Cambridge A Level Biology 9700 — 2014 May/June Paper 4 · Variant 2

This document consists of 22 printed pages and 2 lined pages. DC (CW/CGW) 79179/4 © UCLES 2014 [Turn over Cambridge International Examinations Cambridge International Advanced Level * 5 0 0 3 4 2 7 5 8 8 * BIOLOGY 9700/42 Paper 4 A2 Structured Questions May/June 2014 2 hours Candidates answer on the Question Paper. Additional Materials: Answer Paper available on request. READ THESE INSTRUCTIONS FIRST Write your Centre number, candidate number and name in the spaces provided at the top of this page. Write in dark blue or black pen. You may use a pencil for any diagrams, graphs or rough working. Do not use staples, paper clips, highlighters, glue or correction fluid. DO NOT WRITE IN ANY BARCODES. Section A Answer all questions. Section B Answer one question. Circle the number of the Section B question you have answered in the grid below. Electronic calculators may be used. At the end of the examination, fasten all your work securely together. The number of marks is given in brackets [ ] at the end of each question or part question. For Examiner’s Use Section A 1 2 3 4 5 6 7 8 Section B 9 or 10 Total

2 9700/42/M/J/14 © UCLES 2014 Section A Answer all the questions. 1 (a) Fig. 1.1 is an electron micrograph of a chloroplast from a maize leaf cell. X Y Z Fig. 1.1 Indicate below which of X, Y or Z contains: transport proteins … pigments … [2] (b) A chloroplast also contains DNA. Suggest the functions of DNA in this organelle. … … … … … …[2]

3 9700/42/M/J/14 © UCLES 2014 [Turn over (c) Changes in the atmospheric carbon dioxide concentration, light intensity and temperature alter the rate of photosynthesis. These three factors directly affect different stages of photosynthesis. Complete the table below using a tick (✓) if the factor directly affects the stage or a cross (✗) if it does not affect the stage. factor stage ✓ or ✗ carbon dioxide concentration Calvin cycle … photolysis … light intensity Calvin cycle … photolysis … temperature Calvin cycle … photolysis … [3] [Total: 7]

4 9700/42/M/J/14 © UCLES 2014 2 Myostatin is a protein that is produced in mammalian skeletal muscle cells. It circulates in the blood and acts on muscle tissue to slow down further differentiation and growth. In thoroughbred racehorses, a mutation involving the substitution of a single nucleotide has been identified in the MSTN gene which codes for myostatin. At the site of this mutation, the DNA nucleotide has either a cytosine (C) base or a thymine (T) base, giving race horses three possible genotypes for this mutation: CC, CT or TT. (a) At two years of age, racehorses with the MSTN CC genotype have greater muscle mass than those with the TT genotype. Suggest an explanation for this difference. … … … … …[2] (b) Racehorses that had won races of different distances were tested to determine their MSTN genotype. The results are shown in Fig. 2.1. 100 80 60 genotype CC percentage of winners with each genotype 1.0 1.2 1.4 1.6 1.8 race distance / km 2.6 2.0 2.2 2.4 40 20 0 CT TT Fig. 2.1

5 9700/42/M/J/14 © UCLES 2014 [Turn over With reference to Fig. 2.1, describe the effect of the MSTN genotype on the ability of racehorses to win races of different lengths. … … … … … … … … …[4] (c) Modern thoroughbred racehorses are the result of many years of artificial selection. Explain: (i) what is meant by artificial selection … … … … …[2] (ii) how genetic tests for the MSTN genotype can help in the selective breeding of racehorses. … … … … …[2] [Total: 10]

6 9700/42/M/J/14 © UCLES 2014 3 A group of membrane proteins which transport sugars out of cells have been identified and called SWEETs. They are found in the cell surface membranes of both animal and plant cells, including mammalian liver cells and rice mesophyll cells. Each SWEET is a protein with seven coiled regions which together make a pore through a membrane bilayer as shown in Fig. 3.1. coiled region Fig. 3.1 (a) (i) Explain why, to enter or leave a cell, sugars need molecules such as SWEETS. … … … … … …[2] (ii) Suggest how a SWEET is held within the membrane bilayer. … … … … … … …[3]

