RBSE 12th Biology Model Paper Set 4 with Answers in English

Students must start practicing the questions from RBSE 12th Biology Model Papers Set 4 with Answers in English Medium provided here.

RBSE Class 12 Biology Model Paper Set 4 with Answers in English

Time : 2.45 Hours
Maximum Marks : 56

General Instruction to the Examinee:

1. Candidate must write first his/her Roll No. on the question paper compulsorily.
2. All the questions are compulsory.
3. Write the answer to each question in the given answer book only.
4. For questions having more than one part the answers to those parts are to be written together in continuity.
5. If there is any error/difference/contradiction in Hindi & English version of the question paper, the question of the Hindi version should be treated valid.

Section – A

(1 Mark)

Question 1.
Select the correct option of the following questions and write in notebook. (1 × 9 = 9)
(i) Conidia are produced by which of the following? [1]
(a) Aspergillus
(b) Chlamydomonas
(c) Yeast
(d) Albugo
Answer:
(a) Aspergillus

(ii) The bisexual flower which never open represents : [1]
(a) allogamy
(b) autogamy
(c) cleistogamy
(d) None of these
Answer:
(c) cleistogamy

(iii) Trisomy of 21st chromosome results in: [1]
(a) Down’s syndrome
(b) Sickle cell anaemia
(c) Turner’s syndrome
(d) Klinefelter’s syndrome
Answer:
(a) Down’s syndrome

(iv) DNA length per turn of helix is : [1]
(a) 3.8 Å
(b) 3.4 Å
(c) 6.8 Å
(d) 34 Å
Answer:
(d) 34 Å

(v) Black lung disease is common in: [1]
(a) farmers
(b) refinery workers
(c) petrochemical industry
(d) coal mines
Answer:
(d) coal mines

(vi) Radiation is health hazard because it causes : [1]
(a) haemophilia
(b) leukemia
(c) pneumonia
(d) leucopenia
Answer:
(b) leukemia

(vii) Murrah is the productive breed of: [1]
(a) chick
(b) goat
(c) cow
(d) buffalo
Answer:
(d) buffalo

(viii) Firstly biotechnology word used by: [1]
(a) Karl Mullis
(b) Karl Erkey
(c) Watson and Crick
(d) Alec Jeffery
Answer:
(a) Karl Mullis

(ix) dsRNA in RNAi is cut by: [1]
(a) dicer protein
(b) endonuclease
(c) aniline
(d) ligase
Answer:
(a) dicer protein

Question 2.
Fill in the blanks. (1 × 4 = 4)
(i) Milk yielding breeds of cattle are called ………………. [1]
(ii) About ………………. percent of India’s geographical land area is covered by forest. [1]
(iii) DNA molecule takes a complete turn at a distance of ………………. nm. [1]
(iv) The pyramid of energy is always ………………. [1]
Answer:
(i) milch breed
(ii) 23
(iii) 3.4
(iv) upright.

Question 3.
Give the answers of the following questions in one word or one line. (1 × 8 = 8)
(i) What is sickle cell anaemia? [1]
Answer:
Sickle cell anaemia is an autosomal-linked recessive trait (disorder) that can be transmitted from parents to the offspring when both the partners are carrier for the gene.

(ii) What is the exception to Mendel’s law of independent assortment? [1]
Answer:
Linked genes are exceptions to the law of independent assortment because two genes are located on the same chromosome, but this is generally mitigated when chromosomes cross over.

(iii) Name two viral diseases and their causal viruses. [1]
Answer:
(a) Chicken Pox — Varicella herpes virus
(b) AIDS — HIV virus.

(iv) What is blue-revolution. [1]
Answer:
Inereasement in production of fish and other aquatic animals is known as blue revolution.

(v) Basic difference between plasmid and chromosome. [1]
Answer:
Plasmids are circular extrachromosomal DNA while chromosomes are the type of genomic DNA having important factors for growth, reproduction etc.

(vi) What are ‘cry gene’? In which organism they are present? [1]
Answer:
Cry genes are Bt toxic genes, which produce toxic protein crystals.

(vii) Name the pioneer organisms in hydrosere. [1]
Answer:
Phytoplanktons

(viii) More than 70 percent of all the species recorded on earth are. [1]
Answer:
Animals

Section – B

(1.5 Marks)

Question 4.
What is test tube baby? Why is the term test tube baby a misnomer? [1.5]
Answer:
Test tube Baby : This is a technique of in vitro fertilization of an ovum with sperm carried out outside the body of mother in a glass container in the laboratory. The term test-tube baby is a misnomer because the baby is not developed in the test tube; only fertilization is carried out in such devices in the laboratory conditions (in vitro). The fertilized egg (zygote) or early embryo is then transferred into the fallopian tube or uterus of the mother where it develops and a normal baby is born.

Question 5.
Within what age group sexually transmitted diseases (STDs) are reported to very high. Mention three practices to avoid them. [1.5]
Answer:
In the age group of 15-24 years, STDs are reported to be very high. Following are the three practices to avoid them:

• Always use condom during sexual intercourse.
• Avoid coitus with unknown partners or multiple partners.
• In case of any doubt, medical help should be taken for early detection.

