a level bio questions Flashcards
1
Q
Give the two types of molecule from which a ribosome is made.
A
rna and amino acids
2
Q
Describe the role of a ribosome in the production of a polypeptide. Do not include
transcription in your answer.
A
1)Mana binds to ribosomes
2)two codons
3)tuna with anticpdpms to bind
4)catalysis formation of peptide bond between amino acids
5)moves along Mrna to next codon
3
Q
Describe the structure of glycogen.
A
Polysaccharide of α-glucose;
(Joined by) glycosidic bonds
4
Q
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.
A
Hydrolysed (to glucose);
2. Glucose used in respiration;
5
Q
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.
A
Membrane folded so increased/large surface
area;
Large number of protein channels/carriers (in
membrane) for facilitated diffusion;
3. Large number of protein carriers (in membrane)
for active transport;
4. Large number of protein (channels/carriers in
membrane) for co-transport;
6
Q
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.
A
- 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
7
Q
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.
A
(Higher salt) results in lower water potential of
tissue fluid;
2. (So) less water returns to capillary by osmosis
(at venule end);
8
Q
Describe how bacteria divide.
A
- Binary fission;
- Replication of (circular) DNA;
- Division of cytoplasm to produce 2 daughter
cells; - Each with single copy of (circular) DNA;
9
Q
Suggest one advantage to a bacterium of secreting an extracellular protease in its
natural environment.
Explain your answer.
A
- To digest protein;
- (So) they can absorb amino acids for
growth/reproduction/protein synthesis/synthesis
of named cell component;
10
Q
Describe the action of these membrane-bound dipeptidases and explain their
importance.
A
- Hydrolyse (peptide bonds) to release amino
acids; - Amino acids can cross (cell) membrane;
11
Q
Give three environmental variables that should be controlled when growing the
plants before treatment with the different sprays.
A
- Concentration of mineral ion/named mineral
ion in soil; - Soil pH;
- Temperature;
- Light intensity/wavelength/duration;
- Distance between seeds/plants;
- Volume of water given;
- CO2 concentration;
- Humidity;
12
Q
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.
A
so no change in
shape/structure/denaturation of membrane
proteins;
So all surfaces of the leaf discs are
exposed (to water)/so all submerged;
13
Q
Describe how phagocytosis of a virus leads to presentation of its antigens
A
- Phagosome/vesicle fuses with lysosome;
- (Virus) destroyed by lysozymes/hydrolytic
enzymes; - Peptides/antigen (from virus) are displayed on
the cell membrane;
14
Q
State three comparisons of genetic diversity that the scientists used in order to
generate Classification Y.
A
- The (base) sequence of DNA;
- The (base) sequence of mRNA;
- The amino acid sequence (of proteins);
15
Q
Suggest how the immune response to this viral protein can result in the
development of RA.
A
16
Q
Explain three ways in which an insect’s tracheal system is adapted for efficient gas
exchange
A
- Tracheoles have thin walls so short
diffusion distance to cells; - Highly branched/large number of
tracheoles so short diffusion distance to
cells; - Highly branched/large number of
tracheoles so large surface area (for gas
exchange);
17
Q
Explain how the presence of gills adapts the damselfly to its way of life.
A
- Damselfly larvae has high(er)
metabolic/respiratory (rate); - (So) uses more oxygen (per unit time/per
unit mass);
18
Q
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.
A
- Use random sample of seeds (from each
population); - Use (large enough) sample to be representative
of whole population; - Indication of what size was measured eg mass;
- Calculate a mean and standard deviation (for
each population); - Use the (Student’s) t-test;
- Analyse whether there is a significant difference
between (the means of) the two populations;
19
Q
Give two pieces of evidence from Figure 1 that this cell was undergoing mitosis.
Explain your answers.
A
The (individual) chromosomes are
visible because they have condensed
chromosome is made up of two
chromatids because DNA has
replicated
20
Q
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.
A
- Water moves into the cells/cytoplasm by
osmosis; - Cell/cytoplasm gets bigger;
21
Q
Suggest one way the structure of the chromosome could differ along its length to
result in the stain binding more in some areas.
A
Differences in base sequences
OR
Differences in histones/interaction with
histones
OR
Differences in condensation/(super)coiling;
22
Q
What is a homologous pair of chromosomes?
