12th biology case study questions

Case Study Based Questions for CBSE Class 12 Biology Board Exam 2024: Read this article for Last Minute Revision

Cbse class 12 biology important case study questions : practise important case study based questions for class 12 biology board exam. these case study based questions are important for the upcoming cbse class 12 biology board exam 2024 on march 19, 2024..

CBSE Class 12th Biology Board Exam 2023-24 Pattern

The paper will be of 70 marks and the time duration for completing the paper will be 3 hours.

The paper will have 33 questions divided into 5 sections.

Section–A 16 questions of 1 mark each,

Section–B 5 questions of 2 marks each;

Section–C 7 questions of 3 marks each;

Section–D 2 case-based questions of 4 marks each,

and Section–E 3 questions of 5 marks each.

CBSE Class 12 Biology Important Case Study Based Questions

Case Study 1: Nondisjunction is the failure of homologous chromosomes to disjoin correctly during meiosis. It leads to the formation of a new cell with an abnormal amount of genetic material. A number of clinical conditions are the result of this type of chromosomal mutation. This results in the production of gametes containing a greater or lesser chromosomal amount than normal ones. Consequently, the individual may develop a trisomy or monosomal syndrome. Nondisjunction can occur in both Meiosis I and Meiosis II of the cellular division. It is also the main cause of many genetic disorders; however, its origin and process remain vague. Although it results in the majority of cases from errors in maternal meiosis II, both paternal and maternal meiosis I do influence it. Maternal age is considered a risk factor for trisomy, as well as recombination alterations and many others that can affect chromosomal segregation.

It is the presence of an extra chromosome in a diploid cell.
An aneuploid cell differs from other cells only in size.
It can be less number of chromosomes in a diploid cell.
Aneuploidy always affects female individuals.
both i and iii
both ii and iii
i, iii and iv
Errors in meiosis I is the only cause of aneuploidy
Aneuploidy always affects sex chromosomes.
Most of the aneuploidy results from errors in cell division involved in egg formation.
Nondisjunction in meiosis I can lead to more abnormal cells than disjunction in meiosis II.
both I and iii
both iii and iv
I, iii and iv
Aneuploidy is not influenced by the mother’s age.
Delivery before 30 years of age can decrease the incidence of aneuploidy in most cases
The chance of aneuploidy increases up to 22 years of age.
There is a dramatic increase in aneuploidy if the maternal age exceeds 30

both ii and iv
Chromosomal disorders
Mendelian disorders
Incomplete dominance
All the above

Q5: Assertion: All types of genetic disorders are caused by chromosomal nondisjunction.

Both assertion and reason are correct and the reason is the correct explanation of assertion
Both assertion and reason are correct but the reason is not the correct explanation of the assertion
Assertion is correct but the reason is incorrect
Both assertion and reason are incorrect

Case Study 2: A Representative Diagram of the Human Genome Project:

Biotechnology
Biomonitoring
Bioinformatics
Biosystematics

Q2: Name a free living, non-pathogenic nematode, the DNA of which has been completely sequenced.

Answer: Caenorhabditis elegans

Q3: Summarize the methodology adopted in the Human Genome Project.

Answer: Expressed Sequence Tags (ESTs) : The approach focused on identifying all the genes that are expressed as RNA.

Sequence Annotation : The other took the blind approach of simply sequencing the whole set of genome that contained all the coding and non-coding sequence, and later assigning different regions in the sequence with functions.

Q4: What are SNPs’? How are they useful in human genomics?

Identify disease-causing genes in humans
Can be used to understand the molecular mechanisms of sequence evolution.

Q5: Mention at least four salient features of the Human Genome Project.

Human genome contains 3164.7 million bp.
Average gene consists of 3000 bases, but sizes vary greatly.
Almost all (99.9 percent) nucleotide bases are exactly the same in all people.
Less than 2 percent of the genome codes for proteins.

Case Study 3: Two blood samples of suspects ‘A’ and ‘B’ were sent to the Forensic Department along with sample ‘C’ from the crime scene. The Forensic Department was assigned the responsibility of running the samples and matching the samples of the suspects with that of the sample from the scene of the crime and thereby identifying the culprit.

A radioactively labelled double stranded RNA molecule.
A radioactively labelled double stranded DNA molecule.
A radioactively labelled single stranded DNA molecule.
A radioactively labelled single stranded RNA molecule.

Q3: What does ‘minisatellite’ and ‘microsatellite’ mean in relation to DNA Fingerprinting?

Answer: Minisatellite: the repeating unit consists of 10-100 base pairs.

Microsatellite: the repeating unit consists of 2-6 base pairs.

Q3: How does polymorphism arise in a population?

Answer: Polymorphism (variation at the genetic level) arises due to mutations.

Q4: State the steps involved in DNA Fingerprinting in a sequential manner.

DNA isolation
DNA digestion with restriction enzymes.
DNA fragment separation by electrophoresis.
Hybridization
DNA visualization under UV light.

Case Study 4: Bacteria like Streptococcus pneumoniae and Haemophilus influenzae are responsible for the disease pneumonia in humans which infects the alveoli (air-filled sacs) of the lungs. As a result of the infection, the alveoli get filled with fluid leading to severe problems in respiration. The symptoms of pneumonia include fever, chills, cough, and headache. In severe cases, the lips and fingernails may turn gray to bluish in colour. A healthy person acquires the infection by inhaling the droplets/aerosols released by an infected person or even by sharing glasses and utensils with an infected person. Dysentery, plague, diphtheria, etc., are some of the other bacterial diseases in man. Many viruses also cause diseases in human beings. Rhinoviruses represent one such group of viruses that cause one of the most infectious human ailments – the common cold. They infect the nose and respiratory passage but not the lungs.

The common cold is characterized by nasal congestion and discharge, sore throat, hoarseness, cough,

headache, tiredness, etc., which usually lasts for 3-7 days. Droplets resulting from the cough or sneezes of an infected person are either inhaled directly or transmitted through contaminated objects such as pens, books, cups, doorknobs, computer keyboards or mice, etc., and cause infection in a healthy person.

By exhaling droplets of a non-infected person.
By headache or leg pain.
By eating fast food.
By inhaling droplets of an infected person.

Q4: How long does the common cold last?

Answer: 3-7 days

Q5: Write any two symptoms of the common cold and pneumonia.

Answer: Cough and nasal congestion.

Case Study 5: When you insert a piece of alien DNA into a cloning vector and transfer it into a bacterial, plant, or animal cell, the alien DNA gets multiplied. In almost all recombinant technologies, the ultimate aim is to produce a desirable protein. Hence, there is a need for the recombinant DNA to be expressed. The foreign gene gets expressed under appropriate conditions. The expression of foreign genes in host cells involves understanding many technical details. After having cloned the gene of interest and having optimised the conditions to induce the expression of the target protein, one has to consider producing it on a large scale. Can you think of any reason why there is a need for large-scale production? If any protein encoding gene is expressed in a heterologous host, it is called a recombinant protein. The cells harbouring cloned genes of interest may be grown on a small scale in the laboratory. The cultures may be used for extracting the desired protein and then purifying it by using different separation techniques.

