Paper 1 Questions I got wrong Flashcards
1
Q
Start of 2019:
A
2
Q
What’s equation to find magnification, image length or actual length?
What are units conversions?
A
M= I/A
I= MA
A=I/M
m–> cm x 100
cm–> mm x 10
mm –> um/micrometer
3
Q
Specific question:
What’s the uncertainty of calculated maximum length? (1 mark)
A
+- 2 micrometer
4
Q
Why are there no large lipid drops are visible under optical microscope in samples of suspension A.
Suspension A includes Bile salts, lipids, no pectin(non-competitive inhibitor), no lipase.( 2 marks)
A
1) Bile salts emulsify lipids
2) Increase surface area to volume ration
3) small lipid droplets
4) Low resolution (wavelength of light too long )
5
Q
How does a non-competitive inhibitor do reduce rate of reaction?( 3 marks)
A
-Non-competitive inhibitor binds to allosteric site
changes tertiary structure of enzyme/ changes shape of active site
-Active site is no longer complementary to substrate so substrate can’t bind /fit
-no/less enzyme-substrate complexes form
6
Q
Evaluate question:
A student concluded from Figure 3 that eating an extra 10 g of fibre per day would
significantly lower his risk of cardiovascular disease.(4 marks)
Evaluate his conclusion.
2019paper 1
A
For:
- Negative correlation (between fibre eaten per day and
risk of cardiovascular disease)
-.(Idea of) significance linked to (2x) standard deviation
overlap (at 10 g day-1 change)
- If current intake between 5 and 30 (g day-1) then (eating
10g more results in a significant) decrease in risk
Against:
-don’t know initial mass of fiber in diet
-no stats test to show significant decrease
-if current intake is between 30-50 g , the adding 10g would result in NO significant decrease in risk of cardiovascular disease
-correlation doesn’t equal causation
-large standard deviations at high or low fibre diet shows less precise data.
-Little evidence/data for higher mass of fibre per day;
7
Q
Food frequency questionaire(FFQ) (over year)Positives and negatives compared to nurse asking detailed question about diet for last 2 hrs
A
Pros:
-long term so representative
-diet may vary sayto day /over the year
-More cost effective because fewer
people/nurses required;
Cons:
Recall of 24 hr diet likely to be more accurate
Person may be more honest when being
interviewed;
8
Q
Why does biodiversity decrease in larger fields (figure 4 paper 1 2019)
A
-more centre
-less edge
-less hedge
-fewer species - less habitat-less niche
9
Q
Pros of farmer replanting hesges on farmland
A
- Greater (bio)diversity so increase in predators of
pests
-Increase in predators of pests so more
yield/income/less pesticides/less damage to
crops
-Increase in pollinators so more yield/income
-May attract more tourists/subsidies to their farm
so more income (from diversification);
10
Q
Cons of farmer replanting hedges
A
- Reduced land area for crop growth/income
-Greater (bio)diversity so increase pest
population so less yield/income/profit
-Increased (interspecific) competition so less
yield/income
-more difficult to farm(can’t use bigger machinery) so less income
11
Q
Where is a specific allele of a gene found on a chromosome
A
Gene locus/loci
12
Q
Fill in the blank.
mRNA is then —- into a polypeptide 304 amino acids long
A
Transcribed NOT transcripted
13
Q
Polypetide is modified in which organelle
A
Golgi apparatus
RER
14
Q
If no bonds between quaternary structure is mentioned. what’s the structure?
A
Tertiary
15
Q
What stats test test the null hypothesis:
‘The presence of KIR2DS1 in the mother’s genome does not affect the frequency of
births above 4500 g plus data in table 2 Paper 1 2019
A
Chi squared
16
Q
what’s the opposite statement to this null hypothesis:
‘The presence of KIR2DS1 in the mother’s genome does not affect the frequency of
births above 4500 g’
A
Presence of KIR2DS1/allele does
(significantly) affect the frequency of high
birth mass;
17
Q
What does a virus/viral load do?
