1.Carbon Fixation: CO₂ is fixed to
ribulose-1,5-bisphosphate (RuBP) by the enzyme RuBisCO. This step does not
require ATP or NADPH directly.
2.Reduction: The intermediate
3-phosphoglycerate (3-PGA) is converted into glyceraldehyde-3-phosphate (G3P).
This step requires 2 NADPH and 2 ATP per CO₂
molecule.
3. Regeneration of RuBP: RuBP is
regenerated from G3P, which requires 1 ATP per CO₂ molecule.
This
energy input ensures the Calvin cycle can continuously fix carbon dioxide and
produce carbohydrates].
1.
A. Nucleolus - III. Site of active ribosomal
RNA synthesis
2.
B. Centriole - II. Organization like the
cartwheel
3.
C. Leucoplasts - IV. For storing nutrients
4.
D. Golgi apparatus - I. Site of formation of
glycolipid
Answer- The correct answer is (3) C
Reasoning:
· In a seed embryo, the embryonic root is called
the radicle. When the seed germinates, the radicle is the first
part to emerge and develops into the primary root.
· In the diagram, "C" appears to point
to the lowermost part of the embryo, which is typically where the radicle is
located.
· Therefore, (3) C is the most
likely answer.
129.
Spindle fibers attach to kinetochores of chromosomes during
(1)
Metaphase (2) Anaphase (3) Telophase (4) Prophase
Answer- The
correct answer is: (1) Metaphase
Explanation:
Here's
the breakdown:
1.
Kinetochores: These are protein
structures on chromosomes where spindle fibers attach.
2.
Spindle fibers: These are microtubules
that help separate chromosomes during cell division.
During
the different phases of mitosis:
· Prophase: Chromosomes condense,
and the spindle apparatus begins to form, but spindle fibers do not yet attach
to kinetochores.
· Metaphase: This is the stage
when the spindle fibers attach to the kinetochores of the chromosomes. The
chromosomes then align along the metaphase plate.
· Anaphase: The sister chromatids
separate and move to opposite poles of the cell, guided by the spindle fibers.
· Telophase: The chromosomes
arrive at the poles, and the nuclear envelope reforms.
Thus,
spindle fibers attach to the kinetochores of chromosomes during Metaphase.
130.
Given below are two statements:
Statement
I: Chromosomes become gradually visible under light microscope during leptotene
stage
Statement
II: The beginning of diplotene stage is recognized by dissolution of
synaptonemal complex
In the
light of the above statements, choose the correct answer from the options given
below:
(1) Both Statement
I and Statement II are false
(2)
Statement I is true but Statement II id false
(3)
Statement I is false but Statement II is true
(4) Both
Statement I and Statement II are true
Answer- The
correct answer is (4) Both Statement I and Statement II are true
Here's a more detailed explanation:
·
MMeiosis:
This is a type of cell division that
reduces the number of chromosomes in a cell, resulting in the production of
gametes (sperm and egg cells).
·
PProphase
I:
The first phase of meiosis I, which is
further divided into several substages, including leptotene.
·
LLeptotene:
During this stage, the chromosomes, which
were previously in a diffuse, thread-like form (chromatin), start to condense
and become visible under a light microscope.
·
CChromatin
Condensation:
As the cell prepares for division, the
chromatin within the nucleus coils and condenses, becoming increasingly compact
and visible as distinct chromosomes.
·
SSister
Chromatids:
Each chromosome at this
stage consists of two identical sister chromatids joined at the
centromere.
131.
Given below are two statements:
Statement
I: Parenchyma is living but collenchymas is dead tissue
Statement
II: Gymnosperms lack xylem vessels but presence of xylem vessels is the characteristic
of angiosperms.
In the
light of the above statements, choose the correct answer from the options given
below”
(1) Both
Statement I and Statement II are false
(2)
Statement I is true and Statement II is false
(3)
Statement I is false and Statement II is true
(4) Both
Statement I and Statement II are true
Answer- The
correct answer is: (3) Statement I is false and Statement II is true
Justification:
· Statement I: Parenchyma is living but
collenchymas is dead tissue.
o This
statement is false. Parenchyma cells are indeed living and
have thin cell walls. However, collenchyma cells are also living cells,
although they have unevenly thickened cell walls that provide support to young
plant parts.
· Statement II: Gymnosperms lack xylem
vessels but the presence of xylem vessels is the characteristic of angiosperms.
o This
statement is true. Gymnosperms typically lack xylem vessels;
their water transport is mainly through tracheids. Angiosperms, on the other
hand, have xylem vessels, which are more efficient in water transport. This is
a distinguishing characteristic between these two groups of plants.
Thus,
Statement I is incorrect, and Statement II is correct, making option (3) the
right choice.
132.
These are regarded as major causes of biodiversity loss:
A. Over
exploitation, B.
