Diagram of AQA A-Level Biology Paper 2 | Quizlet

Alleles that are neither dominant nor recessive to one another, so both alleles are always expressed in the phenotype.

Codominant

The inheritance of a single gene

Monohybrid inheritance

Physical, behavioural, biochemical expression of an organisms genotype

Phenotype

The type of genes an individual has

Genotype

Alles that is always expressed in the phenotype

Dominant

Only expressed in the phenotype when homozygous

Recessive

both alleles are the same

Homozygous

Both alleles for a specific gene are different

Heterozygous

Position of a gene on a chromosome

Loci

A set of instructions for a specific polypeptide

Gene

Different forms of a gene

Allele

1. Expected ratios are probability

2. Sexual reproduction is random due to random fusion of gametes and random assortment homologous chromosomes.

3. Small sample size

4. Linked genes

Suggest four reasons why observed ratios are not the same as expected ratios (4).

1. Homologous chromosomes pair up

2. Crossing over / chiasmata form;

3. Produces new combination of alleles

4. Chromosomes separate at random

5. This produces varying combinations of genes

6. Chromatids separated at meiosis II

Meiosis results in cells that have the haploid number of chromosomes and show genetic variation. Explain how. (6)

1. Refer to the specific individuals (using their number)

2. Explain what happened with the genes (passed on recessive/dominant)

3. Describe the genotype of your examples and mention their phenotype (homozygous/heterozygous etc)

Pedigree Questions Mark Scheme (3)

Group of organisms of the same species occupying a particular space at a particular time that can potentially interbreed to produce fertile offspring.

Populations

The total number of genes of every individual in an interbreeding population.

Gene pool

How often an allele appears in a population.

desired allele/total alleles = allele frequency (decimal form)

Allele frequency [definition][Equation] (2)

No emigration or immigration

No mutations

Mating is random

No natural selection

Hardy-Weingberg Assumptions (4)

1. Selection pressure exists in an environment (name it)

2. Variation exists in stated phenotype of organism/ mutation occurs creating new alleles (name the allele if applicable)

3. Some individuals have the selective advantage (describe it)

4. Produces differential survival/ organisms with successful alleles more likely to survive

5. Natural selection occurs via directional selection

6. Survivors breed and pass on alleles to offspring

7. Over time, there is a change in allele frequency

Natural selection MS (7)

Favours the mean phenotype. (Normal distribution becomes narrower)

Stabilising selection

Favours one extreme end. (Normal distribution translates that way)

Directional Selection

Favours both extreme phenotypes. Mean is at disadvantage and dies. Can create two new species

Disruptive Selection

1. Natural disaster

2. Few survive

3. New population develops with different allele

4. Frequency to the original

Genetic Drift: Bottleneck Effect (4)

Genetic drift that occurs after a small number of individuals colonize a new area. Allele frequency is different to the original

Founder effect

1. Populations geographically separated (formation of a river) (Allopatric Speciation)

2. Separated populations now unable to reproduce

3. Different environments have different selective pressures so each population will accumulate different beneficial mutations over time to help them survive so change in allele frequency

4. Two populations become so genetically different that they cannot reproduce to form fertile offspring, so are now classed as different species

Allopatric speciation (4)

Populations live in the same region but occupy different habitats

Reproductive isolation: Ecological

Same area but are sexually mature at different times

Reproductive isolation: Seasonal

Different species ensure successful mating by specific courtship (bird dances). As this is genetic, mutation can change this

Reproductive Isolation: Behavioural

Plants exhibit pheromones to any mate with their own species

Reproductive isolation: Incompatibility

Lack of fit between sexual organs in insects

Reproductive Isolation: Incompatibility in Arthropods

1. Occurs in the same habitat so sympatric speciation

2. Mutation causes different flowering times

3. Seasonal Reproductive separation

4. No gene flow between the organisms so different alleles passed on.

5. Disruptive natural selection

6. Eventually, different species cannot interbreed to produce fertile offspring;

Lord Howe Island in the Tasman Sea possesses two species of palm tree which have arisen via sympatric speciation. The two species diverged from each other after the island was formed 6.5 million years ago. The flowering times of the two species are different.

