Q1. The transverse section of a plant shows the following
anatomical features:
(a) Large number of scattered vascular bundles surrounded by bundle sheath.
(b) Large conspicuous parenchymatous ground tissue.
(c) Vascular bundles conjoint and closed.
(d) Phloem parenchyma absent.
Identify the category of
plant and its part:
(1) Dicotyledonous root
(2) Monocotyledonous stem
(3) Monocotyledonous root
(4) Dicotyledonous stem
Answer: (2) Monocotyledonous stem
Explanation:
1. Scattered
vascular bundles with bundle sheaths:
Monocot stems are characterized by vascular bundles scattered throughout the
ground tissue, each surrounded by a sclerenchymatous bundle sheath.
This contrasts with dicot stems, where vascular bundles are arranged in a ring.
2. Parenchymatous
ground tissue:
The presence of a large, prominent parenchyma-rich ground tissue (as seen in
plants like maize) is typical of monocot stems.
3. Conjoint
and closed vascular bundles:
Conjoint bundles (xylem and phloem in the same bundle) that are closed (lacking
cambium, meaning no secondary growth) are a hallmark of monocots.
4. Absence
of phloem parenchyma:
Monocot vascular bundles lack phloem parenchyma, which is replaced by
sclerenchyma in the bundle sheath for structural support.
Conclusion: These features collectively confirm
the plant as a monocotyledonous stem.
Q2. Q2. Which
of the following would help in prevention of diuresis?
1.
Decrease in secretion of renin by JG cells
2.
More water reabsorption due to under-secretion of ADH
3. Reabsorption
of Na⁺ and water from renal tubules due to aldosterone
4. Atrial
natriuretic factor causes vasoconstriction
The question asks which factor helps prevent diuresis (excessive
urine production). Here's the analysis:
Answer: (3) Reabsorption of Na+ and water from renal
tubules due to aldosterone
Explanation:
1. Aldosterone's
Role:
Aldosterone, a hormone produced by the adrenal glands, enhances sodium (Na⁺)
reabsorption in the distal tubules and collecting ducts of the kidneys. Increased Na⁺ reabsorption is followed by
passive water reabsorption, reducing urine output and preventing diuresis. This mechanism counteracts fluid loss and
maintains blood pressure.
2. Why
Other Options Are Incorrect:
o
(1) Decrease in renin secretion:
Reduced renin lowers angiotensin II and aldosterone levels, leading to less Na⁺/water reabsorption
and increased
diuresis.
o
(2) Under-secretion of ADH:
Antidiuretic hormone (ADH) promotes water reabsorption. Low ADH causes water
loss (e.g., diabetes insipidus), increasing diuresis.
o
(4) Atrial natriuretic factor (ANF):
ANF promotes
Na⁺/water excretion (natriuresis) and causes vasodilation, not
vasoconstriction. This increases diuresis.
Key Mechanism: Aldosterone directly opposes diuresis by retaining Na⁺ and water,
making option (3) the correct choice.
Q3. Which of the following statements is not correct?
1. Genetically
engineered insulin is produced in E. coli.
2. In
humans, insulin is synthesised as proinsulin.
3. Proinsulin
has an extra peptide called C-peptide.
4. Functional
insulin has A and B chains linked together by hydrogen bonds.
Answer: (4) Functional insulin has A and
B chains linked together by hydrogen bonds.
Explanation:
Correct Statements:
1. Genetically
engineered insulin in E. coli:
Recombinant human insulin is predominantly produced in E. coli (and
yeast) using engineered genes. For example, E.
coli expresses insulin precursors as inclusion bodies, which are later
refolded into functional insulin.
2. Proinsulin
synthesis in humans:
Human insulin is initially synthesized as proinsulin, a precursor molecule
containing the A and B chains connected by a C-peptide.
3. C-peptide
in proinsulin:
The C-peptide is a connecting peptide in proinsulin that is enzymatically
removed during maturation to yield active insulin.
Incorrect Statement:
4. A
and B chains linked by hydrogen bonds:
Functional insulin’s A and B chains are covalently linked by disulfide
bonds (not hydrogen bonds). These bonds are critical for insulin’s
structural integrity and biological activity. Hydrogen bonds are
weaker interactions that stabilize protein folding but do not directly link the
chains.
Conclusion: Statement (4) is incorrect because
disulfide bonds, not hydrogen bonds, connect insulin’s A and B chains.
Q4. Which of the following statements is not correct?
1. Genetically
engineered insulin is produced in E. coli.
2. In
humans, insulin is synthesized as proinsulin.
3. Proinsulin
has an extra peptide called C-peptide.
