Molecular Basis of Inheritance MCQ

C. Thymine. Adenine and Guanine are purines, while Uracil is a pyrimidine found in RNA, not DNA.

C. N-glycosidic bond. Phosphodiester bonds link nucleotides together, hydrogen bonds connect base pairs, and peptide bonds join amino acids in proteins.

B. The 5′ end has a free phosphate group, while the 3′ end has a free hydroxyl group. This directionality is crucial for DNA replication and transcription.

C. Contains uracil. Uracil is found in RNA; DNA contains thymine instead.

C. They help to coil and condense DNA into chromosomes. Histones are positively charged proteins that bind to negatively charged DNA, allowing it to be tightly packed.

A. Nucleosome. Chromatin is a complex of DNA and proteins, including histones, that make up chromosomes. The nucleoid is the region where DNA is located in prokaryotes.

B. Euchromatin is loosely packed and transcriptionally active, while heterochromatin is tightly packed and transcriptionally inactive.

C. A bacteriophage has more base pairs than E. coli. E. coli has a larger genome than a bacteriophage.

A. In large loops held by proteins in a region called the nucleoid. Prokaryotes lack a defined nucleus.

A. Lysine and arginine. These amino acids have positively charged side chains that help histones bind to negatively charged DNA.

D. It should be composed of both DNA and RNA. Genetic material can be either DNA or RNA, but not necessarily both.

B. Mice infected with the heat-killed rough strain die from pneumonia. The rough strain is non-virulent, and heat-killing the virulent smooth strain makes it non-lethal unless mixed with live rough strain.

C. DNA is the genetic material. Their experiments showed that DNA, not protein or RNA, was responsible for transformation in bacteria.

B. ³²P and ³⁵S. ³²P labels DNA specifically (due to phosphate groups), while ³⁵S labels proteins (due to sulfur-containing amino acids).

C. DNA is more chemically and structurally stable than RNA. The presence of deoxyribose sugar (lacking the 2′-OH group) makes DNA less reactive and less prone to degradation compared to RNA.

D. It can directly code for proteins. Genetic information in DNA is transcribed into RNA, which is then translated into proteins. RNA is more directly involved in protein synthesis.

C. The genetic material that is transferred between bacteria. Griffith discovered the phenomenon, and Avery, MacLeod, and McCarty identified it as DNA.

B. To separate the viral coats from the bacteria. Agitation in a blender detached the phage particles (protein coats) attached to the outside of the bacteria, allowing separation by centrifugation.

B. High rate of mutation. A good genetic material requires stability and the ability to undergo slow mutations (evolution) but not a high rate, which would compromise genetic integrity.

D. It is more reactive and easily degradable. The presence of the 2′-OH group in ribose makes RNA chemically less stable and more susceptible to hydrolysis than DNA.

B. It can act as both a carrier of genetic information and a catalytic enzyme. This dual role (genetic storage and catalysis, like ribozymes) supports the RNA world hypothesis.

B. DNA is more stable and less prone to mutations. The chemical stability of DNA (double helix, deoxyribose sugar) makes it a better repository for genetic information.

D. Proteins were the first catalysts. The RNA world hypothesis proposes that RNA molecules (ribozymes) were the primary catalysts before the evolution of protein enzymes.

C. RNA can act as both a carrier of genetic information and a catalyst. The discovery of ribozymes and RNA’s central role in protein synthesis (mRNA, tRNA, rRNA) support this hypothesis.

B. DNA is more stable and less prone to degradation. DNA’s double-helical structure and deoxyribose sugar provide greater chemical stability, making it better suited for long-term storage of genetic blueprints.

B. Semiconservative replication. They suggested that the DNA double helix unwinds, and each strand serves as a template for the synthesis of a new complementary strand.

D. Meselson and Stahl. Their experiments using heavy (¹⁵N) and light (¹⁴N) nitrogen isotopes traced the distribution of parental DNA strands during replication in E. coli.

