Genetics

Step 1: Understanding the statements.
- Statement i) is true; alleles of a gene are typically located on different loci on homologous chromosomes.
- Statement ii) is false; an allele does not individually control a trait, but the interaction of alleles controls traits.
- Statement iii) is false; an allelic pair typically controls a single trait, not two traits.
- Statement iv) is true; an allelic pair controls a single trait, with one allele from each parent.
Step 2: Conclusion.
The correct answer is (B) Two, as only two statements are true.

Step 1: Understanding the genetics.
The question refers to the inheritance of two genes controlling wheat kernel colour. The F2 generation typically results from a dihybrid cross, where the genes segregate independently, leading to four possible combinations of alleles. The question asks about the parental combinations, which are those combinations that involve no recombination. In a typical dihybrid cross, the parental combinations occur in 2 out of the 16 total offspring.
Step 2: Analyzing the options.
(A) Three: This is incorrect because parental combinations are less frequent in this case, occurring only in two of the offspring.
(B) Four: Incorrect. Although four combinations are possible, only two represent the parental combinations.
(C) One: Incorrect. The parental combinations occur more frequently than just one.
(D) Two: Correct. Parental combinations of the two genes appear in two offspring in a typical F2 generation.
Step 3: Conclusion.
The correct answer is (D) Two, as only two offspring typically have the parental gene combinations in a dihybrid cross.

Step 1: Understanding human ovum.
The human ovum is the largest cell in the human body and is visible to the naked eye under suitable conditions.

Step 2: Size of ovum.
The diameter of an unfertilized human ovum is approximately 100 micrometres.

Step 3: Conclusion.
Thus, the correct diameter of the human female gamete is 100 m.

Step 1: Understanding exalbuminous seeds.
Exalbuminous seeds are those in which the endosperm is completely consumed during seed development and food is stored in cotyledons.

Step 2: Analyzing the given plants.
Pea, bean, gram, and groundnut store food in cotyledons and lack endosperm at maturity.

Step 3: Eliminating incorrect sets.
Castor and coconut are albuminous seeds as they retain endosperm.
Maize and wheat also possess endosperm.

Step 4: Conclusion.
Therefore, the correct set of exalbuminous seeds is Pea, Bean, Gram, and Ground-nut.

Step 1: Understanding fungal classification.
Fungi are classified based on their mode of sexual reproduction and spore formation.

Step 2: Special feature of Deuteromycetes.
Deuteromycetes are also known as fungi imperfecti because sexual reproduction has not been observed in them.

Step 3: Eliminating other options.
Phycomycetes, Ascomycetes, and Basidiomycetes all show sexual reproduction.

Step 4: Conclusion.
Thus, sexual reproduction is absent in Deuteromycetes.

Step 1: Understanding biofortification.
Biofortification is a plant breeding strategy aimed at improving the nutritional quality of food crops, especially their vitamin and mineral content.

Step 2: Purpose of biofortification.
It focuses on increasing essential nutrients such as iron, zinc, vitamin A, and protein content to combat malnutrition.

Step 3: Eliminating other options.
Disease resistance, pest resistance, and yield improvement are achieved through other breeding strategies, not biofortification.

Step 4: Conclusion.
Therefore, biofortification is used to increase the nutritional value of crop plants.

Step 1: Identifying endangered species.
From the given list, endangered species include tiger, musk deer, one-horned rhinoceros, and great Indian bustard.

Step 2: Excluding non-endangered species.
Lion, camel, red fox, and peacock are not classified as endangered species.

Step 3: Conclusion.
Thus, the total number of endangered species in the list is 4.

Step 1: Understanding dihybrid cross.
A dihybrid cross involves two pairs of contrasting traits, such as seed shape and seed colour in pea plants.
Step 2: F generation outcome.
When heterozygous individuals are crossed, the F generation produces a 9:3:3:1 phenotypic ratio.
Step 3: Counting genotypes and phenotypes.
Total number of possible genotypes in F generation = 16.
Total number of phenotypic combinations = 9.
Step 4: Conclusion.
The correct number of genotypes and phenotypes are 16 and 9 respectively.

Step 1: Understanding the genetic nature of colour blindness.
Colour blindness is a sex-linked recessive trait, carried on the X-chromosome. A male has only one X-chromosome, so if it carries the defective gene, the trait is expressed. Females have two X-chromosomes, so the trait is expressed only if both X-chromosomes carry the defective gene.
Step 2: Writing the genotypes of parents.
Colour blind man:
Normal visioned woman (homozygous):
Step 3: Formation of offspring.
All daughters receive one normal X-chromosome from the mother and the colour blind X-chromosome from the father, so daughters are (carriers but phenotypically normal).
All sons receive the Y-chromosome from the father and a normal X-chromosome from the mother, so sons are (normal).
Step 4: Conclusion.
Since none of the offspring express colour blindness phenotypically, the percentage of colour blind offspring is zero.

