BIOLOGY Control, Genomes and Environment F215 Monday 13 June 2011 Afternoon

Unofficial Mark Scheme. Please note that this is not a comprehensive mark scheme. OCR will probably have many other key points so do not worry if you wrote different points to the ones given here. If this scheme is useful to you please show your appreciation by visiting my Google Blog (http://cambridgeacademic.blogspot.com/) and clicking on an advert or two. Cheers. 1(i) Describe the steps that must occur for plant protein to be converted to animal protein. [3 marks]

Plants are eaten/ingested by animals. Digested/broken down into amino-acids. Amino-acids absorbed by the animal. Amino-acids used in translation to make animal proteins.

(ii) List the processes that contribute to B in the meadow where sheep are grazed. [2 marks]

Death/decay (Plants and animals). Excretion (Animals). Urination (Animals).

(iii) Name the bacteria that carry out processes C and D, and explain the significance for the growth of plants. [3 marks]

C = Nitrosomonas (incorrect spelling in Heinemann OCR book Nitrosomonass) D = Nitrobacter Plants need nitrates to make DNA and amino-acids. Nitrites are toxic to plants.

(iv) Why will the soil nitrate concentration decrease in the cabbage field if it is used to repeatedly grow cabbages every year. [3 marks]

Cabbages harvested each year so very little plant protein returned to the humus/soil by decomposition of dead plant material (B). No animals = no waste material to contribute to humus/soil (B).

Lack of ammonium compounds/urea for nitrification by chemoautotrophic bacteria (C) into nitrites and nitrates.

(v) Suggest a crop that would allow process F and explain how this would add nitrate to the soil. [3 marks]

Leguminous plant (eg clover/beans/peas) Nitrogen fixation by Rhizobium bacteria in root nodules. Gaseous nitrogen fixed into ammonium ions, which are used by host plant.

(b) Why is the continued existence of rare breeds of farm animals desirable? [2 marks]

Ethical responsibility. Maintain genetic diversity/gene pool. Potential use to improve traits/disease resistance of domesticated species. Commercial potential/tourism.

(c) (i) State the two steps that must have occurred for a breed to develop a distinctive metabolism, such as the ability to eat seaweed. [2 marks]

Mutation in metabolic enzyme. Genetic isolation/lack of new genes into the pool. Selection pressure/restricted food suppy.

(ii) Suggest what particular problems make the Ronaldsay sheep one of the most endangered sheep breeds in the United Kingdom. [2 marks]

Isolation/inbreeding reduced genetic diversity so possibly more susceptible to disease. Cannot easily be moved to different location as cannot feed on grass as susceptible to copper poisoning so need supply of seaweed.

2.(i) Innate behaviour [2 marks]

Genetically determined/inherited. No learning required. Rigid and inflexible. Similar responses between individuals (Sterotypical).

(ii) Learned behaviour. [2 marks]

Determined by interaction between genetic and environmental factors. Not inherited. Altered by experience. Variable responses between individuals Basis of inheritance.

(b) Advantages to animals of innate and learned behaviour, with reference to specific examples. [11 marks] [Did not expect this type of question as the “Examiner’s Tip in the OCR book by Heinemann states “You should not try to learn lots of examples of behaviour from the types listed…..it is more important that you can identify what type of learned behaviour ….. given in the question” Innate Behaviour Reflexes: escape behaviour to avoid predation, eg. earthworm recoil into ground. Kineses: non-directional movement to find more favourable conditions eg. woodlice moving from light into dark. Taxes: directional movement to find food/favourable conditions/mates eg pheromones in moths to find mate. Complex innate behaviour to provide information about food sources, eg waggle dance in bees gives direction and distance to flower. Learned Behaviour Habituation: ignoring a neutral stimulus eg birds ignore scarecrows so save energy by not flying away. Imprinting: strong association to another organism usually the mother which allows protection by the mother and learning skills from mother. Conditioning: Association of two stimuli, which may help in obtaining food, eg dog learns that the smell of a rabbit means that it might catch its dinner. Latent (exploratory) learning: Exploring and remembering surroundings may allow escape from prey (eg down burrow) or location of food source. Insight: Ability to solve a problem by reasoning. Will allow individuals to access food that not within easy reach, eg some birds using tools to extract insects from bark, monkey reaching fruit by building tower. Social behaviour allows an individual to establish a place within a group that might provide protection from predators (safety in numbers), enhanced survival of young, opportunity to learn from other members of group.

