some additional questions for +2 board exams …drarvindsbiology.com/content/papers/8c_p25.pdf ·...

1 DR. ARVIND'S BIOLOGY CLASSES

(A Unit of Med-Xel Tutorials)

SOME ADDITIONAL QUESTIONS FOR +2 BOARD EXAMS

REPRODUCTION

SECTION – A

I. Each question carries 1 mark.

1. Cucurbits and papaya plants bear staminate and pistillate flowers. Mention the categories they are put under separately on the basis of the type of flowers they bear.

Ans. Papaya is dioecious because the staminate and pistillate flowers are borne in two different plants, while cucumber is monoecious because it bears both staminate and pistillate flowers in the same plant.

2. Mention the differences between spermiogenesis and spermiation.

Ans. Differences between spermiogenesis and spermiation.

Spermiogenesis Spermiation

It is the process of transforming spermatids into matured spermatozoa or sperms.

It is the process when mature spermatozoa are released from the Sertoli cells into the lumen of seminiferous tubules.

SECTION – C

III. Each question carries 3 marks.

1. Differentiate between perisperm and endosperm giving one example of each.

Ans. Differences between perisperm and endosperm

Perisperm Endosperm

1. It represents persistent remains of

nucellus (of ovule) in the seed

2. It is a part that belongs to seed.

3. It is usually dry.

Example – Black pepper

1. It develops from Primary Endosperm

Nucleus (PEN).

2. It contains reserve food materials

3. It is usually in fluid form or is soft.

Example – Water of coconut, maize

2. Fertilization is essential for the production of seed, but in some angiosperms seeds develop

without fertilization.

(a) Give an example of an angiosperm that produces seeds without fertilization. Name the

process.

(b) Explain the two ways by which seeds develop without fertilization.

Ans. (a) Example, the members of Asteraceae like sunflower. The process is called apomixis.

(b) (i) A diploid egg cell is formed without meiosis and it develops without fertilization into an

embryo in some cases.

(ii) In some cases, some of the cells of nucellus around the embryo sac develop into embryo.

For example - mango and citrus.

3. Write the function of each one of the following:

(a) Fimbriae (Oviducal) (b) Coleoptile (c) Oxytocin

Ans. (a) Fimbriae are the feathery finger-like processes present at the end of Fallopian tubes and it

collects the ovum after it is released from the ovary into the Fallopian tube.

(b) Coleoptile is a conical sheath present in the monocot. Seeds, its function is to protect the

developing plumule.



(c) Oxytocin is a hormone released by the posterior pituitary and it stimulates the contraction of uterine

muscles during child birth (parturition).

4. Write the function of each of the following:

(a) Middle piece in human sperm

(b) Tapetum in anthers

(c) Luteinizing hormone in human males

Ans. (a) Middle piece in human sperm. It contains several mitochondria which produce energy for the

motility of the sperm.

(b) Tapetum in anthers is the innermost layer of the anther. The main function of tapetum is to

provide nourishment to the developing pollen grains.

(c) Leuteinizing hormones in human males stimulate the Leydig cells to produce testosterone.

5. Write the function of each of the following:

(a) Seminal vesicle (b) Scutellum (c) Acrosome of human sperm

Ans. (a) Seminal vesicle. It secretes an alkaline fluid that helps in neutralizing the acidity of the

vaginal tract thereby increasing the life span of sperms. Secretion of seminal vesicle is also a

source of nutrition for the sperms.

(b) Scutellum. The single cotyledon present in seed (e.g., Maize) is called scutellum. It occupies

the major part of the embryo. Its outmost part is called epithelial layers. It is both secretory and

absorptive. The epithelial layer secretes hormones into the endosperm for the synthesis of

enzymes required for solubilisation of food. The solubilised food is absorbed by it and then

transferred to the embryo axis.

(c) Acrosome. It is the cap-like covering or structure that is present at the tip of the sperm (male

gamete). The acrosome contains enzymes which help the sperm enter into the cytoplasm of the

ovum and thus helps in fertilization.

SECTION – D

IV. Each question carries 5 mark.

1. The following is the illustration of the sequence of ovarian events (a – i) in a human female.

(a) Identify the figure that illustrates ovulation and mention the stage of oogenesis it

represents.

(b) Name the pituitary hormone that has caused the above mentioned event.

(c) Explain the changes that occur in the uterus simultaneously in anticipation.

(d) Write the differences between ‘c’ and ‘h’.

(e) Draw a labelled sketch of the structure of a human ovum prior to fertilization.