Step 1: Understanding genome manipulation.
Genome manipulation refers to deliberate alteration of the genetic material of a cell to introduce, remove, or modify specific genes.
Step 2: Role of recombinant DNA (r-DNA) technology.
Recombinant DNA technology allows insertion of foreign genes into a host genome using vectors such as plasmids. This enables genetic modification at the DNA level.
Step 3: Analysis of options.
(A) Cloning: Produces identical copies but does not alter genome content.
(B) Transcription: Involves synthesis of RNA from DNA, not genome modification.
(C) r-DNA technology: Correct, as it directly manipulates genetic material.
(D) Reverse transcription: Converts RNA into DNA but does not itself manipulate the genome.
Step 4: Conclusion.
Genome manipulation is achieved through recombinant DNA technology.

Step 1: Understanding sickle cell anaemia.
Sickle cell anaemia is a genetic disorder caused by abnormal haemoglobin (HbS).
Step 2: Role of oxygen.
Under low oxygen conditions, HbS polymerizes, causing RBCs to deform into a sickle or half-moon shape.
Step 3: Effect on RBC shape.
Oxygen deficiency leads to reduced solubility of HbS, resulting in sickling of RBCs.
Step 4: Conclusion.
Therefore, RBCs become sickle-shaped due to oxygen deficiency.

Step 1: Understanding jaundice.
Jaundice is a condition characterised by yellowing of skin and eyes due to accumulation of bilirubin in blood.

Step 2: Role of liver.
Bilirubin is produced from haemoglobin breakdown and is normally processed and excreted by the liver.
Excessive haemolysis and impaired liver function lead to increased bilirubin levels.

Step 3: Conclusion.
Hence, jaundice is associated with abnormal function of the liver.

Step 1: Applying Hardy–Weinberg principle.
Frequency of homozygous dominant genotype is given by .

Step 2: Calculation.
Here, .
.

Step 3: Conclusion.
Therefore, the frequency of homozygous dominant genotype is 0.49.

Step 1: Understanding the law of dominance.
Mendel’s law of dominance states that when two alleles are present, one allele may mask the expression of the other. However, in some organisms, incomplete dominance occurs, where neither allele completely dominates the other.
Step 2: Analyzing the options.
(A) completely dominant: Incorrect. Completely dominant alleles follow the law of dominance.
(B) both dominant and recessive: Incorrect. The law of dominance does not typically apply in such cases.
(C) completely recessive: Incorrect. Recessive alleles are masked by dominant alleles, and this is consistent with Mendel’s laws.
(D) incompletely dominant: Correct. In incomplete dominance, neither allele completely dominates, leading to intermediate phenotypes.
Step 3: Conclusion.
The correct answer is (D) incompletely dominant, as this is the condition where the law of dominance does not fully apply.

Step 1: Understanding the role of bile salts.
Bile salts are secreted by the liver and released into the duodenum through bile. They do not digest fats chemically but help in their physical breakdown.
Step 2: Process of emulsification.
Emulsification is the process by which large fat globules are broken into smaller droplets. This increases the surface area of fats, making them more accessible to the action of lipase enzymes.
Step 3: Analysis of options.
(A) Assimilate: Refers to utilization of nutrients by body cells, not a digestive action in the duodenum.
(B) Absorb: Absorption of fats occurs mainly in the intestine via lacteals, not by bile salts.
(C) Emulsify: Correct, as bile salts break fats into small droplets in the duodenum.
(D) Regurgitate: Unrelated to digestion of fats.
Step 4: Conclusion.
Bile salts emulsify fats in the duodenum to aid efficient digestion.

Step 1: Understanding commensalism.
Commensalism is an interaction where one organism benefits while the other is neither harmed nor benefited.
Step 2: Evaluating the given pairs.
Sea-anemone and crab, shark and remora, and barnacles and whale are classic examples of commensalism.
In cow–bacteria interaction, both organisms benefit: bacteria get shelter and nutrients, while the cow benefits from cellulose digestion.
Step 3: Analysis of options.
(A) Commensalism.
(B) Commensalism.
(C) Commensalism.
(D) Mutualism, not commensalism.
Step 4: Conclusion.
The pair that does not show commensalism is cow – bacteria.

