BIO PAPER 1 2018 Flashcards
0 1 Figure 1 shows all the chromosomes present in one human cell during mitosis. A
scientist stained and photographed the chromosomes. In Figure 2, the scientist has
arranged the images of these chromosomes in homologous pairs.
Figure 1 Figure 2
0 1 . 1 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;
0 1 box . 2 Tick ( ) one box that gives the name of the stage of mitosis shown in Figure 1. [1 mark] Anaphase Interphase Prophase Telophase
prophase
1 . 3 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;
0 1 box . 4 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
Question 01.4 tested Assessment Objective 2 (application of knowledge) and there were several
parts of the specification from which students could select material to support their answer. 42.2%
did this successfully; those who did not often did not describe sufficiently how a feature would be
different along the length of the chromosome to result in the striped appearance. For example,
mentioning “histones” or “bases” alone was not creditworthy.
1 . 5 In Figure 2 the chromosomes are arranged in homologous pairs.
What is a homologous pair of chromosomes?
[1 mark]
(Two chromosomes that) carry the same
genes;
Reject ‘same alleles’
Accept ‘same loci’ (plural) or
‘genes for the same
characteristics
In question 01.5, only 36.9% of students could define the term ‘homologous chromosome’. It is
likely that more students could have written about independent segregation or crossing over of
homologous chromosomes, but this question revealed that they did not fully understand this
biological term. Many students only referred to the origin of the chromosomes as paternal and
maternal.
0 1 . 6 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;
Max 1 if prokaryotic DNA only found as plasmids OR if prokaryotic DNA is single stranded. Ignore references to nucleus, exons, introns or length of DNA. Do not credit converse statements. Ignore descriptions of eukaryotic DNA alone.
For question 01.6, most students (83.2%) scored at least one mark, but there were many good
answers limited to one out of two by demonstration of fundamental misunderstanding. The most
common of these was that only eukaryotic DNA is a double helix and that prokaryotic DNA is
single-stranded.
0 2 box A student investigated the effect of surface area on osmosis in cubes of potato.
• He cut two cubes of potato tissue, each with sides of 35 mm in length.
• He put one cube into a concentrated sucrose solution.
• He cut the other cube into eight equal-sized smaller cubes and put them into a
sucrose solution of the same concentration as the solution used for the large cube.
• He recorded the masses of the cubes at intervals.
His results are shown in Figure 3
0 2 . 1 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;
This question was based on a variant of required practical activity 3 – “Production of a dilution
series of a solute to produce a calibration curve with which to identify the water potential of plant
tissue”.
Although question 02.1 was based on a more straightforward investigation of osmosis, very similar
principles apply to those used when carrying out the required practical. When asked for a method,
students must write out what actions need to be carried out as if instructing a fellow student.
Generic statements, such as “keep the temperature the same” or “record the mass at set intervals”,
were not credited. Instead, students should give specific instructions about how the results would
be obtained, for example using a water bath set to 30°C, removing the potato chips every 10
minutes to record their mass, and blotting the potato chips dry with a paper towel.
0 2 box . 2 The loss in mass shown in Figure 3 is due to osmosis. The rate of osmosis between
0 and 40 minutes is faster in B (the eight small cubes) than in A (single large cube).
Is the rate of osmosis per mm2 per minute different between A and B during this time?
Use appropriate calculations to support your answer.
[3 marks]
Allow 1 mark for calculation of surface area of two
(sets of) cubes 7350 (mm2
) and 14700 (mm2
)
Allow 1 mark for calculation of both rates of
osmosis shown in first 40 minutes – between 0.12
and 0.13 and between 0.22 and 0.23
If surface area and/or rate of osmosis is incorrect
then, allow 1 mark for (their) calculated rate divided
by (their) calculated surface area
Question 02.2 was the first question on the paper testing mathematical skills (specifically MS 3.5
and MS 4.1). Students needed to calculate rate of change from a graph showing a linear
relationship; this was the most successfully completed part of the calculation, although using the
wrong part of the graph or incorrectly reading from the graph was not uncommon. Students also
needed to calculate the surface area of the one large cube and the eight smaller cubes, but only a
small number of students could do this correctly. Only 5% could correctly combine these to come
up with two correct rates per unit surface area. There was a small but sizeable minority of students
who did not show the steps in their calculations, but just gave the final (incorrect) answer, meaning
it was not possible to award any marks. A small minority wrote descriptions of the change in mass
shown by the graph, or explained why the cubes with large surface area would lose mass more
quickly, without any calculations, thereby ignoring the question stem. Disappointingly, 56.4% of
students failed to gain even one mark for this question.
0 3 Bees are flying insects that feed on nectar made in box 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.
Figure 4
0 3 . 1 What is meant by ‘species richness’?
