paper 2 exam Qs Flashcards
(172 cards)
1
Q
Describe the process of glycolysis. [4 marks]
A
• phosphorylation of glucose using ATP
• oxidation of triose phosphate to pyruvate
• net gain of ATP
• NAD is reduced
2
Q
Malonate inhibits a reaction in the Krebs cycle.
Explain why malonate would decrease the uptake of oxygen in a respiring cell.
[2 marks]
A
• less reduced NADP
• oxygen is the final electron acceptor
3
Q
Explain how a resting potential is maintained across the axon membrane in a neurone. [3 marks]
A
• potassium ions diffuse out, while sodium ions diffuse in
• membrane is more permeable to potassium ions
• sodium ions actively transported out, and potassium ions move in
4
Q
Explain why the speed of transmission of impulses is faster along a myelinated axon than along a non-myelinated axon. [3 marks]
A
• myelination provides insulation
• in myelinated axons, saltatory conduction occurs (depolarisation at the nodes of Ranvier)
• in non-myelinated, depolarisation occurs along the whole of the axon
5
Q
A scientist investigated the effect of inhibitors on neurones. She added a respiratory inhibitor to a neurone. The resting potential of the neurone changed from –70 mV to 0 mV. Explain why. [3 marks]
A
• no ATP produced
• no active transport (of sodium ions)
• electrochemical gradient isn’t maintained
6
Q
A
• tip produces IAA
• IAA diffuses into shoot
• more elongation of cells on one side than the other
7
Q
Using the procedure in Figure 2 and the calibration curve in Figure 3, describe how you could compare the IAA concentration in shoot tips from two different plant species.
In your answer you should refer to all the variables that should be controlled to produce a valid comparison. [5 marks]
A
• size of shoot tip
• number of shoot tips
• type/size of agar block
• shoots at the same stage of development
• time tips are kept in agar
• temperature
• repeat several times and calculate a mean
• compare degree of curvature on calibration curve to determine IAA concentration
8
Q
State two conclusions about IAA that you can make from the results shown in Table 1. [2 marks]
A
• light/dark doesn’t affect IAA concentration
• IAA moves to shaded side (away from light)
9
Q
In fruit flies, males have the sex chromosomes XY and the females have XX.
In fruit flies, a gene for eye colour is carried on the X chromosome. The allele for red eyes, R, is dominant to the allele for white eyes, r.
Male fruit flies are more likely than female fruit flies to have white eyes. Explain why.
A
• males only have one allele
• females must be homozygous recessive (to have white eyes)
10
Q
A
second box
11
Q
A
• the genes are linked (autosomal linkage)
• no crossing over occurs
• no Gl and no gL gametes are produced
12
Q
A
• 400 offspring per generation = 200 females per generation
• 50 x 200^4 = 8x10^10
13
Q
Neonatal diabetes is a disease that affects newly born children. The disease is caused by a change in the amino acid sequence of insulin.
This change prevents insulin binding to its receptor. Explain why this change
prevents insulin binding to its receptor.
[2 marks]
A
• changed tertiary structure
• no longer complementary to receptor
14
Q
A
• less AKT activated
• fewer vesicles move to the membrane (so, fewer channel proteins in membrane)
• less glucose diffuses into cell (so, high blood glucose)
15
Q
Using your knowledge of the kidney, explain why glucose is found in the urine of a person with untreated diabetes.
[3 marks]
A
• high concentration of glucose in filtrate
• not all the glucose is reabsorbed at the proximal convoluted tubule
• as carrier proteins are saturated
16
Q
A
• produces healthy blood cells
• no cancerous blood cells
• stem cells replicate
17
Q
A
• AZA reduces methylation of DNA
• tumour suppressor gene is expressed
• prevents uncontrolled cell division
18
Q
A
• effect of AZA can be compared
• unethical to not treat the control group
19
Q
One method of transferring RNAi molecules into cells involves combining these molecules with a lipid. Suggest why this increases uptake of RNAi molecules into cells. [1 mark]
A
cell membrane has a phospholipid bilayer
20
Q
Using all the information provided, evaluate the use of the two types of RNAi in treating hepatitis B in humans.