Question 6.
What is pisciculture? Write the name of common diseases of fishes. [1.5]
Answer:
Pisciculture : The process of fish farming in isolated water bodies is known as pisciculture. It involves proper utilization of freshwater, brackish water and coastal areas. Common diseases of fishes :

• Viral Haemorrhagic Septicemia (VHS).
• Bacterial Infectious Pancreatic Necrosis (IPN)
• Gill rot (Blackning of gills).
• Dropsy (Swollen belly).

Question 7.
Explain three examples of mutation breeding for high-yielding cultivated varieties. [1.5]
Answer:

• Resistance to yellow mosaic virus Bhindi (Abelmoschus esculentus) was transferred to a wild species and resulted in a new variety A. esculentus called Parabhani Kranti.
• Sharbati Sonora and Pusa Lerma (high-yielding dwarf wheat varieties) are two amber-grain coloured mutants produced from the red-grained Sonora-64 and Maxican Lerma Rojo 64A respectively.
• Resistance to yellow mosaic virus and powdery mildew in mung bean were induced by mutation.

Question 8.
What is the role of plasmid in DNA technology?
Or
Why plasmid is selected as a vector? [1.5]
Answer:
Plasmid are small, circular, extrachromosomal, double stranded DNA, present in bacterial cell. They are low molecular weight and carry ‘origin of replication’ restriction sites and selectable markers, as they are ideal vectors to inserted into host cell because they can easily incorporate with the host DNA to form recombinant DNA.

Question 9.
Why E. coli is most frequently used in genetic engineering? [1.5]
Answer:
Because it can grow and cultivate easily in vitro → it is not expensive to use.
E.coli adapted itself very easily in laboratory and grown at high rate of replication.
Plasmid of E.coli can well replicate by using host’s DNA polymerase enzyme.

Question 10.
What was the conventional method to obtained insulin and what kind of problems had to be faced to produced genetically engineered insulin? [1.5]
Answer:
Insulin is naturally secreted by pancreas through P-cells of islets of langerhans. Insulin control the blood glucose level. When pancreas failed to produce enough insulin, it causes diabetes. This problem could only be controlled by giving insulin to the body.

Initially, insulin gives to a diabetic patient wax extracted from pancreas by killing animals such as cow and pig. This insulin injected to human create allergies other kind of problems.

Naturally synthesised pro-insulin made by mammals contain a C-peptide that insulin undergoes maturation and C-peptide separate out when blood glucose level increases. When insulin was produced using rDNA techniques, this separation of C-peptide become a major challange to get mature insulin artificially.

Question 11.
What is gene therapy? Name the first clinical case in which it was used. [1.5]
Answer:
Gene therapy, is the therapy of defective genes which are replaced by healthy gene. In the child, get corrects the defect of malfunctioning gene before it grow older then the disease will no longer grow in body. So “Gene therapy is a medical field which focuses on the utilization of the therapeutic delivery of nucleic acid (gene) into a patient’s cell as a drug to treat disease.”

Very first experiment of gene therapy performed on 4 years old girl with adenosine deaminase (ADA) deficiency.

Question 12.
What are the differences between standing state and standing crop? [1.5]
Answer:
Differences between standing state and standing crop are as follows:

	Standing State	Standing Crop
1.	Amount of nutrients such as nitrogen, phosphorus, calcium etc. present in the soil of an ecosystem at any given point of time.	It is the amount of living biomass available at a given trophic level at a given point of time.
2.	It is an abiotic component.	It is a biotic component.

Question 13.
Write the relationship between productivity, gross primary productivity, net primary productivity and secondary productivity. [1.5]
Answer:
The rate of biomass production is called productivity in terms of gm^-2yr^-1 for an ecosystem, it can be divided in gross primary productivity (GPP) and net primary productivity (NPP). GPP of an ecosystem is the rate of production of organic matter during photosynthesis. NPP is the available biomass, for the consumption of heterotrophs i.e., herbivores and decomposers. Secondary productivity is defined as the rate of formation of new organic matter by consumers.

Question 14.
What is the significance of the slope of regression in a species-area relationship? [1.5]
Answer:
The slope is very helpful in finding the species-area relationship. An analysis of the species-area relationship in smaller regions reveals that the values of slopes of regression is the same irrespective of the taxonomic assemblage or the region. But, when the same analysis is carried out in large areas, the slope of regression was found to be much steeper.

Question 15.
What is ecological diversity? Explain. [1.5]
Answer:
It is diversity at ecosystem level. For example, in case of India there are deserts, rain forests, mangrooves, coral reefs, wet lands, estuaries and alpine meadows having a greater ecosystem diversity than a Scandinavian country like Norway.

It has taken millions of years of evolution, to accumulate this rich biodiversity in nature. Biodiversity and its conservation are now vital environmental issues of international concern as more and more people around the globe started recognising the importance of biodiversity for our survival and well being of our planet.

