HS 2nd Year Biology Question Paper Solution
H.S. (Final year) 2013
Sub: Biology
Full Marks: 70
Part-I (BOTANY)
1. Fungal association with roots of higher plants is called _______.
Ans: Mycorrhiza
2. Write true or false:
(a) Apomixis is a phenomenon in which ovary develops into a fruit without fertilization.
Ans: False. Apomixis refers to the formation of seeds without fertilization, not the development of the ovary into a fruit.
(b) Aquatic fern 'Marsilea' is an excellent biofertilizer.
Ans: False. Marsilea is not a biofertilizer; it is an aquatic fern used for ornamental purposes or food in some regions.
3. What is 'hidden hunger'?
Ans: Hidden hunger is a condition caused by the deficiency of essential micronutrients like iron, zinc, and vitamins in a person's diet. Although the individual consumes enough calories, the lack of these vital nutrients leads to health problems, reduced immunity, and developmental issues.
Or
What is carrying capacity?
Ans: Carrying capacity is the maximum number of individuals in a population that an environment can sustainably support without degrading its resources. It depends on the availability of food, water, shelter, and other ecological factors.
4. 'Archaeopteryx' is considered a connecting link between reptiles and birds – Explain.
Ans: Archaeopteryx is a fossil organism that shares characteristics with both reptiles and birds, making it a transitional species. It possesses reptilian traits like a long bony tail, teeth in its jaws, and clawed fingers. At the same time, it has bird-like features, including feathers and wings, which are essential for flight. These mixed traits support the theory of evolution, showing the transition from reptiles to birds.
5. An orchid plant is growing on the branch of a mango tree. How do you describe this interaction between them?
Ans: This interaction is an example of commensalism. The orchid benefits from the mango tree by using it as a support to grow and access sunlight. The orchid does not derive nutrients from the tree, nor does it harm the tree. The mango tree remains unaffected, as it neither gains nor loses anything from the presence of the orchid.
6. How is DNA isolated in purified form from a bacterial cell?
Ans: DNA is isolated using the following steps:
Cell wall breakdown: The bacterial cell wall is broken down using enzymes like lysozyme.
Cell membrane disruption: Detergents or surfactants are added to dissolve the lipid bilayer of the cell membrane, releasing cellular contents.
Protein removal: Proteins are removed using proteases, and RNA is degraded using RNase enzymes.
DNA precipitation: Salt is added to neutralize the charge on DNA, and alcohol (e.g., ethanol or isopropanol) is used to precipitate the DNA. The DNA is then spooled out or collected by centrifugation.
Or
Define antigen and antibody with a suitable example for each.
Ans:
Antigen: A molecule or substance that induces an immune response in the body. For example, the spike protein of the SARS-CoV-2 virus acts as an antigen.
Antibody: A specialized protein (immunoglobulin) produced by the immune system in response to an antigen. For example, monoclonal antibodies are artificially produced to target specific antigens like those on cancer cells.
7. Write the scientific names and their utility of the following plants (any two):
(a) Neem (Azadirachta indica) – Used for its antibacterial, antifungal, and insecticidal properties.
(b) Cinchona (Cinchona officinalis) – The source of quinine, an effective treatment for malaria.
(c) Teak (Tectona grandis) – Renowned for its durable and water-resistant timber, widely used in furniture and construction.
8. Give the classification of biofertilizers with one example from each group.
Ans:
(i) Free-living nitrogen-fixing bacteria – Example: Azotobacter
(ii) Free-living nitrogen-fixing cyanobacteria – Example: Nostoc
(iii) Symbiotic nitrogen-fixing bacteria – Example: Rhizobium
(iv) Symbiotic nitrogen-fixing cyanobacteria – Example: Anabaena azollae
(v) Mycorrhiza:
Ectomycorrhiza – Example: Rhizopogon
Endomycorrhiza – Example: Glomus
Or
“Give me a living cell of any plant, and I will give you a thousand plants of the same type.” Is it scientifically possible? Justify.
Ans: Yes, it is possible through plant tissue culture, a technique in biotechnology. Plant cells or tissues are grown on a nutrient medium under sterile conditions, allowing them to develop into whole plants. This process helps produce thousands of genetically identical plants from a single parent cell, useful in agriculture and horticulture.
9. Discuss the Modern Synthetic Theory of evolution.
Ans: The Modern Synthetic Theory of evolution combines Darwin’s theory of natural selection with Mendelian genetics and molecular biology. It explains evolution as a gradual process driven by:
Genetic variation: Caused by mutations, genetic recombination during reproduction, and gene flow between populations.