Do not refer in your answer to how RNAi reduces hepatitis B virus replication. [5 marks]
A
• both are more effective than the control
• significant difference in effectiveness between both types
• as SDs don’t overlap
• replication is reduced, but isn’t at zero
• not investigated in humans
• shRNA is more effective as 7.5% of control compared with 50% for lhRNA
• no indication of sample size
21
Q
Describe and explain how the polymerase chain reaction (PCR) is used to amplify a DNA fragment. [4 marks]
A
• requires DNA fragment, DNA polymerase, DNA nucleotides, and primers
• heat to 95°C to break hydrogen bonds to separate strands
• reduce temperature (to 55°C), so primers bind to DNA
• increase temperature (to 72°C - optimum for DNA polymerase), and DNA polymerase joins nucleotides (and repeat method)
22
Q
A
• (DNA produced) doubles each cycle to result in an exponential increase
• plateaus as no more nucleotides/primers
23
Q
A
• use a random number generator to generate random coordinates for quadrat positions
• large number of quadrats
• divide total percentage by number of quadrats
24
Q
Explain how succession results in a wide variety of fish living on coral reefs. Do not describe the process of succession in your answer. [2 marks]
A
• increase in variety of species
• provides more habitats/niches
• provides greater variety of food
25
• significant difference with C compared with A
• as less than 5% probability that the difference is due to chance
• no significant difference with B and D compared with A
• algae species is unknown
• only done off coast of Florida
• only done at 16 to 18 meters
• only 34 weeks
• artificial reef could affect growth
• cage may allow other fish to enter
26
• mutation leads to variation
• allopatric speciation
• different lakes have different selection pressures
• reproductive isolation
• different alleles selected (individuals with these alleles survive, reproduce, and pass on alleles, changing frequency of alleles)
• eventually, different species cannot breed to produce fertile offspring
27
Loss of nutrients into Lake Malawi has resulted in a decrease in some fish populations (lines 12–13). Explain why. [4 marks]
• algal bloom blocks light
• no photosynthesis, so plants die
• saprobiotic microorganisms aerobically respire
• less oxygen for fish to respire
28
The mark-release-recapture method can be used to estimate the size of a fish population (lines 13–14).
Explain how. [4 marks]
• collect sample, mark, and release
• ensure marking isn’t harmful to fish
• allow time for fish to randomly distribute before collecting second sample
• population = (number in first sample x number in second sample) / number of marked fish in second sample
29
Suggest why the mark-release-recapture method can produce unreliable results in very large lakes (lines 14–15). [1 mark]
less chance of recapturing fish
30
• circular muscle contracts
• radial muscle relaxes
31
The fovea of the eye of an eagle has a high density of cones. An eagle focuses the image of its prey onto the fovea.
Explain how the fovea enables an eagle to see its prey in detail. Do not refer to colour vision in your answer. [3 marks]
• high visual acuity
• each cone connected to a single neurone
• comes send separate sets of impulses to the brain
32
The retina of an owl has a high density of rod cells. Explain how this enables an owl to hunt its prey at night. Do not refer to rhodopsin in your answer. [3 marks]
• high visual sensitivity (retinal convergence)
• several rods connected to a single neurone
• spatial summation to reach threshold
33
• is lipid soluble
• can diffuse through phospholipid bilayer
34
In the cytoplasm, testosterone binds to a specific androgen receptor (AR). An AR is a protein.
Suggest and explain why testosterone binds to a specific AR. [2 marks]
• specific tertiary structure
• structures are complementary
35
The binding of testosterone to an AR changes the shape of the AR. This AR molecule now enters the nucleus and stimulates gene expression.