A
(Two chromosomes that) carry the same
genes;
23
Q
Give two ways in which the arrangement of prokaryotic DNA is different from the
arrangement of the human DNA in Figure 1
A
- Circular (as opposed to linear);
- Not associated with proteins/histones ;
- Only one molecule/piece of DNA
OR
present as plasmids;
24
Q
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 controll
A
- 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;
25
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.
1. The natural habitat is most favourable for bees.
2. The town is the least favourable for bees.
26
uggest and explain two ways in which the scientists could have improved the
method used for data collection in this investigation
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;
27
Formation of an enzyme-substrate complex increases the rate of reaction.
Explain how.
1. Reduces activation energy;
2. Due to bending bonds
28
The genetic code is described as degenerate.
What is meant by this?
More than one codon codes for a single amino
acid;
29
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.
30
Explain how the treatment with antivenom works and why it is essential to use passive
immunity, rather than active immunity.
Antivenom/Passive immunity) antibodies
bind to the toxin/venom/antigen and
(causes) its destruction;
2. Active immunity would be too slow/slower
31
A mixture of venoms from several snakes of the same species is used.
Suggest why.
1. May be different form of antigen/toxin
(within one species
2. Different antibodies (needed in the
antivenom)
32
During the procedure shown in Figure 8 the animals are under ongoing observation
by a vet.
Suggest one reason why.
1. (So) the animal does not suffer from the
venom/vaccine/toxin;
2. (So) the animal does not suffer
anaemia/does not suffer as a result of
blood collection;
3. (So) the animal does not have pathogen
that could be transferred to humans;
33
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
1. B cells specific to the venom reproduce by
mitosis;
2. (B cells produce) plasma cells and memory
cells;
3. The second dose produces antibodies (in
secondary immune response) in higher
concentration and quickly
34
What can you conclude about the movement of Fe3+ in barley plants?
Use all the information provided.
1. 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
35
Describe the role of two named enzymes in the process of semi-conservative
replication of DNA
1. (DNA) helicase causes breaking of hydrogen/H
bonds (between DNA strands);
2. DNA polymerase joins the (DNA) nucleotides;
3. Forming phosphodiester bonds;
36
Describe the gross structure of the human gas exchange system and how we breathe
in and out.
1. Named structures – trachea, bronchi, bronchioles,
alveoli;
2. Above structures named in correct order
OR
Above structures labelled in correct positions on a
diagram;
3. Breathing in – diaphragm contracts and external
intercostal muscles contract;
4. (Causes) volume increase and pressure decrease
in thoracic cavity (to below atmospheric, resulting
in air moving in);
5. Breathing out - Diaphragm relaxes and internal
intercostal muscles contract;
6. (Causes) volume decrease and pressure increase
in thoracic cavity (to above atmospheric, resulting
in air moving out);
37
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.
1. Both contain ester bonds (between glycerol and
fatty acid);
2. Both contain glycerol;
3. Fatty acids on both may be saturated or
unsaturated;
4. Both are insoluble in water;
5. Both contain C, H and O but phospholipids also
contain P;
6. Triglyceride has three fatty acids and
phospholipid has two fatty acids plus phosphate
group;
7. Triglycerides are hydrophobic/non-polar and
phospholipids have hydrophilic and
hydrophobic region;
8. Phospholipids form monolayer (on
surface)/micelle/bilayer (in water) but
triglycerides don’t;
38
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
1. Glucose and galactose;
2. Joined by condensation (reaction);
3. Joined by glycosidic bond;
4. Added to polypeptide in Golgi (apparatus);
39
Describe how a non-competitive inhibitor can reduce the rate of an
enzyme-controlled reaction.
1. Attaches to the enzyme at a site other than the
active site;
2. Changes (shape of) the active site
3. (So active site and substrate) no longer
complementary so less/no substrate can fit/bind;
40
The scientist concluded that pectin is a non-competitive inhibitor of the lipase enzyme.
Use Figure 1 to explain why the scientist concluded that pectin is a non-competitive
inhibitor
(With inhibitor) increase substrate/lipid
(concentration) does not increase/affect/change
rate of reaction
41
No large lipid droplets are visible with the optical microscope in the samples from
suspension A.
Explain why.
Emulsification;
2. (Cannot be seen) due to resolution (of optical
microscope);
42
A student concluded from Figure 3 that eating an extra 10 g of fibre per day would
significantly lower his risk of cardiovascular disease.
Evaluate his conclusion.
