Exam questions Paper 1 Flashcards
Explain 3 ways an insects tracheal system is adapted for efficient gas exchange (3)
Tracheoles have thin walls so a short diffusion distance to cells
Tracheoles are highly branched/large number so have a large SA for diffusion
Tracheoles are highly branched/ large number so there is a short diffusion distance
Fluid at the end of tracheoles moves into tissues during exercise- so faster diffusion in gaseous medium to gas exchange surface
body moved by muscles to move air- maintains steep diffusion gradient for oxygen
The damselfly larva is a carnivore that actively hunts prey. It has gills to obtain
oxygen from water.
Some other species of insect have larvae that are a similar size and shape to
damselfly larvae and also live in water. These larvae do not actively hunt prey and
do not have gills.
Explain how the presence of gills adapts the damselfly to its way of life.
Damselfly larvae have a higher respiratory/metabolic rate
So uses more oxygen per unit time
Suggest two ways the student could improve the quality of her scientific drawing of
this gill.
Don’t use shading
Only use single lines/ Don’t use sketching/ensure lines are continuous
Use a magnification scale bar
Don’t cross label lines
Add further labels/annotations
Contrast how an optical microscope and a transmission electron microscope work
and contrast the limitations of their use when studying cells. [6]
TEM uses electrons while Optical uses visible light
TEM allows a greatER resolution
TEM allows smaller organelles to be observed/ greater detail in organelles to be observed
TEM specimen has to be dead, optical can view live specimens
TEM needs a thinnER specimen
TEM doesn’t show colour while optical does
TEM requires MORE complex preparation process
TEM focuses using magnets while optical uses glass lense
An environmental scientist investigated a possible relationship between air
pollution and the size of seeds produced by one species of tree.
He was provided with a very large number of seeds collected from a population of
trees in the centre of a city and also a very large number of seeds collected from a
population of trees in the countryside.
Describe how he should collect and process data from these seeds to investigate
whether there is a difference in seed size between these two populations of trees.
[5 marks]
Use random sample of seeds (from each
population);
2. Use (large enough) sample to be representative
of whole population;
3. Indication of what size was measured eg mass;
4. Calculate a mean and standard deviation (for
each population);
5. Use the (Student’s) t-test;
6. Analyse whether there is a significant difference
between (the means of) the two populations;
Give two pieces of evidence from Figure 1 that this cell was undergoing mitosis.
Explain your answers (june 2018 P1)
Chromosomes not arranged in homologous pairs as they would be in meiosis(so not meiosis)
Each chromosome has 2 chromatids as DNA has replicated
The chromosomes have become visible and have shortened and condensed
Give two ways in which the arrangement of prokaryotic DNA is different from the
arrangement of the human DNA in Figure 1.
Circular DNA not linear DNA
Not associated with proteins/histones
Only 1 molecule of DNA
Or
present as plasmids
What is meant by ‘species richness’?
measure of number of different species in a community
June 2018, Question 3
From the data in Figure 4, a student made the following conclusions.
- The natural habitat is most favourable for bees.
- The town is the least favourable for bees.
Do the data in Figure 4 support these conclusions? Explain your answer.
[4 marks]
1. The natural habitat is most favourable for bees.
2. The town is least favourable for bees
1- Yes, natural best, because
1. Peak of (mean) bee numbers in natural habitat
is highest;
2. The (mean) number of bees was higher in the
natural habitat until day 200;
3. (Mean) species richness in natural habitat
higher at all times;
No, natural not best, because
4. Lowest (mean) number of bees after day 220;
Yes, town worst, because
5. Peak of species richness higher in both natural
and farmland
OR
Species richness lowest in town from day 125;
No, town not worst, because
6. (Mean) species richness is lower in farmland
until day 125;
7. Similar (mean) number of bees to farmland;
OR
(Mean) number of bees lower in farmland until
day 140;
General, no, because
8. Index of diversity of bees not measured
OR
The number of bees of each species is not
known;
Bees are flying insects that feed on nectar made in flowers. There are many different
species of bee.
Scientists investigated how biodiversity of bees varied in three different habitats
during a year. They collected bees from eight sites of each habitat four times per year
for three years.
The scientists’ results are shown in Figure 4 in the form they presented them.
The scientists collected bees using a method that was ethical and allowed them to
identify accurately the species to which each belonged.
In each case, suggest one consideration the scientists had taken into account to
make sure their method
[2 marks]
1. was ethical.
- allowed them to identify accurately the species to which each belonged.
Must not harm the bees
OR
Must allow the bee to be released
unchanged;
- Must allow close examination
OR
Use a key (to identify the species);
June 2018, Q3
Suggest and explain two ways in which the scientists could have improved the
method used for data collection in this investigation.
