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Q:

Holding a flower, the teacher asked Heena to show the part of a flower which has four-sided (tetragonal) structure consisting of four sporangia, grouped as a pair in each lobe. What would be the part of the flower?

A:
The part of the flower is anther.
Q:

An anther with malfunctioning tapetum often fails to produce viable male gametophytes. Give one reason.

A:

The tapetum is the innermost layer of a microsporangium. It is found in a cluster of two in each lobe of the anther. It nourishes the pollen grains. If it malfunctions, then the pollen grains are provided with adequate nutrition. Hence, the anther is not able to produce mature viable male gametes.

Q:

Bacteria can not be seen with the naked eye, but these can be seen with the help of a microscope. If you have to carry a sample from your home to your biology laboratory to demonstrate the presence of microbes under a microscope, which sample would you carry and why?

A:
A sample of curd may be taken to a lab to demonstrate the presence of microbes. Microbes present in the curd are lactic acid bacteria.
Q:

Why RNAi is considered as an important technique of Biotechnology?

A:

RNAi is considered as an important technique of Biotechnology because:

1. This technique can be used to trace the function of a gene.

2. It can be used to discover new drugs which can suppress or enhance the expression of a gene or set of genes.

Q:

How do conformers cope up with changes in the surrounding environment?

A:
Conformers cope up with changes in the surrounding environment by changing the body temperature.
Q:

What is sere ?

A:
Sere is the entire sequence of communities that successively change in a given area.
Q:

From which organism restriction endonuclease was first extracted?

A:
Escherichia coli
Q:

Fill in the blanks in the different columns of the table given below:

Diseases

Causal Organisms

Medium of transfer

Symptoms

Ascariasis

Ascaris

Through contaminated waters, vegetables and fruits.

d

Amoebiasis

a

House fly

Constipation, abdominal pain, cramps, stool with excess mucous and blood.

Common Cold

b

Droplets from sneezing of infected persons.

Affect nose and respiratory passage, sore throat

Pneumonia

Streptococcus- pneumoniae

c

Fever, chills, cough and headache.

A:
(a) Entamoeba histolytica.
(b) Rhino virus.

(c) By inhaling droplets released by an infected person or by sharing glasses and utensils with an infected person.

(d) Internal bleeding, muscular pain, fever, anemia and blockage of the passage intestinal.

Q:

Identify A, D, E and F in the diagram of an antibody molecule given below:

A:
A: Antigen binding site
D: Light chain
E: Heavy chain
F: Disulphide bond
Q:

Enumerate any four differences between induction and repression.

A:

Induction

Repression

Switch on the genes

Switch off the genes

Needs a inducer

Needs a supressor

Associated with catabolic pathway

Associated with anabolic pathway.

Responsible for starting transcription and translation

Responsible for stopping transcription and translation

Q:

(a) Mention the cause and the body system affected by ADA deficiency in humans.
(b) Name the vector used for transferring ADA-DNA into the recipient cells in humans. Name the recipient cells.

A:

(a) ADA (Adenosine deaminase) is an enzyme crucial for immune system. Its deficiency is caused due to the deletion of the gene coding for adenosine deaminase. Its deficiency impairs the immune system.

(b) In humans, retroviral vector is used to introduce functional adenosine deaminase coding cDNA into the lymphocytes cells of the patient.
Q:

(a) Name the lymphoid organ in humans where all the blood cells are produced.
(b) Where do the lymphocytes produced by the lymphoid organ mentioned above migrate and how do they affect immunity?

A:

(a) Bone marrow

(b) B-lymphocytes migrate to the secondary lymphoid organs like spleen, lymph nodes, etc. T-lymphocytes migrate to thymus for their maturation.

B-lymphocytes produce antibodies and T-lymphocytes help B-cells in production of antibodies. Hence lymphocytes are involved in providing immunity to the body.
Q:

Honey collection improves when beehives are kept in crop-fields during flowering season. Explain.

A:
Keeping beehives in crop fields during flowering period increases the frequency of pollination by bees. Since bees visit the flowers in the crop fields often, they collect more nectar and use this nectar to produce additional honey.
Thus, keeping beehives in crop-fields during flowering season is beneficial both for crop yield and for honey yield.
Q:

For what purpose are air scrubbers and electrostatic precipitators used?

