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biology paper 2 > energy transfers in and between organisms > Flashcards

energy transfers in and between organisms Flashcards

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AP® Biology flashcards Biology 101 flashcards
1
Q

what are the two stages of photosynthesis

A

light dependent and independent reaction (calvin cycle)

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2
Q

where does the ldr occur

A

the thylakoids of chloroplasts (between the thylakoids and the stroma)

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3
Q

why is the ldr not sufficient for a plant without the calvin cycle

A

it doesnt happen at night as its light dependent and doesnt produce glucose or enough atp (some tissues cant produce atp and atp cant be moved or stored as its too reactive)

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4
Q

what does the ldr produce

A

oxygen (waste product), atp and nadph

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5
Q

which ldr products are needed for the lidr

A

atp and nadph

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6
Q

what is ps II

A

photo system II - a transmembrane complex containing chlorophyll

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7
Q

what does the proton pump do

A

actively transports protons into the thylakoid using energy from electrons from chlorophyll

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8
Q

what is ps I

A

photo system I - an electron carrier

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9
Q

what makes up the electron transport chain (ETC)

A

psII, proton pumo, and ps I

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10
Q

what is the first step of the ldr

A
  1. light enters the thylakoid and excites 2 electrons (photoionisation) from the chlorophyll molecule (oxidation of psII). the proton pump gains 2 electrons (reduction). 2 electrons move up in energy level and move from psII to the proton pump.
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11
Q

whats the 2nd step of the ldr

A

the reduced proton pump can actively transport protons (H+) from the stroma into the thylakoid space. this creates an electrochemical gradient between the thylakoid space and the stroma (more H+ in TS than stroma). the proton pump requires the bond with 2 electrons (e-) to work. these 2 e- become unstable

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12
Q

whats the 3rd step of the ldr

A

the e- are transported from the proton pump (oxidation) to the electron carrier psI (reduction)

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13
Q

whats the 4th step of the ldr

A

the protons diffuse down an electrochemical concentration gradient through the ATP synthase channel - this allows the enzyme to catalyse adp + Pi to atp. the channel turns to bring the substrates close together, adp is phosphorylated

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14
Q

whats the 5th step of the ldr

A

2e- and a h+ are transferred to a coenzyme - a molecule that works with an enzyme to help it do its job - called nadp+ (+ means it can carry an e-) to make reduced nadp (nadph) ready to be used in the lidr. this occurs with the help of the enzyme dehydrogenase.

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15
Q

what does dehydrogenase do

A

removes h ions or protons and sticks it to other things. in the ldr it removes proteins from atp synthae ad adds them to nadp+ to make nadp+h+. the 2e- then join to form nadph so the 2e- are at an even lower energy state

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16
Q

6th step of the ldr

A

at this point the reaction will stop because the chlorophyll has given up its electrons but it needs to continuously produce atp and nadph for the lidr

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17
Q

7th step ldr

A

photolysis of h2o occurs in the thylakoid space. this replenishes chlorophylls 2e-, addds to the h+ electrochemical gradient, and produces o2 as a byproduct

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18
Q

what is photolysis

A

‘light breakdown’ - replenishing electrons by adding light energy to h2o to make 2e- 2h+ and oxygen

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19
Q

where does the lidr occur

A

the stroma

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20
Q

what is the lidr

A

a series of reactions that produce the organic compounds a plant needs including glucose

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21
Q

what does nadph do in the lidr

A

supplies protons and electrons

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22
Q

why will the lidr eventually stop without light

A

nadph and atp arent being replaced by the ldr

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23
Q

what are the 3 stages of the lidr

A
  1. carbon fixation
  2. reduction
  3. regeneration
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24
Q

what happens during 1. carbon fixation in the lidr

A

the enzyme RUBISCO catalyses the reaction of ribulose bisphosphate (RuBP) and co2 into 2 x glycerate-3-phosphate (G3P). G3P is a 3 carbon sugar. if it is too hot then RUBISCO denatures and the plant will die