B. To synthesize a new DNA strand from a template strand. DNA polymerase catalyzes the addition of deoxyribonucleotides to the 3′ end of a growing DNA chain, using the template strand as a guide.

C. To join DNA fragments together. DNA ligase forms phosphodiester bonds to seal the nicks between Okazaki fragments on the lagging strand and during DNA repair.

C. The Y-shaped region where DNA is unwound and replicated. It forms as the double helix separates, allowing access for replication machinery.

D. All of the above. DNA polymerase synthesizes only 5′ to 3′. Since strands are antiparallel, one strand (lagging strand) must be synthesized backwards relative to fork movement, in short Okazaki fragments.

B. Short DNA fragments synthesized on the lagging strand. These are later joined together by DNA ligase to form a continuous strand.

A. The point where DNA replication begins. It is a specific DNA sequence recognized by initiator proteins to start replication.

B. S phase. The S phase (Synthesis phase) is dedicated to replicating the cell’s DNA before cell division.

B. Polyploidy. If DNA replicates but the cell fails to divide (cytokinesis), the resulting cell will have multiple sets of chromosomes.

B. The synthesis of RNA from a DNA template. Transcription is the process of copying genetic information from a DNA segment into an RNA molecule.

D. All of the above. Transcribing both strands would produce complementary RNAs, potentially forming double-stranded RNA, and would yield two different protein products from one gene if both were translated, which is generally not how genes function.

A. A sequence of DNA that signals the start of a gene. The promoter is located upstream (towards the 5′ end) of the transcriptional start site and serves as the binding site for RNA polymerase.

A. The template strand is transcribed into RNA, while the coding strand has the same sequence as the RNA (except for T instead of U). The RNA polymerase reads the template strand (also called antisense strand) to synthesize a complementary RNA molecule.

A. A segment of DNA that codes for a polypeptide. It is essentially a gene, a functional unit that specifies the sequence of a single polypeptide chain.

A. Monocistronic genes code for a single polypeptide, while polycistronic genes code for multiple polypeptides. Monocistronic mRNAs are typical of eukaryotes, while polycistronic mRNAs (encoding multiple proteins from one transcript) are common in prokaryotes (e.g., operons).

A. Exons are coding sequences, while introns are non-coding sequences. In eukaryotic genes, introns are intervening sequences that are removed from the pre-mRNA during splicing, while exons are the sequences that are joined together to form the mature mRNA.

C. To carry genetic information from DNA to the ribosome. Messenger RNA (mRNA) contains the codons that specify the amino acid sequence for protein synthesis.

A. To carry amino acids to the ribosome. Transfer RNA (tRNA) acts as an adaptor molecule, matching specific amino acids to their corresponding codons on the mRNA during translation.

B. To catalyze protein synthesis. Ribosomal RNA (rRNA) is a major structural component of ribosomes and also possesses the catalytic activity (peptidyl transferase) responsible for peptide bond formation.

B. RNA polymerase. Bacteria typically have a single type of RNA polymerase responsible for synthesizing mRNA, tRNA, and rRNA.

B. The removal of introns and joining of exons in eukaryotic RNA. Splicing occurs in the nucleus and converts the primary transcript (pre-mRNA) into mature mRNA.

A. The addition of a methyl guanosine triphosphate to the 5′ end of eukaryotic RNA. The 5′ cap (a modified guanine nucleotide added in reverse linkage) protects the mRNA and is important for ribosome binding.

B. The addition of a poly-A tail to the 3′ end of eukaryotic RNA. Polyadenylation (addition of a tail of adenine nucleotides) increases mRNA stability and aids in its export from the nucleus.

D. All of the above. In prokaryotes (bacteria), there is no nuclear membrane separating transcription (DNA to RNA) and translation (RNA to protein). Translation can begin on the nascent mRNA molecule even before transcription is finished.