Negative correlation (between fibre eaten per day and
risk of cardiovascular disease);
2. Original/current fibre intake (of student) not known;
3. (Idea of) significance linked to (2x) standard deviation
overlap (at 10 g day-1 change);
4. If current intake between 5 and 30 (g day-1) then (eating
10g more results in a significant) decrease in risk
43
what data would the students need to collect to calculate their index of diversity in
each habitat?
Do not include apparatus used for species sampling in your answe
Number of species and) number of individuals in
each species (in each habitat)
44
Give two ways the students would have ensured their index of diversity was
representative of each habitat.
1. Random samples;
2. Large number (of samples)
45
Modern farming techniques have led to larger fields and the removal of hedges
between fields.
Use Figure 4 to suggest why biodiversity decreases when farmers use larger fields.
Less hedge
Fewer species;
46
Suggest and explain one advantage and one disadvantage to a farmer of replanting
hedges on her farmland.
1. Greater (bio)diversity so increase in predators of
pests
. Reduced land area for crop growth/income
47
Use Figure 5 to explain how human mass at birth is affected by stabilising selection.
(Most likely to be) transferred to a special care
unit are those under 2800 g
Extreme mass babies least likely to survive (to
reproduce) and so less likely to pass on their
alleles (for extreme mass at birth);
Extreme mass at birth decreases in frequency
48
KIR2DS1 is an (1) of the KIR gene, found at a (2) on
chromosome 19. KIR2DS1 is 14 021 bases long and is (3) into mRNA
that is 1101 bases long. This mRNA is then (4) into a polypeptide 304
amino acids long. The polypeptide is then modified in the organelle, (5) ,
before forming its functional (6) protein structure.
1. Allele
2. Locus/loci
3. Transcribed
4. Translated
5. Golgi (apparatus)/Rough endoplasmic reticulum
6. Tertiary;;;
49
The scientists calculated a P value of 0.03 when testing their null hypothesis.
What can you conclude from this result? Explain your answer.
Probability that difference (in frequency of
births above 4500 g) is due to chance is
less than 0.05
Reject null hypothesis;
50
Describe the structure of the human immunodeficiency virus (HIV).
1. RNA (as genetic material);
2. Reverse transcriptase;
3. (Protein) capsomeres/capsid;
4. (Phospho)lipid (viral) envelope
5. Attachment proteins;
51
The scientists determined the percentage of heart cells undergoing DNA replication
by using a chemical called BrdU. Cells use BrdU instead of nucleotides containing
thymine during DNA replication.
The scientists determined the percentage of heart cells undergoing DNA replication
by using a chemical called BrdU. Cells use BrdU instead of nucleotides containing
thymine during DNA replication.
1. DNA helicase;
2. Breaks hydrogen bonds (between 2 DNA
strands);
3. BrdU complementary to adenine (on template
strand)i98ii
52
Use your knowledge of the ELISA test to suggest and explain how the scientists
identified the cells that have BrdU in their DNA.
1. Add antibody (anti-BrdU with enzyme attached)
to cells/DNA
2. Wash (cells/DNA) to remove excess/unattached
antibody
3. Add substrate to cause colour change;
53
Unlike plants, Ulva lactuca does not have xylem tissue.
Suggest how Ulva lactuca is able to survive without xylem tissue
Short diffusion pathway (to cells)
54
Suggest and explain one reason why successful reproduction between Ulva prolifera
and Ulva lactuca does not happen.
1. They are different species;
2. (So) if fused together they would not
produce fertile offspring
55
Use your knowledge of gas exchange in leaves to explain why plants grown in soil
with very little water grow only slowly.
. Mouse haemoglobin/Hb has a lower
affinity for oxygen
More oxygen can be
dissociated/released/unloaded (for
metabolic reactions/respiration);
56
Mammals such as a mouse and a horse are able to maintain a constant body
temperature.
Use your knowledge of surface area to volume ratio to explain the higher metabolic
rate of a mouse compared to a horse.
Mouse
1. (Smaller so) larger surface area to volume
ratio;
2. More/faster heat loss (per gram/in relation
to body size);
3. (Faster rate of) respiration/metabolism
releases heat;
57
Explain five properties that make water important for organisms.
