Genetics Final Exam

G and U are not present in both DNA and RNA. DNA contains the base Guanine (G), while RNA contains the base Uracil (U). Therefore, the statement is false.

The correct answer is base, sugar, phosphate. Nucleotides are the building blocks of DNA and RNA, and they consist of three main components: a nitrogenous base (adenine, guanine, cytosine, or thymine/uracil), a sugar molecule (deoxyribose in DNA or ribose in RNA), and a phosphate group. These components combine to form a nucleotide, which then join together to form the DNA and RNA strands. The base pairs in DNA (adenine with thymine and guanine with cytosine) and the sequence of nucleotides in RNA determine the genetic information encoded in these molecules.

The statement is true because spliceosomes are responsible for removing introns from precursor mRNA molecules. Spliceosomes are large complexes made up of proteins and small nuclear RNA molecules. They recognize specific sequences at the beginning and end of introns, and then catalyze the removal of the introns through a process called splicing. This results in the production of mature mRNA molecules that only contain the exons, which are the coding regions of the gene. Therefore, the correct answer is true.

Frameshift mutations occur when nucleotides are inserted or deleted in the DNA sequence, causing a shift in the reading frame. This shift alters the codons, which changes the amino acid sequence during translation. As a result, the protein produced is usually nonfunctional or severely impaired. On the other hand, substitutions only replace one nucleotide with another, which may or may not change the amino acid sequence. Therefore, frameshift mutations have more drastic effects on the phenotype as they can completely disrupt the protein's structure and function.

Nondisjunction is the correct answer because it refers to the failure of chromatids to separate and move to opposite poles during anaphase. This can result in an abnormal distribution of chromosomes in the daughter cells, leading to genetic disorders or abnormalities. Transposition, translocation, duplication, and segregation do not specifically refer to the failure of chromatids to separate during anaphase.

Each type of cell within an organism has a specific function and therefore requires specific genes to be expressed. This ensures that the cell can carry out its specialized role effectively. If all genes were expressed in every cell, there would be a lack of specificity and coordination in the organism's functions. Therefore, it is true that only a small subset of genes are expressed in each type of cell within an organism.

The appearance of traits expressed by genes in association with environmental influence is referred to as the organism's phenotype. The phenotype is the observable characteristics of an organism, such as its physical appearance, behavior, and other traits. It is determined by the interaction between an organism's genotype (its genetic makeup) and its environment. The phenotype can be influenced by various factors, including genetic mutations, gene expression, and environmental factors such as diet, temperature, and exposure to toxins.

Helicase is responsible for opening the double helix structure of DNA during replication. It does this by breaking the hydrogen bonds between the DNA strands, allowing the replication machinery to access the DNA template strands and synthesize new complementary strands. Helicase acts as a molecular motor, using energy from ATP hydrolysis to unwind the DNA helix and separate the strands. This process is crucial for DNA replication, as it provides the single-stranded DNA templates needed for DNA polymerases to synthesize new DNA strands.

For any gene with multiple alleles, a diploid organism may have a maximum of 2 alleles for that locus. This is because diploid organisms have two sets of chromosomes, one inherited from each parent. Each set of chromosomes carries one allele for each gene. Therefore, a diploid organism can have a maximum of two different alleles for any given gene locus.

Proto-oncogenes are indeed found in normal, non-cancerous cells. These genes play a crucial role in regulating normal cell processes, such as cell growth, division, and differentiation. However, under certain circumstances, proto-oncogenes can become oncogenes, which are genes that have the potential to cause cancer. This transformation can occur due to genetic mutations or alterations in the regulation of proto-oncogenes. Therefore, the statement that proto-oncogenes are found in normal, non-cancerous cells and are responsible for normal cell processes is true.

The primary structure of a protein refers to the specific sequence of amino acids that make up the protein chain. This sequence is determined by the genetic code and is crucial for the overall structure and function of the protein. The sequence of amino acids determines how the protein folds into its three-dimensional shape and how it interacts with other molecules in the body. Therefore, the correct answer is that the primary structure of a protein is determined by the sequence of amino acids.

If two plants that are heterozygous for flower color (Ww) are mated, the possible genotypes of the offspring are WW, Ww, and ww. This is because the dominant allele (W) can be inherited from one parent, resulting in the genotype WW, while the recessive allele (w) can be inherited from the other parent, resulting in the genotype ww. Additionally, there is a 50% chance that the offspring will inherit the dominant allele (W) from one parent and the recessive allele (w) from the other parent, resulting in the genotype Ww. Therefore, all of the above genotypes are possible outcomes of the mating.