4. Functional
insulin has A and B chains linked together by hydrogen bonds.
Answer: (4) Functional insulin has A and
B chains linked together by hydrogen bonds.
Explanation:
1. Genetically
engineered insulin in E. coli:
Recombinant human insulin is predominantly produced in E. coli (and
yeast) using engineered genes. For example, E. coli expresses insulin
precursors as inclusion bodies, which are later refolded into functional
insulin.
2. Proinsulin
synthesis in humans:
Human insulin is initially synthesized as proinsulin, a precursor molecule
containing the A and B chains connected by a C-peptide.
3. C-peptide
in proinsulin:
The C-peptide is a connecting peptide in proinsulin that is enzymatically
removed during maturation to yield active insulin.
Incorrect Statement:
4. A
and B chains linked by hydrogen bonds:
Functional insulin’s A and B chains are covalently linked by disulfide
bonds (not hydrogen bonds). These bonds are critical for insulin’s
structural integrity and biological activity.
Conclusion: Statement (4) is incorrect because
disulfide bonds, not hydrogen bonds, connect insulin’s A and B chains.
Q5. Which of the following disapproved embryological support
for evolution?
1. Oparin
2. Karl
Ernst von Baer
3. Alfred
Wallace
4. Charles
Darwin
Answer: (2) Karl Ernst von Baer
Explanation:
Karl Ernst von Baer disapproved embryological support for
evolution by rejecting the idea that embryonic development recapitulates
evolutionary history (a concept known as "recapitulation theory"
proposed by Ernst Haeckel). Von Baer argued that embryos of different species
do not resemble each other at all stages of development but instead diverge as
they develop, which contradicted Haeckel's claims that embryology provides
direct evidence for evolution.
Q6. Goblet cells of the alimentary canal are
modified from:
1. Compound
epithelial cells
2. Squamous
epithelial cells
3. Columnar
epithelial cells
4. Chondrocytes
Answer: (3) Columnar epithelial cells
Explanation:
1. Goblet
Cells and Their Function:
Goblet cells are specialized epithelial cells found in the lining of the
respiratory and gastrointestinal tracts. They secrete mucus, which acts as a
protective and lubricating layer.
2. Origin
of Goblet Cells:
Goblet cells are derived from simple columnar epithelial cells.
These columnar cells undergo differentiation to form goblet cells, which are
characterized by their goblet-like shape due to the accumulation of mucus
granules in their apical region.
3. Why
Other Options Are Incorrect:
o
Compound epithelial cells:
These are multicellular structures, not individual cells like goblet cells.
o
Squamous epithelial cells:
These are flat, thin cells that do not differentiate into goblet cells.
o
Chondrocytes: These are
cartilage-producing cells and have no role in the formation of goblet cells.
Conclusion: Goblet cells originate from columnar
epithelial cells, making option (3) the correct answer.
Q7. The QRS complex in a standard ECG represents:
1. Repolarization
of ventricles
2. Repolarization
of auricles
3. Depolarization
of auricles
4. Depolarization
of ventricles
Answer: (4) Depolarization of ventricles
Explanation:
1. What
is the QRS Complex?
The QRS complex is the most prominent waveform in a standard ECG and represents
the depolarization of the ventricles, which initiates
ventricular contraction (systole). It follows the P wave (which represents
atrial depolarization) and precedes the T wave (which represents ventricular
repolarization).
2. Components
of the QRS Complex:
o
Q wave: A small downward
deflection representing depolarization of the interventricular septum.
o
R wave: A sharp upward
deflection representing depolarization of the main ventricular mass.
o
S wave: A downward deflection
following the R wave, representing depolarization of the basal parts of the
ventricles.
3. Why
Other Options Are Incorrect:
o
(1) Repolarization of ventricles:
This is represented by the T wave, not the QRS complex.
o
(2) Repolarization of auricles:
This occurs during the QRS complex but is masked by ventricular depolarization
and is not separately visible on an ECG.
o
(3) Depolarization of auricles:
This is represented by the P wave, not the QRS complex.
4. Clinical
Significance:
The duration and morphology of the QRS complex are crucial for diagnosing heart
conditions such as bundle branch blocks, ventricular hypertrophy, and
arrhythmias.
Conclusion: The QRS complex specifically
indicates ventricular depolarization, making option (4)
correct.
Q8. In light reactions, plastoquinone facilitates the transfer
of electrons from:
1. PS-I
to ATP synthase
2. PS-II
to cytb₆f complex
3. Cytb₆f
complex to PS-I
4.
PS-I to NADP+
Correct Answer: (2)
PS-II to Cyt b₆f complex.
