CBSE previous Year Solved  Papers  Class 12 Biology Outside Delhi 2015

Time allowed : 3 hours                                                                                           Maximum Marks: 70

General Instructions :

  1.  There are a total of 26 questions and five sections in the question paper, All questions are compulsory.
  2. Section A contains question number 1 to 5, Very Short Answer type questions of one mark each.
  3.  Section B contains question number 6 to 10, Short Answer type I questions of two marks each.
  4.  Section C contains question number 11 to 22, Short Answer type II questions of three marks each.
  5.  Section D contains question number 23, Value Based Question of four marks.
  6. Section E contains question number 24 to 26, Long Answer type questions of five marks each.
  7. There is no overall choice in the question paper, however, an internal choice is provided in one question of two marks, one question of three marks and all three questions of five marks. An examined is to attempt any one of the questions out of two given in the question paper with the same question number.



Question.1. How many chromosomes do drones of honeybee possess ?
Name the type of cell division involved in the production of sperms by them.
Answer : Drones of honey bees are haploid and possess 16 chromosomes.
Mitosis is the cell division that is involved in the production of sperms by drones.

Question.2. What is a cistron ?
Answer : Cistron is a section of DNA or RNA molecule that codes for a specific protein or polypeptide.

Question.3. Retroviruses have no DNA However, the DNA of the infected host cell does possess viral DNA. How is it possible ?
Answer : After an attack on the host cell, retro virus enters the micro phages (in case of HIV) where the virus undergoes Teminism or Reverse Transcription to use its RNA genome to form viral DNA in presence of reverse transcriptase enzyme. This viral DNA then incorporates into the host cell DNA, directing the infected cells to prpduce more copies of viruses. In this way, the infected cell possesses viral DNA.

Question.4. Why do children cured by enzyme-replacement therapy for adenosine deaminase deficiency need periodic treatment?
Answer : Children cured by enzyme-replacement therapy for adenosine deaminase (ADA) deficiency need periodic treatment because it is not a completely curative method. In this method, the lymphocytes are retrieved from the blood of the patient. These lymphocytes are then grown in a culture medium. The lymphocytes are then incorporated with functional ADA, i.e. cDNA using a retro viral vector. Since the-lymphocytes have a definite life cycle, there is a need for periodic transfusion of genetically engineered lymphocytes to the patient.

Question.5. List two advantages of the use of unleaded petrol in automobiles as fuel.
Answer : Following are the two advantages of using unleaded petrol as fuel in automobiles :

  1. Pollution caused by unleaded petrol is less as it does not release lead compounds from exhaust fumes into the atmosphere.
  2. It also helps in preventing health issues like anaemia, loss of appetite, damage to erythrocytes & damage to control Nervous system.


Question.6. Why do moss plants produce very large number of male gametes ? Give one reason. What are these gametes called?
Answer : Mosses are bryophytes and they need water for . fertilization. They lay their flagellated male gametes that swim across the water to reach the female gamete. In this process, a large number of male gametes are destroyed or lost. Thus, very large number of male gametes are produced by moss plants so that even if some of the gametes get destroyed, the remaining can fertilise the female gamete. The male gametes are called antherozoids.

Question.7. (a) Select the homologous structures from the combinations given below:
(i) Forelimbs of whales and bats
(ii) Tuber of potato and sweet potato
(iii) Eyes of octopus and mammals
(iv) Thorns of Bougainvillea and tendrils of Cucurbita
(b) State the kind of evolution they represent.
(a) From the above options which forms homologus structures they are:
(i) Forelimbs of whales and bats are similar in structure but perform different functions of swimming and flying, respectively.
(iv) Thorns of Bougainvillea and tendrils of Cucurbita are both modifications of a stem arising from axillary bud but perform different functions like protection and climbing, respectively.
(b) The evolution represented by homologous organs or structures is divergent evolution as their origin is common but have diverged (became dissimilar) with evolution.