A
infect/destroy named T helper cell
18
Q
Percentage of heart cells undergoing mitosis 0.6% vs percentage of heart cells undergoing DNA replication 0%. Explain why(1)
A
DNA replication happens before mitosis
OR
Heart growth slowing until (fully) developed
OR
These cells lost the ability to divide;
19
Q
Ulva lactuca is an alga that lives on rocks on the seashore. It is regularly covered by
seawater.
Unlike plants, Ulva lactuca does not have xylem tissue.Why?(1)
A
Short diffusion pathway (to cells)
20
Q
Circular diagrams for mitosis or meiosis. What’s the pattern for meiosis and mitosis.(2)
A
Meosis:
Diploid –> haploid
Mitosis:
Haploid –> haploid
Diploid –> diploid
Haploid cells - diploid zygote
21
Q
Why doesn’t successful reproduction between Ulva prolifera
and Ulva lactuca happen?(2)
A
- They are different species;
- (So) if fused together they would not
produce fertile offspring so are reproductive isolated - seperate gene pools
22
Q
How do you find the volume of a 1moldm^-3 of sodium chloride solution and volume of distilled water (cm^3) using a known conc and volume of solution you want to achieve?(2)
A
conc x total volume of solution = volume of NaCl solution
Volume of distilled water = total- volume of NaCl.
23
Q
How does the leaf growth of xerophytic plants compared to sunflowers?(2)
A
-less (open) stomata/lower stomatal density
-slower growth
24
Q
Why do plants grown in soil of very little water grow slowly?( 2)
A
-less open stomata/more closed stomata
-less CO2 take up
-lower rate of photosynthesis
25
How the oxygen dissociation curve of a mouse compare to that of a horse to cause mice to have a higher metabolic rate than horses.(2 marks)
-Hemoglobin in mice have lower affinity of oxygen/oxygen less readily associates at same pO2
-Oxygen more readily dissociates (higher rat of respiration more metabolic rate)
26
Why do mice have a higher metabolic rate than horses. Mice and horses must maintain constant body temperature. Use knowledge of surface area to volume ratio.
-mice have a higher surface area to volume ratio than horses
-higher rate of heat loss in mice
-higher rate of respiration releasing more heat /replacing heat lost.
27
Explain how water is cohesive(2)
-Cohesion (between water molecules) so
supports continuous columns of water (in plants)
-Cohesion (between water molecules) so
produces surface tension supporting (small)
organisms;
28
Test for AMYLASE enzyme.(2)
1) Add Biuret reagent
2) colour changes from, blue to mauve/lilac is a positive test for proteins.
AND
Add starch, (leave for a time), test for
reducing sugar/absence of starch using Potassium iodide solution and well test.
29
Definition of a condensation reaction:(1)
A condensation reaction joins monomers
together and forms a NAMED (chemical) bond and
releases water;
30
Definition of hydrolysis reaction.(1)
A hydrolysis reaction breaks a NAMED (chemical)
bond between monomers and uses water;
31
End of 2019 paper 1
32
Start of 2022 paper 1
33
Structure of nucleus(2)
Nuclear envelope and pores
OR
Double membrane and pores;
2. Chromosomes/chromatin
OR
DNA with histones;
3. Nucleolus/nucleoli;
34
Function of nucleotide(2)
1.(Holds/stores) genetic information/material
for polypeptides (production)
OR
(Is) code for polypeptides;
DNA replication (occurs);
2.Production of mRNA/tRNA
OR
Transcription (occurs);
3. Production of rRNA/ribosomes;
35
Suggest one reason the scientists used biomass instead of the number of individuals
of each plant species when collecting data to measure diversity.(1 mark)
-Too time consuming
-Too small/numerous to count individuals
--individual organisms could not be
identified/separated
36
What's the the effect of farming on biodiversity?