A:
Air scrubbers and electrostatic precipitators are used to remove particulate matter from various industrial exhausts as a means to reduce the entry of such pollutants into the atmosphere.
Q:

A cross between two plants heterozygous for a single locus was made. The progeny contained the following:
(i) Round seeds, large starch grains : 1
(ii) Round seeds, intermediate starch grains : 2
(iii) Wrinkled seeds, small starch grains : 1
State the phenomenon exhibited by the above result? Work out the genotype of the parents and offspring using a punnet square.

A:

The gene controls two traits, a phenomenon called pleiotropy.

The gene is completely dominant for one trait, the seed shape, while it shows incomplete dominance for the size of starch grains.

Parent: Bb X Bb

Gamete: B, b B, b

Progeny:

B

b

B

BB

Round seeds & large starch grains

Bb

Round seeds & intermediate starch grains

b

Bb

Round seeds & intermediate starch grains

bb

Wrinkled seeds & small starch grains:

Q:

Cleistogamous flowers are strictly self pollinating,” Explain. What is the disadvantage of this phenomenon?

A:
Cleistogamous flowers remain closed until pollination is accomplished. i.e. anthers and stigma are not exposed to outside, therefore, they are invariable autogamous/self pollinating. The phenomenon does not bring genetic variations and leads to loss of vigour.
Q:

What are exotic species? Explain with the help of two examples how the exotic species disturb the native species of an ecosystem?

A:

A species that has been introduced from another geographic region to an area outside its natural range is known as exotic species.
Their introduction into a particular geographical area leads to invasion by them. They spread very fast as there are no natural predators of the exotic species in the invaded area.For example the prickly pear cactus introduced into Australia caused havoc by rapidly spreading into vast areas of rangeland.
'Terror of Bengal' or water hyacinth introduced in Bengal has spread vastly choking most of the waterways.

Q:

Trace the development of a mature ovule from a megaspore mother cell.

A:
The embryo sac is the female gametophyte plant undergoing the process of meiosis and in this process, the megasporocyte produces four (n) megaspores. Three megaspore nearest the micropyle disintegrate and disappear and the megaspore farthest from the micropyle enlarges and develops into a mature embryo sac that contains (after 3 meiotic divisions) eight nuclei. At the end of the process, an egg cell and two synergid cells are positioned at the micropylar end of the embryo sac (all 3 are often referred to as the egg apparatus), two polar nuclei have migrated to the center of the large central cell and three nuclei at the opposite end of the micropyle are antipodal cells.
Q:

In the adjacent figure of a typical dicot embryo, label the parts (1), (2) and (3). State the function of each of the labelled part.

A:

(1) Plumule - The portion of an embryo that gives rise to the shoot system (as distinct from the root system) of a plants.
(2) Cotyledons - It involves in either storage or absorption of food reserves.
(3) Radicle – The portion of an embryo that gives rise to the root system of a plant.

Q:

(a) Mention the specific geographical region where these organisms are found.
(b) Name and explain the phenomenon that has resulted in the evolution of such diverse species in the region.
(c) Explain giving reasons the existence of placental wolf and Tasmanian wolf sharing the same habitat.

A:
a) Australia
b) Adaptive radiations: The process of evolution of different species in a given geographical area.
c)As both the animals share the same habitat and face the same changes, placental wolf exhibit adaptive radiation and appears to be similar to Tasmanian wolf.
Q:

Below given are the process of manufacturing of two commercial products ‘X’ and ‘Y’ as directed by FAO from soy.

Examine the two flow charts (I and II) and answer the following questions:

  1. Fill the blanks (a), (X), (b) and (Y) in the flow chart
  2. Which one of them is a fermented product? Support your answer with a reason.
  3. Which among the two products ‘X’ and ‘Y’ is more nutritious?

A:
  • (a): Rhizopus oligosporus
  • (X): Tempeh
  • (b): Coagulant (CaSO4.2H2O)
  • (Y): Tofu

  1. Product ‘X’ (Tempeh) is a fermented product as it has required an inoculum of a fungal strain (Rhizopus oligosporus) in order to carry out fermentation.
  1. Product ‘X’ (Tempeh) is more nutritious as compared to that of product ‘Y’ (Tofu) because. it contains more proteins, dietary fibres and vitamins. Moreover, it has less fat and is less processed.
Q:

Emma’s mother got eye bags around her eyes. She thought that harmful UV rays have affected her mother’s skin and this might symptomise skin cancer.

