A level june 2018 Flashcards
Give two pieces of evidence from Figure 1 that this cell was undergoing mitosis.
Explain your answers.
[2 marks]
1. The (individual) chromosomes are visible because they have condensed; 2. (Each) chromosome is made up of two chromatids because DNA has replicated; 3. The chromosomes are not arranged in homologous pairs, which they would be if it was meiosis;
When preparing the cells for observation the scientist placed them in a solution that
had a slightly higher (less negative) water potential than the cytoplasm. This did not
cause the cells to burst but moved the chromosomes further apart in order to reduce
the overlapping of the chromosomes when observed with an optical microscope.
Suggest how this procedure moved the chromosomes apart.
[2 marks]
- Water moves into the cells/cytoplasm by
osmosis; - Cell/cytoplasm gets bigger;
The dark stain used on the chromosomes binds more to some areas of the
chromosomes than others, giving the chromosomes a striped appearance.
Suggest one way the structure of the chromosome could differ along its length to
result in the stain binding more in some areas.
[1 mark]
Differences in base sequences OR Differences in histones/interaction with histones OR Differences in condensation/(super)coiling
What is a homologous pair of chromosomes?
(Two chromosomes that) carry the same
genes;
Give two ways in which the arrangement of prokaryotic DNA is different from the
arrangement of the human DNA in Figure 1.
[2 marks]
(Prokaryotic DNA) is 1. Circular (as opposed to linear); 2. Not associated with proteins/histones ; 3. Only one molecule/piece of DNA OR present as plasmids;
Describe the method the student would have used to obtain the results in Figure 3.
Start after all of the cubes of potato have been cut. Also consider variables he should
have controlled.
[3 marks]
- Method to ensure all cut surfaces of the eight
cubes are exposed to the sucrose solution; - Method of controlling temperature;
- Method of drying cubes before measuring;
- Measure mass of cubes at stated time intervals;
What is meant by ‘species richness’?
[1 mark]
(A measure of) the number of (different) species in
a community
- 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] - The natural habitat is most favourable for bees.
- The town is the least favourable for bees.
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;
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
2. allowed them to identify accurately the species to which each belonged.
1. Must not harm the bees OR Must allow the bee to be released unchanged; 2. Must allow close examination OR Use a key (to identify the species);
Suggest and explain two ways in which the scientists could have improved the
method used for data collection in this investigation.
[2 marks]
1. Collect at more times of the year so more points on graph/better line (of best fit) on graph; 2. Counted number of individuals in each species so that they could calculate index of diversity; 3. Collected from more sites/more years to increase accuracy of (mean) data;
Three of the bee species collected in the farmland areas were Peponapis pruinosa,
Andrena chlorogaster and Andrena piperi.
What do these names suggest about the evolutionary relationships between these
bee species? Explain your answer.
[2 marks]
- A. chlorogaster and A. piperi are more
closely related (to each other than to
P. pruinosa); - Because they are in the same genus;
Formation of an enzyme-substrate complex increases the rate of reaction.
Explain how.
[2 marks]
1) Enzyme provides an alternative pathway with a lower activation energy.
2) Due to the bending of bonds.
OR
Without enzyme, very few substrates have
sufficient energy for reaction;
Calculate by how many times the rate of reaction is greater with the enzyme present.
Give your answer in standard form.
[2 marks]
578/3.0 × 10 -9 =1.93x10 11
Lyxose binds to the enzyme.
Suggest a reason for the difference in the results shown in Figure 5 with and without
lyxose.
[3 marks]
1. (Binding) alters the tertiary structure of the enzyme ; 2. (This causes) active site to change (shape); 3. (So) More (successful) E-S complexes form (per minute) OR E-S complexes form more quickly OR Further lowers activation energy;
The genetic code is described as degenerate.
What is meant by this? [2 marks]
- More than one codon codes for a single amino
acid; - Suitable example selected from Table 1
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]
- (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?
[2 marks]
- There is a less than 0.05/5% probability
that the difference(s) (between observed
and expected) occurred by chance; - Calculated value is greater than critical
value so the null hypothesis can be
rejected; - (The scientists can conclude that) the
proportion of plants that produce 2n
gametes does change from one breeding
cycle to the next;
Use your knowledge of directional selection to explain the results shown in Table 3.
[3 marks]
- The scientists selected/used for breeding plants
that produced 2n gametes; - (So these plants) passed on their alleles (for
production of 2n gametes to the next
generation); - The frequency of alleles for production of 2n
gametes increased (in the population).
When a person is bitten by a venomous snake, the snake injects a toxin into the
person. Antivenom is injected as treatment. Antivenom contains antibodies against
the snake toxin. This treatment is an example of passive immunity.
Explain how the treatment with antivenom works and why it is essential to use passive
immunity, rather than active immunity.
[2 marks]
- (Antivenom/Passive immunity) antibodies
bind to the toxin/venom/antigen and
(causes) its destruction; - Active immunity would be too slow/slower;
A mixture of venoms from several snakes of the same species is used.
Suggest why.
[2 marks]
1. May be different form of antigen/toxin (within one species) OR Snakes (within one species) may have different mutations/alleles; 2. Different antibodies (needed in the antivenom) OR (Several) antibodies complementary (to several antigens);
Horses or rabbits can be used to produce antivenoms.
When taking blood to extract antibody, 13 cm3 of blood is collected per kg of the
animal’s body mass.
The mean mass of the horses used is 350 kg and the mean mass of the rabbits used
is 2 kg
Using only this information, suggest which animal would be better for the production of
antivenoms.
Use a calculation to support your answer.
[2 marks]
1. Horses because more antivenom/antibodies could be collected (as more blood collected); 2. 4550 (cm3 ) v 26 (cm3 ) (blood collected);
During the procedure shown in Figure 8 the animals are under ongoing observation
by a vet.
Suggest one reason why.
[1 mark]
- (So) the animal does not suffer from the
venom/vaccine/toxin; - (So) the animal does not suffer
anaemia/does not suffer as a result of
blood collection; - (So) the animal does not have pathogen
that could be transferred to humans;
During vaccination, each animal is initially injected with a small volume of venom.
Two weeks later, it is injected with a larger volume of venom.
Use your knowledge of the humoral immune response to explain this vaccination
programme.
[3 marks]
- B cells specific to the venom reproduce by
mitosis; - (B cells produce) plasma cells and memory
cells; - The second dose produces antibodies (in
secondary immune response) in higher
concentration and quickly
OR
The first dose must be small so the animal
is not killed;
The scientists concluded that this heat treatment damaged the phloem.
Explain how the results in Figure 9 support this conclusion.
[2 marks]
- The radioactively labelled carbon is converted
into sugar/organic substances during
photosynthesis; - Mass flow/translocation in the phloem
throughout the plant only in plants that were
untreated/B/control
OR
Movement of sugar/organic substances in the
phloem throughout the plant only in plants that
were untreated/B/control;
OR - Movement in phloem requires living
cells/respiration/active transport/ATP; - Heat treatment damages living cells so
transport in the phloem throughout the plant
only in plants that were untreated/B/control
OR
Heat treatment stops respiration/active
transport/ATP production so transport in the
phloem throughout the plant only in plants that
were untreated/B/control;
