paper 1 exam Qs Flashcards
Describe the induced-fit model of enzyme action and how an enzyme acts as a catalyst. [3 marks]
- substrate binds to active site of enzyme to form an enzyme-substrate complex
- active site shape is altered, so it is complementary to the substrate
- lowers activation energy
adenosine diphosphate
- boil
- denatures ATP synthase
Explain the change in ATP concentration with increasing inorganic phosphate concentration. [2 marks]
- with increasing inorganic phosphate concentration, more enzyme-substrate complexes are formed
- above 40mmoldm-3, all active sites are occupied
Explain the advantage for larger animals of having a specialised system that facilitates oxygen uptake. [2 marks]
- larger organisms have a smaller surface area to volume ratio
- allows faster diffusion
- water has a lower oxygen concentration than air
- so, the system on the outside provides a larger surface area that is in contact with water
Explain how the counter-current principle allows efficient oxygen uptake in the fish gas exchange system. [2 marks]
- blood and water flow in opposite directions
- maintains concentration gradient across the whole length of the lamella
Describe how one amino acid is added to a polypeptide that is being formed at a ribosome during translation. [3 marks]
- tRNA brings specific amino acid to ribosome
- anticodon on tRNA binds to codon on mRNA
- amino acids join by peptide bonds, formed via condensation reactions, using ATP
Suggest two ways the student could improve the quality of his scientific drawing of the blood vessels in this dissection. [2 marks]
two of:
1. only use single lines
2. add labels
3. add magnification
4. draw all parts to the same scale
5. do not shade
Identify the type of blood vessel labelled as X and the type of blood vessel labelled as Y in Figure 4.
Describe one feature that allowed you to identify the blood vessels. [2 marks]
X – artery
Y – vein
difference in lumen size OR difference in wall thickness
Describe two precautions to be taken when clearing away after a dissection. [2 marks]
- carry sharp instruments by holding handle
- disinfect instruments/surfaces
- disinfect hands
- put organ in a separate bin to dispose
Describe how a sample of chloroplasts could be isolated from leaves. [4 marks]
- homogenise to break open cells and filter to remove debris
- in cold, isotonic, buffered solution
- centrifuge and remove nuclei
- centrifuge at higher speed, chloroplasts settle out
Some proteins found inside the chloroplast are synthesised inside the chloroplast.
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. [2 marks]
- DNA
- chloroplast DNA is circular, but nuclear DNA is linear
OR chloroplast DNA is shorter than nuclear DNA
OR chloroplast DNA is not associated with histones, but nuclear DNA is
OR
- ribosomes
- chloroplast ribosomes are smaller than cytoplasmic ribosomes
- less thylakoid membrane
- less chlorophyll
- (less chlorophyll, so) reduced light absorption
- so, slower rate of photosynthesis
Name the three phases of mitosis shown by C, D and E on Figure 7.
Describe the role of the spindle fibres and the behaviour of the chromosomes during each of these phases. [5 marks]
- C = prophase, D = metaphase, E = anaphase
- in prophase, chromosomes condense
- in prophase OR metaphase, centromeres attach to spindle fibres
- in metaphase, chromosomes line up at the equator of the cell
- in anaphase, centromeres divide
- in anaphase, sister chromatids are pulled apart to opposite poles of the cell and spindle fibres shorten
- cell engulfs ADC and engulfs it in a phagosome
- lysosome fuses with phagosome containing ADC
- lysozymes digest ADC, releasing the drug
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. [2 marks]
- ADC binds to healthy cells
- causes death of healthy cells
Suggest one reason why there are no data for Group G and Group H after day 8. [1 mark]
mice died OR not ethical to continue
Suggest and explain two further investigations that should be done before this ADC is tested on human breast cancer patients. [2 marks]
- tested on other mammals to check for side effects
- tested on healthy humans to check for side effects
- in group J, see if repeat doses stop the the tumours regrowing
- investigate different concentrations of ADC to find suitable dosage
Describe how a triglyceride molecule is formed. [3 marks]
- one glycerol and three fatty acids
- join by condensation reactions, releasing three molecules of water
- forming ester bonds
- increase accuracy/resolution as lengths are small
- increase accuracy as it reduces the risk of human error
- increase accuracy as roots are less likely to be damaged
- reduce error/uncertainty as lengths are small
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. [1 mark]
same species OR can breed to produce fertile offspring OR it is the genus and species name
name of the type of DNA replication shown in Q?
