Practice Questions

Plant cells possess an extracellular, rigid cell wall made primarily of cellulose that protects the cell from mechanical stress and osmotic lysis.

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1. The dynamic outermost boundary protecting the living protoplasm of a typical animal cell is the

A. Cell wall
B. Plasma membrane
C. Tonoplast
D. Glycocalyx only

Animal cells lack a cell wall entirely; their outermost living boundary is the selectively permeable phospholipid bilayer known as the plasma membrane.

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Triple-stranded H-DNA forms when a third single strand winds into the major groove of a duplex, binding via alternative Hoogsteen hydrogen bonds.

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99. A chemical mutagen modifies an adenine base within a gene via deamination, converting it into hypoxanthine. During subsequent rounds of DNA replication, hypoxanthine preferentially pairs with cytosine instead of thymine, resulting in a permanent

A. Transition mutation from an A-T pair to a G-C pair
B. Transversion mutation from an A-T pair to a T-A pair
C. Frameshift mutation via single-nucleotide deletion
D. Nonsense mutation that halts transcription

Hypoxanthine pairs with cytosine, meaning the original template A-T pair becomes a G-C pair after a few rounds of replication, causing a transition mutation.

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98. A mutation alters the conserved Shine-Dalgarno sequence upstream of a bacterial structural gene. When this mutated polycistronic mRNA interacts with prokaryotic translation machinery, the immediate molecular outcome is the

A. Failure of the small (30S) ribosomal subunit to correctly align with the start codon
B. Accelerated transcription of downstream regulatory genes
C. Spontaneous excision of the mutated sequence by the spliceosome
D. Direct methylation of the structural gene's promoter region

The Shine-Dalgarno sequence base-pairs with the 16S rRNA of the 30S subunit, aligning the bacterial ribosome with the start codon to initiate translation.

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97. An analysis of the spatial orientation of the B-DNA double helix shows that the glycosidic bonds linking the bases to the sugar rings project at unequal angles. This structural asymmetry means that the phosphodiester backbones are

A. Arranged directly opposite each other, forming a perfectly symmetrical cylinder
B. Arranged closer together on one side of the helix than the other, creating distinct grooves
C. Linked covalently across the core via disulfide bridges
D. Restricted to a rigid left-handed zigzag spatial orientation

Because the glycosidic bonds project unevenly, the backbones wind asymmetrically around the axis, creating alternating major and minor grooves.

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NMD is a surveillance mechanism that detects premature stop codons on transcripts and degrades them, preventing the accumulation of toxic, truncated proteins.

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95. A researcher utilizes a chemical toxin that selectively prevents the hydrolysis of pyrophosphate (PPi​) during nucleic acid synthesis. The immediate impact of this drug on gene replication and transcription is the

A. Acceleration of nucleotide polymerization rates
B. Thermodynamic arrest of phosphodiester bond formation
C. Spontaneous conversion of ribose sugars into deoxyribose variants
D. Uncontrolled duplication of upstream promoter sequences

Polymerization relies on the cleavage of pyrophosphate (PPi​→2Pi​) to provide the forward driving force; blocking this halts the reaction.

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94. A point mutation occurs within the consensus sequence of a 5′ splice donor site of a crucial structural gene. During pre-mRNA processing, this molecular defect will most likely cause the spliceosome to

A. Skip the downstream exon entirely or retain the entire mutated intron
B. Add an exceptionally long poly-A tail to the 5' end of the transcript
C. Convert the mature mRNA back into a double-stranded DNA template
D. Shift the promoter sequence into the coding exon

Mutilating a splice site prevents the spliceosome from recognizing the intron-exon boundary, leading to intron retention or exon skipping during splicing.

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93. A segment of double-stranded DNA is analyzed and found to have a melting temperature (Tm​) of 85∘C, while a second fragment of identical length melts at 72∘C. The physical explanation for the higher Tm​ of the first fragment is

A. A higher density of negative charges along its outer phosphate rails
B. A higher proportion of Guanine-Cytosine (G-C) base pairs
C. The presence of long non-coding intron loops within its sequence
D. Its unique organization into nucleosome core filaments

G-C pairs are bound by three hydrogen bonds, meaning G-C rich fragments require higher temperatures to denature than A-T rich sequences of the same length.

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