forest as 10kg C/m2 year and the amount of CO2 given off into the atmosphere as 5kg C/m1 year, what is the GPP.

A

2) 10 = GPP –5 = 15C/m2rr

186
Q

In the patch of a forest in the previous problem, how much energy is available in the primary producer level for herbivore consumption. Assume 1KG of carbon produces 10,000KJ

3) GPP = change in dark = change in light

A

GPP = 5 = 1 = 6

187
Q

reducing energy loss

A

-energy is lost from a food chain due to urine/faeces, heat loss or respiration (higher ). Some parts of organisms are not consumed like bores or feathers

-Net production of consumers = I - (F + R) –> respitory losses to the environment

I = energy store of ingested food

F = energy lost in faeces/urine

R = chemical energy used in respiration

During respiration almost half of GPP is lost as heat. This is respiratory loss (R)

188
Q

energy passing between trophic levels

A

As energy passes from one trophic level to the next, energy is lost. Secondary production allows us to study how much energy consumers store in their biomass. It is also called consumer production

producer –> primary consumer –> secondary consumer etc

189
Q

primary productivity

A

Primary productivity refers to the rate of primary production by green plants.

190
Q

secondary productivity

A

Secondary productivity refers to the rate of secondary production by consumers (e.g. herbivores and carnivores).

Herbivores have a lower secondary productivity than carnivores.

This is because herbivores eat plant material which is higher in cellulose content than meat. Cellulose is hard to digest so more energy is lost in excretion.

191
Q

steps for nitrogen cycle

A

1) Nitrogen in the atmosphere (N2) is fixated with nitrogen fixing bacteria in the soil to form ammonium ions

2) Ammonium ions cannot be assimilated so are converted to nitrate ions (NO3-) via the process of nitrification using nitrifying bacteria

3) Nitrate ions are assimilated into ammonium containing molecules in producers e.g amino acids

4) Through consumption consumers gain amine groups

5) Consumers are decomposed and saprobionts hydrolyse ammonium containing molecules

6) Ammonification to reform ammonium ions

7) However, during anaerobic conditions denitrifying bacteria carry out deniftrification to reform nitrogen in the atmosphere. This is prevented via plouing, maintaining aerobic conditions etc

192
Q

tRNA

A

transfer RNA

193
Q

calorimetry

A

-in order to estimate how much organic matter has been built up by a plant, it is necessary to take a sample, weight it to get the fresh weight then remove the water. The sample is placed in an oven set at 105 degrees. Then is weighed and returned to the oven again. This is repeated to constant mass to give the dry mass

-The plant material is killed but not destroyed by this process and therefore scientists will often take a sample else all the organic matter will be killed e.g may choose to measure leaft biomass then scale to provide an average

194
Q

mass of carbon in biomass

A

-organic compounds all contain carbon and with varying concentrations of hydrogen, oxygen and nitrogen in proteins

195
Q

using calorimetry to measure chemical energy

A

1) collect a sample from a given area which has been allowed to grow for a given time

2) Dry the sample by heating and extracting the water in a drying room until at constant mass

3) Record the starting mass of the sample and then burn completely. The energy released from the burning should be used to heat a known mass of water

4) q = m x c x change in T

1 calorie is the amount of heat needed to raise the temperature of 1g of water by 1 degree

4.81 joules = 1 calorie

It is very inefficient attempting to measure heat by simply burning a container of water -> the standard equipment to do this is a bomb calorimeter

-the biomass sample is burned in an atmosphere of oxygen inside a strong metal container, surrounded by an insulated water jacket

196
Q

describe another process carried out by microorganisms which adds ammonium ions to soil

A

-decomposition/hydrolysis of ammonium containing molecules like amino acids using saprobionts
-this then turns to ammonification where NH4+ ions are reformed

197
Q

explain how ploughing would affect the fertility of the soil

A

-fertility of the soil would increase
-more nitrate ions (nitrification) and less denitrification

198
Q

suggest 2 ways in which crop rotation may lead to high crop yields

A

1) grow crops with nitrogen fixing bacteria
2) different crops have different pathogens

199
Q

explain how farming practices increase the producitivty of agricultural crops

A

-ploughing = increases rate of nitrification due to aeration
-fertilises are nitrate for the synthesis into amino acids
-crop roation reduces pests
-selective breeding produces variety of crops

200
Q

2 variables controlled when preparing liquid medium cultures

A

1) volume of bacteria culture
2) time the culture is left to divide

201
Q

describe how the ecologists could have determined whether the water was all removed from the sample

A

-record mass and reheat until a constant mass has been achieved

202
Q

what does the graph show about the relationship between diameter of tree and biomass (scatter graph)

A

-positive correlation
-non-linear

203
Q

ho wto estimate mass of carbon of wood for particular species

A

-calculate mean daimeter of tree
-use percentage water content to find dired biomass
-use dried density to calculate mass of the tree
-estimate the number of trees and calculate mean carbon content

204
Q

biological molecules containing phosphorus

A

-ATP
-phospholipids
-DNA/ RNA
-TP/GP

205
Q

abiotic source of phosphorus

A

-rocks and sediments

206
Q

where are nitrogen fixing bacteria found

A

-root nodules
-legumes e.g peas

-bacteria have a symbiotic relationship with plants as bacterua provide nitrogen containing compounds and plants provide organic compounds like glucose or carbohydrates

207
Q

Applying very high concentrations of fertiliser to the soil can reduce plant growth. Use your knowledge of water potential to explain why.