Question.8. (a) Why are the plants raised through micropropagation termed as somaclones ?
(b) Mention two advantages of this technique.
(a) The plants obtained by micropropagation are called somaclones because they are genetically identical to the original plant from which they are produced.
(b) The advantages of micropropagation are as follows :
(i) It helps in the propagation of a large number of plants in a short span of time.
(ii) It helps in the production of plants which are disease and pest resistant plants.

Question.9. Explain the different steps involved during primary treatment phase of sewage.
Answer : The primary phase of sewage treatment involves physical removal of particles through filtration andsedimentation Initially, floating debris is removed by sequential Alteration.
Then the grit are removed by sedimentation. All solids that settle form the Primary sludge, and the supernatant forms the effluent. The effluent from the primary setting tank is taken for secondary treatment.

Question.10. What is mutualism ? Mention any two examples where the organisms involved are commercially exploited in agriculture.
List any four techniques where the principle of ex-situ conservation of biodiversity has been employed.
Answer : Mutualism is a relationship between two organisms of different species in which both organisms are benefited.Examples of the organisms involved that are commercially exploited in agriculture are as follows :

  1. Commercial exploitation of Rhizobium in agriculture : Continuous growth of crops leads to the nutrient deficiency in soil. Farmers, then, grow leguminous crops containing Rhizobium in its root nodules to replenish the lost nutrients (especially nitrogen) in the soil.
  2. Commercial exploitation of Mycorrhiza in agriculture.
    Mycorrhiza is an association of the soil fungus with the roots of the higher plants. Farmers use Mycorrhiza commercially in agriculture as it improves the soil quality and reduces soil erosion by improving plant rooting capacity. Fungal symbiont absorbs and store N2(nitrogen), calcium, phosphorus, and potassium in the fungal mantel, and overall increase in plant growth and development.

Ex-site conservation is the preservation of biological diversity outside their natural habitats. This involves conservation of genetic resources, as well as wild and cultivated or species, and draws on a diverse body of techniques and facilities. Some of these include :

  1. Gene banks, e.g. seed banks, sperm and ova banks, field banks.
  2.  In vitro plant tissue and microbial culture collections.
  3. Captive breeding of animals and artificial propagation of plants, with possible reintroduction into the wild, and
  4.  Collecting living organisms for zoos, aquaria, and botanical gardens for research and public awareness.


Question.11. State what is apomixis. Comment on its significance. How can if be commercially used ?
Answer : Apomixis is the mechanism of formation of new individuals without involving the process of meiosis and syngamy. In this process, gamete- formation is absent. Apomixis is observed in certain plants like grasses, citrus plants, conifers like pine etc. It occurs through agamospermy, parthenogenesis appogamy.Significance of Apomixis : Apomixis provides a better alternative for raising hybrid varieties from hybrid seeds. Hybrid seeds have to be produced every year as these cannot be collected from hybrid plants. The hybrid characters segregate during meiosis. If apomixis is introduced in the hybrid seeds there will be no need of producing seeds every year.
Commercial applications of apomixis are :

  1.  By apomixis, hybrid varieties of seeds can be produced, which will provide higher and better yield.
  2.  It prevents the loss of specific characteristics in the hybrid plants.
  3. This method is a cost-effective method of producing seeds.

Question.12. During a monohybrid cross involving a tall pea plant with a dwarf pea plant, the offspring populations were tall and dwarf in equal ratio. Work out a cross to show how it it possible.
Answer : The given condition is possible only when the tall pea plant is heterozygous and dwarf (small) pea plant is homozygous. This cross can be represented as follows :
Alleles can be similar as in the case of homozygotes (TT) and (tt) can be dissimilar as in the case of the heterozygote it. Since the TT plant is heterozygous for gene controlling one character, it is a monohybrid cross.

