BIO PAPER 1 Flashcards
0 1 . 1 Give the two types of molecule from which a ribosome is made.
[1 mark]
One of RNA/ribonucleic acid(s)/nucleotide(s)/nucleic
acid(s)/rRNA/ribosomal RNA/ribosomal ribonucleic
acid
and
one of protein(s)/polypeptide(s)/amino acid(s)/
peptide(s)/ribosomal protein;
01.1 Only 22% of students could correctly name the types of molecule that make up a ribosome.
0 1 . 2 Describe the role of a ribosome in the production of a polypeptide. Do not include
transcription in your answer.
[3 marks]
- mRNA binds to ribosome;
- Idea of two codons/binding sites;
- (Allows) tRNA with anticodons to bind/associate;
4. (Catalyses) formation of peptide bond between amino acids (held by tRNA molecules);
- Moves along (mRNA to the next
codon) /translocation described;
01.2 This question was answered much more successfully than 01.1, with 48% of students
gaining all three marks.
1 . 3 Table 1 shows the base sequence of part of a pre-mRNA molecule from a
eukaryotic cell.
Complete the table with the base sequence of the DNA strand from which this
pre-mRNA was transcribed.
[1 mark]
TGCGTAATA;
Any errors = 0 marks
01.3 This question proved to be very accessible, with 96.3% of students gaining the mark.
0 1 . 4 In a eukaryotic cell, the base sequence of the mRNA might be different from the
sequence of the pre-mRNA.
Explain why.
[2 marks]
- Introns (in pre-mRNA);
- Removal of sections of (pre-mRNA)/splicing;
01.4 Although 83% of students scored 2 marks for this question, there were some incorrect
answers, e.g., indicating that the introns removed from the pre-mRNA were made of DNA.
0 2 In mammals, in the early stages of pregnancy, a developing embryo exchanges
substances with its mother via cells in the lining of the uterus. At this stage, there
is a high concentration of glycogen in cells lining the uterus.
0 2 . 1 Describe the structure of glycogen.
[2 marks]
1. Polysaccharide of α-glucose; OR polymer of α-glucose; 2. (Joined by) glycosidic bonds OR Branched structure;
02.1 The majority of students scored mark point 2 for the idea that glycogen is a branched
molecule or that it contains glycosidic bonds. Although many students knew that glycogen
is a ‘chain’ of alpha glucose, fewer could use the correct terminology to state that it is a
polysaccharide or a polymer. Students must take care to specify alpha glucose or αglucose rather than a-glucose. It was not uncommon for students to describe glycogen as
if it were identical to starch and to include references to amylose and amylopectin in their
answers
2 . 2 During early pregnancy, the glycogen in the cells lining the uterus is an important
energy source for the embryo.
Suggest how glycogen acts as a source of energy.
Do not include transport across membranes in your answer.
[2 marks]
- Hydrolysed (to glucose);
- Glucose used in respiration;
02.2 This question required a little application of biological principles, with students’ knowledge
of glycogen. Many students started their answers with reference to branching molecules
providing many ‘ends’ for enzymes – this was not relevant and is indicative of students
rushing to write all they know, rather than considering what is relevant to answer the
specific question asked. As with question 02.1, students often knew the basic principle of
breaking down glycogen to release the glucose, but did not use the appropriate term of
‘hydrolysis’ to gain the first mark point. Many students suggested that this hydrolysis of
glycosidic bonds releases energy, rather than thinking about glucose as the substrate for
respiration in all human cells.
0 2 . 3 Suggest and explain two ways the cell-surface membranes of the cells lining the
uterus may be adapted to allow rapid transport of nutrients.
