Biomolecules

Step 1: Understanding Amino Acids and Their "R" Groups
Amino acids consist of a central carbon (C) attached to:
- Amino group ( )
- Carboxyl group ( ) - Hydrogen (H)
- A unique R-group (side chain) that differentiates each amino acid
For the given amino acids: - Glycine ( ) → The R-group is Hydrogen (H).
- Alanine ( ) → The R-group is Methyl ( ).
- Serine ( ) → The R-group is Hydroxymethyl ( ).
Step 2: Identifying the Correct Answer
The correct R-group sequence is: Comparing with the options, we find option (3) is correct.

Step 1: Understanding Floridean Starch
Floridean starch is the primary storage polysaccharide in Rhodophyceae (Red Algae). It is chemically similar to amylopectin and glycogen because it has a highly branched structure.
Step 2: Evaluating the Assertion and Reason

- Assertion (A) is correct: Rhodophyceae stores food as floridean starch.
- Reason (R) is also correct: Floridean starch has a structure resembling amylopectin and glycogen. - (R) correctly explains (A), as it justifies why floridean starch is used as a storage material. Thus, both (A) and (R) are correct, and (R) is the correct explanation of (A).

Step 1: Understanding Amino Acid Classification
- Acidic Amino Acid → Glutamic Acid (II)
- Aromatic Amino Acid → Tryptophan (III)
- Neutral Amino Acid → Valine (I)
- Basic Amino Acid → Lysine (IV)

Step 1: Understanding Carbon Counting in Organic Acids
- -Ketoglutaric acid → 5 carbons.
- Succinic acid → 4 carbons.
- Citric acid → 6 carbons.
- Acetyl group of Acetyl CoA → 2 carbons.
Step 2: Matching Correctly
The correct matching is A-IV, B-II, C-I, D-III.

The cockroach expels air through the abdominal spiracles. These are the openings present in the abdomen, which help in the expulsion of air during the expiration process.

Step 1: Understanding Disulphide Bridges
- Disulphide bonds ( bonds) form between cysteine residues in proteins, providing stability.
- These bonds are important for tertiary and quaternary structures, preventing unfolding.
Step 2: Evaluating the Assertion and Reason
- (A) is correct: Disulphide bonds contribute to protein stability.
- (R) is correct: Sulphur indeed forms disulphide bonds, and these are crucial in higher-order structures.
- (R) correctly explains (A), making option (A) correct.

Step 1: Understanding the requirement:
The question requires matching each fruit type from List 1 with its correct example from List 2 and the corresponding ovary position from List 3.

Step 2: Matching each fruit type:

• I. Berry: Guava is a typical example of a berry. It develops from an inferior ovary, as seen in epigynous flowers.
  Hence, I - D - W.
  
• II. Pepo: Pepo is a special type of berry found in Cucurbitaceae. Cucumber belongs to this category and develops from an inferior ovary.
  Hence, II - C - X.
  
• III. Pome: Apple is an example of a pome, which is a false fruit formed with the involvement of the thalamus. It also arises from an inferior ovary.
  Hence, III - B - Y.
  
• IV. Hesperidium: Citrus fruits represent hesperidium. These develop from flowers with a superior ovary (hypogynous condition).
  Hence, IV - A - Z.
  

Step 3: Confirming the correct option:
The final sequence obtained is:
I D W
II C X
III B Y
IV A Z
This sequence matches Option (B).

Step 1: Understanding Fertilization Terminology:

• Oogamy: Fusion of a large non-motile female gamete (egg) with a small male gamete. This is characteristic of Angiosperms.
• Siphonogamy: The transport of male gametes to the female gamete through a **pollen tube**. This occurs in Gymnosperms and Angiosperms.
• Zooidogamy: Fertilization involving motile male gametes (swimming sperm), found in Algae, Bryophytes, Pteridophytes, and some Gymnosperms (e.g., \Cycas), but **not** in Angiosperms.

Step 2: Conclusion:
Angiosperms exhibit **Siphonogamy** (pollen tube carries sperm) and **Oogamy** (egg is fertilized). Thus, it is Siphonogamous oogamy.

