Respiration q's Flashcards
w, z, x, w
why does aerobic respiration yield fewer molecules of ATP than the theoretical maximum?
- some ATP used to actively transport pyruvate into the mitochondrian
- some ATP used to actively transport H+ from reduced NAD formed in glycoloysis
- some energy released in ETC, is releases as heat
- not all the H+ movement back across membrane is used to generate ATP
- not all the reduced NAD is used to feed into the ETC
Explain why the incomplete breakdown of glucose in anaerobic respiration produces less ATP than aerobic respiration
in anaerobic
- glycolysis/ conversion of glucose into pyruvate occurs
- produces 2 molecules of ATP (net)
- only substrate level phosphorylation ocurrs
- O2 not available as final electron acceptor
- pyruvate used to regenerate NAD for glycolysis to continue
- krebs cycle + ETC / chemosmosis / oxidative phosphorylation doesnt occur
describe the pattern of respiration shown by cauliflower at increasing storage temperatures of 0 celcius to 20c (table)
as temp incr, respiration rate incr
respiration rate double w/ a 10 c temp increase
discuss what the data in table 5.1 indicate about the best conditions for staorage of fruits + vege
- best conditions are lower temps bc respiration rate is low
- 0c/ freezing could be the best
- 0c/ might be too low as food cells might be damaged at 0c
- some food storage temo doesnt seem to matter
- data is incompletet for potato / parsnip so no conclusions can be made
- if product needs to ripen during storgae then a higher temp / not above 20c will be ideal
identify w/ reasons which fruit or vege listened in table 5.1 is least likely to spoile during storage
- onion
- has lowest respiration rate
- across all temps
- can be kept at room temp
which fruit / vege listed in table 5.1 is likely to be the most difficult to keep fresh during storgae? give a reason for ur ans
- asparagus
- has a high respiration rate across all temps
certain parasites live in blood of mammals
suggest why even though blood carries O2, these parasites are adapted to respire anaerobically
- parasites have little access to O2
- inaccesible bc little O2 dissolved in plasma
- inaccesible bc little O2 combined w/ HB
- Hb has greater affinity for O2 than parasite pigment
- bc the O2 is bound to Hb the parasite is unable to use it
the anaerboic respiration in animal cells can be reversed, but the anaerobic repiration pathway in yeast cells cannot be reversed.
explain why, using ur knowledge of the differences between the 2 pathways
in animals
- pyruvate is converted to lactate
- can be reversed as no other product formed
- lactate dehydrogenase available to reverse the reaction
in plants
- pyruvate converted to thanol ( in 2 steps) + co2
- cannot be revered as CO2 is lost
- dehydrogenase enzyme cannot reverse the reaction
state precisely where the in the cell glycolysis occurs
cytoplam of cell
outline the process of glycolysis
- phosphoryation of glucose
- so forming hexose 1, 6 bisphosphate
- then forming of 2 triose phosphate
- for formation of pyruvate oxidation occurs
- pyruvate produced from TP
- total production 4 ATP/ net production 2 ATP
W - Ethanal
X - CO2
Y - reduced NAD
Z - NAD+
w/ ref to table 5.1 describe the effect of the diff treatments on the alcohol conc of the treated samples compared w/ the control samples
a : produced less alcohol than the control at all times
v: had the same alcohol as the control at 15 hrs
c: produced more alcohol than the control at 15+ 60 hrs
suggest a reason for the dif in alcohol conc at 60 hrs between the 2 bark treatments v and c
at 60 hrs
- C has more yeast cells
- only a small no. of bacteria in V are producing alcohol
to be used as a source of comercial sugar , the sap needs to be as uncontaminated as possible. suggest w/ reason , which of the treatment shown in table 5.1 would be the best for use w/ sap so that it is suitable as a source of commercial sugar
A,
has the least contamination as it has very few bacteria + little alcohol