[2 marks]
- Membrane folded so increased/large surface
area;
OR
Membrane has increased/large surface area
for (fast) diffusion/facilitated diffusion/active
transport/co-transport; - Large number of protein channels/carriers (in
membrane) for facilitated diffusion; - Large number of protein carriers (in membrane)
for active transport; - Large number of protein (channels/carriers in
membrane) for co-transport;
02.3 Although this question gave the context of the adaptation of the cell-surface membranes of
cells lining the uterus, making it an AO2 question, this should have been fairly
straightforward application of knowledge from 3.2.3. Many students did not gain full marks
because they seemed not to read the question sufficiently carefully. It required reference to
the cell-surface membranes only, so answers relating to “one cell thick”, “many
mitochondria”, or “a good blood supply” were not relevant. The question also required a
suggestion and an explanation to gain each mark. It was not uncommon to see reference
to “thin membranes”, indicating a clear confusion between the phospholipid bilayer
structure of a cell-surface membrane and the epithelial layer of an exchange surface.
Reference to ‘villi’ also demonstrated similar confusion and prevented a student from
obtaining mark point 1. It is important to note that, when referring to protein
carriers/channels, only carriers can be used for active transport; either can be involved in
facilitated diffusion. Also, for a membrane to be adapted for rapid transport, it must have
more of these proteins, rather than them just being present.
2 . 4 In humans, after the gametes join at fertilisation, every cell of the developing
embryo undergoes mitotic divisions before the embryo attaches to the uterus lining.
• The first cell division takes 24 hours.
• The subsequent divisions each take 8 hours.
After 3 days, the embryo has a total volume of 4.2 × 10–3 mm3
.
What is the mean volume of each cell after 3 days? Express your answer in
standard form.
Show your working.
[2 marks]
3.3 x 10–5 OR 3.28 x 10–5 OR 3.281 x 10–5
;;
1 mark for
Evidence of 128 (cells)
Correct numerical calculation but not in standard
form gains 1 mark (0.00003281 OR 0.0000328 OR
0.000033);
02.4 As with all calculations that meet the national requirement to be at the standard of higher
tier GCSE Mathematics, this required a multi-step process. Firstly, students needed to
work out that there would be 128 cells after 3 days: 2 cells after the first day, then another 6
divisions. This was where many students struggled. It was pleasing to see how many
students could correctly use standard form to express their final answer. It should be noted
that the convention for standard form to be correct is to have a number greater than 1 and
less than 10, multiplied by 10 to a power, i.e. the decimal point should be after the first
digit.
0 3 . 1 Sodium ions from salt (sodium chloride) are absorbed by cells lining the gut. Some
of these cells have membranes with a carrier protein called NHE3.
NHE3 actively transports one sodium ion into the cell in exchange for one proton
(hydrogen ion) out of the cell.
Use your knowledge of transport across cell membranes to suggest how NHE3
does this.
[3 marks]
- Co-transport;
- Uses (hydrolysis of) ATP;
- Sodium ion and proton bind to the protein;
- Protein changes shape (to move sodium ion
and/or proton across the membrane);
03.1 There were some excellent answers here, but many students did not gain marks because
their answers failed to focus on the NHE3 carrier protein. Many tried to explain why the
ions would move, often describing how movement of one ion would generate an
electrochemical gradient allowing the other to move (by facilitated diffusion), rather than
sticking with the question of how the active transport (as stated in the stem of the question)
of these two ions could work.
0 3 . 2 Scientists investigated the use of a drug called Tenapanor to reduce salt absorption
in the gut. Tenapanor inhibits the carrier protein, NHE3.
The scientists fed a diet containing a high concentration of salt to two groups of
rats, A and B.
• The rats in Group A were not given Tenapanor (0 mg kg–1).
• The rats in Group B were given 3 mg kg–1 Tenapanor.
One hour after treatment, the scientists removed the gut contents of the rats and
immediately weighed them.
Their results are shown in Table 2.
Table 2
Concentration of Tenapanor
/ mg kg–1
Mean mass of contents of the gut
/ g
0 2.0
3 4.1
The scientists carried out a statistical test to see whether the difference in the
means was significant. They calculated a P value of less than 0.05.
