Principle of inheritance and variation practice · Medium

Option D (Round seeds with yellow cotyledons and wrinkled seeds with yellow cotyledons) is correct because the F1 generation will be RrYy (round, yellow). Selfing the F1 will produce round yellow, round green, wrinkled yellow, and wrinkled green phenotypes. The question asks what phenotypes are expected, not the ratio.

Option B (Purity of gametes) is correct because the law of segregation states that alleles separate during gamete formation, ensuring each gamete carries only one allele for each gene.

Option D (11th) is correct because in $\mathrm{\beta }$-thalassaemia, the affected chromosome is the 11th.

Option C (B and D) is correct because these crosses (TtRr x ttrr and Ttrr x ttRr) result in a 1:1:1:1 phenotypic ratio in the progeny.

Option A (Foetus will transmit antigen to mother blood) is correct because the Rh+ foetus has Rh antigens on its red blood cells, which can enter the mother's bloodstream.

Option C is correct because if both parents are ii (constricted), the F1 progeny cannot have the II or Ii genotype needed for the inflated phenotype.

Option C (16) is correct because the number of gamete types produced by a heterozygous diploid organism is 2^n, where n is the number of heterozygous loci. In this case, n=4, so 2⁴ = 16.

Option C (Down’s syndrome – 45+XY) is correct because individuals with Down syndrome have an extra chromosome 21, resulting in a total of 47 chromosomes (45 autosomes + XY).

Option B (Codominance) is correct because both red and white coat colors are expressed simultaneously in the roan offspring, without blending.

Option B ($Aa \times aa$) is correct because this cross will produce 50% heterozygous offspring (Aa) expressing the dominant character and 50% homozygous recessive offspring (aa) expressing the recessive character.

Option A (a ,c, b, d) is correct because the gene order is determined by minimizing the distances between adjacent genes based on the given recombination frequencies.

Option D (Eugenics) is correct because eugenics is defined as the improvement of the human race through hereditary qualities.

Option C (Alleles) is correct because alleles are defined as the alternate forms of a gene.

Option D (Both 1 and 2) is correct because both Sutton and Boveri independently proposed the chromosomal theory of inheritance, linking Mendel's laws to the behavior of chromosomes during meiosis.

Option C (4) is correct because the cross ABcDE × AbcdE results in the genotype AABbcDdEE. Only the heterozygous loci (Bb and Dd) contribute to the number of gamete types. The number of gamete types is 2^n, where n is the number of heterozygous loci. In this case, n=2, so 2² = 4.

Option C (Four) is correct because with two unlinked heterozygous genes (PpQq), the possible gametes are PQ, Pq, pQ, and pq.

Option D (No antigens A and B on RBCs) is correct because the 'O' blood type signifies the absence of both A and B antigens on the surface of red blood cells.

Option D (1 and 3) is correct because the unit of a linkage map is both a map unit and a centimorgan, which are equivalent.

Option C (Mutated genes are frequently recessive) is correct because recessive mutations are less likely to be immediately expressed in heterozygotes, allowing them to persist and potentially affect future generations.

Option B (Alleles shows complete dominance) is correct because Mendel's laws are based on the principle of complete dominance, where one allele masks the expression of the other.

Option B (4) is correct because in a dihybrid cross with independent assortment, there are typically 4 different phenotypes.

Option D ($\mathrm{RrYy} \times \mathrm{rryy}$) is correct because this cross represents a test cross, where a dihybrid heterozygous individual (RrYy) is crossed with a homozygous recessive individual (rryy), resulting in a 1:1:1:1 phenotypic ratio.

Option B is correct because a color-blind girl needs to inherit the color-blind allele from both parents, meaning her father must be color-blind and her mother must carry the allele, which she inherited from her maternal grandfather.

Option B is correct because chromatids separate during mitosis, allele pairs replicate and are passed to daughter cells, maintaining the genetic composition.

Option D ($7 :9$) is correct because the cross involves two genes (tall/dwarf and yellow/green pods), and the parental combinations are homozygous dominant and recessive. The F2 generation will have a 9:3:3:1 ratio, and the parental combinations will be less frequent than the new combinations, resulting in a 7:9 ratio.

Option B (Haploid - diploid sex - determination system) is correct because in haplodiploidy, males are haploid (have no father) and females are diploid (have a father and can have sons).

Option C (Both (a) and (b)) is correct because codominance is observed in both plants and animals.

Option C ($9\mathrm{ }:\mathrm{ }3\mathrm{ }:\mathrm{ }3\mathrm{ }:\mathrm{ }1$) is correct because even though the genes are linked, some crossing over will still occur, leading to a modified dihybrid ratio that approximates 9:3:3:1, especially if the genes are not extremely tightly linked.

Option C (Provides immunity against malaria) is correct because individuals with sickle cell trait (heterozygous) are more resistant to malaria, providing a selective advantage in malaria-prone regions.

Option B (heterozygous and green) is correct because the ratio of 760 green pod and 240 yellow pod plants (approximately 3:1) suggests a heterozygous parent (Gg) with green pods, where green is dominant over yellow (gg).