Molecular Basis of Inheritance MCQs for NEET — Botany Questions with Answers

If one base is coding one amino acid then (12)¹ = 12, but we have 96 amino acids. Suppose two bases form one codon, then (12)² = 144. It is sufficient to code 96 or 20 amino acids and hence two, bases form one codon.

Euchromatin replicate in early S-phase and heterochromatin in late S + G₂-phase and hence, both incorporate radioactive thymidene.

If in a gene encoding a polypeptide of 50 amino acids, 25^th codon (UAU) is mutated to UAA. A polypeptide of 24 amino acids will be formed. Since, UAA is a termination codon, no further polypeptide will be formed.

Okazaki fragments are the small DNA fragments, which are synthesized from DNA polymerase III on the lagging strand. It is plays an important role to understand the non-conservative nature of DNA replication.

The understanding of genetic transformation in bacteria was made by Frederick F Griffith.

In a DNA segment amount of adenine is equal to thymine. According to given information adenine and thymine constitutes 45% of DNA. So, guanine and cytosine will constitute 55%. Amount of guanine is equal to cytosine so amount of guanine in DNA will be 55/2 = 27.5%.

AUG is a initiation codon which codes for methionie

Valine → GUU

Histidine → CAU

Phenylalanine → UUU or UUC

A recessive mutant is one which is expressed only in homozygous or hemizygous stage.

Adenine + Thymine = 60%

The amount of guanine + cytosine = 40%

So, the amount of guanine in this DNA will be = 20%

$\frac{\mathrm{A}+\mathrm{T}}{\mathrm{G}+\mathrm{C}}$ is constant (Chargaff’s rule) for a species and also A = T and G = C.