A level june 2019 Flashcards
Describe how a non-competitive inhibitor can reduce the rate of an
enzyme-controlled reaction.
[3 marks]
- Attaches to the enzyme at a site other than the
active site; - Changes (shape of) the active site
OR
Changes tertiary structure (of enzyme); - (So active site and substrate) no longer
complementary so less/no substrate can fit/bind;
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.
[1 mark]
(With inhibitor) increase substrate/lipid
(concentration) does not increase/affect/change
rate of reaction
OR
(With inhibitor) increase substrate/lipid
(concentration) does not increase/affect/change
lipase activity
OR
High substrate (concentration) does not overcome
inhibition
OR
High substrate (concentration) does not meet
maximum rate of reaction/lipase activity;
No large lipid droplets are visible with the optical microscope in the samples from
suspension A.
Explain why.
- Emulsification;
- (Cannot be seen) due to resolution (of optical
microscope) ;
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.
[4 marks]
- Negative correlation (between fibre eaten per day and
risk of cardiovascular disease); - Original/current fibre intake (of student) not known;
- (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
OR
If current intake between 30 and 50 (g day-1
) then
(eating 10g more results in) no significant decrease in
risk; - Correlation does not mean causation
OR
Another named factor may be involved; - Little evidence/data for higher mass of fibre per day;
- Large (2x) standard deviation at high/low mass of fibre
makes (mean) less precise
OR
Large (2x) standard deviation at high/low amounts of
fibre means there is a greater uncertainty; - No statistical test (to show if differences are significant)
Suggest one advantage of using the FFQ method and one disadvantage of using the
FFQ method compared with the alternative method.
[2 marks]
(Advantage)
1. Over longer period so more representative
OR
Diet over 24 hr may not be representative
OR
Diet may vary during the year/from day to day
OR
Person more likely to be honest on questionnaire
(rather than speaking to nurse)
OR
More cost effective because fewer
people/nurses required;
(Disadvantage)
2. Relies on (long term) memory so may not be
accurate
OR
Recall of 24 hr diet likely to be more accurate
OR
Estimation (from FFQ) may be less accurate
(than details of last 24hrs)
OR
Person may be more honest when being
interviewed;
What data would the students need to collect to calculate their index of diversity in
each habitat?
[1 mark]
(Number of species and) number of individuals in each species (in each habitat)
Give two ways the students would have ensured their index of diversity was
representative of each habitat.
[2 marks]
- Random samples;
- Large number (of samples)
OR
(Continue sampling) until stable running mean;
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.
[1 mark]
(Larger fields have relatively) More centre OR Less edge OR Less hedge OR Fewer species;
Farmers are now being encouraged to replant hedges on their land.
Suggest and explain one advantage and one disadvantage to a farmer of replanting
hedges on her farmland.
[2 marks]
Advantage -
1. Greater (bio)diversity so increase in predators of
pests
OR
Increase in predators of pests so more
yield/income/less pesticides/less damage to
crops
OR
Increase in pollinators so more yield/income
OR
May attract more tourists/subsidies to their farm
so more income (from diversification);
Disadvantage -
2. Reduced land area for crop growth/income
OR
Greater (bio)diversity so increase pest
population
OR
Increase pest population so less yield/less
income/(more) need for pesticides/(more)
damage to crops
OR
Increased (interspecific) competition so less
yield/income
OR
More difficult to farm so less income;
Use Figure 5 to explain how human mass at birth is affected by stabilising selection.
[3 marks]
- (Most likely to be) transferred to a special care
unit are those under 2800 g
OR
(Most likely to be) transferred to a special care
unit are those over 4200 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
(in the population)
OR
Alleles (for extreme mass at birth) decrease in
frequency (in the population);
The scientists studied the effect of one form, KIR2DS1, of the human KIR gene on
mass at birth.
Write the correct biological term beside each number below, that matches the space in
the passage.
[3 marks]
- Allele
- Locus/loci
- Transcribed
- Translated
- Golgi (apparatus)/Rough endoplasmic reticulum
- Tertiary
The scientists calculated a P value of 0.03 when testing their null hypothesis.
What can you conclude from this result? Explain your answer.
[3 marks]
- Probability that difference (in frequency of
births above 4500 g) is due to chance is
less than 0.05
OR
Probability that difference (in frequency of
births above 4500 g) is due to chance is
0.03; - Reject null hypothesis;
- Presence of KIR2DS1/allele does
(significantly) affect the frequency of high
birth mass;
Describe the structure of the human immunodeficiency virus (HIV).