Using this information, suggest how these two species of palm tree arose by sympatric speciation. (6)

1. Colonisation by pioneer species

2. Change in environment by organisms present

3. Enables other species to survive as the environment becomes less hostile

4. Increase in biodiversity

5. Stability increases

6. Leading to a climax community

Succession occurs in natural ecosystems. Describe and explain how succession occurs. (6)

1. Collect sample, mark and release

2. Ensure marking does not affect survival of the fish

3. Allow time for fish to randomly distribute before collecting the second sample (decide time based on question)

4. Population = (first sample x second sample )/ total recaptured in 2nd sample

The mark-release-recapture method can be used to estimate the size of a fish population (4)

1. Same species present over long time

2. Populations stable (around carrying capacity)

Give two features of a climax community (2)

Maximum population size that a particular environment can support due to abiotic and biotic factors.

Carrying capacity (k)

All the different organisms interacting in a particular place at a particular time

Community

The biotic and abiotic environment of a specific area and their interactions

Ecosystem

Position an organism occupies in an ecosystem

Ecological Niche

The niche species could potentially occupy.

Fundamental Niche

Due to interactions with other organisms where they are outcompeted, species occupy a smaller area than the fundamental niche

Realised Niche

1. Use tapes placed at right angles to produce a grid

2. Use a random number table to generate pairs of numbers to be used as coordinates for areas on the grid.

3. Sample in those areas using a quadrat

4. Use at least 30 quadrats so that sample is representative

5. Count all squares occupied by each species being sampled (percentage cover) in total and divide by 30 to get mean percentage cover.

Random Sampling MS (same abiotic factors, like the sand dunes in Wales) (5) [Make your answer specific to the question]

1. Use tape laid across the ecosystem- a transect

2. Sample through all the areas that show differences

3. Take samples at regular intervals along the transect (e.g quadrants every 5 metres)

4. Count all squares occupied by each species being sampled (percentage cover)

5. Kite diagrams

Systematic sampling MS (Different abiotic factors, Rocky shore at Wales) (5) [Make answer specific to the question]

1. Reduce sample distance in half

2. If sample pattern of zonation remains, then sample interval is okay

How do you know if your sample distance is okay ? (2)

1. Useful for areas with dense vegetation

2. Vertical zonation

Point quadrats (2)

No soil

primary succession

Succession following a disturbance that destroys a community without destroying the soil (contains seeds often)

Secondary succession

1. A community that remains stable only because human activity prevents succession.

2. Animals grazing the land preventing woodlands establishing

3. Conservation programmes.

Define deflected succession and provide 2 examples. (3)

1. Chlorophyll absorbs light which excites electrons

2. And causes them to be lost

Describe what happens during photoionisation in the LDR. (2)

1. Reduced transfer of protons across thylakoid membrane

2. Less ATP produced

3. Less NADPH produced

4. Light-independent reactions slows

Atrazine binds to proteins in the electron transfer chain in chloroplasts of weeds, reducing the transfer of electrons down the chain.

Explain how this reduces the rate of photosynthesis in weeds. (4)

1. Phosphorylation of glucose using ATP;

2. Oxidation of triose phosphate to pyruvate

3. Net gain of ATP

4. NAD reduced

Describe the process of Glycolysis (4)

1. Pyruvate —> Lactate whilst oxidising NADH so it can enter glycolysis.

Pyruvate —> ethanol in plants

Anaerobic respiration [Animals] [Plants] (2)

1. Excessive algae growth due to leached nitrates.

2. Plants below cannot photosynthesise and die.

3. Aerobic bacteria feed on fish in water. They respire and use up oxygen in water.

4. Fish and other aquatic organisms die due to lack of oxygen in water so they cannot respire.

Eutrophication (4)

I - (F+R)

Net production

Nitrogen fixing bacteria found in soil or in leguminous plants convert N2(g) —> NH4+ (in soil)

Nitrogen Fixation (1)

Organic N compounds decomposed to NH4+ by saprobionts.

