Block 1 DNA Technology

The percentage of guanine in the genome can be determined by subtracting the percentage of uracil (15%) from 100%. Since uracil and guanine pair up in RNA, if there is 15% uracil, there must be 15% guanine as well. Therefore, the percentage of guanine in this genome is 35%.

DNA is a double-stranded molecule consisting of two strands that run in opposite directions. This is known as the antiparallel nature of DNA. The two strands are held together by hydrogen bonds between complementary nitrogenous bases, forming base pairs. The phosphate groups are located on the exterior of the DNA molecule, while the nitrogenous bases are stacked along the interior. The sugar component in DNA is deoxyribose, not ribose. Therefore, the correct answer is "Two strands that run in the opposite direction."

The UCA → UCU mutation would not produce a change in the protein translated from an mRNA because both UCA and UCU code for the amino acid serine. This mutation is a silent mutation, meaning it does not result in a change in the amino acid sequence of the protein.

When a radioactive double-stranded DNA molecule undergoes replication in a solution free of radioactive label, the resulting DNA molecules will have one radioactive strand and one non-radioactive strand. This is because during replication, the two strands of the original DNA molecule separate and each strand serves as a template for the synthesis of a new complementary strand. Since the solution is free of radioactive label, the newly synthesized strands will not be radioactive. Therefore, half of the resulting four double-stranded DNA molecules will contain no radioactivity.

The correct answer is RNA polymerase because the hepatotoxic octapeptide a-amanitin, which is present in the death-cap mushroom, inhibits RNA polymerase. This inhibition of RNA polymerase disrupts the process of transcription, leading to the dysfunction of protein synthesis and ultimately causing the symptoms of nausea, vomiting, diarrhea, and abdominal pain. Supportive care is provided as there is no specific antidote for the toxin.

The number of amino acids in a protein is determined by the number of codons in the mRNA sequence. Each codon codes for a specific amino acid. The mRNA in this question is 336 nucleotides long, which includes the initiator and termination codons. Since there are 3 nucleotides per codon, we can divide 336 by 3 to get the number of codons. 336 divided by 3 equals 112 codons. However, we need to subtract 1 for the termination codon, so the final answer is 111 amino acids.

When synthesis of segment C begins, segment E is also being synthesized.

The LD50 of amanitin is the dose at which 50% of the recipients die. In this case, the LD50 is 0.1 mg per kg body weight. The average mushroom contains 7 mg of amanitin. To calculate how many mushrooms must be consumed by someone who is 100 kg to be above the LD50, we need to divide the LD50 dose by the amount of amanitin in one mushroom. 0.1 mg/kg divided by 7 mg/mushroom equals approximately 0.014 mushrooms. Since you can't consume a fraction of a mushroom, the person would need to consume at least 2 mushrooms to be above the LD50.

The 1,000-kilobase fragment of DNA has 10 evenly spaced and symmetric replication origins. This means that each replication origin is 100 kilobases apart from each other. Since DNA polymerase moves at a rate of 1 kilobase per second, it will take 100 seconds to replicate the entire 1,000-kilobase fragment of DNA. However, the question asks for the time it takes to produce two daughter molecules. Since there are 10 replication origins, each origin will produce one daughter molecule. Therefore, it will take 50 seconds to produce two daughter molecules.

The start AUG codon is located at the beginning of the coding sequence, which is preceded by a 5'-untranslated region of 100 bp. The final termination codon is located at the end of the coding sequence, which is followed by a 3'-untranslated region of 200 bp. Therefore, the total number of bases between the start AUG codon and the final termination codon is 100 (5'-untranslated region) + 150 (coding sequence) + 200 (3'-untranslated region) = 450.

The correct answer is "-CCA-CCC-TAG-GTT-CAG-". This sequence is a frame-shift mutation because it shifts the reading frame by deleting one base pair (C) from the original sequence. This causes a shift in the codons and results in a premature stop codon (TAG) being introduced earlier in the sequence. As a result, the encoded protein is terminated prematurely.

The most likely outcome of the patient's deficiency in Histone Acetyltransferase (HAT) activity is reduced transcription rates because nucleosomes occupy promoters. HAT is responsible for adding acetyl groups to histones, which helps in loosening the chromatin structure and allowing transcription factors to access the DNA and initiate gene expression. Without HAT activity, the chromatin remains tightly packed around the DNA, making it difficult for transcription factors to bind and initiate transcription. As a result, transcription rates are reduced.

The 3' to 5' exonuclease activity of DNA polymerase δ is critical in determining the accuracy of nuclear DNA replication and the base substitution mutation rate in human chromosomes. This activity allows DNA polymerase δ to proofread and correct errors that may occur during DNA replication. By removing misincorporated bases, DNA polymerase δ ensures that the newly synthesized DNA strand is accurate and free from mutations. This helps maintain the integrity of the genetic information and reduces the occurrence of genetic diseases caused by single nucleotide substitutions.