They concluded that Tenapanor did reduce salt absorption in the gut.
Use all the information provided and your knowledge of water potential to explain
how they reached this conclusion.
[4 marks]
1. Tenapanor/(Group)B/drug causes a significant increase; OR There is a significant difference with Tenapanor/drug/between A and B;
- There is a less than 0.05 probability that the
difference is due to chance; - (More salt in gut) reduces water potential in gut
(contents) ; - (so) less water absorbed out of gut (contents)
by osmosis
OR
Less water absorbed into cells by osmosis
OR
Water moves into the gut (contents) by
osmosis.
OR
(so) water moves out of cells by osmosis;
03.2 This question was the first test of AO3 skills in this paper and gave students their first
opportunity to interpret and evaluate scientific information. There were two parts to the
mark scheme. The first required students to explain why an increase in mean mass of
contents in the gut would allow scientists to reach the conclusion that salt absorption had
been reduced. Many students successfully achieved these two marks, although some
showed confusion between the contents in the lumen of the gut and in the cells lining the
gut wall. The second required students to interpret the statistical test; this was often not
attempted by students and those who did were rarely completely successful. Even though
the stem contained a statement to help students (The scientists carried out a statistical test
to see whether the difference in the means was significant), many referred to significant
‘results’, and very few could correctly describe what the P value showed with appropriate
use of the words ‘probability’ and ‘chance’.
0 3 . 3 High absorption of salt from the diet can result in a higher than normal
concentration of salt in the blood plasma entering capillaries. This can lead to a
build-up of tissue fluid.
Explain how.
[2 marks]
- (Higher salt) results in lower water potential of
tissue fluid; - (So) less water returns to capillary by osmosis
(at venule end);
OR - (Higher salt) results in higher blood
pressure/volume; - (So) more fluid pushed/forced out (at arteriole
end) of capillary;
03.3 This question required students to apply their knowledge of topic 3.3.4.1 and only 15%
could do this with sufficient detail to score both marks. The question was carefully worded
to lead students to the high salt concentration in the blood entering the capillaries, with the
hope that an explanation linked to decrease in water potential of the tissue fluid would be
given. Those who went down this route often scored this mark, but then referred to less
‘tissue fluid’ being reabsorbed at the venule end, rather than less water reabsorbed by
osmosis. Students are not expected to have knowledge of how salt affects the blood, but
may have some knowledge of the fact that a high salt diet can lead to an increase in blood
pressure. Any explanation of how a high salt concentration increased blood pressure was
ignored, and only the part of the answer focusing on what happens at the capillary bed to
cause a build-up of tissue fluid was marked. Consequently, many students scored one
mark for salt increasing the blood pressure (although it had to be clearly higher than
normal; statements referring to ‘high’ pressure at the arteriole end were insufficient to gain
credit). Only a few could go on to suggest that, as a result, more fluid would move out at
the arteriole end of the capillary.
4 . 1 Bacteria are often used in industry as a source of enzymes. One reason is
because bacteria divide rapidly, producing a large number of them in a short time.
Describe how bacteria divide.
[2 marks]
- Binary fission;
- Replication of (circular) DNA;
- Division of cytoplasm to produce 2 daughter
cells; - Each with single copy of (circular) DNA;
4.1 There were many excellent answers to this question, with over 50% of students achieving
both marks. The majority of students knew that bacteria divide by binary fission, but often
did not include sufficient, or specific enough, detail in their description of what happens
during this process to score the second mark. There were several statements about
mitosis and some descriptions of viral replication
0 4 . 2 Washing powders often contain enzymes from bacteria. These enzymes include
proteases that hydrolyse proteins in clothing stains.
Figure 1 shows the effect of temperature on a protease that could be used in
washing powder.
Figure 1
11
11
Turn over ►
IB/M/Jun17/E5
Do not write
outside the
box
Explain the shape of the curves at 50 °C and 60 °C.