semi-conservative replication
Describe the structure of DNA. [5 marks]
- polymer of nucleotides
- each nucleotide is formed from a deoxyribose, a phosphate group, and an organic base
- phosphodiester bonds formed between nucleotides via condensation reactions, using ATP
- double helix structure held by hydrogen bonds
- hydrogen bonds formed between the complementary base pairs: adenine with thymine, and cytosine with guanine
Name and describe five ways substances can move across the cell-surface membrane into a cell. [5 marks]
- simple diffusion of small, non-polar molecules down a concentration gradient
- facilitated diffusion down a concentration gradient via carrier proteins or protein channels
- osmosis of water down a water potential gradient
- active transport against a concentration gradient via carrier proteins, using ATP
- co-transport of two different substances using a carrier protein
Contrast the structure of the two cells visible in the electron micrographs shown in Figure 14. [5 marks]
- magnification figures shows A is bigger than B
- A has a nucleus, but B has free DNA
- A has mitochondria, but B doesn’t
- A has no cell wall, but B has a murein cell wall
- A has a Golgi body, but B doesn’t
- A has no capsule, but B does
- A has DNA associated with histones, but B has DNA that isn’t associated with histones OR A has linear DNA, but B has circular DNA
- A has larger ribosomes than B
- hydrolyse ATP to ADP and Pi, releasing energy
- this energy allows the active transport of ions
The movement of Na+ out of the cell allows the absorption of glucose into the cell lining the ileum.
Explain how. [2 marks]
- maintains a concentration gradient for Na+
- Na+ moves in by co-transport, bringing glucose with it
Describe and explain two features you would expect to find in a cell specialised for absorption. [2 marks]
- microvilli/folded membrane so large area for absorption
- large number of carrier proteins for facilitated diffusion/active transport/fast rate of absorption
- large number of mitochondria so more ATP
- membrane-bound digestion enzymes to maintain a concentration gradient for fast absorption
Describe how amino acids join to form a polypeptide so there is always NH2 at one end and COOH at the other end. You may use a diagram in your answer. [2 marks]
- the amine group of one amino acid joins to the carboxyl group on another amino acid to form a peptide bond
- so in the chain, there is a free amine group on one end and a free carboxyl group at the other
To study lipid digestion, a scientist placed a tube into the gut of a healthy 20-year-old man. The end of the tube passed through the stomach but did not reach as far as the ileum.
After collecting the samples, the scientist immediately heated them to 70°C for 10 minutes.
Explain why. [2 marks]
- denature lipase
- so no further digestion occurred
Describe the role of micelles in the absorption of fats into the cells lining the ileum. [3 marks]
- micelles include bile salts and fatty acids
- make fatty acids more soluble in water
- carry fatty acids to the lining of the ileum
- maintain the higher concentration of fatty acids to the lining of the ileum
- fatty acids absorbed by diffusion
- pressure in the aorta is higher than in the ventricle
- semi-lunar valve is closed
At Q on Figure 3 there is a small increase in pressure and in rate of blood flow in the aorta.
Explain how this happens and its importance. [2 marks]
- elastic recoil
- maintains the rate of blood flow
A student correctly plotted the right ventricle pressure on the same grid as the left ventricle pressure in Figure 3.
Describe one way in which the student’s curve would be similar to and one way it would be different from the curve shown in Figure 3. [2 marks]
- similarity – similar pattern
- difference – lower pressure
Anthocyanins are coloured pigments found in the cell vacuole of some plant cells. Anthocyanins cannot move across undamaged cell membranes.