A continuous culture system
A stirred-tank bioreactor without in-lets and out-lets
Laboratory flask of the largest capacity
None of the above
upstream processing
downstream processing
bioprocessing
postproduction processing
Human insulin
Growth hormone
cleaving and joining of DNA segments with endonuclease
cleaving DNA segments with endonuclease and re-joining with ligase
cleaving and re-joining DNA segments with ligase
cleaving DNA segments with ligase and re-joining with endonuclease

Case Study 6: Gene Therapy

Read the following and answer the questions that follow:

Replacing a disease-causing gene with a healthy copy of the gene
Inactivating a disease-causing gene that is not functioning properly
Introducing a new or modified gene into the body to help treat a disease
Adenosine deaminase
phenylketonuria
Phenylalanine
Bone marrow transplantation
Southern blotting

Q4 Introduction of gene isolate from bone marrow producing ADA should be introduced at what age to

acute diseases
physiological diseases
hereditary diseases
infectious diseases
CBSE Class 12 Biology Previous Year Question Papers with Solutions PDF Download
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CBSE 12th Standard Biology Subject Case Study Questions With Solution 2021 Part - II

By QB365 on 21 May, 2021

QB365 Provides the updated CASE Study Questions for Class 12 Biology, and also provide the detail solution for each and every case study questions . Case study questions are latest updated question pattern from NCERT, QB365 will helps to get more marks in Exams

QB365 - Question Bank Software

12th Standard CBSE

Final Semester - June 2015

Case Study Questions

Read the following and answer any four questions from (i) to (v) given below: A woman unable to conceive after many years of regular unprotected coitus went to specialised infertility clinic. On complete examination, woman was found to be normal while male partner was diagnosed with infertility. Male partner is unable to copulate the female. Couple was advised to opt for assisted reproductive technology (ART). (i) Which ART could have been useful for this couple?

(ii) In which of the following infertility issues could this technique be used?

(iii) Which among the following techniques is useful to conceive when there is very low sperm count?

(iv) Assertion: Artificial insemination (AI) is intra-uterine insemination. Reason: In AI, sperms collected from donor are introduced into the uterus.

(v) Success rate of artificial insemination technique is

Read the following and answer any four questions from (i) to (v) given below : Riya studies in II standard in a government schooL She belongs to a backward family and her parents did not get her properly vaccinated according to immunisation programme. Once while playing in school playground she fell down due to weakness and developed high fever, headache and stiffness in her neck. Identify the illness she could be suffering from and answer the following questions. (i) The microbe responsible for Riyas illness could be

(ii) Which vaccine, if administered earlier, would have saved Riya from the illness she unfortunately contracted?

(iii) The disease that Riya has contracted spreads through

(iv) Riya can spread her illness to other children through

(v) Assertion: Polio produces inflammation of the nervous system. Reason: Stiffness of the neck, paralysis of particular skeletal muscle is an important symptom of polio.

(ii) The given graph indicates that person Y is suffering from

(iii) Which of the following conditions are common in person Y?

(iv) A person suffering from diabetes mellitus becomes weak because

(v) Assertion: Type I diabetes involves failure of insulin to facilitate the movement of glucose into cells. Reason : Type II diabetes is caused by failure of beta cells to produce adequate amount of insulin due to beta cell depletion.

Read the following and answer any four questions from (i) to (v) given below : Turner's syndrome is an example of mono somy. It is formed by the union of an allosome free egg and a normal 'X' containing sperm or a normal egg and an allosome free sperm. The individual has 2n = 45 chromosomes (44 + X0) instead of 46. Such individuals are sterile females who have rudimentary ovaries, under developed breasts, small uterus, short stature, webbed neck and abnormal intelligence. They may not menstruate or ovulate. This disorder can be treated by giving female sex hormone to the women from the age of puberty to make them develop breasts and have menstruation. This makes them feel more normal. (i) Number of Barr body present in a female with Turner's syndrome is

(ii) Turner's syndrome is an example of

(iii) Turner's syndrome is a/an

(iv) Which of the following statements regarding Turner's syndrome is incorrect?

(v) Assertion : Turner's syndrome is caused due to absence of anyone of the X and Y sex chromosome. Reason : Individuals suffering from Turner's syndrome show masculine as well as feminine development

Read the following and answer any four questions from (i) to (v) given below: Aditya went to his hometown located in countryside along with his parents during his summer vacations. His grand parents' house is surrounded by farmland from all sides. Lots of crops were growing nearby and Aditya was very excited to visit the crop fields. He seeked permission from his mother to play in farmland along with his friends and then went to play in the fields. On returning back he had running nose, watering eyes and continuous sneezing which was very frequent. The symptoms worsened with time. Based on the above information, answer the following questions. (i) What could be the possible reason for Adityas condition?

(ii) How can allergy be diagnosed in a person?

(iii) The symptoms which Aditya developed on account of being allergic are consequence of

(iv) Name the type of allergy that Aditya developed.

(v) Assertion: Hay fever is the form of allergy due to pollens of grasses and other plants. Reason : Hay fever symptoms are due to release of histamines and often respond well to treatment with antihistamines.

Read the following and answer any four questions from (i) to (v) given below: A group of teenagers was involved in drug abuse. They used syringes and needles to inject drugs. They indulged in this habit when they became adults. Administration of drug through needles became a piece of cake for them. Raj was the most active drug abuser amongst them and used to take drugs in high profile parties. In a span of time he started losing weight and suffered persistent diarrhoea. He developed constant low grade fever and used to catch opportunistic infection. When he consulted a doctor, he got himself tested for HIV in his blood and finally diagnosed with AIDS. Based on the above information, answer the following questions. (i) Select the incorrect statement.

(ii) How do you think Raj got AIDS infection?

(iii) How AIDS can be diagnosed?

(iv) How can AIDS be prevented?

(v) Select the correct statement for AIDS virus.

Read the following and answer any four questions from (i) to (v) given below: Kangaroo rat seldom drinks water. It has thick coat to minimise evaporative desiccation. The animal seldom comes out of its comparatively humid and cool burrow during the day time. 90% of its water requirement is met from metabolic water (water produced by respiratory breakdown of fats) while 10% is obtained from its food. Loss of water is minimised by producing nearly solid urine and faeces. As the animal faces acute water scarcity, it develops two types of adaptations: reducing water loss and ability to tolerate desert conditions. (i) Kangaroo rat is a

(ii) Metabolic water refers to

(iii) Desert animals minimise water loss by

(iv) Assertion : Kangaroo rat can tolerate and thrive in wide temperature range and is known as stenothermal. Reason: Kangaroo rats go into hibernation during winter to escape cold weather.