Suggest how the AR could stimulate gene expression. [2 marks]
• AR is a transcription factor
• binds to promoter
• stimulates RNA polymerase
36
• less than or equal to 16 repeats is significant
• as there is a less than 5% probability that the difference is due to chance
• so, with less than or equal to 16, reject the null hypothesis
37
• stirrer distributes heat
• air (provides insulation, so) reduces heat loss
• water has a high specific heat capacity
38
• light is the wrong wavelength
• light is reflected
• light misses chlorophyll
• CO2 concentration or temperature is a limiting factor
39
The light absorbed by chlorophyll is used in the light-dependent reaction. Name the two products of the light-dependent reaction that are required for the light-independent reaction. [2 marks]
• ATP
• NADPH
40
• I band (white band) is only actin
• H band (grey band) is only myosin
• overlapping region (the darkest region) is actin and myosin
41
42
Describe how you would determine the concentration of creatinine in a urine sample using your calibration curve. [2 marks]
• add creatine-detecting solution to urine
• read off creatine concentration against absorption value obtained
43
Describe the sequence of events involved in transmission across a cholinergic synapse.
Do not include details on the breakdown of acetylcholine in your answer. [5 marks]
• depolarisation of presynaptic membrane
• calcium channels open and calcium ions enter
• calcium ions cause presynaptic vesicles to fuse with the presynaptic membrane, releasing acetylcholine
• acetylcholine attaches to receptors on the postsynaptic membrane
• sodium ions enter the postsynaptic membrane, leading to depolarisation
44
Mutation is one cause of genetic variation in organisms. Give two other causes of genetic variation. [2 marks]
• crossing over
• independent segregation
• random fusion of gametes
45
codominance
46
In a species of flowering plant, the T allele for tallness is dominant to the t allele for dwarfness. In the same species, two alleles CR (red) and CW (white) code for the colour of flowers.
47
Using all the information, evaluate the effect on plant growth of adding the different fertilisers to the soil. [5 marks]
48
The scientist determined the dry mass of the spinach plants. First, he heated each sample at 80 °C for 2 hours.
Suggest what the scientist should do to ensure that he has removed all the water from the sample. [2 marks]
• weigh and heat
• until mass is constant
49
Alport syndrome (AS) is an inherited disorder that affects kidney glomeruli of both men and women. Affected individuals have proteinuria (high quantities of protein in their urine).
Suggest how AS could cause proteinuria. [2 marks]
• damages basement membrane / more protein channels in basement membrane
• proteins can pass into the glomerular filtrate
50
box 4
51
52
[wild type mice aren’t affected by AS, an inherited disorder that affects kidney glomeruli of both men and women. Affected individuals have proteinuria (high quantities of protein in their urine).]
• transplanted stem cells differentiate
• reduce the loss of protein at the glomerulus
53
• restriction endonuclease enzyme cuts plasmid (to produce sticky ends)
• ligase joins sticky ends (joins the gene and plasmid together)
54
• cell division has occurred (before gene is added)
• (cells producing) gametes don’t receive the gene
55
• no overlap in SDs
• significant difference in growth (so isn’t due to chance)
56
• large sample size, so representative
• 12 months, so can assess growth
• control present for comparison
57
• Fat (store) used in respiration
• Less energy is required due to low respiration/less movement
• Gluconeogenesis
• Low surface area to volume reduces heat loss
• Fat (layer/insulation) reduces heat loss
• Long loop of Henle so less water lost
• Water provided from respiration
• No urination
• Less evaporation
58
During hibernation, the heart rate and the metabolic rate of black bears decrease (lines 3−5).
Use your knowledge of the nervous control of heart rate to describe how these are linked. [4 marks]
• (lower metabolism, so) low levels of CO2 in the blood
• detected by chemoreceptors
• chemoreceptors located in the aorta
• fewer impulses to medulla
• more impulses along parasympathetic nerve
• to SAN
59
In many mammals, ‘uncoupling proteins’ help to maintain a constant body temperature during hibernation (lines 6−7).
