independent study questions Flashcards
where precisely on a cell does the Calvin cycle take place?
stroma
Explain the decrease in the amount of GP after the carbon dioxide concentration was reduced (1 mark)
less RUBP combined with CO2
the scientist carried out a similar experiment but increased the carbon dioxide concentration from 1% to 2%. The relative amounts of GP and RuBP remained the same.
Suggest two reasons why (2 marks)
temperature is a limiting factor
limited by RuBP
Some bacteria use hydrogen sulfur, H2S to produce organic compounds. The hydrogen sulfide has a similar role to that of water in photosynthesis (2 marks)
hydrogen sulfide is used as an oxidising agent
they add a source if H+ ions to reduce the CO2
energy enters most ecosystems through the light dependent reaction of photosynthesis. Describe what happens during the light independent reaction. (5 marks)
chlorophyll absorbs light energy
excites electrons
electrons move along carriers releasing energy
energy used to join ADP and Pi to form ATP
photolysis of water produces protons, electrons and oxygen
changes in ecosystems can lead to speciation. A high concentration of copper in soil is toxic to most plants. In some areas where the soil is polluted with copper, populations of grasses are found to be growing. These populations of grass belong to a species also found growing on unpolluted soils.
It has been suggested that a new species of grass may evolve on soil that has been polluted with copper. Explain how this new species might evolve
(5 marks)
variation
mutation
some plants have allele to survive in polluted soils
adapted organisms reproduce
increase in frequency of allele
During the light independent reaction of photosynthesis, carbon dioxide is converted into organic substances. Describe how ( 6 marks)
carbon dioxide combines with RuBP
produces 2 glycerate 3 phosphate
GP reduced to triose phosphate
Using reduced NADP
Using energy from ATP
TP converted to glucose
Beech trees have two types of leaves called sun leaves and shade leaves. Sun leaves grow on branches exposed to direct sunlight, shade leaves grow on branches exposed to light that has passed through leaves. An ecologist collected sun leaves and shade leaves from beech trees and determines the mean mass of each photosynthetic pigment, in both types of leaf.
Describe how you would present the data in the table as a graph (2 marks)
bar chart
error bars to represent standard deviation
the ecologist collected shade leaves at random from a branch. Suggest a method he could have used to collect shade leaves at random from a branch ( 2 marks)
Number the leaves on the branch
use random number generator to determine which leaf to pick
the ecologist concluded that there is a significant difference of chlorophyll b in sun leaves and shade leaves of beech trees (2 marks)
no stats test carried out
95% confidence
each type of chlorophyll is produced by a specific enzyme controlled pathway. Use this information to suggest how the same beech trees can produce more chlorophyll b in some leaf cells than others (2 marks)
in shade leaves greater amount of enzyme activity for production of chlorophyll b
more mRNA produced
The solution that the student used to produce the chloroplast suspension had the same water potential as the chloroplasts. Explain why it was important that these water potentials were the same ( 2 marks)
osmosis does not occur
chloroplast does not burst
explain why the student set up tube 1 (2 marks)
to show light does not affect DCPIP
to show chloroplasts are required
explain the results in tube 3 (2 marks)
reduction of DCPIP by electrons
from light dependent reaction
explain the advantage of the student using the IC50 in this investigation (1 mark)
provides a reference point
explain how chemicals which inhibit the decolorisation of DCPIP could slow the growth of weeds (2 marks)
less ATP produced
less NADP produced
less GP converted to TP
calculate the total uptake of carbon dioxide between 11:40 and 13:15 in trees exposed to full sun in a forest that is 12 000 m2 in area ( 3 marks )
6.0 x 10^8
Figure 1 shows there is a small difference in the mean uptake of carbon dioxide between 08.30 and 09.40 hours by trees in full sun and by trees in the shade. When the scientists performed a statistical test on these data, they calculated P>0.5 (2 marks)
difference is not significant
there is a greater than 0.5 probability that this difference is due to chance
In this species of tree, very high light intensities can inhibit the release of electrons from chlorophyll
suggest how this could explain the results shown in figure 1 for 11.40 to 13.15 hours ( 4 marks)
less ATP and reduced NADP produced
less GP converted to triose phosphate
less triose phosphate to regenerate RuBP
less RuBP to react with carbon dioxide
mature leaves from slow growing shade tolerant plants produce poisonous chemicals that are a defence against being eaten by herbivores
suggest how this benefits slow growing, shade tolerant plants (2 marks)
less photosynthesis
slow growing so would take a long time to replace mature leaves
A journalist published the following summary of these results
the more recently a shade- tolerant plant species evolved, the greater the concentration of poisonous chemicals it produces
Do the data in figure 2 support this summary? Justify your answer (2 marks)
the highest concentration was not in the most recently evolved species
the lowest concentration was not in the least recently evolved species
Heat stress decreases the light dependent reaction of photosynthesis
Explain why this leads to a decrease in the light independent reaction (2 marks)
less ATP
less reduced NADP produced
A decrease in the activity of the enzyme rubisco would limit the rate of photosynthesis
Explain why (2 marks)
less CO2 reacts with RuBP
less GP
where precisely is rubisco found in a cell? (1 mark)
stroma
The scientists concluded that heat stress reduces the activity of rubisco in plant leaves by affecting rubisco activase
Use all the information to evaluate their conclusion ( 4 marks)
rubisco activity increase with temperature
rubisco activase optimum temperature 30°
no stats test done
results are only for cotton
other substances are produced in the Krebs cycle in addition to the carbon compounds shown in the diagram. Name three of these other products (3 marks)
reduced NAD
reduced FAD
ATP
Where in a cell does stage 1 occur? (1 mark)
cytoplasm
explain how stage 2 enables stage 1 to continue. (2 marks)
Regenerates NAD
NAD reduced in stage 1
what would be the RQ for aerobic respiration of glucose? (1 mark)
1
A student calculated the RQ of germinating seeds was 1.8
Use the information to provided to explain this result (2 marks)
aerobic and anaerobic respiration occurring
more carbon dioxide produced than oxygen uptake
aerobic respiration produces more ATP per molecule of glucose than anaerobic respiration
Explain why (2 marks)
oxygen is final electron acceptor
oxidative phosphorylation
anaerobic respiration only glycolysis occurs
Describe the process of glycolysis (4 marks)
phosphorylation of glucose using ATP
oxidation of triose phosphate to pyruvate
net gain of ATP=4
NAD reduced
malonate inhibits a reaction in the Krebs cycle
explain why malonate would decrease the uptake of oxygen in a respiring cell (2 marks)
fewer electrons removed and passed to electron transfer chain
oxygen is the final electron acceptor
describe the part played by the inner membrane of a mitochondrion in producing ATP (3 marks)
electrons transferred down electron transport chain oxygen
provides energy to take protons into space between membranes
protons pass back through membrane
why was the solution in which the mitochondria were suspended isotonic? (1 mark)
prevent damage to mitochondria caused by osmosis
explain why the scientist did not use glucose as the respiratory substrate (2 marks)
glucose is broken down during glycolysis in cytoplasm
glucose cannot cross the mitochondrial membrane
explain why the oxygen concentration would change during this investigation (1 mark)
oxygen is final electron acceptor and used to make water
Explain why a layer of oil is required in this investigation (1 mark)
prevents oxygen being absorbed
suggest why the rate of gas production decreased between 50 and 60 minutes (1 mark)
glucose is a limiting factor