(v) The adaptations in an organism are meant for

(ii) Total number of individuals of a species per unit area per unit time is called

(iii) Which of the following is correct regarding age sex pyramid A and B?

(iv) A population with a large proportion of older individuals than younger ones will likely to

(v ) Assertion: Bell shaped age pyramid represents a stable population. Reason : In a stable population, proportion of individuals in reproductive age group is higher than the individuals in pre-reproductive age group.

(ii) Which of the following associations is an example of competitions?

(iii) If '+ ' sign is assigned to beneficial interaction, '-' sign to detrimental and 0 sign to neutral interaction, then the population interaction of competition refers to

(iv) Intraspecific competition is more severe due to

(v) Assertion: Two members of a competing species may co-exist. Reason: Different individuals of a species have different resource requirements.

Read the following and answer any four questions from (i) to (v) given below: The Kakapo is the world's largest and heaviest parrot, found only in New Zealand. It is unusual in that it is nocturnal, flightless and ground-dwelling. It is an excellent climber of trees, has strong legs that allow it to "jog" several kilometres in a single trip, and has mossy green plumage mottled with brown and yellow. The Kakapo is also critically endangered as of now, there were only few known living individuals left. (i) Which could be the possible reason for Kakapo to be well-adapted to its environment prior to the arrival of humans in New Zealand?

(ii) When humans started to settle in New Zealand, they took with them non-native animals, including mammals such as cats, dogs and stoats. By which of the following ways, human settlement likely contributed to a near decimation of Kakapo populations in New Zealand?

(iii) All known survived Kakapo have been relocated by the New Zealand government to three predator-free islands, where they are monitored year round by staffs and volunteers to ensure that the birds are safe, healthy and well-fed. The extremely low population of Kakapo is a hurdle to the species becoming viable in the long term, despite such dedicated conservation efforts. This is because

(iv) The reasons behind conserving biodiversity have been grouped into which of the following categories?

(v) One of the ex situ conservation methods for endangered species is

Read the following and answer any four questions from (i) to (v) given below: Non-native or alien species are often introduced in advertently for their economic and other uses. They often become invasive and drive away-the local species. Exotic species have proved harmful to both aquatic and terrestrial ecosystems. For example, water hyacinth (Eichhornia crassipes) was introduced in Indian waters to reduce pollution. It was clogged water bodies including wetlands at many places resulting in death of several aquatic plants and animals. (i) Island water ecosystem are the most vulnerable due to

(ii) Which of the following is not an alien species?

(iii) Second major cause of species extinction is

(iv) Assertion: Eichhornia crassipes drains off oxygen from water and can be seen growing in standing water. Reason: Eichhornia crassipes is an indigenous species of India.

Read the following and answer any four questions from (i) to (v) given below: Golden rice was engineered from normal rice by Potrykus and Beyer in the 1990s. The typical golden colour is due to the production of \(\beta\) carotene a precursor of vitamin - A. Golden rice differs from its parental strain by the addition of three \(\beta\) carotene genes. These included two genes from daffodil plant and third from a bacterium. The incorporation of these genes allows the rice plant to modify certain metabolic pathways in its cells to produce \(\beta\) -carotene. (i) Due to genetic modification golden rice plants produce and store \(\beta\) - carotene in

(ii) Transfer of genes to produce golden rice is achieved by

(iii) In golden rice two genes were taken from

(iv) Golden rice is helpful to fight against disease caused by the deficiency of

(v) Golden rice was genetically engineered by

Read the following and answer any four questions from (i) to (v) given below: Transgenic cows have extra gene or genes inserted into their DNA. Firstly the genes for the desired product is identified and sequenced. Then a gene construct containing this desired gene is introduced into female cow cells. Transgenic bovine cells are selected and fused with bovine oocytes that have had all of their chromosomes removed. Once fused with the oocyte, the transgenic cells chromosomes are reprogrammed to direct development into an embryo which can be implanted into a recipient cow. The resulting transgenic cow only express the trans gene in her milk. This is because expression of the trans gene is controlled by a promoter specific to lactating mammary cells. The first transgenic cow was 'Rosie: (i) The gene construct with desired gene is introduced into female cow cells by

(ii) Production of transgenic cow fulfill the objective of

(iii) The name of first transgenic cow is

(iv) Transgenic cow is produced through the implantation of _____containing transgene into recipient cow.

(v) Assertion : Transgenes only express in the mammary glands of transgenic cow. Reason : Transgenes are present in chromosomes of every cell in transgenic cow

(ii) Equation for relationship (A) between species richness and area is

(iii) What is the value of slope of line or regression coefficient Z for frugivorous birds?

(iv) The shape of curve for relationship between species richness and areas for wide variety of taxa is

(v) Who gave this concept of increase in species richness with increasing offered area?

(ii) The following is a list of some stages involved in producing human insulin from genetically engineered bacteria. 1. The bacteria are cultured in a fermenter for large scale production. 2. Recombinant insulin is extracted from the bacterial cells that expresses insulin gene. 3. The same restriction enzyme is used again to cut the bacterial plasmid for insertion of the human insulin gene. 4. Bacteria take up the plasmid carrying the insulin gene. 5. A restriction enzyme is used to cut human DNA to extract the insulin gene. Select the correct order of these stages.

(iii) To insert the insulin gene into bacterial DNA, both the bacterial plasmid and the human chromosome containing the insulin gene are treated with the same restriction enzyme. Using the same restriction enzyme ensures that

(iv) Why is the fermentor important for the production of human insulin by transgenic bacteria

(v) A bacteriologist carries out his first attempt at engineering E.coli with the gene for human insulin. During the process, he realises that his stock of DNA ligase has depleted but decides to continue anyway. What is a likely consequence of his decision?

*****************************************

Cbse 12th standard biology subject case study questions with solution 2021 part - ii answer keys.

(i) (c) : Artificial Insemination (AI) is done in infertility cases either due to inability of male partner to copulate the female or due to very low sperm count in the semen of male partner. (ii) (b) : Oligospermia is very low sperm count. (iii) (d): ICSI is intra cytoplasmic sperm injection. (iv) (a) (v) (a)

(i) (b) : Riya is suffering from polio that is caused by an Enterovirus, called Poliovirus. (ii) (a) (iii) (c) : Polio virus enters the body via alimentary canal (faecal oral route) (iv) (a): Riya has contracted polio which is transmitted through faecal oral route, as urine and faeces of the patient contain polio virus. (v) (b)

(i) (c) : Blood sugar level fluctuations in person X indicate that sugar level never exceed the normal limit and sufficient secretion of insulin at required times removes any extra sugar from blood and converts it into glycogen for future use. This implies that person X is normal and healthy. (ii) (a) : Elevated blood sugar levels in person Y indicate that he is suffering from diabetes mellitus. (iii) (d): (iv) (d): (v) (d) : Type I diabetes or insulin dependent diabetes mellitus or juvenile diabetes is an autoimmune disorder caused by failure of beta cells to produce adequate amount of insulin. Type II diabetes or non insulin dependent diabetes mellitus involves failure of insulin to facilitate the movement of glucose into body cells.