-Plant (bio)diversity is lower on (previously used)
crop land
- Farming reduces (bio)diversity of fungi
37
Explain how the use of antibiotics has led to antibiotic-resistant strains of bacteria
becoming a common cause of infection acquired when in hospital.(3 marks)
-Some bacteria have alleles
for resistance
-2. (Exposure to) antibiotics is the
selection pressure
-More antibiotics used in
hospital (compared with
elsewhere)
-Patients have weakened
immune systems
-high frequency of
resistance allele (in bacterial
population)
38
What do you sterilize instead of of inoculating loop.
-Sterlise pipette
-sterilise gas syringe
-sterlise spreader
-sterlise neck of bottle using flame after dipping in alcohol
to kill microorganism
39
How to transfer bacteria to agar plate?(2)
-Lift lid of (agar) plate at an
angle
-for minimal time to prevent contamination
40
Use Figure 2 to evaluate whether more trehalose in the diet could be a factor in the
increased number of antibiotic-resistant C. difficile infections.(3marks)
For
1. Resistant bacteria grow faster with trehalose;
2. Resistant bacteria (likely to) increase in
frequency in the population/people;
3. Resistant bacteria (likely to) outcompete
non-resistant bacteria
Against:
-In laboratory not in people
-Other disaccharides (in the diet) might affect
bacteria
-Other bacterial species (in the body) might
affect bacteria
-No stats test to see if difference/increase is
significant
-No data for both resistant and non-resistant
bacteria growing together
-No data for different concentrations of trehalose
41
What's 2 features of prokaryotic cells that aren't in eukaryotic cell?
(3 marks)
-circular DNA (reject plasmid)
-Murein and peptidoglycan in cell wall
-DNA not associated with proteins/histones
42
Assess why the APs do not damage the eukaryotic cells of the organisms that
produce them.
Prokaryotic cell membranes do not contain cholesterol.( 2 marks)
- Cholesterol stabilises,makes it less flexible (the membrane)
-(So) APs do not make channels in
(eukaryotic) membranes
43
They stained the APs using a monoclonal antibody with gold attached
to it. TEM
-Antibody binds to AP
-As antibody is
complementary to AP
-Gold interacts with electrons (in TEM)
-TEM has a high
resolution;
44
How does a non-dysfunctional mutation occurs in a chromosome?
the sister chromatids separate unevenly (3 in one an 1 in another cell)
45
What should scientist put in the control tubes Q5 2022 Paper 1
1. Same volume of (each) buffer/pH solution;
2. Same concentration/mass of substrate (at start);
3. Same concentration/mass of denatured
enzyme;
46
Give three conclusions you can make from Figure 8.
From pH7.5 and 8.4. Enzyme P and Q.
2022 paper 1
1. Both/P and Q (are) active at pH 8.4
2. P is (equally/most) active at both pHs
3. All reactions reach same end (point)
4.Q is denatured/not active at pH 7.5
47
Explain the rate of transpiration between 5:00am to midday.
-More stomata open (at sunrise/after 5 am) allowing
carbon dioxide to enter;
-Rate of transpiration/evaporation increases due
to increased temperature and light intensity
- (So) increased kinetic energy (causing more
water loss
-Some)stomata close at midday/after 11 am
(reducing transpiration
48
Describe an experiment that you could do to investigate whether the mangrove root
cells have a lower water potential than sea water.
You are given:
* a piece of fresh mangrove root
* sea water
* access to laboratory equipment.
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.
49
3 Differences in DNA and tRNA molecules.
DNA vs tRNA:
Deoxyribose v ribose
Thymine v uracil
Double-stranded v single-stranded
Linear v clover leaf
Does not bind to amino acid v does bind to
amino acid
-No exposed bases v anticodon;
50
Describe how the scientists would remove large organelles from this suspension of
cell contents after cell is homogenised
Use centrifuge/centrifugation at
slow/low/increasing (sequence of) speed(s)
-Large/dense organelles (removed) in (first/early)
pellet
51
Position of bands in ribosome in tube A and tube B Fig 10 paper 1 2022
Tube A:
Ribosomes bound to) rough endoplasmic
reticulum so (Are) denser/heavier so move further
Tube B:Only free ribosomes because)
membrane/phospholipids/endoplasmic reticulum
dissolved (by detergent); hence 1 band.