Ans. (a) Figure ‘f’ illustrates ovulation. It represents the ovulatory stage of oogenesis.

(b) Luteinizing Hormone (LH) is the pituitary hormone released during ovulation.

(c) In anticipation of receiving the fertilized egg, the endometrium of the uterus gets thickened and also

the blood supply to the endometrium increases.

(d) In the figure (c) stage represents the secondary follicle and the (h) stage represent the

degenerating corpus luteum.

Secondary Follicle Corpus Luteum

It is surrounded by layers of granulosa cells.

Presence of theca layer.

Layers of granulosa cells absent.

Theca layer is absent.

(e)

GENETICS

SECTION – B


1. Write the full form of VNTR. How is VNTR different from ‘Probe’?

Ans. VNTR – Variable Number Tandem Repeat

Differences between VNTR and Probe

VNTR Probe

It is a class of satellite DNA, where a small

sequence is arranged tandemly in many copy

numbers.

It is a radioactivity labelled VNTR, used for

hybridization with DNA segments.

SECTION – C


1. Study the given pedigree chart showing the pattern of blood group inheritance in a family



a) Given the genotype of the following:

(i) Parents

(ii) The individual ‘X’ in second generation

b) State the possible blood groups of the individual ‘Y’ in third generation.

c) How does the inheritance of this blood group explain codominance?

Ans. (a) (i) Parents IA IO IB IO

A B

(ii) X individual IOIO or IOIA

(b) Individual ‘Y’ blood groups IA, IO, IO, IO genotype

(c) IA and IB when stay together, show the phenomena of codominance and express themselves

in pressure of each other. In heterozygous hybrid, when both alternative alleles coexist, both

the alleles show codominance.

2. (a) Construct a complete transcription unit with promoter and terminator on the basis of

hypothetical template strand given below

(b) Write the RNA strand transcribed from the above transcription unit along with its

polarity.

Ans. (a) Transcription unit

(b)

SECTION – D

IV. Each question carries 5 marks.

1. (a) Why is haemophilia generally observed in human males? Explain the conditions under which a

human female can be haemophilic.



(b) A pregnant human female was advised to undergo MTP. It was diagnosed by her doctor that

the foetus she is carrying has developed from a zygote formed by an XX-egg fertilized by

Y-carrying sperm. Why was she advised to undergo MTP?

Ans. (a) The genes for haemophilia are present on the X-chromosome. A male has only one X-

chromosome and bears only one allele for the trait. He is hemizygous for the trait as Y-

chromosome does not have a corresponding allele.

A female contains two X-chromosomes. She has to be homozygous recessive. It means her

father must be a sufferer and mother either a carrier or sufferer to be affected by the disease.

(b) The zygote will have XXY-chromosomes. It will develop into a male with Klinefelter’s syndrome

such males are sterile and show feminine characters. That is why the female is advised to

undergo MTP.

2. (a) State the central dogma in molecular biology. Who proposed it? Is it universally

applicable? Explain.

(b) List any four properties of a molecule to be able to act as a genetic material.

Ans. (a) Francis Crick proposed the ‘central dogma’ in molecular biology with states that the genetic

information flows from -

In some viruses, the flow of information is in reverse direction, that is from RNA to DNA.

(b) Properties in a molecule to act as a genetic material:

It should be able to generate its replica.

It should chemically and structurally be stable.

It should provide scope for slow changes (mutation) that are required for evolution.

It should be able to express itself in the form of Mendelian characters.

EVOLUTION

SECTION – D

IV. Each question carries 5 marks.

1. Explain the three different ways the natural selection can affect the frequency of a heritable trait in

a population show in the graph given below:

Ans. Natural selection can lead to Stabilization (more individuals acquire mean character value),

Directional change (more individuals acquire a value other than the mean character value) or



Disruption (more individuals acquire peripheral character value at both ends of the distribution

curve).

APPLICATIONS

SECTION – A

I. Each question carries 1 mark.

1. Name the following:

(i) The semi-dwarf variety of wheat which is high-yielding and disease resistant.

(ii) Any one inter specific hybrid mammal.

Ans. (i) Sonalika and Kalyan Sona are the semi-dwarf varieties of wheat that are high-yielding and

disease resistant.

(ii) Mule is an interspecific hybrid mammal.

2. Write the names of the semi-dwarf and high yielding rice varieties developed in India after

1966.

Ans. Jaya and Ratna are two semi-dwarf and high yielding rice varieties developed in India after 1966.