Question.13. Explain the significance of satellite DNA in DNA fingerprinting technique.
Answer: Satellite DNA are short sequences of DNA repeated again and again to form long sequences and generally non-coding DNA. It is very significant in DNA fingerprinting for the following reasons:

  1.  It is specific for every species
  2. It is a highly polymorphic DNA, an example of which is Variable Number of Tandem Repeats (VNTR). This means number of repeats vary in different individuals and these differences can be used for identification purpose.
  3. VNTR sequences are inheritable and an individual inherits it from both parents. This can be used for paternity testing.
  4. Regions of crossing over can also be easily marked, making identification easier.

Question.14. What does the following equation represent ? Explain f + 2pq + q2 = 1
Answer : The Hardy-Weinberg equation is a mathematical equation that can be used to calculate the genetic variation of a population at equilibrium. The Hardy-Weinberg equation is expressed asp + q = 1 where p is the frequency of the “A” allele and q is the frequency of the “a” allele in the population. In the equation, p2 represents the frequency of the homozygous genotype AA, q2 represents the frequency of the homozygous genotype cut, and 2pq represents the. Creoency of the. heterozygous Aa. Hence, p1 + 2pq + q2 = 1, which is the expression of (p + q)2.
Any deviation in the values of frequency of alleles from the usual value, indicate that there is an evolutionary change in the population.

Question.15. A heavily bleeding and bruised road accident victim was brought to a nursing home. The doctor immediately gave him an injection to protect him against a deadly disease.
(a) Write what did the doctor inject into the patients body.
(b) How do you think this injection would protect the patient against the disease ?
(c) Name the disease against which this injection was given and the kind of immunity it provides.
(a) In the patients body, the doctor has injected antiserum containing preformed antibodies against the causative organism or toxin produced by it.
(b) The solution injected by the doctor had antibodies; hence, the injection provide humoral immunity to patient and would protect the patient against the disease.
(c) The disease against which this injection was given is tetanus caused by Clostridium tetani, which usually exists in environment as spores and may gain access to the body through wound.
The kind of immunity that the injection containing
antiserum provides is passive immunity as performed antibodies are used because fast action is required in this emergency case.

Question.16. Enumerate any six essentials of good, effective Dairy Farm Management Practices.
Answer : Six important ways of good and effective dairy farm management practices are as follows :

  1.  Choosing the breeds that have high yielding potential is essential.
  2. Proper accomodation (including the aspects of hygiene) and adequate water are essential for the care of catde.
  3.  Periodic visit by a veterinary doctor should be compulsory for the good health of the cattles.
  4.  The fodder given to the cattle should be adequate quantity and good quality (including grains and protein concentrates).
  5.  The feeding of cattle should be carried out in a scientific
    manner and become mechanised which reduces chance of direct contact with the handler.
  6.  Regular inspection of dairy farms should be done by appointed officials to ensure that all the instructions are being strictly followed.

Question.17. State the medicinal value and the bioactive molecules produced by Streptococcus, Monascus and Trichoderma.
What are methanogens ? How do they help to generate biogas ?
Streptococcus : It produces Streptokinase, also called Tissue plasminogen activator (TPA) is used as clot buster to dissolve blood clots in patients who have undergone heart attacks.
Monascus : This fungal species produce statins that help in lowering blood cholesterol, statins, act as inhibitors of enzyme HMG CoA reductase of liver which form mevalonate, required . for cholesterol synthesis.
Trichoderma : It produces cyclosporin A which act as immunosuppressive agent by inhibiting the activation of T-cell response to transplanted organs.
Methanogens are the bacteria found in catde dung (gobar) and in anaerobic sludge during sewage treatment. They grow anaerobically on cellulosic material and produce a large amount of methane (main constituent of bio gas) along with C02 and H . Thus, methanogens are used in biogas production.