[4 marks]
- Both denatured (by high temperature);
- Denaturation faster at 60 °C due to more
(kinetic) energy; - Breaks hydrogen/ionic bonds (between amino
acids/R groups); - Change in shape of the active site/active site no
longer complementary so fewer enzymesubstrate complexes formed/substrate does not
fit
04.2 This question required an explanation and so there were no marks for a description of the
graph. Many students gave lengthy descriptions and, even if they went on to write
creditworthy statements after this, they had wasted much time and energy. The first
requirement was for students to realise that, since the 30 ºC line stays at 100%, both lines
at 50 ºC and 60 ºC show denaturation of the enzymes. Many students got this far, but their explanations of what happens to the protein during denaturation were often incomplete or
imprecise. For example, some stated that ‘bonds’ would be broken or that ‘hydrogen, ionic,
disulfide and peptide bonds’ would be broken, so not achieving mark point 3. The mark
least often awarded was for the explanation that a temperature of 60 ºC would denature the
enzyme more quickly because of the increased (kinetic) energy at this temperature.
4 . 3 Some proteases are secreted as extracellular enzymes by bacteria.
Suggest one advantage to a bacterium of secreting an extracellular protease in its
natural environment.
Explain your answer.
[2 marks]
- To digest protein;
- (So) they can absorb amino acids for
growth/reproduction/protein synthesis/synthesis
of named cell component;
OR
(So) they can destroy a toxic substance/protein;
V04.3 / 04.4 Both of these questions required careful reading of the question stems to access the
mark points; many students got half-way through their answers, but gave insufficient detail
to score full marks. For question 04.3, many students appreciated that the extracellular
proteases would digest protein and this might protect the bacteria in some way, but their
answers were not specific enough to gain credit for this idea. Similarly, many appreciated
that extracellular protein digestion would provide useful products for the bacteria but did not
go on to state that these products would be amino acids that could be used for
growth/protein synthesis within the bacteria.
4 . 4 Mammals have some cells that produce extracellular proteases. They also have
cells with membrane-bound dipeptidases.
Describe the action of these membrane-bound dipeptidases and explain their
importance.
[2 marks]
- Hydrolyse (peptide bonds) to release amino
acids; - Amino acids can cross (cell) membrane;
OR
Dipeptides cannot cross (cell) membrane;
OR
Maintain concentration gradient of amino acids
for absorption;
OR
Ensure (nearly) maximum yield from protein
breakdown;
For question 04.4, it was pleasing to see how
many students could state that the dipeptidases would hydrolyse bonds in dipeptides, but
many were not confident what these bonds were, or what the products of this hydrolysis
would be. Many did not continue their answer to state the importance of this hydrolysis with
reference to the passage of amino acids across the cell-surface membrane into the cells for
absorption.
0 5 Scientists investigated treatment of a human bladder infection caused by a species
of bacterium. This species of bacterium is often resistant to the antibiotics currently
used for treatment.
They investigated the use of a new antibiotic to treat the bladder infection. The new
antibiotic inhibits the bacterial ATP synthase enzyme.
0 5 . 1 Place a tick () in the appropriate box next to the equation which represents the
reaction catalysed by ATP synthase.
[1 mark]
ADP + Pi —- ATP + H2O
0 5 . 2 The new antibiotic is safe to use in humans because it does not inhibit the ATP
synthase found in human cells.
Suggest why human ATP synthase is not inhibited and bacterial synthase is
inhibited.
[1 mark]
Human ATP synthase has a different tertiary
structure to bacterial ATP synthase
OR
Human ATP synthase has a different shape
active site to bacterial ATP synthase
OR
Antibiotic cannot enter human cells/mitochondria
OR
Antibiotic not complementary (to human ATP
synthase);
05.2 Some good understanding was demonstrated of how this universal enzyme could be
different in different species. Some students demonstrated understanding but their
responses were insufficiently precise to fulfil the marking criteria; they failed to reference
the ‘tertiary’ structure or the ‘shape’ of the active site. Some students successfully
suggested that the human enzyme would be found in the mitochondria and so be
inaccessible to the antibiotic.