Step 1: Understanding leukocyte distribution.
Different types of white blood cells vary in percentage and function in human blood.
Step 2: Matching Column-I and Column-II.
Neutrophils constitute about 54–62% of total leukocytes.
Eosinophils are present in about 3%.
Lymphocytes constitute about 25–33%.
Basophils are present in 0–5%.
Step 3: Matching Column-III (functions).
Neutrophils are phagocytic.
Eosinophils secrete antihistamine.
Lymphocytes are involved in synthesis of antibodies.
Basophils secrete heparin.
Step 4: Conclusion.
The correct matching corresponds to option (A).

Step 1: Understanding seed adaptations.
Many seeds and fruits develop special structural modifications to spread away from the parent plant. These adaptations reduce competition for resources like light, water, and nutrients.
Step 2: Role of wings, pappus, and hooks.
Wings and pappus help seeds to be carried by wind.
Hooks help seeds attach to animal fur or clothing and get transported to distant places.
Step 3: Analysis of options.
(A) Protection: These structures are not mainly for protection.
(B) Dispersal: Correct, as these structures assist in seed dispersal by wind or animals.
(C) Attraction: Attraction is mainly related to flowers and pollination.
(D) Germination: Germination depends on internal seed conditions and environment, not such structures.
Step 4: Conclusion.
These specialized structures help seeds in dispersal.

Step 1: Understanding Mendelian traits in {Drosophila. Wing size in {Drosophila melanogaster} is a classical example used in genetic studies.
Step 2: Dominant and recessive traits.
Normal wings are dominant, whereas vestigial, notched, and nicked wings are mutant recessive traits.
Step 3: Conclusion.
Therefore, the dominant wing type in {Drosophila} is Normal wings.

Step 1: Understanding dihybrid cross.
In a dihybrid cross, two pairs of contrasting characters are studied simultaneously, and each pair segregates independently.
Step 2: Statistical principle applied.
Mendel calculated the dihybrid ratio (9:3:3:1) by multiplying the probabilities of two independent monohybrid ratios (3:1 × 3:1).
Step 3: Conclusion.
Thus, Mendel applied the principle of probability.

Step 1: Understanding ABO blood group inheritance.
Blood group A phenotype requires at least one A allele. Blood group O has genotype ii and cannot contribute an A allele.
Step 2: Analyzing option (B).
In O × B cross, possible offspring blood groups are B or O only, as neither parent contributes allele A.
Step 3: Conclusion.
Therefore, parents with blood groups O and B cannot have a child with blood group A.

Step 1: Understanding Mendel’s choice of garden pea.
Garden pea plants were selected by Mendel due to clear contrasting characters and true-breeding nature.
Step 2: Evaluating the options.
Garden peas show discrete traits, are self-pollinating, annual, and have contrasting characters.
Step 3: Conclusion.
They do not show intermediate characters, hence option (C) is correct.

Step 1: Understanding the cells and their functions.
- Lymphocytes are responsible for producing antibodies, which help in immune response.
- Basophils secrete heparin, a substance that prevents blood clotting and is involved in inflammatory responses.
- Eosinophils are associated with the defense against parasitic infections but do not secrete globulin.
- Neutrophils are involved in phagocytosis, not the production of anti-histamine. Step 2: Analyzing the options.
(A) Lymphocyte : antibodies: This is true, but it’s not the correct match for this question.
(B) Basophil : heparin: Correct. Basophils secrete heparin.
(C) Eosinophil : globulin: This is incorrect. Eosinophils do not produce globulin.
(D) Neutrophil : anti-histamine: This is incorrect. Neutrophils do not produce anti-histamine. Step 3: Conclusion.
The correct answer is (B) Basophil : heparin.

Step 1: Understanding the genetic cross.
In the case of human skin color, a cross between pure black and pure white parents would follow Mendelian inheritance patterns. The F2 generation from this cross will show a variety of genetic combinations. Multihybrid individuals would be those showing multiple traits (multihybrid).
Step 2: Analyzing the options.
(A) One: Incorrect — The F2 generation will have more than one multihybrid.
(B) Six: Incorrect — Six multihybrid individuals is not the expected outcome.
(C) Twenty: Correct — In the F2 generation of a dihybrid cross, about 20% will be multihybrids.
(D) Fifteen: Incorrect — Fifteen multihybrids are not the expected number for this genetic cross.
Step 3: Conclusion.
The correct answer is (C) Twenty, as the expected number of multihybrids is twenty among the 64 offspring.