Suggest and explain how. [2 marks]
• allows passage of protons
• energy released as heat
60
Climatic change has reduced the survival rate of snowshoe hares in mountain habitats (lines 11−13).
Suggest and explain how. [2 marks]
• less snow, so less camouflage
• more hares are seen and killed by predators
61
Snowshoe hares within a population moult at different times (line 15).
Explain how this could ensure the survival of snowshoe hare populations in these mountain habitats. [4 marks]
• hares that moult earlier are more likely to survive
• hares that moult earlier are more likely to reproduce
• and pass on advantageous allele
• frequency of allele increases in future populations
62
Succession occurs in natural ecosystems. Describe and explain how succession
occurs. [4 marks]
• colonisation by pioneer species
• species change the environment
• environment becomes less hostile for new species
• increase in diversity
• to climax community
63
• change in primary structure
• change in hydrogen/ionic/disulfide bonds
• alters tertiary structure
64
• produces healthy red blood cells
• no SCD red blood cells
• stem cells replicate
• less chance of rejection (since from brother/sister)
65
66
• tip produces IAA
• affects concentration of IAA / affects shoot length/growth/elongation
• mitosis occurs in shoot tips
• affects shoot length/growth/elongation
67
Explain why the student added glucose solution to each Petri dish. [2 marks]
• for respiration
• to provide ATP
68
Explain why the lids were kept on the Petri dishes. [2 marks]
• to prevent evaporation
• which alters water potential / concentration of IAA solution
69
Describe and explain the results shown in Table 1 and suggest how the results might have differed if lengths of root had been used. [3 marks]
• as IAA concentration increases, the higher the mean change in length
• IAA stimulates cell elongation
• in roots, growth is inhibited (by IAA)
70
A student looked at the results and concluded that a decrease in pH does cause a decrease in the force of muscle contraction.
Use Figure 3 to evaluate this conclusion.
[4 marks]
71
• less tropomyosin moved from binding site
• shape of tropomyosin not changed, so binding site isn’t exposed
• fewer actinomyosin bridges formed
• myosin head doesn’t move / doesn’t pull actin / less ATP hydrolase activation
72
In muscles, pyruvate is converted to lactate during prolonged exercise.
Explain why converting pyruvate to lactate allows the continued production of ATP by anaerobic respiration. [2 marks]
• regenerates NAD
• so, glycolysis continues
73
Describe the role of glucagon in gluconeogenesis.
Do not include in your answer details on the second messenger model of glucagon action. [2 marks]
• binds to receptors on target cells and stimulates enzymes
• to convert glycerol/amino acids/fatty acids into glucose
74
• more insulin binds to receptors
• stimulates uptake of glucose via channel proteins
• activates enzymes for glycogenesis (converting glucose to glycogen)
75
Explain how inhibiting adenylate cyclase may help to lower the blood glucose concentration. [3 marks]
• less ATP is converted to cyclic AMP (cAMP)
• less kinase activated
• less glycogenolysis
76
Give one piece of evidence from Figure 4 to show that the allele for grey body colour is dominant. [1 mark]
3 and 4 produce 9
77
Give the full genotype of the fly numbered 6 in Figure 4. [1 mark]
78
Explain one piece of evidence from Figure 4 to show that the gene for body colour is not on the X chromosome. [2 marks]
• 3 and 4 produce 9
• 3 would pass dominant allele to 9
79
fourth box
80
Describe what happens during photoionisation in the light-dependent reaction. [2 marks]
• chlorophyll absorbs light
• light excites electrons in chlorophyll
• electrons are lost
81
ink and leaf pigments would mix
82
Describe the method the student used to separate the pigments after the solution of pigments had been applied to the origin. [2 marks]
• level of solvent is below the origin
• stop before solvent reaches the top
83
The pigments in leaves are different colours. Suggest and explain the advantage of having different coloured pigments in leaves. [1 mark]
absorb more wavelengths of light for photosynthesis
84
What is a DNA probe? [2 marks]
• short, single strand of DNA
• bases are complementary with gene
85
• restriction endonuclease enzyme
• cuts DNA at recognition site
86
The DNA on the nylon membrane is treated to form single strands. Explain why. [1 mark]
so the DNA probe anneals
87
• lane 1 has DNA fragments of known lengths
• (can use lane 1 to) compare (the positions of viral fragments)
88
The sundew is a small flowering plant, growing in wet habitats such as bogs and marshes. The soil in bogs and marshes is acidic and has very low concentrations of some nutrients. The sundew can trap and digest insects.
