Spoiler
1.
a. Hydrophilic = glucose, glucose enters cells for the generation of energy (ATP) water, for use in metabolic rxns
Hydrophobic = carbon dioxide, carbon dioxide diffuses into a plant cell for photosynthesis
b. Outline the process of endocytosis or exocytosis. e.g. exocytosis is the energy dependent mode of bulk transport out of a cell, involving the fusing of a secretory vesicle with the plasma membrane and the subsequent release of its contents into the extracellular environment.
2.
a. Note that this question asks us to give an example of a type of protein, so I assume anything such as transport, enzyme etc would be accepted. If you did have a specific example that would be even better. One suggested response could simply be – enzymes are proteins that reduce the activation energy required for reactions, thereby increasing their rate. Enzymes, such as those responsible for anaerobic respiration, may increase in environments requiring high energy use (during replication) or stress to increase the production of energy or ATP.
b. Pre-mRNA, mRNA, Polypeptide chain, Amino acid
c. Regulatory genes influence the expression of other genes, therefore they will be ADR1 which acts as a transcription factor influencing the transcription of that particular gene, and GCN4 which controls the activity of RNA polymerase, influencing transcription of that gene.
d.Exons, coding regions of DNA.
3.
a. Glucose is an input of cellular respiration to produce ATP, therefore, by measuring the change in glucose we are essentially measuring the rate of cellular respiration, allowing us to measure metabolic activity.
b. Higher, increased demand for glucose if you decrease oxygen. i.e. you decrease oxygen = decrease ATP output, cell needs to compensate by using more glucose via anaerobic respiration.
4.
a. Talk about the entire immune response. Phagocytosis, antigen presentation, T helper cell, activation of B cells, clonal selection differentiation into plasma B and memory B cells, which persist in the body. Plasma B cells = antibodies. On subsequent exposure to the pathogen, activation of memory B cells provides a rapid, large response against the pathogen before it can cause disease, allowing long-term protection.
b. Second dose can increase memory B cells present. On subsequent exposure, even greater, more rapid response.
c. Different = active involves generation of an immune response by the body (e.g. production of memory, antibodies), passive = immunity received from an external source (e.g. antibodies injected). Active = long lasting, passive = short lasting. Similarities = both can potentially involve the use of antibodies.
5.
a. Cell signalling involves the use of a cell signalling molecule that binds to a complementary receptor on a target cell to initiate signal transduction, which involves the conversion of an external stimulus into a response within the cell via the use of secondary messengers that amplify the signal.
b. X = cytokine, macrophage, Y = neurotransmitter = muscle cell, neuron
c. Pheromones, fan their wings to spread it further into the air
6.
a. Standard speciation question here. Introduce geographical barrier, preventing gene flow (here can be distance between the two populations), populations are subjected to different environmental selection pressures, genetic differences allowed to accumulate. New species are formed when they can no longer interbreed to form viable and fertile offspring.
b. This question tells us that the two populations are basically genetically similar. This can be achieved through the presence of gene flow – the idea that the populations are still migrating with one another and interbreeding with each other, preventing the accumulation of genetic differences. Another way could be through having similar environmental selection pressures/ living in similar environments.
7.
a. Opposable thumb, forward facing eyes
b. Fast mutation rate, not subjected to recombination
c. Most recent common ancestor between L. catta, E. coronatus, E. flavifrons was around 21mya, most recent common ancestor between E. coronatus, E. flavifrons was around 7.5mya.
d. DNA hybridisation involving denature, mixing and anneal, then reheat and measure the melting temperature. Higher melting = high degree of similarity as more bonding needs to be broken. Lower melting = less degree of similarity. Temperature of hybrid DNA between E. coronatus, E. flavifrons will be higher than that between L. catta, E. coronatus and L catta and E. flavifrons.
8.
a. Cave paintings showing that passage of knowledge down between generations, showing customs and traditions etc with differences in paintings between generations also illustrating that cultural evolution.
b. Drawings going to be older than the dated deposits (65,000yrs), as the deposits covered the drawings i.e. drawings came first, and then the mineral deposit formed over the top.
c. First appearance of sapiens was 45,000 yrs ago which is too early for the cave paintings, which in the previous question we said arose later than 65,000 years ago.
9.
a. GM cotton, GM canola both transgenic = contain DNA from external source (bacteria). GM safflower not transgenic – only silencing no external DNA.
b. Dig it up? lol Use a different insecticide?
c. Social = increase health of wider population; biological = is it safe for consumption.
10.
a. The idea of forming DNA from different sources or something along these lines
b. Knew the amino acid sequence = can reverse engineer and determine a DNA sequence that codes for the amino acid sequence. Differences with the natural gene is that it won’t contain introns (non coding regions), it will only contain exons.
c. Standard outline how to use recombinant plasmids question. Restriction enzymes to produce complementary sticky ends in both plasmid and gene of interest, DNA ligase joins them together, chuck plasmid back into bacteria (e.g. through electroporation or some other method), then grow the bacteria.
d. Ethical response? Wat…? Ensure safety before human use? Share research with others to advance medical care? Such a broad question….
11.
a. Type of bacteria – E coli, Temperature, time of incubation
b. If the bacteria were cultured on 0 ug/mL of ampicillin on day 0 then there will be less bacteria at the end of day 3 on 6ug/mL of ampicillin? Something along the lines of that or anything that relates IV and DV. IV = antibiotic concentration of day 0, DV = bacterial growth.
c. Analyse the results, you’ve got 3 marks, so just systematically talk about the results. E.g. identify trends, did growth go down? Did growth go up? support your answer. Comparisons between the two groups would’ve been good as well.
d. Results of the experiment don’t support the predicted. Students predicted that in the control group you got no bacterial for 6ug/mL of ampicillin but you still have colonies in the actual experiment. Then for experimental group, you have that initial increase in bacterial growth between day 0 and day 1 which they did not draw. Something else could’ve been the idea that they only predicted the presence of distinct colonies, but in day 1 you had a bacterial lawn for the control group in the experiment.