AQA Biology Non-included Questions Flashcards
1
Q
C14 - Diagram: part of the retina in the human eye
when light falls on cells 1 and 2, only one spot of light is seen. But when light falls on cells 2 and 3, two spots of lights are seen.
A
- cells 1 and 2 share neurone but 2 and 3 have separate neurones
2
Q
C14 - why is it that when cell 3 receives one unit of light energy no light is seen, but when one unit of light energy is received by 3, cells light is seen (3)
A
- 1 unit of light energy on one cell is below the threshold
- 3 units of light energy is above the threshold so gives sufficient depolarisation
- no impulses in sensory neurone (3>impulses)
- through spatial summation a sufficient neurotransmitter is released
3
Q
C14 - cells of the same type as cells 6 and 7 are found in large numbers at the fovea, this results in colour vision with high visual acuity
A
- cone cells are sensitive to different wavelengths/different frequencies of these waves cause different colours
- Impulses along separate neurone from each receptor cell/each receptor cell connects to separate neurones
4
Q
C14 - explain why an increased cardiac output is an advantage during exercise (3)
A
during exercise:
- more energy is released
- more respiration
- more actively respiring muscles
A higher cardiac output:
- increases O2 supply to muscles
- increases glucose supply to muscles
- lactate removal
- increases heat removal from muscles (for cooling)
5
Q
C14 - Explain how nervous control in human can cause increased cardiac output during exercise (4)
A
- coordination via medulla located in the cardiac centre
- Increased impulses along sympathetic nerve
- to the SA node
- more impulses= increased rate of SA node to increase rate of firing
6
Q
C14 - Points A and B show when the atria and ventricles contract at which point was the ventricle contracting and why
A
B
- 2nd contraction
- occurs immediately after the atrium contraction
- larger more force, more pressure
7
Q
C14 - Addition of acetylcholine in the experiment mimics the effect of which branch of the autonomic nervous system
A
parasympathetic nervous system
8
Q
C14 - If acetylcholine reduces the heart rate and has no effect on the stroke volume. What effect will it have on the cardiac output
A
- Cardiac output is reduced
- Q= HR x SV
- if heart rate is reduced then so is the cardiac output
9
Q
C4 - DNP prevents Oxidative phosphorylation. when treated with DNP the sodium potassium pump no longer works. Why is this
A
- lack of ATP produced from oxidative phosphorylation
- pump = is active transport so requires energy from ATP
- if there is a lack of it then sodium ions cannot be transported out and the concentration gradient remains the same
10
Q
C4 - Why is it that sodium ions and amino acids are not absorbed from the lumen of the small intestine in the presence of DNP
A
- concentration of Na ions inside the cell aren’t transported so there is no concentration gradient
- as a result no facilitated diffusion of sodium ions from the lumen
- amino acid absorption requires diffusion of sodium ions so therefore cannot enter via co-transport
11
Q
C4 - By what mechanism would amino acids leave the epithelial cell
A
facilitated diffusion
12
Q
C7 - use the cohesion-tension theory of water movement through a plant and why the diameter of the trunk is smallest at midday (6)
A
- diameter of trunk minimal at warmest/brightest time of day - (midday is the warmest and brightest)
- Stomata open in light = more water loss
- Water evaporates even more when warm
- Hydrogen bonds occur between water molecules (cohesion)
- Then adhesion between water molecules and walls of xylem vessels
- Xylem is pulled inwards by the faster flow of water and pulled in by tension
13
Q
C7 - Three ways in which the leaves of xerophytic plants may be adapted to reduce water loss (3)
A
- thick cuticle (wax layer)
^waterproof and impermeable - reduced number of stomata
^reduced surface area for water loss - small leaves
^reduced surface area for water loss
14
Q
C15 - Why are the arrangement of actin and myosin filaments visible under the electron microscope but not under the optical microscope
A
electron microscope has greater resolution
- able to tell two close objects apart better
- electrons have a shorter wavelength
15
Q
C16 - When would glucagon and insulin be released
A
Glucagon: blood glucose levels are below norm
Insulin: blood glucose levels are above the norm
16
Q
C21 - Many diabetics require regular injections of insulin.