Describe how you could estimate the size of a population of sundews in a small marsh. [5 marks]
• use a grid
• obtain random coordinates using a random number generator
• count frequency in a quadrat
• obtain a large sample and calculate a mean
• total number of sundews = mean number per quadrat x number of quadrats in marsh
89
The sundew is a small flowering plant, growing in wet habitats such as bogs and marshes. The soil in bogs and marshes is acidic and has very low concentrations of some nutrients. The sundew can trap and digest insects. Suggest and explain how digesting insects helps the sundew to grow in soil with very low concentrations of some nutrients. [2 marks]
• digestion of proteins
• provides amino acids
90
• no saltatory conduction, so depolarisation occurs along the whole length of the axon
• nerve impulses are slowed
• at the neuromuscular junction / sarcolemma
91
Sometimes Guillain–Barré syndrome causes heart rate irregularities (lines 4–5).
Suggest and explain why. [3 marks]
• fewer impulses from the medulla
• along the parasympathetic/sympathetic pathway
• to the SAN
92
The first successful drug trial to reduce concentrations of huntingtin in the brain used single-stranded DNA molecules (lines 13-14).
Suggest and explain how this drug could cause a reduction in the concentration of the protein huntingtin. [3 marks]
• this DNA is complementary to the mRNA for huntingtin
• binds to the mRNA for huntingtin
• prevents translation
93
Scientists from the first successful drug trial to reduce concentrations of
huntingtin (lines 9-11) reported that the drug is not a cure for Huntington's disease.
Suggest two reasons why the drug should not be considered a cure.
Do not include repeats of the drug trial in your answer. [2 marks]
• small sample size
• only four months
• huntingtin is still produced
• gene for Huntington’s is still present
• brain is already damaged / brain damage may continue
94
Suggest two reasons why people had the drug injected into the cerebrospinal fluid (lines 12-13) rather than taking a pill containing the drug. [2 marks]
• drug will directly/quickly reach the brain
• cerebrospinal fluid bathes the brain
• drug isn’t destroyed by acid / digested by enzymes
95
Suggest and explain one way epigenetics may affect the age when symptoms of
Huntington's disease start. [2 marks]
• decreased methylation of DNA
• stimulates transcription
• decreased acetylation of histones
• inhibits transcription
• increased acetylation of histones
• stimulates transcription
96
• less ATP
• less NADPH
97
A decrease in the activity of the enzyme rubisco would limit the rate of photosynthesis.
Explain why. [2 marks]
• less carbon dioxide reacts with RuBP
• less GP
98
Where precisely is rubisco found in a cell? [1 mark]
stroma
99
100
second box
101
Describe how alterations to tumour suppressor genes can lead to the development of tumours. [3 marks]
• increased methylation of tumour suppressor gene
• or mutation in tumour suppressor gene
• tumour suppressor genes are not transcribed (or primary structure is altered)
• results in uncontrolled cell division
102
so oxygen is used
103
During her investigation, the coloured liquid moved to the right. Explain why it moved to the right. [2 marks]
• anaerobic respiration produces carbon dioxide
• increase in pressure
104
large range in numbers
105
Many yeast cells die during the death phase. Suggest one reason why. [1 mark]
• decrease in glucose/substrate
• increase in carbon dioxide/ethanol/acidity
106
Describe how you would determine the mean percentage cover for beach grass on a sand dune. [3 marks]
• obtain random coordinates for quadrat positions using a random number generator
• obtain a large number of quadrats
• divide total percentage by number of quadrats
107
The scientists concluded that the results shown in Figure 5 were due to succession taking place.