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25
Q

what happens during 2. reduction in the lidr

A

reduced nadp (nadph) from the ldr reduces (gains electrons) g3p into 2 x triose phosphate (TP) using energy from atp

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26
Q

what happens during 3. regeneration in the lidr

A

some TP is converted into useful organic compounds e.g glucose. some is regenrated into ribulose bisphosphate using energy from atp to maintain its supply

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27
Q

how many times does the calvin cycle occur per glucose molecule

A
  1. 6 x RuBP = 30 carbon atoms. + (6 x co2) = 36 carbon atoms. - 6 carbon atoms to make glucose (C6H12O6) = 30 carbon atoms. these 30 carbon atoms are regenerated into 6 x RuBP (5 carbon sugar)
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28
Q

limiting factors of photosynthesis and which stage do they affect

A
  • co2 concentration (lidr)
  • light intensity (ldr)
  • temperature (both)
  • chlorophyll (ldr)
  • water availability (ldr - photolysis)
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29
Q

how can we measure photosynthesis

A
  • presence of starch (using iodine)
  • volume of o2 released
  • volume of co2 used up (hydrogencarbonate indicator can measure the acidity (co2) of water (pH) - as co2 is used up, the water becomes less alkaline)
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30
Q

how can we measure rate of photosynthesis

A
  • volume of o2 produced per minute (cm^3/min^-1)
  • count bubbles of o2 released per minute
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31
Q

what is the law of limiting factors

A

at any given moment, the rate of photosynthesis is limited by the factor that is least favourable

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32
Q

what are the 4 stages of aerobic respiration

A
  1. glycolysis
  2. link reaction
  3. krebs cycle
  4. oxidative phosphorylation
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33
Q

where does glycolysis occur

A

cytoplasm

34
Q

where does the link reaction occur

A

matrix

35
Q

where does the krebs cycle occur

A

matrix

36
Q

where does oxidative phosphorylation occur

A

through the cristae of the intermembrane space of mitochondria

37
Q

where does aerobic respiration take place

A

mitochondria

38
Q

what does glycolysis do

A

breaks down glucose into pyruvate whilst producing atp and harvesting protons (h+) and electrons (e-)along the way. it removes an e- to break the covalent bond between H and another molecule in glucose (carbon)

39
Q

step 1 glycolysis

A

glucose is phosphorylated into glucose phosphate using 2 x atp

40
Q

step 2 glycolysis

A

glucose phosphate breaks down into 2 x triose phosphate

41
Q

step 3 glycolysis

A

2 x TP are oxidised into 2 x pyruvate using 2 x nad+

42
Q

step 4 glycolysis

A

oxidation of 2 x tp into 2 x pyruvate using 2 x nad+ forms 2 x nadh and 4 x atp

43
Q

what is the yield of 1 glucose molecule undergoing aerobic respiration

A
  • 2 x (net) atp –> 4 are produced but 2 are invested
  • 2 x nadh
  • 2 x pyruvate (3 carbon sugar) (gets into the matrix using active transport crossing the double membrane)
44
Q

what does the link reaction happen to

A

2 x pyruvate

45
Q

step 1 link reaction

A

pyruvate is oxidised (with nad) to remove hydrogen ions and released co2 (decarboxylation). an enzyme removes a h+ and an e- making the molecule unstable, causing it to drop a carbon with 2 o2 attached. this forms acetate (2 carbon sugar)

46
Q

step 2 link reaction

A

coenzyme A is added to acetate to produce acetyl CoenzymeA (ACoA)

47
Q

how many times does the krebs cycle occur for each glucose molecule

A

2 x for every glucose molecule (same as the link reaction)

48
Q

step 1 krebs cycle

A

acetyl CoA combines with a 4 carbon molecule to form a 6 carbon molecule

49
Q

step 2 krebs cycle

A

the 6 carbon molecule is oxidised into a 5 carbon molecule using nad+ (decarboxylation occurs)