[1 mark]
(A measure of) the number of (different) species in a community; For ‘community’ accept ‘habitat/ecosystem/one area/environment’ Reject ‘in a population’
67.1% of students could answer question 03.1, a test of recall from section 3.4.6 of the
specification. Once again, many students confused the terms population and community, and the
terms species richness and an index of diversity.
0 3 box . 2 From the data in Figure 4, a student made the following conclusions.
1. The natural habitat is most favourable for bees.
2. 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.
- The town is the least favourable for bees.
(4)
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;
Question 03.2 was the first opportunity on the paper for students to interpret scientific evidence
and demonstrate Assessment Objective 3 skills. The question, ‘Do the data in Figure 4 support these conclusions?’, should have demonstrated that careful use of these data would be required in
the answer. Both conclusions were comparative statements, natural habitat being the most
favourable and town habitat being the least favourable, and so the marking points were linked to
comparative statements between the habitats. A description of one habitat alone was, thus, not
sufficient to gain credit. Numbers were provided on the axes and so quoting days of the year was
expected in the answers. This question provides a good example to demonstrate the importance
of using and quoting the correct evidence from the data when trying to support the conclusions. It
is also a good example of how times from the x-axis of the graph should be used to illustrate where
observed trends begin and end. Some 76.6% of students were able to gain at least one mark; only
7.9% scored four marks.
0 3 box . 3 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]
- 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);
In part 03.3, the majority of students appreciated that the bees should not be harmed and so
gained the first mark. Many students got confused with mark-release-recapture techniques but,
since no population estimate was being made in this investigation, this was not relevant.
3 . 4 Suggest and explain two ways in which the scientists could have improved the
method used for data collection in this investigation.
[2 marks]
- 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;
0 3 box . 5 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;
03.5 was answered well, with most students (62.1%) understanding the binomial system for
naming of species and linking it to their evolutionary relationship. Some students suggested that
the two Andrena species were unrelated to Peponapis pruinosa and that they did not share a
common ancestor – this was not creditworthy
0 4 . 1 Formation of an enzyme-substrate complex increases the rate of reaction.
Explain how.
[2 marks]
- Reduces activation energy;
- Due to bending bonds
OR
Without enzyme, very few substrates have
sufficient energy for reaction;
In question 04.1, although many students could describe that the enzyme lowers the activation
energy, only 18% could explain that this was due to the enzyme bending the bonds in the
substrate. Some students gave lengthy descriptions of how the E-S complex forms, losing focus
on the question of how this results in an increase in the rate of reaction.
0 4 . 2 A scientist measured the rate of removal of amino acids from a polypeptide with and
without an enzyme present. With the enzyme present, 578 amino acids were
released per second. Without the enzyme, 3.0 × 10−9 amino acids were released per
second.
Calculate by how many times the rate of reaction is greater with the enzyme present.
Give your answer in standard form.
[2 marks]
1.93 x 1011;; Allow 1 max for 578/3.0 × 109 1.93 x 10x when x ≠11 Correct answer with incorrect standard form e.g. 19.3 x 1010
Questions 04.2 and 04.3 assessed mathematical skills MS 0.2, MS 0.3 and MS 3.1. It was
pleasing to see standard form being used successfully, with 71.9% of students achieving both
marks for 04.2.
Another scientist investigated an enzyme that catalyses the box following reaction. ATP → ADP + Pi The scientists set up two experiments, C and L. Experiment C used • the enzyme • different concentrations of ATP. Experiment L used • the enzyme • different concentrations of ATP • a sugar called lyxose. The scientists measured the rate of reaction in each experiment. Their results are shown in Figure 5.
0 4 box . 3 Calculate the rate of reaction of the enzyme activity with no lyxose at 2.5 mmol dm−3 of
ATP as a percentage of the maximum rate shown with lyxose.
[2 marks]
Students found part 04.3 more challenging. Many thought they needed to
calculate a rate, not appreciating that the y-axis already gave the rate of reaction. Many did not
read the correct parts of the graph – at 2.5 mmol dm-3 for curve C and at the highest point for curve L. Credit of a single mark was given if a correct calculation was completed with readings from
incorrect parts of the graph, but all readings had to be accurate and this was often not the case.
4 box . 4 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 key to answering question 04.4 successfully was to use the information provided – that the
lyxose binds to the enzyme and that the graph shows that lyxose increases the rate of reaction.
Many students started their answer with the idea of lyxose being an inhibitor and reducing the rate
of reaction – this limited them to one mark out of three. A significant number of students inferred
that lyxose is a respiratory substrate (rather than using the information that it binds to the enzyme),
so there would be an increase in ATP levels, so more substrate and hence a faster rate. It was a
shame that some students who worked out what was going on did not achieve full marks due to
imprecise use of language, for example referring to changing the 3-D shape of the enzyme rather
than changing the tertiary structure of its active site. Fewer than half the students scored any
marks on this question.