3. BOD of two samples of water A and B were 120 mg/L and 400 ms/L respectively. Which

sample is more polluted?

Ans. Sample B (BOD 400 mg/L) is more polluted.

SECTION – B

II. Each question carries 2 marks.

1. Due to undue peer pressure, a group of adolescents started using opioids intravenously.

What are the serious problems they might face in future?

Ans. Serious problems due to opioids taken intravenously:

(i) AIDS and hepatitis – B may occur. The viruses for these diseases can be transmitted through

sharing infected needles and syringes.

(ii) Due to regular use, tolerance of the receptors increase, so they will become addicted to these

drugs.

2. Name the blank spaces a, b, c and d in the table given below:



Name of the drug Plant Source Organ System Affected

‘a’ Poppy plant ‘b’

Marijuana ‘c’ ‘d’

Ans. a – Morphine

b – Central nervous system

c – Cannabis sativa

d – Cardiovascular system

3. Study the flow chart given below

(a) Identify the events that take place at stages ‘a’ and ‘b’ respectively.

(b) State the importance of the technology explained above.

Ans. (a) (i) The hormone induces follicular maturation and super ovulation, i.e., production of 6 – 8

eggs.

(iii) They are transferred to surrogate mother.

(b) The technology called MOET (Multiple Ovulation Embryo Transfer Technology) is used to

increase the herd size by breeding high milk yielding breeds of females with high quality meat

yielding bulls.

SECTION – C


1. (a) Name the drug used (i) as an effective sedative and pain killer (ii) for helping patients to

cope with mental illness like depression but often misused.

(b) How does moderate and high dosage of cocaine affected the human body?

Ans. (a) (i) Morphine

(ii) Lysergic acid diethylamide (LSD) or Barbiturates.

(b) Moderate dose of cocaine has a stimulating action on central nervous system.

Cow is administrated with FSH hormone

‘a’

6 – 8 eggs per cycle are derived

Artificially inseminated

Fertilized eggs at 8 – 32 cells are recovered

‘b’



It produces a sense of euphoria and increased energy.

High dosage of cocaine causes hallucinations.

2. Describe the functions of anaerobic sludge in a sewage treatment plant.

Ans. Function of activated sludge in sewage treatment plant:

(i) A small part of activated sludge is pumped back into aeration tank to serve as aeration tank.

(ii) The remaining part of activated sludge is pumped back into large tanks called anaerobic sludge

digesters. All kinds of bacteria grow anaerobically in it, which digest the bacteria and the fungi in

the sludge. During this digestion, bacteria produce a mixture of gases, such as methane,

hydrogen sulphide and carbon dioxide. These gases form biogas and can be used as source of

energy.

3. Name the genes responsible for making Bt cotton plants resistant to bollworm attack. How

do such plants attain resistance against bollworm attacks? Explain.

Ans. The Bt toxin is encoded by the cry gene. The cry IIAc and cry IIAb control cotton bollworms,

while the cry IAb controls corn borer. Bt gene produces Bt toxin. This toxin provides resistance to

plants against lepidopteron, coleopteron and dipterans pests.

Specific BT toxin genes are isolated from B. thuringiensis and incorporated into the crops. Since,

these toxins are insert specific, they do not harm the crops or humans.

BIOTECHNOLOGY

SECTION – B


1. Study the diagram given below and answer the following questions.

(a) Why have DNA fragments in band ‘d’ moved farther away in comparison to those in band

‘c’?

(b) Identify the anode end in the diagram.

(c) How are these DNA fragments visualized?

Ans.

(a) In band ‘d’, DNA fragments are smaller than those on band ‘c’. The fragments separate according

to their size through the sieving effect provided by the gel. So, the smaller fragments move farther

away than the larger ones.

(b) End towards ‘b’.

(c) Gel containing DNA fragments is stained with ethidium bromide and exposed to UV radiation.

Orange-colour bands of DNA are visible.



2. Name the source organism that possesses taq polymerase. What is so special about the

function of this enzyme?

Ans. taq polymerase – from thermophilic bacterium, Thermus aquaticus. It remains active during the high

temperature induced denaturation of double-stranded DNA. Thus is helpful in PCR technique.

3. (a) Mention the cause and the body system attacked by ADA deficiency in humans.

(b) Name the vector used for transferring ADA-DNA into the recipient cells in humans.

Name the recipient cells.

Ans. (a) ADA is caused due to deletion of gene for adenosine deaminase.

Immune system of body is affected due to this.