52
How does uncontrolled cell devision change the gills? (3 marks)
Thicker lamellae as there are more cells
Longer diffusion distance
- smaller surface area for gas exchange.
-Slower rate of gas exchange
53
Compare fish vs mammal circulation system.
-single circulation vs double circulation
-2 chambers vs 4 chamber
-1 atrium vs 2 atria
-1 ventricle vs 2 ventricle
Heart contains
deoxygenated
blood vs Heart contains
oxygenated
and
deoxygenated
blood;
-one vein vs 2 veins
-Blood reaching
body capillaries
at low(er)
pressure vs Blood reaching
body capillaries
at high(er)
pressure;
54
Transport of carbohydrate in plants:
09.1
1. Sucrose actively transported into phloem (cell);
OR
Sucrose is co-transported/moved with H+ into
phloem (cell);
2. (By) companion/transfer cells;
3. Lowers water potential (in phloem) and water
enters (from xylem) by osmosis;
4. ((Produces) high(er) (hydrostatic) pressure;
OR
(Produces hydrostatic) pressure gradient;
5. Mass flow to respiring cells
OR
Mass flow to storage tissue/organ;
6. Unloaded/removed (from phloem) by active
transport;
55
End of 2022 paper 1
56
Start of 2020 paper 1
57
What are the adaptations of a cell specialised for absorption?
-High number of carrier and channel proteins for a high rate of facilitated diffusion
-high number of mitochondria to produce more ATP in RESPIRATION for ATP hydrolysis to release more energy for a faster rate of active transport moving a substance against concentration gradient.
-membrane bound digestive enzymes maintain steep concentration gradient - higher rate of facilitated diffusion
-Folded membrane of microvilli increases surface area for absorption, simple diffusion ,Facilliated diffusion
58
Describe how amino acids join to form polypetides with NH2 on one end and COOH on the other end? (2)
Draw diagram
-amine group and carboxyl group of 2 different amino acids joining by peptide bond
- free NH2 group
at one end and a free COOH group
at the other
- amino acids are always orientated in the same direction in a chain.
59
Use knowledge of lipid digestion to explain differences in sample A and sample B in table 1 shown. 2020 paper 1 Q2.
-Triglycerides decrease because of lipase activity
-triglycerides are hydrolysed by lipase
-lipase hydrolyzing ester bonds between glycerol and fatty acids in triglycerides causing fatty acids concentration to increase.
60
Describe the role of micelles in the absorption of fats into the cells lining the ileum. (3 marks)
1. Micelles include bile salts and fatty acids.
2. Make the fatty acids (more) soluble in water;
3. Carry fatty acids to cell lining ileum
4. Fatty acids absorbed by simple diffusion diffusion.
5.This Maintain higher concentration of fatty acids to
cell lining ileum.
61
Which valve is between atrium and ventricle
Left or right atrioventricular valve.
62
Which valve is between ventricle and aorta
Left or right semilunar valve.
63
At P on Figure 3,(2020 paper 1 ) the pressure in the left ventricle is increasing. At this time, the rate
of blood flow has not yet started to increase in the aorta. use fig 3
- Semi-lunar valves is closed;
-Because pressure in aorta HIGHER than in
ventricle.
64
Explain how and why there is a small increase in pressure AND in rate of blood flow in the aorta Use fig 3 2020 paper1.
-Elastic recoils
-smooths the blood flow
65
Describe 1 similarity and one difference between pressures of left and right ventricle (fig 3 2020)
Similarities:
-same trend/pattern of increasing and decrease peaks
-peak at same/similar time.