1. A metabolite in condensation/hydrolysis/
photosynthesis/respiration;
2. A solvent so (metabolic) reactions can occur
3. High heat capacity so buffers changes in
temperature;
4. Large latent heat of vaporisation so provides a
cooling effect (through evaporation);
5. Cohesion (between water molecules) so
supports columns of water (in plants);
6. Cohesion (between water molecules) so
produces surface tension supporting (small)
organisms;
58
Describe the biochemical tests you would use to confirm the presence of lipid,
non-reducing sugar and amylase in a sample.
Lipid
1. Add ethanol/alcohol then add water and
shake/mix
OR
Add ethanol/alcohol and shake/mix then
pour into/add water;
2. White/milky emulsion
OR
emulsion test turns white/milky;
Non-reducing sugar
3. Do Benedict’s test and stays blue/negative;
4. Boil with acid then neutralise with alkali;
5. Heat with Benedict’s and becomes
red/orange (precipitate);
Amylase
6. Add biuret (reagent) and becomes
purple/violet/mauve/lilac;
7. Add starch, (leave for a time), test for
reducing sugar/absence of starch;
59
Describe the chemical reactions involved in the conversion of polymers to monomers
and monomers to polymers.
Give two named examples of polymers and their associated monomers to illustrate
your answer.
1. A condensation reaction joins monomers
together and forms a (chemical) bond and
releases water;
2. A hydrolysis reaction breaks a (chemical)
bond between monomers and uses water;
3. A suitable example of polymers and the
monomers from which they are made;
4. A second suitable example of polymers and
the monomers from which they are made;
5. Reference to a correct bond within a named
polymer
60
Explain the function of this ATP hydrolase.
(ATP to ADP + Pi ) Releases energy;
2. (energy) allows ions to be moved against a
concentration gradient
OR
(energy) allows active transport of ions;
61
The movement of Na+ out of the cell allows the absorption of glucose into the
cell lining the ileum.
Explain how.
1. (Maintains/generates) a
concentration/diffusion gradient for Na+ (from
ileum into cell);
2. Na+ moving (in) by facfacilitated diffusion,
brings glucose with it
62
Describe and explain two features you would expect to find in a cell specialised for
absorption.
1. Folded membrane/microvilli so large surface
area (for absorption);
2. Large number of
co-transport/carrier/channel proteins so fast
rate (of absorption)
63
Describe how amino acids join to form a polypeptide so there is always NH2 at
one end and COOH at the other end.
1. One amine/NH2 group joins to a
carboxyl/COOH group to form a peptide
bond;
2. (So in chain) there is a free amine/NH2 group
at one end and a free carboxyl/COOH group
at the other
64
After collecting the samples, the scientist immediately heated them to
70 °C for 10 minutes.
Explain why.
1. To denature the enzymes/lipase;
2. So no further digestion/hydrolysis/catalysis
occurred;
65
Describe the role of micelles in the absorption of fats into the cells lining the ileum.
. Micelles include bile salts and fatty acids;
2. Make the fatty acids (more) soluble in water;
3. Bring/release/carry fatty acids to cell/lining (of the
ileum);
4. Maintain high(er) concentration of fatty acids to
cell/lining (of the ileum);
5. Fatty acids (absorbed) by diffusion
66
The student kept constant:
* the mass of fresh blueberries
* the volume of extraction solvent
* the time for the mixture to stand.
Name two other variables the student should have kept constant during this
investigation.
1. Temperature;
2. Agitation/mixing/stirring;
3. Source/age/type of blueberries;
4. Crushing of the blueberries;
5. Rinsing of the blueberries prior to mixing;
6. Concentration of ethanol/acid
67
A different student did this investigation. He did not have a colorimeter.
Describe a method this student could use to prepare colour standards and use them
to give data for the total anthocyanin extracted.
1. Use known concentration of blueberry
juice/extract
2. Prepare dilution series;
3. Compare (results) with colour standards to give
score/value/concentration;
68
Describe the role of DNA polymerase in the semi-conservative replication of DNA
1. Joins (adjacent DNA) nucleotides;
2. (Catalyses) condensation (reactions);
3. (Catalyses formation of) phosphodiester bonds
(between adjacent nucleotides);
69
Cyclin D stimulates the phosphorylation of DNA polymerase, which activates the
DNA polymerase.
Describe how an enzyme can be phosphorylated.
1. Attachment/association of (inorganic) phosphate
(to the enzyme);
2. (Released from) hydrolysis of ATP
70
Some tumour cells contain higher than normal concentrations of cyclin D.