50
Q

step 3 krebs cycle

A

the 5 carbon molecule is oxidised into a 4 carbon molecule using the coenzyme fad+ (fad+ holds 2 x H+ creating fadh2) (decarboxylation occurs). ATP is also generated during this step (the enzyme contains a Pi that turns adp to atp)

51
Q

what does oxidative phosphorylation involve the use of from the other stages of aerobic respiration

A

nadh and fadh2 from glycolysis, the link reaction, and the krebs cycle to start a series of redox reactions - energy is harvested from these reactions

52
Q

step 1 oxidative phosphorylation

A

nadh or fadh2 is oxidised into nad+ or fad+. this releases a proton into the matrix and 2e- are transferred to carrier a in the phospholipid bilayer (reduction). once carrier a is reduced it can transport a proton into the intermembrane space. the energy from these electrons allows carrier a to pump a proton into the intermembrane space as it changes the tertiary structure of the carrier slightly

53
Q

step 2 oxidative phosphorylation

A

the electrons are lost from carrier a (oxidation) and transported to carrier b (reduction)., this provides energy to pump another proton into the intermembrane space from the matrix

54
Q

step 3 oxidative phosphorylation

A

the electrons are lost from carrier b (oxidation) and transported to carrier c (reduction). this provides energy to pump another proton into the intermembrane space from the matrix

55
Q

step 4 oxidative phosphorylation

A

the electrons in the intermembrane space from an electrochemical gradient. they diffuse through an atp synthase channel protein to allow catalysis of adp + Pi to atp. the enzyme rotates in such a way that allows the protons to pass through. the diffusion of molecules down an electrochemical gradient is called chemiosmosis

56
Q

step 5 oxidative phosphorylation

A

the 2 electrons combine with 1/2 o2 - the final electron acceptor - in the matrix which reacts with 2 h+ to make h20. the o2 delivered to mitochondria through inhalation is missing 2 electrons (so theres room for the 2e- from carrier c). this is why h2o is a product of respiration. the role of o2 in this process is critical - in its absence the ETC and the transport of hydrogen would grind to a halt

57
Q

what is biomass

A

the mass of living material in a specific area at a given time. units: gm^-2 (land), gm^-3 (water)

58
Q

why do we measure biomass

A

for farming (measuring crop/meat yield), monitoring ecosystem health/productivity, and studying food chains

59
Q

how do we prepare dry mass

A

heat a sample up to 100 degrees celsius to evaporate the contained water (dont exceed 100 degrees to avoid burning the sample as this would take carbon from its biomols for the co2 in the sample therefore losing dry mass). weigh the sample until the mass is constant i.e no more water is being lost. to prepare dry mass you need to kill the organism so usually small samples are used

60
Q

how to find the chemical energy in dry biomass

A
  • found using calorimetry (bomb calorimeter)
  • burn the sample to release its energy and measure the temperature rise of the surrounding water
  • a fuse wire lights the sample ; dont use a wooden splint as wood is biomass, using a fuse wire means nothing additional is being burnt
  • the water in the bomb calorimeter acts as an insulator to ensure no thermal energy is lost or gained
  • the stirrer ensures a uniform temperature change
  • the amount of energy needed to raise kg of wate by 1 degree celsius (the specific heat capacity) is 4184 J (4.184 for 1g)
  • we use the increase in water temperature to calculate the energy contained in the sample
61
Q

how much light energy from the sun is taken in by producers

A

1-3%

62
Q

why is only 1-3% of light energy from the sun taken in by producers

A
  • over 90% is reflected by clouds back into space
  • not all wavelengths of light can be absorbed by pigments
  • light may not fall on a chlorophyll pigment molecule
  • limiting factors may prevent energy storage e.g co2 concentration
63
Q

what is gross primary production (GPP)

A

th total amount of energy stored by a producer in a given time (kJ m^-2 year^-1). plants use 20-50% of gpp in respiration, the remaining energy is stored as biomass