Section – C

(3 Marks)

Question 16.
(a) Higher organisms have resorted to sexual reproduction in spite of its complexity. Why?
(b) Explain why meiosis and gametogenesis are always interlinked? [3]
Or
What do you understand by binary fission? Explain binary fission with suitable example and labelled diagram.
Answer:
(a) Most organisms reproduce sexually. Although sexual reproduction involves more time and energy, higher organisms have resorted to sexual reproduction in spite of its complexity. This is because this mode of reproduction helps in introducing new variations in offsprings through the combination of the DNA from two different individuals. These variations allow the individual to cope with various environmental conditions and thus, make the organism better suited for the environment. Variations also lead to the evolution of better organisms and therefore, provide better chances of survival. On the other hand, asexual reproduction does not provide genetic differences in the offsprings produced.

(b) Gametogenesis is the formation of both types of gametes i.e., male and female. These are always haploid. A haploid parent produces gametes by mitosis. But in majority of organisms, parent body is diploid. Thus, diploid organisms produce haploid gametes by meiosis. At the end of meiosis, only one set of chromosomes is present in each gamete.

Question 17.
Define and design a test cross. [3]
Or
Using a punnet square, workout the distribution of phenotypic features in the first filial generation after a cross between a homozygous female and a heterozygous male for a single locus. [3]
Answer:
A test cross can be defined as a cross of an F₁ individual that has a dominant phenotype with its homozygous recessive parent. Test cross can be used to determine if an individual displaying dominant character is homozygous or heterozygous.

Example : When a cross is made between tall (TT) pea plant and dwarf (tt) pea plant, the F₁ offsprings show all tall plants. If the F₁ plants crossed with dwarf parent plants producing tall and dwarf plants in the equal proportion i.e., 1 : 1 ratio is observed.

Question 18.
From which plant is cocoa alkaloid or cocaine obtained ? What are its effect on the human body? Write the chemical structure of morphine. [3]
Or
Which of the following Sets of diseases is caused by bacteria ?
(a) Tetanus and mumps
(b) Herpes and influenza
(c) Cholera and tetanus
(d) Typhoid and smallpox [3]
Answer:
Cocoa alkaloid or cocaine is obtained from the young leaves and twigs of a shrub Erythroxylum coca found in South America.

Effect on Human body :

• Cocaine is a CNS stimulant.
• It increases mental alertness and physical strength.
• It gives a delayed feeling.
• It reduces sleep and appetite.
• It’s addiction ultimately leads to irritational behaviour, sweating, dilation of pupils, flushed face, blurred vision and faster heart beat.

Section – D

(4 marks)

Question 19.
Differentiate between microsporogenesis and megasporogenesis. Which type of cell division occur during these events? Name the structure formed pt the end of these two events. [4]
Or
Explain the structure of angiospermic pollen grain with diagrams. [4]
Answer:
Differences between Microsporogenesis and Megasporogenesis

	Microsporogenesis	Megasporogenesis
1.	It is the process in which a diploid microspore mother cell undergoes meiosis to form haploid microspores.	It is the process of formation of haploid Mega spores from the diploid megaspore mother cell.
2.	It occurs inside pollen sac in anther.	It occurs inside ovule in ovary.
3.	Pollen grains are produced by microsporogenesis.	Embryo sacs are produced by megasporogenes is.
4.	The arrangement of microspores is tetrahedral.	The arrangement of megaspores is linear.
5.	All the four microspores produced by microsporogenesis, are functional.	Only one, out of the four megaspore formed by megasporogenes is, is functional.

Meiotic cell division occurs during megasporogenesis and microsporogenesis. It is also called reduction division that leads to the production of haploid gametes.
The structures formed at the end of these events are :
Microsporogenesis : Pollen grains. Megasporogenesis : Embryo sac.

Question 20.
Differentiate between translation in prokaryotes and eukaryotes. [4]
Or
What do you understand by DNA fingerprinting? Write its four applications. [4]
Answer:
Differences between Translation in Prokaryotes and Eukaryotes

	Translation in Prokaryotes	Translation in Eukaryotes
1.	In this ribosomes are of 70S type.	In this ribosomes are of 80S type.
2.	mRNA is polycistronic.	mRNA is monocistronic.
3.	Three initiation factors IF1, IF2 and IF3 are required for initiation.	About 10 initiation factors are required.
4.	First amino acid for the initiation of synthesis is formylated methionine (F-mat).	First amino acid for initiation of synthesis is methionine (met).
5.	Elongation factors are EF-Tu and EF-Ts.	Elongation factors is eEF1.
6.	Transcription coupled translation occur in prokaryotes.	Translation occurs only after transcription of m-RNA is completed and it comes out of the nucleus into the cytoplasm.
7.	It is speedy, about 20 amino-acids are added per second.	Translation is slow, only one amino acid is added per second.
8	m-RNA has short life span, say only a few seconds.	Eukaryotic mRNA has a comparatively longer life span, say a few hours to a few days.