Difference:
Pressure in left ventricle is higher than pressure in right ventricle.
66
How to find beats per minute using fig 3 and points Q (2020 )
Find time between points Q and divide by 60 seconds ( 1 minute).
67
Use knowledge of membrane structure to explain why (ethanol water and acid have a higher absorbance than ethanol and water which have a higher absorbance than just water which has a higher absorbance than 0. 2020 Q4.3
Blueberry practical and anthocyanin pigment.
General trend :1. Higher absorbance indicates more anthocyanin
as more membrane damaged increasing membrane permeability.
OR
-G/water only not zero because) some membrane damage
when blueberries are crushed;
2. More membrane damage/permeability results in
more anthocyanin release
3. (E(with acid) and F (without acid) greater than water because)
phospholipids dissolve in ethanol;
4. (E greater than F because) acid denatures
membrane proteins;
68
Control variables student should keep constant in blueberry and pigment practical.
1.Temperature of extraction solvent.
2. Concentration of ethanol/acid;
3.age/type of blueberries
4. Mixing
5. Degree of Crushing of the blueberries;
6. Rinsing of the blueberries prior to mixing;
69
How to produce colour standards for blueberry practical with no colorimeter.
-serial dilution/prepare dilution series
-Use known concentration of
anthocyanin pigment
-compare results with colour standards to give score/value.
70
How is an enzyme phosphorylated?(2)
-Hydrolysis of ATP to produce ADP and and Pi
-Attachment/association of Pi to enzyme to changes the enzyme's tertiary structure so it's more reactive (make it's active site more complementary to substrates increasing n.o of enzyme-substrate complexes produced per second)
71
Some tumour cells contain higher than normal concentrations of cyclin D. (2)
Use Figure 5 to suggest why higher than normal concentrations of cyclin D could
result in a tumour. 2020 question 5.
- DNA replication occurs earlier and shorterns interphase
-faster and uncotrolled mitosis and cell division
-produces a mass of abnormal cells-tumour forms.
72
How can one antibody be complementary to both a tick protein and alpha gal.(2)
- (part of) tick protein and alpha gal have similar tertiary structures
-both complementary to specifically shaped antigen -binding site of antibody.
73
Define ‘non-coding base sequences’ and describe where the non-coding multiple
repeats are positioned in the genome.(2)
-They are introns that code for a specific sequence of DNA nucleotide bases that code don't code for a polypeptide
-positioned between genes.
74
The scientists studied five individuals from each species. Within the five individuals of
species T they found a percentage similarity of 66%. (2)
Use Table 3 to evaluate how this information affects the validity of the
phylogenetic tree.
T and C = 59.7% similarity
T and L =53.7% similarity
T and R=36.6% similarity.
1. FOR - more similar than with any other
species;
2. Against high intraspecific variation in
species T (compared with variation between T
and C);
3. Small sample size of 5 people so not representative.
75
If a graph looks like a line graph with with a line of best fit and no mean what is the stats test to be used?(1)
Correlation coefficient
76
A journalist saw Figure 10 (2020) and suggested that future increases in atmospheric
carbon dioxide concentration could result in less transpiration. (4)
Evaluate his suggestion.
1. Increasing carbon dioxide concentration shows
decreased stomatal density;
2. Fewer stomata means less rate of transpiration
Against:
3. less stomata can absorb same volume of CO2.
4. Don’t know the size of the stomata;
5. Don’t know whether leaf size has changed;
6. Don’t know if this is true for all species (of plant);
7. Don’t know how long the stomata are open for;
8. Don’t know if this trend will continue beyond the
concentrations of carbon dioxide shown in Figure
10- extrapolation
9. Other factors affect transpiration (rate) like light intensity, humidity and temperature.
77
Define ‘gene mutation’ and explain how a gene mutation can have(4 marks)
1. Change in the DNA nucleotide base sequence
2.New allele forms
No effect:
Degenerate is genetic code so different sequence of DNA base triplet can form the same sequence of amino acid and the tertiary structure.