Question.18. Rearrange the following in the current sequences to accomplish an important biotechnological reaction :
(a) In vitro synthesis of copies of DNA of interest
(b) Chemically synthesized oligonucleotides
(c) Enzyme DNA-polymerase
(d) Complementary region of DNA
(e) Genomic DNA template
(f) Nucleotides provided
(g) Primers
(h) Thermostable DNA-polymerase (from Thermus aquaticus)
(i) Denaturation of ds-DNA.
Answer : The steps given are the steps involved in PCR (Polymerase chain reaction) correct sequence of the steps are :
(i) Denaturation of ds-DNA
(d) Complementary region of DNA
(c) Enzyme DNA polymerase
(b) Chemically synthesized oligonucleotides
(g) Primers
(h) Thermostable DNA-polymerase (from Thermus acquticus)
(f) Nucleotides Provided
(a) In vitro synthesis of region of DNA of interest

Question.19. Describe any three potential applications of genetically modified plants.
Answer: Three potential applications of genetically modified (GM) plants are as follows :

  1.  Nutrient Enrichment
    GM plants provide essential nutrients to people through the consumption of main staple crop.
    Example: Golden rice is a variety of rice that is genetically engineered by inserting genes to produce beta carotene, which is a precursor of Vitamin A. Thus, consumption of this crop helps in the prevention of Vitamin A deficiency diseases.
  2. Insect/Pest Resistance
    G.M. Crops Such as BT Cotton are insect|pest resistant Crops. It is a genetically modified variety of cotton that produces a protein called Bt toxin. The Bt toxin act as insecticide & kills cotton balloworms and Larvae of Lepidopterans & Dipterans.
  3.  Abiotic Stress Resistance
    GM crops are tolerant to stress conditions such as high and low temperature, salinity and drought.

Question.20. How did an American Company, Eli Lilly use the knowledge of r-DNA technology to produce an insulin ?
Answer : Insulin Consists of 51 amino acids forming two short polypeptide chains chain A having 21 amino acids and chain B with 30 amino acids. The two chains are linked by disulfide bond. In animals including humans, insulin occurs as proinsulin is made of chain A, chain B and chain C (30 amino acids). As the insulin matures, chain C is removed.
The American company, Eli Lily used the following technique to produce synthetic human insulin or humulin. The steps followed were:

  1.  The genetic engineering of insulin begins with identification & separation of DNA sequences coding for chain A and chain B.
  2. These sequences were then introduced into plasmid (pBR322) of E.coli bacterium. It is referred as factory used in genetic engineering of insulin.
  3. In E.coli, p-galactosidase controls the transcription of these genes, therefore, insulin gene needs to be tied to this enzyme. The protein formed by E-coli consists partly of P-galactosidase joined to either A or B chain of insulin.
  4.  These are then extracted from P-galactosidase fragment & purified.
  5. The two chains are mixed & reconnected in a reaction that forms disulfide bridges resulting in pure synthetic human insulin (humulin).

Question.21. How do snails, seeds, bears, zooplanktons, fiingi and bacteria adapt to conditions unfavourable for their survival ?
Answer: Snails adapt to unfavourable conditions by producing epiphragm during hibernation that covers the opening of its shell and thus prevent desiccation.
Seeds adapt to unfavourable conditions by getting into the state of dormancy.
Bears adapt to unfavourable conditions by hibernating and reducing their body metabolic activities by 75%. Zooplanktons adapt to unfavourable conditions by entering into diapause (stage of suspended development).
Fungi adapt to unfavourable conditions by reducing their metabolic rate and by forming thick-walled spores.
Bacteria adapt to unfavourable conditions by forming endospores.

Question.22. With the help of a flow chart, show the phenomenon of biomagnification of DDT in an aquatic food chain.
Answer: Biomagnification is the increase in the concentration of pollutants or harmful chemicals within each step of the food chain. The levels of biomagnification will be different at different trophic levels. For example, in a pond of water, DDT was sprayed and the producers were found to have 0.04 ppm concentration of DDT. Since many types of planktons are eaten by some fishes and clams, their body accumulates
0.23 ppm of DDT. Sea gull that feeds on clams accumulates- more DDT as one sea gull eats many clams. Hawk, the top carnivore, has the highest concentration of DDT.


