Genetics

Step 1: Understanding RNA as Genetic Material
Frankel Conrat demonstrated that RNA, rather than protein, carries genetic information in viruses. He conducted experiments on Tobacco Mosaic Virus (TMV) by separating its protein and RNA components and reconstituting the virus, proving that RNA was the genetic material.
Step 2: Identifying the Correct Answer
- Option (1) Boveri (Incorrect – Known for chromosomal inheritance).
- Option (2) Hugo de Vries (Incorrect – Proposed mutation theory).
- Option (3) Frankel Conrat (Correct – Discovered RNA as genetic material).
- Option (4) Khorana (Incorrect – Known for genetic code deciphering).

Given that the diploid chromosome number of A is 4. Thus, the haploid number of A will be .
The haploid number of B is twice that of A, so the haploid number of B will be . Therefore, the diploid number of B is .
The diploid number of C is thrice that of A, so the diploid number of C is .
Thus, we have the following:

Step 1: Understanding Dihybrid Cross and Phenotypes

- Dihybrid cross follows the 9:3:3:1 ratio for phenotypic expression.
- The number of genotypes determines dominant or recessive traits in the F generation.
- The correct genotype matches are I and IV.

To arrange the ratios, we need to consider the phenotypic ratios for different types of crosses. Step 1: Incomplete dominance For incomplete dominance, the typical phenotypic ratio is as seen in the ratio . This happens when the heterozygote phenotype is distinct from both homozygous phenotypes. Step 2: Monohybrid cross In a standard monohybrid cross with complete dominance, the phenotypic ratio is , which corresponds to . Step 3: Dihybrid cross In a dihybrid cross, where two genes are involved, the phenotypic ratio is , which corresponds to . Thus, the correct order is:

Step 1: Understanding Linkage and Recombination

- Morgan’s experiments with Drosophila demonstrated genetic linkage between the white-eye and miniature-wing genes.
- The expected recombination percentage was 37.2%, based on observed crossover frequencies.

Step 1: Understanding the given statements:
- A) ABO blood groupings: The ABO blood groupings are inherited based on co-dominance, where both alleles are expressed. So, this matches with IV.
- B) Starch synthesis in pea seeds: This is an example of pleiotropy, where one gene affects multiple traits. Hence, it matches with III.
- C) Point mutation: A point mutation can cause a genetic disorder like Sickle cell anemia, which is a result of a point mutation in the hemoglobin gene. This matches with II.
- D) Deletion of DNA base pairs: A deletion of DNA base pairs leads to frame-shift mutations, which result in a change in the reading frame of the gene. This matches with I.
Thus, the correct matching is:

Step 1: Understanding Restriction Enzymes:
Restriction endonucleases are molecular scissors that cut DNA at specific recognition sites.
Step 2: Mechanism of Action:
DNA strands are made of nucleotides linked by a sugar-phosphate backbone. The bond connecting the sugar of one nucleotide to the phosphate of the next is the phosphodiester bond. Restriction enzymes catalyze the hydrolysis of this specific bond to break the DNA backbone. (Note: While sticky ends involve breaking hydrogen bonds between bases effectively separating the strands, the enzymatic \cutting action is on the phosphodiester backbone).
Step 3: Evaluating Options:
(A) Hydrogen bonds: Connect base pairs; broken by helicase or heat.
(B) Glycosidic bonds: Connect base to sugar.
(C) Phosphodiester bonds: Connect nucleotides in the strand; cut by endonucleases.
(D) Peptide bonds: Connect amino acids in proteins.
Step 4: Final Answer:
The correct option is (C).

Step 1: Analyze Missing Data:

• A (Cell Wall of Green Algae): Typically composed of an inner layer of **Cellulose** and an outer layer of **Pectose/Pectin**.
• B (Pigments of Brown Algae): Chlorophyll a, c, and **Fucoxanthin** (gives brown colour).
• C (Stored Food of Brown Algae): Complex carbohydrates like **Laminarin** or **Mannitol**.
• D (Pigments of Red Algae): Chlorophyll a, d, and **Phycoerythrin** (gives red colour).