0 5 . 3 The scientists tested the new antibiotic on mice with the same bladder infection.
They divided these mice into three groups, C, R and A.
• Group C was the control (untreated).
• Group R was treated with an antibiotic currently used against this bladder
infection.
• Group A was treated with the new antibiotic.
They removed samples from the bladder of these mice after treatment and
estimated the total number of bacteria in the bladder.
Their results are shown in Figure 2.
Figure 2
The antibiotics were given to the mice at a dose of 25 mg kg–1 per day.
Calculate how much antibiotic would be given to a 30 g mouse each day.
Show your working.
[2 marks]
0.75;;
One mark for showing 30 g = 0.03 kg;
One mark for showing 0.025 mg g05.3 67% of students scored 2 marks. Those who did not, usually could not convert 30g to
0.03kg.-1
0 5 . 4 Calculate the percentage difference in actual numbers of bacteria in group A
compared with group R. The actual number of bacteria can be calculated from the
log10 value by using the 10× function on a calculator.
Show your working.
[2 marks]
Answer in range 97.0 – 97.8%;;
OR
Answer in range 3288 – 4368%;;
4 There were two ranges of correct answer here, crediting a percentage difference compared
with either group A or group R. Ranges were calculated to allow for a tolerance of half a
2x2 mm square when reading from the graph, and for rounding at different stages. It was
pleasing to note that 46% of students could correctly complete this calculation. Some
clearly were unfamiliar with using the 10x function on their scientific calculators. 14% of
students could correctly read from the graph and convert these values into actual numbers
of bacteria, but then could not calculate the percentage difference.
0 5 . 5 The scientists suggested that people newly diagnosed with this bladder infection
should be treated with both the current antibiotic and the new antibiotic.
Explain why the scientists made this suggestion.
Use information from Figure 2 and your knowledge of evolution of antibiotic
resistance in bacteria in your answer.
[3 marks]
- (From Fig 2) New/old antibiotic does not kill
all bacteria;
OR
(From Fig 2) Some bacteria are resistant to
the new/old antibiotic; - Resistant bacteria will reproduce to produce
(more) resistant bacteria; - (Use of both) one antibiotic will kill bacteria
resistant to the other antibiotic;
OR
Unlikely that bacteria are resistant to both the
new and the old antibiotic;
OR
Use of both antibiotics (likely to) kill all/most
bacteria;
6 2,4-D is a selective herbicide that kills some species of plants but not others. 2,4-D
disrupts cell-surface membranes but the extent of disruption differs in different
species.
Scientists investigated the effect of 2,4-D on wheat plants (a crop) and on wild oat
plants (a weed).
They grew plants of both species in glasshouses. They put plants of each species
into one of two groups, W and H, which were treated as follows:
• Group W – leaves sprayed with water
• Group H – leaves sprayed with a solution of 2,4-D.
After spraying, they cut 40 discs from the leaves of plants in each group and
placed them in flasks containing 10 cm3 de-ionised water. After 5 minutes, they
calculated the disruption to cell-surface membranes by measuring the
concentration of ions released into the water from the leaf discs.
Their results are shown in Table 3.
