Subject: Marketing Strategies Q1. Choosing a sales compensation plan is an important decision. However, there is no one-size-fits-all approach. What are some of the factors involved in the compensation process?….
Biology for International Science
|Lifelong Learning Centre|
- Supplementary booklet available on request
- You are allowed to use a non-programmable calculator which has an approval sticker issued by the Biological Sciences Department
- You are not allowed to use a dictionary in this exam.
- There are two sections to this exam: Section A and Section B.
- Answer all questions in Section A.
- Section A earns 85 marks.
- You are advised to spend 2 hours on this section.
- Answer one question in Section B.
- Section B earns 15 marks.
- You are advised to spend 30 minutes on Section B.
- There are 22 pages to this exam.
- There will be 2.5 hours to complete this exam
Please do not remove this paper from the exam venue.
Any supplementary booklets must be attached with a treasury tag to the back of this answer booklet.
Answer ALL questions from this section.
This section earns 85 marks.
You are advised to spend 2 hours on this section.
- Plant and animal cells are both eukaryotic cells, yet have several structural and functional differences.
- State TWO cell features present in the plant cell, but not in an animal cell and explain their functions in the plant cell.
|Any of cell wall – support/structural role chloroplasts – photosynthesis large vacuole – store of water and minerals (1/2 mark naming each organelle and 1 mark each for correct function)|
- Exocytosis is a process of vesicular transport in eukaryotic cells.
Describe this process after vesicular transportation from the rough endoplasmic reticulum and also how mitochondria are required.
|Transport to golgi apparatus/body (1/2) for processing/packaging (1/2) Bud off golgi as vesicle to cell/plasma membrane (1/2) Fuse/merge with CM (1/2) and release contents (½)|
|Energy produced by mitochondria for this process (1/2)|
- Phototropism is an example of a response by a plant to change in its external environment. Light is the stimulus which is detected by the plant. The drawing shows the shoot tip of a plant. A light is switched on.
- Draw the next ~2cm of growth as accurately as possible in the space above
(see diagram) 1 mark
- What is the molecule which is responsible for the response to light?
|Indoleacetic Acid (IAA)|
- Describe how the distribution of molecule referred to in 2b is affected by the stimulus
|The IAA moves away from the light source causing an increase in concentration on the shadow side |
- Explain how the change in distribution of the molecule (referred to in 2b) brings about the response
|IAA cause cell elongation .|
|Given the higher concentration on the shadow side there is more cell elongation here than at the lit side |
|This differential in growth/cell elongation cause the bending of the tip towards the light |
- In Figure 1 below the actual length of the mitochondrion is 8 µm.
- What is the magnification of this image? Show your calculations.
Actual length is 8um
Observed (image) approx. 12cm = 12000um (1) (allow 0.5cm either way)
Mag = 12,000/8 = 15000 (1)
- Calculate how long a 5 µm scale bar would be on this electron micrograph
Observed (scale bar/image) = mag x actual/specimen size (1)
15000 x 5 = 75000um / 10000 = 7.5cm (1)
- Determine the width of the mitochondrion in Figure 1
Actual width = measured/observed / mag
Approx. 2cm 20000um /15000 = 1.33um
No line on image to show where measured – 0.5 marks
- Use the DNA sequence below for this question
5’ CCTCATTGTACAACT 3’
- What RNA sequence would be produced from this DNA molecule?
5’ AGU UGU ACA AUG AGG 3’
1 mark each for correct base pairings, U instead of T in mRNA and correct orientation
- Using the codon table below determine the protein sequence which would be produced from this DNA sequence?
- The Llama is a goat-like mammal which lives at altitudes of 5000 m in the Andes. The table below shows the effect of varying the partial pressure of oxygen on the percentage saturation of haemoglobin for a llama and a goat which lives at sea level.
|Partial pressure of oxygen (mm Hg)||% saturation of haemoglobin|
|1 mark for Accurate plotting Reasonable scaling Reasonable curves drawn and id All label and units present|
Using the same axis plot oxygen dissociation curves for the haemoglobin of both of these animals on the graph paper below.