Step 2: Conclusion:
Option (A) accurately lists all these characteristics.

Step 1: Match the Phases to Events:

• A. Diakinesis: The final stage where the chiasmata move to the ends of chromosomes. This is called **Terminalisation** (II).
• B. Pachytene: Crossing over occurs here, characterized by the appearance of **Recombination nodules** (IV).
• C. Zygotene: Homologous chromosomes pair up (synapsis), facilitated by the **Synaptonemal complex formation** (I).
• D. Leptotene: Chromosomes start condensing and appear as **thin threads** (III).

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

Classify the given amino acids by side-chain properties:
Glutamic acid (Glu) is an acidic amino acid (negatively charged at physiological pH).
Lysine (Lys) is a basic amino acid (positively charged at physiological pH).
Valine (Val) is a neutral, non-polar (hydrophobic) amino acid.
The requested order "Acidic → Basic → Neutral" corresponds to Glutamic acid → Lysine → Valine, which is option (1).

Primers for DNA replication and transcription are short nucleic acid sequences (oligonucleotides) that provide a free 3′-OH group for DNA polymerase or RNA polymerase to extend. Polypeptides or sugars cannot act as primers.

1. Papaya Roundup ready → Herbicide tolerant (Z).
2. Tomato → Fungal pathogen resistant (B), pathogen X → Pseudomonas.
3. Potato → Bacterial pathogen resistant (C), pathogen Y → Phytophthora.
4. Soyabean → Papaya ring spot resistance (D), pathogen W → Virus.
5. Correct matching is (1) I-A-Z, II-B-Y, III-C-X, IV-D-W.

1. Recombination frequency ∝ distance between genes.
2. Arrange genes to satisfy given distances: a-d = 29%, d-b = 9%, b-c = 15%, a-c = 5%, c-d = 24%, a-b = 20%.
3. The linear sequence satisfying all distances is a, d, b, c.
4. Correct answer is (1) a, d, b, c.

1. Unisexual flowers borne on different plants (dioecious condition) mean an individual plant bears only male or only female flowers.
2. Autogamy (self-pollination within the same flower) is impossible because there is no bisexual flower on a plant.
3. Geitonogamy (transfer of pollen between different flowers of the same plant) is also prevented because all flowers on a plant are of the same sex — a female plant has no pollen to donate and a male plant has no ovules to receive pollen.
4. Xenogamy (cross-pollination between different plants) is still possible and, in fact, required for fertilization in dioecious species.
5. Hence, unisexual flowers on different plants prevent both autogamy and geitonogamy. Therefore the correct answer is (1) Autogamy and Geitonogamy.

Sickle cell anemia arises due to a single amino acid substitution in the β-globin chain of hemoglobin: Glutamic acid (hydrophilic) is replaced by Valine (hydrophobic) at position 6. This leads to abnormal hemoglobin (HbS) and sickling of RBCs.

1. Papaya Roundup ready → Herbicide tolerant (Z).
2. Tomato → Fungal pathogen resistant (B), pathogen X → Pseudomonas.
3. Potato → Bacterial pathogen resistant (C), pathogen Y → Phytophthora.
4. Soyabean → Papaya ring spot resistance (D), pathogen W → Virus.
5. Correct matching is (1) I-A-Z, II-B-Y, III-C-X, IV-D-W.

1. Water potential , where = osmotic potential, = pressure potential.
2. In hypertonic solution, cell loses water → turgor pressure drops → .
3. Thus, MPa.
4. Hence the water potential is (2) -0.5 MPa.

1. Recombination frequency ∝ distance between genes.
2. Arrange genes to satisfy given distances: a-d = 29%, d-b = 9%, b-c = 15%, a-c = 5%, c-d = 24%, a-b = 20%.
3. The linear sequence satisfying all distances is a, d, b, c.
4. Correct answer is (1) a, d, b, c.

Water is absorbed by root hairs from the soil via osmosis due to water potential difference. Casparian strips in the endodermis are suberized to block apoplastic movement and force water to move symplastically. While both statements are true, Casparian strips do not directly cause water uptake by root hairs; they regulate water movement inside the root. Hence, option 2 is correct.