3. (a) (i) Suggest a technique to provide molecular evidence that all Enqlish Elms are a clone. [1 mark]

DNA sequencing. (ii) Why is the English Elm genetically isolated from other types of elm. [1 mark]

Lack of seed production/sexual reproduction will not allow gene mixing from other Elms. (iii) What is the process in which plants reproduce asexually by means such as suckers. [1 mark]

Vegetative propagation (b) Explain why there was such a rapid loss of elm trees in Britain as a result of this disease. [4 marks]

All English Elm trees are clones and genetically identical. All are equally susceptible/lack of resistance to the fungus. No genetic variation to allow natural selection of resistant types.

(c) (i) Explain why the plugging of xylem vessels will result in the leaves of the upper branches turning yellow. [2 marks]

Lack of water supply to leaves. Water stress/stomata closure Less CO2 for photosynthesis Less sugars/amino-acids/lipids made Less ATP production. Reduced chlorophyll production. Chloroplast breakdown

(ii) Explain why the loss of leaves from the tree may result in the death of the tree's roots. [2 marks]

Less sugar/nutrition from leaves. Less substrate for ATP production by roots. Cell death.

(d) Describe the process of cloning plants by tissue culture. [7 marks] Sterile conditions/aseptic techniques. Plant tissue containing meristem cells removed.

Explant tissue. Growth on nutrient medium/agar. Callus formation. Plant growth hormone to stimulate shoots (cytokinin). Plant growth hormone to stimulate roots (auxin). Incubation in light. Plantlets subdivided. Handling, medium/sterile soil.

(e) List two advantages and two disadvantages of cloning plants by tissue culture. [4 marks] Advantages

Many plants generated quickly. Genetically identical so all have desired, characteristics / genotypes / phenotypes ; No need for (artificial) selection. Easy to transport/store. Easy to genetically engineer. Disease/virus, free.

Disadvantages

Genetically identical so may be susceptible to disease. Loss in genetic diversity (if cloned plants are grown exclusively). Farmers have to buy plants from suppliers. Expensive and may not be available in for developing countries. Patented technology.

4(a) (i) Percentage decrease in the number of breeding pairs of snipe in area 2. [2 marks]

Answer = 57.25% 655-280 = 375 375/655 x 100 = 57.25%

(ii) Use the data in Table to describe and explain the effect of the introduction of hedgehogs on the number of breeding pairs of waders in area 2. [6 marks]

General reduction in the number of waders since introduction of hedgehogs. Probably due to hedgehogs eating wader eggs. Reduced wader reproduction/offspring. Without hedgehogs, increase in lapwing and redshank (in area 1).

Reduction in dunlin and snipe in area 2 may not only be from hedgehogs since they have also decreased in area 1, although not as much. Some waders show less reduction (lapwing, 31% and redshank 41%). Perhaps eggs better hidden/camouflaged or at higher level so hedgehogs cannot eat as many.

iii) Suggest and explain two factors that has led to an increase in the hedgehog population. [4 marks]

Lack of hedgehog predators allows high survival of offspring and increase in numbers. Lack of any disease in founder hedgehogs allows high survival rate. Abundant food supply with no interspecies competition allows good nutrition and survival. Small number of original hedgehogs means very little intraspecies competition. Food supply not limiting allowing high survival rate.