7 9700/42/M/J/14 © UCLES 2014 [Turn over (b) The bacterium, Xanthomonas oryzae (Xoo), causes the disease known as bacterial leaf blight in rice plants. It infects the intercellular spaces of the leaves of the host plant. Then, by switching on SWEET genes in the mesophyll cells, it stimulates the secretion of glucose into the intercellular spaces. Several different recessive alleles have been found, in rice plants from different countries, which give resistance to bacterial leaf blight. All these alleles have a mutation in the promoter of the SWEET gene. The effect of Xoo on wild type and resistant rice plants is compared in Fig. 3.2. wild type rice Xoo secretes a chemical into the rice mesophyll cells chemical binds with promoter region of SWEET gene SWEET protein produced and inserted into cell surface membrane switches on SWEET gene glucose secreted into intercellular spaces Xoo multiplies resistant rice Xoo secretes a chemical into the rice mesophyll cells chemical cannot bind with mutated promoter region of SWEET gene number of Xoo cells remains low Fig. 3.2 (i) Using the information in Fig. 3.2, explain this resistance of rice plants to Xoo. … … … … … … …[3]

8 9700/42/M/J/14 © UCLES 2014 (ii) Explain why it would be difficult to transfer this resistance into susceptible rice plants by genetic engineering. … … … … …[2] (iii) Explain why the presence of large numbers of Xoo in the intercellular air spaces of rice plants affects the ability of the plants to grow with their roots submerged in water. … … … … … … … … …[4] [Total: 14]

9 9700/42/M/J/14 © UCLES 2014 [Turn over 4 Collared lizards, Crotaphytus collaris, show variation in body length and running speed. Fig. 4.1 shows a collared lizard. Fig. 4.1 An investigation was carried out to find out whether body length and running speed affected the number of offspring that a male lizard sired (fathered). The lizards reproduce sexually, and females lay eggs after mating with males. A large number of male lizards was captured during the breeding season. For each lizard: • body length was measured • it was chased down a small race track and its fastest running speed over 1 m recorded • a blood sample was taken and DNA sequences at 10 different loci were analysed. The lizards were then released back into the place where they were captured, to allow mating to occur. Several weeks later, after the lizards’ eggs had hatched, as many young hatchlings as possible were captured. DNA testing was carried out on each hatchling to determine which male was the father. The results were used to determine the number of offspring sired by each of the male lizards in the first sample. (a) Outline how DNA analysis can be carried out and then used to establish which male lizard sired each hatchling. … … … … … … …[3]

10 9700/42/M/J/14 © UCLES 2014 (b) Fig. 4.2 shows the number of offspring sired plotted against the body length of the adult male lizards. Fig. 4.3 shows the number of offspring sired plotted against the fastest running speed (sprint speed) of the adult male lizards. 12 number of offspring sired 2.00 2.01 2.02 2.03 2.04 2.05 2.06 body length (log scale) 8 4 10 6 2 0 Fig. 4.2 12 number of offspring sired 0.40 0.45 0.50 0.55 0.60 0.65 0.70 sprint speed (log scale) 8 4 10 6 2 0 Fig. 4.3

11 9700/42/M/J/14 © UCLES 2014 [Turn over (i) With reference to Fig. 4.2 and Fig. 4.3, describe the relationships between • body length and the number of offspring produced • sprint speed and the number of offspring produced. body length … … … … … sprint speed … … … … …[4] (ii) Research has also shown that, in a population of collared lizards with varying leg lengths, those with longer hind legs are able to run faster. With reference to the results shown in Fig. 4.3, explain how, over time, this could lead to a change in the mean hind leg length in a population of collared lizards. … … … … … … … … …[3]

12 9700/42/M/J/14 © UCLES 2014 (c) Small islands often contain species of lizards that are not found on other islands or on the mainland. Explain how a population of collared lizards that became isolated on an island could evolve to form a new species. … … … … … … … … … …[5] [Total: 15]

13 9700/42/M/J/14 © UCLES 2014 [Turn over Question 5 starts on page 14

14 9700/42/M/J/14 © UCLES 2014 5 The hormone FSH (follicle stimulating hormone) plays important roles in the reproductive cycles of mammals. (a) State the precise site of secretion of FSH. …[1] (b) Certain cells in the ovaries and testes have receptors for FSH in their cell surface membranes. FSH can bind with these receptors, which triggers the cells to respond to the hormone. Mice were genetically modified so that they lacked functioning alleles of the gene that codes for the production of FSH receptors. (i) Female mice without FSH receptors were sterile. They were found to have normal primary and secondary follicles in their ovaries, but no Graafian (ovarian) follicles or corpora lutea. Explain these observations. … … … … … … … … …[4]

