As stated in the text, bacteriophages have been discovered with the following base substitutions in their DNA: (a) dUMP completely substituting for 'TMP (b) 5-hydroxymethyl-dUMP completely substituting for d'TMP (c) 5-methyl-dCMP completely substituting for dCMP For any one of these cases, formulate a set of virus-coded enzyme activities that could lead to the observed substitution. Write a balanced equation for each reaction you propose.
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- The DNA of a deletion of alpha bacteriophage has a length of 15 micrometers instead of 17 micrometers? How many base pairs are missing from this mutant?Researchers have been determining the nucleotide sequences of variant forms of SARS-CoV-2, looking for versions of the virus that might be more easily transmitted between humans or that might be more deadly. (a) For example, one mutation in a viral gene changed a GAU codon to a GGU codon. How does this change affect the sequence of the polypeptide encoded by that gene? (b) In another variant form of the virus, a gene is missing six consecutive nucleotides. How would this change affect the sequence of the polypeptide encoded by that gene? (c) In another coronavirus variant, the spike protein (the prominent protein on the surface of the virus) contains a histidine residue where an aspartate (aspartic acid) residue should be. Describe a point mutation in the coronavirus genome that could have caused this change in the spike protein.As stated in the text, bacteriophages have been discovered with the followingbase substitutions in their DNA:(a) dUMP completely substituting for dTMP(b) 5-hydroxymethyl-dUMP completely substituting for dTMP(c) 5-methyl-dCMP completely substituting for dCMPFor any one of these cases, formulate a set of virus-coded enzyme activitiesthat could lead to the observed substitution. Write a balanced equation foreach reaction you propose.
- The DNA of a deletion mutant of λ bacteriophage has a length of 15.4383 μm instead of 19.6356 μm. How many base pairs are missing from this mutant? *A mixture of four a-[32P]–labeled ribonucleoside triphosphates was added to permeabilized bacterial cells undergoing DNA replication in the presence of an RNA polymerase inhibitor, and incorporation into high-molecular-weight material was followed over time, as shown in the accompanying graph. After 10 minutes of incubation, a 1000-fold excess of unlabeled ribonucleoside triphosphates was added, with the results shown in the graph. Incorporation of radioactivity, cpm Excess (a) Why was the excess of unlabeled rNTPs added?(b) How could you tell that radioactivity is being incorporated as ribonucleotides rather than as an alternative such as reduction to deoxyribonucleotides, followed by incorporation?(c) What does this experiment tell you about the process of DNA replication?A mixture of four a-[32P]–labeled ribonucleoside triphosphates was added to permeabilized bacterial cells undergoing DNA replication in the presence of an RNA polymerase inhibitor, and incorporation into high-molecular-weight material was followed over time, as shown in the accompanying graph. After 10 minutes of incubation, a 1000-fold excess of unlabeled ribonucleoside triphosphates was added, with the results shown in the graph. (a) Why was the excess of unlabeled rNTPs added?(b) How could you tell that radioactivity is being incorporated as ribonucleotides rather than as an alternative such as reduction to deoxyribonucleotides, followed by incorporation?(c) What does this experiment tell you about the process of DNA replication?
- The anti-viral drug Acyclovir is a nucleotide analog that is lacking the 3’ OH group which is required to form a 3’→5’ phosphodiester bond. This drug is ineffective against DNA polymerases with proofreading abilities, which is why human DNA polymerases are not targeted. Acyclovir can be used to treatsevere cases of Epstein-Barr viral (EBV) infection, but has little to no effect under non-severe infections. Based on this information, EBV will use ________ DNA polymerase during severe infections and __________ DNA polymerase during non-severe infections. Human; Human EBV; Human EBV; EBV Human; EBVA plasmid vector pBS281 is cleaved by the enzymeBamHI (5′ G^GATCC 3′), which recognizes only onesite in the DNA molecule. Human DNA is digestedwith the enzyme MboI (5′ ^GATC 3′), which recognizes many sites in human DNA. These two digestedDNAs are now ligated together. Consider only thosemolecules in which the pBS281 DNA has been joinedwith a fragment of human DNA. Answer the following questions concerning the junction between thetwo different kinds of DNA. a. What proportion of the junctions between pBS281and all possible human DNA fragments can becleaved with MboI?b. What proportion of the junctions between pBS281and all possible human DNA fragments can becleaved with BamHI?c. What proportion of the junctions between pBS281and all possible human DNA fragments can becleaved with XorII (5′ C^GATCG 3′)?d. What proportion of the junctions between pBS281and all possible human DNA fragments can becleaved with BstYI (5′ R^GATCY 3′)? (R and Ystand for purine and pyrimidine, respectively.)e.…Which of the following set(s) of primers a–d couldyou use to amplify the following target DNA sequence, which is part of the last protein-coding exonof the CFTR gene?5′ GGCTAAGATCTGAATTTTCCGAG ... TTGGGCAATAATGTAGCGCCTT 3′3′ CCGATTCTAGACTTAAAAGGCTC ... AACCCGTTATTACATCGCGGAA 5′a. 5′ GGAAAATTCAGATCTTAG 3′;5′ TGGGCAATAATGTAGCGC 3′b. 5′ GCTAAGATCTGAATTTTC 3′;3′ ACCCGTTATTACATCGCG 5′c. 3′ GATTCTAGACTTAAAGGC 5′;3′ ACCCGTTATTACATCGCG 5′d. 5′ GCTAAGATCTGAATTTTC 3′;5′ TGGGCAATAATGTAGCGC 3′
- Let’s suppose you make a transposon library of the cellulose-secreting bacterium Komagataeibacter xylinus, with the goal of finding mutants that produce higher than normal amounts of cellulose, which would be useful industrially. However, despite your best efforts you are unable to isolate any transposon mutants that make more cellulose than the wild-type strain.Why might this have failed? List as many reasons as you can think of.The principal genomic component isolated from equine influenza virus is 22% C, 23% A, 22% G and 33% U. Is this sufficient data to determine whether the genomic material is DNA/RNA and is double- or single-stranded? Why or why not? Explain, using a diagramRNA was extracted from certain virus and found to contain 35% cytosine. With this information, is it possible to predict what percentage of the bases in virus are adenine? If so, what percentage? If not, why not?