(i) (a) : Barr body is a structure consisting of a condensed X chromosome that is found in nondividing nuclei of female mammals. The presence of Barr body is used to confirm the sex of athletes in sex determination tests. It is named after the Canadian anatomist M.L. Barr, who identified it. The number of Barr bodies is one less than total number of X chromosomes. In Turner's syndrome genotype is 45 + X0,so, the number of Barr body is O. (ii) (a) : Failure of segregation of chromatids during cell division result in the gain or loss of a chromosomes called aneuploidy. For example, Turner's syndrome results due to loss of X chromosome in human females. (iii) (d) : Turner's syndrome is a chromosomal disorder that occurs due to absence of one chromosome. (iv) (c) : In Turner's syndrome individual lacks one X chromosome. This situation is known as monosomy. (v) (d): Turner's syndrome occurs due to absence of X chromosome. Individuals having a single X chromosome 22A + X0 (45) have female sexual differentiation but ovaries are rudimentary. Other associated phenotypes of this condition are short stature, webbed-neck, broad chest, lack of secondary sexual characteristics and sterility. Thus, any imbalance in the copies of the sex chromosomes may disrupt the genetic information necessary for normal sexual development.

(i) (a) : Aditya most probably had developed some sort of allergy due to pollens of grasses, trees and other plants. (ii) (a) : Allergies mainly involve IgE antibodies and chemicals like histamine and serotonin from mast cells. IgE antibodies are produced in response to an antigen, coat mast cells and basophils. (iii) (a) : The allergy which Aditya developed is characterised by inflammation of membrane lining the nose and conjunctiva leading to running nose, watering eyes and constant sneezing. (iv) (c) (v) (b)

(i) (d) : Every year, December 1 is designated as World AIDS Day. (ii) (d) (iii) (a) : ELISA test, also called EIA for enzyme immunoassay is used to detect HIV antibody. It checks for certain proteins that the body makes in response to HIY. (iv) (d) (v) (c) : HIV is spherical with a diameter of 90-120 nm. Its genome consists of single stranded RNA. The envelope consists of a lipid bilayer derived from host cell membrane and projection knob like glycoproteins. It contains two protein coats.

(i) (c) : Kangaroo rat is a regulator that performs homeostasis through thermoregulation osmoregulation by physiological adjustments and behavioural changes. (ii) (c) (iii) (d) (iv) (d) : Organisms that can tolerate and thrive in wide temperature range are known as eurythermal. Kangaroo rats do not go into hibernation rather, they stay if}cool moist burrows during summer days. (v) (d): Adaptation is an attribute of an organism (morphological, physiological and behavioural) that enables it to survive and reproduce in its habitat. Adaptations lead to the formation of some specialised and peculiar features which have evolved over a long period of time through natural selection.

(i) (b) (ii) (b) : Population density is the number of individual present per unit area at a given time. (iii) (d) : A represents nearly stable population whereas B represents rapidly growing population. (iv) (c) : A population with large number of older individuals than younger ones is likely to decline since older individuals do not take part in reproduction. (v) (c): In a bell-shaped age pyramid, the number of pre-reproductive and reproductive individuals is almost equal. Post-reproductive individuals are comparatively fewer. It represents a stable population.

(i) (b): Competition is a sort of rivalry among two or more organisms for obtaining the same resources. The competition among individuals of the same species is called intraspecific competition and among members of different species is called interspecific competition. Intraspecific competition is more severe than interspecific competition due to similar needs. Now, according to the given graph, competition for water in a population leads to more root weight (mg) per leaf area (crrr'). This is because competition causes more root growth so that each sapling can derive more water from the pot. (ii) (b) : The association or interactions of Cuscuta and hedge plant is parasitism, cactus and moth is predation and orchid and mango is commensalism. (iii) (b) (iv) (d) : Intraspecific competition is more severe because of common resource, similar needs and similar adaptations. (v) (d)

(i) (d) : Since the Kakapo is nocturnal, it was active only in the night when its potential predators would not be out for hunting. With its greenish plumage, the Kakapo could likely be well camouflaged among the forest foliage in the daytime when it is resting hence evading detection by its predators. It was able to effectively hunt for food in the night given its ability to climb trees and travel significant distances over land despite lacking the ability to fly. (ii) (d): As humans settled in New Zealand, they would have cleared the land to make way for their own needs e.g., farmland, hence shrinking the natural habitats of the Kakapo. The new mammals that were introduced into the Kakapos habitats might have out-competed the Kakapo for the limited food resources available. The new mammals that were introduced might also have easily preyed on the Kakapo (e.g., by using their sense of smell), as the KakapoIikely lacked the necessary adaptations to defend itself given that such predators were never present in the past. For instance, many mammals such as cats are nocturnal and hence would prey on Kakapo when the latter are also active at night. (iii) (d) : The small populations results in a yery small gene pool, i.e., a very limited variety of alleles / traits among surviving individuals of the species. Even if the existing birds manage to breed and multiply significantly, there would be very limited genetic diversity among the resulting offspring. This would lead to reduce capacity of the species to adapt to and survive changes in the environment. There may also be reduced fitness in the offspring given increased likelihood of homozygosity of recessive harmful! deleterious alleles, which would result in these alleles being expressed to bring about unfavourable phenotypes. (iv) (d) : We should conserve biodiversity. The reason for this can be broadly divided into three categories: (i) Narrowly utilitarian (Humans derive a major part of their requirement from organisms). (ii) Broadly utilitarian (Biodiversity is fundamental to ecosystem services of nature). (iii) Ethical (Every living species has an intrinsic value, it is our moral duty not to destroy them). (v) (c): Ex situ (off site) conservation is conservation of selected rare plants/animals in places outside their natural homes. It is a desirable approach to save threatened or endangered plant or animal species from extinction. Ex situ conservation includes offsite collections, gene banks, in vitro fertilisation, cryopreservation techniques and tissue culture.

(i) (d) (ii) (d) : Pronuba yuccasella shows obligatory mutualistic relationships with Yucca. (iii) (c) : Alien species invasion is considered to be second major cause of extinction of species. (iv) (c): Eichhornia crassipes is an aquatic plant, native to Amazon basin. This plant was introduced into India for its beautiful flowers and shape of leaves. (v) (a)

(i) (b) : Golden rice is a transgenic variety of rice (Oryza sativa) which contains good quantities of \(\beta\) -carotene (provitamin A - inactive state of vitamin A). \(\beta\) -carotene is a principal source of vitamin A. Since the grains (seeds) of the rice are yellow in colour due to \(\beta\) -carotene, the rice is commonly called golden rice. (ii) (a) : Foreign genes were inserted into the genome of a tempera,te rice variety by using Agrobacterium tumefaciens as the vector. (iii) (a) : In golden rice, two genes for \(\beta\) -carotene were taken from daffodil (Narcissus sp.) and one gene from a bacterium Erwinia. (iv) (c) : Golden rice is rich in \(\beta\) -carotene which is a precursor of vitamin A. (v) (b) : Prof. Ingo Potrykus and Peter Beyer produced genetically engineered rice by introducing three gene associated with synthesis of carotene. The grains (seeds) of transgenic rice are rich in provitamin.