Step 1: Understanding sex chromosomes in birds.
In birds, the sex chromosomes are opposite of mammals. Male birds have ZZ chromosomes and female birds have ZW chromosomes.
Step 2: Analyzing the options.
(A) XY: Incorrect — This is the chromosomal pattern for males in mammals, not birds.
(B) ZW: Incorrect — This is the chromosomal pattern for females in birds.
(C) XX: Incorrect — This is the chromosomal pattern for females in mammals.
(D) ZZ: Correct — Male birds have ZZ chromosomes.
Step 3: Conclusion.
The correct answer is (D) ZZ, as male birds have ZZ chromosomes.

Step 1: Understanding the third kingdom.
The third kingdom, Protista, was proposed by the German biologist Ernst Haeckel. He introduced the kingdom to include unicellular organisms, distinguishing them from plants, animals, and fungi.
Step 2: Analyzing the options.
(A) R.H. Whittaker: R.H. Whittaker proposed the five-kingdom classification system but did not suggest the third kingdom Protista.
(B) Carl Linnaeus: Carl Linnaeus is known for developing the binomial nomenclature system and the classification of organisms, but he did not propose the third kingdom.
(C) Haeckel: Correct — Ernst Haeckel proposed the kingdom Protista.
(D) Carl Woese: Carl Woese is known for the three-domain system of classification but did not propose the third kingdom Protista.
Step 3: Conclusion.
The correct answer is (C) Haeckel, as he proposed the third kingdom Protista.

Step 1: Understanding genetic expression.
In the case of AB blood group, both A and B alleles are expressed when a person is heterozygous (having one A allele and one B allele). This is an example of codominance, where both alleles are equally expressed.
Step 2: Analyzing the options.
(A) diabetes: Incorrect — Diabetes is not an example of codominance.
(B) AB blood group: Correct — Both alleles (A and B) are expressed in AB blood type.
(C) colour blindness: Incorrect — Colour blindness follows X-linked recessive inheritance, and the trait is expressed only when the individual has two recessive alleles.
(D) haemophilia: Incorrect — Haemophilia also follows X-linked inheritance, and it is typically expressed in males with a single recessive allele.
Step 3: Conclusion.
The correct answer is (B) AB blood group, where both alleles are expressed in a heterozygous individual.

Step 1: Understanding co-dominance.
Co-dominance occurs when both alleles contribute equally and visibly to the organism's phenotype. In this case, both alleles are dominant and neither allele is recessive.
Step 2: Analyzing the options.
(A) one is recessive: Incorrect — Co-dominance involves both alleles being dominant.
(B) both are dominant: Correct — In co-dominance, both alleles express their traits simultaneously.
(C) one is dominant: Incorrect — Both alleles are dominant in co-dominance.
(D) both are recessive: Incorrect — Recessive alleles do not show their effect when co-dominance occurs.
Step 3: Conclusion.
The correct answer is (B) both are dominant, as co-dominance involves both alleles contributing equally to the phenotype.

Step 1: Understanding Mendelian crosses.
A dihybrid cross involves two traits, such as height and flower position in pea plants. The 16:1 phenotypic ratio (315:108:101:32) corresponds to a dihybrid cross, where two traits are inherited independently.
Step 2: Analyzing the options.
(A) Test: Incorrect — A test cross is used to determine the genotype of an individual with a dominant phenotype.
(B) Trihybrid: Incorrect — A trihybrid cross involves three traits, which is not the case here.
(C) Dihybrid: Correct — A dihybrid cross is the correct representation of the given data, involving two traits (height and flower position).
(D) Monohybrid: Incorrect — A monohybrid cross involves only one trait, not two.
Step 3: Conclusion.
The correct answer is (C) Dihybrid, as the cross involves two traits.

Step 1: Understanding point mutation.
A point mutation involves a change in a single nucleotide base in the DNA sequence of a gene.
Step 2: Evaluating the options.
(A) sudden change in the chemical make up of a gene: Correct. Point mutation alters the nucleotide sequence of a gene.
(B) alteration in allele frequency: Incorrect. It relates to population genetics.
(C) changes in gene arrangement in chromosome: Incorrect. This refers to chromosomal mutation.
(D) gradual, continuous change in genetic make up: Incorrect. This describes evolution.
Step 3: Conclusion.
The correct answer is (A) sudden change in the chemical make up of a gene.

Step 1: Human skin color is determined by multiple genes, with each gene contributing to the final phenotype. This type of inheritance is called polygenic inheritance, where multiple genes influence a single trait.

Step 2:
- (A) Mendelian inheritance involves single-gene traits, such as flower color in pea plants.
- (C) Incomplete dominance involves a blending of traits, such as flower color in certain plants.
- (D) Codominance occurs when both alleles are expressed equally, such as in AB blood type.
Conclusion: Skin color is an example of polygenic inheritance.