15 9700/42/M/J/14 © UCLES 2014 [Turn over (ii) Spermatozoa were collected from male mice with and without FSH receptors. The mean number of spermatozoa per mouse was estimated. The percentage of spermatozoa that could swim actively was calculated. Table 5.1 shows the results. Table 5.1 male mice with FSH receptors male mice without FSH receptors mean number of spermatozoa per mouse 5.6 × 106 3.6 × 106 percentage of spermatozoa that could swim actively 62 47 Discuss what these results suggest about the role of FSH in the development of spermatozoa in male mice. … … … … … …[3] [Total: 8]

16 9700/42/M/J/14 © UCLES 2014 6 (a) In mammals, the water potential of the blood is constantly monitored by osmoreceptor cells in the hypothalamus of the brain. When the water potential of the blood decreases, ADH (antidiuretic hormone) is produced by cells in the hypothalamus and released into the blood via an endocrine gland. (i) Explain what is meant by the term water potential. … … …[1] (ii) Describe the effect on water potential of adding solute to a solution. … …[1] (iii) State precisely where ADH is released into the blood. …[1] (iv) The decrease in the water potential of the blood is sometimes due to the loss of water from the body of a mammal. List two ways by which water may be lost from the body. … …[1]

17 9700/42/M/J/14 © UCLES 2014 [Turn over (b) Describe the action of ADH on the kidney. … … … … … … … … … … … … … … … … …[5] (c) Diabetes insipidus is a condition caused by an inability to produce ADH or by the kidneys being unable to respond to ADH. Suggest symptoms that may be experienced by a person who has diabetes insipidus. … … … … … …[2] [Total: 11]

18 9700/42/M/J/14 © UCLES 2014 7 Neurofibromatosis (NF) is a genetically inherited condition in humans where tumours grow in the nervous tissue. One symptom, which can develop around the age of 20 years, is loss of sight due to tumours on the optic nerve. (a) If one parent has NF, there is at least a 50% chance that his or her children will develop the condition, even if the other parent is unaffected. Complete the genetic diagram below to show how NF may be transmitted from parent to child. key to symbols … … parental phenotypes parent with NF unaffected parent parental genotypes gametes offspring genotypes offspring phenotypes [3] (b) Suggest how a person may develop NF when there is no family history of the condition. … … … … …[2]

19 9700/42/M/J/14 © UCLES 2014 [Turn over (c) Suggest how a tumour on the optic nerve could prevent the transmission of nerve impulses to the brain. … … … … … …[3] [Total: 8]

20 9700/42/M/J/14 © UCLES 2014 8 Adipose tissue is specialised connective tissue that functions as the major storage site for fat in the form of triglycerides. The human body contains two types of adipose tissue: white adipose tissue (WAT) and brown adipose tissue (BAT). • WAT is more common and is found under the skin and around some internal organs. • BAT is found in infants around the back and shoulders. • BAT is also found in adults but in relatively smaller quantities. • BAT cells contains more mitochondria than WAT cells. • BAT is involved in the maintenance of a constant blood temperature when the external environment is cold. (a) (i) Blood temperature in humans is maintained by a process called homeostasis. With reference to blood temperature, outline the main principles of homeostasis. … … … … … … … … …[4] (ii) Suggest why infants have relatively more BAT than adults. … … … … …[2]

21 9700/42/M/J/14 © UCLES 2014 [Turn over (b) Mitochondria in BAT cells function differently from those in other cells during periods of cold environmental conditions. Fig. 8.1 shows part of a mitochondrion in a BAT cell. cytosol outer membrane intermembrane space uncoupling protein 1 heat energy ATP ADP+Pi electron transport chain H+ H+ H+ H+ H+ H+ H NAD reduced NAD e– process 3 process 2 matrix process 1 B A Fig. 8.1 (i) Name structures A and B. A … B … [2] (ii) Draw an arrow on Fig. 8.1 to indicate the direction of the proton gradient that exists between the matrix and the intermembrane space. [1] (iii) State the two processes, shown in Fig. 8.1, that will be more active during periods of cold external environmental conditions. …[1] (iv) State the by-product that is obtained as a result of processes 1 and 2. …[1] (v) Suggest the main respiratory substrate for BAT cells. …[1] [Total: 12]