Describe how bacteria can be genetically modified to produce human insulin
A
1) cut out insulin gene and cut open plasmid with restriction enzyme
2) use same restriction enzyme on second DNA
3) when cut there are complimentary sticky ends
4) DNA ligase used to join two DNA molecules
5) modified plasmid taken up by bacteria
17
Q
C21 - The fragment of DNA containing the normal allele and the fragment containing the beta-thalassaemia allele moved the same distance on the gel
Explain why
A
- cut at the same base sequence
- same enzyme used
- fragments are the same length/size and have same charge
18
Q
C4 - by what process do potassium ions normally enter a bacterial cell
A
- rate of movement/diffusion proportional to concentration gradient
- High concentration of potassium ions inside cell compared to outside
19
Q
C1 - if you was to draw a peptide bond where the COOH group of one amino acids joins to the NH2 group of another
A
peptide bond between the carbon and nitrogen molecule
OH from COOH is taken away
H from NH2 is taken away
20
Q
C1 - Describe how you would use a bio mechanical test to show that a solution contained protein
A
- Biuret solution
- should turn into a purple colour
21
Q
C1 - What is the R group attached to in the amino acid structure
A
- Carbon molecule
22
Q
C1 - What is the bond between two amino acids
A
peptide bond
23
Q
C1 - What does the term hydroxylating refer to
A
- addition of hydroxyl
- OH group
24
Q
C6 - Describe the part played by the diaphragm in causing air to enter the lungs during breathing
A
- diaphragm contracts
- increases thoracic cavity volume of chest
- Decreases thoracic cavity pressure
- allows air to enter
25
C6 - Table shows the flow of blood to the lungs and to the diaphragm in a seal when it is on land and when it is under water
Explain why the figures in the table would be given per gram of tissue
- allows a comparison in results
- as organs differ in size, larger organs will need more blood
26
C6 - Calculate the percentage by which blood flow to the lungs is reduced when the seal is swimming under water
Lungs:
On land - 0.88
Under water - 0.52
0.88-0.52 = 0.36
0.36/0.88 x100 = 40.9
27
C6 - Explain the advantage to a diving seal if blood is continuing to flow to the lungs
- some oxygen still in lungs
- enters the blood
- helps remove carbon dioxide from the blood
28
C6 - Explain the advantage to a diving seal if a large reduction in blood flow to the diaphragm
- Diaphragm muscles not contracting means that it does not require as much oxygen/glucose
- more blood available for other organs
29
C3 - What is the process by which the protein thyroglobulin is secreted from the cell
exocytosis
30
C3 - function of the rough endoplasmic reticulum
- Joining together of amino acids
- protein synthesis
31
C3 - Explain why the rough endoplasmic reticulum can be seen with an electron microscope and not an optical microscope
- electron microscope has a higher resolution
- uses electrons which have a smaller wavelength
32
C3 - Explain how the shape of a red blood cell allows it to take up a large amount of oxygen in a short time (2)
- large surface area to volume ratio
- short diffusion pathway
- happens quicker
33
C1 - Explain what is meant by the proteins tertiary structure
- way in which the whole protein/polypeptide is folded
34
C1 - What type of chemical reaction is involved in breaking down chitin
hydrolysis
35
C1 - an unusual protein in found in a grain. its function is to act as an enzyme and break down chitin which is found in the walls of the fungus
breakdown of chitin leads to death by osmosis of fungi attacking the grain
Explain how this happens
- water enters the fungus by osmosis which increases the pressure inside
- cell wall no longer strong enough (cell bursts)
36
C1 - an unusual protein in found in a grain. its function is to act as an enzyme and break down chitin which is found in the walls of the fungus
Explain why the protein does not break down the cell walls of the job tears plant
- cell wall of plant is not made of chitin its made out of cellulose
- enzyme is specific to chitin so will not break down cellulose
37
C1 - Explain how heating an enzyme leads to it being denatured
- more kinetic energy
- bonds holding the tertiary structure together break
- Hydrogen, ionic and disulphide bridges
38
C1 - How can protein engineering make enzymes. more stable and less prone to heat denaturation
- change amino acids
- allowing formation of more hydrogen bonds and disulphide bridges
39
C1 - an unusual protein in found in a grain. its function is to act as an enzyme and break down chitin which is found in the walls of the fungus
Describe how the sequence of amino acids in part of the protein from Job's Tears could enable this protein to act as an enzyme inhibitor (6)
- Sequence of amino acids gives the shape
- this is the tertiary structure
- has similar shape to substrate
- competes for active site (competitive)
- fits at site other than the active site (non-competitive)
- distorting active site
- therefore substrate will not fit (active site)
40
C6 - in each cardiac cycle, the arterial pressure has a maximum value. Explain the link between this maximum value and the events of the cardiac cycle
- corresponds to ventricular systole
41
C6 - Graph based on artery
Describe two ways in which you would expect blood pressure in a vein to differ from that in an artery (refer to a graph)
- more constant
- little difference between the maximum and minimum values
42
C6 - Plasma proteins Albumin and Globulin are both not permeable.