Co-extinction, C.
Mutation,
D.
Habitat loss and fragmentation, E.
Migration
Choose
the correct option:
(1) A, B,
C and D only, (2) A, B
and E only
(3) A, B
and D only, (4) A, C
and D only
Answer- The
correct answer is:(3) A, B, and D only
Justification:
Let's
analyze each option:
· A. Overexploitation: This is a
major cause of biodiversity loss. Overuse of resources can lead to the
depletion or extinction of species.
· B. Co-extinction: This is a
significant factor in biodiversity loss. When one species goes extinct, other
species that are dependent on it (e.g., parasites, mutualists) may also face
extinction.
· C. Mutation: While mutation is
a fundamental process in evolution, it is not a primary cause of
biodiversity loss in the current ecological context. Mutations occur naturally
and are a source of genetic variation, but they don't directly drive
large-scale species extinctions.
· D. Habitat loss and fragmentation:
This is a leading cause of biodiversity loss. Destruction or division of
habitats reduces the space and resources available for species to survive,
often leading to population declines and extinctions.
· E. Migration: Migration, a
natural process of species relocation, is not a primary cause of biodiversity
loss itself. It can be affected by habitat loss and climate change, but it is
not a direct driver of species extinction.
Therefore,
A, B, and D are the most significant and directly linked causes of biodiversity
loss among the options provided.
The correct answer is (3) A, B, and D only.
133. The
lactose present in the growth medium of bacteria is transported to the cell by
the action of:
(1)
Acetylas, (2)
Permease, (3)
Polymerase, (4)
Beta-galactosidase
Answer- The
correct answer is: (2) Permease
Justification:
1.
Permease: This is a specific membrane
protein that facilitates the transport of lactose across the bacterial cell
membrane. In the lac operon system, lactose permease is essential for
bringing lactose into the cell, where it can then be metabolized.
2.
Other options:
o Acetylase:
This is an enzyme involved in acetylation reactions and not directly involved
in lactose transport.
o Polymerase:
Polymerases are enzymes involved in DNA or RNA synthesis and are not
responsible for transporting lactose.
o Beta-galactosidase:
This enzyme is responsible for breaking down lactose into glucose and galactose
inside the cell, but it doesn't transport lactose into the
cell.
Thus,
permease is the correct enzyme responsible for transporting lactose into the bacterial
cell.
134.
Bulliform cells are responsible for
(1)
Protecting the plant from salt stress
(2)
Increase photosynthesis in monocots
(3)
Providing large spaces for storage of sugars
(4)
Inward curling of leaves in monocots.
Answer- The
correct answer is: (4) Inward curling of leaves in monocots.
Justification:
· Bulliform cells are large,
empty, bubble-shaped epidermal cells that occur in groups on the upper surface
of leaves, particularly in monocots. They are involved in the mechanism that
causes leaves to curl inward to minimize water loss during drought stress.
· When water is scarce, bulliform cells lose
turgidity (water pressure), causing the leaf to fold or curl inward, reducing
the surface area exposed to the environment and thus decreasing transpiration.
Here's
why the other options are incorrect:
(1) Protecting the plant from salt
stress: While some plants have adaptations to deal with salt stress,
bulliform cells are not directly involved in this process. Halophytes, for
example, have other mechanisms for salt tolerance.
(2) Increase photosynthesis in monocots:
Bulliform cells do not directly increase photosynthesis. Photosynthesis
primarily occurs in mesophyll cells.
(3) Providing large spaces for storage
of sugars: While some plant cells are specialized for storage, this is
not the primary function of bulliform cells.
Therefore,
the correct answer is that bulliform cells are responsible for the inward
curling of leaves in monocots during water stress.
135.
The cofactor of the enzyme carboxylpeptidase is:
(1)
Niacin (2) Flavin (3) Haem (4) Zinc
Answer- The
correct answer is (4) Zinc
Explanation
Carboxypeptidase
is a proteolytic enzyme that requires zinc as an inorganic
cofactor for its activity.
Cofactors are non-protein molecules essential for enzyme function, categorized
as:
· Inorganic ions (e.g., zinc,
magnesium)
· Organic coenzymes (e.g., NAD,
FAD derived from vitamins).
Zinc acts
by binding to the enzyme's active site, stabilizing its structure, and
facilitating substrate binding or catalysis12.
The other options are organic coenzymes:
· Niacin (option 1) is a
precursor for NAD, involved in redox reactions.
· Flavin (option 2) forms
FMN/FAD, also used in redox processes.
· Haem (option 3) is an
iron-containing prosthetic group in proteins like hemoglobin.
Since
carboxypeptidase relies on a metal ion cofactor rather than an
organic coenzyme, zinc is the correct answer.
Botany:
Section – B (Q. No. 136 to 150)
136.