The lowest significant difference (LSD), is the smallest difference between two
means that would be significant at P≤0.05
Table 3
Table 3 Group Treatment Mean concentration of ions in water / arbitrary units Wheat Wild oats W Water 26 45 H 2,4-D 27 70 Lowest significant difference (LSD) 7 10 0 6 . 1 Give three environmental variables that should be controlled when growing the plants before treatment with the different sprays. [2 marks]
1. Concentration of mineral ion/named mineral ion in soil; 2. Soil pH; 3. Temperature; 4. Light intensity/wavelength/duration; 5. Distance between seeds/plants; 6. Volume of water given; 7. CO2 concentration; 8. Humidity;
1 and 2. Allow ‘growing solution’ for ‘soil’. 2. pH alone is insufficient. 3. Allow ‘colour of light’ Reject ‘amount’ for mps 1, 4, 6 and 7. Ignore O2 concentration Three correct = 2 marks Two correct = 1 mark One or none correct = 0 marks
06.1 Some good understanding was demonstrated here, but students often failed to gain credit
due to poor use of language. “Amount” and “level” are not accepted units; pH and light
must be qualified, e.g. ‘soil pH’ and ‘light intensity’, and ‘nutrients’ is not an acceptable
alternative to mineral ion concentration at this level.
0 6 . 2 Evaluate the use of 2,4-D as a herbicide on a wheat crop that contains wild oats as
a weed. Use all the information provided.
[4 marks]
- 2,4-D causes an increase in release of ions
from wild oat cells and 2,4-D does not
affect/has little effect on the release of ions
from wheat cells; - (For wheat) Difference is less than LSD/7 so
difference is not significant;
OR
(For wild oats) Difference is more than LSD/10
so difference is significant; - Loss of ions from cells (likely to) lead to
cell/plant death/damage;
OR
Disruption of cell membrane (likely to) lead to
cell/plant death/damage; - No evidence here about death of plants as a
result of this ion loss; - No evidence here of other
ecological/environmental impact;
06.2 A very small number of students achieved 4 marks here. The most commonly awarded
mark was for identifying that 2,4-D increased the release of ions from wild oats but had very
little effect on wheat (mark point 1). There was quite a lot to read and understand in the
stem of the question, and some students did not appreciate that disruption of the cellsurface membranes was linked to loss of ions, and so damage to the plant itself. Some
students confused these ions from leaves with ions in the soil and consequently suggested
greater ion concentration in the water would allow more ion uptake, to be used for growth.
The lowest significant difference (LSD) was a novel context, and many students did not
read/understand the explanation, and assumed the LSD was a form of standard deviation
0 6 . 3 The scientists incubated the flasks containing the leaf discs at 26 °C and gently
shook the flasks.
Suggest one reason why the scientists ensured the temperature remained constant
and one reason why the leaf discs were shaken.
[2 marks]
Temperature
Shaken
- (Maintain temperature) so that the rate of
diffusion (of ions out of cells) remains constant
OR
(Maintain temperature) so no change in fluidity
of phospholipids/kinetic energy of
phospholipids;
OR
(Maintain temperature) so no change in
shape/structure/denaturation of membrane
proteins; - (Shaking) So all surfaces of the leaf discs are
exposed (to water)/so all submerged;
OR
To maintain diffusion/concentration gradient
(for ions out of leaf discs);
06.3 Unfortunately, many students did not fully link the context of this question to the practical
they had carried out, and so were not confident in their suggestions here. Many mistakenly
thought that the 2,4-D was in the water at this point. Those who did appreciate that the
temperature must remain constant, to avoid further changes in membrane permeability
(rather than enzyme activity), were often too vague in their responses to gain credit. Better
students often appreciated that temperature would affect the rate of diffusion of ions out of
the leaf discs. Why the leaf discs were shaken was generally better answered, although
those who showed some idea of maintaining the diffusion gradient for ions often gave
vague, imprecise answers that could not be given credit
0 7 . 1 Describe how phagocytosis of a virus leads to presentation of its antigens.
[3 marks]
- Phagosome/vesicle fuses with lysosome;
- (Virus) destroyed by lysozymes/hydrolytic
enzymes; - Peptides/antigen (from virus) are displayed on
the cell membrane;
07.1 Many students did not refer to the antigen being presented on the surface membrane of the
phagocyte, so could not be awarded mark point 3.