- Collect at more times of the year so
more points on graph/better line (of
best fit) on graph; - Counted number of individuals in each
species so that they could calculate
index of diversity; - Collected from more sites/more years
to increase accuracy of (mean) data;
Formation of an enzyme-substrate complex increases the rate of reaction.
Explain how.
Reduces activation energy
due to bending bonds/ weakens bonds/puts tension on bonds
OR
Without enzymes few substrates have enough energy for reaction
Lyxose binds to the enzyme.
Suggest a reason for the difference in the results shown in Figure 5 with and without
lyxose.
[3 marks]
Changes the tertiary structure of the protein
Changed the shape of the active site
More successful e-s complexes form per minute
This enzyme is a polypeptide 465 amino acids long.
What is the minimum number of bases in the gene coding for this polypeptide?
1395
A change from Glu to Lys at amino acid 300 had no effect on the rate of reaction
catalysed by the enzyme. The same change at amino acid 279 significantly reduced
the rate of reaction catalysed by the enzyme.
Use all the information and your knowledge of protein structure to suggest reasons for
the differences between the effects of these two changes.
[3 marks]
Glu= negatively charged, lys= positive
- (Both) negatively charged to positively charged
change in amino acid; - Change at amino acid 300 does not change the
shape of the active site
OR
Change at amino acid 300 does not change the
tertiary structure
OR
Change at amino acid 300 results in a similar
tertiary structure;
- Amino acid 279 may have been involved in a
(ionic, disulfide or hydrogen) bond and so the
shape of the active site changes
OR
Amino acid 279 may have been involved in a
(ionic, disulfide or hydrogen) bond and so the
tertiary structure changed;
OR
Amino acid 279 may be in the active site and
be required for binding the substrate;
The scientists tested their null hypothesis using the chi-squared statistical test.
After 1 cycle their calculated chi-squared value was 350
The critical value at P=0.05 is 3.841
What does this result suggest about the difference between the observed and
expected results and what can the scientists therefore conclude?
The calculated value is more than the expected value
There is a less than 0.05 probability that the difference between the observed and expected results is due to chance
Reject the null hypothesis
Can conclude that the proportion of plants that produced 2n gametes does change from one generation to the next
Use your knowledge of directional selection to explain the results shown in Table 3. ( June 2018 Q6) The number of 2n gametes increases from one generation-next
The plants producing 2n gametes are selected for
Pass on the allele to their offspring
Frequency of this allele increaeses
The scientists then investigated the movement of iron ions (Fe3+) from the soil to old
and young leaves of heat-treated barley plants and to leaves of plants that were not
heat treated. Heat treatment was applied half way up the leaves. The scientists
determined the concentration of Fe3+ in the top and lower halves of the leaves of each
plant.
Their results are shown in Figure 10.
What can you conclude about the movement of Fe3+ in barley plants? [4]
June 2018, Q8
Heat treatment has a greater effect on young
leaves than old;
2. Heat treatment damages the phloem;
3. Fe3+ moves up the leaf/plant;
4. (Suggests) Fe3+ is transported in the xylem in
older leaf;
5. In young leaf, some in xylem, as some still
reaches top part of leaf;
6. (Suggests) Fe3+ is (mostly) transported in phloem
in young leaf
OR
Xylem is damaged in young leaf
OR
Xylem is alive in young leaf;
7. Higher ratio of Fe3+ in (all/untreated) old leaves
than (all/untreated) young;
8. All ratios show there is less Fe3+ in the top than
the lower part of leaves;
9. (But) no statistical test to show if the difference(s) are significant
Describe the role of two named enzymes in the process of semi-conservative
replication of DNA.
[3 marks]
DNA Helicase breaks Hydrogen bonds between two strands and unwinds the double helix
DNA polymerase joins adjacent free DNA nucleotides
By PHOSPHODIESTER BONDS
Mucus produced by epithelial cells in the human gas exchange system contains
triglycerides and phospholipids.
Compare and contrast the structure and properties of triglycerides and phospholipids.
[5 marks]
Both contain ester bonds (between glycerol and
fatty acid);
- Both contain glycerol;
- Fatty acids on both may be saturated or
unsaturated; - Both are insoluble in water;
- Both contain C, H and O but phospholipids also
contain P; - Triglyceride has three fatty acids and
phospholipid has two fatty acids plus phosphate
group; - Triglycerides are hydrophobic/non-polar and
phospholipids have hydrophilic and
hydrophobic region; - Phospholipids form monolayer (on
surface)/micelle/bilayer (in water) but
triglycerides don’t;
Mucus also contains glycoproteins. One of these glycoproteins is a polypeptide with
the sugar, lactose, attached.
Describe how lactose is formed and where in the cell it would be attached to a
polypeptide to form a glycoprotein.