Use Figure 5 to explain why the scientists reached this conclusion. [4 marks]
• beach grass is the pioneer species
• pioneer species changes the environment
• so, makes the environment less hostile for shrub-bunchgrass
• conifer/hardwood trees represent climax community
108
trees block sunlight
109
Using evidence from Figure 6, what can you conclude about the net primary productivity (NPP) in the sand dunes that are older than 1000 vears?
Explain your answer. [2 marks]
• NPP remains constant
• photosynthesis/GPP and respiration is constant
• NPP is low
• less light, so less photosynthesis/GPP
110
• affects water potential of blood
• affects volume of urine produced
111
Using Table 1, what can you conclude about the effectiveness of furosemide and CVT in the removal of excess fluid from the body? [2 marks]
• furosemide and CVT are more effective than the placebo
• significant difference as SDs don’t overlap
112
Furosemide and CVT are drugs used to remove excess fluid from the body. Furosemide is sometimes used to treat high blood pressure.
Suggest how furosemide would cause a decrease in blood pressure. [1 mark]
lower volume of blood
113
Furosemide inhibits the absorption of sodium and chloride ions from the filtrate produced in the nephrons.
Explain how furosemide causes an increase in the volume of urine produced. [3 marks]
• water potential of filtrate is decreased
• less water is reabsorbed by osmosis from the filtrate
• at the collecting duct
114
In genetic crosses, the observed phenotypic ratios obtained in the offspring are often not the same as the expected ratios.
Suggest two reasons why. [2 marks]
• small sample size
• fusion of gametes is random
• linked genes
• epistasis
• lethal genotypes
115
In tomato plants, the genes for height and for the type of leaf are on the same homologous pair of chromosomes. The allele T, for a tall plant, is dominant to the allele t, for a dwarf plant. The allele M, for normal leaves, is dominant to the allele m, for mottled leaves.
What would be the genotype of the offspring with dwarf plants and mottled leaves? [1 mark]
ttmm
116
Use the information provided to explain the results in Table 3. [3 marks]
• genes are linked
• so, produces fewer tall mottled and dwarf normal offspring
• crossing over has occurred
117
In tomato plants, the genes for height and for the type of leaf are on the same homologous pair of chromosomes. The allele T, for a tall plant, is dominant to the allele t, for a dwarf plant. The allele M, for normal leaves, is dominant to the allele m, for mottled leaves. These genes are linked.
118
• dopamine diffuses across synaptic cleft
• attaches to receptors on the postsynaptic membrane
• stimulates entry of sodium ions and depolarisation
119
• morphine attaches to opioid receptors
• more dopamine released to provide pain relief
120
GABA is a neurotransmitter released in some inhibitory synapses in the brain. GABA causes negatively charged chloride ions to enter postsynaptic neurones.
Explain how this inhibits postsynaptic neurones. [3 marks]
• inside of postsynaptic neurone becomes more negative
• more sodium ions required to reach threshold
• for depolarisation (to lead to an action potential)
121
What is meant by a genome? [1 mark]
all the DNA in a cell
122
Explain why the antibody binds to the transcription factor. [2 marks]
• antibody has a specific tertiary structure
• that is a complementary structure (to the transcription factor)
123
Use Figure 8 to explain what ‘precipitated DNA’ consists of. [1 mark]
DNA, transcription factor, and antibody
124
binds to P34 mRNA
125
Describe the roles of two named types of enzymes used to insert DNA fragments into plasmids. [2 marks]
• restriction endonuclease to cut plasmid
• ligase to join gene to plasmid
126
• charge
• mass / number of amino acids/polypeptides
• R groups differ
127
• LP arises due to mutation
• milk provides glucose
• individuals with LP are more likely to survive, reproduce (and pass on alleles)
• directional selection
• frequency of allele increases
128
Use the information provided to explain why the number of people showing LP would rapidly increase once selection for this condition had been established. (2 marks]
• dominant allele
• always expressed (when present in the phenotype)
129
Lactase persistence is caused by a mutation in DNA. This mutation does not occur in the gene coding for lactase.