A student investigated how to extract anthocyanins from blueberries.
She mixed 10 g of crushed, fresh blueberries with 100 cm3 of extraction solvent for 1 hour.
She investigated three different extraction solvents:
• E – Ethanol, water and acid
• F – Ethanol and water
• G–Water
- higher absorbance indicates more membrane damage
- more membrane damage results in more anthocyanin release
- E and F are greater than G, water, as the phospholipid bilayer dissolves in ethanol
- E is greater than F as acid denatures membrane proteins
Anthocyanins are coloured pigments found in the cell vacuole of some plant cells. Anthocyanins cannot move across undamaged cell membranes.
A student investigated how to extract anthocyanins from blueberries using different solvents.
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. [3 marks]
- use known concentrations of anthocyanin solution
- prepare dilution series
- compare results with colour standards to give values
Describe the role of DNA polymerase in the semi-conservative replication of DNA. [2 marks]
catalyses the formation of phosphodiester bonds (1) via condensation reactions (2) to join adjacent nucleotides (3)
Cyclin D stimulates the phosphorylation of DNA polymerase, which activates the
DNA polymerase.
Describe how an enzyme can be phosphorylated. [2 marks]
attachment of inorganic phosphate to the enzyme (1) released from the hydrolysis of ATP (2)
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. [2 marks]
- faster DNA replication
- uncontrolled cell division
- results in a mass of excessive cells
If alveolar epithelium cells die inside the human body they are replaced by non-specialised, thickened tissue.
Explain why death of alveolar epithelium cells reduces gas exchange in human lungs. [3 marks]
- reduces surface area
- increased distance for diffusion
- reduced rate of gas exchange
Some people eat red meat for many years without having any reaction, then have an allergic reaction to the alpha-gal in red meat.
An allergic reaction is caused by an immune response.
Draw a labelled diagram of an antibody and identify the specific alpha-gal binding site. [3 marks]
(3rd mark for either disulfide bridge, constant region, or variable region labelled)
Suggest how one antibody can be specific to tick protein and to alpha-gal. [2 marks]
- similar structure
- so, antibody is complementary to both and can form antigen-antibody complexes with both
Define ‘non-coding base sequences’ and describe where the non-coding multiple repeats are positioned in the genome. [2 marks]
- DNA that does not code for protein
- positioned between genes
Describe how mRNA is formed by transcription in eukaryotes. [5 marks]
- DNA helicase breaks hydrogen bonds between DNA bases
- only one DNA strand acts as a template
- free RNA nucleotides align by complementary base pairing
- in RNA, uracil pairs with adenine in DNA, instead of thymine
- RNA polymerase joins adjacent RNA nucleotides
- by phosphodiester bonds between adjacent nucleotides formed via condensation reactions, using ATP
- pre-mRNA is spliced to remove introns to form mRNA
Describe how a polypeptide is formed by translation of mRNA. [6 marks]
- mRNA attaches to ribosome
- tRNA anticodons bind to complementary mRNA codons
- tRNA brings specific amino acids
- amino acids join by peptide bonds formed by condensation reactions, using ATP (5)
- tRNA released
- ribosome moves along the mRNA (and the process repeats until the stop codon is reached) to form the polypeptide
Define ‘gene mutation’ and explain how a gene mutation can have:
• no effect on an individual
• a positive effect on an individual.