Question.23. Your school has been selected by the Department of Education to organize and host an interschool seminar on “Reproductive Health – Problems and Practices”. However, many parents are reluctant to permit their wards to attend it. Their argument is that the topic is “too embarrassing”.
Put forth four arguments with appropriate reasons and explanation to justify the topic to be very essential and timely.
Answer: Reproductive health is the total well-being in all aspects of reproduction. It includes physical, emotional, behavioural and social well-being of an individual. Therefore, there is an urgent need to educate and discuss topics related to the reproductive health.
Following are the topics., about reproductive health that should be discussed with the students :

  1.  Sexually transmitted diseases, such as AIDS and Gonorrhoea, are transferred from one individual to another through sexual contact. Therefore, making the students aware about these diseases will help to jprevent their spread.
  2. Lack of knowledge about the reproductive status may lead to unwanted pregnancies. Hence, it is necessary to create awareness among people, especially the youth.
  3.  Learning about one’s sexuality at a proper age may help the students to know about the different changes happening in their body; thereby, leading to a better mental and physical state of health.
  4.  Counselling and creating awareness about reproductive health also help to curb the problems of infertility, birth control, mortality, etc.


Question.24. (a) Plan an experiment and prepare a flow chart of the steps that you would follow to ensure that the seeds are formed only from the desired sets of pollen grains. Name the type of experiment that you carried out.
(b) Write the importance of such experiments.
Describe the roles of pituitary and ovarian hormones during the menstrual cycle in a human female.
Answer : The experiment or technique that is carried out to obtain seeds from a desired set of pollen grains is referred as Artificial Hybridization. In this technique, only desired pollen grains are used in pollination and the stigma of the desired plant is kept protected from contamination from the unwanted pollen. This involves two processes :
(a) Emasculation and (b) Bagging.
(a) Emasculation : In this, the anthers are removed by fine forceps in their bud condition from the bisexual flowers of plants selected as female parent Such flowers are called emasculated flowers.
(b) Bagging : This involves covering the stigma of both emasculated and non-emasculated flowers with butter paper of polythene in their bud condition to prevent contamination from unwanted pollen.
When the stigmas of emasculated flowers mature, the bags are removed for a while. The stigmas are dusted with pollen grains fo desired male plants by means of a brush. The flowers are rebagged till the fruits develop. This is how desired seeds are obtained.
(b) Artificial hybridization is important for the following reasons :
It ensures that the crops produced have the desired characteristics.
It helps to improve the crop yield.
It helps to yield commercially superior varieties.
The cycle of events, starting from one menstruation till the next is called the menstrual cycle. Following are the changes that  are about by ovarian and pituitary hormones in the phases of menstrual cycle.

  1.  Menstrual phase : It lasts for 3-5 .days. The unfertilised ovum alongwith ruptured uterine epithelium, some blood & mucus is discharged through the vaginal orifice & this is called Menstruation. The decrease in the level of progesterone & estrogen in the event of failure of fertilization stimulates the hypothalmus & anterior pituitary to release GnRH and FSH, to start the phase of next menstrual cycle.
  2. Follicular/proliferative phase : It exhibits following changes:
    (a) Under the stimulation of GnRH of hypothalamus there is increased secretion of FSH from the anterior pituitary. The FSH stimulates the ovary and bring about the change| maturation of 1° follicle (6-12 in number) into a Graffian follicle. The follicular cells of Grafian follicle secrete estrogen.
    (b) This estrogen produced, stimulates the growth, maintenance & normal functioning of secondary sex organs.
    (c) The estrogen also aids in repairing the endometrial lining for 2 days & later this lining becomes more vascular in 7-9 days.
    (d) The end of phase marks the inhibition of FSH secretion & stimulation of LH secretion.
  3.  Ovulatory phase : Process of release of mature egg cell is called as ovulation. Ovuiationis controlled by increased level of LH. LH starts the change of empty Graffian follicle into corpus luteum and secretion of progesterone from corpus luteum which is responsible for maintenance of endometrium. Ovulatory phase is characterised by high levels of estrogen & progesterone.
  4. Luteal/Secretory phase : It secretes progesterone by suppressing the secretion of FSH & inhibits the maturation of follicle & ovulation. The uterine endometirum gets ready for implantation. In absence of GnRH, levels of LH & FSH falls, thus decreasing progesterone levels & prepare to start the menstrual phase.