Natural selection: The differential survival and reproduction of individuals based on their traits.
Genetic drift: Random changes in allele frequencies in small populations.
Speciation: Accumulation of genetic changes over generations leads to the emergence of new species.
This theory emphasizes the role of populations, gene pools, and gradualism in evolutionary changes.
10. What do you mean by speciation? Name different forms of speciation. 1+2=3
Ans: Speciation is the formation of one or more new species from an existing species.
The different forms of speciation are-
(i) Allopatric speciation: In this type a part of the popu- lation becomes geographically isolated from the main population.
(ii) Sympatric Speciation: In this type a small segment of the original population becomes isolated reproductively.
(iii) Parapatric Speciation: It takes place when a popu- lation of a species enters a new niche or habitat.
(iv) Quantum speciation: It is the rapid and abrupt mode of species formation.
11. What is sewage? In which way sewage can be treated? (1+2=3)
Ans:
Sewage is municipal wastewater that contains human excreta, household waste, and other organic pollutants.
Sewage can be treated in two stages:
Primary treatment:
This is a physical process that removes large and small floating or suspended solids.
It involves filtration to trap larger particles and sedimentation to separate heavier particles from the liquid.
Secondary treatment:
This is a biological process where microorganisms decompose organic matter in the sewage.
Microbial activity forms flocs, which settle as sludge. The sludge is further digested to reduce waste. This process helps in purifying the sewage further.
Or
What is a haploid plant? Mention the purpose of production of haploid plants in modern cultivation. (1+2=3)
Ans:
A haploid plant originates from an unfertilized egg cell and contains only one set of chromosomes (n).
Purpose of production of haploid plants:
Haploid plants are highly sterile but useful in modern agriculture.
They can be converted into homozygous diploids in 2–3 years using colchicine treatment or through culture-induced diploidy.
Homozygous plants are extensively used in crop improvement programs, mutation studies, and allele analysis.
Superior homozygous lines are selected for breeding programs to develop new and improved crop varieties.
12. Explain any three methods to force recombinant DNA into a host cell. (3)
Ans:
Three methods to introduce recombinant DNA into host cells are:
Transformation:
Host cells are made competent by treating them with calcium chloride to uptake foreign DNA.
The DNA enters the cell via pores in the cell membrane.
Electroporation:
An electrical pulse creates temporary pores in the cell membrane.
Recombinant DNA is then forced into the host cell through these pores.
Microinjection:
Direct injection of DNA into the nucleus of a host cell using a fine micro-needle or micropipette.
This method is commonly used in animal cells.
Or
What do you mean by GEAC? What are its roles? (1+2=3)
Ans:
GEAC stands for Genetic Engineering Approval Committee.
Roles of GEAC:
It evaluates and approves genetically modified organisms (GMOs) for research, production, and commercialization.
Ensures that GMOs and their products are safe for human consumption and the environment.
Regulates the import, export, and field trials of GMOs.
13. Describe the stages of embryo sac formation in angiosperms with a suitable diagram. (3+2=5)
Ans:
The process of embryo sac formation is called megagametogenesis. It involves the following stages:
Functional megaspore formation: The diploid megaspore mother cell undergoes meiosis to produce four haploid megaspores. Out of these, only one functional megaspore survives.
Nuclear division: The functional megaspore enlarges and undergoes three mitotic divisions to produce eight nuclei.
Polar nuclei movement: Four nuclei move to each pole. One nucleus from each pole migrates to the center, forming two polar nuclei.
Cellular organization:
At the micropylar end, three cells form the egg apparatus (1 egg cell and 2 synergids).
At the chalazal end, three cells form antipodal cells.
The central cell contains the two polar nuclei.
This 7-celled, 8-nucleate structure is the mature embryo sac.
14. What are molecular scissors? Explain their role in recombinant DNA technology. (1+4=5)
Ans:
Molecular scissors are enzymes, also called restriction endonucleases, that cut DNA at specific sequences called recognition sites.
Role in recombinant DNA technology:
Recognition and cutting: They identify specific palindromic sequences in DNA and cut them to produce sticky or blunt ends.
Gene insertion: The sticky ends facilitate the insertion of foreign DNA fragments into a vector (e.g., plasmid).
Creation of recombinant DNA: The foreign DNA and vector are joined using DNA ligase, forming recombinant DNA.
Precision and specificity: Molecular scissors ensure that only the desired DNA fragments are cut and inserted, making the process highly accurate.
Applications: They are crucial in genetic engineering for cloning, gene therapy, and the production of genetically modified organisms (GMOs).