[4 marks]
[Definition]
1. change in the base sequence of DNA
2. results in the formation of new allele
[no effect as]
3. genetic code is degenerate OR mutation is in an intron, so amino acid sequence may not change
4. does change the amino acid sequence, but this has no effect on the tertiary structure
5. new allele is recessive, so does influence the phenotype
[positive effect as]
6. results in change in polypeptide that positively changes the properties of a protein
7. may result in increased reproductive succession OR survival chances
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 (the allosteric site)
- changes tertiary structure of the enzyme, so changes the shape of the active site
- active site and substrate are no longer complementary, so no substrate binds to form enzyme-substrate complexes
• with the inhibitor, increasing the substrate doesn’t affect the rate of the reaction OR doesn’t affect lipase activity
• high substrate concentration doesn’t overcome inhibition OR doesn’t meet maximum rate of reaction/lipase activity
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 answer. [1 mark]
number of species and number of individuals in each species
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
Use Figure 5 to explain how human mass at birth is affected by stabilising selection. [3 marks]
- most likely to be transferred to special care unit if under 2800g or over 4200g
- extreme mass babies are less likely to survive to reproduce, so less likely to pass on their alleles for extreme mass at birth
- alleles for extreme mass at birth decreases in frequency in the population
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 the difference is due to chance is less than 0.05
- reject null hypothesis
- presence of KIRDS1 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 capsid
- phospholipid viral envelope
- 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]
- have more CD4 cells
- lower viral load that can destroy CD4 cells
- continued activation of B cells, cytotoxic T cells, and phagocytes
- with B cells still being activated, there is continued production of antibodies
- more able to destroy pathogens
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.
Describe how BrdU would be incorporated into new DNA during semi-conservative replication. [5 marks]
- DNA helicase
- breaks hydrogen bonds between complementary base pairs, separating the two strands of DNA
- free RNA nucleotides align by base pairing, and BrdU is complementary to adenine, so forms hydrogen bonds with adenine on the template strand
- DNA polymerase joins adjacent RNA nucleotides together, incorporating BrdU into the new DNA strand
- phosphodiester bonds form between nucleotides
Cells with BrdU in their DNA are detected using an anti-BrdU antibody with an enzyme attached.
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 the antibody to the cells
- wash cells to remove excess antibody
- add substrate to cause a colour change
Scientists grew sunflower plants. They supplied different plants with different volumes of water.
After two days, they determined the water potential in the leaf cells.
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]
- slower growth
- due to smaller number of stomata
OR
- growth may continue at lower water potentials
- due to adaptations in enzymes involved in photosynthesis
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, so less photosynthesis
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]
- a mouse is smaller, so has a larger surface area to volume ratio
- so, faster heat loss
- faster rate of respiration
Explain five properties that make water important for organisms. [5 marks]
- a metabolite in photosynthesis, respiration, condensation, and hydrolysis
- a solvent, so metabolic reactions can occur and allows the transport of substances
- high specific heat capacity, so buffers changes in temperature
- large specific latent heat of vaporisation, so provides a cooling effect via evaporation
- cohesion between water molecules, so supports columns of water in plants
- cohesion between water molecules, so produces surface tension, supporting small organisms
Describe the biochemical tests you would use to confirm the presence of lipid, non-reducing sugar and amylase in a sample. [5 marks]
LIPID
1. add ethanol, then add water and mix
2. white emulsion
NON-REDUCING SUGAR
3. do Benedict’s test and expect a negative result
4. boil with acid, then neutralise with alkali
5. heat with Benedict’s and a red precipitate will be formed
AMYLASE
6. add biuret solution and becomes lilac
7. add starch, leave for some time, and then test for the absence of starch
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. [5 marks]
- condensation reactions join monomers together and forms chemical bonds, releasing water
- hydrolysis reactions breaks chemical bonds between monomers in a polymer, releasing water
EXAMPLES
• amino acids join via peptide bonds to form polypeptides
• alpha glucose molecules join via glycosidic bonds to form glycogen or starch
• beta glucose molecules join via glycosidic bonds to form cellulose
• nucleotides join via phosphodiester bonds to form polynucleotides, RNA or DNA
- the individual chromosomes are visible because they have condensed
- each chromosome is made up of two sister chromatids because DNA has replicated
- the chromosomes aren’t arranged in homologous pairs, which they would be if it was meiosis
When preparing 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. [2 marks]
- water moves into cell by osmosis
- cell gets bigger
The dark stain used on the chromosomes binds more to some areas of the chromosomes than others, giving the chromosomes a striped appearance.