Question.26. (a) List the different attributes that a population has and not an individual organism.
(b) What is population density ? Explain any three different ways the population density can be measured, with the help of an example each.
“It is often said that the pyramid of energy is always upright. On the other hand, the pyramid of biomass can be both upright and inverted”. Explain with the help of examples and sketches.
Answer : (a) Following are the attributes that a population
has but an individual organism does not have :
Birth rate : per capita Births.
Death rate : per capita deaths.
Sex ratio : Ratio of number of males to females in a population.
(b) Population density means numbers of individuals present per unit area. Population density can be measured by determining the population size.
Different ways to measure population density are :

  1.  Direct method : It involves the counting of organisms in the given area, e.g., in order to determine the number of bacteria growing in a Petri dish, their colonies are counted.
  2. Quadrat method : It is the method that involves the use of square of particular dimensions to measure the number of organisms, e.g., the number of Parthenium plants in a given area can be measured using the quadrat method.
  3.  Indirect method : In this method, there is no need to count the organism individually, e.g., the number of fishes caught per trap gives the measure of their total density in a given water body.

Pyramid of Energy
The graphical representation of the amount of energy trapped at different trophic levels of food chain in an ecosystem is called pyramid of energy. It represents the total amount of energy consumed by each trophic level. The amount of energy transferred through food to successive higher levels decreases with each level. This means that a greater amount of energy is available at the producer level than at the primary consumer level (herbivores). This is because some amount of energy is lost at each trophic level. The energy that is lost is almost 80-90% (by the second law of thermodynamics) of the energy available at the lower trophic level. Thus, this pyramid is always upright or straight. The law states that only 10% of chemical energy is retained at each trophic level and this law is known as the 10 percent law. This 10% of energy is only available to the next trophic level.

Pyramid of Biomass
Biomass is the renewable organic (living) material or the energy contained inside plants and animals. It is measured as both fresh weight and dry weight. It is measured in gram/ (meter)2 or calories/(meter)2. A graphical representation that shows the amount of biomass present at each tropic level in a unit area at any point of time is called pyramid of biomass. The biomass pyramids are of two types : upright and inverted. Pyramids of biomass in a terrestrial ecosystem is an upright or straight pyramid. In this type of pyramid, the combined weight of producers is larger than the combined weight of consumers.



Question.3. State the cause of adenosine deaminase enzyme deficiency.
Answer : Adenosine deaminase (ADA) enzyme deficiency is caused by the mutation in the ADA gene. This gene is present on chromosome number 20 and provides instruction for producing the enzyme adenosine deaminase. ADA deficiency is inherited as an autosomal recessive trait.


Question.8. Explain the process of secondary treatment given to the primary effluent up to the point it shows significant change in the level of biological oxygen demand (BOD) in it.
Answer: Secondary treatment refers to those treatment process that use biological process to convert dissolved suspended and colloidal organic wastes to more stable solids that can either be removed by settling or discharged to the environment without causing harm. Supernatant from the primary treatment is passed into large aeration tanks during secondary treatment. In these tanks, the effluent is agitated mechanically and air is pumped into it. This causes vigorous growth of the bacteria that lead to the formation of floes, containing bacteria and fungal filaments in a mesh-like structure. While growing, these microbes consume the major part of organic matter in the effluent; it decreases the biological oxygen demand (BOD), the BOD of sewage or waste water is reduced significantly, the effluent if then passed into a setding tank where the bacterial ‘floes’ are allowed to sediment, called activated sludge.