Why is it that the water potential of the plasma at the venule end of the capillary is more negative than the water potential at the arteriole end
- plasma proteins are too large to leave the capillary
- therefore water lost and increase in concentration of proteins in the blood plasma
43
C6 - Although the capillary walls are slightly permeable to haemoglobin molecules, there is no haemoglobin in the tissue fluid
explain what causes the absence of haemoglobin in tissue fluid
- haemoglobin is in the red blood cells
- too large to pass through the membrane of the red blood cells
- red blood cells are to large to pass through the wall
44
C3 - in what part of the cell does this part of protein synthesis take place
'Two strands of DNA molecule separate'
nucleus
45
C8 - Order the following events of protein synthesis
A - tRNA molecules bring specific amino acids to the mRNA molecule
B - mRNA nucleotides join with exposed DNA bases and form a molecule of mRNA
C - The two strands of a DNA molecule separate
D - peptide bonds form between the amino acids
E - the mRNA molecule leaves the nucleus
F - a ribosome attaches to the mRNA molecule
C > B > E > F > A > D
46
C8 - Which of these steps takes place in translation
A - tRNA molecules bring specific amino acids to the mRNA molecule
B - mRNA nucleotides join with exposed DNA bases and form a molecule of mRNA
C - The two strands of a DNA molecule separate
D - peptide bonds form between the amino acids
E - the mRNA molecule leaves the nucleus
F - a ribosome attaches to the mRNA molecule
A, D and F
47
C8 - Threonine has the mRNA codon of ACC
What is the DNA base sequence that codes for threonine
TGG
48
C3 - What event takes place during the metaphase of the cell cycle
chromosomes attach to equator (middle of the cell)
49
C3 - Chromosomes coil and shorten
what part of mitosis is this
- prophase
50
C3 - Daughter chromosomes move to the poles of the cell
what part of mitosis is this
Anaphase
51
C3 - What event takes place during the S-phase of the cell cycle
- DNA replication/synthesis
^chromosome copying
52
C3 - Nuclear envelope re-forms
what part of mitosis is this
Telophase
53
C3 - If a Haploid organism has 32 chromosomes present
How many chromosomes will the male gamete have
32
- because the haploid goes through meiosis so there is 64 then split between the female and male gamete
54
C21 - Human DNA cut into separate pieces using a restriction enzyme which produced a staggered cut. Inserted these pieces into plasmids using the same restriction enzyme.
Why will the pieces of human DNA would be able to join to the cut DNA of the plasmids?