product: urea
organ the product is transported to : liver
-hepatocytes can tolerate lacatate which would otherwise be toxic
- hepatocytes have enzymes to metabolise lactate
- conversion of lactate require O2 + muscle cells dont have enough O2
explain what might happen to a person if the liver did not break down insulin
- blood glucose conc woud fall too low (causes hypoglycaemia )
- glucose continued to be taken up by cells results in low blood glucose
- mitochondria eventually cannot generate enough ATP as less avaibake glucose in blood
- death
- build up of lactate kills liver cells
- disruption of enzymes as a reuly of low pH
- lack of fatty acids not avaiable for respiration
- lack of oxidised NAD for metaboilc reactions
- some deamination cannot occur
- build up of fatty acids present in fat deposits
state precisiely where in the liver cells the excess reduced NAD can be re oxidised
cristae
suggest the type of reaction that removes a phosphate group form an ATP molecule
hydrolysis
1
none
2/3
state the precise location of the ETC in the cell
cristae
describe the differences observed in the seeds in groups a b and c
A has more stain than B
and C has none
suggest reasons for the results observed in the seeds in group A
- shaded areas in A are respiring
- 22c is suitable temp for respiration
- reduced NAD produced in glycolysis/ link reaction/ krebs cycle
- more electron transfer to TTC (electron acceptor)
suggest reasons for the diff in the amount of staining observed in the seeds in groups B and C when compared to those in group A
- respiration uses enzyme
- group b: not enough energy for ESC formation
- group c : enzymes denatured by high temperature
i) ethanal
ii) ethanal
iii) Co2 and ethanol
explain why this pathway (anaerobic pathways after glycolysis) is important for plant cell
- release NAD to accept more H+
- some ATP available for active transport
a) krebs cycle, link reaction
b) oxidative phosphorylation
suggest why at each temp , respirator B contained some glass beads
- to make the vol of contents the same in the respirometer
- the vol of peas in A is greater than the vol of peas in B
suggest how the student determined the quantity of glass beads to place in repirometer B at each temp
- determined by finding diff in vol between soaked peas to dry peas
- the diff represents the vol of glass beads required
- calc vol of 1 bead to determine no. of beads equivalent to vol required
explain why there is an inceased rate of respiration in soaked seeds at 25 c compared w/ soaked seeds at 15 c
- at high temp 25 c , incr kinetic energy
- named respiratory enzymes involved
suggest a reason for diff in ROR between soaked + dry pea seeds
- reaction need to take place in solution
- enzyme + substarte can move in soaked seeds
- soaked seeds need more ATP
- for protein synthesis
mammal : pyruvate , no , lactate
yeast: ethanal, yes, ethanol
suggest one benefit of anaerobic respiration to an organism
- ATP produced
- recycles NAD
- allows glycolysis
W: glycolysis
X: calvin cycle
Y: krebs cycle
- takes place in diff parts of the cell
- glycolysis in cytoplasm
- calvin cycle in cholorplasts / stroma of choloplast
- krebs cycle in martix/ mitochondrion
- substrate respired changes over time
- initially respires carbohydrate
- decrease in RQ indicates lipid as substrate
- carbohydrate used up overtime
- proteint not likely to be used as substrate/ proteins only used as a last resort
i) glycolysis
ii) cytoplasm
iii) D- ATP, E- NAD, F- pyruvate
describe the pattern of growth of the yeast population in this fermenter ( overall trend , shape : curve straight etc, fig)
- slow incr during day 1
- esponential increase
- rate of increase less steep during day 1
- stationary phase
- day 2 , yeast pop % 24
fig 5.1 shows that as the sugar conc decreases the ethanol conc increase. explain this relationship
- sugar converted to ethanol
- in anerobic respiration
- sugar undergoes glycolysis
- pyruvate loses CO2
- reduced NAD gives hydrogen to ethanal