(i) (c) (ii) (d) : The two chief objectives of transgenic cow production are as follows: (i) increased milk and meat production and (ii) molecular farming. (iii) (c) ( iv) (b) : Transgenic bovine cells are selected and fused with bovine oocytes that have had all its chromosomes removed. Once fused with oocyte, the transgenic cells chromosomes are reprogrammed to direct development into embryo which is implanted into recipient cow. (v) (b)

(i) (b) (ii) (a) : On a logarithmic scale, the relationship between species richness and area is a straight line. (iii) (b) : The regression coefficient Z have a value of 1.15 for frugivorous birds and mammals of tropical forests of different continents. (iv) (c) (v) (b) : While exploring the wilderness South American jungles, Alexander von Humboldt found that within a region, the species richness increased with increasing area but upto a certain limit

(i) (c) : A represents polypeptide chain C which is removed prior to insulin formation. (ii) (d) (iii) (d) : Each particular restriction enzyme produces unique sticky ends. Using the same enzyme for both the bacterial and human DNA will produce complementary sticky ends that can bind together by complementary base pairing. This would allow the human insulin gene to be inserted into the plasmid. (iv) (a) : The optimal temperature, pH, oxygen and nutrient conditions in the fermenter allow the bacteria containing the insulin gene to reproduce quickly and produce large quantities of it. (v) (c) : DNA ligase forms strong hydrogen bonds between the DNA bases on the human insulin gene and the bacterial plasmid, producing a continuous double stranded DNA loop. Without DNA ligase, the human insulin gene, despite being able to undergo complementary base pairing with the bacterial DNA at the sticky ends would not be securely inserted into the plasmid. Thus, the resulting E.coli bacteria would receive plasm ids that lack the human insulin gene.

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Class 12 Biology Case Study Questions

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As we know that CBSE will now ask case study questions in each subject. In most of the cases, we have noticed that these case-based questions are high-scoring. A little effort on these case study questions can help you get good marks in your board exams. You can download CBSE Class 12 Biology Case Study Questions from the myCBSEguide App or from our Student Dashboard .

Let’s understand what type of case study questions CBSE asks in class 12 Biology. If you analyze the latest class 12 Biology sample papers , you will find that there are two types of case study questions in the Biology question papers.

Case Studies with objective questions
Case studies with subjective questions

As per the latest circular issued by CBSE on Assessment and Evaluation Practices of the Board for the Session 2022-23 , CBSE has clearly mentioned that competency-based questions including case studies will be different from subjective questions. Hence, we expect that CBSE will ask only objective questions in CBSE class 12 Biology case study questions too.

Biology Competency Based Questions

As discussed earlier too, the competency-based questions promote learning development for our students and test higher-order skills, such as analysis, critical thinking and conceptual clarity. Case study questions are actually competency-based questions. The very purpose of including such questions in the curriculum is to emphasise on development of problem-solving ability and the ability to apply knowledge in real-life situations.

Even in CBSE Class 12 Biology case studies, you will find some text input like paragraphs, pictures, data etc followed by some objective-type questions. You should read the given information carefully and then answer the questions.

CBSE 12th Biology Case Study MCQs

Here is one example question on subjective type case study questions. This was given in the term-2 sample paper in 2022.

Some restriction enzymes break a phosphodiester bond on both the DNA strands, such that only one end of each molecule is cut and these ends have regions of single-stranded DNA. BamH1is one such restriction enzyme which binds at the recognition sequence, 5’-GGATCC- 3’and cleaves these sequences just after the 5’- guanine on each strand.

What is the objective of this action?
Explain how the gene of interest is introduced into a vector.
You are given the DNA shown below. 5’ ATTTTGAGGATCCGTAATGTCCT 3’ 3’ TAAAACTCCTAGGCATTACAGGA 5’ If this DNA was cut with BamHI, how many DNA fragments would you expect? Write the sequence of these double-stranded DNA fragments with their respective polarity.
A gene M was introduced into E.coli cloning vector PBR322 at the BamH1 site. What will be its impact on the recombinant plasmids? Give a possible way by which you could differentiate non-recombinant to recombinant plasmids.

Let’s take another example from MCQ type question:

To answer the questions, study the graphs below for Subject-1 and 2 showing different levels of certain hormones.

The peak observed in Subject-1 and 2 is due to

progesterone
luteinizing hormone
follicle stimulating hormone

Subject 2 has higher level of hormone B, which is

If the peak of Hormone A does not appear in the study for Subject 1, which of the following statement is true?

Peak of Hormone B will be observed at a higher point in the graph
Peak of Hormone B will be observed at a point lower than what is given in the graph
There will be no observed data for Hormone B
The graph for Hormone B will be a sharp rise followed by a plateau

Which structure in the ovary will remain functional in subject 2?

Corpus Luteum
Tertiary follicle
Graafian follicle
Primary follicle

For subject 2 it is observed that the peak for hormone B has reached the plateau stage. After approximately how much time will the curve for hormone B descend?

Which of the following statements is true about the subjects?

Subject 1 is pregnant
Subject 2 is pregnant
Both subject 1 and 2 are pregnant
Both subject 1 and 2 are not pregnant

Another example of a class 12 Biology case study question

We use microbes or products which are derived from them every day. A common example is the production of curd from milk. Micro-organisms such as Lactobacillus and others commonly called lactic acid bacteria (LAB) grow in milk and convert it to curd. The dough, which is used for making foods such as dosa and idli is also fermented by bacteria. A number of traditional drinks and foods are also made by fermentation by microbes. ‘Toddy’, a traditional drink in some parts of southern India is made by fermenting sap from palms. The ‘Roquefort cheese’ is ripened by growing specific fungi on them, which gives them a particular flavour. Different varieties of cheese are known by their characteristic texture, flavour and taste, the specificity coming from the microbes used.

Penicillium notatum
Saccharomyces cerevisiae
Aspergillus niger
Clostridium butylicum
thermal vents
polluted water
all of these
None of the above
production of a large amount of CO 2
production of O 2
due to the presence of water
none of these
Both Assertion and Reason are true and Reason is the correct explanation of the Assertion
Both Assertion and Reason are true but Reason is not the correct explanation of the Assertion
Our Assertion is true but the Reason is false
Both the statements are false

Download 12 Biology Case Study Questions

In this article, we have given you a few examples of class 12 Biology case study questions. We advise you to download the myCBSEguide App or access our Student Dashboard to get more case study questions for CBSE class 12 biology. We have hundreds of questions on case studies related to CBSE Class 12 Biology. As CBSE is now focusing more on the understanding of the concepts, it is a must for students to practice such questions regularly.