Suggest and explain how this mutation causes LP. [2 marks]
• mutation in promoter region for transcription factor/lactase gene OR mutation in gene for transcription factor
• lactase gene continues to be transcribed
130
• no cone cells
• cones are connected to a single neurone
• cones send separate sets of impulses to the brain
131
Red-green colour blindness affects more men than women (lines 7–8).
Explain why. [2 marks]
• gene is on the X chromosome
• females require two alleles, while males require one allele
132
People with red-green colour blindness are unable to distinguish between red and green, and also between other colours (lines 8–10).
Explain why. [3 marks]
• green sensitive cones are non-functional
• there are three different types of cones
• different colours are seen due to stimulation of more than one cone
133
Current research into the treatment of red-green colour blindness involves the use of induced pluripotent stem cells (iPS cells) (lines 17–19).
Suggest how iPS cells could correct red-green colour blindness. [2 marks]
• iPS cells divide
• iPS cells differentiate into green sensitive cones
134
The use of iPS cells could have advantages over the use of gene therapy to correct red-green colour blindness (lines 19–20).
Using the information from the passage, suggest and explain reasons why. [3 marks]
use of iPS cells is…
• long-term
• less chance of rejection/immune response
• single treatment
• side effects could occur from using viruses in gene therapy
135
Exercise causes an increase in heart rate.
Describe the role of receptors and of the nervous system in this process. [4 marks]
• chemoreceptors detect a rise in carbon dioxide
• and send impulses to medulla
• to the SAN
• by the sympathetic nervous system
136
• less malonyl-CoA
• more fatty acids moved into mitochondria
• respiration of fatty acids provides ATP
137
• intraspecific competition
• don’t breed (so, reducing the population of mosquitoes)
138
Describe how the mark-release-recapture method could be used to determine the population of A. aegypti at the start of the investigation. [3 marks]
• collect sample, mark and release
• leave time for mosquitos to randomly disperse before second collection
• population = (number in first sample x number in second sample) / number of marked in second sample
139
The release of radiation-sterilised A. aegypti has not been very successful in controlling the transmission of dengue.
Suggest one reason why. [1 mark]
radiation affects survival
140
to maintain numbers as they die / have a short life span
141
The release of transgenic males proved successful in reducing the number of A. aegypti.
Describe how the results in Figure 2 support this conclusion. [2 marks]
number of mosquitoes in treated area is lower after 12/13/14/15/16 weeks
142
• increase in aerobic respiration
• more mitochondria
• more slow muscle fibres
143
The scientists then compared the length of time that the control mice and the trained mice could carry out prolonged exercise. The trained mice were able to exercise for a longer time period than control mice.
Explain why. [3 marks]
• more aerobic respiration produces more ATP
• anaerobic respiration is delayed
• less lactate
144
145
• osmosis doesn’t occur
• chloroplast doesn’t burst/shrink
146
Explain why the student set up Tube 1. [2 marks]
• to show light doesn’t affect DCPIP
• to show chloroplasts are required
147
Explain the results in Tube 3. [2 marks]
• reduction of DCPIP by electrons
• from light dependent reaction / chlorophyll
148
• provides a standard/reference point
• can compare different chemicals/weed killers
• can compare different concentrations of chemicals/weed killers
149
Explain how chemicals which inhibit the decolourisation of DCPIP could slow the
growth of weeds. [2 marks]
• less ATP produced
• less NADPH produced
• less GP reduced to triose phosphate
150
Arbuscular mycorrhiza fungi (AMF) are fungi which grow on, and into, the roots of plants. AMF can increase the uptake of inorganic ions such as phosphate.
Suggest one way in which an increase in the uptake of phosphate could increase
plant growth. [1 mark]
used to make ATP/ADP/phospholipids/DNA etc.