Ammonification (2)

NH4+ --> NO2- --> NO3- by nitrifying bacteria.

Nitrification (3)

Anaerobic denitrifying bacteria converts nitrogen compounds back into N2(g). Prevent this by aerating soil.

Denitrification (4)

1. CO2 reacts with Ribulose bisphosphate

2. Produces 2 Glycerate 3-phosphate molecules using rubisco.

3. Glycerate 3 phosphate reduced to triose phosphate

4. Using reduced NADP

5. Using energy from ATP

6. Out of 6 Triose phosphate molecules, 5 are recycled into the calvin cycle (light independent reaction) and 1 is saved to form glucose.

Describe the light-independent reaction of photosynthesis. (6)

1. Energy is lost between trophic levels

2. Energy is lost via excretion and respiration

Farming cattle for humans to eat is less efficient than farming crops because of energy transfer. Explain why. (2)

1. Constant volume of oxygen

2. Constant concentration of respiratory substrate

3. Constant concentration of bacteria

4. Time must be maintained as constant for each dish to incubate

Apart from temperature and pH, give two variables the scientist would have controlled when preparing the liquid medium cultures. (2)

1. Cells that desired protein is expressed in are selected and should contain a large amount of mRNA for the protein

2. Identify and extract mRNA for desired proteins

3. Use reverse transcriptase to convert mRNA —> cDNA (complementary to the mRNA template and single stranded)

4. DNA polymerase used to make cDNA double stranded

5. Insert gene into plasmid using restriction and ligase enzymes

6. Insert plasmid into bacterium and clone it.

Using reverse transcriptase to make a specifically selected protein (6)

1. Order of animo acids in protein sequenced allowing mRNA to be determined, therefore cDNA to be sequenced

2. Feed cDNA into computer

3. Check sequence passes international standard biosafety

4. Computer designs series of small overlapping strands of single stranded DNA called oligonucleotides which are assembled into desired gene

5. Oligonucleotides joined together to make gene and gene replicated using PCR

6. PCR also makes copied genes double stranded

7. Gene inserted into bacterial cell or other organism (in vivo cloning)

8. Genes in transformed organisms checked using genetic sequencing. Those containing errors are rejected

Outline the method used to manufacture genes (Gene machine) (8)

1. DNA fragments - to be copied

2. DNA polymerase - obtained from bacteria Thermus Aquaticus that lives in hot springs making it thermostable

3. Primers - Short sequences of nucleotides called oligonucleotides that have a set of bases that are complementary to those at each end of the DNA fragments

4. Nucleotides - Containing all four bases

Requirements for PCR (4)

A short strand of DNA that is complementary to a known DNA sequence (e.g a mutated allele) labelled with a fluorescent or radioactive tag that makes it identifiable.

DNA Probe

1. Thermal cycler heats to 95 degrees celsius

2. 2 separated strands now used as templates for building complementary strands

In Vitro Step 1 (Template DNA double stranded —> single stranded) (2)

1. Cycler cools the mixture to 55 degrees celsius, allowing hydrogen bonds to reform.

2. Primers attach to ends of desired sequence in DNA fragments. Primers allow the attachment of DNA polymerase

In Vitro Step 2 (Annealing the Primer) (2)

1. Cycler raises the temperature of DNA to 72 degrees celsius (optimum temp for Thermus Aquaticus DNA polymerase)

2. DNA polymerase attaches to primer and synthesises new strands that are complementary to template DNA

3. Two new double stranded DNA fragments are formed.

4. PCR is repeated around 25 times creating over 50 million copies of template DNA

In Vitro Step 3: Synthesis (4)

2^x where x = number of cycles

Number of DNA fragments produced =

1. Easily manipulated

2. Restriction sites well known

3. Easily taken up by cells in culture as bacteria absorb plasmids

Why are bacterial plasmids used in In Vivo cloning ? (3)

1. Target DNA (spliced prior so introns removed) is cut using restriction endonucleases to create fragments with sticky ends.