- NAD being recycled so glycolysis continues
- pyruvate decarboxylase
- glucose enters glycolysis w/o being broken down
- maltose must be hydrolysed
- enzyme only made when glucose is running out
- enzyme induced
- transcription + translation / protein synthesis takes time
- enzymes to bread down malotriose made list
adv of using yeast
- less energy required
- doesnt need incr temp / incr pressure
- can use waste material as a substrate
- substrate is sustainable
- process doesnt use up fossil fuels
- product is carbon neutral
- yeast is easily accessible
dis of using yeast
- time consuming
- needs purification of product
- is killed by product
- can only use batch method
- sterile conditions required
- less ventiliation
- less O2 for aerobic respiration
- so less ATP produced
- increased acidity as CO2 build up interfering w/ enzymes
- less glucose uptake into cells
- less glucose for ATP production
- glucose not converted to glycogen
i) mitochondria
ii) - molecules are not lipid soluble to require protein channels to cross membrane
- AT ornithine+ citrulline need to be moved into + out of mitochondrion more quickly than would be met by diffusion
- facilitates diffusion - conversion of ornithine ino citrullline creates conc gradients
identify the compound labelled X
ATP
give one limitation of using this method to investigate respiration rate
- measurements only taken every 30 sec
- difficult to read the meniscus (may be subjective)
experiment:
- mass of invertebrate + mass of the beans is the same
- safe+ethical use of invertebrate eg, ass screen so that animal cannot touch the muslin bag
- bigger syringe needed (5-10cm3)
- keept temp+ light constatnt
- use same mass of soda lime in both assays
- measuring distance moved by coloured red liquid at regular time intervals
- repeat experiments
results+ conclusion
- invertebrates ROR is expected to be higher than the ROR of the beans bc invertebrates are moving around
- metabolic processe requires energy/ generate heat
- increased vol of water aded to seedling , leads to lower survival of seedlings
- larger decrease in survival for added water from 30cm3
- vol of water has no effect on no. of seedlings surviving up to the first 3 days
- quote data points to support any point
aerobic respiration
- no O2 so no aerobic respiration occurs
- no link reaction/ krebs cycle/ ETC/ oxidative phosphorylation
- no O2 to ac as the final e- acceptor
anaerobic respiration respiration
- plant has to switch anaerobic respiration
- only glycolysisi occrus
- NAD regenerated for glycolysisi
- pyruvate to ethanal to ethanol
- pyruvate decarboxylase
- only 2 ATP
consequences
- ethanol is toxic
- alcoholic fermentation is irreversible
- less ATP procduced
- no AT
- Root hair cells cannot take up mineral ions by AT
- so plant cannot make proteins
- cannot generate WP gradient into roots
- water cannot be absorbed (so cell cannot remain turgid)
- no PS
U : matrix
W : crista
Z : inter membrane space
cyanide: prevents aerobic respiration
fluoride prevents anaerobic respiration
alanine : both have 3 carbons
akutanic acids : both have 5 carbon atoms
aspartic acid: highest proportion of O2 atoms in its structure
outline the reactions that must occur to convert alpha ketoglutaric acid to oxaloacatate
- decarboxylation/ CO2 produced
- dehydrogenation / hydrogen removal
- ATP prodcued
- 4c intermediate produced
i) 16 c atoms in fatty acids ; 2 c atoms in acetyl coA which enters krebs cycle
ii) acetyl coA (2c) is produced form pyruvate ( 3 C atom) in link reaction
K: acetyl group of coA
L: Citrate
M:co2
N:oxaloacetate
coenzyme A + ADP/ATP
coenyme A:
- transfers acetly from link reaction to krebs cycle
ADP/ATP
- phosphorylation (addition of) of glucose to form hexose 1 6 bisphosphate in glycolysis
- dephosphorylation (removal of) of TP in glycolysis
- dephosphorylation of intermediate in krebs cycle
- formation from substrate level phosphorylation
- formation from oxidative phosphorylation, harnessing chem energy from chemiosmosis