When she took her to the doctor, the doctor prescribed her a drug which was extracted from yeast strain.

Based on the given information, answer the following questions:

  1. Was Emma’s assumption correct?
  2. The drug prescribed to her mother is possibly extracted from which microorganism?
  3. Write the possible mechanism involved in the mode of action of the drugs.

A:
  1. No, Emma’s assumption was wrong. Her mother had high level of cholesterol in her blood, due to which doctor had prescribed a drug with statins.
  2. Statins are extracted from yeast Monascus purpureus.
  3. Statins act by competitively inhibiting the enzyme responsible for cholesterol synthesis. Statins resemble mevalonate and is a competitive inhibitor of ß-hydroxy-ß-methylglutaryl CoA reductase.
Q:

(i) Name the organism in which the vector shown is inserted to get the copies of the desired gene.
(ii) Mention the area labelled in the vector responsible for controlling the copy number of the inserted gene.
(iii) Name and explain the role of selectable marker in the vector shown.

A:
(i) The organism in which the vector is inserted to get the copies of the desired gene is E.coli.
(ii) Origin of replication (ori) is responsible for controlling the copy number of the inserted gene.
(iii) The selectable markers in the vector are Ampicillin resistant gene and Tetracycline resistant gene. The role of selectable markers is that they help in identifying and eliminating non-transformants and selectively permitting the growth of the transformants.
Q:

Name any two types of intra uterine contraceptive devices and the mechanism by which they prevent conception.

A:
a) Copper releasing IUDs, e.g., Cu T, Cu 7, multiload 375 releases Cu which suppress sperm motility and the fertilising capacity of sperms.
b) Hormone releasing IUDs, e.g., Progesteron, make the uterus unsuitable for implantation and the cervix hostile to the sperms.
Q:

Give the technical term for:
(a) Region of DNA that helps in initiating the process of transcription and serves as the binding site for RNA polymerase.
(b) A small RNA that reads the genetic code present on mRNA.
(c) Coding sequences of DNA in eukaryotes that transcribe for proteins.

A:
(a) Promoter
(b) tRNA
(c) Exons
Q:

a. Why did Meselson and Stahl grow E.coli in a medium containing 15NH4Cl for many generations?
b. In which medium were they transferred later on?
c. What were their observations?

A:
a. They grew E.coli in a medium containing 15NH4Cl to get E.coli with all its nitrogen as heavy isotope i.e. 15N (heavy isotope of nitrogen).
b. They were transferred into a medium with normal 14NH4Cl.
c. They observed that the DNA which was extracted had a hybrid or intermediate density.

Q:

(a) What is the programme called that is involved in improving success rate of production of desired hybrid and herd size of cattle?
(b) Explain the method used for carrying this programme for cows.

A:

(a) To improve chances of successful production of desired hybrids and herd size of cattle, Multiple Ovulation Embryo Transfer Technology (MOET) is being used.

(b) In this method, a cow is administered hormones, to induce follicular maturation and super ovulation. Instead of one egg per cycle, they produce 6-8 eggs. The animal is either mated with an elite bull or artificially inseminated. The fertilised eggs at 8-32 cells stages, are recovered non-surgically and transferred to surrogate mothers. The genetic mother is available for another round of super ovulation. This technology has been demonstrated for cattle, sheep, rabbits, buffaloes, mares, etc. High milk-yielding breeds of females and high quality (lean meat with less lipid) meat-yielding bulls have been bred successfully to increase herd size in short time.

Q:

How will the second amendment of the Indian Patents Bill help Indian bio-resources?

A:

Second amendment of the Indian Patents Bill will help Indian bio-resources as follows:

• It is in the area of genetically modified plants and seeds. Methods of agriculture and plants were excluded from patentability in the Indian patent act to ensure that seed can be held as a common property resource in the public domain and hence there will not be any violation of farmers' right to save, exchange and improve seed.

• It takes into consideration issues like exploitation of bio-resources and traditional knowledge related to bio-resources including patent terms, emergency provisions and research and development initiative.

Q:

A patient was complaining of frequent urination, excessive thirst, hunger, and tiredness. His fasting blood glucose level was found higher than 130 mg/dl on two occasions.
(i) Name the disease.
(ii) Give the root cause of this disease.
(iii) Explain why the blood glucose level is higher than’ 130 mg/dl.

