Genetics and Plant Breeding

All three are correct. (d) is incorrect. A single pea pod has many seeds in it. This is not the reason due to which Mendel used pea plant in his experiments.

Green colour of unripe pods: Dominant Axillary position of flowers: Dominant Round shape of seeds: Dominant

Siblings usually share many genetic traits, but they do not have identical sets of alleles unless they are monozygotic (i.e. identical) twins.

Law of independent assortment states that when two pairs of independent alleles enter into combination in the F2 generation, they exhibit independent dominant effects. It means that during the formation of gametes, the law of segregation operates but the factors assort themselves independently at random and freely. So, for explaining this principle a monohybrid cross will be insufficient. We will need a dihybrid cross to depict this. Thus, this is the correct answer.

The genes for MHC-I and MHC-II are located closely on chromosome-6 of humans, and these are transferred as a set to the next generation in order to maintain the MHC compatibility.

This is the correct option as the colour of leaves of the pea plant's transmission to progeny was not investigated by Mendel.

This is the correct option. If the genotype of a person is I^AI^B, then AB will be his blood group.

This is the correct option as this statement is false. Blonde hair is a phenotype produced by the genotype hh.

This is the correct option. Gene is a unit of heridity that transfers characteristics from parents to their offsprings during reproduction. Genotype is the description of genes present in an organism. Phenotype is the trait that is visible in an organism.

In X-linked recessive inheritance, males are primarily affected because they have only one X chromosome. Affected females require an affected father and a carrier or affected mother to inherit two recessive alleles. Sons of affected females always receive the X chromosome with the recessive allele and will be affected. These traits can skip generations through carrier females, so S is false.

For a heterozygous female (X^H X^h) crossed with a normal male (X^H Y): each child has 1/4 chance of being unaffected (X^H X^H female or X^H Y male). So P (unaffected) = 3/4. For 5 offspring, the probability exactly 2 are affected is calculated using binomial distribution: C(5,2) × (1/4)^2 × (3/4)^3 = 10 × 1/16 × 27/64 ≈ 26.3%.

For autosomal dominant traits, every affected individual must have an affected parent (except new mutations), males and females are equally affected, and there's a 50% transmission risk to each offspring. However, both sons AND daughters of an affected male have equal 50% chance - not all daughters and no sons. This sex-specific pattern describes X-linked inheritance, not autosomal dominant.

Allele frequencies always sum to 1, and if there are two alleles, knowing one frequency gives the other. A missing allele has frequency 0. However, two populations can have identical allele frequencies but different homozygote proportions because the frequencies themselves don't determine homozygosity - you need to know whether the population is in Hardy-Weinberg equilibrium to make that calculation.

Classical Mendelian traits are qualitative and show discontinuous variation (distinct categories like tall vs short). They are typically monogenic (controlled by one or few genes), not polygenic. Polygenic traits show continuous variation and are called quantitative or metric traits, not qualitative.

For test cross of AaBb: P (unlinked) - parental aabb is 1/4 (25%). Q (completely linked) - aabb requires no crossover, so 50% parental types. R (10 map units) - 10% recombinants, so aabb = 45% parental. S (24 map units) - 24% recombinants, so aabb = 38% parental. The further apart the genes, the more crossing over, so fewer parental aabb progeny.