Suggest one way the structure of the chromosome could differ along its length to result in the stain binding more in some areas. [1 mark]
• differences in base sequences
• differences in histones
• differences in condensation / coiling
What is a homologous pair of chromosomes? [1 mark]
two chromosomes that carry the same genes
What is meant by species richness? [1 mark]
a measure of the number of different species in a community
Three of the bee species collected in the farmland areas were Peponapis pruinosa, Andrena chlorogaster and Andrena piperi.
What do these names suggest about the evolutionary relationships between these bee species? Explain your answer.
[2 marks]
Andrena chlorogaster and Andrena piperi are more closely related to each other than to Peponapis pruinosa (1) as they are in the same genus (2)
Formation of an enzyme-substrate complex increases the rate of reaction. Explain how. [2 marks]
- reduces activation energy
- due to bending bonds
Draw the general structure of an amino acid. [1 mark]
The genetic code is described as degenerate. What is meant by this?
more than one codon codes for a single amino acid
The scientists used the following null hypothesis:
‘The proportion of plants that produce 2n gametes will not change from one breeding cycle to the next.’
The scientists tested their null hypothesis using the chi-squared statistical test. After 1 cycle their calculated chi-squared value was 350
The critical value at P=0.05 is 3.841
What does this result suggest about the difference between the observed and
expected results and what can the scientists therefore conclude? [2 marks]
- calculated value is greater than the critical value, so the null hypothesis can be rejected
- there is a less than 5% probability that the difference is due to chance
- the scientists can conclude that there proportion of plants that produce 2n gametes changes from one breeding cycle to the next
- the scientists selected for breeding plants that produced 2n gametes
- these plants passed on their alleles or the production of 2n gametes to the next generation
- the frequency of alleles for production of 2n gametes increased in the population
When a person is bitten by a venomous snake, the snake injects a toxin into the person. Antivenom is injected as treatment. Antivenom contains antibodies against the snake toxin. This treatment is an example of passive immunity.
Explain how the treatment with antivenom works and why it is essential to use passive immunity, rather than active immunity. [2 marks]
- antivenom antibodies bind to the toxin to cause its destruction
- active immunity would be too slow
- may be different form of toxin within the same species
- different antibodies would be needed in the antivenom OR several antibodies are complementary to several antigens
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. [3 marks]
- 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 OR the first dose must be small so the animal isn’t killed
The scientists concluded that this heat treatment damaged the phloem. Explain how the results in Figure 9 support this conclusion. [2 marks]
- radioactively labelled carbon is converted to sugars during photosynthesis
- translocation in phloem throughout the plant only occurs in B
OR
- movement in phloem requires respiration to provide ATP
- heat treatment stops respiration, so transport in phloem throughout the plant only occurs in B
The scientists concluded that this heat treatment did not affect the xylem.
Explain how the results in Table 4 support this conclusion. [2 marks]
- water content of the leaves isn’t different as standard deviations overlap
- water is still being transported in the xylem to the leaf OR movement in xylem is passive, so is unaffected by heat treatment
Describe the role of two named enzymes in the process of semi-conservative replication of DNA. [3 marks]
- DNA helicase causes hydrogen bonds between DNA strands to break
- DNA polymerase joins DNA nucleotides
- to form phosphodiester bonds
Suggest explanations for the results in Table 5. [3 marks]
- In D, the antibody binding to cyclin A cannot bind to initiate DNA replication
- in E, RNA interferes with polypeptide production, so cyclin A isn’t made
- in F, added cyclin A can bind to the enzyme to initiate DNA replication
Describe the gross structure of the human gas exchange system and how we breathe in and out. [6 marks]
- trachea, bronchi, bronchioles, alveoli (second mark for having them in this order)
When breathing IN,
3. diaphragm contracts and external intercostal muscles contract
4. causes volume to increase and pressure to decrease in the thoracic cavity, resulting in air to move in
When breathing OUT,
5. diaphragm relaxes and internal intercostal muscles contract
6. causes volume to decrease and pressure to increase in the thoracic cavity, resulting in air to move 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. [5 marks]
- both contain ester bonds between fatty acids and glycerol
- both contain glycerol
- fatty acids in both may be saturated or unsaturated
- both are insoluble in water
- both contain carbon, hydrogen, and oxygen, but phospholipids also contain phosphorus
- triglycerides contain 3 fatty acids, but phospholipids contain 2 and a phosphate group
- triglycerides are hydrophobic, but phospholipids have hydrophilic phosphate heads and hydrophobic fatty acid tails
- phospholipids form a 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. [4 marks]
- glucose and galactose
- joined by glycosidic bond
- formed by condensation reaction
- lactose is added to a polypeptide in the Golgi apparatus
The enzymes DNA helicase and DNA polymerase are involved in DNA replication.