Read the following statements and choose the set of correct statements:
In the
members of Phaeophyceae,
A.
Asexual reproduction occurs usually by biflagellate zoospores.
B. Sexual
reproduction is by oogamous method only.
C. Stored
food is in the form of carbohydrates which is either mannitol or laminarian
D. The
majorpigments found are chlorophyll a, c and carotenoids and xanthophylls.
E.
Vegetative cells have a cellulosic wall, usually covered on the outside by
gelatinous coating of algin
Choose
the correct answer from the options given below:
(1) B, C,
D and E only
(2) A, C,
D and E only
(3) A, B,
C and E only
(4) A, B,
C and D only
Answer- The
correct answer is (2) A, C, D and E only
Read
carefully following explanation to enhance your understanding.
Phaeophyceae,
or brown algae, exhibit specific characteristics related to their reproduction,
pigments, and cell structure. Let's analyze each statement to determine the
correct set:
A.
Asexual reproduction occurs usually by biflagellate zoospores.
Correct. Phaeophyceae reproduce asexually via biflagellate zoospores, which
have one anterior tinsel and one posterior whiplash flagellum.
B.
Sexual reproduction is by oogamous method only.
Incorrect. Sexual reproduction in brown algae can be isogamous, anisogamous, or
oogamous. The term "only" makes
this statement false.
C.
Stored food is in the form of carbohydrates, either mannitol or laminarin.
Correct. Laminarin (a β-1,3-glucan) and mannitol are the primary storage
carbohydrates.
D.
Major pigments include chlorophyll a,
c, carotenoids, and
xanthophylls.
Correct. Pigments are chlorophyll a,
c, fucoxanthin (a xanthophyll), and other carotenoids. While xanthophylls are a subset of
carotenoids, the statement is accurate.
E.
Vegetative cells have a cellulosic wall with an outer gelatinous algin coating.
Correct. Cell walls contain cellulose and are coated with algin, a gelatinous
polysaccharide.
Correct
Answer: Option (2) A, C, D, and E only
Justification:
·
B is excluded because sexual
reproduction in Phaeophyceae is not exclusively oogamous.
·
A, C, D, and E align with the
characteristics confirmed by multiple sources.
137.
Match List I with List II
List
I
List II
A. Robert
May I.
Species-Area relationship
B.
Alexander von Humboldt II.
Long term ecosystem experiment using out door plots
C. Paul
Ehrlich III. Global
species diversity at about 7 million
D. David
Tilman IV.
River popper hypothesis
Choose
the correct answer from the options given below:
(1)
A-III, B-I, C-IV, D-II
(2) A-I,
B-III, C-II, D-IV
(3)
A-III, B-IV, C-II, D-I
(4) A-II,
B-III, C-I, D-IV
Answer- The correct answer is (1) A-III, B-I, C-IV, D-II
Read
the following and find the answer
To match
the names in List I with their corresponding contributions in List II, let us
analyze each individual:
Analysis:
1. Robert May: Known for his work on the
relationship between species diversity and ecosystem stability. He also
contributed to the understanding of species-area relationships, which describe
how species richness increases with area siz
o Match:
I. Species-Area Relationship
2. Alexander von Humboldt: He is credited
with pioneering the concept of the species-area relationship, where larger
areas support more species due to greater habitat diversity
o Match:
I. Species-Area Relationship
3. Paul Ehrlich: Proposed the Rivet
Popper Hypothesis, which compares species in an ecosystem to rivets in an
airplane wing, emphasizing the importance of each species in maintaining
ecosystem stability
o Match:
IV. Rivet Popper Hypothesis
4. David Tilman: Conducted long-term
ecosystem experiments using outdoor plots to study biodiversity and its impact
on ecosystem productivity and stability
o Match:
II. Long-term ecosystem experiment using outdoor plots
Justification:
·
Robert May (A) is associated with global species
diversity estimates (~7 million).
·
Alexander von Humboldt (B) is linked to the
species-area relationship.
·
Paul Ehrlich (C) developed the Rivet Popper
hypothesis.
·
David Tilman (D) is known for long-term
biodiversity experiments using outdoor plots.
138.
Given below are two statements:
Statement I: In C3 plants, some O2
binds to RuBisCO, hence CO2 fixation is decreased.
Statement II: In C4 plants, mesophyll cells show
very little photorespiration while bundle sheath cells do not show
photorespiration.
In the light of the above statements, choose the correct
answer from the options given below:
(1) Both Statement I and Statement II are false
(2) Statement I is true but Statement II are false
(3) Statement I is false but Statement II is true
(4) Both Statement I and Statement II are true
Answer- The correct answer is (4) Both Statement I and Statement II are true
Logical analysis-
Statement I: In C3 plants, some O2
binds to RuBisCO, hence CO2 fixation is decreased.