A:
(i) Diabetes mellitus
(ii) Hyposecretion of insulin
(iii) Because glucose is not being converted into glycogen due to hyposecretion of insulin hormone by B-cells of islets of langerhans in pancreas.

Q:

Huntington’s disease, a disease of the nervous system, is autosomal dominant. The pedigree below shows the inheritance of the disease in three generations of a family. Observe the pedigree carefully and answer the questions that follow:

(a) What is the probable genotype of individual D? How do you know?

(b) What is the probability that individual N will not have Huntington’s disease?

A:

(a) Individual D is an affected person because individual K has inherited the dominant allele from him.

Now the question arises whether individual D is homozygous (HH) or heterozygous (Hh) for the dominant allele. For this, we have to look at the cross between his parents, i.e., individual A and B:

Parent: Hh X hh

(Individual A) (Individual B)

Gamete: H,h h,h

Progeny:

H

h

h

Hh

Affected

hh

Normal

h

Hh

Affected

hh

Normal

Therefore, genotype of individual D is Hh.

(b) The genotype of individual N entirely depends on the genotype of Individual H and I.

Genotype of individual H: Individual O is homozygous (HH) for the dominant trait points out that both of his parents are affected individuals. Now, looking at the genotype of individual A and B (parents of individual H), the genotype of any of their affected offspring must be Hh. Thus, the genotype of individual H is Hh.

Genotype of individual I: Individual O is homozygous (HH) for the dominant trait points out that both of his parents are affected individuals. Therefore, individual I can be homozygous (HH) or heterozygous (Hh) for the trait.

So, the probability that individual N will not express the trait depends on the genotype of her parents.

In case individual I is homozygous (HH) for the trait:

Parent: Hh X HH

(Individual H) (Individual I)

Gamete: H,h H,H

Progeny:

H

h

H

HH

Affected

Hh

Affected

H

HH

Affected

Hh

Affected

In such case, all the offspring will express the trait or the probability that individual N will not express the trait is 0%.

In case individual I is heterozygous (Hh) for the trait:

Parent: Hh X Hh

(Individual H) (Individual I)

Gamete: H,h H,h

Progeny:

H

h

H

HH

Affected

Hh

Affected

h

Hh

Affected

hh

Normal

In such case, 1/4th of the offspring will not express the trait or the probability that individual N will not express the trait is 25%.
Q:

A variety of wild beetles have been observed to lay their eggs in dead animals and then they their eggs in the ground until they hatch. Assuming that the preference for fresh meat (F) and tendency to bury the meat shallow (S) are dominant traits.
a) What will be the genotype of the offspring if a female carrion beetle homozygous dominant for both traits mates with a male homozygous recessive for both traits.
b) What will be the expected genotypic ratio of the F2 generation (FFSS : FFSs : FFss : FfSS : FfSs : Ffss : ffSS : ffSs : ffss)?

A:

a) Female - FFSS

Male - ffss

FS

FS

FS

FS

fs

FfSs

FfSs

FfSs

FfSs

fs

FfSs

FfSs

FfSs

FfSs

fs

FfSs

FfSs

FfSs

FfSs

fs

FfSs

FfSs

FfSs

FfSs

Phenotype – Fresh meat/shallow

Genotype – FfSs

b) FfSs x FfSs

FS

Fs

fS

fs

FS

FFSS

FFSs

FfSS

FfSs

Fs

FFSs

FFss

FfSs

Ffss

fS

FfSS

FfSs

ffSS

ffSs

fs

FfSs

Ffss

ffSs

ffss

FFSS : FFSs : FFss : FfSS : FfSs : Ffss : ffSS : ffSs : ffss
1:2:1:2:4:2:1:2:1

Q:

Study the population growth curve shown above.
(i) Identify curve 'a' and 'b'.
(ii) mention the conditions responsible for the curve 'a' and 'b' respectively.
(iii) Give the necessary equation for the curve 'b'.

A:
(i) a is a exponential curve and b is a logistic curve.
(ii) a when responses are not limiting the growth, plot is exponential
 b when responses are limiting the growth, plot is logistic,
(iii) dN/dt =rN(K-N)/k
Q:

How does radioactive pollution harm living organisms? Suggest ways to control radioactive pollution.