Use Figure 5 to suggest why higher than normal concentrations of cyclin D could
result in a tumour.
1. Shortens interphase
2. Fast(er) cell cycle/division/multiplication/mitosis
71
Explain why death of alveolar epithelium cells reduces gas exchange in human lungs.
1. Reduced surface area;
2. Increased distance for diffusion;
3. Reduced rate of gas exchange
72
Alpha-gal is a disaccharide found in red meat.
Alpha-gal is made of two galactose molecules. Galactose has the chemical formula
C6H12O6
Give the chemical formula for the disaccharide, alpha-gal, and describe how it is
formed from two galactose molecules.
1. C12H22O11;
2. Condensation reaction
73
Suggest how one antibody can be specific to tick protein and to alpha-gal.
(Part of tick protein and alpha-gal) have a similar
shape/structure;
2. Antibody is complementary to both (tick protein
and alpha-gal)
74
Define ‘non-coding base sequences’ and describe where the non-coding multiple
repeats are positioned in the genome.
. DNA that does not code for protein/polypeptides
2. (Positioned) between genes
75
Describe how mRNA is formed by transcription in eukaryotes.
1. Hydrogen bonds (between DNA bases) break;
2. (Only) one DNA strand acts as a template;
3. (Free) RNA nucleotides align by complementary
base pairing;
4. (In RNA) Uracil base pairs with adenine (on
DNA)
7. Pre-mRNA is spliced (to form mRNA)
76
Describe how a polypeptide is formed by translation of mRNA.
1. (mRNA attaches) to ribosomes
2. (tRNA) anticodons (bind to) complementary
(mRNA) codons;
3. tRNA brings a specific amino acid;
4. Amino acids join by peptide bonds;
5. (Amino acids join together) with the use of ATP;
6. tRNA released (after amino acid joined to
polypeptide);
7. The ribosome moves along the mRNA to form
the polypeptide;
77
Define ‘gene mutation’ and explain how a gene mutation can have:
* no effect on an individual
* a positive effect on an individual
1. Change in the base/nucleotide (sequence of
chromosomes/DNA);
2. Results in the formation of new allele;
(Has no effect because)
3. Genetic code is degenerate (so amino acid
sequence may not change);
(Has positive effect because)
6. Results in change in polypeptide that positively
changes the properties (of the protein)
78
Describe the induced-fit model of enzyme action and how an enzyme acts as a
catalyst.
1. Substrate binds to the active site/enzyme
2. Active site changes shape (slightly) so it is
complementary to substrate
3. Reduces activation energy
79
Suggest and explain a procedure the scientists could have used to stop each reaction
1. Boil
2. Denatures the enzyme/ATP synthase
3. Put in ice/fridge/freezer;
4. Lower kinetic energy so no enzyme-substrate
complexes form;
5. Add high concentration of inhibitor;
6. Enzyme-substrate complexes do not form
80
Explain the change in ATP concentration with increasing inorganic phosphate
concentration.
1. (With) increasing Pi concentration, more
enzyme-substrate complexes are formed;
2. At or above 40 (mmol dm-3) all active sites
occupied
81
Explain the advantage for larger animals of having a specialised system that
facilitates oxygen uptake
1. Large(r) organisms have a small(er) surface
area:volume (ratio);
2. Overcomes long diffusion pathway
82
Explain how the counter-current principle allows efficient oxygen uptake in the
fish gas exchange system.
1. Water has low(er) oxygen partial
pressure/concentration (than air);
2. So (system on outside) gives large surface area
(in contact with water
3. Water is dense(r) (than air);
4. (So) water supports the systems/gills
1. Blood and water flow in opposite directions;
2. Diffusion/concentration gradient (maintained)
along (length of) lamella/filament;
83
Describe how one amino acid is added to a polypeptide that is being formed at a
ribosome during translation
1. tRNA brings specific amino acid (to ribosome);
2. Anticodon (on tRNA) binds to codon (on
mRNA);
3. Amino acids
84
Use information in Table 2 to suggest why this amino acid replacement changes the
properties of crystallin.
. Hydrogen bonds form instead of ionic bonds;
2. Changes the tertiary structure (of the crystallin);
85
Suggest two ways the student could improve the quality of his scientific drawing of the
blood vessels in this dissection
1. Only use single lines/do not use sketching
(lines)/ensure lines are continuous/connected;
2. Add labels/annotations/title;
3. Add magnification/scale (bar);
4. Draw all parts to same scale/relative size;
5. Do not use shading/hatching;
86
Describe two precautions the student should take when clearing away after the
dissection.