Describe the function of each of these enzymes. [2 marks]
DNA helicase – unwind DNA and break hydrogen bonds between strands
DNA polymerase – joins adjacent nucleotides
Adenosine triphosphate (ATP) is a nucleotide derivative.
Contrast the structures of ATP and a nucleotide found in DNA to give two
differences. [2 marks]
- ATP has ribose, but DNA nucleotide has deoxyribose
- ATP has 3 phosphate groups, but DNA nucleotide has 1 phosphate group
- ATP base is always adenine, but DNA nucleotide base varies
- [boiling the agar] so no contamination
- [transferring the same volume of liquid culture onto each agar plate] so same number of bacteria transferred to allow comparison
Explain the advantages of lipid droplet and micelle formation. [3 marks]
- droplets increase surface areas for lipase action
- so, faster hydrolysis of lipids
- micelles carry fatty acids and monoglycerides to the membrane
Name structure Q in Figure 2 and suggest how it is involved in the absorption of lipids. [4 marks]
- Golgi apparatus
- modifies and processes triglycerides
- combined triglycerides with proteins
- forms vesicles for exocytosis
Explain the role of the heart in the formation of tissue fluid. [2 marks]
- contraction of ventricle produces high hydrostatic pressure
- this forces water out of blood capillaries
- t-test
- looking for differences between two means
- difference is significant, so not due to chance, as the P value is less than 0.05
68% of all the fish caught in this investigation came from sample A.
A student thought this showed that sample A had a greater index of diversity than any of the other samples.
It is not possible to draw this conclusion from the given data. Give reasons why. [3 marks]
- number of individuals of each species isn’t known
- almost all of sample A could be the same species
- other samples have a higher species richness but a lower number of individuals
- other samples may have more individuals of each species
- person has HIV DNA in their DNA
- new HIV particles are still made
- AZT inhibits reverse transcriptase
- AZT prevents these new HIV particles from forming new HIV DNA (OR AZT stops the replication of HIV)
- stops the destruction of more T cells
- so, immune system continued to work, and AIDS doesn’t develop
Suggest and explain two advantages of using HAART (lines 7–9). [4 marks]
- stops development of AIDS
- as AZT-resistant HIV is prevented from replicating by other drugs
- AZT continues to work as a drug
- as HAART prevents the spread of AZT-resistant HIV to the rest of the human population
- no new HIV particles are made
- as HAART may interfere with viral protein synthesis
Suggest why high doses of AZT lead to muscle wastage (lines 10–11). [2 marks]
- fewer mitochondria, so less aerobic respiration
- muscles receive less ATP, so waste
FOR
1. appears to be no HIV-1, so could be effective
2. no CCR5, so can’t get HIV-1 in the future
3. only one BSCT is needed, shown by patient Q
4. wouldn’t need to take ART daily 16 months after BSCT
AGAINST
5. don’t know if chemotherapy or radiotherapy is needed
6. it is only for HIV-1
7. only worked in two cases
8. might not be long term
9. HIV-1 could mutate and be able to bind on a different receptor on helper T cells
10. might be a lack of suitable donors
Whey as…
1. it is absorbed quicker
2. still stimulates protein synthesis, even it is lower than casein
3. inhibits the breakdown of body proteins
4. becomes body protein significantly more
- antisense mRNA is completely to sense mRNA
- antisense mRNA would bind to sense mRNA
- ribosome wouldn’t be able to bind to sense mRNA
- results in less translation of sense mRNA, so less production of SUT1 protein
SAME
1. age, BMI, ethnicity, sex, other/previous medications, diet, exercise, health issues
DIFFERENT
2. no sodium in the same medicine
[physiological change]
1. increase in breathing rate
[explanation]
2. similar pCO2 per breath, but more breaths
3. increase in tidal volume
4. similar pCO2 per breath, but increased volume per breath
- valve A = atrioventricular valve
- chamber B = left ventricle
substitution
Use the information provided to suggest how Trexall slows cell division. [3 marks]
- Trexhall acts as a competitive inhibitor, and is able to bind to the active site of DR
- fewer enzyme–substrate complexes form
- fewer nucleotides available for DNA replication
correlation coefficient as looking for correlation between two variables
From these data, what can you conclude about the effectiveness of Rituximab in treating patients with CLL?