(b) Retroviral vector is used to transfer ADA-DNA into the recipient cells of human.

Recipient cells – Lymphocytes.

4. List four applications of genetically modified plants.

or

Highlight any four advantages of Genetically Modified Organism (GMOs).

Ans. Advantages of GMOs

(i) Tolerant against biotic stresses, such as cold, drought, salt, heat

(ii) Reduces dependence on chemical pesticides.

(iii) Reduces post harvest losses.

(iv) Increased efficiency of mineral usage by plants.

ECOLOGY SECTION – A

III. Each question carries 1 mark.

1. What is an interaction called when an orchid grows on a mango plant?

Ans. An orchid growing on the branch of a mango tree is an epiphyte. Epiphytes are plants growing

on other plants which however, do not derive nutrition from them. Hence, the relationship between

a mango tree and an orchid is an example of commensalism.

2. Mention the unique feature with respect to flowering and fruiting in bamboo species.

Ans. Bamboo plants flower only once in their life time, generally after 50-100 years, produce large

number of fruits and then die.

3. Why are cattle and goats not seen browsing on Calotropis growing in the fields?

Ans. Calotropis plant produces poisonous cardiac glycosides. Therefore, cattle or goat do not graze

these plants.

4. Why do predators avoid eating Monarch butterfly? How does the butterfly develop this

protective feature?

Ans. The monarch butterfly is highly distasteful to its predators (birds) because of a special chemical present

in its body. It acquires this chemical during the caterpillar stage by feeding on a poisonous weed.

5. Give one example where population estimation of an organisms is done indirectly without

actually counting the organism

Ans. The number of fish caught per trap is a population estimation method indirectly without actually

counting them.

6. List any two ways of measuring the standing crop of a trophic level.

Ans. Standing crop can be measured by

(i) Biomass (ii) Number in a unit area

7. Name the interaction between sucker fish and shark.

Ans. Commensalism; as sucker fish is benefitted by attaching itself to the shark and the shark is not

affected.



SECTION – B


1. Some organisms suspend their metabolic activities to survive in unfavourable condition.

Explain with the help of any four examples.

Ans.

(i) Bacteria, fungi and lower plants. They form thick-walled spores which help them to survive in

unfavourable conditions. Spores germinate on return of favourable conditions.

(ii) Higher plants. Seeds and some other vegetative reproductive structures serve as means to tide

over periods of stress. They reduce their metabolic activity and undergo dormancy.

(iii) Animals. They undergo hibernation or aestivation, if unable to migrate. For example, some

snails and fishes.

(iv) Zooplanktons. They enter diapause (suspended development) under unfavourable conditions.

2. Humming birds live among the bushes in tropics while penguins live on icebergs. They

cannot survive if their habitats are reversed justify.

Ans. Humming birds are natural habitants of tropics. They have large surface area relative to their

volume. So, they tend to lose heat very fast, when it is cold outside.

Penguins live on icebergs (natural habitat). They reduce the surface area to volume ratio. The

lesser the ratio, more effective the thermoregulation. Also, they huddle in groups to remain warm.

3. (a) What is ‘r’ in the population equation given below:

dN/dt = rN

(b) How does the increase and the decrease in the value of ‘r’ affect the population size?

Ans. (a) ‘r’ is an intrinsic factor assessing impacts of biotic and abiotic factor on population growth.

(b) When ‘r’ increases, population size increases while a decrease in ‘r’ decreases the population

size.

4. Explain the function of ‘reservoir’ in nutrient cycle. List the two types of nutrient cycles in

nature.

Ans. ‘Reservoir’ in an ecosystem meets the deficit that arises due to imbalance in the influx and efflux

of nutrients. The two types of nutrient cycles are:

(i) Gaseous cycle (ii) Sedimentary Cycle

5.

The above graph shows species area relationship. Write the equation of the curve ‘a’ and

explain.

Ans. S = CA2

Where, S = Species richness

A = Area

C = Y – Intercept

(i) Alexander Von Humboldt observed that within a region, species richness increased with

increasing explored area, but only up to a limit.



(ii) The relation between species richness and area for a wide variety of taxa like angiosperms,

birds, fishes, etc., turns out to be a rectangular hyperbola.

6. The figure given inside, shows the relative contribution of four greenhouse gases to global

warning.

(i) Identify the gases a and c

(ii) Why are these four gases called green house gases?