22 9700/42/M/J/14 © UCLES 2014 Section B Answer one question. 9 (a) Describe the main features of an organism belonging to the plant kingdom. [8] (b) Describe the methods used to conserve endangered animal species. [7] [Total: 15] 10 (a) Describe the production of penicillin using the batch culture method. [8] (b) Mycoprotein is produced using a continuous culture method. Describe the advantages of the batch culture method and the continuous culture method. [7] [Total: 15] … … … … … … … … … … … … … … … … … … …

23 9700/42/M/J/14 © UCLES 2014 [Turn over … … … … … … … … … … … … … … … … … … … … … … … … … … … …

24 9700/42/M/J/14 © UCLES 2014 … … … … … … … … … … … … … … … … … … … … … … … Copyright Acknowledgements: Fig. 1.1 Dr. George Chapman, Visuals Unlimited / Science Photo Library Fig. 4.1 George H. H. Huey / Alamy Permission to reproduce items where third-party owned material protected by copyright is included has been sought and cleared where possible. Every reasonable effort has been made by the publisher (UCLES) to trace copyright holders, but if any items requiring clearance have unwittingly been included, the publisher will be pleased to make amends at the earliest possible opportunity. Cambridge International Examinations is part of the Cambridge Assessment Group. Cambridge Assessment is the brand name of University of Cambridge Local Examinations Syndicate (UCLES), which is itself a department of the University of Cambridge.

CAMBRIDGE INTERNATIONAL EXAMINATIONS GCE Advanced Subsidiary Level and GCE Advanced Level MARK SCHEME for the May/June 2014 series 9700 BIOLOGY 9700/42 Paper 4 (A2 Structured Questions), maximum raw mark 100 This mark scheme is published as an aid to teachers and candidates, to indicate the requirements of the examination. It shows the basis on which Examiners were instructed to award marks. It does not indicate the details of the discussions that took place at an Examiners’ meeting before marking began, which would have considered the acceptability of alternative answers. Mark schemes should be read in conjunction with the question paper and the Principal Examiner Report for Teachers. Cambridge will not enter into discussions about these mark schemes. Cambridge is publishing the mark schemes for the May/June 2014 series for most IGCSE, GCE Advanced Level and Advanced Subsidiary Level components and some Ordinary Level components.

Page 2 Mark Scheme Syllabus Paper GCE AS/A LEVEL – May/June 2014 9700 42 © Cambridge International Examinations 2014 Mark scheme abbreviations: ; separates marking points / alternative answers for the same point R reject A accept (for answers correctly cued by the question, or by extra guidance) AW alternative wording (where responses vary more than usual) underline actual word given must be used by candidate (grammatical variants accepted) max indicates the maximum number of marks that can be given ora or reverse argument mp marking point (with relevant number) ecf error carried forward I ignore AVP alternative valid point (examples given as guidance)

Page 3 Mark Scheme Syllabus Paper GCE AS/A LEVEL – May/June 2014 9700 42 © Cambridge International Examinations 2014 1 (a) transport proteins – Y ; pigments – X ; [2] (b) DNA codes for, proteins / polypeptides / enzymes ; one example of protein or enzyme ; e.g. rubisco / electron acceptor / ATP synthase / transport ref. transcription / mRNA ; [max 2] (c) factor stage
or  carbon dioxide concentration Calvin cycle
photolysis  light intensity Calvin cycle  photolysis
temperature Calvin cycle
photolysis  ;;; all 6 correct = 3 marks 4 or 5 correct = 2 marks 2 or 3 correct = 1 mark [3] [Total:7] 2 (a) presence of C base in DNA (code) changes amino acid (in myostatin) ; myostatin in CC horses, is inactive / not produced ; ora for TT in CC horses muscle, differentiation / growth, has not been slowed ; ora for TT [max 2]