A Barr body is a condensed, inactive X chromosome that is typically found in the nuclei of female cells. In females, one of the two X chromosomes is randomly inactivated during early development to ensure dosage compensation between males and females. This inactivated X chromosome becomes highly compacted and forms a dense structure known as a Barr body. The presence of a Barr body can be used to identify the sex of an individual's cells, as males typically have one X chromosome and females have two, with one being inactivated.

Not all RNAs are translated. While messenger RNA (mRNA) is translated into proteins, other types of RNA such as transfer RNA (tRNA) and ribosomal RNA (rRNA) are involved in the process of translation but are not themselves translated. Additionally, non-coding RNAs do not undergo translation. Therefore, the statement that all RNAs are translated is false.

Variance is the statistical measure used to describe sample variability. It quantifies how spread out the data points in a sample are from the mean. A higher variance indicates that the data points are more scattered, while a lower variance suggests that the data points are closer to the mean. Variance is calculated by taking the average of the squared differences between each data point and the mean.

A purine's being replaced by a different purine is called a transition mutation. This means that if a purine base (adenine or guanine) is replaced by another purine base, it is considered a transition mutation. This is true because transition mutations involve the substitution of a nucleotide with a different nucleotide of the same type (purine to purine or pyrimidine to pyrimidine).

The regulation of the lac operon of E. coli by lactose is both negative and inducible. This means that the presence of lactose acts as an inducer, causing the lac operon to be transcribed and translated into enzymes that can break down lactose. However, the regulation is also negative, meaning that in the absence of lactose or the presence of glucose, the lac operon is repressed and not transcribed. Therefore, the statement "Regulation of the lac operon of E. coli by lactose is both negative and inducible" is true.

The given information states that there are 81 red flowers (R1R1), 18 pink flowers (R1R2), and 1 white flower (R2R2) in the population. In incomplete dominance, the heterozygous genotype (R1R2) displays a phenotype that is intermediate between the two homozygous genotypes (R1R1 and R2R2). Since the frequency of the R1 allele is stated to be 0.9, it means that the R1 allele is present in a majority of the population. Therefore, the statement "True" is the correct answer.

The correct answer is genotype. Genotype refers to the complete set of genes that an organism possesses. It includes all the genetic information encoded in an organism's DNA. This term is used to describe the genetic makeup of an individual, which determines the traits and characteristics that they may exhibit.

At the conclusion of meiosis, there are four haploid cells. Meiosis is a type of cell division that occurs in sexually reproducing organisms to produce gametes (sperm and egg cells). During meiosis, a single diploid cell undergoes two rounds of division, resulting in the formation of four haploid cells. Haploid cells contain half the number of chromosomes as the original diploid cell, allowing for genetic diversity in offspring during sexual reproduction. Therefore, the correct answer is four haploid cells.

Epistasis refers to the interaction between two or more genes where the expression of one gene masks or modifies the expression of another gene. In this case, the gene in question is masking the expression of a different, nonallelic gene, indicating an epistatic relationship. This means that the gene is exerting control over the expression of the other gene, overriding its normal function or expression pattern. Therefore, the correct answer is epistatic.

DNA Polymerase I is the correct answer because it is an enzyme that is involved in DNA replication and repair. It is responsible for removing the RNA primers during DNA replication and replacing them with DNA nucleotides. DNA Polymerase I also has a 5' to 3' exonuclease activity, which allows it to proofread and correct errors in the newly synthesized DNA strand. Additionally, DNA Polymerase I has a role in DNA recombination and the repair of damaged DNA.

Nonsense mutations actually do alter DNA sequence information. These mutations introduce premature stop codons in the coding region of a gene, causing the synthesis of a truncated and usually nonfunctional protein. This alteration in the DNA sequence disrupts the normal functioning of the gene and can lead to various genetic disorders or diseases. Therefore, the correct answer is False.

During the S phase of the cell cycle, the genetic material of eukaryotic cells is duplicated. This phase is also known as the synthesis phase, where DNA replication occurs. The cell prepares itself for division by creating an identical copy of its DNA. This ensures that each daughter cell will have a complete set of genetic material. The S phase is followed by the G2 phase, where the cell prepares for cell division, and then the M phase, where the actual cell division takes place. Therefore, the correct answer is S.

Chromatin is a complex structure found in the nucleus of cells that contains DNA and proteins. DNA is the genetic material that carries the instructions for the development and functioning of all living organisms. Proteins, on the other hand, play various roles in the organization and regulation of DNA. Therefore, the correct answer is D) both A and C, as chromatin contains both DNA and protein.

In a double-stranded RNA molecule, the percentage of uracil and cytosine should be equal since they pair together. If the percentage of uracil is 30%, then the percentage of cytosine should also be 30%. Therefore, the answer of 20% is incorrect.