Or
Soil bacleruim Bacillus thuringiensis possesses a gene fam- ily cry gene over its plasmid which synthesises an endotoxic protein, cry protein capable of killing larvae of certain insects. As the toxic Cry protens are obtained from Bacillus thuringiensis, they are called by toxins. The genes have been cloned from bac- teria and incorporated in many plants. The genetically modified crop plants having Bt toxin genes do not require protection of insecticides as they themselves act as bioinsecticides. The insec- ticide protains are produced by the bacterial genes in inactive protoxin crystalline state. As an insect feeds over the plant the inactive protoxin crystals pass into its gutwhere alkaline PH and digestive enzymes sdubilise the crystals and convert protoxin into toxin. The toxin binds to epithelial cells midgut of insects and produces pores which result in swelling and lysis of cells and as a result the insect dies. Thus by applying techineques of biotech- nology insect resistant plants can be developed.
Part-II (ZOOLOGY)
1. Fill in the blank: (any two) 1×2=2
(a) Kaseeru is______food plant of eri silkworm.
(b) Filariasis is caused by_______worm
(c) Acrosome of sperm is formed from the________.
(d) Phenylketonuria causes________.
Ans: (a) Host
(b) Wuchereria
(c) Golgi complex of spermatid
(d) Mental retardation.
2. Answer any two:
(a) What is a transgenic animal?
Ans: A transgenic animal is an organism whose genome has been altered by the introduction of foreign DNA using genetic engineering techniques. These animals are created to study gene function, develop models for diseases, or produce therapeutic proteins.
(b) Express the genotype of AB blood group.
Ans: The genotype of the AB blood group is IAIB. This is an example of codominance, where both A and B alleles are expressed equally.
(c) Why should motor vehicles fitted with catalytic converters use unleaded petrol?
Ans: Motor vehicles fitted with catalytic converters should use unleaded petrol because lead in fuel can poison the catalysts, rendering the catalytic converter ineffective at reducing harmful emissions.
(d) Write the equation of species-area relationships in a logarithmic scale.
Ans: The equation for species-area relationships in a logarithmic scale is:
Where:
S = Number of species
C = Constant
A = Area
Z = Slope of the line (species-area exponent)
3. Write answers of any four of the following: 2×4=8
(a) How is placenta formed?
Ans: The placenta is formed when the trophoblast cells of the blastocyst invade the endometrial lining of the uterus. This results in the formation of a vascular connection between the mother and the fetus, allowing the exchange of nutrients, gases, and waste products.
(b) Why does Cu-T act as an effective contraceptive?
Ans: Cu-T is an effective contraceptive as it releases copper ions in the uterus, which:
Inhibit sperm motility and viability.
Create a hostile environment for fertilization and implantation.
(c) Explain the chromosomal theory of inheritance.
Ans: The chromosomal theory of inheritance states that genes are located on chromosomes, and their segregation and independent assortment during meiosis explain inheritance patterns. This theory was proposed by Sutton and Boveri and correlates Mendel's laws with chromosomal behavior.
(d) What are the four features of genetic material?
Ans: The four features of genetic material are:
Replication: It must be able to copy itself accurately.
Storage of information: It should store all the necessary information for cellular functions.
Expression of information: It must direct the synthesis of proteins and other molecules.
Variation: It should be capable of mutations for evolutionary adaptability.
(e) State the symptoms of typhoid.
Ans: Symptoms of typhoid include:
Sustained high fever (up to 104°F).
Weakness and fatigue.
Abdominal pain.
Loss of appetite.
Constipation or diarrhea.
(f) Write the names of two primary and two secondary host plants of muga silkworm.
Ans:
Primary host plants: Som (Machilus bombycina) and Soalu (Litsaea polyantha).
Secondary host plants: Dighloti (Litsaea citrate) and Mejankori (Litsea salicifolia
4. State the difference between (any two) 2×2= 4
(a) Passive immunity and active immunity.
Ans:- Passive immunity and active immunity
Passive Immunity:
Passive immunity occurs when antibodies are directly transferred to an individual, rather than being produced by their immune system.
It provides immediate, short-term protection. Examples include antibodies passed from mother to child through breast milk or injected during treatments (e.g., antiserum for snake bites).
Active Immunity:
Active immunity develops when an individual's immune system produces its own antibodies in response to an antigen, either through infection or vaccination.
It provides long-term protection, as memory cells are created.
(b) Down's syndrome and Turner's syndrome.
Ans:- Down's syndrome and Turner's syndrome
Down's Syndrome:
Caused by the presence of an extra chromosome 21 (trisomy 21).
Symptoms include intellectual disability, characteristic facial features (flat face, almond-shaped eyes), and developmental delays.