Do not include considerations of statistical analyses in your answer. [3 marks]
- the more CD20 on B cells, the higher the percentage of B cells destroyed
- in no cases are all B cells destroyed
- don’t know proportion of cancer cells killed
- won’t cure CLL
- little effect below 5 CD20 on cells
Complete Figure 3 by putting A or a in the boxes. One box has been completed for you with A.
lowercase a in both boxes
Name the process that produced the combination of alleles on the chromosome in the first polar body in Figure 3.
crossing over
there are fewer white blood cells, so no need to dilute further to see enough
Explain how the stain allowed the doctor to count the white blood cells amongst all the red blood cells. [1 mark]
white blood cells have a nucleus that stains, but red blood cells don’t
With some samples, the scientists decided they needed to carry out a series of dilutions of the sample before counting the bacteria.
Use evidence from Figure 3 to explain why dilutions were necessary for some
samples but not for others. [2 marks]
- log scale on graph shows big range in number of bacteria
- Some samples too many to count (so dilute) but some countable (so don’t dilute)
- significant increase in species richness on Islay and Colonsay, and significant fall on Harris
- change in diversity on Islay is not significant
- as greater than 0.05 probability that the difference is due to chance (on Islay)
- changed tertiary structure of enzyme
- active site becomes complementary to substrate
- ATM won’t bind to broken DNA
- DNA isn’t repaired
- cell division continues
- tumour suppressor gene isn’t activated
- may have no effect in diploid organism
- that has a functional ATM gene
- similar size, so can fit the channel
- similar shape, so can bind to the channel
complementary base pairs – hydrogen bond
adjacent nucleotides in a DNA strand – phosphodiester bond
A – attachment protein
B – capsid
- resolution is too low
- as wavelength of light is too long
- chromosomes are becoming visible
- because still condensing
OR
- chromosomes arranged randomly
- because no spindle activity OR because not attached to spindle fibres
A
- endopeptidase
- 3
- vein
- wider lumen
The student wanted to use the data from plot 1 to estimate the total number of the beetle species in the meadow.
