a level bio questions Flashcards
Give the two types of molecule from which a ribosome is made.
rna and amino acids
Describe the role of a ribosome in the production of a polypeptide. Do not include
transcription in your answer.
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
Describe the structure of glycogen.
Polysaccharide of α-glucose;
(Joined by) glycosidic bonds
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.
Hydrolysed (to glucose);
2. Glucose used in respiration;
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.
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;
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.
- 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
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.
(Higher salt) results in lower water potential of
tissue fluid;
2. (So) less water returns to capillary by osmosis
(at venule end);
Describe how bacteria divide.
- Binary fission;
- Replication of (circular) DNA;
- Division of cytoplasm to produce 2 daughter
cells; - Each with single copy of (circular) DNA;
Suggest one advantage to a bacterium of secreting an extracellular protease in its
natural environment.
Explain your answer.
- To digest protein;
- (So) they can absorb amino acids for
growth/reproduction/protein synthesis/synthesis
of named cell component;
Describe the action of these membrane-bound dipeptidases and explain their
importance.
- Hydrolyse (peptide bonds) to release amino
acids; - Amino acids can cross (cell) membrane;
Give three environmental variables that should be controlled when growing the
plants before treatment with the different sprays.
- 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;
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.
so no change in
shape/structure/denaturation of membrane
proteins;
So all surfaces of the leaf discs are
exposed (to water)/so all submerged;
Describe how phagocytosis of a virus leads to presentation of its antigens
- Phagosome/vesicle fuses with lysosome;
- (Virus) destroyed by lysozymes/hydrolytic
enzymes; - Peptides/antigen (from virus) are displayed on
the cell membrane;
State three comparisons of genetic diversity that the scientists used in order to
generate Classification Y.
- The (base) sequence of DNA;
- The (base) sequence of mRNA;
- The amino acid sequence (of proteins);
Suggest how the immune response to this viral protein can result in the
development of RA.
Explain three ways in which an insect’s tracheal system is adapted for efficient gas
exchange
- 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);
Explain how the presence of gills adapts the damselfly to its way of life.
- Damselfly larvae has high(er)
metabolic/respiratory (rate); - (So) uses more oxygen (per unit time/per
unit mass);
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.
- 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;
Give two pieces of evidence from Figure 1 that this cell was undergoing mitosis.
Explain your answers.
The (individual) chromosomes are
visible because they have condensed
chromosome is made up of two
chromatids because DNA has
replicated
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.
- Water moves into the cells/cytoplasm by
osmosis; - Cell/cytoplasm gets bigger;
Suggest one way the structure of the chromosome could differ along its length to
result in the stain binding more in some areas.
Differences in base sequences
OR
Differences in histones/interaction with
histones
OR
Differences in condensation/(super)coiling;
What is a homologous pair of chromosomes?
(Two chromosomes that) carry the same
genes;
Give two ways in which the arrangement of prokaryotic DNA is different from the
arrangement of the human DNA in Figure 1
- Circular (as opposed to linear);
- Not associated with proteins/histones ;
- Only one molecule/piece of DNA
OR
present as plasmids;
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
- 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;
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.
- The town is the least favourable for bees.
uggest and explain two ways in which the scientists could have improved the
method used for data collection in this investigation
- Collect at more times of the year so
more points on graph/better line (of
best fit) on graph; - Counted number of individuals in each
species so that they could calculate
index of diversity; - Collected from more sites/more years
to increase accuracy of (mean) data;
Formation of an enzyme-substrate complex increases the rate of reaction.
Explain how.
- Reduces activation energy;
- Due to bending bonds
The genetic code is described as degenerate.
What is meant by this?
More than one codon codes for a single amino
acid;
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.
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
A mixture of venoms from several snakes of the same species is used.
Suggest why.
- May be different form of antigen/toxin
(within one species - Different antibodies (needed in the
antivenom)
During the procedure shown in Figure 8 the animals are under ongoing observation
by a vet.
Suggest one reason why.
- (So) the animal does not suffer from the
venom/vaccine/toxin; - (So) the animal does not suffer
anaemia/does not suffer as a result of
blood collection; - (So) the animal does not have pathogen
that could be transferred to humans;
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
- B cells specific to the venom reproduce by
mitosis; - (B cells produce) plasma cells and memory
cells; - The second dose produces antibodies (in
secondary immune response) in higher
concentration and quickly
What can you conclude about the movement of Fe3+ in barley plants?
Use all the information provided.
- Heat treatment has a greater effect on young
leaves than old; - Heat treatment damages the phloem;
- Fe3+ moves up the leaf/plant;
- (Suggests) Fe3+ is transported in the xylem in
older leaf; - In young leaf, some in xylem, as some still
reaches top part of leaf; - (Suggests) Fe3+ is (mostly) transported in phloem
in young leaf
Describe the role of two named enzymes in the process of semi-conservative
replication of DNA
- (DNA) helicase causes breaking of hydrogen/H
bonds (between DNA strands); - DNA polymerase joins the (DNA) nucleotides;
- Forming phosphodiester bonds;
Describe the gross structure of the human gas exchange system and how we breathe
in and out.
- Named structures – trachea, bronchi, bronchioles,
alveoli; - Above structures named in correct order
OR
Above structures labelled in correct positions on a
diagram; - Breathing in – diaphragm contracts and external
intercostal muscles contract; - (Causes) volume increase and pressure decrease
in thoracic cavity (to below atmospheric, resulting
in air moving in); - Breathing out - Diaphragm relaxes and internal
intercostal muscles contract; - (Causes) volume decrease and pressure increase
in thoracic cavity (to above atmospheric, resulting
in air moving out);
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.
- Both contain ester bonds (between glycerol and
fatty acid); - Both contain glycerol;
- Fatty acids on both may be saturated or
unsaturated; - Both are insoluble in water;
- Both contain C, H and O but phospholipids also
contain P; - Triglyceride has three fatty acids and
phospholipid has two fatty acids plus phosphate
group; - Triglycerides are hydrophobic/non-polar and
phospholipids have hydrophilic and
hydrophobic region; - Phospholipids form monolayer (on
surface)/micelle/bilayer (in water) but
triglycerides don’t;
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
- Glucose and galactose;
- Joined by condensation (reaction);
- Joined by glycosidic bond;
- Added to polypeptide in Golgi (apparatus);
Describe how a non-competitive inhibitor can reduce the rate of an
enzyme-controlled reaction.
- Attaches to the enzyme at a site other than the
active site; - Changes (shape of) the active site
- (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
(With inhibitor) increase substrate/lipid
(concentration) does not increase/affect/change
rate of reaction
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);
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
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)
Give two ways the students would have ensured their index of diversity was
representative of each habitat.
- Random samples;
- Large number (of samples)
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;
Suggest and explain one advantage and one disadvantage to a farmer of replanting
hedges on her farmland.
- Greater (bio)diversity so increase in predators of
pests
. Reduced land area for crop growth/income
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