Genetics

Step 1: Understanding the Concept:
The question asks to identify the scientist who discovered the process of transformation in bacteria. Transformation is the genetic alteration of a cell resulting from the direct uptake and incorporation of exogenous genetic material from its surroundings through the cell membrane.
Step 2: Detailed Explanation:


F. Griffith (1928): Conducted experiments with two strains of {Streptococcus pneumoniae}, a virulent (S) strain and a non-virulent (R) strain. He observed that when heat-killed S-strain bacteria were mixed with live R-strain bacteria and injected into mice, the mice died. He isolated live S-strain bacteria from the dead mice. He concluded that some "transforming principle" from the heat-killed S-strain had been taken up by the live R-strain, transforming them into the virulent S-strain. This was the discovery of bacterial transformation.

Lederberg and Tatum (1946): Discovered bacterial conjugation, a process of genetic transfer between bacterial cells through direct contact or a bridge-like connection.

Avery, McCarty, and Macleod (1944): Expanded on Griffith's work to identify the "transforming principle." Through a series of experiments using enzymes to destroy different types of molecules (proteins, RNA, DNA), they proved that DNA was the genetic material responsible for transformation.

Therefore, while Avery and McCarty identified the molecule responsible for transformation, it was F. Griffith who first discovered the phenomenon itself.
Step 3: Final Answer:
Based on the historical context of these discoveries, Frederick Griffith is credited with the discovery of bacterial transformation.

Step 1: Understanding the Concept:
The question asks for the correct experimental procedure for a Mendelian dihybrid cross, which is designed to study the inheritance of two different traits simultaneously and demonstrate the Law of Independent Assortment.
Step 2: Detailed Explanation:
Let's outline the logical steps of the experiment:

A. Selection of two traits in the parent generation: The experiment begins with the scientist choosing two distinct, contrasting traits to study (e.g., seed shape and seed color in peas).
B. Crossing of pure - breeding parents: The next step is to perform the parental (P generation) cross. This involves crossing two parents that are pure-breeding (homozygous) for the selected traits, one dominant for both and the other recessive for both (e.g., RRYY x rryy).
D. Observations of gamete combinations in the F₁ generation: The offspring of the P cross are the F₁ generation. They are all heterozygous for both traits (e.g., RrYy). Mendel then self-crossed these F₁ individuals. The key to independent assortment is observing that the F₁ generation produces four different types of gametes (RY, Ry, rY, ry) in equal proportions. Step D, "Observations of gamete combinations in the F₁ generation," refers to this step of generating and understanding the gametes from the F₁ that will create the F₂ generation.
C. Analysis of phenotypic ratios in the F₂ generation: The offspring of the F₁ self-cross are the F₂ generation. The final step is to count the offspring and analyze their phenotypic ratios. If the genes assort independently, this ratio will be approximately 9:3:3:1. This analysis is the ultimate test of the law.
Step 3: Final Answer:
The correct experimental sequence is: Select traits (A) Cross pure parents (B) Observe F₁ and its gametes for the next cross (D) Analyze F₂ results (C). This corresponds to the order A, B, D, C.

Step 1: Understanding the Concept:
The question asks about the role of DNA methylation, a key epigenetic modification, in regulating gene expression. Specifically, it focuses on the effect of hypermethylation (an excess of methylation) in the promoter region of a gene.
Step 2: Detailed Explanation:


DNA methylation is the addition of a methyl group (-CH₃) to the DNA molecule, typically at CpG sites (a cytosine nucleotide followed by a guanine nucleotide).
The promoter is a region of DNA where transcription is initiated. It's the binding site for RNA polymerase and transcription factors.
Hypermethylation of a gene's promoter region is a well-established mechanism for gene silencing. It works in two main ways:

The methyl groups can physically block the binding of transcription factors and RNA polymerase to the promoter, thus preventing transcription from starting.
Methylated DNA can be recognized and bound by specific proteins called methyl-CpG-binding domain proteins (MBDs). These proteins, in turn, recruit histone deacetylases (HDACs) and other chromatin-remodeling proteins. This leads to the compaction of chromatin into a condensed, inactive state (heterochromatin), making the gene inaccessible for transcription.

Therefore, hypermethylation of the promoter is strongly associated with the repression or silencing of transcription.
Step 3: Final Answer:
The typical effect of hypermethylation of promoter regions is the silencing of transcription.

Step 1: Understanding the Concept:
This question tests the knowledge of modified Mendelian dihybrid ratios that result from different types of gene interactions, specifically epistasis. Epistasis occurs when the effect of one gene is masked or modified by another gene at a different locus. The standard dihybrid ratio is 9:3:3:1, which gets modified in epistasis.
Step 2: Detailed Explanation:
Let's match each type of gene interaction with its characteristic phenotypic ratio:

A. Duplicate dominant epistasis: The dominant allele at either of the two loci (A_B_, A_bb, or aaB_) produces the same dominant phenotype. Only the double recessive genotype (aabb) produces the recessive phenotype. The ratio becomes (9+3+3) : 1, which is 15:1. Thus, A matches with II.

B. Duplicate recessive epistasis (Complementary genes): The dominant alleles at both loci are required to produce the dominant phenotype (A_B_). All other genotypes (A_bb, aaB_, aabb) produce the recessive phenotype. The ratio becomes 9 : (3+3+1), which is 9:7. Thus, B matches with I.