Suggest how the student should use the data from plot 1 and other information
provided to estimate the total number of the beetle species in the meadow. [4 marks]
- determine the area of plot 1
- calculate total area of meadow
- divide area of meadow by area of plot
- multiply this by the number of beetles per plot
table – independent variable in first column
data – same number of decimals in final column
• ribosomes / rough endoplasmic reticulum
• Golgi apparatus/vesicles
Explain the biological advantage to athletes of injecting synthetic EPO (lines 7−8). [2 marks]
- more red blood cells, so more haemoglobin, so more oxygen transport
- more aerobic respiration to produce more ATP, so can exercise for longer
Describe how mice injected with human EPO produce anti-human EPO antibody (line 14). [3 marks]
- human EPO antigen displayed on antigen-presenting cells
- specific helper T cell simulates specific B cell to clone by mitosis
- plasma cells secrete antibodies
Describe the roles of anti-human EPO antibody and anti-mouse antibody with enzyme attached (lines 14−16) in producing a positive result for EPO in the ELISA test. [3 marks]
[Anti-human EPO antibody]
1. binds to EPO in plastic well
[Anti-mouse antibody]
2. binds to anti-human antibody
3. Substrate is added, and enzyme causes colour change that indicates a positive result
- unsaturated
- double bond between carbons
[similarity]
1. both have a phospholipid bilayer OR both have protein
[differences]
2. no channel or carrier proteins, but the fluid mosaic model does
3. cholesterol is not present, but it is present in the fluid mosaic model
4. glycoprotein is not present, but it is present in the fluid mosaic model
5. glycolipid is not present, but it is present in the fluid mosaic model
- non-disjunction
- in meiosis
- chromosomes have not separated
- amino acids are used in protein synthesis
- so, more enzymes produced for plasmid replication
- circular DNA is heavier
- as the band is lower in the tube
What can you conclude from Figure 6 about the effect of antibiotic X on plasmid replication and on circular DNA replication? Explain your answer. [2 marks]
- plasmid replication increases with X as the band is wider
- circular DNA replication doesn’t increase with X as the band is identical
- amylase hydrolyses starch
- into maltose
- different primary structure
- different tertiary structure, so different shape of active site
- enzyme-substrate complexes are more likely with the enzyme from ADF allele
- flies with ADF allele have selective advantage in the presence of alcohol
- so, insects with ADF are more likely to survive and reproduce
- and pass on ADF allele
- so, allele frequency increases
Using Figure 8 and the information provided, what can you conclude about amino acid uptake by G cells and by H cells? [3 marks]
- amino acid uptake by active transport
- cyanide reduces amino acid uptake
- ATP production is stopped on membranes
- ATP production continues in the cytoplasm
What can you conclude from Figure 10 about the effect of trapping the enzymes GOx and HRP inside cages? [3 marks]
- trapping increases enzyme activity
- difference is significant
- as the standard deviations don’t overlap
- higher affinity for oxygen, so dissociates oxygen less readily
- allows aerobic respiration when diving, so when at a lower partial pressures of oxygen
- first oxygen binds to Hb, causing change in shape
- cooperative binding - the change in shape allows more oxygen to bind more easily
K = bronchiole
L = artery
- tissue doesn’t contain starch
- makes nucleus visible as it contains DNA
- no, as if it was addition, there would be a frameshift
- would affect more than one amino acid
OR
- no, as G to C in the second codon
- so, substitution has occured, not an addition mutation
- expression of gene from another species
- so, new protein
Explain how an increase in the rate of transcription of the PIP1b gene (lines 6–7) will affect the permeability of tobacco plant cell membranes to water. [2 marks]
- more aquaporin made
- increased water permeability
Suggest and explain one advantage and one disadvantage of increased stomatal
density on the growth of tobacco plant leaves (lines 8–9). [4 marks]
[advantage]
1. more carbon dioxide uptake
2. more photosynthesis, so faster growth
[disadvantage]
1. more water loss
2. less photosynthesis, so slower growth
What can you conclude from the appearance of valves in Figure 1 about heart muscle activity and blood movement between:
1. the ventricles and arteries? [2 marks]
2. atria and ventricles? [2 marks]
(1)
1. ventricles relaxed
2. no backflow of blood into the ventricles
(2)
3. atria muscle contracted
4. blood movement from atria into ventricles
Give two variables that the student did not control in her procedure. [2 marks]
- length and diameter (OR surface area / volume / mass/weight of cylinder)
- time in solution
Phloem pressure is reduced during the hottest part of the day. Use information in Figure 6 along with your understanding of transpiration and mass flow to explain why. [3 marks]
- high rate of transpiration
- water lost through stomata
- causes less water movement from xylem to phloem