A:

Radioactive substances can cause deleterious mutations in genes, impairment of various biological functions. In higher doses, these can cause immediate death of the organism.

Ways for controlling radioactive pollution:
ª Storage of radioactive wastes after sufficient pretreatment in shielded containers.
ª Prevention of leakage of radioactive substances from nuclear reactors, labs, etc.
ª Banning the use of nuclear weapons.

Q:

State true or false and rewrite the false statement correctly.
i.The ‘Chernobyl’ disaster resulted into the release of uranium fuel into the environment.
ii.Yukka mountain is a site of nuclear waste storage.
iii.Radioactive pollution results in temporary changes.
iv.Radioactive pollution may cause cancers.
v.The Yukka mountain is located in southeast of Las Vegas.

A:
i.False; The ‘Three Mile Island’ disaster resulted into the release of uranium fuel into the environment.
ii.True
iii.False; Radioactive pollution results in permanent changes.
iv.True.
v.False; The Yukka mountain is located in northwest of Las Vegas.
Q:

Wolves have been seen with black coats and blue eyes. If normal coat colour (N) is dominant to black (n) and brown eyes (B) are dominant to blue (b). Suppose the breeding male and female are black with blue eyes and normal coloured with brown eyes respectively and female is also heterozygous for both traits. How many of the offspring (assume 16) living in the pack will have each of the following genotypes?
a) 1)nnBB 2)Nnbb
b)What percent of the offspring will be normal coloured with blue eyes?
c)What percent of the offspring will be brown with brown eyes?

A:

a) Male-nnbb

Female-NnBb

NB

Nb

nB

nb

nb

NnBb

Nnbb

nnBb

nnbb

nb

NnBb

Nnbb

nnBb

nnbb

nb

NnBb

Nnbb

nnBb

nnbb

nb

NnBb

Nnbb

nnBb

nnbb

1) nnBB-0 2) Nnbb-4

b) Normal coloured with blue eyes – Nnbb – 25% (4/16x100)

c) Brown with brown eyes – NnBb – 25% (4/16x100)
Q:

(a) A colour blind man marries a woman with normal vision whose father was colour blind. Work out a cross to demonstrate the genotype of the new couple and their future sons?(b) Answer the following questions with reference to the given pedigree.

i) Is the trait autosomal dominant, autosomal recessive or sex-linked? Justify your answer.
ii) Give the genotypes of the parents (individual 1 and 2).
iii) Give the genotype of the daughter in the first generation and the son and the daughters in the second generation.

A:

(a) Since the father of the normal women is colour blind, her genotype will be XcX (carrier)

Colour blind man X Normal woman

XcY XcX

Gamete: Xc, Y Xc, X

Progeny:

Xc

Y

X

XcX

Normal (carrier)female

XY

Normal male

Xc

XcXc

Colour Blind female

XcY

Colour blind male

50% of their sons will be colour blind.

(b) (i) The trait is sex linked (recessive) as it has appeared in male child in the first generation pointing at the possibility that the female parent is a carrier as his father is normal.

(ii) Genotype of individual 1 => Male => XY

Genotype of individual 2 => Female => XcX (Carrier)

(iii)
Genotype of individual 3 in generation I => XcY
Genotype of individual 4 in generation I => XcX

Genotype of individual 6 in generation II => XcX Genotype of individual 7 in generation II => XcXc

Q:

Study the following flow chart. Name the hormones involved at each stage. Explain their function.

A:

Gonadotropin releasing hormone secreted by hypothalamus causes release of follicle stimulating hormone (FSH) and luteinizing hormone (LH) from pituitary.
• Luteinising hormone (LH) acts on the Leydig cells and stimulates synthesis and secretion of androgens.
• FSH acts on the sertoli cells and stimulates secretion of some factors which help in the process of spermiogenesis.
• Androgens(Testosterone) stimulate the process of spermatogenesis.

Q:

Give a schematic representation of oogenesis in humans. Mention the number of chromosomes at each stage. Correlate the life phases of the individual with the states of the process.

A:

At oogonia- 46 chromosomes per cell
At primary oocyte- 23 chromosomes per cell
At secondary oocyte- 23 chromosomes per cell
The flow diagram above shows the correlation between the life phases of the individual with the states of the process.

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