2. Promoter and terminator regions (STOP codon) are added allowing transcription.

3. Same restriction endonuclease is used to cut open plasmid (vector) to allow complementary sticky ends.

4. DNA ligase is used to incorporate the DNA fragment into the plasmid, forming recombinant DNA

5. Recombinant DNA is taken up by bacteria in Ca2+ solution to increase membrane permeability, forming a transformed cell.

Preparation and Insertion of DNA fragments (5)

1. Genes for ampicillin and tetracyline resistance added before human gene is added.

2. Human gene inserted in the middle of tetracycline gene, omitting tetracycline resistance.

3. Bacteria that can grow on ampicillin but not tetracycline have recombinant plasmid.

Identifying Transformed Bacteria Cells (Replica plating) (3)

1. Human gene is inserted in the middle of GFP gene (isolated from jellyfish)

2. GFP no longer made in transformed cells but have ampicillin resistance so is grown on agar plate

3. So transformed cells are not fluorescent

Identifying transformed Bacteria Cells (Green Fluorescent Protein) (3)

1. Human gene inserted between B-galactosidase gene.

2. B-galactosidase + lactose analogue —> blue

3. Transformed colonies are not blue

Identifying transformed Bacteria Cells (B-galactosidase and lactose analogue) (3)

1. Nucleotides with radioactive phosphate.

2. Probe identified using photographic plate.

Radioactively Labelled Probes (2)

1. Emit light under specified conditions

Fluorescent Labelled Probes

1. Gel plate is prepared to hold gel matrix. Gel comb is used to create pores in the agarose gel.

2. Gel tray placed in electrophoresis chamber and is filled with a buffer (allowing electric current to flow through).

3. This separates the DNA fragments, as DNA is negatively charged.

4. DNA fragments migrate towards positive electrode.

5. Gel has fine pores, so small fragments move faster than large ones.

6. Results in separated fragments in the form of bands, with the smallest fragments nearest to the positive electrode.

Gel Electrophoresis (6)

1. After gel electrophoresis, bands are transferred to nylon membrane as gel is fragile.

2. DNA probe is made complementary to target gene by DNA sequencing techniques (known genotypes)

3. Quantity of DNA probe increased by PCR

4. DNA tested is turned into single stranded DNA

5. Strands mix with copies of DNA probe causing DNA probes to bind to complementary bases on strands causing DNA hybridisation.

6. DNA is then washed clean of any unattached probes, and marked DNA is identified.

Southern Blots (6)

Fluorescent insitu hybridisation probes can be used to identify mutations of nearly any human gene.

FISH Probes

1. Heritable changes in gene function

2. Without changes to the base sequence of DNA

Define what is meant by epigenetics (2)

1. Methyl groups added to a tumour suppressor gene;

2. The transcription of tumour suppressor genes is inhibited

3. Leading to uncontrolled cell division.

Explain how increased methylation could lead to cancer. (3)

1. EGCG binds to active site of DNMT

2. DNMT cannot methylate promoter region of tumour suppressor gene

3. Transcription factors can bind to promoter region

4. RNA polymerase stimulated

Increased methylation of the promoter region of a tumour suppressor gene causes one type of human throat cancer. In this type of throat cancer, cancer cells are able to pass on the increased methylation to daughter cells. The methylation is caused by an enzyme called DNMT. Scientists have found that a chemical in green tea, called EGCG, is a competitive inhibitor of DNMT. EGCG enables daughter cells to produce messenger RNA (mRNA) from the tumour suppressor gene.

Suggest how EGCG allows the production of mRNA in daughter cells. (3)

1. Chemoreceptors in aorta and carotid artery detect decrease in pH.

2. Send impulses to medulla oblongata

3. More impulses to SAN

4. By sympathetic nerve

5. Resulting in higher heart rate

Exercise causes an increase in heart rate. Describe the role of receptors and of the nervous system in this process. (4)

1. Electrical activity only through Bundle of His

2. Wave of electrical activity passes through both ventricles at the same time

When the heart beats, both ventricles contract at the same time.