Question.10. In mosses sexual reproduction cannot complete without water. A moss plant is unable to complete its life-cycle in a dry environment. State two reasons.
Answer : Mosses cannot complete their life cycle in a dry environment because of the following reasons :
1. Water acts as a medium for flagellated sperm to reach the egg and undergo fertilization.
2. Mosses cannot absorb water because their roots are rudimentary, thus, for their survival, they need to grow in moist environment.


Question.14. Two independent monohybrid crosses were carried out , involving a tall pea plant with a dwarf pea plant. In the first . cross, the offspring population had equal number of tall and dwarf plants, whereas in the second cross it was different. Work out the crosses, and explain giving reasons for the difference in the off sping populations.
Answer : First Cross (Case I)
In the first cross between the tall plant and the dwarf plant, equal number of tall and dwarf plants are produced. This is because the tall plant is heterozygous dominant (Tt) and dwarf plant is homozygous recessive (tt). The progenies will be 50 % tall and 50 % dwarf.
Second Cross (Case II)
In the second cross between the tall plant and the dwarf plant, all the offsprings will be tall. This is because the tall plant is  homozygous dominant (TT) and dwarf plant is homozygous recessive (tt).
The two monohybrid crosses can be represented as follows :

Question.19. Explain co-evolution with reference to parasites and their hosts. Mention any four special adaptive features evolved in parasites for their parasitic mode of life.
Answer : Co-evolution is a term used to describe the mutual changes in two or more species, usually one following the other, that affect their interactions. In terms of the relation of host and parasite, it can be explained as follows :
A parasite is an organism that derives nourishment and protection from other living organism known as host, but in doing so, it also harms the host. The host evolves over a long period of time to protect itself from parasite, while parasite evolves so that it can find another way to derive nutrition from the host and hence, the cycle continues.
Four special adaptive features evolved in parasites for their parasitic mode of life are as follows :

  1.  Parasites have suckers as organs for attachment such that help them to firmly attach to the host body and help in deriving nutrition from them. For example suckers and hooks in Taenia solium.
  2. Parasites are covered by protective body covering, that is tegument (in case of Taenia solium) and cuticle (in case of Ascaris lumbricoides) to protect them from harmful effects of digestive enzymes of the host.
  3.  All parasitic organisms usually lack locomotory structures, they do not require to move in search of food.
  4.  High reproductive capacity of parasitic organisms ensures the continuation of parasitic race.

Question.21. With the help of a flow-chart exhibit the events of eutrophication.
Answer : Enrichment of water body by excessive nutrients is known as eutrophication.


