In some members of which of the following pairs of
families, pollen grains retain their viability for months after release?
AI-generated explanation. Check against your textbook.
Explanation:
Months-long pollen viability is seen in some members of Rosaceae and Leguminosae (Fabaceae). Thus, the pair Rosaceae; Leguminosae (option 4) is correct. In NCERT, Poaceae (grasses) pollen is typically short-lived and loses viability quickly, while Solanaceae pollen also has limited longevity. Some Rosaceae and Fabaceae pollen grains are highly durable under dry conditions, enabling them to remain viable for extended periods.
🧠 Did You Know?
Dry, well-protected pollen grains can stay viable for months in certain plant families, aiding plants with delayed fertilization strategies.
💡 NCERT Memory Line:
Pollen grains in some plants can remain viable for months under dry storage, notably in Rosaceae and Leguminosae.
AI-generated explanation. Check against your textbook.
Explanation:
Correct option: 2. In bacteria, transcription termination can be Rho-dependent, where the Rho factor binds to the RNA and helps terminate transcription. This mechanism is described in NCERT biology.
1 is wrong because the 5' end of hnRNA receives a 7-mmG cap (7-methylguanosine cap), not at the 3' end. 3 is wrong because the template strand, not the coding strand, is used to synthesize mRNA; the mRNA sequence is similar to the coding strand (with U replacing T). 4 is wrong because split genes (introns) are a feature of eukaryotes, not prokaryotes.
🧠 Did You Know?
Rho-dependent termination requires the Rho protein to catch up with the RNA polymerase and unwind the RNA-DNA hybrid.
💡 NCERT Memory Line:
Rho factor-dependent termination occurs in some bacteria; transcription in prokaryotes uses a simpler termination mechanism than in eukaryotes.
Plasmid pBR322 has PstI restriction enzyme site
within gene ampR that confers ampicillin resistance. If this enzyme is used for inserting a gene for β-galactoside production and the recombinant plasmid is inserted in an E.coli strain
AI-generated explanation. Check against your textbook.
Explanation:
In pBR322, the ampR gene provides ampicillin resistance. If PstI cuts within ampR to insert a β-galactoside gene, ampR is disrupted, so the plasmid can no longer confer ampicillin resistance. Therefore, transformed E. coli would not survive ampicillin unless ampR remains functional. Hence, option 1 is correct. Options 2 and 4 assume both resistance and new function, which is unlikely when ampR is inactivated by insertion. Option 3 about lysis is not a general consequence of such recombination.
🧠 Did You Know?
In blue-white screening, insertion of a foreign gene into lacZ (β-galactosidase gene) often disrupts beta-galactosidase production, helping identify recombinant colonies.
💡 NCERT Memory Line:
Restriction enzymes create cuts at specific sites; inserting foreign DNA into a plasmid can disrupt a gene (e.g., ampR) and alter antibiotic resistance.
Now a days it is possible to detect the mutated gene causing cancer by allowing radioactive probe to
hybridise its complimentary DNA in a clone of cells,
followed by its detection using autoradiography
because:
In the exponential growth equation Nt = N0ert, e represents:
AI-generated explanation. Check against your textbook.
Explanation:
In the exponential growth equation N_t = N_0 e^{rt}, e is the base of natural logarithms. It makes growth continuous and smooth, as the rate of change is proportional to the current amount. This is why e is called the base of natural logarithms and appears in continuous growth/decay models in NCERT biology.
🧠 Did You Know?
Option 1 refers to base 10 (common logarithms). Option 2 and 4 describe non-standard phrases; they are not used in this biology context.
💡 NCERT Memory Line:
In exponential growth, N_t = N_0 e^{rt}; e is the base of natural logarithms.