(b) Three methods to reduce the number of hedgehogs were propsed: trapping and moving to the mainland. trapping and keeping them in captivity. trapping followed by killing. Comment on the ethical issues involved in choosing the last option. [3 marks]

Unethical to trap and kill when other options that do not kill the hedgehogs exist. Unethical to move hedgehogs to the mainland where they may be at a disadvantage for survival in competition with other species/disease exposure. Unethical to move to mainland as may have a negative impact on existing species/ecosystem there. Unethical to keep in captivity as these are wild creatures and would be placed under stress.

5.(a) Identify the amino acids in the sequence. [2 marks]

Methionine Arginine (Lysine) Threonine Tryptophan

(b) What is the stage of protein synthesis and name the organelle in the cell where this takes place. [2 marks]

Translation Ribosome

(c) Identify the type of nucleic acid [2 marks]

mRNA (d) Give the three triplet codons that would cause termination of the polypeptide (stop codons) and explain why these codons have this effect. [2 marks]

UAA UAG UGA No corresponding tRNA for these codons so polypeptide synthesis stops.

(e) What name would be given to a UUU to UUC mutation that resulted in a change of the codon. [1 mark]

Silent Describe the differences between: (a) somatic gene therapy and germ line gene therapy [2 marks]

Somatic cell gene therapy involves delivery/replacement/expression of a gene in somatic/body cells with no transmission to offspring. Germ line cell gene therapy involves gene delivery to the sperm/eggs so that the gene is passed onto the offspring (transgenic animals).

(b) the central nervous system and the peripheral nervous system. [4 marks]

The CNS consists of the brain and spinal cord. PNS consists of neurons that carry impulses/signals/action potentials to and from the CNS. In the PNS, sensory neurons carry signals from receptors to CNS. In the PNS, motor neurons carry signals from CNS to effector/target organs. Motor neurons subdivided into somatic (voluntary control of effects, eg skeletal muscle) and autonomic (not under voluntary control, eg smooth muscle).

(c) prophase 1 of meiosis and prophase 2 of meiosis. [2 marks]

Prophase 1: Cell carries homologous pairs of chromosomes (4n) that form a bivalent and allow crossing over. Prophase 2: Cell carries chromosomes with no homologous partner/pair (2n). No bivalent or crossing over.

7(a) (i) Suggest a technique by which the student might have collected his samples along this transect. [1 mark]

Sampling using a quadrat to count ladybirds.

(ii) It was suggested he should make several transects up the hill. Explain why this is good experimental design. [1 mark]

Less chance of sampling errors/more accurate data by taking multiple measurements. Allows better statistical analysis. Might identify differences between different parts of the hill.

(b) (i) Suggest a method of processing this data to make comparisons more valid and explain why your method is an improvement. [2 marks]

Express the number of ladybirds as a % of the total number for each height above sea level. This will allow an easier comparison of differences between groups.

[Processing the data suggests that statistical test which analyse the data may not be credited]. (ii) Was the student correct to say: “Data showed a positive correlation between increasing altitude and the frequency of the black ladybird. I concluded that high altitude causes the black form to survive better." [3 marks]

The student is correct in the statement about a positive correlation since number of black ladybirds increases from 7% to 20.5% with altitude. BUT conclusion is not correct since other factors (eg reduced predation) rather than altitude may contribute to better survival. Correlations do not prove causality. The black ladybirds do survive better at altitude but this does not prove that the high altitude is the cause.

(c)(i) What is meant by the term recessive. [1 mark]

Characteristic in which the phenotype is only present in the absence of a dominant allele. (ii) Total number of the red ladybirds is 296, and the total number of the black ladybirds is 50. Use the Hardy-Weinberg principle and the figures given above to calculate the frequency of the dominant allele, p, and the recessive allele, q, in the two-spot ladybird population. [3 marks]

Red (q) is recessive Total ladybirds 296 + 50 = 346 q2 = 296/346 = 0.8555 q = 0.92 Since p + q = 1 p = 1-0.92 p = 0.08