Ans. (i) a – Carbon dioxide (60%)

c – Chlorofluorocarbons (14%)

For additional information

b – Methane (20%) and d is N2O (6%)

(ii) These are called greenhouse gases, as they absorb infrared radiations emitted by the Earth’s

surface.

SECTION – C


1. Study the graph below and answer the questions which follow:

a) The curve ‘a’ is represented by the equation rN dt

dN . What does ‘r’ represent in the

equation and what is its importance?

b) Which one of the two curves is considered a more realistic one for most of the animal

population?

c) Which curve would depict the population of a species of deer if there are no predators in

the habitat? Why is it so?

Ans.

(a) r is intrinsic rate of natural increase. It is an important parameter for assessing the impact of

any abiotic or biotic factors on the population growth.

(b) Curve ‘b’ is more realistic.



(c) Curve ‘b’. When the predators are absent, there will be competition in large prey population for

resources.

2. Draw and complete the following model of carbon cycle filling a, b, c, d, e and f.

Ans.

3. By the end of 2002, the public transport of Delhi switched over to a new fuel. Name the fuel.

Why is this fuel considered better? Explain.

Ans. Delhi had been categorized as the fourth most polluted city of the world in a list of 41 cities.

Burning of fossil fuels has added to the pollution of air in Delhi. So, in year 2002, it is switched over

to new fuel-CNG (Compressed Natural Gas).

Reasons of CNG being better fuel:

(i) It is a clean fuel that produces very little unburnt particles. Hence, it is ecofriendly.

(ii) It burns most efficiently, unlike petrol of diesel, very little is left unburnt.

(iii) It is cheaper than petrol or diesel and cannot be siphoned off by thieves and adulterated like

petrol or diesel.

4. How is ozone formed in the stratosphere? Why is it called good ozone? CFCs contribute to

ozone-hole formation. Explain.

Ans. Formation of ozone is stratosphere :

(i) Nascent oxygen combines with molecular oxygen (O2) to form ozone by the action of UV-rays.

UV O2 [O] + [O] Molecular oxygen Nascent oxygen

UV O2 + [O] O3 Ozone

(ii) Ozone is degraded into molecular oxygen in the stratosphere by the action of UV rays to maintain its balance between production and degradation

UV O3 O2 + [O]

UV O2 + [O] O3

Good ozone is formed in the stratosphere and absorbs harmful UV radiations from the sun. CFCs react with UV in stratosphere to release chloride atoms. Chloride atoms act as catalyst to degrade ozone and release molecular oxygen. CFCs have permanent and continued effect as



chloride atoms are not consumed. This leads to thinning of ozone layer that has resulted in the formation of ozone hole as observed over the Antarctica region.

SECTION – D


1. What is the association between the bumble bee and its favourite orchid Ophrys? How

would extinction or change of one affect the other?

Ans. Mutualism is an association seen between the bumble bee and the orchid Ophrys. In this

association both species are benefitted. One petal of its flower bears an uncanny resemblance to

the female of the bee in size, colour and markings. The male bee is attracted to what it perceives

as a female, ‘pseudocopulates’ with the flower and during that process, it is dusted with pollen from

the flower. When this same bee ‘pseudocopulates’ with another flower, it transfers pollen to it and

thus cross-pollinates the flower.

2. (a) Draw a ‘pyramid of numbers’ of a situation where a large population of insects feed upon

a very big tree. The insects in turn, are eaten by small birds which in turn are fed upon by

big birds.

(c) Differentiate giving reasons, between the pyramid of biomass of the above situation and

the pyramid of numbers that you have drawn.

Ans. (a) Pyramid of numbers showing interaction between trees; insects, birds and big birds.

(b) (i) Pyramid of number is spindle-shaped as the number of insects is maximum. The number

of trees and birds are less than the insects. The number is gradually decreasing at each

trophic level.

(iii) The pyramid of biomass in this ecosystem is erect because the biomass decreases at each

trophic level.



3. (a)

(i) Name the biogeochemical (nutrient) cycle shown above.

(ii) Name an activity of the living organisms not depicted in the cycle by which this nutrient

is returned to the atmosphere.

(b) How would the flow of nutrient in the cycle be affected due to large scale deforestation?

Explain giving reasons.

Ans.

(a) (i) It is carbon cycle.

(ii) Decomposition of organic wastes by microbes.

(b) Deforestation leads to increase in carbon dioxide levels in the air. Because the CO2 present is not

being utilized for photosynthesis in the absence of plants.

some additional questions for +2 board exams …drarvindsbiology.com/content/papers/8c_p25.pdf ·...

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