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Case Study Questions Class 12 Biology Organisms and Populations

Case study questions class 12 biology chapter 13 organisms and populations.

CBSE Class 12 Case Study Questions Biology Organisms and Populations. Term 2 Important Case Study Questions for Class 12 Board Exam Students. Here we have arranged some Important Case Base Questions for students who are searching for Paragraph Based Questions Organisms and Populations.

At Case Study Questions there will given a Paragraph. In where some Important Questions will made on that respective Case Based Study. There will various types of marks will given 1 marks, 2 marks, 3 marks, 4 marks.

CBSE Case Study Questions Class 12 Biology Organisms and Populations

Case study 1

What are the key elements that lead to so much variation in the physical and chemical conditions of different habitats? The most important ones are temperature, water, light and soil. We must remember that the physico-chemical (abiotic) components alone do not characterise the habitat of an organism completely; the habitat includes biotic components also – pathogens, parasites, predators and competitors – of the organism with which they interact constantly. We assume that over a period of time, the organism had through natural selection, evolved adaptations to optimise its survival and reproduction in its habitat.

Each organism has an invariably defined range of conditions that it can tolerate, diversity in the resources it utilises and a distinct functional role in the ecological system, all these together comprise its niche.

Que. 1) Organisms has defined many that they can …………………………………………………………………………… range of conditions.

(a) Destroy

(b) Cannot tolerate

(d) Reproduce

Que. 2) When the organism plays role in the ecological system then it is known as ……………………………………………………………………………………. .

(a) Ecology

(d) Biotic

Que. 3) For the survival and reproduction, organisms are ………………………………………………………………………………………… .

(a) Living free

(b) Interacted

(d) Adapted

Que. 4) What are the main reason behind so much variation in the physical and chemical conditions?

Que. 5) Write examples of components which characterise habitat of an organism.

Que. 1)(c) Tolerate.

Que. 2) (b) Niche.

Que. 3) (d) Adapted .

Que. 4) Answer: The main reasons behind so much variation in physical and chemical conditions are temperature, soil, water and light.

Que. 5) Answer: Abiotic components such as temperature, soil, water, light and biotic components such as predators, pathogens, parasites and competitors.

Case study 2

Water is another the most important factor influencing the life of organisms. In fact, life on earth originated in water and is unsustainable without water. Its availability is so limited in deserts that only special adaptations make it possible for organisms to live there. The productivity and distribution of plants is also heavily dependent on water. You might think that organisms living in oceans, lakes and rivers should not face any water-related problems, but it is not true. For aquatic organisms the quality (chemical composition, pH) of water becomes important. The salt concentration (measured as salinity in parts per thousand), is less than 5 in inland waters, 30-35 in the sea and > 100 in some hypersaline lagoons. Some organisms are tolerant of a wide range of salinities (euryhaline) but others are restricted to a narrow range (stenohaline).

Many freshwater animals cannot live for long in sea water and vice versa because of the osmotic problems, they would face.

Que. 1) Many ocean living organisms cannot live for long in fresh water due to …………………………………………………………………………………. .

(a) Butterfly

(b) Thermal insulation

(d) None of them

Que. 2) On the earth, life is unsustainable without ………………………………………………………………………………………… .

(a) One tree

(b) A virus

Que. 3) In inland water, salt concentration is less than ………………………………………………………………………………………… .

Que. 4) Which factor can affect productivity and distribution of the plants.

Que. 5) Give reason:

Aquatic organisms do face water related issues.

Que. 1)(c) Osmotic issue.

Que. 2) (c) Water.

Que. 3) (d) 5.

Que. 4) Answer: Water is main factor that can affect productivity and distribution of the plants

Que. 5) Answer: The water include chemical composition and pH. Water quality is very important for the aquatic organisms. Also, Salt concentration is different for different water habitat. Hence, aquatic organisms do face water related issues.

Case study 3

Since plants produce food through photosynthesis, a process which is only possible when sunlight is available as a source of energy, we can quickly understand the importance of light for living organisms, particularly autotrophs. Many species of small plants (herbs and shrubs) growing in forests are adapted to photosynthesise optimally under very low light conditions because they are constantly overshadowed by tall, canopied trees. Many plants are also dependent on sunlight to meet their photoperiodic requirement for flowering. For many animals too, light is important in that they use the diurnal and seasonal variations in light intensity and duration (photoperiod) as cues for timing their foraging, reproductive and migratory activities. The availability of light on land is closely linked with that of temperature since the sun is the source for both. But, deep (>500m) in the oceans, the environment is dark and its inhabitants are not aware of the existence of a celestial source of energy called Sun.

What, then is their source of energy? The spectral quality of solar radiation is also important for life. The UV component of the spectrum is harmful to many organisms while not all the colour components of the visible spectrumare available for marine plants living at different depths of the ocean.

Que. 1) Small plants include many species which as adapted to …………………………………………………………………..……………………… light conditions.

(d) Constant

Que. 2) Light is very important for ………………………………………………………………………………………. .

(a) Autotrophs

(b) Heterotrophs

(d) Non-living organisms

Que. 3) In the photosynthesis process, sunlight is required as a source of ……………………………………………………………………………………. .

Que. 4) Which organisms are unaware of sun?

Que. 5) Which factor can affect source of light to the organisms?

Que. 1)(a) Low.

Que. 2) (a) Autotrophs.

Que. 3) (d) Energy.

Que. 4) Answer:Organisms that live deep in the oceans with dark environment are unaware of sun.

Que. 5) Answer: Temperature is the factor that can affect source of light because sun is the source of light.

Case study 4

Some organisms are able to maintain homeostasis by physiological (sometimes behavioural also) means which ensures constant body temperature, constant osmotic concentration, etc. All birds and mammals, and a very few lower vertebrate and invertebrate species are indeed capable of such regulation (thermoregulation and osmoregulation). Evolutionary biologists believe that the ‘success’ of mammals is largely due to their ability to maintain a constant body temperature and thrive whether they live in Antarctica or in the Sahara desert. The mechanisms used by most mammals to regulate their body temperature are similar to the ones that we humans use. We maintain a constant body temperature of 370C. In summer, when outside temperature is more than our body temperature, we sweat profusely. The resulting evaporative cooling, similar to what happens with a desert cooler in operation, brings down the body temperature. In winter when the temperature is much lower than 370C, we start to shiver, a kind of exercise which produces heat and raises the body temperature. Plants, on the other hand, do not have such mechanisms to maintain internal temperature.

Que. 1) Homeostatic of an organism is maintained by …………………………………………………………………………………… means.