- Sticky ends from the cut
- complementary base pairing
55
C21 - other than the restriction enzyme what is the other enzyme the scientist must have used to form recombinant plasmids
ligase
56
C21 - A plasmid may be used as a vector
Explain what is meant by a vector
carrier of DNA
carries into the cell/host/other organism
57
C21 - Molecular Biologists often use plasmids which contain antibiotic resistance genes
Explain the reason for this
- act as marker gene
- allows detection of cells containing plasmid
58
C21 - Explain why the DNA for each species of a shark shows differences in base sequence
- base sequences determine protein sequence
- different species have different protein structure
59
C21 - Each primer is specific to the DNA of one shark species
Explain why a particular primer will only bind to the DNA of one species
- DNA primers have specific DNA sequences
- DNA sequences have specific complementary base pairing
60
C21 - The length of the replicated DNA fragment is different for each species
Explain why this is important in identifying the shark species involved
- Electrophoresis separates DNA
- so they can be identified by position on gel
- smaller/shorter fragments travel further
61
C21 - describe the polymerase chain reaction
- heat DNA
- breaks hydrogen bonds / separates strands
- add primers
- add nucleotides
- cool
- allow binding of nucleotides/primers
- DNA polymerase
- Role of polymerase
- repeat cycle many times
62
C5 - What is an antigen
- protein on the surface of the membrane of a cell
- causes immune response/triggers antibody production
63
C5 - What is a monoclonal antibody
- antibodies all the same/ from one type of plasma
- specific to/ complimentary to antigen
64
C5 - An antigen called PSA is present in the blood of men in the early stages of prostate cancer. Blood test for PSA is done
Why is it that this test detects prostate cancer, but not any other disease?
- antibodies only binds to PSA
- PSA only associated with prostate cancer/ not any other diseases
65
C21 - Describe how altered DNA may lead to cancer
- DNA is altered by mutation
- mutation changes the base sequence of the gene which controls cell growth of the tumour suppressor gene
- changes the protein structure so it's not functional
- tumour suppressor genes produce proteins that inhibit cell division
- mitosis
- uncontrolled abnormal growth
- maligant tumour
66
C21 - Explain why fragments of DNA from cancer cells may be present in blood plasma
- cancer cells die/ break open
- release DNA
67
C21 - Fragments of DNA can now be detected by a new test in which a test strip containing nucleic acid binds to sections of altered DNA.
Explain why the nucleic acid on the test strip will only bind to altered DNA
- Normal DNA and changed DNA have different sequences
- DNA only binds to complementary sequence
68
C21 - explain why cancer is more likely to be treated successfully if the disease is detected at a very early stage
- fewer cancerous cells
- less cell damage
69
C21 - another technique involves removing a tissue sample from a patient. the tissue sample is used to obtain mRNA
Explain how examining mRNA enables scientists to discover wether cancer is present
- mRNA base sequence has changed
- DNA structure is different
- cancer gene active/ tumour suppressor gene inactive
70
C15 - What is the biological importance of reducing the chromosomes number when the cell divides by meiosis
- Later fertilisation
- restoring diploid to original number
- not doubling chromosomes number
71
C15 - Three Differences between meiosis and mitosis
- genetically different (meiosis), genetically identical (mitosis)
- Two divisions - 4 offspring cells (meiosis), one division - 2 offspring cells (mitosis)
- Homologous chromosomes associate in pairs (meiosis), Homologues are independent (do not pair)
72
C19 - Mayfly nymphs were collected by disturbing the gravel of a stream bed.
How would you have collected the samples in order to ensure they were representative of the habitats being investigated and could be compared with each other
- samples collected at random
- method for choosing random sites (coordinates ...)
other methods
- same size of net
- use of quadrat
- quadrat of the same size
73
C19 -
Family Caenidae: deep water (SD - 7.92
Family Caenidae: shallow water (SD - 1.51)
Family Baenidae: deep water (SD - 1.51)
Family Baenidae: shallow water (SD - 6.72)
Which one of the four samples showed the greatest variation within the sample
- Family Caenidae: deep water
- highest standard of deviation
74
C19 - What is meant by the term ecological niche
- an organism's role/ in the ecosystem/ community
75
C19 -
Family Caenidae: deep water (mean amount - 12.88)
Family Caenidae: shallow water (mean amount - 2.38)
Family Baenidae: deep water (mean amount - 6)
Family Baenidae: shallow water (mean amount - 24.5)
Using this evidence describe why the two families of mayflies occupy different ecological niches and the advantages
- Caeridae found mainly in deep water
- Baetidae found in shallow water
+reduces competition for food/shelter
+ensure both types survive otherwise the better adapted type will displace the other
76
C11 - In which part of a chloroplast does the light dependant reaction occur
- grana / thylakoids / lamellae
77
C11 - The role of the chlorophyll in photolysis
- Absorbs light energy
- OXIDISED: loses electrons and becomes positively charged
- accepts electrons from water which causes more water to dissociate
78
C11 - How the energy of light is converted into chemical energy in the light dependent reactions
- Electrons raised to a higher energy level (electrons excited)
- use of electron carriers
- for production of ACT
79
C18 - How can resistance suddenly appear in an isolated population of rats which has never before been exposed to wafarin
- spontaneous mutation
- change in gene and DNA sequence
80
C18 - Give two assumptions that must be made when using the Hardy-Weinberg equation
- no mutation
- no immigration
- large population
- random mating
81
C7 - Haemoglobin is a protein with a quaternary structure.