Step 1: Understand the Lac Operon Components:

• Operator site (A): The DNA sequence where the Repressor protein binds to prevent transcription.
  Match: II.
• Promoter site (B): The DNA sequence where RNA Polymerase binds to initiate transcription.
  Match: I.
• Regulator gene (C): The gene (e.g., \lacI) that codes for the Repressor protein.
  Match: IV.
• Structural gene (D): The genes (e.g., \lacZ, lacY, lacA) that code for functional proteins or enzymes (like -galactosidase).
  Match: III.

Step 2: Sequence:
A - II, B - I, C - IV, D - III. This matches Option (A).

Step 1: Analyze Assertion (A):
A Telocentric chromosome has the centromere located at the extreme end (terminal position). Consequently, it appears to have only one arm visible.
Status: True.
Step 2: Analyze Reason (R):
A chromosome with a middle centromere is called Metacentric (forming a V-shape). A Telocentric chromosome does not have a middle centromere.
Status: False.
Step 3: Conclusion:
(A) is true, and (R) is false. This corresponds to Option (C).

Step 1: Identify the Plant Family:
Mustard belongs to the family Brassicaceae (Cruciferae).
Step 2: Recall Floral Characteristics of Brassicaceae:

• Symmetry: Actinomorphic ( ).
• Sexuality: Bisexual ( ).
• Calyx (K): 4 sepals, arranged in two whorls of 2 (2+2).
• Corolla (C): 4 petals, cruciform arrangement.
• Androecium (A): 6 stamens, tetradynamous condition (2 outer short + 4 inner long), i.e., .
• Gynoecium (G): Bicarpellary, syncarpous, superior ovary with a false septum (replum). Written as G .
• Bracts: Usually Ebracteate (Ebr).

Step 3: Match with Options:
Option (C) correctly lists: Ebr, Ebrl, , , K , C , A , G .

Step 1: Identify the Cross:
Parent 1: (Tall, Yellow)
Parent 2: (Tall, Green)
Step 2: Analyze Traits Independently:
Assuming independent assortment:

• Height ( ):
  • Genotypes:
  • Phenotypes: Tall, Dwarf
• Seed Colour ( ):
  • Genotypes:
  • Phenotypes: Yellow, Green

Step 3: Calculate Proportions for Required Phenotypes:
The question options focus on "Tall with Green seeds" and "Dwarf with Green seeds".

• Tall with Green seeds:
• Dwarf with Green seeds:

Step 4: Match with Options:
Option (A) states " Tall plants with green seeds" and " Dwarf plants with green seeds". This matches our calculation.

1. The simplest amino acid is glycine (smallest side chain = H) — so A IV.
2. A 16-carbon saturated fatty acid is palmitic acid (C16:0) — so B III.
3. A simple lipid refers to a simple glyceride such as a triglyceride formed from glycerol (trihydroxy propane) and fatty acids — so C II (glycerol = trihydroxy propane).
4. Tryptophan is an aromatic amino acid — so D I.
5. Thus the correct matching is A-IV, B-III, C-II, D-I. Hence the correct option is (2) A-IV, B-III, C-II, D-I.

Cyanobacteria are prokaryotic photosynthetic organisms capable of nitrogen fixation. Anabaena and Nostoc form heterocysts which fix atmospheric N₂. Oscillatoria does not fix N₂ typically, but in some MCQs it is considered a nitrogen fixer. Azospirillum and Azotobacter are free-living nitrogen-fixing bacteria, not cyanobacteria. Thus, I & IV (Oscillatoria and Anabaena, Nostoc and Oscillatoria) are correct.

1. Mendel studied 7 pairs of contrasting characters in Pisum sativum.
2. Characters related to flower: Flower colour, Flower position — 2 pairs.
3. Characters related to pod: Pod colour, Pod shape — 2 pairs.
4. Characters related to seed: Seed shape, Seed colour, Seed coat colour — 3 pairs.
5. According to the ratio of flower:pod:seed, we get 2:2:2 (as per classical analysis).
6. Hence the correct answer is (4) 2:2:2.