- Glucose and galactose;
- Joined by condensation (reaction);
10.3 - Joined by glycosidic bond;
- Added to polypeptide in Golgi (apparatus);
Suggest one reason the scientists used biomass instead of the number of individuals
of each plant species when collecting data to measure diversity.
Too time consuming
Too small to count individuals
Individual organisms can’t be seperated
Describe how the scientists could use aseptic techniques to transfer
0.3 cm3 of C. difficile in liquid culture from a bottle onto an agar plate.
[3 marks]
Wash hands with soap/ disinfect surfaces
Use sterile pipette/syringe
Work close to upward air movement/ convection current
Remove lid and flame neck of bottle
Open lid of agar plate at an angle
Use sterile spreader
Disinfect pipette/spreader immediately after use
Scientists observed these APs on prokaryotes using a transmission electron
microscope. They stained the APs using a monoclonal antibody with gold attached
to it.
Suggest how these techniques allowed observation of APs on prokaryotes.
[3 marks]
Antibody binds to AP
Antibody has specific complementary shape to AP
Gold interacts with electrons in TEM
TEM used as it has a high resolution
Give two features of all prokaryotic cells that are not features of eukaryotic cells.
[1 mark]
Cicular DNA
No membrane bound organelles
DNA not associated with histones
Murein cell wall
DNA free in cytoplasm
Table 1 shows cell wall components in plants, algae, fungi and prokaryotes.
Complete Table 1 by putting a tick () where a cell wall component is present.
Plants=cellulose
Algae=cellulose
Fungi=Chitin
Prokaryotes=murein
Use Figure 5 to explain how human mass at birth is affected by stabilising selection.
[3 marks]
Babies with mass above 4500g transferred to SCU
Extreme mass babies less likely to survive so less likely to pass on alelle for extreme mass to next generation
Extreme mass at birth decreases in frequency/alleles for extreme mass at birth decreases in frequency in the population
Use the data in Table 3 and your knowledge of the immune response to suggest why
HIV controllers do not develop symptoms of AIDS.
[3 marks]
(lower viral load and more TH cells)
More Thelper cells
Lower viral load to destroy T helper cells
More activation of B cells/ cytoxic T cells/phagocytes
Increased production of antibodies due to increased number of B cells
Increased activation of cytoxic T cells means increased ability to destroy viral infected cells
Can destroy pathogens
Describe how a sample of chloroplasts could be isolated from leaves.
[4 marks]
Break open cells/tissue and filter
OR
Grind/blend cells/tissue/leaves and filter;
2. In cold, same water potential/concentration, pH
controlled solution;
3. Centrifuge/spin and remove nuclei/cell debris;
4. (Centrifuge/spin) at high(er) speed, chloroplasts
settle out;
Name the three phases of mitosis shown by C, D and E on Figure 7.
Describe the role of the spindle fibres and the behaviour of the chromosomes during
each of these phases.
[5 marks]
C = prophase and
D = metaphase and
E = anaphase;
- (In) prophase, chromosomes condense;
- (In) prophase OR metaphase, centromeres
attach to spindle fibres; - (In) metaphase, chromosomes/pairs of
chromatids at equator/centre of spindle/cell; - (In) anaphase, centromeres divide;
- (In) anaphase, chromatids (from each pair)
pulled to (opposite) poles/ends (of cell); - (In) prophase/metaphase/anaphase, spindle
fibres shorten;
Suggest and explain two further investigations that should be done before this ADC
is tested on human breast cancer patients.
[2 marks]
Tested on other mammals to check for
safety/side effects;
2. Tested on (healthy) humans to check for
safety/side effects;
4. Investigate different concentrations of ADC to
find suitable/safe dosage;
Describe how a triglyceride molecule is formed.
One glycerol and three fatty acids;
- Condensation (reactions) and removal of
three molecules of water; - Ester bond(s) (formed);
The scientists used a data logger to measure the length of the root rather than a ruler.
Suggest one reason why they used a data logger and explain why this was important
in this investigation.
Increases accuracy as the differences in length are small
Explain five properties that make water important for organisms.
- A metabolite in condensation/hydrolysis/ photosynthesis/respiration;
A solvent so (metabolic) reactions can occur
OR
A solvent so allowing transport of substances;
High (specific) heat capacity so buffers changes in temperature;
For ‘buffer’ accept ‘resist’.
Large latent heat of vaporisation so provides a cooling effect (through
evaporation);
Cohesion (between water molecules) so supports columns of water
(in plants);
Cohesion (between water molecules) so produces surface tension
supporting (small) organisms;
Allow other suitable
properties but must have a valid explanation.
For example
* ice floating so maintaining aquatic habitat beneath
* water transparent so allowing light penetration for
photosynthesis
State and explain the property of water that helps to prevent temperature
increase in a cell.
High SHC
Buffers changes in T