Turner's Syndrome:
Occurs in females due to the absence of one X chromosome (45, X karyotype).
Symptoms include short stature, webbed neck, lack of secondary sexual development, and infertility.
(c) Primary succession and secondary succession.
Ans:- Primary succession and secondary succession
Primary Succession:
Occurs in areas where no life existed previously, such as bare rock or lava flows.
Starts with pioneer species like lichens or mosses that create soil for other organisms over time.
Secondary Succession:
Occurs in areas where life existed but was disturbed (e.g., after a forest fire or farming).
Soil is already present, so the recovery process is faster compared to primary succession.
(d) Follicular phase and Luteal phase of Menstrual cycle.
Ans:
Down's syndrome
1. The cause of this genetic disorder is the presence of an additional copy of the chromosome number 21 (21 trisomy)
2. The affected individual is short statured with small round head, furrowed tongue and partially open mouth. Palm is broad with character- istic palm crease. Physical, psychomotor and mental development is retarded.
Turners syndrome
1. It is caused due to the absence of one of the X chromosomes, i.e. 45 with XO.
2. They are sterile female with rudimentary ovaries have shield shaped thorax webbed neek, poor development of breasts, short stature, small uterees and puffy fingers.
c) Primary Seccession
1. It occurs in a bare area.
2. Soil is absent at the time of beginning.
3. Humus or organic matter is absent in the early stages
4. Seral communities are many.
5. No reproductive structures of any living being is present.
Secondary Succession
1. It occurs in an area which has been denuded recently.
2. Soil is present.
3. Humus or organic matter is present from the beginning.
4. Seral communities are few.
5. Reproductive structures such as seeds, spores, etc. are present.
d) Follicular phase
1. It extends for about 10-12 days usually from 6th day to 13th day in a 28 day cycle.
2. During this period one ova- rian follicle is changes to graafian follicle.
3. Oestrogens are secreted.
4. Endometrium is about 2-3 mm thick.
Luteal Phase
1. It lasts for 13-14 days after ovulation (between 17-28 days)
2. Empty graafian follicle converts into corpus luteum.
3. Progesterone is secreted.
4. Endpmetrium becomes 5mm thick.
5. Draw a neat labeled diagram of sperm. 3
Ans:
Or
What do you understand by fishery? Discuss its impor- tance. 1+2=3
Ans: A fishery is an industry or practice related to catching, processing, or selling fish and other aquatic organisms for economic purposes.
Importance:
Economic growth: Fisheries contribute to the economy by providing employment and generating revenue through exports.
Food source: They provide a rich source of protein and essential nutrients.
Livelihood: Millions of people, especially in coastal areas, depend on fisheries for their livelihood.
Or
6. What is co-dominance? Explain with example.1+2=3
Ans:- Co-dominance is a genetic condition where both alleles of a gene in a heterozygous individual are equally expressed, resulting in a phenotype that shows characteristics of both alleles.
Example:
In humans, the AB blood group is an example of co-dominance. The IA and IB alleles are co-dominant. Individuals with genotype IAIB express both A and B antigens on the surface of red blood cells.
Or
Discuss the lac-operon model of gene-expression regula- tion. 3
Ans: When two alleles of a gene are equally dominant and express themselves in the presence of the other the phenomenon is called codominance.
e.g. ABO blood grouping in humans ABO blood groups are controlled by gene I Gene I has three alleles I^, I and i. I^ and IB produce RBC surface antigens A and B, respectively whereas i does not produce any antigen. I^ and I are the dominant alleles whereas i is the recessive allele. When I and I are present to- gether both express equally and produce the surface antigens A and B. Thus alleles for blood group A(I^) and blood group B (IB) are codominant so that when they come together in an individual they produce blood group AB.
Or
7. What are the consequences of loss of biodiversity?
Ans: Biodiversity refers to the variety and variability of life forms on Earth. The loss of biodiversity has serious consequences, affecting ecosystems, human society, and the planet as a whole. Here are some key consequences:
Ecosystem instability: Biodiversity is crucial for ecosystem stability. When species are lost, ecosystems become more vulnerable to disturbances such as climate change, diseases, or natural disasters. For example, the extinction of a species that plays a key role in pollination or water filtration can disrupt the entire ecosystem.
Loss of ecosystem services: Biodiversity supports ecosystem services such as pollination, water purification, soil fertility, and carbon sequestration. Without these services, food production, clean water, and air would become scarce, impacting human well-being.
Economic losses: Many industries, such as agriculture, forestry, and fisheries, rely on biodiversity. Loss of species can lead to reduced crop yields, depletion of fish stocks, and the collapse of forest ecosystems, affecting the livelihoods of millions of people.