OR Does change amino acid but no effect on tertiary
structure;
5. New allele is recessive so does not influence
phenotype
Positive effect:
6. Results in change in polypeptide that positively
changes the properties of the protein
7. May result in increased reproductive success
OR
May result in increased survival (chances) -advantageous allele passed on - topic 4.
78
End of 2020 paper 1
79
Start of 2021 paper 1
80
Describe the induced-fit model of enzyme action and how an enzyme acts as a
catalyst.(3 marks)
1. Substrate binds to the active site/enzyme
OR
Enzyme-substrate complex forms;
2. Active site changes shape (slightly) so it is
complementary to substrate by distorting and straining bonds in substrate.
3. Reduces activation energy;
81
How to stop reaction of ATP being made (2 marks)
1st way:
Boil to Denatures the ATP synthase;
2nd way:
3. Put in ice/fridge/freezer;
4. Lower kinetic energy so no enzyme-substrate
complexes form
3rd way:
- Add high concentration of inhibitor;
-Enzyme-substrate complexes do not form;
82
Explain the change in ATP concentration with increasing inorganic phosphate
concentration. ( increases and plataeus) fig 1 2021
1. (With) increasing Pi concentration, more
enzyme-substrate complexes are formed;
2. At or above 40 (mmol dm-3) all active sites
occupied and ATP synthase concnetration is the limiting factor.
83
Suggest how the environmental conditions have resulted in adaptations of systems
using Model A rather than Model B.(2)
fig 2 2021 A outward fish gill and B inward fish gill
- Water has low(er) oxygen partial
pressure/concentration (than air);
-so (system on outside) gives large surface area
(in contact with water)
84
figure 3 2021
A student studied Figure 3 and concluded that the fish gas exchange system is more
efficient than the human gas exchange system.
1.. In fish, blood leaving (V) has more oxygen than
water leaving (E);
2. (But) in humans, blood leaving (V) has less
oxygen than air leaving (E);
3. Difference in oxygen (concentration) between
artery and vein is greater in fish than in humans;
4. (So) fish remove a greater proportion from the
oxygen they take in;
85
Give two evidences for figure 3 of a cell undergoing mitosis.
Start of 2024
-chromosomes condense and supercoil around histone proteins
-The chromosomes are not arranged in
homologous pairs, which they would be
if it was meiosis SO MITOSIS
-each chromosomes is made of 2 chromatids because DNA is replicated
86
How do chromosomes move apart 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 (1)
1. Water moves into the cells/cytoplasm by
osmosis;
2. Cell/cytoplasm gets bigger
87
Suggest one way the structure of the chromosome could differ along its length to
result in the stain binding more in some areas. (1)
Differences in histones/interaction with
histones,
Differences in condensation/(super)coiling
88
What is a homologous pair of chromosomes?
(Two chromosomes that) carry the same
genes, same gene loci but different alleles
89
Give two ways in which the arrangement of prokaryotic DNA is different from the
arrangement of the human DNA in Figure 1.2018
1. Circular (as opposed to linear);
2. Not associated with proteins/histones ;
3. Only one molecule/piece of DNA
90
Describe the method the student would have used to obtain the results in Figure 3. 2018
Start after all of the cubes of potato have been cut. Also consider variables he should
have controlled.
1.Method to ensure all cut surfaces of the eight
cubes are exposed/FULLY SUBMERGE to the sucrose solution;
2. Controlling temperature using waterbath
3. Drying cubes before measuring by blotting paper
4. Measure mass of cubes at stated time intervals-10 MINS;
91
Species richness meaning
(A measure of) the number of (different) species in
a community/habitat/ecosystem at one area at one time.
92
From the data in Figure 4, a student made the following conclusions.
1. The natural habitat is most favorable for bees.
Do the data in Figure 4 support these conclusions? Explain your answer.
[2018
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;
93
From the data in Figure 4, a student made the following conclusions.