What is meant by a quaternary structure
- structure resulting from aggregation of several polypeptide chains (tertiary structures)
- haemoglobin has 4 polypeptide chains
82
C7 - Explain how oxygen in a red blood cell is made available for respiration in active tissues
- lower pH due to increased CO2 because of an increased respiration
- increases dissociation of haemoglobin
- oxygen diffuses from red blood cell to tissues
83
C7 - Haemoglobin is broken down in the liver. One product of this breakdown is amino acids.
give one use in the body of these amino acids
- use in respiration/ used in protein synthesis
84
C14 - Explain how the release of acetylcholine at an excitatory synapse reduces the membrane potential of the postsynaptic membrane
- binds to receptor (opens sodium ion channels)
- sodium ions enter and make membrane potential less negative/depolarised
85
C14 - Explain what causes transmission at a synapse to occur in only one direction
- vesicles containing neurotransmitter only in presynaptic membrane neurone
- receptor only located in postsynaptic membrane
86
C14 - GABA is a neurotransmitter which inhibits the production of action potentials
When the postsynaptic membrane is stimulated by acetylcholine, an action potential is less likely if GABA is released at the same time
Explain why
- GABA opens potassium and chloride channels so potassium passes out and chloride passes in
- membrane potential more negative (hyperpolarised)
- requires increased stimulation/ must open more sodium ion channels to allow more sodium ions to enter
^in order to reach the threshold
87
C16 - The graph shows changes in plasma glucose concentration that occurred in a person who went without food for some time
Explain the role of negative feedback in the control of plasma glucose concentration
- Deviation of value from norm initiates corrective mechanisms
- Fluctuations in plasma glucose concentration detected by hypothalamus/ islet cells in pancreas
- initial decrease (because no food given) stimulates increased secretion of glucagon
- increases in plasma glucose stimulate increased secretion of insulin from beta cells
-interconversion of glycogen increases or decreases uptake of glucose by cells (change in membrane permeability)
88
C16 - How does maintaining a constant body temp allow metabolic reactions in cells to proceed with maximum efficiency
- 37 degrees body temp is optimum temp for enzymes
- excess heat denatures enzymes by altering the shape of the tertiary structure so the enzyme cannot bind to the active site
- reactions slowed down
- too little reduces kinetic energy of molecules so they move slowly
- fewer collisions so fewer enzyme substrate complexes formed
89
C1 - Lactose is a disaccharide found in milk. In the small intestine, it is digested into glucose and galactose by the enzyme lactase
- Hydrolysis of the disaccharide lactose forms a monosaccharide galactose
90
C1 - Name one other digestive enzyme that is located in the plasma membranes of cells lining the small intestine
- maltase
- dipeptidase
90
C1 - What is an advantage of lactase and other digestive enzymes being located in the plasma membranes of cells lining the small intestine, rather than being secreted into the lumen of the small intestine
- So enzymes are not lost with gut contents
- more effective absorption of products formed by these enzymes
91
C1 - the absorption of galactose from the small intestine is reduced if the absorbing cells are treated with a respiratory inhibitor, such as cyanide
Explain this
- No ATP formed, no energy released by respiration
- ATP is used for active transport of galactose into cells
92
C15 - Explain the role of calcium ions and ATP in bringing about a contraction of muscle fibre
Calcium:
-bind to troponin
- remove blocking action of tropomyosin
- expose myosin binding sites
ATP:
- Allows myosin to detach from actin to break cross bridge
- releases energy to recock myosin head and drive a power stroke
93
C15 - Explain how an action potential passes along the motor neurone to the neuromuscular junction