1. Carry/wash sharp instruments by holding handle
2. Disinfect instruments/surfaces;
3. Disinfect hands
87
Describe how a sample of chloroplasts could be isolated from leaves.
1. Break open cells/tissue 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;
88
Give one feature of the chloroplast that allows protein to be synthesised inside the
chloroplast and describe one difference between this feature in the chloroplast and
similar features in the rest of the cell
1. Less (thylakoid) membrane
OR
Fewer/smaller grana;
2. Smaller surface area (of membrane in
chloroplast)/less chlorophyll;
3. (Less chlorophyll so) reduced light absorption;
4. (So) slower rate of photosynthesis;
89
Use your knowledge of phagocytosis to describe how an ADC enters and kills the
tumour cell.
1. Cell ingests/engulfs the antibody/ADC
2. Lysosomes fuse with vesicle/phagosome
(containing ADC);
3. Lysozymes breakdown/digest the antibody/ADC
to release the drug;
90
Some of the antigens found on the surface of tumour cells are also found on the
surface of healthy human cells.
Use this information to explain why treatment with an ADC often causes side effects.
1. ADC will bind to non-tumour/healthy cells;
2. Cause death/damage of non-tumour/healthy
cells
91
Suggest and explain two further investigations that should be done before this ADC
is tested on human breast cancer patients.
1. Tested on other mammals to check for
safety/side effects;
2. Tested on (healthy) humans to check for
safety/side effects
92
Describe how a triglyceride molecule is formed.
1. One glycerol and three fatty acids;
2. Condensation (reactions) and removal of
three molecules of water;
3. Ester bond(s) (formed);
93
Although these two populations are completely separate and show genetic variation,
they are both called Helianthus annuus.
Explain why they are both given this name
Same species
94
Describe the structure of DNA.
1. Polymer of nucleotides;
2. Each nucleotide formed from deoxyribose, a
phosphate (group) and an organic/nitrogenous base;
3. Phosphodiester bonds (between nucleotides);
4. Double helix/2 strands held by hydrogen bonds;
5. (Hydrogen bonds/pairing) between adenine, thymine
and cytosine, guanine
95
Name and describe five ways substances can move across the cell-surface
membrane into a cell
1. (Simple) diffusion of small/non-polar molecules down a
concentration gradient;
2. Facilitated diffusion down a concentration gradient via
protein carrier/channel;
3. Osmosis of water down a water potential gradient;
4. Active transport against a concentration gradient via
protein carrier using ATP;
5. Co-transport of 2 different substances using a carrier
protein;
96
Contrast the structure of the two cells visible in the electron micrographs shown in
Figure 14.
1. Magnification (figures) show A is bigger than B;
2. A has a nucleus whereas B has free DNA;
3. A has mitochondria whereas B does not;
4. A has Golgi body/endoplasmic reticulum whereas B
does not;
5. A has no cell wall whereas B has a murein/glycoprotein
cell wall;
6. A has no capsule whereas B has a capsule;
7. A has DNA is bound to histones/proteins whereas B
has DNA not associated with histones/proteins
OR
A has linear DNA whereas B has circular DNA;
8. A has larger ribosomes;
97
Describe the structure and function of the nucleus.
1. Nuclear envelope and pores
OR
Double membrane and pores;
2. Chromosomes/chromatin
OR
DNA with histones;
3. Nucleolus/nucleoli;
Function
4. (Holds/stores) genetic information/material
for polypeptides (production)
OR
(Is) code for polypeptides;
5. DNA replication (occurs);
6. Production of mRNA/tRNA
OR
Transcription (occurs);
7. Production of rRNA/ribosomes;
98
Name the main polymer that forms the following cell walls.
Plant cell wall
Fungal cell wall
Cellulose (plants) and
Chitin (fungi);
99
Suggest one reason the scientists used biomass instead of the number of individuals
of each plant species when collecting data to measure diversity.
Individual organisms could not be
identified/separated
100
Sometimes farmers stop growing crops on an area of land to allow the natural
ecosystem to recover. The plant species index of diversity of these areas previously
used to grow crops is different from nearby land that has never been used to grow
crops.
Suggest and explain how the plant species index of diversity would be different in
these areas previously used to grow crops.
