In rice, assume that a recessive gene “a” is responsible for resistance to BLB. This gene is present on chromosome #1. A monosomic resistant line for chromosome #1 is crossed with normal susceptible line. The phenotype of F₁ plants will be

Correct option is A

✅ (a) 50% plants resistant and 50% plants susceptible [Monosomic Genetics] • The resistant line is monosomic for chromosome 1 (i.e., has only one copy with the recessive allele “a”). • The normal susceptible parent is disomic with dominant “A” alleles. • Gametes from the monosomic parent will be 50% with “a” and 50% without chromosome 1. • Crossing yields: • 50% A/a → Susceptible (dominant A) • 50% a/- → Resistant (no A allele present) • Hence, 1:1 ratio of resistant to susceptible.
❌ (b) 75% resistant • Not possible; indicates Mendelian segregation in normal diploids, not monosomics.
❌ (c) All plants resistant • Requires both parents to be homozygous “aa”.
❌ (d) All plants susceptible • Not possible with “a” allele contribution from resistant parent.