- depolarisation of axon membrane (influx of sodium ions establishes local currents)
- Change permeability to sodium ions gates of adjoining region
- adjoining region depolarises
94
C15 - When the action potential arrives at the neuromuscular junction, it results in the secretion of acetylcholine into the synaptic cleft
Explain how
- Depolarisation of presynaptic membrane
- Calcium 2+ ion channels open, increased permeability so there is an influx of calcium 2+ ions
- causes vesicles to move towards and fuse with the presynaptic membrane
95
C15 - Muscle mass is lost as you get older. Most of the muscle lost consists of fast fibres
Explain why explosive exercises, such as sprinting and weightlifting, will be more affected by this muscle loss than aerobic exercises, such as jogging
- Fast fibres used for explosive movements
96
C6 - Is glucose present in the tissue fluid and blood plasma
yes
97
C6 - are sodium ions present un the tissue fluid and blood plasma
- yes
98
C6 - Is haemoglobin present in the tissue fluid and blood plasma
- no
99
C6 - how does the hydrostatic pressure of the blood at the arteriole end of the capillary help to form tissue fluid
Explain how
- Hydrostatic pressure higher than osmotic effect
- forces and pushes out small molecules such as ions (sodium and potassium)
100
C6 - Describe how the increase in pressure of the blood in the artery results from the events in the cardiac cycle
- Caused by blood leaving the heart and entering the artery
- as a result of the ventricles contracting
101
C6 - Elastin fibres in the wall of the artery help smooth out the flow of blood.
What happens to these fibres as the pressure of the blood in the artery changes
- they stretch as pressure increases
- recoil back as pressure drops
102
C6 - Give one way in which the structure of the wall of an artery is similar to the structure of the wall of a capillary
- Both have an epithelial cell and endothelial cell
103
C6 - Why are both the heart and arteries described as organs
- Contain different, more than one tissue/ which is made from similar cells
104
C6 - Give two risk factors associated with coronary heart disease
- smoking
- high salt intake
- high blood cholesterol
- lack of exercise
105
C6 - During a myocardial infarction, areas of heart muscle begin to die
Explain why
- reduced blood flow in coronary artery as it is blocked
- less or not enough oxygen
- respiration drops/stops and cells start to die
106
C3 - Two structures that are adaptations of this cell that make possible the rapid absorption of glucose and explain why
- Microvilli (increases larger surface area - diffusion)
- mitochondria (to supply ATP for active transport)
- Vesicles to bring carrier proteins to membrane
107
C3 - The absorption of glucose into the cell leads to the movement of water into the cell
Explain how
- Lowers the water potential of the cell
- so water moves in by osmosis
108
C7 - Tuberculosis leads to permanent changes in the gas exchange system. these changes include fibrosis
FEV - maximum volume a person can breathe out in 1 second
Explain how fibrosis caused by tuberculosis could have produced the changes in FEV of the miners
Fibrosis causes;
- development of scar tissue
- loss of elasticity in lungs
- restricts inhalation
- breathe out less/reduced FEV due to reduced lung elasticity
109
C19 - what is selective breeding
- Organisms with certain/chosen characteristics/alleles
110
C19 - explain one potential problem that can be caused by selective breeding
- inbreeding (breed closely related organisms
- higher probability of mutations being passed on
- reduced genetic variation/ reduced diversity and smaller gene pool (less variety of alleles)
- So can be killed by a disease, reduces ability to adapt and survive
111
C10 - All primates produce a species-specific type of haemoglobin. An antibody against human haemoglobin could be used to compare the evolutionary relationships between different primates.
Describe and explain how
- antibody and haemoglobin in the blood of different primates
- precipitate complex band formed
- amount of precipitate/ complex thickness of band shows relationship
112