64
Q

what is net primary production (NPP)

A

the chemical energy store of plants (biomass)

65
Q

what is the equation to work out npp

A

npp = gpp (energy from photosynthesis) - respiratory losses

66
Q

how much of the npp stored by producers is transferred to biomass in primary consumers

A

5-10%

67
Q

why is only 5-10% of npp stored by producers transferred to biomass in primary consumers

A
  • the whole organism isnt consumed e.g roots
  • the whole organism isnt digested e.g cellulose/fibre
  • some energy is lost in faeces and urine (inefficiency of digestion and absorption)
  • some energy is lost to the environment as heat from respiration
68
Q

what is the equation to work out the net biomass production of consumers

A

N = ingested energy (I) - (energy lost in faeces/urine (F) = respiratory losses (R))

69
Q

how much energy is transferred between primary consumers and secondary consumers

A

15-20%

70
Q

why is the energy transferred into biomass between primary and secondary consumers higher than that transferred between producers and primary consumers

A

theyre both animals so theres less that ant be utilised/eaten as theyre more similar

71
Q

why do food chains typically only have 4/5 trophic levels

A

energy transferred at each level decreases moving along the food chain. this is also why the biomass of secondary and tertiary consumers (g m^-2) is less than that of producers (e/g in a field there will be more grass than foxes)

72
Q

consequences of agricultural farming

A
  • intensive farming of food has improved food productivity
  • intensive food production places large demand on soil
  • nutrients are removed but not replaced as crops are removed before they can decay
73
Q

how can we replenish nutrients in agricultural farming

A
  • crop rotation gives soil time to recover different nutrients
  • fallow years - e/g grass (less leaching) followed by ploughing
  • allowing animals to grace as faeces/manure is organic fertiliser
  • using organic fertilisers e/g plant/animal remains, manure, slurry, bone meal
  • using inorganic fertilisers e/g NPKfertilisers (a blend of phosphorus, nitrogen, and mined potassium-containing rocks
74
Q

positives of nitrogenous fertiliser

A
  • more amino acids (amine group), atp (adenosine), dna (bases)
  • more growth (protein synthesis) of leaves and roots
  • more photosynthesis leads to improved productivity
75
Q

what is nitrogen burn

A

when fertiliser has been added to such a point that water becomes a limiting factor. the water potential of the soil has decreased meaning water in roots of plants leaves via osmosis which dehydrates the crop. this reduces crop yield

76
Q

negatives of nitrogenous fertiliser

A
  • reduced species diversity - grasses/nettles/weeds favour nitrogen rich soil and outcompete other species
  • leaching - nitrates (NO^3-) easi;y dissolve and are washed away leading to eutrophication
77
Q

describe the process of eutrophication

A
  • nitrogenous fertiliser in ponds/water causes growth of algae/surface plants/algal bloom to block light
  • reduced/no photosynthesis so submerged plants die
  • saprobionts/saprobiotic microorganisms
  • aerobically respire/use oxygen in respiration
  • less oxygen is available for fish to respire/aerobic organisms die
78
Q

advantages of organic fertiliser

A
  • acts as soil conditioner by containing slow release nutrients (needs to decay first)/improves drainage and aerates soil as its fibrous and porous/increases organic content of soil
  • contains a wider range of elements
  • production of artificial fertiliser is energy consuming
  • less leaching/slower release of nutrients as it needs to decay first so less risk of eutrophication
79
Q

what is the nitrogen cycle needed for

A
  • amino acids (amine group)
  • dna/rna (bases)
  • atp (adenine)
  • nad/nadp (adenine)
80
Q

the nitrogen cycle

A
81
Q

the phosphorus cycle

A
82
Q

what is the phosphorus cycle needed for

A
  • atp (triphosphate)
  • dna/rna (phosphodiester bonds/ sugar phosphate backbone)
  • phospholipids

biology paper 2 (4 decks)

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