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2020 paper biological processes

Terms in this set (18)

Q4) Which of the following statements, A to D, does not correctly describe the structure or formation of plant vascular tissues?

A) Companion cells are linked to xylem vessels by plasmodesmata.

B) Mature sieve tube elements do not contain nuclei.

C) Phloem and xylem are formed by differentiation of vascular meristems.

D) Xylem vessels have non-lignified pits to allow movement in and out.
Answer =
Q6) Which of the following statements, A to D, describes the movement of water across plant roots?

A) The Casparian strip blocks movement by the symplast pathway.

B) The symplast pathway requires water to cross partially permeable membranes.

C) Water moves from the soil to the root hair cells up a water potential gradient.

D) Water moves through the leaves only by the symplast pathway and across the roots only by the apoplast pathway.
...
Q7) Which of the following statements about water transport in plants is/are correct?

1) Transpiration happens as a consequence of the need for gas exchange.

2) There are cohesive forces between water molecules because they form hydrogen bonds with
one another.

3) Water is drawn up the stem due to adhesive forces between water molecules.

A) 1, 2 and 3
B) only 1 and 2
C) only 2 and 3
D) only 1
...
8) The graphs, A to D, represent gas exchange in bony fish.
The graphs show the oxygen saturation in blood in the lamellae and water flowing over the lamellae.

Which graph, A to D, shows the relationship between blood oxygen saturation and distance along the lamellae?
...
Questions 9 and 10 refer to the graph below.
An extract of the bark of the Indian fig tree, Ficus racemosa, has antioxidant properties. It has been suggested that this extract could protect against damage to non-cancerous cells during radiotherapy.

Cultures of non-cancerous cells were exposed to increasing doses of radiation and the percentage of damaged cells was measured. The experiments were performed multiple times either with or without the bark extract.

The graph shows the results for the two sets of data. Each data point shows the mean ±1 standard deviation.
...
Q9) A student used Spearman's rank correlation coefficient to test for correlation between radiation
dose and cell damage.
Which of the following statements, A to D, about this correlation shown on the graph is correct?

A) Spearman's rank correlation coefficient, rs, for cells without extract will be between 0 and +1.

B) Spearman's rank correlation coefficient, rs, for cells with extract will be between 0 and −1.

C) There is a negative correlation between radiation dose and cell damage in both cases.

D) There is no correlation between radiation dose and cell damage in either case.
...
Q10) Which of the following statements, A to D, about the graph is correct?
A) The data for the 'with extract' group is more variable than the data for the 'without extract' group.

B) The error bars show the range of data at each radiation dose.

C) The error bars show whether the difference between the two sets of data is significant.

D) The χ2 (chi-squared) test can be used to determine whether there is a significant difference between the two sets of data.
...
Questions 11 and 12 refer to the investigation described below.
The diagram shows a respirometer used to compare respiration in two types of germinating seeds.

- starting position - respiring seeds of the meniscus - soda-lime (absorbs CO2)

A student set up the respirometer to measure oxygen consumption.

• The narrow tube contained coloured water.
• The position of the meniscus was noted at the beginning of the experiment.
• The tube was left for 20 minutes.
• The new position of the meniscus was noted.
• The experiment was repeated with the other type of seed.
...
Q11) Which of the following would be necessary to ensure valid results?

1) Keeping the respirometer in the dark during the experiment.

2) Keeping the respirometer at the same temperature during the experiment.

3) Using the same dry mass of seeds each time.

A) 1, 2 and 3
B) only 1 and 2
C) only 2 and 3
D) only 1
...
Q12) The student investigated two types of seed, pea and sunflower:
• pea seeds store mainly starch
• sunflower seeds store mainly lipid.

Which of the following, A to D, describes the results you would expect with each type of seed?

A) The meniscus would move to the left with pea seeds and further to the left with sunflower
seeds.

B) The meniscus would move to the left with sunflower seeds and to the right with pea seeds.

C) The meniscus would move to the right with pea seeds and further to the right with sunflower seeds.

D) The meniscus would not move.
...
Questions 14 and 15 refer to the diagram below of a synapse.
L, J, K

Q14) Which of the following statements, A to D, describes events occurring at a synapse?

A) Acetylcholine is broken down by enzymes so that it can bind to structure K.

B) An action potential causes structure J to close.

C) Structure J is a voltage gated Ca2+ channel.

D) Structure L is released by exocytosis.
...
Q15) GABA is a neurotransmitter.
GABA reduces the number of action potentials in the postsynaptic neurone by opening chloride ion channels in the post-synaptic membrane.

Which of the following statements, A to D, describes the action of GABA?

A) GABA binds to structure K in competition with acetylcholine.
B) GABA causes hyperpolarisation of the post-synaptic membrane.
C) GABA causes depolarisation of the post-synaptic membrane.
D) GABA inhibits release of neurotransmitter from structure L.
...
The student used a colorimeter to measure the concentration of each substance in the liquid
surrounding the cells.
The colorimeter had an analogue display. The reading was indicated by a needle moving across a scale. The smallest divisions on the scale corresponded to 0.1 absorbance unit.
After the investigation the student suggested some improvements.
Draw a line between each of the improvements to the corresponding justification.

Q16c
DOUBT
Q17a) iii) The solvent used for the separation was non-polar.
Identify the spot corresponding to the least polar pigment. Give a reason for your choice.

Spot -
Reason -
Spot - 5
Reason - It is the most soluble in mobile phase. Non polar solvent will facilitate movement of non polar substance.
17b) The student used the following method for the investigation:

Step 1: Extraction of pigments

• Take 0.5g of fresh spinach and add 1g of sand.

• Grind the mixture until it becomes a fine, light green powder.

• Transfer the powder to a test tube and add 2 cm3 of propanone.

• Stir for about 1 minute then allow to stand for another minute.

• Transfer the dark green upper layer with a pipette to a clean test tube and seal with film when not in use.

Step 2: TLC analysis

• Hold the TLC plate carefully by the edges and avoid damaging the surface of the plate.

• Draw a pencil line across the width of the TLC plate 1 cm from the bottom edge.

• Spot the extract on the pencil line using a pipette, one drop at a time, allowing the spot to dry before adding the next drop.

• Place chromatography solvent in a jar so that it is no more than 0.5 cm deep.

• Lower the TLC plate into the jar and lean the top against the side of the jar. Make sure the plate does not touch the sides of the jar anywhere else.

• Place a cap on the jar and allow the solvent to soak up the plate.

• When the solvent has reached a few mm from the top of the plate, remove the plate from the jar and mark the position of the solvent front with a pencil.
...
17b) i) Explain why the method included the following precautions:

1) Hold the TLC plate carefully by the edges and avoid damaging the surface of the plate

2) Make sure the plate does not touch the sides of the jar anywhere else.
1) to avoid contamination by fingerprints or oils
OR
so that spots aren't affected by damage

2) to prevent spots from traveling in wrong directions
OR
condensation of liquids on wall may affect movement of spots
Q17c) Describe and explain the effect of reducing the CO2 concentration on the concentration of RuBP and GP in the algae. (3m)
1) RuBP concentration increases and then decreases
2) Increases because it is not converted into GP
3) Increases because it is still being produced by TP
4) Decreases because no GP available to regenerate RuBP.
5) GP decreases
6) Decreases because no CO2 to react with RuBP to produce GP
Explain the advantages of using a confocal microscope. (2020 paper 2, q18b)
1) Can be used with living cells/ thick samples
2) Sharp image/ less blurred image
3) High resolution
4) Can be used to see distribution of molecules within cells
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