Page 4 Mark Scheme Syllabus Paper GCE AS/A LEVEL – May/June 2014 9700 42 © Cambridge International Examinations 2014 (b) CC genotype does best in short races ; data quote ; e.g. 75% winners at 1.0 km ; TT genotype does better in longer races ; data quote ; e.g. about 60% of winners at 2.6 km CT genotype has some winners at all distances ; CT does best at 1.8 km ; [max 4] (c) (i) by humans ; individuals with desired features chosen to breed / AW / named example ; [2] (ii) can choose parents genotypes to breed ; CC for racing short distances / TT for racing long distances / CT as all-rounders ; ref. need to keep all three genotypes in population ; [max 2] [Total:10] 3 (a) (i) idea of sugars unable to pass through phospholipid bilayer ; hydrophilic / polar / not lipid-soluble / water soluble ; large ; [max 2] (ii) forms bonds with hydrophilic heads (of phospholipids) ; hydrophobic parts of SWEET ; bond with, fatty acid chains / hydrophobic tails, (of phospholipids) ; ref. hydrogen bonding / ionic bonds / hydrophobic interactions ; [max 3] (b) (i) (SWEET) gene cannot be switched on ; no SWEET (protein) produced ; no, glucose / sugar, secreted (into intercellular spaces) ; (so) Xoo / bacteria, do not multiply / numbers remain low ; (small numbers of Xoo / bacteria) so no disease ; [max 3]

Page 5 Mark Scheme Syllabus Paper GCE AS/A LEVEL – May/June 2014 9700 42 © Cambridge International Examinations 2014 (ii) allele is recessive ; idea of not expressed when dominant allele present ; ref. promoter ; e.g. normal promoter must be inactivated or removed / must transfer mutated promoter [max 2] (iii) prevents diffusion of air (from leaves to roots) ; ref. aerenchyma ; roots respire anaerobically ; (so) less ATP produced (for growth) ; bacteria use of oxygen ; more ethanol produced may be beyond tolerance / AW ; [max 4] [Total:14] 4 (a) details of electrophoresis ;; any 2 from DNA cut by, restriction enzyme(s) / endonuclease(s) loaded (into wells) at, negative end / cathode end, (of gel) ref. buffer / electrolyte (negatively charged) DNA attracted to, anode / positive electrode separation due to, electric field / potential difference short pieces / smaller mass, move further (in unit time) / move faster ora fluorescent / radioactive, DNA probes compare, DNA sequences / bands, (of male lizard and hatchling) ; [3] (b) (i) body length no relationship (between body length and number of offspring) ; small / intermediate, body length produce more offspring ; sprint speed lizards with greater sprint speed sire more offspring ; use of two paired figures from Fig. 4.3 to support relationship ; [4]

Page 6 Mark Scheme Syllabus Paper GCE AS/A LEVEL – May/June 2014 9700 42 © Cambridge International Examinations 2014 (ii) lizards with longer (hind) legs will tend to have more offspring ; so will have a greater chance of passing on their alleles ; (over time) the frequency of alleles (for long hind leg) will increase ; so mean hind leg length will increase ; directional selection ; [max 3] (c) no, breeding / allele flow / gene flow, between (lizard) populations ; different selection pressures / different (environmental) conditions ; mutations occur ; advantageous alleles, selected for / passed on ; change in, allele frequency / gene pool ; genetic drift ; (eventually) unable to interbreed ; allopatric speciation ; [max 5] [Total:15] 5 (a) anterior pituitary ; [1] (b) (i) early follicle development not dependent on FSH ; with no FSH / no FSH receptors, follicle development stops ; ora with no FSH / no FSH receptors, Graafian / ovarian, follicle does not develop ; ora with no FSH / no FSH receptors, there is no ovulation ; no corpora lutea because these form, from Graafian follicle / after ovulation ; [max 4]

Page 7 Mark Scheme Syllabus Paper GCE AS/A LEVEL – May/June 2014 9700 42 © Cambridge International Examinations 2014 (ii) sperm development better when FSH present ; with FSH receptors more sperm produced ; ora sperm more active ; ora males have increased fertility ; ora without FSH receptors some sperm produced ; [max 3] [Total:8] 6 (a) (i) tendency of water molecules to move from one region to another / potential energy of water / ability of water to do work ; [1] (ii) (water potential) becomes, lower / more negative ; [1] (iii) posterior pituitary ; [1] (iv) for one mark ; any 2 from urine sweat water vapour (from exhaled air) faeces bleeding tears [max 1] (b) affects collecting duct, (cells / wall) ; A distal convoluted tubule cells binds to receptors on cell surface membranes ; activates series of enzyme controlled reactions ; (phosphorylase causes), vesicles / aquaporins, to move to cell surface membrane (on lumen side) ; vesicles / aquaporins, fuse with cell surface membrane ; cells / wall, more permeable to water ; water moves out of lumen (of collecting duct) ; down water potential gradient ; [max 5]