2. The town is the least favourable for bees.
Do the data in Figure 4 support these conclusions? Explain your answer
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
94
The scientists collected bees using a method that was ethical. Suggest one consideration a scientist take into account to make sure method was ethical. (1)
-Must not harm the bees
-Must allow the bee to be released
unchanged /doesn't increase chance of predation
95
The scientists collected bees using a method that allowed them to
identify accurately the bee species to which each belonged.
2. Must allow close examination
OR
Use a key (to identify the species)2. Accept ‘use
photographs/specimens (to identify
species)’
96
Suggest and explain two ways in which the scientists could have improved the
method used for data collection in this investigation.2018
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
97
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. 2018
1. (Both) negatively charged to positively charged
change in amino acid;
2. Change at amino acid 300 does not change the shape of the active site or the tertiary structure
3. Amino acid 279 may have been involved in a
(ionic, disulfide or hydrogen) bond and so the
shape of the active site changes which changes tertiary structure.
98
Explain how the treatment with antivenom works and why it is essential to use passive
immunity, rather than active immunity.(2)
Antivenom antibodies
bind to the venom and
DESTROY it. (not neutralize it)
2. Active immunity would be too slow/slower;
99
A mixture of venoms from several snakes of the same species is used to make vaccine. Suggest why?(1)
May be different form of antigen/toxin to produce different anitbodies to venom
100
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.
1. Horses because more
ANTIVENOM/ANTIBODIES could be collected
(as more blood collected);
2. 4550 (cm3) v 26 (cm3) (blood collected);
101
During the procedure shown in Figure 8 the animals are under ongoing observation
by a vet.
Suggest one reason why.
So the animals don't lose too much blood so they don't suffer from anemia
-don't suffer from antigen/venom
-to prevent the animal having a pathogen
that could be transferred to humans;
102
Why did heat Treatment not affect xylem vessel (2018- USE TABLE)(2)
-SD overlap
-so no significant difference
-Passive movement of water in xylem so not affected by heat-treatment
103
The scientists concluded that this heat treatment damaged the phloem.
Explain how the results in Figure 9 support this conclusion.(Q8-2018 USE)
1. The radioactively labelled carbon is converted
into sugar/sucrose/organic compound.
2. Mass flow/translocation in the phloem
throughout the plant only in plants that were
untreated/B/control
104
What can you conclude about the movement of Fe3+ in barley plants?
Use all the information provided. Question 8 2018 USE
TREND SHOWN AND WHY.(1/2 OF IT) (3 marks total of question)
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. Fe3+ is transported in the xylem in older leaf
WHEREAS
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.
105
What can you conclude about the movement of Fe3+ in barley plants?
Use all the information provided. LIMITATIONS(1 MARK) (1/2) 2018
8. All ratios show there is less Fe3+ in the top than
the lower part of leaves
9. (But) no statistical test to show if the difference(s)
is significant;
106
Suggest explanations for the results in Table 5.( table 5 2018) Question 9 (CDEF Treatment and DNA replication)
1. Treatment D Antibody binds to cyclin A so it
cannot bind to DNA/enzyme so DNA replication doesn't start (11% of cells undergoing DNA replication)
2. (Treatment E) RNA interferes with
MRNA,TRNA ribosome/POLYPEPTIDEformation
(so cyclin A not made)
3. In Treatment F added cyclin A can bind to
DNA/enzyme (to initiate DNA replication)D
107
Similairites of triglyercides and fatty acids
1. Ester bonds in both(between glycerol and fatty acids)
2. Glycerol in both
3. Unsaturated and saturated fatty acids
4. Insoluble in water (BOTH)
5. Both contain C, H and O
108
Differences between trigleyricdes and fatty acids
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 whereas
phospholipids have hydrophilic and
hydrophobic region;
8. Phospholipids form monolayer (on
surface)/micelle/bilayer (in water) but
triglycerides don’t;