Conjugation is the correct answer because it refers to the process by which E. coli transfers genetic material to another bacterium through direct cell-to-cell contact. This process allows for the exchange of plasmids, which are small, circular pieces of DNA that can carry beneficial genes such as antibiotic resistance. Conjugation plays a crucial role in bacterial evolution and the spread of antibiotic resistance genes among bacterial populations.

Genes in both prokaryotes and eukaryotes are primarily regulated at the level of transcription. This is because transcription is the process by which the genetic information in DNA is used to synthesize RNA molecules, which then serve as templates for protein synthesis. By controlling the rate of transcription, cells can regulate the amount of RNA produced and therefore the amount of protein that is ultimately synthesized. This allows cells to respond to changing environmental conditions and to regulate gene expression in a precise and coordinated manner.

A eukaryotic promoter does not contain a Pribnow box. The Pribnow box, also known as the -10 element, is a conserved DNA sequence found in prokaryotic promoters. It is recognized by RNA polymerase and plays a crucial role in initiating transcription. In eukaryotes, transcription is initiated by different promoter elements, such as the TATA box and the initiator element (Inr), which are distinct from the Pribnow box. Therefore, the statement that a eukaryotic promoter contains a Pribnow box is false.

In a double-stranded RNA molecule, the two strands are twisted around each other in a helical structure. The number of complete turns refers to the number of times the two strands wrap around each other in a full circle. In this case, the approximate number of complete turns in a 10,000 base pair RNA molecule is 1000. This means that the two strands wrap around each other 1000 times to form the helical structure.

F plasmids are known as plasmids that can confer the ability to conjugate. Conjugation is a mechanism of horizontal gene transfer in bacteria where genetic material is transferred between two bacterial cells through direct cell-to-cell contact. F plasmids contain the Fertility (F) factor which allows them to transfer their genetic material to recipient cells during conjugation. This transfer can lead to the acquisition of new traits or resistance to antibiotics. Therefore, F plasmids are specifically associated with the ability to conjugate.

The ratio of (A+G) to (T+C) in a double-stranded DNA molecule is equal to 1, meaning that the number of adenine (A) and guanine (G) bases is equal to the number of thymine (T) and cytosine (C) bases. This is because in DNA, adenine always pairs with thymine and guanine always pairs with cytosine, forming base pairs. Therefore, if the ratio is equal to 1, it indicates that the DNA molecule is properly paired and balanced.

DNA synthesis in eukaryotes is semi-conservative. This means that during replication, each strand of the original DNA molecule serves as a template for the synthesis of a new complementary strand. As a result, each new DNA molecule formed contains one original strand and one newly synthesized strand. The other options, proceeding in a 3' to 5' direction and involving a single protein, are not accurate statements about DNA synthesis in eukaryotes.

Eukaryotes, which include plants, animals, fungi, and protists, generally have larger amounts of DNA compared to prokaryotes, which are bacteria and archaea. This is because eukaryotes have more complex cellular structures and functions that require a larger amount of genetic information. Prokaryotes, on the other hand, have simpler cellular structures and therefore do not require as much DNA. This difference in DNA content is more typical in eukaryotes than in prokaryotes.

Chromosomal inversions can indeed result in duplications and deletions. During a chromosomal inversion, a segment of the chromosome is flipped in orientation. This can lead to the breakage and rearrangement of genetic material, causing duplications and deletions of certain genes or regions. Therefore, the statement is true.

Crossing over occurs most frequently during prophase I of meiosis. This is because during prophase I, homologous chromosomes pair up and exchange genetic material through a process called crossing over. This exchange of genetic material increases genetic diversity and helps in the formation of genetically unique gametes. In contrast, crossing over does not occur during metaphase II, prophase and telophase of mitosis, or prophase II of meiosis.

Aminoacyl transferase is the enzyme responsible for charging tRNAs. Charging refers to the process of attaching the correct amino acid to the corresponding tRNA molecule, which is necessary for protein synthesis. DNA Polymerase I is involved in DNA replication, not tRNA charging. RNA Polymerase II and RNA Polymerase III are responsible for transcribing DNA into RNA, but they are not involved in charging tRNAs. Therefore, the correct answer is Aminoacyl transferase.

A diploid number (2n) of 18 means the normal somatic cells have 18 chromosomes. Tetraploid organisms have four sets of chromosomes (4n), so their somatic cells contain twice the diploid number:

4 × 9 = 36 chromosomes.