· Explanation: In C3
plants, RuBisCO (Ribulose-1,5-Bisphosphate Carboxylase/Oxygenase) is the enzyme
responsible for fixing CO2 into organic compounds during the Calvin
cycle. However, RuBisCO can also bind to O2, leading to photorespiration.
When O2 binds to RuBisCO, it initiates a process that results in the
loss of CO2 and energy, thereby decreasing the efficiency of CO2
fixation. Thus, Statement I is true.
Statement II: In C4 plants, mesophyll cells
show very little photorespiration while bundle sheath cells do not show
photorespiration.
· Explanation: C4
plants have evolved a mechanism to reduce photorespiration by spatially
separating CO2 fixation and the Calvin cycle. In C4
plants, CO2 is first fixed into a four-carbon molecule in the
mesophyll cells, which then moves to the bundle sheath cells where the Calvin
cycle occurs. This process concentrates CO2 in the bundle sheath
cells, reducing the likelihood of O2 binding to RuBisCO and thus
minimizing photorespiration in these cells. Mesophyll cells, however, do not
perform the Calvin cycle and therefore do not exhibit photorespiration related
to RuBisCO activity in the same way as C3 plants. Thus, Statement II
is true.
Justification:
· Statement I is true because in
C3 plants, O2 binding to RuBisCO leads to
photorespiration, which decreases CO2 fixation efficiency.
· Statement II is true because C4
plants minimize photorespiration by concentrating CO2 in bundle
sheath cells, where the Calvin cycle occurs, and mesophyll cells do not show
significant photorespiration due to their role in initial CO2
fixation.
139. The
DNA present in chloroplast is:
(1)
Circular, double stranded
(2)
Linear, single stranded
(3)
Circular, single stranded
(4)
Linear, double stranded
Answer- The correct answer is (1) Circular, double stranded
Read
these facts carefully and build subject concept along with the answer
Chloroplast DNA Characteristics:
·
Structure: Chloroplast DNA is
typically circular in structure, similar to bacterial DNA.
This circular structure is a common feature among many organelle DNAs.
·
Strand Type: Chloroplast DNA is
double-stranded, meaning it consists of two complementary
strands of nucleotides.
Justification:
Option (1) Circular, double-stranded:
This is the correct description of chloroplast DNA. It is circular, similar to
bacterial DNA, and double-stranded, allowing for replication and transcription
processes.
Option (2) Linear, single-stranded:
Incorrect. Chloroplast DNA is not linear like some viral genomes, nor is it
single-stranded.
Option (3) Circular, single-stranded:
Incorrect. While chloroplast DNA is circular, it is not single-stranded.
Option (4) Linear, double-stranded:
Incorrect. Chloroplast DNA is not linear.
Justification:
Chloroplast
DNA is circular and double-stranded, reflecting its evolutionary origin from
cyanobacteria, which also have circular DNA. This structure supports the
replication and transcription processes necessary for chloroplast function.
140.
In an ecosystem if the Net Primary Productivity (NPP) of first trophic level is
Answer- The correct answer is :(1) x(kcal m-2)/y-1
Read
the following explanation and learn the facts behind it
Definitions:
·Gross Primary Productivity (GPP):
The total amount of organic matter produced by photosynthesis in an ecosystem,
including the energy used by the plants themselves for respiration.
·Net Primary Productivity (NPP):
The amount of organic matter left after the plants have used some of the GPP
for their own respiration. It is essentially GPP minus the energy used by
plants for respiration.
Relationship Between GPP and NPP:
The
relationship between GPP and NPP can be expressed as:
NPP=GPP−R
where R is the respiration of the plants. Typically,
NPP is about half of GPP, but this ratio can vary.
Trophic Levels and Energy Transfer:
Energy is
transferred from one trophic level to the next with a significant loss at each
step, typically following the "10% rule," where about 10% of the
energy from one trophic level is transferred to the next.
Given Information:
· The NPP of the first trophic level (primary
producers) is 100x (kcal m−2) yr−1
Calculating GPP of the First Trophic Level:
If we
assume that NPP is roughly half of GPP (though this can vary), then:
GPP=2×NPP=2×100x=200x (kcal m−2) yr−1
Energy Transfer to Higher Trophic Levels:
Applying
the 10% rule:
·From First to Second Trophic Level:
10%×200x=20x (kcal m−2) yr−1 (GPP of the second trophic level, assuming
it's all used by the next level).
·From Second to Third Trophic Level:
10%×20x=2x (kcal m−2) yr−1 (GPP
of the third trophic level).
However,
the question asks for the GPP of the third trophic level, which would be the
energy available for that level to produce biomass, not necessarily what is
transferred to the next level. Since the GPP of a trophic level includes the
energy used by organisms in that level for their own metabolic processes, the
GPP of the third trophic level would be slightly higher than the energy
transferred to it, but in this simplified model, we focus on the energy transferred.