- The oxygen dissociation curve for the haemoglobin of the goat is typical of most mammals that live at low altitudes. Explain the advantage to the llama in having the different oxygen dissociation curve for its haemoglobin
|Partial pressure of O2 is less at higher altitudes |
|Llama curve shifted to the right |
|More effective at absorbing O2 at lower pressure |
|Curve needs to be steeper so that OxyHb still dissociates in body tissues |
- Describe how vaccination can result in actively acquired immunity in an individual.
|Vaccination involves introducing a substance to activate their immune response|
|Follows the process – recognition of an antigen by certain lymphocytes – activation of these lymphocytes – differentiation and proliferation into effector cells – effector cells eliminate antigen – development of memory cells – memory cells elicit a rapid and long-term response to re-exposure|
|Any 6 of the above|
- Describe the problems associated with using a vaccine to control influenza in the UK population
|Influenza rapidly mutates|
|Need for herd immunity – misconception among the public reduces uptake|
|Side effects, believe already immunised, government distrust|
5. The diagram below shows the position of the diaphragm at times A and B.
- Air moves out of the lungs between times A and B. Explain how this happens including reference to the volume and pressure changes involved.
|Diaphragm relaxes and domes reduces volume of thorax (1) Results in increase in pressure (1) Air moves from high to low pressure / down pressure gradient out lungs (1)|
- Gaseous exchange in mammals involves the exchange of oxygen for carbon dioxide at the interface between the respiratory and circulatory system. Name structure A labelled below where this occurs
……………………………………………….. [1 mark]
- State and explain 2 features that enable structure A labelled above to be structurally adapted to perform its function
|Any of below (1 mark point made and 1 mark for appropriate explanation) Large surface area – allowing efficient diffusion of gases Good blood supply – blood flow creates diffusion gradient and allows gases to move down conc gradient Alveoli/capillary cell wall one cell thick – creates ‘short’ diffusion pathway’|
- With the aid of a diagram describe how an action potential message is transmitted to the postsynaptic membrane. Include in your answer reference to the key structures and molecules as well as how the synapse recovers.
|An action potential causes calcium channels to open and Ca2+ enters the synaptic knob |
|Influx of Ca2+ ions causes synaptic vesicles to fuse with presynaptic membrane  and release neurotransmitter [½] /acetylcholine  other named neurotransmitter |
|Neurotransmitter /acetylcholine diffuses across synaptic cleft and fuses/attaches to active sites on Na+ channels on post synaptic gap |
|Na+ ions diffuse rapidly along conc gradient generating a new action potential in post synaptic neurone |
|Acetylcholinesterase hydrolyses acetylcholine into choline and ethanoic (acetyl) acid which diffuse back across the synaptic cleft  ATP released by the mitochondria is used to regenerate acetylcholine from choline and ethanoic (acetyl) acid |
- The Heart
- On the diagram (Figure 10) label the following:
- The biscupsid valve
- The left atrium
- The aorta
- Label the chambers of the heart with 1, 2, 3, and 4 in order of blood flow, starting with deoxygenated blood returning from the body.
1=LA. 2=LV, 3=RA, 4=RV order RA, RV, LA, LV
- Describe the part played by each of the following in myofibril contraction
|Moves out of the way (or changes shapes) when Ca2+ bind|
|This allows myosin to bind (to actin)/ or cross bridge formation|
|Head of myosin binds to actin and moves/pulls/slides actin past |
|Myosin detaches form actin and re-sets/moves further long |
|(answer must refer to myosin head or cross bridges and give an indication of driving the slide)|
- The table below shows the mean thickness of the wall of some mammalian blood vessels
|Blood vessel||Mean thickness of wall / mm|
- Which of the blood vessels A to E is:
- a capillary? ………C………
Explain why in terms of the vessels function
|Very thin wall – for exchange of gases/nutrients.|
- the aorta? ……B…………
Explain why in terms of the vessels function
|Thick wall to withstand high blood pressure.|
- The diagram shows a simplified version of a reaction which occurs as part of respiration
- Complete the table by naming stages A and B and giving the location of each stage in a cell such as a liver cell
|Stage||Name of Stage||Location in cell|
|B||Krebs cycle/Citric acid cycle/ TCA cycle||Matrix (of mitochondria)|
- How many carbon atoms are there in each pyruvate ion?
- What happens to the H+ ions and electrons released in Stage C?
|React with oxygen  to produce water |
- The diagram below shows the structure of two amino acids, alanine and serine
- In the space below, draw a diagram to show the dipeptide formed when these two molecules bond together.
Peptide bond drawn correctly  Rest of both amino acids correctly drawn 
- Computer-generated models were used to measure the length in nanometres (nm) of a polypeptide containing a large number of amino acids. Measurements were made of the primary and secondary structure of the polypeptide at 25 °C and at 55 °C. The results are shown in the table below
|Length of polypeptide/nm|
|Structure of polypeptide||at 25 °C||at 55 °C|
- Explain what is meant by the term secondary structure of a protein
|Reference to α-helix or β-pleated sheet|
|Reference to hydrogen bonds|
- Suggest why the increase in temperature has this effect on the length of the secondary structure of the polypeptide
|Reference to increase in kinetic energy at 55 °C|
|Leading to more vibrations within a molecule|
|Resulting in breaking of the hydrogen bonds holding the α-helix or β-pleated sheets together|