0 5 box . 1 Draw the general structure of an amino acid.
[1 mark]
H
H2N —–c—COOH
R
64.5% of students could answer question 05.1, a test of recall from section 3.1.4.1 of the
specification.
5 . 2 The genetic code is described as degenerate.
What is meant by this? Use an example from Table 1 to illustrate your answer.
[2 marks]
- More than one codon codes for a single amino
acid; - Suitable example selected from Table 1;
A scientist investigated changes in the amino acid sequence of a human enzyme box
resulting from mutations. All these amino acid changes result from single base
substitution mutations.
This enzyme is a polypeptide 465 amino acids long.
Table 2 shows the result of three of the base substitutions.
Table 2
Amino acid
number
Correct amino
acid
Amino acid inserted as a result of
mutation
203 Val Ala
279 Glu Lys
300 Glu Lys
0 5 . 3 What is the minimum number of bases in the gene coding for this polypeptide?
[1 mark]
1395
05.3 was an interesting question as it could be interpreted as a straightforward 465 amino acids
multiplied by three to give the 1395 bases of the gene. Students could also add three or six bases
to this, for a start and/or stop codon. Some students appreciated that a gene would be made up of
double-stranded DNA so would comprise double this number of bases – this was also awarded
credit. 67% of students could complete this calculation.
0 5 box . 4 Use information from Table 1 to tick ( ) one box that shows a single base substitution
mutation in DNA that would result in a change from Val to Ala at amino acid number
203
CAA → CGA GUU → GCA GUU → GUC CAC → CGG [1 mark]
CAA → CGA
Question 05.4 showed that multiple-choice questions are not always simplistic or based on recall.
This question required students to interpret Table 1, convert the mRNA codon to a DNA base
triplet, and also notice in the question that this was a single base substitution. Only 9.6%
successfully achieved all of this. The majority of students ticked box 2 that did show a Val to Ala
change, but with a mutation in the mRNA (not DNA) and of 2 bases
0 5 . 5 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 answers to question 05.5 revealed much misunderstanding of the principles of the link
between DNA, polypeptides and proteins, as many students discussed the changes in amino acid
in terms of silent mutations or frameshift mutations. The instruction to use all the information
required the students to look back at Table 1 and use the key showing the properties of the R
group of the amino acid. Very few students could apply this information and link the reduction in
rate of reaction with the R group of amino acid 279 changing from being negatively-charged to
being positively-charged. Imprecise biological language was also often seen, for example omitting
to reference the ‘shape’ of the active site and discussing the 3-D structure of the protein rather than
its tertiary structure. Just over half of the students (54.2%) managed to score at least one mark
here; only 3.2% gained maximum credit.
0 6 box Figure 6 shows a faulty form of meiosis that can occur in some plant
s.
0 6 . 1 Complete Figure 7 to show the chromosome content of the cells that would result
from a normal meiotic division of the diploid parent cell shown in Figure 6.
[2 marks]
- 1 long and 1 short chromosome, each
made up of 2 chromatids held (by
centromere), in each cell of 1st division; - 1 long and 1 short (separate) chromosome
in each cell of 2nd division;
2. Allow ECF for correct chromosomes shown in each cell from candidate’s 1 st division cells. 2. Ignore drawing of centromere
0 6 box . 2 If two diploid (2n) gametes fuse at fertilisation, it can result in the growth of a tetraploid
plant which has 4 copies of each chromosome.
Red clover is a plant grown to produce cattle feed. Tetraploid red clover plants
produce a higher yield than diploid red clover plants.
Whether a red clover plant produces 2n gametes is genetically controlled.
Scientists investigated the possibility of breeding red clover plants that only produced
2n gametes.
• In breeding cycle 0, they grew red clover plants and identified plants that produced
2n gametes.
• In breeding cycle 1, they used the plants producing 2n gametes to produce
offspring.
• In breeding cycles 2 and 3, they identified plants producing 2n gametes and used
these to produce offspring.
Their results are shown in Table 3.
Table 3
The scientists used the following null hypothesis.
‘The proportion of plants that produce 2n gametes will not change from one breeding
cycle to the next.’
Complete Table 3 to show the expected number of plants that did not produce 2n
gametes and the expected number of plants that did produce 2n gametes after 1
cycle.
Give each answer to the nearest whole number.
[2 marks]
52 4
Allow 1 mark for numbers totalling 56 except
14/42 - repetition of observed values.
If table is blank, award 1 mark for evidence of
56.
Question 06.2 tested students’ understanding of the chi-squared test. The null hypothesis stated
‘the proportion of plants will not change from one breeding cycle to the next’. Students, therefore,
needed to maintain the proportions from breeding cycle 0 into breeding cycle 1, taking into account
that the total number of plants had changed from 54 to 56. 22.3% of students could successfully
combine this understanding of chi-squared with their maths skills, using proportions to achieve
both marks.