Justification:
· The energy transferred from the second to the
third trophic level is 2x, but considering the simplified nature of the question and typical assumptions, the closest match is
x (kcal m-2)/yr-1, acknowledging
that the exact calculation might slightly differ based on specific ecosystem
efficiencies and assumptions about energy transfer. However, given the options
provided and the typical understanding of energy transfer in ecosystems, the
closest match is x.
Note:
The exact calculation provided suggests 2x for the
energy transferred to the third trophic level, but none of the options directly
match this. The closest interpretation under typical assumptions and given the
options is x, acknowledging that the question might be
simplifying or assuming different efficiencies in energy transfer.
141.
Which of the following are fused in somatic hybridization involving two
varieties of plants?
(1)
Somatic embryos, (2)
Protoplasts
(3)
Pollens, (4)
Callus
Answer- The Correct Answer is (2) Protoplasts
Explanation:
Somatic
hybridization involves the fusion of protoplasts, which are
plant cells that have had their cell walls removed. This process allows for the
combination of genetic material from two different plant varieties, species, or
genera to create a hybrid plant with desirable traits.
Key Points:
1. Protoplasts
in Somatic Hybridization:
o Protoplasts are isolated by enzymatic digestion
of the cell wall using enzymes like cellulase and pectinase.
o These protoplasts can then be fused using
chemical agents (like polyethylene glycol, PEG) or physical methods (like
electrofusion).
2. Fusion
Process:
o The fusion of protoplasts results in a
heterokaryon (a cell containing nuclei from both parent cells).
o The fused protoplast is then cultured to
regenerate a new cell wall, divide, and develop into a callus, which can be
grown into a hybrid plant.
3. Why
Not Other Options?:
o (1) Somatic embryos: These are
formed later during tissue culture but are not involved in the fusion process.
o (3) Pollens: Pollens are
involved in sexual reproduction, not somatic hybridization.
o (4) Callus: Callus is an
undifferentiated mass of cells formed after protoplast fusion but is not
directly involved in the fusion process.
The correct answer is (2) Protoplasts, as they
are the cellular components fused during somatic hybridization to create
somatic hybrids.
142.
Match List I with List II
List
I
List II
A. Citric
acid cycle
I. Cytoplasm
B.
Glycolysis II.
Mitochondrial matrix
C.
Electron Transport system
III. Intermembrane space of mitochondria
D. Proton
gradient
IV. Inner mitochondrial membrane
Choose
the correct answer from the options given below:
(1) A-II,
B-I, C-IV, D-III
(2)
A-III, B-IV, C-I, D-II
(3) A-IV,
B-III, C-II, D-I
(4) A-I,
B-II, C-III, D-V
Answer- The correct answer is (1) A-II, B-I, C-IV, D-III
Explanation:
Let us analyze the locations of the cellular processes listed in List
I and match them with their corresponding locations in List II:
1. A.
Citric Acid Cycle (Krebs Cycle)
o
The citric acid cycle occurs in the mitochondrial
matrix, where acetyl-CoA is oxidized to produce NADH, FADH2,
and ATP/GTP.
o
Match: II.
Mitochondrial matrix.
2. B.
Glycolysis
o
Glycolysis is the first step of cellular
respiration and occurs in the cytoplasm, where glucose is
broken down into pyruvate.
o
Match: I. Cytoplasm.
3. C.
Electron Transport System (ETS)
o
The ETS is located in the inner
mitochondrial membrane, where electrons are transferred through
complexes to generate a proton gradient.
o
Match: IV. Inner
mitochondrial membrane.
4. D.
Proton Gradient
o
The proton gradient is established in the intermembrane
space of mitochondria, as protons are pumped from the matrix into this
space during electron transport.
o
Match: III.
Intermembrane space of mitochondria.
The correct matches are:
·
A-II: Citric Acid Cycle → Mitochondrial Matrix
·
B-I: Glycolysis → Cytoplasm
·
C-IV: Electron Transport System → Inner
Mitochondrial Membrane
·
D-III: Proton Gradient → Intermembrane Space
143.
Match List I with List II
List
I
List II
A.
Frederick Griffith I. Genetic
code
B.
Francisco Jacob & Jacque Monod
II. Semi-conservative mode of DNA replication
C. Har
Gobind Khorana
III. Transformation
D.
Meselson & Stahl IV. Lac operon
Choose
the correct answer from the options given below:
(1)
A-III, B-IV, C-I, D-II
(2) A-II,
B-III, C-IV, D-I
(3) A-IV,
B-I, C-II, D-III
(4)
A-III, B-II, C-I, D-IV
Answer- The correct answer is (1) A-III, B-IV, C-I, D-II
Explanation:
Let us analyze each scientist and their corresponding discovery
or contribution:
1. Frederick
Griffith (A)
o
Griffith's experiment in 1928 demonstrated the
phenomenon of bacterial transformation, where genetic material
from dead virulent bacteria transformed non-virulent bacteria into virulent
ones. This is referred to as the "transforming principle".
o
Match: III.