. Plant (bio)diversity is lower on (previously used)
crop land
Farming reduces (bio)diversity of fungi
101
Explain how the use of antibiotics has led to antibiotic-resistant strains of bacteria
becoming a common cause of infection acquired when in hospital.
1. (Some bacteria have) alleles
for resistance;
2. (Exposure to) antibiotics is the
selection pressure
3. More antibiotics used in
hospital (compared with
elsewhere)
102
Name another disaccharide formed from two glucose molecules.
Maltose;
103
Give two features of all prokaryotic cells that are not features of eukaryotic cells.
Prokaryotes have
No membrane-bound organelles/correct
example
OR
(Single,) circular/loop DNA (in cytoplasm)
OR
DNA free in cytoplasm
OR
DNA not associated with proteins/histones
OR
Murein/peptidoglycan (in) cell wall;
104
Tick () the box to show which type of bond maintains the helical structure of the
polypeptide
hydrogen
105
Suggest how these properties of the APs allow them to become positioned across the
membrane (as shown in Figure 3) and make a channel through which ions can pass.
1. Hydrophobic side next to/in/face fatty
acids/tails
2. Hydrophilic sides allow ion movement
through membrane
106
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.
1. Antibody binds to AP
OR
Gold (present) where AP located;
2. (As antibody/tertiary structure is)
complementary (to AP);
3. Gold interacts with electrons (in TEM);
4. (T)EM (used as it) has a high
resolution;
107
Describe viral replication.
1. Attachment proteins attach to receptors;
2. (Viral) nucleic acid enters cell;
3. Nucleic acid replicated in cell
108
Define the quaternary structure of a protein
More than 1 polypeptide;
109
Explain how two enzymes with different amino acid sequences can catalyse the
same reaction.
1. (Both) active sites have similar/identical tertiary
structures
2. (So) form enzyme-substrate complexes (with the
same substrate)
110
Describe what the scientists should place in the control tubes in this investigation.
1. Same volume of (each) buffer/pH solution;
2. Same concentration/mass of substrate (at start);
3. Same concentration/mass of denatured
enzyme;
111
Explain the rate of transpiration between 5 am and midday
1. (Rate of) transpiration/evaporation increases due
to increased temperature
. (So) increased kinetic energy
3. Stomata open (at sunrise/after 5 am) allowing
gas exchange
4. (Some) stomata close at midday/after 11 am
(reducing transpiration)
112
Describe an experiment that you could do to investigate whether the mangrove root
cells have a lower water potential than sea water
1. Record mass/length before and after;
2. Place in sea water for (specified/equal) time;
3. Method to remove surface water;
4. Increase in mass/length shows water has been
absorbed by osmosis
113
Complete Table 3 to give three differences between DNA molecules and
tRNA molecules.
1. Deoxyribose v ribose;
2. Double-stranded v single-stranded;
3. Many nucleotides v few ;
4. Thymine v uracil;
5. Linear v clover leaf (structure)
114
Describe how the scientists would remove large organelles from this suspension of
cell contents.
1. Use centrifuge/centrifugation at
slow/low/increasing (sequence of) speed(s);
2. Large/dense organelles (removed) in (first/early)
pellet
115
Other than surface area:volume ratio, describe one way this uncontrolled cell division
changes the gills, as shown in Figure 11.
Explain how this difference would affect gas exchange.
1. More cells (between water and capillary/ blood)
2. Longer diffusion pathway
116
Describe the transport of carbohydrate in plants.
1. (Red blood cells) do not have a nucleus/DNA;
2. Haemoglobin;
117
Compare and contrast the structure of starch and the structure of cellulose.
1. Both polysaccharides
OR
Both are glucose polymers
OR
Both are made of glucose monomers;
2. Both contain glycosidic bonds (between
monomers);
3. Both contain carbon, hydrogen and oxygen/C, H
and O;
4. Starch has α-glucose and cellulose has β-
glucose;
5. Starch (molecule) is helical/coiled and cellulose
(molecule) is straight;
6. Starch (molecule) is branched and cellulose is
not/unbranched;
7. Cellulose has (micro/macro) fibrils and starch
does not;
118
Describe the complete digestion of starch by a mammal
1. Hydrolysis;
2. (Of) glycosidic bonds;
3. (Starch) to maltose by amylase;
4. (Maltose) to glucose by disaccharidase/maltase;
5. Membrane-bound (disaccharidase/maltase);