(a) Behavioural

(b) Modificational

(d) Both (a) and (b)

Que. 2) A constant body temperature of ……………………………………………………………………………. °C is maintained by humans.

Que. 3) According to the evolutionary biologist, mammals are successful because they have ability to maintain ………………………………………………………………………………………. of body.

(a) Zero temperature

(b) Low temperature

(d) High temperature

Que. 4) What will happen if outside temperature will increase more than our body temperature?

Que. 5) Why our body will shiver in winters?

Que. 1)(d) Both (a) and (b)

Que. 2) (a) 37.

Que. 3) (c) Constant temperature.

Que. 4) Answer:Constant human body temperature is 37° C. If outside temperature will increase more than our body temperature then the body will sweat.

Que. 5) Answer: Our body will shiver in the winters because outside temperature will be less than our body temperature.

Case study 5

An overwhelming majority (99 per cent) of animals and nearly all plants cannot maintain a constant internal environment. Their body temperature changes with the ambient temperature. In aquatic animals, the osmotic concentration of the body fluids change with that of the ambient air, water osmotic concentration. These animals and plants are simply conformers. Considering the benefits of a constant internal environment to the organism, we must ask why these conformers had not evolved to become regulators. Recall the human analogy we used above; much as they like, how many people can really afford an air conditioner? Many simply ‘sweat it out’ and resign themselves to suboptimal performance in hot summer months. Thermoregulation is energetically expensive for many organisms. This is particularly true for small animals like shrews and humming birds. Heat loss or heat gain is a function of surface area. Since small animals have a larger surface area relative to their volume, they tend to lose body heat very fast when it is cold outside; then they have to expend much energy to generate body heat through metabolism. This is the main reason why very small animals are rarely found in polar regions.

During the course of evolution, the costs and benefits of maintaining a constant internal environment are taken into consideration. Some species have evolved the ability to regulate, but only over a limited range of environmental conditions, beyond which they simply conform. If the stressful external conditions are localised or remain only for a short duration, the organism has two other alternatives for survival.

Que. 1) The conformers are ……………………………………………………………………………………… to become regulators.

(a) Not evolved

(b) Evolved

(d) None of above

Que. 2) In the …………………………………………………………………………… , temperature of body changes with the ambient temperature.

(d) Insects

Que. 3) Which of the following is unable to maintain a constant internal environment?

(a) Mammals

Que. 4) In the polar region, very small animals rarely found. What will be the main reason?

Que. 5) Which region has very small animals rarely?

Que. 1)(a) Not evolved.

Que. 2) (b) Plants.

Que. 3)(d) Plants.

Que. 4) Answer: Very small animals can lose body heat fast when they have larger surface area. These animals generate body heat through metabolism when there is cold outside. Hence, They found rarely in the polar region.

Que. 5) Answer: Polar region has very small animals rarely.

Case study 6

The size of a population for any species is not a static parameter. It keeps changing with time, depending on various factors including food availability, predation pressure and adverse weather. In fact, it is these changes in population density that give us some idea of what is happening to the population – whether it is flourishing or declining. Whatever might be the ultimate reasons, the density of a population in a given habitat during a given period, fluctuates due to changes in four basic processes, two of which (natality and immigration) contribute to an increase in population density and two (mortality and emigration) to a decrease.

(i) Natality refers to the number of births during a given period in the population that are added to the initial density. (ii) Mortality is the number of deaths in the population during a given period. (iii) Immigration is the number of individuals of the same species that have come into the habitat from elsewhere during the time period under consideration. (iv) Emigration is the number of individuals of the population who left the habitat and gone elsewhere during the time period under consideration.

Que. 1) For any species, size of ………………………………………………………………………………….. is not a static parameter.

(a) Population

Que. 2) Which of the following are responsible for the fluctuation of density of population?

(a) Mortality

(b) Natality

(d) All of them

Que. 3) In the population, the number of death during a given period is known as ………………………………………………………………………………………. .

(a) Immigration

(b) Mortality

(d) Emigration

Que. 4) Define the following term: Mortality and Immigration

Que. 5) Which two basic process can decrease population density?

Que. 1)(a) Population

Que. 2) (d) All of them

Que. 3) (b) Mortality

Que. 4) Answers:

Natality: In the population, the number of births which occurs during a period of time and which are added to initial density is known as Natality.

Immigration: During the time period under consideration, the individuals number of same species which have come into the habitat from elsewhere is known as Immigration.

Que. 5) Answer: Emigration and Mortality are the two basic processes that can decrease population density.

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Biology Case Study for Class 12

While preparing for the board exams, students are being judged on different levels of skills, such as writing, reading, etc. Biology Case Study for Class 12 questions are one of them that helps in assessing critical thinking.

The Central Board of Secondary Education will be asking the case study questions in the Class 12 board examination. Therefore, here on this page, we have provided the Biology Case Study for Class 12 at free of cost. Our subject matter experts have prepared Case Study questions so that Apart from the basic standard questions, students can have a variety of problems to solve.

Just like MCQs, and other written types questions Biology Case Study for Class 12 questions will impact the overall performance of a student. Therefore for the convenience of the students we have provided the download links here, so that they can easily access the Class 12 Case Study.

Download Chapter Wise Biology Case Study for Class 12 Question and Answers PDF

There are lots of chapters in class 12 Biology from which Class 12 Case Study Questions can be framed. Going through such types of questions help the students to assess their understanding level in the topics discussed in NCERT Class 12 Biology Books. By practicing the Class 12 Case Study Questions for Biology students will be very confident to ace the board exam. Also the Biology case study will be very useful for the NEET exam preparation.

Doing a regular practice of Class 12 Biology Case Study questions is a great way to score higher marks in the board exams as it will help students to develop a grip on the concepts.

Here our subject experts have crafted the Subject Wise Case Study For Class 12. Download Subject Wise CBSE Case Study Class 12 Question and Answers PDF from the below given links.

Case Study Questions Class 12 Chapter 2 Sexual Reproduction in Flowering Plants

Case Study Questions Class 12 Chapter 3 Human Reproduction

Case Study Questions Class 12 Chapter 4 Reproductive Health

Case Study Questions Class 12 Chapter 5 Principles of Inheritance & Variation

Case Study Questions Class 12 Chapter 6 Molecular Basic of Inheritance

Case Study Questions Class 12 Chapter 7 Ecosystem

Case Study Questions Class 12 Chapter Evolution

Case Study Questions Class 12 Chapter 8 Human Health And Diseases

Case Study Questions Class 12 Chapter 10 Microbes in Human Welfare

Case Study Questions Class 12 Chapter 11 Biotechnology - Principles & Processes

Case Study Questions Class 12 Chapter 12 Biotechnology And Its Applications

Case Study Questions Class 12 Chapter 13 Organisms And Populations

Case Study Questions Class 12 Chapter 15 Biodiversity And Conversation

Case study types of questions are generally descriptive that helps to gather more information easily so, it is kinda easy to answer. However, our subject matter experts have given the solutions of all the Biology Case Study for Class 12 Biology questions.