Transformation
2. Francisco
Jacob & Jacques Monod (B)
o
Jacob and Monod proposed the Lac
operon model, which explains the regulation of gene expression in
prokaryotes through an operon system.
o
Match: IV. Lac
operon
3. Har
Gobind Khorana (C)
o
Khorana played a key role in deciphering the genetic
code, determining how nucleotide sequences specify amino acids during
protein synthesis.
o
Match: I. Genetic code
4. Meselson
& Stahl (D)
o
Meselson and Stahl conducted the famous
experiment that demonstrated the semi-conservative mode of DNA
replication, where each daughter DNA molecule consists of one parental
strand and one newly synthesized strand.
o
Match: II.
Semi-conservative mode of DNA replication
Final Matches:
·
A-III: Frederick Griffith → Transformation
·
B-IV: Francisco Jacob & Jacques Monod → Lac
operon
·
C-I: Har Gobind Khorana → Genetic code
·
D-II: Meselson & Stahl → Semi-conservative
mode of DNA replication
144.
Match List I with List II
List
I
List II
(Type of
Stamens) (Example)
A.
Monadelphous I.
Citrus
B.
Diadelphous II. Pea
C.
Polydelphous
III. Lily
D.
Epiphyllous
IV. China-rose
Choose
the correct answer from the options given below:
(1) A-IV,
B-I, C-II, D-III
(2) A-I,
B-II, C-IV, D-III
(3)
A-III, B-I, C-IV, D-II
(4) A-IV,
B-II, C-I, D-III
Answer- The correct answer is (4) A-IV, B-II, C-I, D-III
Explanation:
Let us match the types of stamens in List I with
their corresponding examples in List II:
1. A.
Monadelphous
o
In this type, all the filaments are united into
a single bundle, while the anthers remain free.
o
Example: China-rose (Hibiscus)
belongs to the Malvaceae family and exhibits monadelphous stamens.
o
Match: IV. China-rose
2. B.
Diadelphous
o
In this type, the filaments are united into two
bundles, while the anthers remain free.
o
Example: Pea (Pisum sativum)
exhibits diadelphous stamens, where 9 stamens form one bundle and 1 remains
separate.
o
Match: II. Pea
3. C.
Polydelphous
o
In this type, the filaments are united into more
than two bundles, while the anthers remain free.
o
Example: Citrus exhibits
polydelphous stamens, where the filaments form multiple groups.
o
Match: I. Citrus
4. D.
Epiphyllous
o
In this type, the stamens are attached to the
tepals (undifferentiated perianth).
o
Example: Lily exhibits
epiphyllous stamens attached to its tepals.
o
Match: III. Lily
Final Matches:
·
A-IV: Monadelphous → China-rose
·
B-II: Diadelphous → Pea
·
C-I: Polydelphous → Citrus
·
D-III: Epiphyllous → Lily
145.
Identify the correct description about the given figure:
(1) Water
pollinated flowers showing stamens with mucilaginous covering.
(2)
Cleistogamous flowers showing autogamy.
(3)
Compact inflorescence showing complete autpgamy
(4) Wind
pollination plant inflorescence showing flowers with well exposed stamens.
Answer- The correct answer is (4) Wind pollination plant
inflorescence showing flowers with well-exposed stamens.
Explanation:
The question refers to identifying the correct description of a
plant structure based on the figure. Let us analyze each option in light of the
characteristics of pollination mechanisms:
1. Option
(1): Water-pollinated flowers showing stamens with mucilaginous covering
o
Water-pollinated flowers (hydrophily) typically
have pollen grains protected by a mucilaginous covering to prevent wetting.
However, this is specific to aquatic plants like Vallisneria and Hydrilla,
and there is no mention of such adaptations in the question.
o
Incorrect.
2. Option
(2): Cleistogamous flowers showing autogamy
o
Cleistogamous flowers are closed flowers that do
not open and enforce self-pollination (autogamy). These flowers lack exposed
stamens as they do not rely on external pollination agents.
o
Incorrect.
3. Option
(3): Compact inflorescence showing complete autogamy
o
Compact inflorescences may promote
self-pollination, but this does not align with the description of exposed
stamens, which are typically associated with wind pollination rather than
autogamy.
o
Incorrect.
4. Option
(4): Wind pollination plant inflorescence showing flowers with well-exposed
stamens
o
Wind-pollinated plants (anemophily) have
adaptations such as:
§ Well-exposed
stamens for easy pollen dispersal by wind.
§ Large
feathery stigmas to trap airborne pollen.
