NEET Zoology - MCQ Practice Questions

A gene is the fundamental unit of heredity that carries information for a specific trait. It consists of a segment of DNA coding for a protein or RNA.

Humans have 46 chromosomes (23 pairs) in somatic cells - 22 pairs of autosomes and 1 pair of sex chromosomes (XX or XY).

Mendel's Law of Segregation states that hereditary traits are controlled by discrete units (genes) that separate during gamete formation.

Monohybrid cross Aa × Aa produces genotypic ratio 1 AA : 2 Aa : 1 aa. Probability of aa (homozygous recessive) = $\frac{1}{4}$ or 25%.

Alleles are alternative forms of a gene that occupy the same locus on homologous chromosomes and control different variations of the same trait.

AA × aa produces all Aa offspring (test cross principle). All offspring are heterozygous with dominant phenotype.

The 1:1 ratio indicates a testcross result (Rr × rr). The round seed parent must be heterozygous (Rr) to produce wrinkled (rr) offspring.

The correct hierarchical order is: Nucleotide (basic unit) → DNA (polymer of nucleotides) → Gene (segment of DNA) → Chromosome (condensed DNA with proteins).

In Hardy-Weinberg equation, p and q represent frequencies of two alleles in a population. p² and q² are homozygous frequencies; 2pq is heterozygous frequency.

Recessive phenotype = $\frac{360}{1000}$ = 0.36 = q². Therefore q = √0.36 = 0.60. This represents the frequency of recessive allele.

X-linked recessive traits show more affected males (XᵃY) than females because males need only one copy of recessive allele on their single X chromosome.

Haemophilia A (clotting factor VIII deficiency) is X-linked recessive. Affected males (XᵃY) pass the allele to all daughters but no sons.

Color blindness is an X-linked recessive trait. To find the percentage of color-blind sons, we need to perform a cross between a carrier mother and a normal father, then analyze the male offspring.

Step 1: Identify the genotypes

Mother (carrier): $X^B{X}^b$ — has one normal allele ($X^B$) and one color-blind allele ($X^b$)

Father (normal): $X^{B}Y$ — has the normal allele on his X chromosome

Step 2: Set up the Punnett square

Cross all maternal alleles with all paternal alleles:

Step 3: Identify offspring genotypes and phenotypes

The four possible offspring are:

Step 4: Calculate percentage of color-blind sons

Among all sons (male offspring only):

Answer: 50% of their sons will be color blind (Option C)

Mutation is a permanent change in the DNA sequence that can occur spontaneously or be induced. It is the source of genetic variation.

Silent mutation occurs in the third position of a codon and does not change the amino acid due to degeneracy of genetic code. No phenotypic effect.

Natural selection is the non-random differential reproduction of organisms with traits that increase fitness. Organisms with advantageous traits leave more offspring.

Phenotypic plasticity is the ability of a single genotype to produce different phenotypes in response to environmental variation (e.g., hydrangea flower color).

Multiple lines of evidence support evolution: homologous structures indicate common ancestry, vestigial organs show evolutionary remnants, and fossil records show gradual changes over time.

Directional selection occurs when one extreme phenotype has higher fitness. Dark beetles have selective advantage, shifting population toward darker phenotype.

Heterozygous frequency = 2pq = 2(0.7)(0.3) = 0.42. Expected heterozygous individuals = 0.42 × 500 = 210.