Principle of inheritance and variation practice · Medium

Option B (Type O) is correct because individuals with blood type O are universal donors and can donate blood to individuals with any ABO blood type in emergencies.

Option C (Colour blindness) is correct because Haemophilia and Colour blindness are both X-linked recessive disorders.

Option D (Brown body, white eyes) is correct because y+ represents the wild-type allele for body color (brown), and w represents the mutant allele for eye color (white).

Option B (Yy $\times$ Yy) is correct because a cross between two heterozygotes (Yy) results in a 1:2:1 genotypic ratio (YY:Yy:yy) and, assuming incomplete dominance, a corresponding phenotypic ratio.

Option A (Cretinism) is correct because Cretinism is caused by iodine deficiency and is not a hereditary disease, while the others are.

Option A (Klinefelter’s syndrome –44 autosomes + XXY) is correct because Klinefelter's syndrome is characterized by the presence of an extra X chromosome in males, resulting in an XXY karyotype.

Option D (42370) is correct. This question is flawed as it provides an impossible context (selfing a heterozygous plant and asking for a specific offspring count). The correct approach would involve calculating the probability of the homozygous recessive offspring (1/16) and multiplying by the total population size, but the provided options are not realistic.

Option D (Archibald Garrod) is correct because he is recognized as the father of physiological genetics or biochemical genetics due to his work on inborn errors of metabolism.

Option A (Both Assertion and Reason are true and the reason is the correct explanation of the assertion) is correct because a cross between a heterozygous tall plant (Tt) and a homozygous recessive dwarf plant (tt) would result in 50% tall (Tt) and 50% dwarf (tt) offspring, and the reason correctly identifies the heterozygous nature of one parent.

Option A (Cones) is correct because cones in the retina are responsible for color vision, and defects in these cells lead to color blindness.

Option D (All of these) is correct because in Drosophila, these traits are typically X-linked, and a male progeny receives its X chromosome from its mother.

The correct answer is B (Less than 5 per cent) because in polygenic inheritance with multiple genes, the proportion of progeny resembling either extreme parental type is very low in the F2 generation.

Option C (Alfred Sturtevant) is correct because Alfred Sturtevant developed the concept of using recombination frequencies to measure the distance between genes on a chromosome, creating the first genetic map.

Option D ($I^AI^B$ and $I^oI^o$) is correct because the offspring of these parents can only have blood groups A or B, and cannot have blood group O or AB.

Option C (Law of independent assortment) is correct because this law does not apply to linked genes, which are located close together on the same chromosome and tend to be inherited together.

The correct answer is D (salivary glands of Drosophila) because giant chromosomes, also known as polytene chromosomes, are commonly found in the salivary glands of Drosophila melanogaster.

Option A (Sex chromosome) is correct because sex-linked traits are determined by genes located on the sex chromosomes.

Option B (Wheat and rye) is correct because Triticale is a hybrid of wheat (Triticum) and rye (Secale).

The correct answer is D (0) because daughters inherit one X chromosome from each parent, and in this case, the daughter will inherit a normal X from her father and a carrier X from her mother, making her a carrier but not colorblind.

Option D (34) is correct because with three pairs of polygenes, there are 7 genotypes and 27 phenotypes, and the sum is 34.

Option C (Assertion is true but Reason is false) is correct. Phenylketonuria (PKU) involves the conversion of phenylalanine to tyrosine, not alanine.

Option A (${X^H}{X^h}$ and ${X^H}Y$) is correct because the mother must be a carrier to have a hemophilic son, and the father must be normal to have normal daughters.

Option C (Trillium) is correct because Trillium species are known for having exceptionally large chromosomes, making them ideal for cytological studies.

Option D (Yellow pod) is correct because true breeding plants produce offspring with the same traits as the parents, thus a true breeding pea plant for yellow pods will only produce yellow pods.

Option D (All of these) is correct because fruit flies have a short life cycle, can be easily fed, produce many offspring, have clear sex differentiation, and exhibit easily observable hereditary variations.

Option B ($Mirabilis$) is correct because Mirabilis jalapa (Four O'Clock plant) is a classic example of incomplete dominance, where the heterozygous offspring show an intermediate phenotype.

Option B (A, B, and AB blood types) is correct because parents with AB blood type can produce offspring with A, B, or AB blood types, but not O.

Option A (Turner’s syndrome) is correct because Turner's syndrome is characterized by having only one sex chromosome (XO).

Option D (Chromosome-12, chromosome-4 and chromosome-11) is correct because Phenylketonuria is associated with chromosome-12, Huntington’s disease with chromosome-4 and sickle cell anaemia with chromosome-11.

Option B (Half of the male children are with both colorblindness and hypertrichosis) is correct because it considers the inheritance patterns of both X-linked colorblindness and Y-linked hypertrichosis.