§ Light,
non-sticky pollen grains for efficient wind transport.
o
Examples include grasses, maize, and other
plants with compact inflorescences adapted for wind pollination. This matches
the description given in the question.
o
Correct.
The correct answer is (4) Wind pollination plant
inflorescence showing flowers with well-exposed stamens, as this
aligns with the adaptations seen in anemophilous plants like grasses and maize.
146.
Match List I with List II
List
I List II
A.
GLUT-4 I. Hormone
B.
Insulin II. Enzyme
C.
Trypsin III.
Intercellular ground substance
D.
Collagen IV.
Enables glucose transport into cells
Choose
the correct answer from the options given below:
(1) A-I,
B-II, C-III, D-IV
(2) A-II,
B-III, C-IV, D-I
(3)
A-III, B-IV, C-I, D-II
(4) A-IV,
B-I, C-II, D-III
Answer-The correct answer is (4) A-IV, B-I, C-II, D-III
Explanation:
Let us analyze each item in List I and match it
with its corresponding description in List II:
1. A.
GLUT-4
o
GLUT-4 is a glucose transporter protein that
facilitates the uptake of glucose into cells, particularly in adipose tissue
and striated muscle. It is regulated by insulin and enables glucose transport
into cells.
o
Match: IV. Enables
glucose transport into cells
2. B.
Insulin
o
Insulin is a hormone secreted by the pancreas
that regulates blood glucose levels by promoting the uptake of glucose into
cells through mechanisms like GLUT-4 activation.
o
Match: I. Hormone
3. C.
Trypsin
o
Trypsin is an enzyme involved in protein
digestion. It breaks down proteins into smaller peptides in the small
intestine.
o
Match: II. Enzyme
4. D.
Collagen
o
Collagen is a structural protein found in
connective tissues and serves as an intercellular ground substance, providing
strength and support to tissues like skin, bones, and cartilage.
o
Match: III.
Intercellular ground substance
Final Matches:
·
A-IV: GLUT-4 → Enables glucose transport into
cells
·
B-I: Insulin → Hormone
·
C-II: Trypsin → Enzyme
·
D-III: Collagen → Intercellular ground substance
147.
Identify the step in tricarboxylic acid cycle, which does not involve oxidation
of substrate.
to
(1) Succinic acid Malic acid
to(2) Succinyl – Co-A Succinic acid
to(3) Isocitrate α-ketoglutaric acid
to(4) Malic acid Oxaloacetic acid
Answer-The correct answer is (2) Succinyl-CoA → Succinic acid
Explanation:
The Tricarboxylic Acid (TCA) cycle, also known as the citric acid
cycle or Krebs cycle, involves multiple steps where substrates are oxidized to
generate energy in the form of NADH, FADH₂, and ATP. However, not all steps
involve oxidation. Let us analyze each option:
1. Succinic
acid → Malic acid
o
This step involves the oxidation of succinate to
fumarate and then to malate. FAD is reduced to FADH₂ during this process.
o
Involves oxidation, so this
option is incorrect.
2. Succinyl-CoA
→ Succinic acid
o
In this step, Succinyl-CoA is converted to
succinate by replacing CoA with a phosphate group, which is then transferred to
ADP (or GDP) to form ATP (or GTP). This is a substrate-level phosphorylation
step and does not involve oxidation.
o
Does not involve oxidation, so
this is the correct answer.
3. Isocitrate
→ α-ketoglutaric acid
o
This step involves the oxidation of isocitrate
to α-ketoglutarate, with NAD⁺ being reduced to NADH and CO₂ being released.
o
Involves oxidation, so this
option is incorrect.
4. Malic
acid → Oxaloacetic acid
o
This step involves the oxidation of malate to
oxaloacetate, with NAD⁺ being reduced to NADH.
o
Involves oxidation, so this
option is incorrect.
The correct answer is (2) Succinyl-CoA → Succinic acid,
as it is the only step in the TCA cycle that does not involve oxidation but
instead involves substrate-level phosphorylation.
148.
Spraying sugarcane crop with which of the following plant growth regulators,
increases the length of stem, thus, increasing the yield?
(1)
Gibberellin
(2)
Cytokinin
(3)
Abscisic acid
(4) Auxin
Answer-The correct answer is (1) Gibberellin
Explanation:
Gibberellins (GAs) are plant growth regulators known for their
role in promoting stem elongation and internodal growth. Spraying sugarcane
crops with gibberellins increases the length of the stem, which directly
enhances the yield by increasing the biomass and sucrose accumulation in the
elongated internodes.
Key Points:
1. Role
of Gibberellins:
o
Gibberellins regulate stem elongation by
promoting both cell division and cell elongation.
o
They reduce the activity of DELLA nuclear
repressor proteins, enabling the expression of genes involved in growth and
development.
o
Studies have shown that applying gibberellins
(GA3) increases internode elongation in sugarcane, leading to higher yields.