Proto-oncogenes do not directly cause cancer. They are normal genes that play a role in regulating cell growth and division. However, when proto-oncogenes undergo certain changes or mutations, they can become oncogenes, which have the potential to cause cancer. Therefore, it is incorrect to say that proto-oncogenes cause cancer.

Bacterial recombination requires all of the above options because F pilus is necessary for the transfer of genetic material between bacteria, plasmids are the carriers of the genetic material being transferred, and physical contact between bacteria is required for the transfer to occur. Therefore, all three options are essential for bacterial recombination to take place.

Gyrase is the correct answer because it is an enzyme that relieves torsional stress during DNA replication. Torsional stress occurs when the DNA double helix becomes overwound or underwound during the unwinding process. Gyrase helps to alleviate this stress by introducing negative supercoils into the DNA, allowing for smooth unwinding and replication to occur. This ensures that the DNA strands do not become tangled or damaged during the replication process.

The correct order of these genes is def because in the offspring, the genotype + + + is the most frequent, followed by + e f, d + f, d e f, + e +, and d + +. This indicates that the dominant allele for each gene (represented by +) is present in the majority of the offspring, suggesting that it is the first gene in the order. The second gene can be determined by looking at the offspring with the genotype + e f, which is the second most frequent genotype. This indicates that the heterozygous individual must have the e allele, and the homozygous recessive individual must have the f allele. Therefore, the correct order is def.

HMG (High Mobility Group) is a nonhistone protein found in chromatin. It functions as a structural protein that helps in the organization and stabilization of chromatin structure. HMG proteins are involved in various cellular processes such as transcription, DNA repair, and recombination. They bind to DNA and help in regulating gene expression by facilitating the binding of transcription factors and other proteins to DNA. Unlike histone proteins, which are also found in chromatin, HMG proteins do not participate in nucleosome formation but play a crucial role in maintaining chromatin architecture.

The transfer of bacterial genes from one cell to another requires multiple factors, including the F factor, F plasmids, and F+ cells. The F factor is a piece of DNA that allows for the transfer of genetic material between bacterial cells. F plasmids are self-replicating circular pieces of DNA that contain the F factor and are responsible for transferring the F factor to other cells. F+ cells are bacterial cells that contain the F factor and are capable of transferring genetic material to F- cells. Therefore, all of these factors are necessary for the transfer of bacterial genes.

The lac operon is a group of genes involved in lactose metabolism in bacteria. In the absence of allolactose, which is an inducer molecule, the lac operon is not transcribed constitutively. Instead, it is repressed by a regulatory protein called the lac repressor. The lac repressor binds to the operator region of the lac operon, preventing RNA polymerase from transcribing the genes. When allolactose is present, it binds to the lac repressor and causes a conformational change, releasing the repressor from the operator and allowing transcription of the lac operon. Therefore, the statement that the lac operon is constitutively transcribed in the absence of allolactose is false.

In human males, genes on the X chromosome are hemizygous. This is because males have one X chromosome and one Y chromosome, while females have two X chromosomes. Therefore, males have only one copy of genes on the X chromosome, making them hemizygous.

The binding of the CAP protein to the DNA sequence adjacent to the promoter of the lac operon in the absence of glucose helps RNA polymerase to bind to the promoter, resulting in a high level of transcription of the lac operon. This indicates that the CAP protein positively regulates the lac operon by enhancing the transcription process.

Ion channels are NOT commonly known to include oncogenes. Oncogenes are genes that have the potential to cause cancer when mutated or overexpressed. They are typically involved in cell growth, division, and regulation. Transcription factors, growth factors, signal-transduction proteins, and DNA-repair enzymes are all commonly associated with oncogenes as they play crucial roles in cellular processes that can become dysregulated in cancer. However, ion channels primarily regulate the flow of ions across cell membranes and are not typically implicated in oncogenesis.

The sequence of the template strand of DNA is the complementary sequence to the RNA molecule. In RNA, adenine (A) pairs with uracil (U), cytosine (C) pairs with guanine (G), and vice versa in DNA. Therefore, the template strand of DNA would have the sequence 3’-CCGTAGCTGC-5’, which is the complementary sequence to the given RNA sequence.

The correct answer is "all are different". This means that all of the options listed (splicing, 5' capping, poly-adenylation, and termination) are different between eukaryotic and prokaryotic transcription. In eukaryotes, splicing is required to remove introns from the pre-mRNA, 5' capping is the addition of a modified guanine nucleotide to the 5' end of the mRNA, poly-adenylation is the addition of a poly-A tail to the 3' end of the mRNA, and termination involves the recognition of specific termination sequences. In contrast, prokaryotic transcription does not involve splicing, 5' capping, or poly-adenylation, and termination occurs differently.