Erythromycin is effective in treating respiratory tract infections in legionnaire's disease, whooping cough, and Mycoplasma-based pneumonia because it inhibits translocation by binding to 50S ribosomal subunits. This means that it prevents the movement of the ribosome along the mRNA strand during protein synthesis, effectively stopping the production of proteins in certain bacteria. By inhibiting this essential step in bacterial protein synthesis, erythromycin can effectively kill the bacteria and treat the respiratory tract infections.

The correct answer is Single-strand binding protein (SBP). The question states that a viral-encoded enzyme inactivates a host-cell nuclear protein required for DNA replication. Among the options given, SBP is the only protein that is directly involved in DNA replication. It binds to single-stranded DNA and stabilizes it, allowing for DNA replication to occur. Therefore, it is a potential substrate for the viral enzyme that inactivates the host-cell nuclear protein.

If adenine is the first base on the template strand for segment E, then the corresponding nucleotide on the newly synthesized strand would be uracil, as adenine pairs with uracil in RNA. Therefore, the precursor molecule that would serve as the substrate for the formation of the first nucleotide in segment E is UTP (uridine triphosphate).

Tetracycline, streptomycin, and erythromycin are effective antibiotics because they inhibit protein synthesis in prokaryotes. Prokaryotes, such as bacteria, have different cellular machinery for protein synthesis compared to eukaryotes. These antibiotics specifically target the ribosomes in prokaryotic cells, preventing them from synthesizing proteins effectively. This disruption in protein synthesis weakens the bacteria and inhibits their growth and reproduction, making these antibiotics effective in treating bacterial infections.

The average size of a human gene is 40,000 bp. This is because genes in humans can vary in size, ranging from a few hundred base pairs to over a million base pairs. However, when considering the average size, it is found to be around 40,000 base pairs. This is determined by analyzing the sizes of all known human genes and calculating the average length.

Nucleosomes are subunits of chromatin. Chromatin is the material that makes up chromosomes and is composed of DNA tightly wrapped around histone proteins. Nucleosomes are formed when DNA is wound around a core of eight histone proteins. Therefore, the phrase "Subunits of chromatin" accurately describes nucleosomes.

Histone acetylation is a process that involves the addition of acetyl groups to histone proteins, which are involved in packaging DNA into a compact structure called chromatin. This modification is typically associated with gene activation because it loosens the interaction between DNA and histones, allowing for easier access of transcription factors and other regulatory proteins to the DNA. This facilitates the transcription of genes, leading to increased gene expression and ultimately, gene activation.

The given correct answer for the question is "Translocation". Translocation is the process in protein synthesis where the ribosome moves along the mRNA strand, shifting the tRNA molecules from the A site to the P site and then to the E site. This movement allows for the synthesis of the polypeptide chain. In this case, the antibiotic inhibits the translocation process, resulting in the only product being methionyl-phenylalanyltRNA.

Based on the given information, the base composition of Strand 2 matches the base composition of the transcribed mRNA. This suggests that Strand 2 is the coding strand, serving as a template for mRNA synthesis.

Enhancers are transcriptional regulatory sequences that function by enhancing the activity of RNA polymerase at a single promoter site. They do not bind to the promoter or RNA polymerase directly, but they interact with other transcription factors to increase the efficiency of transcription initiation. By facilitating the binding of RNA polymerase to the promoter and stabilizing the transcription initiation complex, enhancers play a crucial role in regulating gene expression.

The cDNA made from mature stimulin RNA would have a size of 400 bp. This is because the cDNA is synthesized from the mature mRNA, which does not contain the intron. The two exons that encode the protein of 100 amino acids are separated by the intron of 100 bp, which is not present in the mature mRNA. Therefore, the cDNA will only include the two exons and the untranslated regions, resulting in a size of 400 bp.

The sequence of an Okazaki fragment is determined by the direction of DNA replication and the location of the 5' and 3' ends. In this case, the replication fork is moving from left to right, so the new strand being synthesized will be in the 5' to 3' direction. The 5' and 3' ends of the Okazaki fragment are located right at the borders of the shown sequence, which means that the new strand will start with the same nucleotide as the last nucleotide of the shown sequence and continue in the 5' to 3' direction. Therefore, the correct answer is 5’ GGATGTTATCATGCACCT 3’.

Norfloxacin inhibits the action of Topoisomerase II, an enzyme that plays a crucial role in bacterial DNA replication. Topoisomerase II is responsible for untangling and unknotting the DNA strands during replication, making it an essential enzyme for chromosome duplication in rapidly dividing cells. By inhibiting this enzyme, norfloxacin disrupts the replication process and prevents the bacteria from replicating their DNA effectively.

The child in this scenario presents with severe growth failure, accelerated aging, microcephaly, and decreased levels of DNA synthesis. The fact that the addition of protein extract from normal cells restores DNA synthesis suggests that there is a defect in an enzyme involved in DNA replication. Unwinding proteins play a crucial role in DNA replication as they help in unwinding the double helix structure of DNA, allowing for DNA synthesis to occur. Therefore, the likely enzyme that is defective in this child is the unwinding proteins.