2. Other
Plant Growth Regulators:
o
Cytokinin: Primarily involved
in cell division and delaying senescence but does not significantly influence
stem elongation.
o
Abscisic Acid (ABA): Known for
its role in stress responses and inhibiting growth, not promoting stem
elongation.
o
Auxin: While it plays a role in
cell elongation, its effect on internode elongation is secondary compared to
gibberellins.
3. Scientific
Evidence:
o
Research indicates that spraying gibberellins
increases endogenous GA levels, which stimulate internode elongation and
improve sugarcane yield.
Spraying sugarcane crops with gibberellins
effectively increases stem length, enhancing yield. Thus, the correct answer is
(1) Gibberellin.
149.
Match List I with List II
List
I List II
A.
Rose I. Twisted aestivation
B.
Pea II. Perigynous
flower
C.
Cotton III. Drupe
D.
Mango IV. Marginal
placentation
Choose
the correct answer from the options given below:
(1) A-I,
B-II, C-III, D-IV
(2) A-IV,
B-III, C-II, D-I
(3) A-II,
B-III, C-IV, D-I
(4) A-II,
B-IV, C-I, D-III
Answer-The correct answer is (4) A-II, B-IV, C-I, D-III
Explanation:
Let us match the items in List I with their
corresponding descriptions in List II based on their botanical
characteristics:
1. A.
Rose → Perigynous flower (II)
o
Roses are perigynous flowers,
where the ovary is partially embedded in the floral cup (hypanthium), and other
floral parts (sepals, petals, and stamens) are attached at the same level
around the ovary.
o
Match: II. Perigynous
flower
2. B.
Pea → Marginal placentation (IV)
o
In peas, the ovules are attached along the
margin of a single carpel, a characteristic of marginal placentation.
o
Match: IV. Marginal
placentation
3. C.
Cotton → Twisted aestivation (I)
o
Cotton exhibits twisted aestivation,
where one petal overlaps the next petal in a twisted manner in the bud stage.
o
Match: I. Twisted
aestivation
4. D.
Mango → Drupe (III)
o
Mango is a classic example of a drupe,
a type of fleshy fruit with a single seed enclosed in a hard endocarp (stone).
o
Match: III. Drupe
Final Matches:
·
A-II: Rose → Perigynous flower
·
B-IV: Pea → Marginal placentation
·
C-I: Cotton → Twisted aestivation
·
D-III: Mango → Drupe
150.
Which of the following statement is correct regarding the process of
replication in E. coli?
(1) The
DNA dependent RNA polymerase catalyses polymerization in one direction, that is
5’
3’.
(2) The DNA dependent DNA polymerase
catalyses polymerization in 5’ to 3’ as
well as 3’ to 5’ direction
(3) The DNA dependent DNA polymerase
catalyses polymerization in 5’
3’direction.
(4) The DNA dependent DNA polymerase
catalyses polymerization in one direction that is 3’ to 5’
Answer-The correct answer is (3) The DNA-dependent DNA
polymerase catalyzes polymerization in the 5' → 3' direction.
Explanation:
DNA replication in E. coli involves several enzymes,
with DNA-dependent DNA polymerase playing a central role. Let us analyze the
given options based on the process of replication:
1. DNA
Polymerization Direction:
o
DNA polymerases, including the main enzyme in E.
coli (DNA polymerase III), can only add nucleotides to the 3' end of a
growing DNA strand.
o
This means that DNA synthesis always occurs in
the 5' → 3' direction, as new nucleotides are added to the
free hydroxyl group (-OH) at the 3' end of the growing strand.
2. Template
Reading Direction:
o
To synthesize DNA in the 5' → 3' direction, the
template strand is read in the 3' → 5' direction.
3. Analysis
of Options:
o
Option (1): Incorrect. RNA
polymerase is not involved in DNA replication; it is involved in transcription.
o
Option (2): Incorrect.
DNA-dependent DNA polymerase does not catalyze polymerization in both
directions; it works only in the 5' → 3' direction.
o
Option (3): Correct. The
DNA-dependent DNA polymerase catalyzes polymerization exclusively in the 5'
→ 3' direction, which aligns with how nucleotide chains are elongated
during replication.
o
Option (4): Incorrect.
Polymerization does not occur in the 3' → 5' direction.
4. Supporting
Evidence:
o
During replication, one strand (the leading
strand) is synthesized continuously in the 5' → 3' direction toward the
replication fork.
o
The other strand (the lagging strand) is
synthesized discontinuously as Okazaki fragments, also in the 5' → 3'
direction, but away from the fork.
The correct answer is (3) The DNA-dependent DNA
polymerase catalyzes polymerization in the 5' → 3' direction, as this
is a fundamental property of all known DNA polymerases involved in replication.