Explain how this is coordinated in the heart after initiation of the heartbeat by the SAN. (2)

1. Saltatory conduction (depolarisation occurs along whole length of axon)

2. Nerve impulses slowed

3. Less impulses converted into neurotransmitters at neuromuscular junction

Damage to the myelin sheath can cause muscular paralysis. Explain how. (3)

1. SAN releases wave of depolarisation across atria, causing atria to contract

2. Wave of depolarisation reaches AVN

3. AVN releases another wave of depolarisation

4. Bundle of His transmits waves down the septum to the purkyne fibres in the ventricle walls

5. Before ventricles contract, delay occurs to make sure ventricles fill with blood.

SAN —> AVN (5)

1. Detected by baroreceptors in the wall of the aorta and the carotid artery.

2. Impulses sent to the medulla oblongata, then back to the SAN via parasympathetic nerve.

3. Decrease in frequency of electrical signals, so heart rate decreases to ensure blood vessels do not burst

Heart rate control: Increase in Pressure (3)

1. Detected by baroreceptors in the aorta and carotid artery.

2. Impulses sent to the medulla oblongata and back to the SAN via sympathetic nerve

3. Increase in frequency of electrical signals, so heart rate increases.

Heart rate control: Decrease in Pressure (3)

1. High visual acuity

2. Each cone is connected to a single neurone

3. Cones send separate sets of impulses to brain

The fovea of the eye of an eagle has a high density of cones. Explain how the fovea enables an eagle to see its prey in detail. (3)

1. High visual sensitivity

2. Several rods connected to a single neurone

3. Spatial summation to overcome threshold

The retina of an owl has a high density of rod cells. Explain how this enables an owl to hunt its prey at night. (3)

1. High visual sensitivity

2. Can detect light in low light intensity as many rod cells are connected to a single bipolar cell (spatial summation)

3. Low visual acuity (quality of vision)

4. Rhodopsin

Rod Characteristics (The Where)

1. Most concentrated in fovea

2. High visual acuity

3. Can only respond to high light intensity due to each cone being connected to one bipolar cell

4. Different types of iodopsin (red, green and blue) to absorb different wave lengths of light

Cone Characteristics (The What)

1. Membrane more permeable to K+ and less permeable to Na+.

2. 2 K+ pumped in and 3 Na+ actively transported out.

Explain how the resting potential of -70 mV is maintained in the sensory neurone when no pressure is applied.

1. Pressure causes membrane to become stretched

2. Sodium ion channels in membrane open and sodium ions move in

3. Greater pressure means more channels open. so more sodium ions enter.

Explain how applying pressure to the Pacinian corpuscle produces the changes in membrane potential recorded by microelectrode P. (3)

1. Threshold reached

2. Which causes maximal response/ All or nothing principle

Why is membrane potential never higher than it is ? (2)

1. IAA diffuses to the shaded side of the shoot promoting growth and cell elongation

2. Causes the shoot to bend towards the light source (positive phototropism)

Phototropism in Shoots (2)

1. IAA diffuses towards the shaded side of the root and inhibits growth and cell elongation.

2. Root bends away from the light source (negative phototropism)

Phototropism in Roots (2)

1. IAA diffuses towards the bottom due to gravity

2. Increase in IAA concentration promotes growth and cell elongation, so the shoot tends and growths against gravity (negative geotropism)

Geotropism in Shoots (2)

1. IAA diffuses to the bottom due to gravity

2. IAA inhibits cell elongation in the roots so the roots bend and grow downward (positive geotropism)

Geotropsim in Roots (2)

Directional response in movement due to stimuli

Positive = towards stimuli

Negative = Against stimuli

Taxes

Non directional response in movement towards a stimulus. (Humans moving around in the cold to stay warm)

Kineses