C. 3. The genetic code is read in triplets of nucleotides, where each triplet (codon) specifies an amino acid or a stop signal.

B. Multiple codons can code for the same amino acid. There are 64 possible codons but only 20 common amino acids, so most amino acids are specified by more than one codon.

D. Oswald Avery. Avery, MacLeod, and McCarty identified DNA as the transforming principle (genetic material). Gamow proposed the triplet nature, while Nirenberg and Khorana performed key experiments to determine codon assignments.

A. A point mutation in the gene for beta globin chain. Specifically, a single nucleotide substitution (GAG to GTG) changes the codon for glutamic acid to a codon for valine at the sixth position.

D. Both B and C. The insertion or deletion of a number of nucleotides not divisible by three shifts the triplet reading frame, altering all downstream codons and usually resulting in a nonfunctional protein.

C. To act as an adapter molecule between mRNA and amino acids. tRNA molecules read the mRNA codons via their anticodons and bring the corresponding amino acids to the ribosome.

B. A sequence of three nucleotides on tRNA that is complementary to a codon on mRNA. The anticodon loop of tRNA base-pairs with the mRNA codon during translation.

C. It is ambiguous. The genetic code is unambiguous, meaning that a specific codon always codes for the same amino acid (with very few exceptions).

D. AUG. AUG serves as the start codon, signaling the beginning of translation, and it also codes for the amino acid methionine (or formylmethionine in bacteria).

A. UAA, UAG, UGA. These three codons do not code for any amino acid and signal the termination of polypeptide synthesis.

C. The synthesis of protein from an RNA template. Translation decodes the nucleotide sequence of mRNA into the amino acid sequence of a polypeptide chain.

B. To catalyze protein synthesis. Ribosomes are the cellular machinery where mRNA is read, tRNAs deliver amino acids, and peptide bonds are formed.

B. A sequence of mRNA that codes for a single polypeptide. It typically starts with a start codon (AUG), is followed by the codons specifying the polypeptide sequence, and ends with a stop codon (UAA, UAG, or UGA).

A. Untranslated regions located at the 5′ and 3′ ends of mRNA. The 5′ UTR is upstream of the start codon, and the 3′ UTR is downstream of the stop codon. They contain regulatory elements.

D. Initiation, elongation, termination. Initiation involves ribosome assembly on mRNA, elongation is the stepwise addition of amino acids, and termination occurs when a stop codon is reached.

D. All of the above. Eukaryotic gene expression is complex and can be controlled at multiple steps, including transcription initiation, RNA processing, mRNA transport, translation, and protein modification/stability.

B. A unit of gene expression found in prokaryotes. An operon is a cluster of genes under the control of a single promoter and operator, allowing coordinated regulation.

C. To regulate the access of RNA polymerase to the promoter. The operator is a DNA sequence, typically located near or overlapping the promoter, where a repressor protein can bind to block transcription.

A. An operon that regulates lactose metabolism in E. coli. It includes genes encoding proteins needed to transport and break down lactose.

A. To code for beta-galactosidase. This enzyme hydrolyzes lactose into glucose and galactose.

B. To code for permease. This membrane protein transports lactose into the bacterial cell.

C. To code for transacetylase. This enzyme acetylates galactosides, though its precise role in lactose metabolism is less critical than the other two enzymes.

B. It acts as an inducer. Specifically, allolactose (an isomer of lactose) binds to the lac repressor protein, changing its shape so it can no longer bind to the operator, thus allowing transcription.

C. By both positive and negative regulation. It is under negative control by the lac repressor (repressed unless lactose is present) and positive control by CAP-cAMP (activated when glucose is low).

B. It is repressed. High glucose levels lead to low cAMP levels. Without sufficient cAMP, the activator protein (CAP) cannot bind effectively, leading to low levels of lac operon transcription even if lactose is present (this is called catabolite repression).

D. All of the above. The HGP was an international effort to sequence the entire human genome, identify the genes, and make the data accessible for research.