[4 marks]
- RNA (as genetic material);
- Reverse transcriptase;
- (Protein) capsomeres/capsid;
- (Phospho)lipid (viral) envelope
OR
Envelope made of membrane; - Attachment proteins;
Use the data in Table 3 and your knowledge of the immune response to suggest why
HIV controllers do not develop symptoms of AIDS.
[3 marks]
- (All) have more T helper/CD4 cells;
- Lower viral load to infect/destroy helper T/CD4
cells; - (So more/continued) activation of B
cells/cytotoxic T cells/phagocytes; - (With B cells more/continued) production of
plasma cells/antibodies
OR
(With cytotoxic T cells more/continued) ability to
kill virus infected cells; - (More able to) destroy other
microbes/pathogens
OR
(More able to) destroy mutated/cancer cells;
Describe and explain the data in Table 4.
[2 marks]
- (Trend of) slowing growth from before birth to 21
days
OR
(Trend of) decreasing percentage undergoing
mitosis from before birth to 21 days
OR
(Trend of) decreasing percentage undergoing
DNA replication from before birth to 21 days; - DNA replication happens before mitosis
OR
Heart growth slowing until (fully) developed
OR
These cells lost the ability to divide;
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.
0 6 . 2 Describe how BrdU would be incorporated into new DNA during semi-conservative
replication.
[5 marks]
- DNA helicase;
- Breaks hydrogen bonds (between 2 DNA
strands); - BrdU complementary to adenine (on template
strand)
OR
BrdU forms hydrogen bonds with adenine (on
template strand); - DNA polymerase joins (adjacent) nucleotides (to
incorporate BrdU into the new DNA strand); - Phosphodiester bonds form (between
nucleotides);
Use your knowledge of the ELISA test to suggest and explain how the scientists
identified the cells that have BrdU in their DNA.
[3 marks]
- Add antibody (anti-BrdU with enzyme attached)
to cells/DNA
OR
Add cells/DNA to antibody (anti-BrdU with
enzyme attached); - Wash (cells/DNA) to remove excess/unattached
antibody
OR
Wash (immobilised antibody) to remove
excess/unattached cells/DNA; - Add substrate to cause colour change;
Unlike plants, Ulva lactuca does not have xylem tissue.
Suggest how Ulva lactuca is able to survive without xylem tissue.
[1 mark]
Short diffusion pathway (to cells)
OR
It has a surface permeable (to water/ions into cells);
Ulva prolifera also produces haploid, mobile single cells that can fuse to form a zygote.
Suggest and explain one reason why successful reproduction between Ulva prolifera
and Ulva lactuca does not happen.
[2 marks]
- They are different species;
- (So) if fused together they would not
produce fertile offspring
OR
(So) they have named characteristics that
means they are reproductively isolated;
Sunflowers are not xerophytic plants. The scientists repeated the experiment with
xerophytic plants.
Suggest and explain one way the leaf growth of xerophytic plants would be different
from the leaf growth of sunflowers in Figure 9.
[2 marks]
- Low/slow growth;
- Due to smaller number/area of stomata (for
gas exchange);
OR - Growth may continue at lower water
potentials; - (Due to) adaptations in enzymes involved in
photosynthesis/metabolic reactions;
Use your knowledge of gas exchange in leaves to explain why plants grown in soil
with very little water grow only slowly.
[2 marks]
- Stomata close;
- Less carbon dioxide (uptake) for less
photosynthesis/glucose production;
The data in Table 7 show differences between the oxyhaemoglobin dissociation curve
for a mouse and the oxyhaemoglobin dissociation curve for a horse.
Suggest how these differences allow the mouse to have a higher metabolic rate than
the horse.
[2 marks]
1. Mouse haemoglobin/Hb has a lower affinity for oxygen OR For the same pO2 the mouse haemoglobin/Hb is less saturated OR At oxygen concentrations found in tissue mouse haemoglobin/Hb is less saturated; 2. More oxygen can be dissociated/released/unloaded (for metabolic reactions/respiration);
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.
[3 marks]
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;
Explain five properties that make water important for organisms.
[5 marks]
- A metabolite in condensation/hydrolysis/
photosynthesis/respiration; - A solvent so (metabolic) reactions can occur
OR
A solvent so allowing transport of substances; - High heat capacity so buffers changes in
temperature; - Large latent heat of vaporisation so provides a
cooling effect (through evaporation); - Cohesion (between water molecules) so
supports columns of water (in plants); - Cohesion (between water molecules) so
produces surface tension supporting (small)
organisms;