Health risks: Loss of biodiversity can lead to the disappearance of medicinal plants or other natural resources that are important for treating diseases. Furthermore, reduced biodiversity can cause new diseases to emerge, as pathogens can spread more easily in disturbed ecosystems.
Cultural impact: Many indigenous and local communities depend on biodiversity for their culture, traditions, and livelihoods. Loss of species can erode cultural heritage and the knowledge tied to those species.
Or
Write a note on prevention of AIDS.
Ans: AIDS (Acquired Immunodeficiency Syndrome) is caused by the HIV (Human Immunodeficiency Virus), which weakens the immune system by attacking white blood cells. The prevention of AIDS involves various strategies to reduce the transmission of HIV. These include:
Safe sexual practices: The most effective way to prevent HIV transmission is through safe sex practices. This includes the consistent and correct use of condoms, which act as a barrier to prevent the exchange of bodily fluids.
Education and awareness: Educating people about the modes of HIV transmission and the preventive measures is essential. Public health campaigns can raise awareness and promote safe sexual behavior.
Testing and counseling: Regular HIV testing and counseling are important in identifying individuals who are infected with the virus. Early diagnosis allows for treatment, and counseling helps people to understand how to prevent the spread of HIV.
Avoiding sharing needles: HIV can also be transmitted through the sharing of needles among intravenous drug users. Promoting the use of sterile needles and offering needle exchange programs are important steps in preventing the spread.
Antiretroviral therapy (ART): People living with HIV can take antiretroviral drugs that help manage the virus and reduce the viral load, making it less likely to transmit the virus to others.
Mother-to-child transmission prevention: Pregnant women who are HIV-positive can receive treatment to reduce the risk of transmitting the virus to their babies during childbirth or breastfeeding.
8. Discuss the various means of prevention and control of drug and alcohol abuse.
Ans: Drug and alcohol abuse can have devastating effects on health, family, and society. Prevention and control of this abuse require a multi-pronged approach:
Education and awareness: Public education campaigns that explain the dangers of drug and alcohol abuse are crucial. Schools, communities, and media outlets can be utilized to spread awareness and provide accurate information.
Counseling and therapy: Counseling and psychological therapy help individuals understand the root causes of their addiction and develop coping strategies. Therapy may include behavioral therapy, cognitive therapy, and group counseling.
Support groups and rehabilitation: Rehabilitation centers and support groups such as Alcoholics Anonymous (AA) or Narcotics Anonymous (NA) offer support to people recovering from addiction. These groups provide a sense of community and shared experiences that aid in recovery.
Family involvement: Families play a critical role in preventing and controlling drug and alcohol abuse. Family therapy and support can help individuals deal with underlying issues that may lead to substance abuse.
Government and legal policies: Governments can enact laws and policies to control the sale and distribution of alcohol and drugs. Additionally, providing access to rehabilitation programs and enforcing penalties for illegal drug use and trafficking are vital.
Early intervention: Early intervention programs in schools, workplaces, and communities help identify at-risk individuals and offer support before addiction develops.
Promoting healthy lifestyles: Encouraging physical activity, healthy hobbies, and social engagement can help individuals resist the temptation to abuse drugs and alcohol.
9. Why DNA replication is said to be semiconservative? How Meselson and Stahl provided evidence in favor of the semiconservative nature of DNA replication?
Ans: DNA replication is described as semiconservative because, during the replication process, each of the two strands of the original DNA molecule serves as a template for the formation of a new complementary strand. As a result, after replication, each new DNA molecule consists of one original strand (parent strand) and one newly synthesized strand.
Meselson and Stahl's Experiment:
Meselson and Stahl conducted an experiment in 1958 using E. coli bacteria to demonstrate that DNA replication is semiconservative. They grew E. coli in a medium containing heavy nitrogen isotopes (N-15) and then transferred the bacteria to a medium containing lighter nitrogen isotopes (N-14). After one round of replication, they extracted the DNA and analyzed its density using a technique called density gradient centrifugation.
First round of replication: The DNA molecules showed an intermediate density, indicating that each DNA molecule contained one strand with heavy nitrogen (from the N-15) and one strand with light nitrogen (from N-14). This supported the idea that one strand of the original DNA was conserved in each daughter molecule.
Second round of replication: The DNA showed two distinct bands: one with intermediate density and one with light density. This confirmed that the DNA replication was semiconservative, as some of the DNA molecules had both strands composed of light nitrogen, while others had one light and one heavy strand.
Thus, the results of their experiment provided strong evidence in favor of the semiconservative model of DNA replication.
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