C. 3 billion base pairs. The haploid human genome consists of approximately 3.3 billion base pairs.

B. 30,000. Initial estimates were higher, but the HGP revealed fewer protein-coding genes than expected, currently estimated around 20,000-25,000, though the total including non-coding RNA genes is higher (~30,000 is a reasonable approximation given in many texts).

D. 99.9%. Despite individual differences, the vast majority of the DNA sequence is identical between any two humans.

A. Expressed Sequence Tags. ESTs are short subsequences of cDNA sequences, representing portions of transcribed genes (mRNA).

B. Sequencing the whole set of the genome and later assigning different regions in the sequence with functions. Annotation involves identifying genes, regulatory elements, and other features within the raw DNA sequence data.

C. Fruit fly & D. Nematode. Bacteria (E. coli) and Yeast (S. cerevisiae) were the primary hosts for cloning large DNA fragments into BACs and YACs, respectively, which were essential tools for sequencing.

C. Frederick Sanger. The dideoxy chain termination method, developed by Sanger, was the dominant sequencing technology used throughout the HGP.

A. Single Nucleotide Polymorphisms. SNPs are the most common type of genetic variation among people, representing differences at a single nucleotide site in the DNA sequence.

C. They provide information about chromosome structure, dynamics, and evolution. While some repetitive elements can influence gene expression, their primary known roles relate to chromosome structure (e.g., centromeres, telomeres) and genome evolution. They do not code for proteins.

A. Chromosome 1. Being the largest chromosome, Chromosome 1 contains the highest number of genes.

D. Chromosome Y. The Y chromosome is significantly smaller than the X chromosome and contains the fewest genes, mostly related to male sex determination and fertility.

A. Identifying disease-associated sequences. SNPs located near disease genes can serve as genetic markers in genome-wide association studies (GWAS) to help find genes linked to diseases or traits.

B. The use of high-speed computational devices for data storage and retrieval, and analysis of biological data. Bioinformatics applies tools from computer science and statistics to manage and analyze large biological datasets, like genomic sequences.

D. All of the above. The HGP laid the foundation for large-scale biological studies, enabling research across genomics, transcriptomics, proteomics, and systems biology.

B. A technique to identify variations in individuals of a population at the DNA level. It compares specific, highly variable regions of DNA (like repetitive sequences) between individuals.

C. Repetitive DNA. DNA fingerprinting primarily relies on polymorphisms (variations) found within repetitive DNA regions, such as microsatellites (STRs) and minisatellites (VNTRs), which are mostly non-coding.

A. DNA that forms a minor peak during density gradient centrifugation. Satellite DNA consists of highly repetitive sequences and often has a different base composition from the bulk genomic DNA, causing it to separate as distinct bands (‘satellites’) during centrifugation.

A. The presence of two or more alleles of a gene or DNA sequence in a population. DNA polymorphism refers to variations in DNA sequences among individuals, which form the basis for DNA fingerprinting.

C. Sequencing the entire DNA. DNA fingerprinting analyzes specific variable regions, not the entire genome sequence.

B. Variable Number Tandem Repeats. VNTRs are regions in the genome where a short sequence of nucleotides is organized as a tandem repeat, and the number of repeats varies among individuals.

A. Polymerase Chain Reaction (PCR). PCR allows amplification of the specific DNA regions (like STRs) used for fingerprinting, enabling analysis from very small samples.

C. Determining the sequence of a gene. DNA fingerprinting compares the lengths or patterns of variable regions, it doesn’t determine the exact base-by-base sequence of a specific gene.

C. Alec Jeffreys. Sir Alec Jeffreys developed the first DNA profiling techniques using VNTRs in 1984.

A. The presence of repetitive sequences in DNA showing high degree of polymorphism. Variations (polymorphisms) in the number of repeats within these sequences (like VNTRs and STRs) create unique patterns for individuals.