151
Arbuscular mycorrhiza fungi (AMF) are fungi which grow on, and into, the roots of plants. AMF can increase the uptake of inorganic ions such as phosphate.
Suggest one way in which AMF may benefit from their association with plants. [1 mark]
obtain glucose/carbohydrates/amino acids (any organic biological compound)
152
Explain why an increase in shoot biomass can be taken as a measurement of net primary productivity. [2 marks]
• represents dry mass
• represents gross production minus respiratory losses
153
Using the data from Figure 5, evaluate the effect on plant productivity of adding
AMF species and adding phosphate to the soil. [4 marks]
154
Each year, a few people with type I diabetes are given a pancreas transplant. Pancreas transplants are not used to treat people with type II diabetes.
Give two reasons why pancreas transplants are not used for the treatment of type II diabetes. [2 marks]
• type II (usually) produce insulin
• receptors are less sensitive to insulin / faulty insulin receptors
• treated/controlled by exercise/diet
155
fourth box
156
Suggest how transcription factors can reprogramme cells to form iPS cells. [2 marks]
• attach to promoter region
• stimulate/inhibit transcription/RNA polymerase
157
Using all the information provided, evaluate the use of iPS cells to treat type I diabetes in humans. [4 marks]
• Effective as group A is lower than group B
2. Effective as group A is similar to group C
3. Investigation done on mice, not humans;
4. Only shows results for 12 weeks / long-term effects not known
158
What is meant by the term phenotype? [2 marks]
appearance due to genotype and environment
159
epistasis
160
produces cDNA from mRNA
161
Explain the role of DNA polymerase in RT-PCR. [1 mark]
joins nucleotides to produce complementary strands of DNA
162
Any DNA in the sample is hydrolysed by enzymes before the sample is added to the reaction mixture.
Explain why. [2 marks]
• to remove any DNA present
• as this DNA would be amplified
163
Suggest one reason why DNA replication stops in the polymerase chain reaction. [1 mark]
limited number of primers/nucleotides
164
Scientists have used the RT-PCR method to detect the presence of different RNA viruses in patients suffering from respiratory diseases.
The scientists produced a variety of primers for this procedure. Explain why. [2 marks]
• base sequences differ
• different complementary primers needed
165
What is a gene pool? [1 mark]
all the alleles in a population
166
• occurs in the same habitat (not geographically isolated)
• mutations cause different flowering times
• reproductive isolation
• different alleles selected for (individuals with the advantageous allele survive, reproduce, and pass on allele, changing the frequency of allele)
• disruptive selection
• eventually, different species cannot breed to produce fertile offspring
167
Explain how donepezil could improve communication between nerve cells (lines 7–9). [3 marks]
• less acetylcholine is broken down
• acetylcholine attaches to receptors
• more sodium ions enter to reach threshold (to lead to depolarisation)
168
Suggest and explain two reasons why there is a high frequency of the E280A
mutation in Yaramul (lines 13–15). [2 marks]
• isolated, so inbreeding/low genetic diversity/small gene pool
• allele is inherited through generations from a common ancestor
169
Explain why natural selection has not reduced the frequency of the E280A
mutation in the population (lines 16–17). [2 marks]
• symptoms develop late/at 49
• have already reproduced
170
The age at which the E280A mutation is expressed to cause AD can vary (lines 11–12).
Suggest and explain one reason for this.
[2 marks]
• epigenetics
• methylation of genes
171
One scientific study which analysed chromosome 14 involved 102 individuals. The scientists recorded a sample size of 204. In this sample they detected 75 E280A mutations but only 74 potential AD cases (lines 19–21).
Suggest explanations for the figures the scientists recorded. [2 marks]
one person was homozygous dominant
172
Suggest why a DNA probe for the mutated triplet was not considered a suitable method for detection of the E280A mutation (lines 22–23). [2 marks]
• triplet is common
• would not know if it was the mutation / produces false positives