A

-soil has lower water potential than plants
-osmosis from plant

208
Q

Give two examples of biological molecules containing nitrogen that would be removed when a crop is harvested.

A

1) DNA
2) amino acids

209
Q

mycorrhizae

A

Mycorrhizae are a type of symbiotic relationship (where each organism benefits from the other) formed between fungi and the roots of plants.

The benefit to the plant is the increased surface area of the roots – used to increase the absorption of water and mineral ions (phosphate ions) from the soil.
The benefit to the fungi is the organic compounds, such as glucose, that the plant provides.
The fungi consist of hyphae – long, filamentous structures that connect to the plants roots and extend into the soil.

210
Q

natural vs artifical fertilisers

A

Natural fertilisers are organic material – such as applying dead plants, animals or waste to the soil. The decomposition of the material leads to ions being replaced.

Artificial fertilisers are inorganic material – they are concentrated solutions of the lost minerals that are sprayed onto the plants.
Artificial fertilizers are faster but have higher environmental risks (increase risk of leaching)

211
Q

process of eutrophication

A

Chemical fertilisers leach into water sources. Nitrates are soluble and are readily leached into ponds and lakes.
The fertiliser accumulates, and is absorbed and utilised by algae. The excess of nitrates will lead to an algae bloom (a rapidly increasing population of algae).
Light is unable to penetrate past the algae. Algae grows rapidly on the surface of the water, blocking any light.
Plants cannot photosynthesise and will die.
Animal species diversity will decrease. Many organisms rely on the plants for food and shelter, so when these die it disrupts the food chain.
Respiration of decomposers uses up oxygen in the water. Dead organic matter provides more nutrients to the saprobionic bacteria. The bacteria respire aerobically, so use up the oxygen.
This causes fish and other aquatic organisms to die

212
Q

steps for phosphorous cycle

A

1) phosphate ions are found in sedimentary rocks

2) Phosphate ions originate in the sea but are brought to terrestrial environments by the uplifting of rocks

3) Weathering and erosion of rocks helps phosphorus become dissolved and available for plants to incorporate into biomass

Phosphate ions are taken up from the soil by plants through their roots or absorbed from water by algae. Transferred to consumers during feeding.

4) Animals can also excrete phosphorus in waste (urine) and some faeces

5) When plants and animals die, saprobionts hydrolyse the biological molecules which releases phosphate ions into the water or soil

6) Phosphate ions are released into the soil and are transported by streams and rivers into the lake and oceans where they reform sedimentary rocks

213
Q

myccohizae

A

Mycorrhizae are important in facilitating the uptake of water and inorganic ions by plants. These are associations between certain types of fungi and the roots of the vast majority of plants. They increase the surface area and act as a sponge holding water and minerals. As a result a plant can better resist drought and take up inorganic ions more easily.

214
Q

why are not all nutrients recycled in an agricultural ecosystem

A

-crop plants are harvested and removed from the area

-animals are kept in barns or taken for slaughter

-compact or waterlogged soils mean conditions become anaerobic so denitrification occurs

Natural fertilisers are organic material – such as applying dead plants, animals or waste to the soil.

215
Q

nitrogen based fertilisers

A

Nitrogen based fertilisers = leads to leaching and eutrophication

Nitrogen fertilisers greatly increase crop yields and therefore can help to deal with the demands of a growing human population. However they have negative effects on the environment which include reducing biodiversity, leaching and eutrophication.

216
Q

leaching

A

process by which mineral ions, such as nitrate, dissolve in rainwater and are carried from the soil to end up in rivers and lakes

217
Q

eutrophication

A

provides algae in waterways with enough nitrate ions to grow more rapidly than it otherwise would do. As a result this can block out light from other plants, causing decay and the use of oxygen in the water way. This eventually leads to the death of the ecosystem.