Passage Based Class 12 Biology Case Study Questions in PDF

CBSE Class 12 Case studies are known as Passage Based Questions. These types of problems usually contain a short/long paragraph with 4 to 5 questions.

Students can easily solve Passage Based Class 12 Case Study Questions by reading those passages. By reading the passage students will get the exact idea of what should be the answers. Because the passage already contains some vital information or data. However a better understanding of the basic concepts that can be learned from the NCERT Class 12 Textbooks will aid in solving the Case based questions or passage based questions.

How to Download CBSE Case Study of Class 12 Biology ?

Follow the below given simple steps to know how to download CBSE Case Study of Class 12 Biology:-

Open Selfstudys website in your browser
Go to the navigation menu that look like this
Now, click on CBSE and then Case Study respectively
A new page will open, where you have to click on “Class 12”
Now, you are ready to select the subject for which you want to download the case study questions.

How to Solve Biology Case Study Based Questions of Class 12?

There are very simple methods that a student should keep in mind while solving Biology Case Study for Class 12 for any subject:

Read each line of paragraph carefully and pay attention to the given data/numbers. Often questions are framed according to the highlighted data of the passage.
Since case study questions are often framed in Multiple choice questions, students should have the knowledge of elimination methods in MCQs.
Having a good understanding of the topics that are discussed in CBSE Class 12 Books are ideal to Solve Case Study Based Questions of Class 12.

Features Of Class 12 Biology Case Study Questions And Answers PDF

The three most noticeable features of Class 12 Biology Case Study Questions And Answers Pdf are -

It Is Free To Use: Keeping in mind the need of students and to help them in doing Self Study, our team has made all the PDF of Biology Case Study for Class 12 free of cost.
Answers Are Given: Not only the PDFs are free provided but answers are given for all the questions of Biology Case Study for Class 12.
PDF Can Be Downloaded Or Viewed Online: Many students don’t like to download the PDFs on their device due to the shortage of storage. Therefore, CBSE Class 12 Case Study questions are made available here in online format, so that students can view them online. However, through the Selfstudys app, a learner can download the PDFs too.

Benefits of Using CBSE Class 12 Biology Case Study Questions and Answers

The CBSE Class 12 Biology Case Study Questions and Answers can help a student in several ways:

In Exam Preparation: Those who go through the Biology Case Study for Class 12 will find support during the exam preparation as case based questions are also asked in the CBSE Class 12 Board examination.
Help in brushing up the previous learnings: No matter how brilliant you are, you have to revise the studied topics time and again to keep them refreshed. And in this task, the CBSE Class 12 Case Study Questions and Answers can help a lot.
To develop the critical thinking: Able to analyse information and make an objective judgement is a skill that is known as critical thinking. A student of class 12 can use Biology Case Study for Class 12 with answers to develop critical thinking so that they can make better decisions in their life and in the board examination.

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Class 12 Biology Case Study of Chapter 5 Principles of Inheritance and Variation

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In Class 12 Boards there will be Case studies and Passage Based Questions will be asked, So practice these types of questions. Study Rate is always there to help you. Free PDF Downloads of CBSE Class 12 Biology Chapter 5 Principles of Inheritance and Variation Case Study and Passage Based Questions with Answers were Prepared Based on the Latest Exam Pattern. Students can solve Class 12 Biology Case Study Questions Principles of Inheritance and Variation to know their preparation level.

Join our Telegram Channel, there you will get various e-books for CBSE 2024 Boards exams for Class 9th, 10th, 11th, and 12th.

In CBSE Class 12 Biology Paper, There will be a few questions based on case studies and passage-based as well. In that, a paragraph will be given, and then the MCQ questions based on it will be asked.

Principles of Inheritance and Variation Case Study Questions With Answers

Here, we have provided case-based/passage-based questions for Class 12 Biology Chapter 5 Principles of Inheritance and Variation

Case Study/Passage-Based Questions

Answer: (c) A and C only

(ii) Teacher asked to conduct an experiment on Drosophila because

Answer: (d) all of these.

(iii) Genes white-eyed and yellow-bodied located very close to one another on the same chromosome tend to be transmitted together and are called

Answer: (c) linked genes

(iv) Select the correct statement regarding the given experiment.

Answer: (a) The physical distance between two genes determines the strength of linkage

(v) Assertion : When yellow-bodied, white-eyed Drosophila females were hybridized with brown-bodied, red-eyed males; and FI progeny was intercrossed, the F2 ratio deviated from 9: 3: 3: 1. Reason: When two genes in a dihybrid are on the same chromosome, the proportion of parental gene combinations is much higher than in the non-parental type.

(a) Both assertion and reason are true and the reason is the correct explanation of assertion.

(b) Both assertion and reason are true but the reason is not the correct explanation of the assertion

(d) Both assertion and reason are false

Answer: (a) Both assertion and reason are true and the reason is the correct explanation of assertion.

Case Study 2: Turner’s syndrome is an example of mono somy. It is formed by the union of an allosome free egg and a normal ‘X’ containing sperm or a normal egg and an allosome free sperm. The individual has 2n = 45 chromosomes (44 + X0) instead of 46. Such individuals are sterile females who have rudimentary ovaries, under developed breasts, small uterus, short stature, webbed neck and abnormal intelligence. They may not menstruate or ovulate. This disorder can be treated by giving female sex hormone to the women from the age of puberty to make them develop breasts and have menstruation. This makes them feel more normal.

What is Turner’s syndrome an example of? A) Trisomy B) Monosomy C) Polysomy D) Karyotype alteration

How many chromosomes are typically present in an individual with Turner’s syndrome? A) 44 B) 46 C) 45 D) 47

What is the chromosomal configuration in Turner’s syndrome? A) 44 + XX B) 44 + XY C) 44 + X0 D) 44 + XXY

Which of the following is a characteristic of Turner’s syndrome? A) Overdeveloped Breasts B) Tall Stature C) Normal Ovaries D) Webbed Neck

What is the primary treatment approach for women with Turner’s syndrome from the age of puberty? A) Giving Male Sex Hormone B) Administering Insulin C) Providing Female Sex Hormone D) Performing Surgery

Individuals with Turner’s syndrome often experience which of the following reproductive issues? A) Overactive Ovaries B) Frequent Ovulation C) Sterility and Rudimentary Ovaries D) Multiple Pregnancies

Hope the information shed above regarding Case Study and Passage Based Questions for Class 12 Biology Chapter 5 Principles of Inheritance and Variation with Answers Pdf free download has been useful to an extent. If you have any other queries about CBSE Class 12 Biology Principles of Inheritance and Variation Case Study and Passage Based Questions with Answers, feel free to comment below so that we can revert back to us at the earliest possible. By Team Study Rate

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