Biology class XII Question anwser

12/18/2011 CBSE

Biology Class-Xll
2000
(CBSE)

 


Q 1. How does industrial melanism bring out the action of natural selection? (3 marks)

Ans1.
The peppered moth, Biston betularia has two strains :

(i) a dull grey form, and

(ii) a black form: carbonaria

The occurrence of Carbonaria was rare before the Industrial Revolution because
the dark form was more conspicuous on the tree trunks and was killed easily by
the birds. However, a couple of centuries of Industrial Revolution offered
better survivability to the dark form on the tree trunks

blackened with industrial soot due to its reduced conspicuousness. The grey form
now stood out more conspicuously and was selectively predated by the birds. The
frequency of dark form among moths increased manifolds. This phenomenon of
'Industrial Melanism' proves that the frequency of occurrence of a trait is
dependent on its survivability and suitability against the factors of natural
selection.


Q 2. How does intestinal juice
contribute in the digestion of proteins? What provides alkaline pH in the small
intestine? (3 marks)

Ans2.
Intestinal juice contains the protease ‘Enteropeptidase’. Some of the
inactive trypsinogen molecules of the pancreatic juice are first hydrolysed by
enteropeptidase into an inactive peptide and active trypsin. Trypsin then
hydrolyses the remaining trypsinogen molecules into trypsin. It also activates
other pancreatic proteases, viz. chymotrypsin and carboxypeptidases. Trypsin
hydrolyses basic proteins to peptides. Chymotrypsin hydrolyses casein and
paracasein to peptides. Carboxypeptidases hydrolyse the terminal peptide bond of
the peptide chain to release individual amino acids.

The bicarbonates of pancreatic and intestinal juices and bile from liver provide
alkaline pH.


Q 3. An mRNA strand has a series
of codons out of which three are mentioned below.

(i) AUG, (ii) UUU and (iii) UAG

(a) What will these codons be translated into?

(b) What are the DNA codons that would have transcribed these RNA codons? (3
marks)

Ans3.


(a) (i) AUG and (ii) UUU will be translated. (iii) UAG is a stop codon which
will not be translated.

(b)





















RNA codon


DNA codon


AUG


TAC


UUU


AAA


UAG


ATC



 


Q 4. List any four symptoms
shown by a Down's syndrome afflicted child. Explain the cause of this disorder.
(3 marks)

Ans4.
A child suffering with Down’s syndrome will have the following
symptoms :

(i) A prominent forehead

(ii) A flattened nasal bridge

(iii) Habitually open mouth

(iv) Projecting lower lips

Down’s syndrome is caused by trisomy 21, i.e. an extra chromosome 21. Trisomy
arises by non-disjunction during egg cell formation.



Q 5. Answer the following with reference to the anatomy of dicot stem :



(i) Where exactly are the cambial cells located in the vascular bundles?

(ii) What is the name given to such a bundle?

(iii) How are the xylem vessels arranged?

(iv) What type of cells constitute the pith? (3 marks)        
      

Ans.5.

(i) The cambial cells are located in the vascular bundles of the young
dicotyledonous stem, between xylem and phloem.

(ii) It is called Fascicular Cambium.

(iii) Xylem vessels are arranged towards the center, protoxylem innermost and
metaxylem next.

(iv) Parenchymatous cells constitute the pith.



Q 6. Explain any three chemical barriers that offer non-specific defence
mechanism. (3 marks)

Ans6.
The non-specific chemical defence to the body is provided by :

(i) Lysozyme: Lysozyme is an enzyme, which destroys cell wall of many bacteria,
and thus renders them non-pathogenic. It is found on skin, tears, and saliva.

(ii) Oil and sweat: Glands secrete oils and waxes, which make the surface
acidic. This discourages many microorganisms from establishing themselves on the
skin.

(iii) Acidic gastric juice: Some bacteria who survive the salivary secretion are
killed by the highly acidic gastric juice in the stomach.



Q 7. Trace the development of microsporocyte in the anther of a mature
pollen-grain. (3 marks)

Ans7.
The four microsporangia of the anthers have sporogenous cells into
them. The sporogenous cells are characterised by their large size, abundant
cytoplasm, and prominent nuclei.

These sporogenous cells or the microsporocytes undergo several mitotic divisions
to increase in number and then function as Microspore Mother Cells (MMC). The
MMCs are diploid cells. Each MMC divides meiotically to give a tetrad of four
haploid microspores. The individual microspores get separated, enlarge and
undergo a mitotic division to produce a large Vegetative cell (tube cell) and a
small Generative cell. This entire structure representing the male gametophyte
is called Pollen Grain. Normally pollen grain is shed at this stage.

In some plants, the generative cell undergoes mitosis to produce two male
gametes before the pollen grains are shed.


Q 9. Name the hormone that
regulates each of the following and mention the source of it:

(i) Urinary elimination of water.



(ii) Storage of glucose as glycogen.



(iii) Sodium and potassium ion metabolism.



(iv) Basal metabolic rate.



(v) Descent of testes into the scrotum 5 marks.





Ans9.


































 

Hormone


Source


(i)


Vasopressin (ADH)


Posterior pituitary


(ii)


Insulin


Islets of Langerhans in the
Pancreas


(iii)


Aldosterone (Mineralocarticoids)


Adrenal cortex


(iv)


Thyroid hormones


Thyroid gland


(v)


Follicle Stimulating
Hormone (FSH)


Anterior Pituitary



Q 10. Describe the C4
pathway in a paddy plant. How is this pathway an adaptive advantage to the
plant? (5 marks)

Ans.
Phosphoenolpyruvate (PEP) forms a 4-carbon compound oxaloaceticacid (OAA)
by accepting CO2 in outer mesophyll cells. OAA is converted to malic
acid, which is transported to the cells surrounding the vascular bundle namely,
the Bundle sheath cells. Malic acid is converted to

pyruvic acid in bundle sheath cells and CO2 is released in the
cytoplasm. Thus the bundle sheath cells maintain a high concentration of CO2.

The pyruvic acid generated in the bundle sheath cells is transferred back to the
mesophyll and is converted to PEP by the expenditure of an ATP molecule. This
conversion generates AMP, which requires two additional ATPs to regenerate an
ATP. The total additional requirement of ATPs per

six CO2 molecules is thus 12 ATPs in comparison to C3
pathway.

However, the plants having C4 pathway save a lot of glucose from
being wasted by photorespiration.

They can maintain a very high CO2 concentration in bundle sheath
cells where RuBPcarboxylase / oxygenase exists. It helps to avoid RUBISCO to go
on the oxygenase mode and oxidise glucose even at high temperatures of tropics.
Thus C4 pathway is an adaptive advantage.

 

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