Question.26. (a) Why are colour blindness and thalassemia categorised as Mendelian disorders ? Write the symptoms of these diseases seen in people suffering from them.
(b) About 8% of human male population suffers from colour blindness whereas only about 0.4% of human female population suffers from this disease. Write an explanation to show how it is possible.
Explain the process of transcription in prokaryotes. How is the process different in eukaryotes ? 
Answer : (a) Colour blindness and thalassaemia are categoried as Mendelian disorders because they are caused by mutation in a single gene. Their mode of inheritance follows the principles of Mendelian genetics. Mendelian disorders can be:
(i) autosomal dominant (muscular dystrophy)
(ii) autosomal recessive (thalassaemia)
(iii) sex linked (colour blindness)
Symptoms of Thalassaemia
(a) growth problems – not putting on weight or growing in height.
(b) anaemia – red blood cell deficiency, leading to tiredness, weakness and shortness of breath.
(c) jaundice – yellowing of the skin and whites of the eyes.
(d) swollen abdomen (tummy) – this is caused by an enlarged liver of spleen.
Symptoms of Colour Blindness
(a) Difficulty distinguishing between colors
(b) Inability to see shades or tones of the same color
(c) Rapid eye movement (in rare cases)
The events of eutrophication are as follows :
(b) Colour blindness is a X-linked recessive disorder. Males have higher chances of getting affected in comparison to females because males have only one X with Y chromosome while females have two X chromosomes (XX). Thus, for a female to get affected by colour blindness, she has to have the mutant gene on both the X chromosomes while males can be affected if they carry it on the single X chromosome.
From the given table, it can be concluded that females have very less probability of getting this disease as compared to males. Females will be colour-blind only when either both parents are affected or male is affected and female is carrier, while males can be colour-blind even if female is carrier and male is normal.
The transfer of genetic information from DNA to mRNA is known as Transcription. It is done by formation of RNA over the template of DNA. It creates single stranded RNA which has a coded information similar to the sense or coding strand of DNA (with exception of U in place ofT).
The segment of DNA that takes part in transcription is called Transcription Unit. It has 3 components—a promoter, structural

  1. Promoter region : It is the proximal area of transcription unit, which provides sites for attachment of transcription factors (CT factor) & RNA polymerase. It is present upstream at 5’ end of coding & 3’ end of template strand. It has AT rich area called TATA box or pribnow Box.
  2.  Structural gene: It is the area of template strand involved in transcription. It is polycistronic in prokaryotes.
  3.  Terminator region : It is the distal end of transcription unit. It is present downstream at 5’ end of template strand. It provides site for attachement of termination factor (rho-factor) for release of RNA polymerase.
    The main enzyme that takes part in transcription is called RNA polymerase. RNA polymerase has a sigma factor for recogniyng the start signal of promoter region. The remaining part of RNA polymerase is called Core enzyme.

Mechanism of Transcription :

  1.  Activation of Ribonucleotides : 4 types of
    ribonucleotides take part in synthesis of RNA over DNA. These are AMP, GMP, UMP & CMP. These are found in nucleoplasm. These are converted to active triphosphates by energy, phosphate & phosphorylase enzyme.
  2.  DNA Template : The DNA strand which function as a template for RNA synthesis is the Template or Antisense strand. It has 3’ —» 5’ polarity & transcription proceeds in 5’ —» 3’ direction. The separation of template strand does not require specific enzymes as in DNA replication. Ill transcription, only the RNA polymerase travels along the template strand.
  3. Initiation : RNA polymerase enzyme binds to the promoter region. The CT (Sigma) factor recognises the recognises the promoter site and initiates transcription, by using triphosphates.
  4. Elongation : (a) The activated ribonucleotides come to lie opposite to the template strand, complementarily & starts pairing. This releases phosphate radical & energy.
    (b) The core enzyme with the help of energy & Mg2* ions, bonds the adjacent nucleotides to form RNA chain. The RNA remians attached to the enzymes in a small region. As the enzyme moves along the DNA, the RNA chain alongates till terminator region is reached.
  5.  Termination : (a) When enzyme reaches, terminator region, the RNA polymerase, by means of Rho (p) factor & NUS G. This results in termination of transcription.
    (b) p factor has ATPase activity. After RNA separation, the sense & antisense strands of DNA re-establish Fl-bonds.

Difference in Eukaryitic Transcription :

  1.  In Eukaryotes, the promoter site is recognised by presence of specific nucleotide sequence called TATA box (Hogness Box) on template strand.
  2.  In Eukaryotes, RNA polymerase is of 3 types-I, II & III.
  3. Monocistronic structural genes are present which have uninterrupted coding sequences, i.e. the information on gene is split. The coding sequences are called exons & the non-coding sequences occuring as interruption in exons are
    called Introns. All introns have GU at 5’ end and AG at 3’ end.