218
Q

eutrophication and leaching

A

-when there is no leaching of nitrates, nitrate concentration is often a limiting factor to grwoth in an aquatic ecosystem

-the leaching of fertilisers triggers an algal bloom. This absorbs many wavelengths of light (light as a limiting factor)

-the death of plants results in an increase of sabropotic bacteria

-anaerobic conditions results in the growth of anaerobic organisms, some of which release hydrogen sulfide (H2S) making the water putrid. (oxygen as limiting factor)

219
Q

how are ions absorbed

A

active transport

nitrate ions = soluble and easily leach

nitrifcation = oxidation reaction
NH4+ –> NO3-

220
Q

effects of nitrogen containing fertilisers

A

-reduces species diversity –> outcompete
-leaching
-eutrophication

221
Q

oxidation during nitrification
ammonia –> nitrites –> nitrates

A

-require oxygen as it is added to compounds during conversion

222
Q

how can biomass be measured

A

-mass of dry carbon and dry tissue per given area
-chemical energy store using calorimetry

223
Q

percentage efficiency between primary and secondary consumer =

A

secondary / primary x 100

224
Q

units for productivity example

A

KJha-1year-1

Ha = hectres

225
Q

decrease in plant growth productivity

A

-increased competition for light
-reduced photosynthesis

226
Q

why biomass shows little increases after 100 years

A

NPP is GPP minus respiratory losses
-biomass showes little increase due to increased respiration

227
Q

high productivity =

A

more producers which are food sources for consumers

228
Q

is it ammonia or ammonium ions

A

ammonium ions (they are oxidised into nitrites and then nitrates) during nitrification

229
Q

plants assimilate

A

phosphate ions

sabropionts break down phosphorus into inorganic forms

230
Q

what are fungal associations between plant roots and fungi

A

-mycorrhizae

-provides larger surface area for the uptake of water and inorganic ions

231
Q

artficial fertilisers =

A

more water soluble = higher absorption

232
Q

explain how ploughing would affect fertility of soil

A

-less anaerobic conditions
-less dentifircation
-increase in nitrates

233
Q

2 ways crop rotation may lead to high crop yield s

A

-crops use diff nutrients
-crops with nitrogen-fixing bacteria can be grown

234
Q

how is acetylCoA formed in the link reaction

A

-pryruvate is oxidised and decarboxylated
-then coenzyme A is added to acetate

235
Q

advantage of lactate being converted back to pyruavte

A

lacate produces muscle fatigue = bad

236
Q

layer of oil

A

prevents oxygen uptake

237
Q

explain what would happen to volume of gas if yeast only respired aerobically

A

stays the same
-same amount of oxygen uptake and CO2 released

238
Q

if production is measured in 1 hour and the scientists only measured for 15minutes what do you do

A

divide answer/difference by 4

239
Q

why does iron deficncy reduce CO2 uptake

A

less TP is converted into RuBP and CO2 combines with RuBP since RuBP concentration has decreased so has the uptake of CO2.

240
Q

describe how you would maintain quantitive measure of their cloudiness

A

-colorimetry = measure cloudiness
-measure absorbance of light
-zero colorimetry
-same wavelength

241
Q

why is determinining dry mass appropriate

A

water content varies

242
Q

reasons why the respirometer should be left with the cap open for the first 10 minutes

A

-to allow respirometer to equilibrate
-to allow respiration rate to reach a constant
-to allow pressure inside respirometer to stabilise

243
Q

what measurements need to be taken to record the respiration rate in cm3hr-1

A

-distance moved by liquid
-cross sectional area of the tubing
-time

244
Q

why did the colour liquid move to left
(respirometer experiment)

A

-as seeds respire, oxygen is taken in and CO2 is released
-CO2 is absorbed by KOH solution
-this means that pressure inside respirometer decreases causing coloured liquid to move to the left

245
Q

why was the tap open at the beginning of the experiment

A

to allow apparatus to equilbirate

246
Q

why is a syringe good for control

A

press down to release air + calibrate to zero

247
Q

calculate cross sectional area of tubing

A

pi x radius squared x distance

248
Q

purpose of water bath

A

maintain constant temperature that is optimum for enzyme activity

249
Q

what measurements should the studen take to calculate rate of aerobic respiration in mm3 of oxygen per gram per hour

A

-mass of woodlice (grams)
-volume of oxygen
-time
-volume + cross sectional area of tubing

250
Q

calculate percentage efficiency

A

GPP / total x 100

251
Q

diffusion for H+ ions down ATP synthase in oxidative phosphorylation

A

not active transport

252
Q

give 2 reasons why most light falling on producers in natural ecosystems is not used in photosynthesis

A

-green wavelenghts of light are reflected
-light does not directly hit chlorophyll molecule

253
Q

describe the effect nitrate concentration may have in the river at point Y

A

-nitrate conc is no longer limiting factor causing growth of algae
-O2 levels initially increase resultng in agal bloom
-light can no longer be abosrbed
-increase in sapribotic bacteria to decompose dead organic material utilising oxygen
-oxygen levels decrease
-anaerobic conditions kill aquatic life

254
Q

change experiment to measure CO2 levels in respirometer

A

-remove potassium hydroxide
-record distance liquid moves
-use difference in liquid movement with and without KOH

255
Q

adaptions of apparatus in calorimetry

A

-stirrer distributes heat energy
-insulation reduces loss of heat
-water has high specific heat capacity

256
Q

calorimeter

A

instrument to determine the chemical energy store of a known mass of carbon

Aqa a level biology (8 decks)

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