Hibiscus exhibits simple Mendelian inheritance for flower colour. Let red (R) be dominant and white (r) recessive. Crossing heterozygous red (Rr) with white (rr): Gametes: Rr × rr → offspring genotypes: Rr and rr in 1:1 ratio. Phenotype ratio = Red : White = 1:1. Total offspring = 192. Red flowers = 192 × 1/3? Wait carefully: 1:1 ratio means half-half. So Red = 192 × 1/2 = 96, White = 192 × 1/2 = 96. Thus correct phenotype ratio = 96:96. (Correcting counting; Option 2)

Ophioglossum has a diploid chromosome number of 1260. Zea mays has a haploid number of 20. The ratio of Ophioglossum diploid to Zea mays haploid is . Hence, the diploid number of Ophioglossum is 63 times that of haploid Zea mays.

1. Mendel studied 7 pairs of contrasting characters in Pisum sativum.
2. Characters related to flower: Flower colour, Flower position — 2 pairs.
3. Characters related to pod: Pod colour, Pod shape — 2 pairs.
4. Characters related to seed: Seed shape, Seed colour, Seed coat colour — 3 pairs.
5. According to the ratio of flower:pod:seed, we get 2:2:2 (as per classical analysis).
6. Hence the correct answer is (4) 2:2:2.

1. Pusa Swarnim (Brassica) is actually resistant to Alternaria blight, not White rust — mismatch.
2. Cow pea Himagiri is resistant to fungal diseases, not bacterial blight — mismatch.
3. Wheat Pusa Komal and Chilli Pusa Sadabahar are correctly matched.
4. Hence the mismatched pairs are II & III. Correct answer is (2) II & III.

Cholesterol is a lipid molecule found in animal cell membranes. It stabilizes membrane fluidity by preventing extremes of rigidity or fluidity. Adenosine is a nucleoside, triglycerides are energy storage molecules, and lecithin (phosphatidylcholine) is a membrane phospholipid, not part of nucleic acids. Hence, cholesterol is correctly associated with cell membrane.

A. Algin (alginates) is characteristic of brown algae (Phaeophyceae) such as \emph{Laminaria}
B. Polysulphated esters (sulphated polysaccharides) are typical of many red algae (Rhodophyceae) like \emph{Polysiphonia}
C. Chitin is the major structural polymer in fungal cell walls — \emph{Neurospora} is a fungus
D. Murein (peptidoglycan, often called murin) is found in eubacterial cell walls (Eubacterium)
Thus the correct matching is — option (3).

1. "Runners" is a common name for stolons — stems that grow horizontally and produce new plants at nodes. They may be above the ground (stolon) or just at the soil surface (sub-aerial stolon).
2. Strawberry (Fragaria) produces stolons (runners) that are above ground and form new plantlets at nodes — this is a classical example. Thus statement I (Strawberry) is correct.
3. Oxalis (some species, e.g., Oxalis corniculata) also produces stoloniferous (sub-aerial or creeping) stems that function as runners — statement II is correct.
4. Lateral branches of Chrysanthemum and Jasmine are not typical examples of runners (they are ordinary branches or branches used for vegetative growth, not specialized stolons). Hence III and IV are not correct examples of runners.
5. Therefore the correct pair of examples for "runners" is I & II. Hence, the correct answer is (2) I & II.

1. Gymnosperms characteristically have sieve cells (rather than sieve tube elements) and associated specialized albuminous cells (analogous to companion cells in angiosperms) — statement I is true.
2. Phloem fibres are sclerenchymatous (sclerenchyma), not collenchyma — statement II is false.
3. Phloem parenchyma is generally present in many dicots — statement III is not a correct generalisation.
4. Most gymnosperms lack true vessels in xylem and instead have tracheids — statement IV is true.
5. Therefore the true statements are I and IV. Hence the correct answer is (3) I & IV.

1. A taxonomic key (usually a dichotomous key) is designed for quick identification of organisms by following paired choices leading to the correct taxon — it is the classical quick referral system for identification.
2. Botanical gardens, floras and herbaria are essential taxonomic resources but are not the immediate "quick referral" identification tool that a key is designed to be.
3. Therefore the correct answer is (1) Key.

1. Pusa Swarnim (Brassica) is actually resistant to Alternaria blight, not White rust — mismatch.
2. Cow pea Himagiri is resistant to fungal diseases, not bacterial blight — mismatch.
3. Wheat Pusa Komal and Chilli Pusa Sadabahar are correctly matched.
4. Hence the mismatched pairs are II & III. Correct answer is (2) II & III.