In the light-dependent reactions of photosynthesis, plastoquinone
(PQ) plays a critical role in shuttling electrons from Photosystem II
(PSII) to the cytochrome b₆f complex. This lipid-soluble molecule transfers
electrons through the thylakoid membrane as part of the electron transport
chain.
Explanation:
1. Electron
Flow Pathway:
o
Light energy excites electrons in PSII, which
are then passed to plastoquinone.
o
Plastoquinone carries these electrons to the
cytochrome b₆f complex, where proton pumping occurs to establish the
electrochemical gradient used for ATP synthesis.
2. Why
Other Options Are Incorrect:
o
(1) PS-I to ATP synthase: ATP
synthase is involved in ATP production via chemiosmosis, not direct electron
transfer from PSI.
o
(3) Cyt b₆f
complex to PS-I: Electrons move from cytochrome b₆f to PSI
via plastocyanin, not plastoquinone.
o
(4) PS-I to NADP+: PSI
transfers electrons to ferredoxin, which then reduces NADP+ to NADPH;
plastoquinone is not involved here.
Correct Answer: (2) PS-II to Cyt b₆f
complex.
Q9. The product(s) of the reaction catalyzed by nitrogenase
in root nodules of leguminous plants is/are:
1. Ammonia
and hydrogen
2. Ammonia
alone
3. Nitrate
alone
4. Ammonia
and oxygen
Correct Answer: (1) Ammonia and hydrogen
Explanation:
Nitrogenase, the enzyme responsible for nitrogen fixation in root
nodules, catalyzes the reduction of atmospheric nitrogen gas (N₂) into ammonia
(NH₃). However, this reaction also produces hydrogen gas (H₂)
as a byproduct. Here’s why:
1. Key
Reaction Mechanism:
o
Nitrogenase reduces N₂ to NH₃ through a process
requiring ATP and electrons.
o
A side reaction reduces protons (H⁺) to H₂, even
when N₂ is available. This occurs because nitrogenase cannot fully suppress
proton reduction during N₂ fixation.
2. Evidence
from Search Results:
o
Root nodules convert atmospheric N₂ into
ammonia, which is assimilated into amino acids and nucleotides.
o
Hydrogen gas production is inherent to
nitrogenase activity, as observed in studies on legume nodules and other
nitrogen-fixing systems.
o
Oxygen is not a product;
instead, it is strictly regulated (via leghaemoglobin) to protect oxygen-sensitive
nitrogenase.
3. Why
Other Options Are Incorrect:
o
(2) Ammonia alone: Excludes H₂,
a proven byproduct.
o
(3) Nitrate alone: Nitrate is
not produced by nitrogenase; it is generated via nitrification in soil.
o
(4) Ammonia and oxygen: Oxygen
inhibits nitrogenase and is actively excluded from nodules.
Q10. Match the following stages of meiosis with their
corresponding events:
List I List II
(a) Zygotene
(i) Terminalization
(b) Pachytene
(ii) Chiasmata
(c) Diplotene (iii)
Crossing over
(d) Diakinesis (iv) Synapsis
Options:
1. (a)-(ii),
(b)-(iv), (c)-(iii), (d)-(i)
2. (a)-(iii),
(b)-(iv), (c)-(i), (d)-(ii)
3. (a)-(iv),
(b)-(iii), (c)-(ii), (d)-(i)
4. (a)-(i),
(b)-(ii),
(c)-(iv), (d)-(iii)
Correct Answer: (3) (a)-(iv), (b)-(iii),
(c)-(ii), (d)-(i)
Explanation:
Matching Stages and Events:
1. Zygotene
(a) → Synapsis (iv)
o
During zygotene, homologous chromosomes pair up
via synapsis, facilitated by the synaptonemal complex.
2. Pachytene
(b) → Crossing over (iii)
o
Crossing over (exchange of
genetic material between non-sister chromatids) occurs at recombination nodules
during pachytene.
3. Diplotene
(c) → Chiasmata (ii)
o
In diplotene, homologous chromosomes begin to
separate but remain connected at chiasmata (sites of crossing
over).
4. Diakinesis
(d) → Terminalization (i)
o
Diakinesis marks the terminalization of
chiasmata (movement of chiasmata toward chromosome ends) and breakdown
of the nuclear envelope.
Why Other Options Are Incorrect:
·
Option (1): Incorrectly assigns
chiasmata to zygotene and synapsis to pachytene.
·
Option (2): Links crossing over
to zygotene and terminalization to diplotene, which contradicts the sequence of
events.
·
Option (4): Associates
terminalization with zygotene and synapsis with diplotene, reversing key
processes.
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