Page 8 Mark Scheme Syllabus Paper GCE AS/A LEVEL – May/June 2014 9700 42 © Cambridge International Examinations 2014 (c) produce, a lot of urine / dilute urine ; dehydration / thirsty ; cramps / loss of salts ; [max 2] [Total:11] 7 (a) symbols and key ; e.g. A = NF allele and a = normal allele parental genotypes and gametes ; e.g. parental genotypes Aa x aa gametes A a x a a offspring genotypes and phenotypes linked ; e.g. Aa has NF and aa is unaffected [3] (b) spontaneous / random / chance ; mutation of, gene / allele ; AVP ; e.g. named mutagen / detail of mutation / in oocyte / in sperm [max 2] (c) compresses nerve ; damages, myelin sheaths / Schwann cells ; prevents, setting up of local circuits / saltatory conduction ; stops Na+ / K+ pumps from working ; blocks blood supply; qualified ; e.g. effect on, oxygen supply / glucose supply / ATP production AVP ; e.g. may stop ion channels opening [max 3] [Total:8]

Page 9 Mark Scheme Syllabus Paper GCE AS/A LEVEL – May/June 2014 9700 42 © Cambridge International Examinations 2014 8 (a) (i) receptors / hypothalamus, detect change in blood temperature ; brain ; (receptor / brain) sends impulses to effector ; effector carries out response / example of response ; blood temperature returns to normal ; negative feedback ; [max 4] (ii) larger SA : V ratio ; lose (relatively) more heat ; ref. more mitochondria to release heat energy ; cannot carry out behavioural actions to get warm ; infants cannot shiver ; [max 2] (b) (i) A – ATP synthase / ATP synthetase / stalked particles ; R ATPase B – inner membrane / crista ; I phospholipid bilayer [2] (ii) arrow going down from intermembrane space to matrix ; [1] (iii) 1 and 3 ; [1] (iv) water ; [1] (v) fatty acids ; A lipid / fat / triglycerides [1] [Total:12]

Page 10 Mark Scheme Syllabus Paper GCE AS/A LEVEL – May/June 2014 9700 42 © Cambridge International Examinations 2014 9 (a) multicellular ; differentiated cells ; (most) have, vascular tissue / xylem and phloem ; eukaryotic (cells) ; ref. meristems ; (most) are not motile ; motile gametes only in mosses and ferns ; autotrophic nutrition / photosynthesis ; cells have: chloroplasts ; large / central, vacuole ; walls made of cellulose ; [max 8] (b) place in zoos ; protected against, disease / predation ; captive breeding programme ; ref. assisted reproduction / cloning / sperm banks ; released into wild ; ref. national parks / reserves ; rangers patrol parks ; human access restricted ; controlled agriculture ; controlled industry ; visitor centres / education ; habitat / breeding sites, protected ; banning sale of protected animals or their products ; banning hunting ; [max 7] [Total:15]

Page 11 Mark Scheme Syllabus Paper GCE AS/A LEVEL – May/June 2014 9700 42 © Cambridge International Examinations 2014 10 (a) use Penicillium (in batch fermenter) ; (main) nutrients added at start ; penicillin only produced, after growth phase / when running out of nutrients ; (penicillin) is a secondary metabolite ; fermentation is stopped ; penicillin is harvested ; fermenter is cleaned out / ref. sterility ; new culture of Penicillium is put in and started again ; ref. fed batch culture ; carbohydrate / named nutrient, added regularly ; keeps fermentation going longer / produces more penicillin ; detail of fermenter ;; e.g. paddle to mix nutrients / sterilising steam inlet / set at pH 6.5 / aeration / kept at 27 °C [max 8] (b) batch easy to set up culture ; can continue with minimal attention / AW ; environmental conditions easy to control ; fermenter can be used for different process afterwards ; only waste one batch if contaminated ; less chance of blockage in fermenter ; continuous no / less, down time / AW ; small vessels can be used ; productivity high ; cost effective ; downstream processing easier ; good for using immobilised enzymes ; [max 7] [Total:15]