Paper 2 24 Flashcards
name the structures at the neuromuscular junction
A= Mitochondrion
B= Pre synaptic membrane
C= Synaptic vesicle
D= Synaptic cleft/gap
Following the release of acetylcholine into a neuromuscular junction, a muscle contraction occurs.
Describe the sequence of events, following the release of acetylcholine, that leads to stimulation of this contraction. (4)
Do not include release of calcium ions in the myofibril.
- Acetylcholine DIFFUSES (across synaptic cleft)
- Acetylcholine attaches to RECEPTOR on the sarcolemma
- Sodium IONS enter leading to depolarisation/action potential
- Calcium (ions) released by endoplasmic reticulum
Inhibitory synapses causes depolarisation in post synaptic neurones.
Explain how this inhibits synaptic transmission (2)
- (Inside postsynaptic neurone/membrane) is more negative
- SO ONLY More sodium ions (required to enter) for depolarisation/ action potential
What happens when sarcomere contracts? (1)
3rd box: H-zone and I-band decreases and A-band has no change
explain why there is a higher glycogen concentration in fast muscles than slow muscle (4)
- Fast fibres CONTRACT quickly whereas slow fibres CONTRACT slowly
- Fast fibres mainly use anaerobic respiration, slow fibres use aerobic respiration
- Fast fibres provides ATP for energy for short intense activity
- Glycogen hydrolysed to glucose (glycogenolysis)
How are products of light dependent reactions used in light-independent reaction.
(3)
Do not include details about the activity of RuBP.
- Light-dependent reaction uses the reduced NADP and ATP from the light dependent reaction
- To form simple sugars (GP reduced to triose phosphate)
- The hydrolysis of ATP in the dependent reaction provides energy for independent reaction
Conclude and evaluate: how does absorption and colour affect the rate of photosynthesis (2)
- % of light correlates with rate of photosynthesis
- HIGHER rate of photosynthesis in RED **and ** BLUE and lower rate of photosynthesis with green light
Name 3 factors that need to be controlled in photosynthesis (2)
1- CO2 Conc
2- temperature
3- mineral ion conc in soil (water)
4- light intensity
5- humidity
1.Dominant allele for grey body G and recessive allele for black body is g.
Explain how you would determine if the genotype of a grey fly is homozygous or heterozygous for body colour (2)
- In fruit flies, male is XY and females are XX. A cross between a grey bodied male fly and black bodied female fly produced some black bodied females.
explain how this shows that the gene for body colour is not x-linked (1)
- cross with black fly (homozygous recessive)
- Black offspring = heterozygous (Gg)
Grey offspring= homozygous (GG)
_________________________________________ - if sex-linked, female offspring only inherits dominant allele for grey body so will be grey bodied
how can sampling be used to estimate the number of fat hen plants in the whole field (5)
- use random number generator to generate coordinates
- Place at least 10 squares down at each of the random coordinates
- In each quadrat count as many species as you can see and calculate the mean
- Convert this to no of plants per m2and multiply by area of the field
- Repeat until you gather a representative sample of the area
In potato plant fields, fat hen plants can grow up to a height of 2M and absorb large quantities. Name the type of competition occuring between fat hen plants and potato plants and why would there be a decrease in potato competition? (3)
- Interspecific competition
- Less Nitrates/nitrogen to produce amino acids/proteins/ DNA/RNA/ATP (as wild hen plants use most of the ions)
- the fat hen plant is 2M tall so it blocks sunlight SO less/no photosynthesis
describe and explain relationship between IAA concentration growth rate and cell plasticity (2)
- higher conc of IAA = greater growth/ elongation and plasticity
- decrease in pH = cell wall loosens and stretches
(draw tangent to graph?)
Suggest how you could determine the size of the different DNA fragments produced in these experiments? (2)
- Use a “ladder” aka a DNA fragment with known lengths sizes / number of base pairs at known positions.
- The bands separate out, heaviest first, so can compare to “ladder” row to see which bands are similar/different
describe how restriction endonuclease and DNA ligase are used to insert a gene into a plasmid (2)
- Restriction endonuclease/ enzymes cut plasmid
- DNA ligase joins sticky ends and ligase forms phosphodiester binds
Increasing IAA concentration increases cell wall plasticity
Increase in plasticity results in CELL elongation
1 part/cm^3 of 10-1 stock to 9 parts of distilled water, then 1 part/cm^3 of 10^-2 to 9 parts of distilled water
Describe a suitable procedure of the student could follow to investigate the effect of different concentrations of GA on the growth of stem segments (5)
include the variables that need to be controlled
M1 - 1 dish with distilled water (control) **and ** different concs of GA and 10 stem segments in each
M2 - Cut the stems as the same length and measure length of segments at start and end of investigation
M3- Place stems in same volume of GA
M4- leave for same period of time
M5- same temp
M6- same species/ type of plant
M7- same age of plant
M8- same diameter/thickness of plant
Give two reasons why a weight loss programme could be used to treat type II why and not type I diabetes? (2)
Type 1= (β cells in islets of langerhans
in pancreas) produce insufficient insulin
Type 2 = receptors are less sensitive to insulin / insulin ‘resistant’
M3: - weight is **not ** linked TO TYPE 2
Extra info
(● So fewer glucose transport proteins → less uptake of glucose → less conversion of glucose to glycogen)
The design of this investigation helps to support the validity of any conclusions obtained.
Suggest and explain three features of this investigation (weight loss) that justify this statement (3)
Large same size so its reliable and representative sample.
Wide range of ages so age range is representative and age is not a factor
Computer generated list so bias avoided
Control group used socan compare effects of weight loss to no weight loss
Done over 2 years so+ long term effects known - reliable
Describe the myogenic stimulation of the heart and how the regular contraction of the atria and ventricles is coordinated (5)
Do not include the autonomic nervous system in your answer
- SAN releases wave of electrical activity
- SO atria to contract at the same time
- AVN relays/ passes electrical activity after a short delay
- via Purkyne tissue in/and bundle of His
- So ventricles contract (at the same time from bottom upwards)
What does a 0.05 significance suggest? (3)
Below 0.05 means significant increased risk of disease with factor. So more than 0.95 probability that results were significant and not due to chance
Hyperthyroidism, age and the high blood pressure were significant impact.
Suggest two factors that are the causes for phenotypic diversity (2)
- Mutations/ alleles
- Crossing over (during meiosis resulting in new combinations of alleles)
- Independent Segregation ( of homologous chromosomes)
- environment/ habitat
- Epigenetics
- Random fertilisation
Why is it advantageous to be able to camouflage based on surroundings? (2)
Camouflage prevents predation
More likely to survive, reproduce fertile offspring with advantageous allele.
- potential successful mating
The scientists concluded that it was probable that disruptive selection was leading to sympatric speciation in the wall gecko. Evaluate this conclusion (5)
M1 (REQUIRED FOR MAX MARKS)- Geckos in same habitat/ geographical location (no geographical isolation between geckos)
M2- Possible allopatric speciation as different habitats / environment
M3: could lead to separate gene pools
M4: Mutations
M5: Disruptive selection for both of the two extremes
M6: Diurnal geckos are a distinct genetic group
M7: DNA suggests the geckos are the same species
DEF:
- disruptive selection = ● Organisms with alleles coding for either extreme variation of a trait have a selective advantage
- sympatric selection= Population is not geographically isolated so prevents interbreeding