C. Recessive epistasis: A recessive genotype at one locus (e.g., aa) masks the expression of alleles at another locus (B/b). The genotypes A_B_ and A_bb produce two different phenotypes, while aaB_ and aabb produce a third, single phenotype. The ratio becomes 9 : 3 : (3+1), which is 9:3:4. Thus, C matches with III.

D. Dominant \& Recessive Epistasis (or Dominant Inhibitory Epistasis): The question uses a slightly ambiguous term, but the ratio 13:3 is characteristic of a specific interaction where a dominant allele at one locus (A) and the recessive genotype at that same locus (aa) combined with the recessive genotype at the second locus (bb) produce the same phenotype. Essentially, the dominant allele at the second locus (B) is required for the alternative phenotype, but only when the first locus is not dominant (aaB_). A more common name is Dominant Inhibitory Epistasis where a dominant allele at one locus suppresses the expression of the other locus. This leads to a (9+3+1) : 3 or 13:3 ratio. Thus, D matches with IV.

Step 3: Final Answer:
The correct set of matches is A-II, B-I, C-III, D-IV, which corresponds to option (B).

Step 1: Understanding the Concept:
The question requires matching different classes of chemical mutagens with their correct examples. A mutagen is a physical or chemical agent that changes the genetic material, usually DNA, of an organism.
Step 2: Detailed Explanation:
Let's analyze and match each class of mutagen in List-I:

A. Alkylating agent: These agents donate an alkyl group (e.g., -CH₃, -CH₂CH₃) to amino or keto groups in nucleotides. This alters base-pairing properties. Mustard gas is a classic example of an alkylating agent. Thus, A matches with III.

B. Base analog: These are molecules that are structurally similar to the normal purine and pyrimidine bases in DNA. They can be incorporated into DNA during replication and cause mispairing. 5-Bromouracil is an analog of thymine. Thus, B matches with IV.

C. Intercalating agent: These are flat, planar molecules that can insert themselves (intercalate) between the stacked base pairs of DNA. This insertion distorts the double helix and can lead to frameshift mutations during DNA replication. Acridine Orange is a well-known intercalating agent. Thus, C matches with I.

D. Deamination agent: These agents work by removing an amino group (-NH₂) from a nucleotide base. Nitrous acid (HNO₂) deaminates cytosine to uracil and adenine to hypoxanthine, causing incorrect base pairing. Thus, D matches with II.

Step 3: Final Answer:
The correct set of matches is A-III, B-IV, C-I, D-II, which corresponds to option (D).

Step 1: Understanding the Concept:
This question deals with the ploidy levels of different cells in an angiosperm during its life cycle. The key is to know the process of double fertilization which forms the zygote and the endosperm.
Step 2: Key Formula or Approach:


The egg cell in an angiosperm is haploid (n).
The endosperm is formed by the fusion of one male gamete (n) with the central cell, which contains two polar nuclei (n + n).
Therefore, the endosperm is triploid (3n).
Step 3: Detailed Explanation:


We are given that the number of chromosomes in the egg cell is 8. Since the egg cell is haploid, we have:

The endosperm is triploid, meaning its chromosome number is 3n.

We can now calculate the number of chromosomes in the endosperm:

Step 4: Final Answer:
The number of chromosomes in the endosperm would be 24.

Step 1: Understanding the Concept:
This question asks for the specific name of a chromatin structural unit defined by its DNA length and the complement of histone proteins, including the linker histone H1.
Step 2: Detailed Explanation:
Let's define the different levels of chromatin structure:

Histone Octamer: This is the protein core of the nucleosome. It consists of eight histone molecules: two copies each of the core histones H2A, H2B, H3, and H4. It does not include DNA or histone H1.
Nucleosome (or Core Particle): This is the fundamental repeating unit of chromatin. It consists of the histone octamer with approximately 147 base pairs (bp) of DNA wrapped around it. It does not include the linker histone H1. Mild digestion of chromatin with micrococcal nuclease (MNase) yields this structure.
Chromatosome: This is the next level of organization. It consists of the nucleosome core particle {plus} the linker histone H1, which binds to the DNA where it enters and exits the octamer. The addition of H1 protects a longer stretch of DNA from nuclease digestion. The chromatosome comprises the histone octamer, the linker histone H1 (total of 9 histone molecules), and about 168 bp of DNA. This description perfectly matches the question.
Histosome: This term is not standard in the context of chromatin structure.
Step 3: Final Answer:
The structure containing 168 bp of DNA, the histone octamer, and the linker histone H1 is called a chromatosome.

Concept: The Five Kingdom Classification system is a widely accepted system that classifies organisms based on cell structure, body organization, mode of nutrition, and reproduction.
Step 1: Identifying the scientist.}
R.H. Whittaker proposed the Five Kingdom Classification in 1969.
Step 2: Understanding the kingdoms.}
The five kingdoms are Monera, Protista, Fungi, Plantae, and Animalia.
Step 3: Eliminating other options.}

• Linnaeus – Proposed the two-kingdom system.
• Darwin – Known for the theory of evolution.
• Mendel – Father of genetics.

Concept: Velamen is a specialized multilayered tissue found in certain plant roots that helps in absorption and retention of moisture.
Step 1: Understanding velamen.}
Velamen is a spongy tissue present in aerial roots that absorbs water from the atmosphere.
Step 2: Identifying plant type.}
This adaptation is characteristic of epiphytic plants such as orchids, which grow on other plants and rely on atmospheric moisture.
Step 3: Eliminating other options.}

• Xerophytes – Adapted to dry conditions but lack velamen.
• Hydrophytes – Aquatic plants.
• Halophytes – Salt-tolerant plants.